1335vch00.indd I 15.02.2007 11:48:03

Hugo Hens

Building Physics -Heat, Air and Moisture Fundamentals and Engineering Methods with Examples and Exercises

irnst&Sohn A W i l e y C o m p a n y

1335vch00.indd II 15.02.2007 11:48:03

1807-2007 Knowledge for Generations

Each generation has its unique needs and aspirations. When Charles Wiley first opened his small printing shop in lower Manhattan in 1807, it was a generation of boundless potential searching for an identity. And we were there, helping to define a new American literary tradition. Over half a century later, in the midst of the Second Industrial Revolution, it was a generation focused on building the future. Once again, we were there, supplying the critical scientific, technical, and engineering knowledge that helped frame the world. Throughout the 20th Century, and into the new millennium, nations began to reach out beyond their own borders and a new international community was born, Wiley was there, ex-panding its operations around the world to enable a global exchange of ideas, opinions, and know-how.

For 200 years, Wiley has been an integral part of each generation's journey enabling the flow of information and understanding necessary to meet their needs and fulfill their aspirations. Today, bold new technologies are changing the way we live and learn. Wiley will be there, providing you the must-have knowledge you need to imagine new worlds, new possibilities, and new oppor-tunities.

Generations come and go, but you can always count on Wiley to provide you the knowledge you need, when and where you need it!

William J. Pesce Peter Booth Wiley President and Chief Executive Officer Chairman of the Board

1335vch00.indd III 15.02.2007 11:48:03

Hugo Hens

Building Physics -Heat, Air and Moisture Fundamentals and Engineering Methods with Examples and Exercises

I I C I H T I H M U L

1 8 O 7

©WILEY 2 O O 7

B I C E N T E N N I A L A W i l e y C o m p a n y rnst&Sohn

1335vch00.indd IV 15.02.2007 11:48:03

Professor Dr. Ir. Hugo S.L.C. Hens K.U. Leuven Faculty of Engineering Department of Civil Engineering Laboratory of Building Physics Kasteelpark Arenberg, 40 3001 Leuven Belgium

Cover illustration: Thermal bridge, calculated with TRYSCO (Physibel CV, Maldegem, Belgium)

Bibliographic information publishd by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.d-nb.de>

ISBN 978-3-433-01841-5

©2007 Ernst & Sohn Verlag für Architektur und technische Wissenschaften GmbH und Co. KG, Berlin

All rights reserved (including those of translation into other languages). No part of this book may be repro-duced in any form - by photoprinting, microfilm, or any other means - nor transmitted or translated into a machine language without written permission from the publisher. Registered names, trademarks, etc. used in this book, even when not specifically marked as such, are not to be considered unprotected by law.

Typesetting: Manuela Treindl, Laaber Printing: Strauss GmbH, Mörlenbach Binding: Litges & Dopf GmbH, Heppenheim Wiley Bicentennial Logo: Richard J. Pacifico

Printed in Germany

To my wife, children and grandchildren

In remembrance of Professor A. de Grave,who introduced Building Physics as a new disciplineat the University of Louvain, K.U. Leuven, Belgium, in 1952

1335vch00.indd V 15.02.2007 11:48:03

Preface

Overview

Until the � rst energy crisis of 1973, building physics survived as a somewhat dormant � eld in building engineering, with seemingly limited applicability in practice. While soil mechanics, structural mechanics, building materials, building construction and HVAC were seen as essential, designers only demanded advice on room acoustics, moisture tolerance, summer comfort or lighting when really needed or when, after construction of their designs, when problems arose. Energy was even not a concern, while indoor environmental quality was presumably guaranteed thanks to the ever present in� ltration, to opening windows and to the heating system. The energy crises of the seventies, persisting moisture problems, complaints about sick buildings, thermal, visual and olfactory discomfort, the move towards more sustainability, changed it all. The societal pressure to diminish energy consumptions in buildings without degrading usability acted as a trigger that activated the whole notion of performance based design and construction. As a result, building physics and its potentiality to quantify performances was suddenly pushed to the frontline of building innovation.As all engineering sciences, building physics is oriented towards application. Hence, creativity in application demands a sound knowledge of the basics in each of the branches building physics encompasses: heat and mass, acoustics, lighting, energy and indoor environmental quality. Integrating these basics into an up to date text on heat and mass has been the main goal of this book, with mass limited to air, water vapour and moisture. The book is the result of thirty years of teaching architectural, building and civil engineers, coupled to thirty � ve years of experience, research and consultancy in the � eld. Apart of that, where and when needed, inputs and literature from over the world is used, reason why each part is followed by an extended literature list. In an introductory chapter, building physics is presented as a discipline. Then, the � rst part concentrates on heat transport, with conduction, convection and radiation as main topics, followed by concepts and applications, typical for building physics. The second part treats mass transport, with air, water vapour and moisture as most important components. Again, much attention is devoted to the concepts and applications which relate to building physics. The last part � nally discusses combined transport of heat and mass, who in fact act as if they were Siamese twins. The � rst and second part is followed by exemplary exercises, with some of them solved and the others offered to the reader as an element of training.The book is written in SI-units. It should be especially usable for undergraduate and graduate studies in architectural and building engineering, although also mechanical engineers, studying HVAC, and practising building engineers, who want to refresh their knowledge, may bene� t from it. The level of presentation anyhow assumes that the reader has a sound knowledge of calculus and differential equations along with some background in physics, thermodynamics, hydraulics, building materials and building construction.

Acknowledgments

A book of this magnitude re� ects the work of many, not only the author. Therefore, � rst of all, we like to thank the thousands of students we had during our thirty years of teaching building physics. They gave us the opportunity to test the content of the book and helped in upgrading

1335vch00.indd VII 15.02.2007 11:48:21

VIII Preface

it by the corrections they proposed and the experience they offered in learning what parts should be better explained.The book should not have been written the way it is, if not standing on the shoulders of those, who preceded us. Although we started our carrier as a structural engineer, our predecessor, Professor Antoine de Grave, planted the seeds that slowly fed our interest in building physics. The late Bob Vos of TNO, the Netherlands, and Helmut Künzel of the Fraunhofer Institut für Bauphysik, Germany, showed the importance of experimental work and � eld testing for understanding building performance, while Lars Erik Nevander of Lund University, Sweden, taught that solving problems in building physics does not always ask complex modelling, mainly because reality in building construction is ever again much more complex than any model may be.During the four decennia at the Laboratory of Building Physics, several researchers and PhD-students got involved. I am very grateful to Gerrit Vermeir, Jan Carmeliet and Staf Roels, who became colleagues at the university, to Piet Standaert, Filip Descamps, now part-time professor at the Free University Brussels (VUB); Arnold Janssens, now associate professor at the University of Ghent (UG); Dirk Saelens, Hans Janssen, now associate professor at the Technical University Denmark (TUD); Rongjin Zheng and Bert Blocken, now associate professor at the Technical University Eindhoven (TU/e), who all contributed to this book by their work. The experiences gained by leading four Annexes of the IEA, Executive Committee on Energy Conservation in Buildings and Community Systems, further-on forced me to rethink parts of the text over time. The many ideas I exchanged and got in Canada and the USA from Kumar Kumaran, Paul Fazio, Bill Brown, William B. Rose, Joe Lstiburek and Anton Ten Wolde were also of great help in drafting the text A number of reviewers took time to evaluate the book. Although we do not know their names, we like to thank them. Many merits also go to Miss Gerrie Doche from Montreal, Quebec, Canada, who made the � rst translation of the text from Dutch into English.Finally, I like to thank my family. My loving mother who died too early. My father, who at a respectable age, still shows interest in our work. My wife, Lieve, who managed living together with a professor who was so busy, my three children who had to live with that busy father, and my many grandchildren who do not know their grandfather is still busy.

Leuven, February 2007 Hugo S.L.C. Hens

1335vch00.indd VIII 15.02.2007 11:48:21

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . VII

0 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

0.1 Subject of the Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

0.2 Building Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.1 De� nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2.2.1 Comfort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2.2.2 Health . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2.2.3 Architectural and Material Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2.2.4 Economy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2.2.5 Environment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

0.3 Importance of Building Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

0.4 History of Building Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

0.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

0.6 Units and Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1 Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2 Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.2.1 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.2.2 Fourier’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.2.2.1 First Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.2.2.2 Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.2.3 Steady State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.3.1 What Is It? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.3.2 One Dimension: Flat Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.3.3 Two Dimensions: Cylinder Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.2.3.4 Two and Three Dimensions: Thermal Bridges . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.2.4 Transient Regime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.2.4.1 What is Transient? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.2.4.2 Flat Walls, Periodic Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.2.4.3 Flat Walls, Transient Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431.2.4.4 Two and Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

1.3 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481.3.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481.3.1.1 Heat Transfer at a Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481.3.1.2 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

1335vch00.indd IX 15.02.2007 11:48:43

X Contents

1.3.2 Convection Typology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501.3.2.1 Driving Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511.3.2.2 Type of � ow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511.3.3 Calculating the Convective Surface Film Coef� cient . . . . . . . . . . . . . . . . . . . . . 511.3.3.1 Analytically . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511.3.3.2 Numerically . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521.3.3.3 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521.3.4 Values for the Convective Surface Film Coef� cient . . . . . . . . . . . . . . . . . . . . . . 541.3.4.1 Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541.3.4.2 Cavities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561.3.4.3 Pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

1.4 Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591.4.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591.4.1.1 Thermal Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591.4.1.2 Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591.4.1.3 Re� ection, Absorption and Transmission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591.4.1.4 Radiant Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 621.4.2 Black Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 621.4.2.1 Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 621.4.2.2 Radiation Exchange Between Two Black Bodies: The Angle Factor . . . . . . . . . 651.4.2.3 Properties of Angle Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 661.4.2.4 Calculating Angle Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671.4.3 Grey Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 691.4.3.1 Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 691.4.3.2 Radiation Exchange Between Grey Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 701.4.4 Colored Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 721.4.5 Practical Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

1.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741.5.1 Surface Film Coef� cients and Reference Temperatures . . . . . . . . . . . . . . . . . . . 741.5.1.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741.5.1.2 Inside Environment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741.5.1.3 Outside Environment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 781.5.2 Steady-state, One-dimension: Flat Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 811.5.2.1 Thermal Transmittance and Interface Temperatures . . . . . . . . . . . . . . . . . . . . . . 811.5.2.2 Thermal Resistance of a Non-ventilated In� nite Cavity . . . . . . . . . . . . . . . . . . . 851.5.2.3 Solar Transmittance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 861.5.3 Steady State, Cylindrical Coordinates: Pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 891.5.4 Steady-state, Two and Three Dimensions: Thermal Bridges . . . . . . . . . . . . . . . . 901.5.4.1 Calculation by the Control Volume Method (CVM) . . . . . . . . . . . . . . . . . . . . . . 901.5.4.2 Thermal Bridges in Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 911.5.5 Transient, Periodic: Flat Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 941.5.6 Heat Balances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

1.7 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

1335vch00.indd X 15.02.2007 11:48:44

XIContents

2 Mass Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

2.1 In General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1112.1.1 Quantities and De� nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1112.1.2 Saturation Degree Scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1132.1.3 Air and Moisture Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1152.1.4 Moisture Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1172.1.5 Air, Moisture and Durability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1172.1.6 Linkages between Mass- and Energy Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . 1192.1.7 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

2.2 Air Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1202.2.1 In General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1202.2.2 Air Pressure Differences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1212.2.2.1 Wind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1212.2.2.2 Stack Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1222.2.2.3 Fans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1232.2.3 Air Permeances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1242.2.4 Air Transfer in Open-porous Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1272.2.4.1 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1272.2.4.2 Flow Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1282.2.4.3 Air Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1282.2.4.4 One Dimension: Flat Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1292.2.4.5 Two- and Three-dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1312.2.5 Air Flow Through Permeable Layers, Apertures, Joints, Leaks and Cavities . . 1322.2.5.1 Flow Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1322.2.5.2 Conservation of Mass, Equivalent Hydraulic Circuit . . . . . . . . . . . . . . . . . . . . . 1322.2.6 Combined Heat- and Air Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1332.2.6.1 Open-porous Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1332.2.6.2 Air Permeable Layers, Joints, Leaks and Cavities . . . . . . . . . . . . . . . . . . . . . . . 140

2.3 Vapour Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1432.3.1 Water Vapour in the Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1432.3.1.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1432.3.1.2 Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1432.3.1.3 Maximum Vapour Pressure and Relative Humidity . . . . . . . . . . . . . . . . . . . . . . 1442.3.1.4 Changes of State in Humid Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1482.3.1.5 Enthalpy of Moist Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1492.3.1.6 Characterizing Moist Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1492.3.1.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1502.3.2 Water Vapour in Open-porous Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1562.3.2.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1562.3.2.2 Sorption Isotherm and Speci� c Moisture Ratio . . . . . . . . . . . . . . . . . . . . . . . . . 1562.3.2.3 The Physics Behind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1582.3.2.4 Impact of Salts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1602.3.2.5 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1612.3.3 Vapour Transfer in the Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1612.3.4 Vapour Transfer in Materials and Construction Parts . . . . . . . . . . . . . . . . . . . . 1642.3.4.1 Flow Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1642.3.4.2 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

1335vch00.indd XI 15.02.2007 11:48:44

XII Contents

2.3.4.3 Vapour Transfer by ‘Equivalent’ Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1672.3.4.4 Vapour Transfer by (Equivalent) Diffusion and Convection . . . . . . . . . . . . . . . 1832.3.5 Surface Film Coef� cients for Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1902.3.6 Some Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1932.3.6.1 Diffusion Resistance of a Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1932.3.6.2 Cavity Ventilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1942.3.6.3 Water Vapour Balance in a Room in Case of Surface Condensation and

Drying . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

2.4 Moisture Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1972.4.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1972.4.2 Moisture Transfer in a Pore . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1972.4.2.1 Capillarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1972.4.2.2 Water Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2002.4.2.3 Vapour Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2092.4.2.4 Moisture Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2112.4.3 Moisture Transfer in Materials and Construction Parts . . . . . . . . . . . . . . . . . . . 2112.4.3.1 Transfer Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2112.4.3.2 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2142.4.3.3 Starting, Boundary and Contact Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2152.4.3.4 Remark . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2162.4.4 Simpli� ed Moisture Transfer Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2162.4.4.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2162.4.4.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

2.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

3 Combined Heat, Air and Moisture Transfer . . . . . . . . . . . . . . . . . . . . . . . . . 255

3.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

3.2 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

3.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

3.4 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2563.4.1 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2563.4.2 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

3.5 Flow Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2603.5.1 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2603.5.2 Mass, Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2603.5.2.1 Open Porous Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2603.5.2.2 Air Permeable Layers, Apertures, Joints, Cracks, Leaks and Cavities . . . . . . . . 2613.5.3 Mass, Moisture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2613.5.3.1 Water Vapour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2613.5.3.2 Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

3.6 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2623.6.1 Enthalpy/Temperature and Water Vapour Saturation Pressure/Temperature . . 2623.6.2 Relative Humidity/Moisture Content . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2623.6.3 Suction/Moisture Content . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

1335vch00.indd XII 15.02.2007 11:48:44

XIIIContents

3.7 Starting, Boundary and Contact Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2623.7.1 Starting Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2623.7.2 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2623.7.3 Contact Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

3.8 Two Examples of Simpli� ed Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2633.8.1 Heat, Air and Moisture Transfer in Non-Hygroscopic, Non-Capillary

Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2633.8.2 Heat, Air and Moisture Transfer in Hygroscopic Materials at Low Moisture

Content . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

3.9 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266

4 Postscript . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269

1335vch00.indd XIII 15.02.2007 11:48:44

0 Introduction

0.1 Subject of the Book

This is the � rst of a series of books on a Building Physics and Applied Building Physics:

Building Physics: Heat, Air and Moisture TransportApplied Building Physics: Boundary Conditions, Building Performance and Material PropertiesApplied Building Physics and Performance Based Design 1Applied Building Physics and Performance Based Design 2

For the Anglo-Saxon notion, the phrase ‘Building Physics’ is somewhat unusual. ‘Building Science’ is the more usual expression. However the two do not cover the same � eld of knowledge. Building science in fact is broader in its approach. It encompasses all subjects related to buildings that claim to be ‘scienti� c’. This is especially clear when looking at journals that publish on ‘Building Science’. The range of subjects treated is remarkably wide, ranging from control issues in HVAC, to city planning and organizational issues.In this book on ‘Building Physics: Heat, Air and Moisture Transport’, the subject is the physics behind heat, air and moisture transfer in materials and building parts. Applied Building Physics deals with climate and the usability conditions confronting building physics. Furthermore, the performance concept and some performance requirements are discussed, and tables with material properties are given. Applied Building Physics and Performance Based Design 1 and 2 then use the performance concept as a tool in the design and construction of buildings. As such, both integrate the � elds of building construction, building materials, building physics and structural mechanics.

0.2 Building Physics

0.2.1 De� nition

Building Physics is an applied science that studies the hygrothermal, acoustical and light-related properties of building components (roofs, façades, windows, partition walls, etc.), rooms, buildings and building assemblies. The basic considerations are the user requirements for thermal, acoustic and visual comfort, the user’s health requisites and the more-or-less compelling demands and limitations imposed by architectural, material-related, economical and ecological considerations.The term ‘applied’ indicates that Building Physics is directed towards problem solving: the theory as a tool, not as a purpose. From the de� nition, the � eld covered contains three sub-sectors. The � rst, hygrothermal, stands for heat, air and moisture, and deals with heat, air and moisture transport in materials, building components and buildings, and between buildings and the outside environment. Speci� c topics are: thermal insulation and thermal inertia, moisture and temperature induced movements, strains and stresses; the moisture balance (rain, initial moisture, rising damp, sorption, surface condensation, interstitial condensation); salt transport; air-tightness and wind resistance; energy demand and energy consumption; ventilation of

��

��

1335vch00.indd 1 15.02.2007 11:49:06

Building Physics - Heat, Air and Moisture: Fundamentals and Engineering Methods with Examples and Exercises by Hugo Hens

Copyright © 2007 Ernst & Sohn Verlag für Architektur und technische Wissenschaften GmbH und Co. KG, Berlin

2

buildings, indoor air quality, wind comfort, etc. The second sub-sector, building acoustics, studies noise problems in and between buildings and their environment. The main topics are: air and impact noise transmission by walls, � oors, façades and roofs; room acoustics and the abatement of installation and environmental noises. Finally, the third sub-sector, lighting, addresses issues with respect to day-lighting, as well as arti� cial lighting and the impact of both on energy consumption.

0.2.2 Criteria

Building Physics has to deal with a variety of criteria: on the one hand requirements related to human comfort and health, on the other hand, architectural, material-related, economical and ecological facts and restrictions.

0.2.2.1 Comfort

Comfort means a state of mental and physical well-being. Attaining such a condition depends on a number of environmental and human factors. By thermal, acoustic and visual comfort we understand those qualities human beings unconsciously request from their environment in order to feel thermally, acoustically and visually comfortable when performing a given activity (not too cold, not too warm, not too noisy, no large contrasts in luminance, etc.).Thermal comfort is connected to global human physiology. As an exothermal creature maintaining a constant body temperature of about 37 °C (310 K), a human being should be able, under any circumstance, to release heat to the environment, be it through conduction, convection, radiation, perspiration, transpiration and breathing. The heat exchanges by means of these six mechanisms are determined by air temperature, air temperature gradient, radiant temperature, radiant asymmetry, contact temperature, relative air velocity, air turbulence and relative humidity in the direct environment. For a certain activity and clothing, some combinations of these parameters are experienced as being comfortable, others are not.Acoustic comfort is strongly connected to our mental awareness. Physically, young adults perceive sound frequencies between 20 and 16 000 Hz. Humans experience sound intensity logarithmically, with better hearing for higher than for lower frequencies. Consequently, acoustics works with logarithmical units: the decibel (dB). 0 dB stands for the audibility thre-shold, while 140 dB corresponds to the pain threshold. Human are easily disturbed by undesired noises, as the one the neighbour makes or the noise caused by traf� c, industry and aircraft.Visual comfort is both mentally and physically related. Physically, the eye is sensitive to electromagnetic waves with wavelengths between 0.38 and 0.78 �m. The maximum sensitivity lies near a wavelength of � 0.58 �m, the yellow-green light. Besides that, eye sensitivity adapts to the average luminance. For example, in the dark, sensitivity increases 10,000 times compared to daytime sensitivity. This adaptability lets the eye, just like the ear, react logarithmically. Too large differences in brightness are disturbing. Psychologically, lighting helps to create atmosphere.

0.2.2.2 Health

Health does not only mean the absence of illness but also the absence of neuro-vegetative complaints, of psychological stress and of physical unease. Human wellbeing may be menaced by dust, � bres, VOC’s, radon, CO, bacteria in the air, molds and mites on surfaces, too much noise in the immeadiate environment, local thermal discomfort, etc.

0 Introduction

1335vch00.indd 2 15.02.2007 11:49:06

3

0.2.2.3 Architectural and Material Facts

Building physics has to operate in an architectural and material framework. Floor, façade and roof form, aesthetics and the choice of materials are all elements which shape the building and whose design among others should be based on the performance requirements building physics imposes. Con� icting structural and physical requirements may make it dif� cult to � nd an ideal solution in every case. The need for a thermal cut, for example, may interfere with the strength and the need for stiffness-based interconnection. Waterproof-ness and vapour permeability are not always compatible. Acoustical absorption is opposed to vapour tightness. Certain materials may not remain wet for a long time, etc.

0.2.2.4 Economy

Not only must the investment in the building remain within the budget limits, also the total present value of all annual expenses should be as low as possible. In that respect, energy consumption, maintenance and building life expectancy play a role. A building which has been designed and constructed in accordance with the performance requirements of building physics, could generate a much lower total present value than buildings built without much consideration for usability.

0.2.2.5 Environment

Societal concern about environmental impact has increased substantially over the past years. Buildings have such impact locally, country-wide as well as globally. Locally their use produces solid, liquid and gaseous waste. Countrywide, building construction and occupancy accounts for 35–40% of the yearly primary energy use. A major part of that use is fossil fuel related, which means that the CO2-emissions by buildings closely follow that percentage. Globally, CO2-release is the most important of all gas emissions which are responsible for a global warming.That striving for more sustainability in building construction and usage is also re� ected in the increasing importance of life cycle analysis. In such analysis, buildings are evaluated in terms of environmental impact from ‘cradle to grave’, i.e. from the preliminary stage, through the construction and occupancy stage until demolition. For each stage, all material, energy and water in� ows as well as the polluting solid, liquid and gaseous out� ows are quanti� ed and their impact on human wellbeing and environment assessed.

0.3 Importance of Building Physics

Building physics is bound by the necessities of building, which means the creation of a comfortable indoor environment which protects human beings against the vagaries of the outside climate. As a consequence, the separation between the inside and the outside, i.e. the envelope of the building (� oors, façades, roofs) is submitted to numerous climatologic loads and climate differences (sun, rain, wind and outside noise; differences in temperatures, partial water vapour pressure and air pressure). An appropriate envelope design along with an appropriate detailing must care of these loads, attenuate them where possible or use them if possible in order to maintain the desired indoor climate with a minimum of technical means, with the least possible energy consumption and with a minimum of discomfort.

0.3 Importance of Building Physics

1335vch00.indd 3 15.02.2007 11:49:06

4

In earlier days, experience was the main guide. Former generations had only a limited range of construction materials (wood, straw, loam, natural stone, lead, copper and cast iron, blown glass), for which the know-how to use them had increased over the centuries. Standard details for attics, eaves and walls were used. From the size and orientation of the windows to the overall lay-out, buildings were constructed as to limit heating during winter and overheating during summer. Because noise sources were scarce, sound annoyance outside urban centres was unknown, and our ancestors saved on expensive energy (wood) by a lifestyle adapted to the seasons.The industrial revolution of the 19th century brought an end to this era. New materials inundated the marketplace: steel, reinforced and later pre-stressed concrete, nonferrous metals, synthetics, bitumen, insulation materials, etc. New technologies created new possibilities for existing materials: cast and � oat glass, rolled metal products, pressed bricks, etc. Better knowledge of structural mechanics and the related calculation methods allowed the construction of any form and span. Energy became cheap, � rst coal, then petroleum and � nally natural gas. Construction volume increased drastically, and buildings turned into a supply-demand market. The consequence was mass-building with a minimum of quality and, in the early 20th century, a ‘modern school’ of architects, who experimented with new structural solutions and new materials. These experiments had nothing to do with former knowledge and experience. Architects designed buildings without any concern either for energy consumption or comfort, nor any understanding of the physical parameters of the new façade and roof solutions they proposed. Typical for these buildings was the profuse application of steel, concrete and glass, which are all dif� cult materials from a hygrothermal point of view. The results were, and are, obvious cases of failure which could have been avoided through a better knowledge of building physics but instead led to severe damage as well as premature restoration of many twentieth century landmark buildings.Building physics is essential if we want to achieve high quality buildings that � t our purposes. The � eld should replace the time consuming practice of learning by experience, for which the evolution in technology and the subsequent changes in architectural fashion have become too rapid.

0.4 History of Building Physics

Building physics originates from a merger of three application-oriented disciplines: applied physics, building services and building construction.Already in the early 20th century, some circles of physicists showed a lot of interest in the application of noise control to building construction. In 1912, Berger submitted a Ph.D. thesis to the Technische Hochschule München entitled ‘Über die Schalldurchlässigkeit’ (about sound transmission). In the thirties and forties, Lothar Cremer was responsible for a break-through in understanding sound transmission. He recognized that the coincidence effect between sound waves in the air and bending waves on a wall played a major role in sound transmission and studied impact noise in detail. In 1920, Sabine published his well-known formula regarding the reverberation time in rooms.The application of lighting to buildings and civil engineering constructions came later. In 1931 a study was completed at the Universität Stuttgart. It dealt with ‘Der Ein� u� der Besonnung auf Lage und Breite von Wohnstra�en’ (The in� uence of solar irradiation on the location and width of residential streets).

0 Introduction

1335vch00.indd 4 15.02.2007 11:49:06

5

In the case of heat and mass transport, during the � rst half of the past century attention was mainly focused on thermal conductivity. In the thirties, measuring the diffusion resistance was added. In the U.S.A. Teesdale of the Forest Products Laboratory published a study in 1937 on ‘Condensation in Walls and Attics’. In 1952 an article appeared in the German journal ‚Der Gesundheitsingenieur’ from J. S. Cammerer on ‚Die Berechnung der Wasserdampfdiffusion in der Wänden’ (Calculation of Water Vapor Diffusion in Walls). Towards the end of the � fties, H. Glaser described in a series of articles in ‘Der Gesundheitsingenieur’ a calculation method for vapour transport by diffusion and interstitial condensation in cold storage walls. Others applied the method to building parts. In the sixties, more researchers got engaged in the study of combined heat and moisture transport. Some well-known names are O. Krischer, J. S. Cammerer and Helmut Künzel in Germany, A. De Vries and B. H. Vos in the Netherlands, L. E. Nevander in Sweden and A. Tveit in Norway.

In the 19th century, engineers were especially concerned with housing and urban hygiene. Max von Pettenkofer (1818–1901) was the � rst to perform research on the relationship between ventilation, CO2-concentration inside and indoor air quality. Later, the notion of “breathing materials” was derived from his work, the result of an erroneous explanation of the link between the air permeability of wall materials and the healthiness of the indoor environment. In the same period, building service technicians were searching for methods to calculate the heating and cooling load of buildings. For that purpose they took advantage of the knowledge developed by physicists who provided concepts such as the ‘thermal transmittance or U-value of a wall’. Organizations such as ASHVE, later ASHRAE, and VDI created technical committees, which dealt with the topics of heat loss and heat gain. An active member of ASHVE was W. H. Carrier (1876–1950), recognized in the U.S.A. as the ‘father’ of air conditioning, who was the � rst to publish a usable psychometric chart. In Germany, Herrmann Rietschel, Professor at the Technische Universtät Berlin and the author of a comprehensive book on ‘Heizungs und Lüftungstechnik’ (Heating and Ventilation Techniques), was also a pioneer in the � eld. Heat loss and heat gain through ventilation was one of his concerns. He and others learned by experience that well-designed ventilation systems did not function properly because of a lack of air-tightness of the building envelope. This sharpened the interest in the overall subject of air transport.

The interest in moisture by HVAC-engineers arose around the period that air conditioning (HVAC) became popular. But the subject was already one of concern, mainly because moisture appeared to be very damaging to the insulation quality of some materials and could cause health problems. They also included sound insulation in their work mainly because the HVAC installations could be quite noisy. Lighting was added because more and more HVAC engineers received contracts for lighting designs and lighting advice. From 1973 on, energy consumption became a concern. The fact that HVAC and building physics are tightly linked can still be observed in the U.S.A where ‘building science’ is often seen as a branch of mechanical engineering.

Finally, building physics helped construction to understand and avoid failures that arose when using solutions which were designed and built according to the existing “state-of-the art” rules. In the thirties, peeling and blistering of paints on insulated walls was an example (insulation materials were rather new at that time). This was precisely the problem which motivated Teesdale to make his study on condensation. Later, ventilated constructions became a research topic. Towards the end of the thirties, the � rst experimental work of its kind on roofs was performed by F. Rowley, Professor in Mechanical Engineering at the University of Minnesota, U.S.A. In Germany, the Freiland Versuchsstelle Holzkirchen, which was created in 1951, did similar work. When, after 1973, insulation became a necessity, the accrued

0.4 History of Building Physics

1335vch00.indd 5 15.02.2007 11:49:06

6

knowledge was very useful for the development of high quality, well insulated building parts. The demand for global quality and durability became part of the performance approach which is so widespread today.At the University of Louvain, Belgium, lectures in building physics were � rst given in 1952. This made it a pioneering establishment in Belgium. In those early years the courses were compulsory for architectural engineers and optional for civil engineers. A. de Grave, a civil engineer and head of the building department at the Ministry of Public Works, became the � rst professor. He taught until 1975, the year of his death. In 1957 he published the book ‘Building Physics’ (in Dutch), followed by ‘Oil Heating in the House’ (in Dutch) in the early seventies. Professor de Grave was a practical man, not a researcher. Former students still remember his enthusiastic way of teaching.

0.5 References

[0.1] CIB-W40 (1975). Quantities, Symbols and Units for the description of heat and moisture transfer in Buildings: Conversion factors, IBBC-TNP, Report No. BI-75-59/03.8.12, Rijswijk.

[0.2] Winkler Prins Technische Encyclopedie, deel 2 (1976). Article on Building Physics. Uitgeverij Elsevier, Amsterdam, pp. 157–159 (in Dutch).

[0.3] ISO-BIN (1985). Standards series X02-101 – X023-113.

[0.4] Donaldson, B., Nagengast, B. (1994). Heat and Cold: Mastering the Great Indoors. ASHRAE Publication, Atlanta.

[0.5] Kumaran, K. (1996). Task 3: Material Properties. Final Report IEA EXCO ECBCS Annex 24. ACCO, Louvain.

[0.6] Künzel, H. (2001). Bauphysik. Geschichte und Geschichten. Fraunhofer IRB-Verlag (in German).

0.6 Units and Symbols

The book uses the SI-system (internationally mandatory since 1977). Base units: the meter (m); the kilogram (kg); the second (s); the Kelvin (K); the ampere (A); the candela. Derived units, which are important when studying building physics, are:

Unit of force: Newton (N); 1 N = 1 kg · m · s–2

Unit of pressure: Pascal (Pa); 1 Pa = 1 N/m2 = 1 kg · m–1 · s–2

Unit of energy: Joule (J); 1 J = 1 N · m = 1 kg · m2 · s–2

Unit of power: Watt (W); 1 W = 1 J · s–1 = 1 kg · m2 · s–3

For the symbols, the ISO-standards (International Standardization Organization) are followed. If a quantity is not included in these standards, the CIB-W40 recommendations (International Council for Building Research, Studies and Documentation, Working Group ‘Heat and Moisture Transfer in Buildings’) and the list edited by Annex 24 of the IEA, ECBCS (International Energy Agency, Executive Committee on Energy Conservation in Buildings and Community Systems), are applied.

�

�

0 Introduction

1335vch00.indd 6 15.02.2007 11:49:06

7

Symbols and Units

Symbol Meaning Units

a

a

b

c

c

e

f

g

g

g

h

h

h

k

l

l

m

n

p

q

r

s

t

u

v

w

x, y, z

Acceleration

Thermal diffusivity

Thermal effusivity

Speci� c heat capacity

Concentration

Emissivity

Speci� c free energy

Temperature ratio

Speci� c free enthalpy

Acceleration by gravity

Mass � ow rate, mass � ux

Height

Speci� c enthalpy

Surface � lm coef� cient for heat transfer

Mass related permeability

Length

Speci� c enthalpy of evaporation or melting

Mass

Ventilation rate

Partial pressure

Heat � ow rate, heat � ux

Radius

Speci� c entropy

Time

Speci� c latent energy

Velocity

Moisture content

Cartesian co-ordinates

m/s2

m2/s

W/(m2 · K · s0.5)

J/(kg · K)

kg/m3, g/m3

–

J/kg

–

J/kg

m/s2

kg/(m2 · s)

m

J/kg

W/(m2 · K)

s

J/kg

kg

s–1, h–1

Pa

W/m2

m

J/(kg · K)

s

J/kg

m/s

kg/m3

m

0.6 Units and Symbols

1335vch00.indd 7 15.02.2007 11:49:06

8

Symbol Meaning Units

A

A

B

D

D

E

F

G

G

H

I

K

K

K

L

M

P

P

P

Q

R

R

S

T

T

U

U

V

W

X

Z

Water sorption coef� cient

Area

Water penetration coef� cient

Diffusion coef� cient

Moisture diffusivity

Irradiation

Free energy

Free enthalpy

Mass � ow (mass = vapour, water, air, salt)

Enthalpy

Radiation intensity

Thermal moisture diffusion coef� cient

Mass permeance (mass may be air, water …)

Force

Luminosity

Emittance

Power

Thermal permeance

Total pressure

Heat

Thermal resistance

Gas constant

Entropy, saturation degree

Absolute temperature

Period (of a vibration or a wave)

Latent energy

Thermal transmittance

Volume

Mass resistance (mass may be air, water …)

Moisture ratio

Diffusion resistance

kg/(m2 · s0.5)

m2

m/s0.5

m2/s

m2/s

W/m2

J

J

kg/s

J

J/rad

kg/(m · s · K)

s/m

N

W/m2

W/m2

W

W/(m2 · K)

Pa

J

m2 · K/W

J/(kg · K)

J/K, –

K

s, days, etc.

J

W/(m2 · K)

m3

m/s

kg/kg

m/s

0 Introduction

1335vch00.indd 8 15.02.2007 11:49:06

9

Symbol Meaning Units

�

�

�

�

�

�

�

�

�

�

�, , �

�

�

�

�

Thermal expansion coef� cient

Absorptivity

Surface � lm coef� cient for diffusion

Volumetric thermal expansion coef� cient

Dynamic viscosity

Temperature

Thermal conductivity

Vapour resistance factor

Kinematic viscosity

Density

Re� ectivity

Surface tension

Transmissivity

Relative humidity

Angle

Speci� c moisture capacity

Porosity

Volumetric moisture ratio

Heat � ow

K–1

–

s/m

K–1

N · s/m2

°C

W/(m · K)

–

m2/s

kg/m3

–

N/m

–

–

rad

kg/kg per unit of moisture

potential

–

m3/m3

W

Currently Used Suf� xes

Indices Meaning

a

c

e

h

i

cr

CO2, SO2

Air

Capillary, convection

Outside, outdoors

Hygroscopic

Inside, indoors

Critical

Chemical symbol for gasses

0.6 Units and Symbols

1335vch00.indd 9 15.02.2007 11:49:06

10

Indices Meaning

m

r

sat

s

rs

v

w

Moisture, maximal

Radiant, radiation

Saturation

Surface, area, suction

Resulting

Water vapour

Water

Relative humidity

Notation

[ ], bold Matrix, array, value of a complex numberdash Vector, e.g: a

0 Introduction

1335vch00.indd 10 15.02.2007 11:49:06

1 Heat Transfer

1.1 Overview

A � rst de� nition of heat is found in thermodynamics. Thermodynamics divides the world into systems and environments. A system can be anything: a volume of material, a building part, a building construction, a part of the heating system, etc. We can even look at a city as a system. ‘Heat’ indicates how energy is transferred between a system and its environment. Whereas ‘work’ is purposeful and organized, heat is diffuse and unorganized. A second de� nition can be found in particle physics in which heat denotes the statistically distributed kinetic energy of atoms and free electrons. In both cases, heat is the least noble, or most diffuse form of energy, to which each nobler form degrades according to the second law of thermodynamics.The ‘quality’ of heat is determined by its potential: the temperature. Higher temperatures mean a higher quality, which means higher mechanical energy of atoms and free electrons and the possibility to convert more heat into power via a cyclic process. That in turn is synonymous to a higher exergy. Lower temperature gives heat a lower quality, which means less mechanical energy on the atomic scale and less exergy. Higher temperatures are obtained by warming up the system, i.e. by adding heat, lower temperature are obtained by cooling down the system, i.e. by removing heat. Like any other potential, temperature is a scalar.Heat and temperature cannot be measured directly. Yet, temperature can be sensed and is indirectly measurable because a great deal of material properties depends on it:

Thermal expansion: for a mercury thermometer, we use the volumetric expansion of mercury as a measure for temperature.Change of electrical resistance: for a Pt100 resistance thermometer, the electrical resistance of a platinum wire functions as a measure for temperature.Change of contact potentials: this is the basis for the measurement of temperature with a thermocouple.

The SI-system uses 2 scales for temperature:

EmpiricalDegree Celsius (indicated as °C, symbol �). 0 °C is the triple point of water, 100 °C the boiling point of water at 1 Atmosphere.

ThermodynamicDegree Kelvin (indicated as K, symbol: T). 0 K is absolute zero, 273.15 K is the triple point of water.

The relation between both is:T = � + 273.15. Temperatures are given in °C or K, temperature differences in K.

Heat exists in sensitive form, which means temperature-related, or in latent form, which means as transformation heat. Sensitive heat is transferred by:

1. Conduction. Conduction refers to the energy transferred when vibrating atoms collide and free electrons move collectively. Heat is transferred that way between solids at different temperature in contact with each other and between points at different temperature within the same solid.

�

�

�

�

�

1335vch01.indd 11 12.02.2007 11:36:07

Building Physics - Heat, Air and Moisture: Fundamentals and Engineering Methods with Examples and Exercises by Hugo Hens

Copyright © 2007 Ernst & Sohn Verlag für Architektur und technische Wissenschaften GmbH und Co. KG, Berlin

12 1 Heat Transfer

The mode also intervenes when heat is transferred in gases and liquids and in the contacts between gases and liquids at one side and solids at the other. Conduction always occurs from points at a higher temperature to points at a lower temperature (2nd law of thermodynamics). The mode needs a medium. No observable macroscopic movement is linked to it.

2. Convection. Convection stands for the displacement of molecule groups at a different temperature. Convection is by nature a consequence of movement (transfer of enthalpy) and occurs in a pronounced way close to the contact between liquids and gases at one side and solids at the other. We distinguish forced, natural and mixed convection depending on whether or not the movement is caused by an external force, a difference in � uid density or both. In forced convection, work exerted by an exterior source may compel heat to � ow from low to high temperatures. Convection needs a medium. Actually, in liquids and gases, convection always includes conduction and even in convection mode, heat transfer between molecules occurs by conduction.

3. Radiation. Radiation refers to heat transfer, caused by the emission and absorption of electromagnetic waves. At temperatures above 0 K, each surface emits electromagnetic energy. Between surfaces at different temperature, that emission results in heat exchanges. Heat transfer through radiation does not need a medium. On the contrary, it is least hindered in vacuum and follows physical laws which diverge strongly from conduction and convection.

Latent heat moves along with a carrier, independent of temperature. Each time the carrier undergoes a change of state, related latent heat is converted into sensitive heat or vice versa. For example, when water evaporates, it absorbs sensitive heat in a quantity equal to its latent heat of evaporation. The water vapour then diffuses to a cooler spot (= transfer) where it may condense with reemission of the latent heat of evaporation as sensible heat. These conversions in� uence the temperature course as well as the � ow of sensitive heat in solid materials and building parts.

Why are heat and temperature so important in building physics? Heat � ow means energy consumption. Thermal comfort obliges to keep the operative temperature in buildings at the right level. That requires heating and cooling. Both are obtained by combusting fossil fuels, by converting work into heat or by transforming thermal electricity, hydraulic electricity or electricity generated by windmills and PV-cells into heat. In every developed country, heating and cooling buildings has a substantial share in the primary energy consumption. Fossil fuels are scarce and their combustion is responsible for environmental problems. Thermal electricity is energy-intensive to generate. Consequently, the design and construction of buildings with low heating and cooling demands is a challenge and a necessity. A low demand presumes a restricted heat � ow traversing the envelope.Temperatures in turn in� uence thermal comfort and durability. Suf� ciently high surface temperatures contribute to thermal comfort and a feeling of well-being. On the other hand, summer indoor temperatures which are too high affect a building’s usability. The higher the temperature differences between the layers in a building parts, the larger are the differential movements, the larger the thermal stresses and the higher the risk of crack formation. Temperatures below freezing may damage moist porous materials. High temperatures accelerate the chemical breakdown of synthetic materials. Differences in temperatures are responsible for displacement of moisture and salts in porous materials, etc. Whether or not we can control these temperature effects depends on how the building parts are designed and constructed.This chapter deals with heat transfer by conduction, convection and radiation and, the typical building physics related applications. But � rst some de� nitions:

1335vch01.indd 12 12.02.2007 11:36:08

13

HeatQuantity which indicates the energy exchange in the form of heat. Because energy is a scalar, heat is also a scalar.Symbol Q; unit [J] (Joule)

Heat � owThe heat per unit of time. Heat � ow is a measure for ‘power’. Similar to heat, heat � ow is a scalar.Symbol �; units [J/s] = [W] (Joule per second = Watt)

Heat � ow rateThe heat per unit of time that � ows through a unit of surface perpendicular to the � ow di-rection. The heat � ow rate is a vector with the same direction as the surface. Components: qx, qy, qz in Cartesian co-ordinates, qR, q�, q� in polar co-ordinates.Symbol q; unit [W/m2] (Watt per m2)

Solving a heat transfer problem means the determination of two � elds: that of temperatures and that of heat � ow rates. Consequently, we have two unknown quantities: T (x, y, z, t) and q (x, y, z, t). One is a scalar (T), the other a vector (q). Their computation demands a scalar and a vector equation.

1.2 Conduction

1.2.1 Conservation of Energy

For a � rst relation between heat � ow rate q and temperature T we go back to one of the axioms of classic physics: conservation of energy. In the case of pure conduction, no mass motion occurs between the system and the environment. As system we take an in� nitely small part dV of a material volume V, with the remaining volume forming the environment. Hence, for the energy equilibrium in the system dV, we have:

d d d dU WΦ + Ψ = + (1.1)

where d� is the resulting heat � ow by conduction between dV and the environment and d� is the dissipation energy released or absorbed per unit of time in dV. Dissipation can take on many forms: heat production as a result of an exothermic chemical reaction, heat absorption as a result of an endothermic chemical reaction, the Joule effect coupled to an electric current; the release or absorption of latent heat, etc. The symbol dU represents the change per unit of time of the internal energy in dV. Finally, dW is the labour exchanged per unit of time between the system dV and the environment (dW = P d (dV) = P d2V with P pressure). The equation states that the heat � owing between the environment and the system (= d�) plus the heat released or absorbed in the system is responsible for a change in internal energy in the system and exchange of labor between the system and the environment. If the process is isobaric (P = Cte), than equation (1.1) can be written as d (U + P dV) = dQ + dE. In this equation U + P dV represents the enthalpy H in the system dV. If we replace d� by

(div ) dV− q , dH by p[ ( ) / ] dc T t V∂ ρ ∂ , cp being the speci� c heat capacity at constant pressure and � being the density of the material (in kg/m3), and d� by �� dV, �� being the dissipation per unit of time and unit of volume, positive for a source and negative for a sink, than the law of conservation becomes:

�

�

�

1.2 Conduction

1335vch01.indd 13 12.02.2007 11:36:08

14 1 Heat Transfer

p( )div d 0

c TV

t

∂ ρ⎡ ⎤+ Φ + =′⎢ ⎥∂⎣ ⎦

q (1.2)

In the case of solids and liquids, the speci� c heat capacity at constant pressure (cp) differs very little from the speci� c heat capacity for other changes of state. This is why we call cp simply ‘speci� c heat capacity’. Symbol c, units: [J/(kg · K)]. The product � c is the volumetric speci� c heat capacity, unit: [J/(m3 · K)]. For gases instead, the speci� c heat capacity depends on the change of state. The speci� c heat capacity at constant pressure and speci� c heat capacity at constant volume are coupled by the following relationship: cp = cv + R with R the speci� c gas constant, [Pa · m3/(kg · K)]. As equation (1.2) holds for each in� nitely small volume of material (here ‘the system’), the equation turns into:

( )div

c T

t

∂ ρ= − − Φ′∂

q (1.3)

i.e, the scalar relationship between q and T that we need.

1.2.2 Fourier’s Laws

1.2.2.1 First Law

As vector relation between q and T, we have the empirical conduction law which was � rst formulated by the French physicist Fourier in the nineteen century:

T= −λ = −λ θq grad grad (1.4)

The equation states that the conductive heat � ow rate at a point in a solid, a liquid or a gas is proportional to the temperature gradient at that point. The proportionality factor is called the thermal conductivity or, in short, the -value of the solid, liquid or gas, units: [W/(m · K)].

Figure 1.1. Lines of equal temperature (called isotherms) and equal heat � ow rate (called iso� ux-lines).

1335vch01.indd 14 12.02.2007 11:36:08

15

The minus symbol in the equation (1.4) indicates that heat � ow rate and temperature gradient oppose each other. Thermodynamics teaches that without external work, heat always � ows from high temperature to low temperature areas. In the opposite case, entropy would decrease without energy supply from the environment, which is physically impossible. Since a gradient is positive from high to low, the symbol is minus.Equation (1.4) is known as Fourier’s � rst law. It is supported by the following experimental observation. If the surfaces linking all points at the same temperature in a material are constructed, then the heat � ow develops perpendicular to these surfaces and increases as the surfaces lie closer to each other (Figure 1.1). This increase happens proportionally to a property which is material-speci� c, the thermal conductivity. For reasons of simplicity, this property is supposed to be constant and scalar, even though with real materials none or only one of the two is true. For all porous materials, thermal conductivity is a function of temperature and moisture content. In anisotropic material the property becomes a tensor instead of a scalar. Fortunately, for � rst order calculations, the assumption ‘scalar and constant’ suf� ces.In right-angled Cartesian coordinates [x, y, z], equation (1.4) gives as heat � ow rates along each of the axes:

xT

qx x

∂ ∂θ= = −λ = −λ∂ ∂x xq u

yT

qy y

∂ ∂θ= = −λ = −λ∂ ∂y yq u

zT

qz z

∂ ∂θ= = −λ = −λ∂ ∂z zq u

The elementary heat � ow through a surface dA with direction n becomes:

2d d dn n n nA u An n

∂θ ∂θΦ = = −λ = −λ∂ ∂nq dA

Along the 3 axes that equation turns into:

d dx xAx

∂θΦ = −λ∂

d dy yAy

∂θΦ = −λ∂

d dz zAz

∂θΦ = −λ∂

1.2.2.2 Second Law

The system, formed by the energy balance (1.3) and the conduction law (1.4) allows calculating the temperature � eld. Combining both gives:

( )div ( )

c TT

t

∂ ρλ = − Φ′∂

grad (1.5)

If the thermal conductivity and volumetric speci� c heat capacity � c are constant, then the equation simpli� es to:

1.2 Conduction

1335vch01.indd 15 12.02.2007 11:36:08

16 1 Heat Transfer

2 c TT

t

ρ ∂ Φ′⎛ ⎞∇ = −⎜ ⎟⎝ ⎠λ ∂ λ (1.6)

with 2 as the Laplace operator. In Cartesian coordinates:

2 2 22

2 2 2x y z

∂ ∂ ∂∇ = + +∂ ∂ ∂

(1.7)

Equation (1.6) is known as Fourier’s second law.The further study focuses on solving the system formed by Fourier’s two laws for a series of cases that are relevant in building physics.

1.2.3 Steady State

1.2.3.1 What Is It?

Steady state means that temperatures and heat � ow rates are independent of time. In this respect, we need constant boundary conditions, time independent material properties and constant energy dissipation. All these, but especially the constant boundary conditions, distort reality. In the case that thermal conductivity (), volumetric speci� c heat capacity (�c) and energy dissipation (��) are set constant by de� nition, then a necessary and suf� cient condition for steady state is: / 0T t∂ ∂ = . Fourier’s second law then becomes (� instead of T):

2 Φ′∇ θ = −λ

(1.8)

1.2.3.2 One Dimension: Flat Walls

In one dimension, (1.8) converts into (temperature changes extending parallel to the x-axis):

2

2

d

dx

θ Φ′= −λ

(1.9)

which without dissipation becomes: 2

2

d0

dx

θ = .Solution:

1 2C x Cθ = + (1.10)

where C1 and C2 are the integration constants, de� ned by the boundary conditions. The linear equation (1.10) lies at the basis of the study of conduction through � at walls with the two wall surfaces at constant, although different temperature. Buildings contain numerous � at walls: � at roofs, the pitches of sloped roofs, facades, � oors, inner walls, glass surfaces, etc. In cross-section, � at walls can be single-layered, i.e. made out of one material, or composite, i.e. consisting of several material layers.

Single-layered Walls

Assume that the thermal conductivity of the material () and the thickness of the wall (d) are known. As boundary conditions we have: x = 0: � = �s1; x = d: � = �s2 with �s1 < �s2, of which �s1 and �s2, the temperatures on both wall surfaces, are given. As a result we obtain: C2 = �s1,

1 s2 s1( ) /C d= θ − θ . (1.10) then becomes:

1335vch01.indd 16 12.02.2007 11:36:08

17

s2 s1s1x

d

θ − θθ = + θ (1.11)

According to Fourier’s First Law the heat � ow rate is:

s2 s1d

dx d

θ − θθ= −λ θ = −λ = −λq grad (1.12)

which means that the heat per square meter � ows in the negative x-direction, from high to low temperature. In absolute terms, the heat � ow rate becomes:

s2 s1qd

θ − θ= λ (1.13)

According to the equations (1.11) to (1.13), the temperature in a � at, single-layered wall, which is in steady state and contains no heat source or sink, varies linearly between the temperatures on both wall surfaces (Figure 1.2). The heat � ow rate is a constant, directly proportional to the thermal conductivity of the material () and the temperature difference between the two surfaces, and inversely proportionate to the thickness of the wall. For a given thickness and a given temperature difference, a lower -value decreases the heat � ow rate, which means lower heat loss or heat gain. This is the reason why materials with a low thermal conductivity are called thermal insulation materials.We can rewrite equation (1.13) as:

qdΔθ=

λ

(1.14)

where d / is the speci� c thermal resistance of the single-layered wall, in short called the thermal resistance. Symbol R, units m2 · K/W. The higher the thermal resistance, the smaller the heat � ow rate for a given temperature difference, or, which is the same, the better the wall insulates. A higher thermal resistance can be obtained by increasing the wall thickness or by using an insulation material, the latter being the most economical option. The inverse of the thermal resistance is called the thermal permeance. Symbol P, units W/(m2 · K). The thermal permeance tells how much heat per unit of time and unit of surface passes through a single-layered wall at a temperature difference of 1 K between both wall surfaces.

Figure 1.2. Single-layered wall.

1.2 Conduction

1335vch01.indd 17 12.02.2007 11:36:08

18 1 Heat Transfer

Composite Walls

Each wall made up of two or more plane-parallel layers is called ‘composite’ (see Figure 1.3). From an architectural point of view, composite walls are the reference. The simplest example is a plastered wall: 3 layers, consisting of two plaster layers and one brickwork layer.

Figure 1.3. Composite wall.

Heat Flow Rate

In steady state, without heat sources or sinks, the heat � ow rate must be identical in each layer. If not, conservation of energy would allow for thermal storage or discharge, and thus the regime would become time-dependant, that is, non steady state. Suppose the temperature is �s1 on wall surface 1 and �s2 on wall surface 2. If the thermal conductivity i of the materials and the thickness di of all layers are known, then we may write per layer (�s1 < �s2):

layer 1 1 s11

1

qd

θ − θ= λ

layer 2 2 12

2

qd

θ − θ= λ

…

layer n – 1 1 21

1

n nn

n

qd

− −−

−

θ − θ= λ

layer n s2 1nn

n

qd

−θ − θ= λ

where �1, �2, …, �n – 1 are the unknown interfaces temperatures between layers and q the constant heat � ow rate in the wall. Rearrangement and summation give:

1335vch01.indd 18 12.02.2007 11:36:09

19

1

1

dqλ

=

1 1sθ − θ

2

2

dqλ

=

2 1θ − θ

…

n

n

dqλ

= s2 1n−θ − θ

Sum: 1

ni

i i

dq

=

⎛ ⎞⎜ ⎟λ⎝ ⎠

∑

= s2 s1θ − θ

or: s2 s1

1( / )

n

i ii

qd

=

θ − θ=

λ∑ (1.15)

We call the sum � (di / i) the total thermal resistance of the composite wall. The symbol is RT, units m2 · K/W. The ratio di / i represents the thermal resistance of layer i, symbol Ri, units m2 · K/W. The higher the total thermal resistance, the lower the steady state heat � ow rate and the better the wall insulates. A high thermal resistance is obtained by incorporating a suf� ciently thick insulation layer in the wall. This procedure is known as ‘insulating the wall’. Since the energy crisis of 1973–1979 and the fear of a global warming, a high performing thermal insulation of all walls in the building envelope is seen as a very ef� cient means to lower energy consumption for heating and cooling.Wall design � xes the thermal resistance. Because the total thermal resistance is the sum of the thermal resistances of the separate layers, the commutation property applies: the sequence of the terms, which means the sequence of the layers, does not change the value of the sum. As a result, inside insulation should be equivalent to outside insulation or to insulation within the wall. From a building physics point of view, this mathematical conclusion is not correct. Indeed, for the same insulation thickness, there is a difference in overall performance between inside and outside insulation. However, a one dimensional steady state model does not allow highlighting these differences.At this point, the analogy between the equations (1.14) and (1.15) and the equation for the current in an electrical circuit with one or more electric resistances in series, which are subjected to a voltage difference �V (i = �V / � Rei) is quite instructive. In fact, temperature replaces volt-age, conductive heat � ow rate the electric current and thermal resistance the electrical resistance. This allows converting thermal problems into an electrical analogy (Figure 1.4). Indeed, the conduction equation is mathematically identical to the equation for the electric � eld.

Figure 1.4. The electrical analogy.

1.2 Conduction

1335vch01.indd 19 12.02.2007 11:36:09

20 1 Heat Transfer

Temperatures

When both surface temperatures are known, then temperature in all interfaces can be calculated. In fact, as for a single-layered wall, temperatures in a layer form a straight line between the temperatures in the interface with both adjacent layers. These in turn are found through a rearrangement of the preceding equations (1.15):

1 11 s1 s1 s2 s1

1 T

( )d R

qR

θ = θ + = θ + θ − θλ

2 1 22 1 1 2 s1 s2 s1

2 T

( )( )

d R Rq q R

R

+θ = θ + = θ + = θ + θ − θ

λ

…

11 s1 s2 s1

T

( )

i

ii i

i ii

Rd

qR=

−θ = θ + = θ + θ − θλ

∑

1

1 11 2 s1 s2 s1

1 T

( )

n

in i

n nn

Rd

qR

−

− =− −

−θ = θ + = θ + θ − θ

λ

∑

So, if we know the thermal conductivity i and thickness di of all layers and the temperature at both wall surfaces, then temperature in each interface can be calculated. Reshuf� ing

1 /i i i iq d−θ = θ + λ into 1( ) /i i iq−θ − θ = λ , shows that the temperature gradient in each layer is inversely proportional to the thermal conductivity. Hence, gradients are large in an insulating material (i small) and small in a conductive material (i large).

The equation s1 s2 s1 T1

( ) /i

i ii

R R=

θ = θ + θ − θ ∑ can be rewritten as:

s1 s1x

x q Rθ = θ + (1.16)

with s1xR the thermal resistance between surface s1 and the interface x in the wall. If the

calculation starts at surface s2, then equation (1.16) becomes: s2 s2x

x q Rθ = θ − . In the [R, �] coordinate system (abscissa R, ordinate �, further called the [R, �]-plane or [R, �]-graph), both relations represent a straight line through the points (0, �s1) and (R, �s2) with the heat � ow rate q as the slope. Or, in a [R, �]-plane, a composite wall behaves as a single-layered wall. Consequently, the graphical construction of the temperature line is an easy job. Draw the wall in the said plane with the layers as thick as their thermal resistance, mark on wall surface s1 temperature �s1, on wall surface s2 temperature �s2 and trace a straight line in between. The correct geometric temperature curve is then obtained by transposing the intersections between the straight line and the successive interfaces into the wall’s thickness graph and linking the successive points with line sections. During the whole operation, the layers have to keep their correct sequence (Figure 1.5).

Heat Flow

For single-layered as well as composite walls, heat � ow is given by the product of the heat � ow rate through the wall with the wall’s surface (A):

q AΦ = (1.17)

1335vch01.indd 20 12.02.2007 11:36:09

21

Single-layered Wall with Variable Thermal Conductivity

Each time we have a temperature difference over, or a moisture content pro� le in a wall, thermal conductivity () becomes a function of temperature (� or the ordinate (x) (Figure 1.6).In case varies linearly with temperature ( = o + a � in a single-layered wall�� the heat � ow rate follows from (steady state, x = 0, � = �s1; x = d, � = �s2)

s2

s1

00

( ) d dd

a q xθ

θλ + θ θ =∫ ∫

giving as a solution:

2 2s2 s1

0 s2 s1( )

( )2

aq d

θ − θλ θ − θ + =

or:

s2 s1m( )q

d

θ − θ= λ θ (1.18)

with m( )λ θ the thermal conductivity at the average temperature in the single-layered wall. The thermal resistance becomes m/ ( )d λ θ . As a temperature curve we obtain a parabola of the form 2

0/ 2a q x Cθ + λ θ = + .

Figure 1.5. Graphical construction of the temperature line in a composite wall.

1.2 Conduction

1335vch01.indd 21 12.02.2007 11:36:09

22 1 Heat Transfer

If the wall is moist with a moisture content pro� le of the form w(x), so that ( ( )) ( )F w x f xλ = = , then the heat � ow rate becomes (x = 0: � = �s1; x = d: � = �s2)

s2

s10

dd

( )

d xq q

x

θ

θ=

λ∫ ∫

If moisture is distributed in such way that thermal conductivity increases proportional to x ( = 0 + a x), that equation gives as a solution:

s2 s1

0

0

1ln

qa d

a

θ − θ=

⎛ ⎞λ +⎜ ⎟λ⎝ ⎠

(1.19)

As for a wall with constant thermal conductivity, the denominator represents thermal resistance. The temperature curve becomes:

0s1

0

1lnx

a xq

a

⎡ ⎤⎛ ⎞λ +θ = θ + ⎢ ⎥⎜ ⎟λ⎝ ⎠⎣ ⎦

(1.20)

Single-layered Wall with Heat Source or Sink Spread Over the Thickness

In building parts, processes may take place that dissipate heat. Hydration in concrete, carbonation in lime plaster, condensation of water vapour, drying, etc., are examples of that. Suppose that �� joule is released or absorbed per unit of time and unit of volume in a single-layered wall. If �� is a constant (which is not the case for most processes), the equation for the steady state heat transfer becomes:

2

2

d

dx

θ Φ′= −λ

Figure 1.6. Wet wall, moisture pro� le turns thermal conductivity into a function of the ordinate x.

1335vch01.indd 22 12.02.2007 11:36:10

23

with as boundary conditions: x = 0: � = �s1; x = d: � = �s2 (�s1 < �s2).

The solution is:

21 22

x C x CΦ′θ = − + +λ

(1.21)

i.e. a parabolic temperature curve. For a heat source, this parabola is convex, for a sink it is concave (Figure 1.7).The heat � ow rate becomes:

1d

dq x C

x

θ= −λ = Φ − λ′ (1.22)

Opposite to the case with no heat dissipation, now the heat � ow rate changes from point to point in the wall. The boundary conditions give as the value for the integration constants:

s2 s11 2 s1,

2

dC C

d

θ − θ Φ ′= + = θλ

Composite Wall with Local Heat Source or Sink

If condensate is deposited in the interface between two layers in a wall, heat of evapouration is released there. Drying of that condensate causes the same heat of evapouration to be absorbed again. As a result, a local heat source and sink is activated. Assume that q� W/m2 heat is released and that surface temperature �s1 is higher than surface temperature �s2 (Figure 1.8).The temperature line in the wall is then found by writing a steady state heat balance at the interface x which contains the source or sink. Heat is supposed to � ow from the environment to x. Hence, according to conservation of energy, the sum of all heat � ow rates in x should be zero.

Figure 1.7. Uniform steady state heat source and heat sink in a single-layered wall: temperature lines.

1.2 Conduction

1335vch01.indd 23 12.02.2007 11:36:10

24 1 Heat Transfer

As heat � ow rates we have:

Conduction between surface s1 and x: s1s1

s1

x xx

qR

θ − θ=

Conduction between surface s2 and x: s2s2

s2

x xx

qR

θ − θ=

Dissipated heat q� in x. In case of drying, q� is negative, in case condensate is deposited, q� is positive.

In the two equations, �x is the unknown temperature in the interface x. Sum zero gives:

s2 s1 s1 s2 s1 s2

s1 s2

x x x x

x x x

R R q R R

R R

θ + θ + ′θ =

+ (1.23)

We obtain the conductive heat � ow rates between the surface s1 and x and the surface s2 and x by introducing this result into the equations for the heat � ow rates s1

xq and s2xq :

s1 s2 s2s1

T T

xx q R

qR R

⎛ ⎞θ − θ ′= −⎜ ⎟⎝ ⎠

s1 s2 s1s2

T T

xx q R

qR R

⎡ ⎤⎛ ⎞θ − θ ′= − +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦ (1.24)

Here RT s1 s2( )x xR R= + is the total thermal resistance of the wall. If the dissipated heat q� is positive, the ingoing heat � ow rate decreases and the outgoing heat � ow rate increases compared to no dissipation. If the dissipated heat q� is negative, then ingoing is larger and outgoing smaller.

1.2.3.3 Two Dimensions: Cylinder Symmetry

In cylinder co-ordinates, cylinder symmetric problems behave one-dimensionally. Pipes are an example of that.

�

�

�

Figure 1.8. Local heat source.

1335vch01.indd 24 12.02.2007 11:36:10

25

Questions to answer with pipes are: heat loss per meter run, pipe temperature, and how ef� cient is pipe insulation? Consider a pipe with inner radius r1 and outer radius r2. The temperature on the inner surface is �s1, the temperature on the outer surface �s2. The heat loss per meter run follows from conservation of energy (Figure 1.9): in steady state, in the absence of dissipation, the same heat � ow will pass each cylinder concentric to the pipe. With the centre of the pipe as origin, that balance per meter run gives: te(2 ) d / dr r CΦ = −λ π θ = . Integration turns that expression into:

s22

1 s1

d2 d

r

r

r

r

θ

θΦ = − π λ θ∫ ∫ ,

or:

s1 s2

2 1ln( / )

2

r rθ − θ

Φ =

π λ

(1.25)

In that equation, 2 1ln ( / ) / 2r r π λ is the equivalent of the thermal resistance R of a � at wall. We call the quantity the thermal resistance of a pipe per meter run. Units: m · K/W. For a composite pipe (for example an insulated pipe) a reasoning analogous to that of a composite � at wall allows the heat � ow per meter run to be written as:

s1 s2

1

1

ln ( / )

2

ni i

i i

r r+

=

θ − θΦ =

⎡ ⎤⎢ ⎥π λ⎣ ⎦

∑ (1.26)

The temperature line follows from:

11 s1

1

ln ( / )

2

ii i

ii i

r r++

=

⎡ ⎤θ = θ + Φ ⎢ ⎥π λ⎣ ⎦

∑ (1.27)

1.2.3.4 Two and Three Dimensions: Thermal Bridges

Overview

When looking in detail at the heat transfer through facades, roofs, � oors, inner walls etc., it must be clear that the hypothesis ‘� at’ does not apply everywhere. What about the reveals around a window; the corner between two outer walls; the corner point between two outer walls and

Figure 1.9. The pipe problem.

1.2 Conduction

1335vch01.indd 25 12.02.2007 11:36:11

26 1 Heat Transfer

a roof, etc. (Figure 1.10). Also, the two surfaces of � at walls are not necessarily isothermal (Figure 1.11). To study steady state heat transfer in these cases, we have to return to:

2 2 2

2 2 2x y z

∂ θ ∂ θ ∂ θ Φ′+ + = ±λ∂ ∂ ∂

or, without dissipation:

2 2 2

2 2 20

x y z

∂ θ ∂ θ ∂ θ+ + =∂ ∂ ∂

Figure 1.10. Locations in the envelope, where heat transfer develops in two- and three dimensions.

Figure 1.11. Wall with non-isothermal inner surface.

1335vch01.indd 26 12.02.2007 11:36:11

27

For some very simple cases (one single material, easy geometry, simple boundary conditions), this partial differential equation can be solved analytically. For the majority of building details, however, a numerical solution is the only possibility.

Control Volume Method (CVM)

Many details in building construction are composed of rectangular shaped material volumes. To apply CVM in such case we mesh the detail in elementary cube or beam-like volumes. Energy conservation is written per volume. In steady state: the sum of heat � ows from all adjacent volumes to each central volume zero (Figure 1.12). When constructing the mesh, we can select either of two options with respect to the interfaces between the various materials:

Meshing coincides with the interfaces between materials. In that case, all volumes are homogenous. However the calculation results do not directly give the interface temperatures.Meshing is done in such a way that volume centers lie on the interfaces. Now the volumes are not material homogenous but the calculation directly gives the interface temperatures.

In practice, the second method is applied. Along the x-, y- and z-axis, the side of the control volume around a mesh point, called hereafter a point, is equal to the sum of half the distances in the negative and positive directions on the axis considered between that point and the adjacent points (in three dimensions six in total: 2 in x-, 2 in y- and 2 in z-direction/in two dimensions four in total: 2 in x- and 2 in y-direction).

Figure 1.12. Control volumes (CVM).

Some Examples

Central Volume and Adjacent Volumes Situated in the Same Material

If the mesh width of the raster equals a along the three axes, then the equations is:

Heat � ow of (l – 1, m, n) to (l, m, n) (= central point):

21, , , ,, ,

1, , 1, , , ,

( )( )l m n l m nl m n

l m n l m n l m n

aa

a−

− −θ − θ

Φ = λ = λ θ − θ

�

�

1.2 Conduction

1335vch01.indd 27 12.02.2007 11:36:11

28 1 Heat Transfer

Heat � ow from the � ve remaining adjacent points:

21, , , ,, ,

1, , 1, , , ,

( )( )l m n l m nl m n

l m n l m n l m n

aa

a+

+ +θ − θ

Φ = λ = λ θ − θ

2, 1, , ,, ,

, -1, , 1, , ,

( )( )l m n l m nl m n

l m n l m n l m n

aa

a−

−θ − θ

Φ = λ = λ θ − θ

2, 1, , ,, ,

, 1, , 1, , ,

( )( )l m n l m nl m n

l m n l m n l m n

aa

a+

+ +θ − θ

Φ = λ = λ θ − θ

2, , 1 , ,, ,

, , 1 , , 1 , ,

( )( )l m n l m nl m n

l m n l m n l m n

aa

a−

− −θ − θ

Φ = λ = λ θ − θ

2, , 1 , ,, ,

, , 1 , , 1 , ,

( )( )l m n l m nl m n

l m n l m n l m n

aa

a+

+ +θ − θ

Φ = λ = λ θ − θ

Adding and sum zero gives:

1, , 1, , , 1, , 1, , , 1 , , 1 , ,6 0l m n l m n l m n l m n l m n l m n l m n− + − + − +θ + θ + θ + θ + θ + θ − θ =

i.e. a linear equation with 7 unknowns: the temperature in the central point and the temperatures in the 6 adjacent points. In two dimensions we obtain a linear equation with 5 unknowns:

1, 1, , 1 , 1 ,4 0l m l m l m l m l m− + − +θ + θ + θ + θ − θ =

Central Volume Spanning the Interface Between Two Materials

If we assume that the interface is parallel with the [x, y] plane and that thermal conductivity is represented by 1 and 2 then for a mesh width a along the three axes, the following equations are obtained:

Heat � ow from (l – 1, m, n) to (l, m, n) (= central point):

2 21, , , , 1, , , ,, ,

1, , 1 2

( ) ( )

2 2l m n l m n l m n l m nl m n

l m n

a a

a a− −

−θ − θ θ − θ

Φ = λ + λ

or:

1 2 1, , , ,, ,1, ,

( ) ( )

2l m n l m nl m n

l m n

a −−

λ + λ θ − θΦ =

Heat � ow from the � ve remaining adjacent points:

1 2 1, , , ,, ,1, ,

( ) ( )

2l m n l m nl m n

l m n

a ++

λ + λ θ − θΦ =

1 2 , 1, , ,, ,, 1,

( ) ( )

2l m n l m nl m n

l m n

a −−

λ + λ θ − θΦ =

1 2 , 1, , ,, ,, 1,

( ) ( )

2l m n l m nl m n

l m n

a ++

λ + λ θ − θΦ =

, ,, , 1 1 , , 1 , ,( )l m n

l m n l m n l m na− −Φ = λ θ − θ

, ,, , 1 2 , , 1 , ,( ) l m n

l m n l m n l m na+ +Φ = λ θ − θ

1335vch01.indd 28 12.02.2007 11:36:11

29

Adding and sum zero gives:

1 2 1, , 1, , , 1, , 1,

2 , , 1 1 , , 1 1 2 , ,

( ) ( )

23 ( ) 0

l m n l m n l m n l m n

l m n l m n l m n

− + − +

− +

λ + λ θ + θ + θ + θ

+ λ θ + λ θ − λ + λ θ =

i.e. again a linear equation with 7 unknowns: the temperature in the central point and the temperatures in the 6 adjacent points. In two dimensions a linear equation with 5 unknowns is obtained:

1 2 1, 1,2 , 1 1 , 1 1 2 , ,

( ) ( )2 ( ) 0

2l m l m

l m l m l m n− +

− +λ + λ θ − θ

+ λ θ + λ θ − λ + λ θ =

Central Volume Lying on the Intersection Between Three Materials

Suppose that the interfaces are parallel to the [x, y] and [y, z]-plane. The thermal conductivity of the three materials is 1, 2 and 3. The following sum is obtained:

1, , 1, , , 1,2 3 2 3 3 , 1, 1 2( ) ( ) ( )

2 2 2l m n l m n l m n

l m n− + +

−θ θ θ

λ + λ + λ + λ + λ θ + λ + λ

, , 1 , , 1 , ,1 2 3 1 2 3( ) (3 3 6 ) 0

4 2l m n l m n l m n− +θ − θ θ

λ + λ + λ − λ + λ + λ =

i.e. a linear equation with 7 unknowns. In two dimensions, the result is a linear equation with 5 unknowns:

1, , 1 , 13 1, 1 2 2 3 1 3 1 2 3 ,( ) ( ) ( ) ( ) 0

2 2 2l m l m l m

l m l m+ − +

−θ θ θ

λ θ + λ +λ + λ +λ + λ +λ − λ +λ +λ θ =

A solution for all other cases is found the same way. In three dimensions, a control volume may include a maximum of eight materials. In two dimensions that maximum is four. Per calculation point, we obtain one linear equation with 7 (three dimensions) or 5 unknowns (two dimensions). In the case of p points, the � nal result is a system of p equations with p unknowns: the p temperatures in these p points. Solving the system gives the temperature distribution. The above equations then allow calculating the heat � ow rate along the x and y or the x, y and z-axis. That way, we get the components of the heat � ow rate vector.CVM leads to the following general algorithm. Suppose Ps is the surface-linked thermal permeance between two adjacent points (units W/K). If the two points lie in the same material, then s ( / )P d A= λ . If they lie in different materials, then that surface-linked thermal permeance becomes a serial and/or parallel circuit of permeances (Figure 1.13). Thus, for the heat � ow from an adjacent point to the central point, the following applies:

, , , ,1, , s, 1, , 1, , , ,( )l m n l m n

l m n l m ns l m n l m nP− − −Φ = θ − θ (1.28)

The sum over all adjacent points becomes:

s, , , s,, , , ,1 1

[ ] 0i j i j l m n i ji l m n i l m nj j

P P+ + += ==± =±

θ − θ =∑ ∑ (1.29)

Consequently, writing the equations only demands calculating the surface-linked thermal permeances Ps. Points with known temperatures act as known terms and are transferred to the

1.2 Conduction

1335vch01.indd 29 12.02.2007 11:36:12

30 1 Heat Transfer

right. In such a manner, each building detail is converted into a system of equations having the form [Ps]p,p [�]p = [Ps,i,j,k �i,j,k]p with [Ps]p,p the p rows, p columns permeance matrix, [�]p the column matrix of the p unknown temperatures and [Ps,i,j,k �i,j,k]p the column matrix of the p known temperatures.The accuracy of a CVM-calculation depends on the mesh width. The smaller it is, the closer the numerical solution approaches the exact solution. However, a smaller mesh width results in a larger system of equations. An in� nitely � ne mesh gives the exact solution. The price, however, is an in� nite number of equations. Consequently, a compromise between accuracy and calculation work has to be sought. An adequate choice of the mesh (large widths where small temperature gradients are expected and small mesh widths where large temperature gradients are expected) minimizes the differences between the numerical and exact solution. How to apply CVM is the subject of the standard EN-ISO 10 211-02. During the past decades, the method has been translated into powerful software packages allowing thermal evaluation of complex building details (Trysco®, Heat®, etc.).

1.2.4 Transient Regime

1.2.4.1 What is Transient?

Transient means that temperatures and heat � ow rates change with time. Varying material properties, a time-dependent heat dissipation and changing boundary conditions may be responsible. In the latter case, Fournier’s second law becomes:

2 c

t

ρ ∂θ∇ θ =λ ∂

Boundary conditions (surface temperature and heat � ow rate) may vary periodically or non-periodically, periodically when a regular � uctuation occurs, non-periodically when a once-only change occurs. For example, the outside temperature changes over a period of 1 year, n days, 1 day and 1/x days (Figure 1.14). Heating-up instead is responsible for a non periodic

Figure 1.13. Thermal permeances around a point.

1335vch01.indd 30 12.02.2007 11:36:12

31

increase of the inside temperature (Figure 1.15). Such a ‘jump’ is one of the non-periodical classics, the other being the Dirac impulse: a sudden increase, instantaneously followed by a decrease of the same size.

1.2.4.2 Flat Walls, Periodic Boundary Conditions

For a � at wall, Fournier’s second law simpli� es to

2

2

c

tx

∂ θ ρ ∂θ=λ ∂∂

or, with ac

λ=ρ

:

2

2a

tx

∂ θ ∂θ=∂∂

a is called the thermal diffusivity of the material. Units: m2/s. The thermal diffusivity indicates how easily a local temperature change spreads through a material. The larger it is, the faster that happens. A large thermal diffusivity requires a light material with high thermal conductivity

Figure 1.14. Periodical changes in temperature.

Figure 1.15. Non-periodical change in temperature.

1.2 Conduction

1335vch01.indd 31 12.02.2007 11:36:12

32 1 Heat Transfer

or a heavy material with low speci� c heat capacity. None of them exist. Light materials have a low thermal conductivity, while the speci� c heat capacity does not differ substantially between light and heavy materials. The majority of materials therefore have similar thermal diffusivities. Metals are the only exception.If we substitute the ordinate x by the thermal resistance R (= x / )�, �which means multiplying both terms with 2, then Fourier’s second law becomes:

2

2 tRc∂ θ ∂θ= ρ λ

∂∂ (1.30)

with a heat � ow rate of:

qR

∂θ= −∂

Boundary Conditions

Periodic means that temperature and heat � ow rate on both wall surfaces � uctuates with time. Each periodic signal with base period T can be transformed into a Fourier series. For the surface temperature:

s0s s s

1

2 2sin cos

2 n nn

B n t n tA B

T T

∞

=

⎡ ⎤π π⎛ ⎞ ⎛ ⎞θ = + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∑

with

sn s0

2 2( ) sin d

T n tA t t

T T

π⎛ ⎞= θ ⎜ ⎟⎝ ⎠∫

s0

2 2( ) cos d

T

n sn t

B t tT T

π⎛ ⎞= θ ⎜ ⎟⎝ ⎠∫

Bs0 /2 is the average value over the base period T. As1, As2…, Asn, Bs1, Bs2…, Bsn are the harmonics of 1st, 2nd, …, nth order. We recall that the suf� x s refers to ‘surface’. Euler’s formulas allow writing the harmonics in complex form:

exp( ) exp( )sin ( )

2

i x i xx

i

− −=

exp ( ) exp ( )cos( )

2

i x i xx

+ −=

or, for 2 /x n t T= π :

s1 2

( ) exp 2s n

n

i n tt

T

∞

=−∞

⎡ ⎤π⎛ ⎞θ = α ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∑ (1.31)

where �sn = Bsn – i Asn. �sn is the nth complex surface temperature. The amplitude and phase shift are:

Amplitude 2 2s sn nB A+

Phase shift s

s

atan⎛ ⎞−⎜ ⎟⎝ ⎠

n

n

A

B

The amplitude gives the size of the complex temperature, while the phase shift represents the time delay with respect to a cosine function with period / (2 )T n π (in radians).

1335vch01.indd 32 12.02.2007 11:36:13

33

Example: The Air Temperature at Ukkel, Belgium

The average monthly air temperature in Ukkel, Belgium, is approximated by (Figure 1.16):

e2

9.45 7.18 cos ( 2.828)365.25

tπ⎛ ⎞θ = + − −⎜ ⎟⎝ ⎠

where t is the time in days (365.25 is a period of 1 year, corrected for the leap years), 9.45 the average annual temperature �em, 7.18 the annual amplitude �e1 and 2.829 the related phase shift �e1 (in radians). Written as a Fourier series, the equation becomes:

e2 2

9.45 2.21 sin 6.83 cos365.25 365.25

t tπ π⎛ ⎞ ⎛ ⎞θ = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

where: B0 = 18.9; A1 = –2.21 and B1 = –6.83.

Amplitude and phase shift are deduced as follows:

e1 e1 e1

2 22.21 sin 6.83 cos

365.25 365.25

2 2sin ( ) sin cos ( ) cos

365.25 365.25

π π⎛ ⎞ ⎛ ⎞− − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎡ ⎤π π⎛ ⎞ ⎛ ⎞θ φ + φ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

t t

t t

for �e1 sin(�e1) = –2.21 and �e1 cos(�e1) = –6.83.

Solution of this system:

e1 e1( ) 0.335 or 2.828 radt g φ = φ = −

2 2e1

ˆ ( 2.21) ( 6.83) 7.18 °Cθ = − + − =

Figure 1.16. Monthly average air temperature in Ukkel, Belgium. The black circles show the measured averages for the period 1901–1930. The thick line gives the result of a Fourier analysis with 1 harmonic. The thinner lines represent the annual mean and the sine and cosine terms.

1.2 Conduction

1335vch01.indd 33 12.02.2007 11:36:13

34 1 Heat Transfer

Graphically, 7.18 cos [2 365.25 ( 2.828)] tπ − − is a rotating vector with value 7.18 (= �e1), which starts rotating from an angle –2.828 (= �e1) with the real axis. As a complex number this vector becomes:

e1 e1 e1 1 e1 1 1cos( ) sin ( ) 6.83 2.21α = θ φ + θ φ = + = − −ei B i A i

We call �e1 the annual complex outside air temperature for Ukkel. Such complex temperature contains all information on the annual � uctuation: an amplitude 7.18 °C and a phase shift –2.828 rad.

Solution

Suppose a periodic temperature or a periodic heat � ow rate is applied at time zero on one or both wall surfaces. The wall responds in a twofold way: by a transient, which dies away slowly, and by a periodic. The latter includes the same harmonics as the boundary signal (a wall cannot compress (smaller period) nor extend (larger period) a thermal signal). However, the amplitudes of both the temperature and the heat � ow rate decrease as the signal travels through the wall and dampening takes place. At the same time, while progressing in the wall, both signals gradually drop behind the boundary signal: a phase shift develops. Or, the two signals in the wall behave as complex numbers. The solution can be represented by a Fourier series, in which the complex temperature and complex heat � ow rate are functions of thermal resistance:

s

1 2( , ) ( ) exp

2

d ( , ) 1 2( ) ( ) exp

d 2

nn

nn

i n tR t R

T

R t i n tq t R

R T

∞

=−∞

∞

=−∞

⎡ ⎤π⎛ ⎞θ = α ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎡ ⎤θ π⎛ ⎞= − = α′ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∑

∑

(1.32)

The accent in snα′ is a reminder that the complex heat � ow rate equals the � rst derivative of the complex temperature to thermal resistance. The amplitude and phase shift of the heat � ow rate are:

Amplitude 2 2s sn nB A′ + ′

Phase shift s

s

atan⎛ ⎞′−⎜ ⎟′⎝ ⎠

n

n

A

B

Single-layered Wall

Inserting the temperature equation (1.32) into equation (1.30) gives:

2

2

d ( ) 2 2( ) exp 0

dn

nn

R c i n i n tR

T TR

∞

=−∞

⎧ ⎫⎡ ⎤α ρ λ π π⎪ ⎪⎛ ⎞− α =⎢ ⎥⎨ ⎬⎜ ⎟⎝ ⎠⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭∑ (1.33)

Sum zero is a fact if all coef� cients of the exp-functions are zero, or:

2

2

d ( ) 2( ) 0

dn

nR c i n

RTR

α ρ λ π− α = (1.34)

Thus, the partial differential equation of second order breaks up into 2 � + 1 differential equations of second order, with the complex temperature �n(R) as the dependent and the

1335vch01.indd 34 12.02.2007 11:36:14

35

thermal resistance R as the independent variable. These equations must be ful� lled for all values of n between –� and +�. Because the solutions between 0 and –� and 0 and +� mirror each other, we are left with � + 1 equations. Because n intervenes as a parameter only and the equations are otherwise identical, we only have to know the solution for one value of n. That solution is:

1 2( ) exp( ) exp( ) n n nR C R C Rα = ω + −ω (1.35)

where 2 2n

c i n

T

ρ λ πω = and 0 n≤ ≤ ∞.

The quantity � is called the thermal pulsation. Conversing (1.35) using the formulas for sinh(x) and cosh(x)

1sinh ( ) [exp( ) exp( )]

2x x x= − − ,

1cosh ( ) [exp( ) exp( )]

2x x x= + −

gives as complex temperature and complex heat � ow rate:

1 2 1 2( ) ( ) sinh ( ) ( ) cosh ( ) n n nR C C R C C Rα = − ω + + ω

1 2 1 2d

( ) [( ) cosh ( ) ( ) sinh ( )] dn n n nR C C R C C R

R

αα = = ω − ω + + ω′

The integration constants (C1 – C2) and (C1 + C2) follow from the boundary conditions:

On wall surface R = 0 the complex temperature is s (0)nα , or:

s 1 2 1 2 1 2(0) ( ) 0 ( )n C C C C C Cα = − + + = +

On wall surface R = 0 the complex heat � ow rate is s 0nα′ = , or:

s 1 2 1 2 1 2(0) [( ) ( ) 0] ( ) n n nC C C C C Cα′ = ω − + + = ω −

That result converts the complex temperature and complex heat � ow rate into:

s sm

s s

sinh ( )( ) (0) cosh ( ) (0)

( ) (0) sinh ( ) (0) cosh ( )

nn n n

n

n n n n n n

RR R

R R R

ωα = α ω + α′

ωα = α ω ω + α ω′ ′

(1.36)

Equation (1.36) forms a system of 2 equations, allowing for the solution of 2 unknowns. Inclusion into the Fourier series (1.32) gives the time-related equations for the temperature and heat � ow rate in a single-layered wall.In general, most of the attention is on the relationship between the complex quantities on both wall surfaces. Suppose R = 0 coincides with the inside surface. Hence, for outside walls, R = RT is the outside surface, for inside walls the surface at the other side (Figure 1.17). In general, for R = RT the system equation becomes:

s T s[ ( )] [ (0)]n n nA R W A= (1.37)

where [Asn(0)] is the column matrix of the unknown complex quantities on the one wall surface, and [Asn(RT)] the column matrix of the known complex quantities on the other. Wn, the coef� cient matrix of the system, is called the wall matrix of the single-layered wall for the nth harmonic. This matrix depends solely on the wall properties (thickness and material data), the base period and the order of the harmonic considered. The wall matrix replaces so to say

�

�

1.2 Conduction

1335vch01.indd 35 12.02.2007 11:36:14

36 1 Heat Transfer

the steady state thermal resistance. However, it includes much more information. In order to interpret that information, we examine some speci� c cases.

Zero Harmonic (n = 0)

For n = 0, only the terms s0α and s0α′ are left, where s0 / 2α and s0 / 2α′ represent the average temperature and average heat � ow rate on the inner surface over the period T.

The thermal pulsation for the average is 20

2 0( 0) 0

i cn

T

ρ λ πω = = = , turning (1.36) into:

s0 0 s00

( ) (0) (0)0sRα = α + α′

s0 s0 s0 s0( ) (0) 0 (0) (0) Rα = α + α′ = α′

The second equation tells us that in a periodic regime the average incoming and outgoing heat � ow rates are identical. For a random value R, i.e. in the wall, we arrive at the same conclusion. Consequently the heat � ow rate s0α′ is a constant, as it is in steady state. The unde� ned ratio 0/0 in the � rst equation is solved using the de l’Hôpital’s rule:

0 00 0 0

0 0lim [sinh ( ) / ] lim [ cosh ( )]R R R R

ω → ω →ω ω = ω =

giving: s0 s0 s0( ) (0) (0)R Rα = α + α .For the thermal resistance at an interface R in the wall, we get the same result. Consequently, in the [R, �] plane, the average temperatures in the wall all lie on a straight line with the heat � ow rate as a slope. This again concurs with steady state. Both results extend the steady state concept from ‘invariable’ to ‘concerns the average over a suf� ciently long period of time’.

Order of the Harmonic Different from Zero (n � 0)

(1) Suppose the inside surface temperature is constant. Than all complex inside surface temperatures are zero and the � rst equation of (1.36) turns into:

ss s

s

sinh ( ) ( ) sinh ( )( ) (0)

(0)n n n

n nn n n

R R RR or

ω α ωα = α =′

ω α ω′ (1.38)

Figure 1.17. The wall surfaces (R = 0 and R = RT).

1335vch01.indd 36 12.02.2007 11:36:14

37

The function sinh ( ) /n nRω ω consequently represents the ratio between the complex temperature on the outside surface or the surface at the other side and the complex heat � ow rate on the inside surface. In steady state, we call that ratio thermal resistance. Analogically, the complex figure sinh ( ) /n nRω ω is named the dynamic thermal resistance (for the nth harmonic). Symbol q

nD , units: m2 · K/W. The amplitude gives the size, the argument q arg[sinh ( ) / ]n

n nRφ = ω ω the time shift between the complex temperature at the one and the complex heat � ow rate at the other surface. The hypothesis behind the de� nition of dynamic thermal resistance may seem theoretical. However, a transfer to effective temperatures indicates that the case is applicable to a building which is kept at constant temperature.For the dynamic thermal resistance we can easily determine the limit values for a zero and in� nitely long period:

The period becomes in� nite:An in� nite period means a thermal pulsation zero. Thus:

q0 0

sinh (0)[ ] lim lim [ cosh (0)]

0n

n nD R R

→ →

⎡ ⎤= = =⎢ ⎥⎣ ⎦

q0

sinh (0)lim arg

0n

n→

⎡ ⎤φ = = ∞⎢ ⎥⎣ ⎦

or the dynamic thermal resistance equals the thermal resistance in that case. An in� nite period stands for steady state.

The period becomes zero:A zero period means a thermal pulsation becoming in� nite. Thus:

q Tsinh ( )

[ ] lim lim [ cosh ( )]n

n nD R

→∞ →∞

∞⎡ ⎤= = ∞ = ∞⎢ ⎥∞⎣ ⎦

qsinh ( )

lim arg 0n

n→∞

∞⎡ ⎤φ = =⎢ ⎥∞⎣ ⎦

or, for ever shorter � uctuations, the dynamic thermal resistance turns in� nite, i.e. the wall dampens the signal completely. The two limit values anyhow show that the dynamic thermal resistance is always larger than the thermal resistance. Thus, it suf� ces to impose a high thermal resistance in order to obtain a high dynamic thermal resistance.

(2) Suppose the heat � ow rate at the inside surface is constant. Then all complex heat � ow rates at that surface go to zero and the � rst equation of (1.36) becomes:

ss s

s

( )( ) (0) cosh ( ) or cosh ( )

(0)n

n n n nn

RR R R

αα = α ω = ω

α (1.39)

The function Tcosh ( )n Rω consequently represents the ratio between the complex temperature on the outside surface or the surface at the other side and the complex temperature on the inside surface of the wall. We call that function temperature damping of the wall (for the nth harmonic). Symbol nDθ , no units. The amplitude again gives the size of that ratio, the argument arg[cosh ( )]n

n RθΦ = ω the time shift between the complex temperatures at both surfaces. The hypothesis behind the de� nition of temperature damping is quite � ctitious. Indeed,

�

�

1.2 Conduction

1335vch01.indd 37 12.02.2007 11:36:15

38 1 Heat Transfer

the property has no equivalent in a steady state, although it is of the utmost importance as a quanti� cation of the natural response, the spontaneous climate control capacity of a wall. In dry, hot regions, temperature damping is an important characteristic. If solar gains are low and ventilation is restricted, then an enclosure with a high temperature damping will create a stable inside climate. Hence, in such regions, speci� cations should be quite demanding with respect to temperature damping. In that sense, it is a performance characteristic. Limit values for a zero and in� nite period are:

The period becomes in� nite:An in� nite period means a thermal pulsation zero. Thus:

0[ ] lim [cosh (0)] 1n

nDθ →

= = 0

lim {arg[cosh (0)]}n

nθ →

φ = = ∞

or, for long lasting � uctuations the temperature damping approaches one. In fact, in steady state, no temperature difference can exist between the inside and the other surface of a wall when no heat � ows through.

The period becomes zero:A zero period means a thermal pulsation going to the in� nite. Thus:

[ ] lim [cosh ( )]n

nDθ →∞

= ∞ = ∞

lim {arg[cosh ( )]} 0n

nθ →∞

φ = ∞ =

or, for ever shorter � uctuations, temperature damping goes to in� nite.

(3) Suppose that the temperature on the outside surface or on the surface at the other side remains constant. Then all complex temperatures on that surface go to zero, converting the � rst equation of the system (1.36) into (after � lling in q

nD and nDθ ):

ss s

s q

(0)0 (0) (0) or cotgh ( )

(0)θ

θα′

= α + α′ = − = −ω ωα

nn n n

n q n n nnn

DD D R

D (1.40)

The ratio q/n nD Dθ thus relates the complex heat � ow rate on the inside surface to the complex temperature. We call it the admittance of the wall (for the nth harmonic). Symbol Adn, units W/(m2 · K). The amplitude gives the size of the ratio, the argument q

n n nAd θφ = φ − φ the time

shift between the complex heat � ow rate and the complex temperature on the inside surface. Also the admittance looks somewhat � ctitious. Anyhow, the property tells us to what extent a wall will absorb heat. The higher the admittance, the larger the absorption caused by a � uctuat-ing surface temperature. Thus, the expression ‘capacitive building’ points to an enclosure with

high admittance. As the thermal pulsation �n can be written as 2 /n i c n Tω = ρ λ π ,

it is obvious that a large admittance requires a large value of cρ λ . This square root of the product of volumetric speci� c heat capacity and thermal conductivity is called the contact coef� cient or effusivity of a material. Symbol b. Units: J/(m2 · s–1/2 · K). The larger the contact coef� cient, the more active is a material as a storage medium. The usage of heavy materials without insulating � nish results in high effusivity as well as a large admittance. Limit values for a zero and in� nite period:

The period becomes in� nite:An in� nite period means a thermal pulsation of zero. Thus:

�

�

�

1335vch01.indd 38 12.02.2007 11:36:15

39

0

1[ ] lim

nn

nnq

DAd

RDθ

→

⎛ ⎞= =⎜ ⎟

⎝ ⎠

As0

q

lim argn

nnn

D

Dθ

→

⎡ ⎤⎛ ⎞⎢ ⎥φ = − = ∞⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

or, for ever longer � uctuations, the admittance approaches the thermal permeance (P) of the wall (1/R). In steady state, the heat � ow rate entering or leaving a wall is indeed equal to P ��s.

The period becomes zero:A zero period means a thermal pulsation going to the in� nite. Thus:

q

[ ] limn

nnn

DAd

Dθ

→∞

⎛ ⎞= = ∞⎜ ⎟

⎝ ⎠

Asq

lim arg 0n

nnn

D

Dθ

→∞

⎡ ⎤⎛ ⎞⎢ ⎥φ = =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

or, for ever faster � uctuations, the admittance turns in� nite.The properties de� ned this way, allow rewriting the complex wall matrix as:

q

2q

n n

n n nn

D DW

D D

θ

θ

⎡ ⎤⎢ ⎥=⎢ ⎥ω⎣ ⎦

From the Complex to a Real Wall Matrix

Thermal pulsation can be written as:

2(1 )

2(1 )

2n

ni b

n nTi b i bT T

π+π πω = = = +

The conversion of i in that formula comes from: 2(1 ) 2i i+ = or 1

2

ii

+= .

With �n redressed that way, the product �n R becomes:

(1 )(1 )n

ni b x

nTR i xa T

π+ πω = = +λ

where x is the ordinate value (equals thickness at the other side of the wall) and a (= �2/b2)

the thermal diffusivity. If we name the product [ / ]π nx n a T X , then the thermal pulsation

�n simpli� es to (1 ) /ni X R+ and the term nDθ in the wall matrix can be written as:

cosh ( ) cosh [(1 ) ]

cosh ( ) cosh ( ) sinh ( ) sinh ( )

nn n

n n n n

D R i X

X i X X i Xθ = ω = += +

or, with cosh ( ) cos( )n ni X X= and sinh ( ) sin ( )n ni X i X= :

cosh ( ) cosh ( ) cos( ) sinh ( ) sin ( )n n n n nR X X i X Xω = +

�

1.2 Conduction

1335vch01.indd 39 12.02.2007 11:36:15

40 1 Heat Transfer

Analogically:

{

}

sinh ( )[sinh ( ) cos( ) cosh ( ) sin ( )]

2

[cosh ( ) sin ( ) sinh ( ) cos( )]

nn n n n

n n

n n n n

R RX X X X

X

i X X X X

ω= +

ω

+ −

{}

sinh ( ) [sinh ( ) cos( ) cosh ( ) sin ( )]

[cosh ( ) sin ( ) sinh ( ) cos( )]

nn n n n n n

n n n n

XR X X X X

R

i X X X X

ω ω = −

+ +

We now de� ne the following functions Gni:

1 cosh ( ) cos( ) n n nG X X=

2 sinh ( ) sin ( ) n n nG X X=

31

[sinh ( ) cos( ) cosh ( ) sin ( )] 2n n n n n

n

G X X X XX

= +

41

[cosh ( ) sin ( ) sinh ( ) cos( )] 2n n n n n

n

G X X X XX

= −

5 [sinh ( ) cos( ) cosh ( ) sin ( )] n n n n n nG X X X X X= −

6 [cosh ( ) sin ( ) sinh ( ) cos( )] n n n n n nG X X X X X= +

which allow us to rewrite the dynamic thermal resistance, temperature damping and the other term as:

1 2cosh ( )n n nR G i Gω = +

3 4sinh ( )

( )nn n

n

RR G i G

ω= +

ω

5 6sinh ( ) n nn n

G i GR

R

+ω ω =

Amplitude and time shift of the dynamic thermal resistance qnD , the temperature damping nDθ

and the admittance Adn of a single-layered wall then become (R = RT):

Amplitude Phase angle

2 2q T 3 4[ ]n

n nD R G G= + 4q

3

atanφ =n n

n

G

G

2 21 2[ ]n

n nD G Gθ = + 2

1

atanθφ =n n

n

G

G

q[ ] [ ] /[ ]n n nAd D Dθ= qn n nAd θφ = φ − φ

1335vch01.indd 40 12.02.2007 11:36:16

41

The nth 2 2 complex wall matrix is herewith converted into a 4 4 real matrix:

1 2 3 4

2 1 4 3

5 61 2

T T

6 52 1

T T

n n T n T n

n n T n T n

n nn nn

n nn n

G G R G R G

G G R G R G

G GG GW

R R

G GG G

R R

⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥− −⎢ ⎥⎣ ⎦

For the phase shift the sign rule of Figure 1.18 applies.

Composite Walls

A composite wall forms a system of series-connected single-layered walls (the composing layers), each with a wall matrix Wn,i, now called a layer matrix, whereas the wall as a whole has a wall matrix WnT, for which the following applies (Figure 1.19):

s T T s[ ( )] [ (0)]n n nA R W A= (1.41)

where [Asn(RT)] is the column matrix of the nth complex temperature and heat � ow rate on the one and [Asn(0)] the column matrix of the nth complex temperature and heat � ow rate on the other wall surface. As relationship between the complex temperature and complex heat � ow rate on the interfaces j and j + 1 we have: , 1 , ,[ ] [ ]n j n j n jA W A+ = . The latter can be written for each layer, starting at the surface R = 0 which, as agreed upon, coincides with the inner surface:

Figure 1.18. Periodic regime: sign rule for the G-functions.

1.2 Conduction

1335vch01.indd 41 12.02.2007 11:36:16

42 1 Heat Transfer

( ),1 ,1 ,

,2 ,2 ,1

, 1 , 1 , 2

s, T , , 1

[ ] [ 0 ]

[ ] [ ]

[ ] [ ]

[ ( )] [ ]

n n s n

n n n

n m n m n m

n n m n m

A W A

A W A

A W A

A R W A

− − −

−

=

=

=

=

� (1.42)

Introducing each of the preceding in each of the subsequent equations gives: s, T[ ( )]nA R =

, , 1 ,2 ,1 s,[ (0)]n m n m n n nW W W W A− … . Combined with (1.41), one thus gets:

T ,1

m

n n jj

W W=

= ∏ (1.43)

or, the wall matrix of a composite wall equals the product of the layer matrixes of the compo-sing layers. Multiplication starts at the inner and ends at the outer layer. Each following layer matrix must be multiplied with the product of the former ones. Because the commutation property does not apply for a product of matrixes, this sequence must be strictly respected. Translated to design and construction practice, contrary to what was applicable for the thermal resistance, the transient response of a composite wall is a function of the layer sequence! The terms in the wall matrix of a composite wall have the same meaning as for a single-layered wall.

Figure 1.19. Composite wall, wall matrix WnT .

1335vch01.indd 42 12.02.2007 11:36:17

43

Temperatures

If two of the four complex boundary conditions (two complex temperatures and two complex heat � ow rates) on the wall surfaces are known, then the two others follow from the matrix equation: [Asn(RT)] = WnT [Asn(0)]. Afterwards the complex temperatures and heat � ow rates in the interfaces are found by ascending or descending equations (1.42) in the correct sequence. For a single-layered wall, equations (1.32) give the temperatures and heat � ow rates in the wall. We may also use the composite wall approach. For that purpose, we divide the single-layered wall into m layers with thickness �x, layer matrix n xW Δ and wall matrix

( )mn n xW W Δ= (Figure 1.20). For the interfaces at a distance �x from each, the system (1.42)

applies, thus allowing for the calculation of the complex temperatures n xα and the complex heat � ow rates n xα′ .

Figure 1.20. Single-layered wall, calculating the periodic temperatures.

1.2.4.3 Flat Walls, Transient Boundary Conditions

Dirac Impulse

With a Dirac impulse, the temperature or the heat � ow rate on a surface is zero between time t = –� and time t = +�, unless for an in� nitesimally short time interval dt, when it has a value of one (Figure 1.21).

Figure 1.21. Dirac impulse.

1.2 Conduction

1335vch01.indd 43 12.02.2007 11:36:17

44 1 Heat Transfer

Response Factors

Fourier’s second law can be solved for a Dirac impulse. Suppose that the temperature �s1 on the wall surface s1 is subjected to an impulse. As a result, a variation qs1(t) of the heat � ow rate on that surface and a variation �s2(t) of the temperature and qs2(t) of the heat � ow rate on the other surface s2 will ensue. In case all material properties are constant, the functions qs1(t), �s2(t) and qs2(t) de� ne the response factors of the heat � ow rate qs1, the temperature �s2 and the heat � ow rate qs2 for a temperature impulse �s1 on surface s1. The symbols are:

1 1 1 2 1 2q qI I Iθ θ θ θ

Analogically we have:

1 1 1 2 1 2

2 1 2 1 1 2

2 1 2 1 2 2

q q q q

q q

q q q q

I I I

I I I

I I I

θ θ

θ θ θ θ

θ θ

Response factors are again a function of the layer sequence. Hence, 1 2

Iθ θ differs from 2 1

Iθ θ , etc.

Convolution Integrals

Consider now a temperature impulse on surface s1 with a value �0. If all response factors are known, then the temperature and heat � ow rate response on surface s2 and the heat � ow rate response on surface s1 become:

Impulse �0 op s1 Impulse q0 op s1

1 1s1 0 qq Iθ= θ1 1s1 0 qq I θθ =

1 2s2 0 Iθ θθ = θ1 2s2 0 qq I θθ =

1 2s2 0 qq Iθ= θ1 2s2 0 q qq q I=

Figure 1.22. The convolution principle.

1335vch01.indd 44 12.02.2007 11:36:18

45

An impulse on surface s2 results in analogue relationships. That way any random signal �s1(t) can be split into a continuous series of successive impulses 0 ( )t tθ Δ . For the response of �s2 on surface s2 on a signal �s1 on surface s1, the following applies (Figure 1.22):

0t = s2 (0) 0θ =

t t= Δ 1 2s2 s1( ) ( 0) ( )t t I t tθ θθ Δ = θ = = Δ

2t t= Δ 1 2 1 2s2 s1 s1(2 ) ( 0) ( 2 ) ( ) ( )t t I t t t t I t tθ θ θ θθ Δ = θ = = Δ + θ = Δ = Δ

3t t= Δ

1 2

1 2

1 2

s2 s1

s1

s1

(3 ) ( 0) ( 3 )

( ) ( 2 )

( 2 ) ( )

t t I t t

t t I t t

t t I t t

θ θ

θ θ

θ θ

θ Δ = θ = = Δ

+ θ = Δ = Δ

+ θ = Δ = Δ

…

t n t= Δ

1 2

1 2

1 2

s2 s1

s1

s1

( ) ( 0) ( )

( ) [ ( 1) ]

( 2 ) [ ( 2) ]

n t t I t n t

t t I t n t

t t I t n t

θ θ

θ θ

θ θ

θ Δ = θ = = Δ

+ θ = Δ = − Δ

+ θ = Δ = − Δ +…

or: 1 2

1

s2 s10

( ) ( ) [( ) ]n

jn t j t I n j t

−

θ θ=

θ Δ = θ Δ − Δ∑

In integral form:

1 2s2 s10

( ) ( ) ( ) dt

t I tθ θθ = θ τ − τ τ∫

This integral of the product of the signal and the response factor scanned counter clockwise, is called the convolution-integral of the temperature �s2 for the signal �s1(t). In general, the response of the three other variables on a random signal of the fourth is found by convoluting the signal with the speci� c response factor. Response factors as well as convolution-integrals have to be calculated numerically.

Step Function at the Surface of a Semi-in� nite Medium

Suppose a semi-in� nite medium is at uniform temperature �s0. At time zero, the surface temperature jumps from �s0 to a temperature �s0 + ��s0 (Figure 1.23).The solution of Fourier’s second law with t = 0, 0 � x � �: � = �s0 as initial condition and t � 0, x = 0: �s = �s0 + ��s0 as boundary condition results from a separation of variables. The outcome is:

2s0 s0

2

2( , ) exp( ) d

xq

a t

x t q q∞

=

⎛ ⎞⎜ ⎟

θ = θ + Δθ −⎜ ⎟π⎜ ⎟⎜ ⎟⎝ ⎠

∫ (1.44)

1.2 Conduction

1335vch01.indd 45 12.02.2007 11:36:18

46 1 Heat Transfer

The term between brackets is the inverse error function [1 – erf(q)], with a the thermal diffusivity. As heat � ow rate at the surface (x = 0) we get:

2s0 s0

0

2dexp

d 42x

bxq

x a ta t t=

⎡ ⎤⎛ ⎞Δθ Δθθ= −λ = −λ − = −⎢ ⎥⎜ ⎟π π⎝ ⎠⎢ ⎥⎣ ⎦ (1.45)

where b is the contact coef� cient. The application of the de� nition of response factor on equation (1.45) gives

1 1/qI b tθ = − π . Semi in� nite mediums of course do not exist, but,

below grade or a thick wall are quite close. In any case, equation (1.45) illustrates in an excellent way the meaning of the contact coef� cient. A high value gives a large heat absorption and a large heat release when the surface temperature suddenly rises or drops, while a low value gives a small heat absorption or heat release in such case. Hence, materials with high contact coef� cient are active as storage media. For so-called passive solar buildings, for which temporary storage of solar gains in the fabric is one of the design concerns, walls and � oors containing suf� ciently thick material layers with high contact coef� cient at the inside are therefore a good choice.The heat, which � ows in or out of the in� nite mass per m2 surface between time zero and time t, is given by:

s0q

0

2d

t bQ q t t A t

Δθ= = =

π∫ (1.46)

The quantity Aq is called the heat absorption coef� cient of the material, units J/(m2 · K · s1/2).If two materials, one at temperature �1 and the other at temperature �2, contact each other, then the heat � ow rate in the interface totals (�1 > �2):

Material 1: 1s1 1 c( )

bq

t= θ − θ

π

Material 2: 2s2 c 2( )

bq

t= θ − θ

π

Figure 1.23. Temperature step at the surface of a semi-in� nite medium.

1335vch01.indd 46 12.02.2007 11:36:19

47

with �c the contact temperature. Since both heat � ow rates must be equal in absolute terms (qs1 = qs2), the contact temperature �c becomes:

1 1 2 2c

1 2

b b

b b

θ + θθ =

+ (1.47)

At the start, the contact temperature between materials apparently depends on the temperature of both materials and their contact coef� cients. This has consequences for our comfort when touching materials. A material with high contact coef� cient (= capacitive materials) feels cold or hot while a material with low contact coef� cient feels comfortably warm. Indeed, in the � rst case the skin temperature moves from 32–33 °C to the material temperature, while in the second case the contact surface adapts to the skin temperature. We therefore experience a material such as concrete as unpleasant and a material such as wood, as pleasant. To indicate that feeling, the terms cold and warm materials are used. The contact coef� cient is an important criterion when choosing a � oor covering or chair � nishes.

1.2.4.4 Two and Three Dimensions

On spots where heat transfer evolves in two or three dimensions, we have to go back to equation (1.6). Analytical solutions for building details do not exist, which is why CVM or FEM (Finite Elements) is used. For the basic principles of CVM we refer to Section 1.2.2.4. In a transient regime, a heat capacity c �V is assigned to each control volume of magnitude �V. According to energy conservation without dissipation, the resulting heat � ow per control volume must equal the change in heat content, or:

m( ) t c VΦ Δ = ρ Δ Δθ∑where �m is the average heat � ow from each adjacent control volume to the central control volume during the time-step �t. If we assume m (1 ) t t tp p+ΔΦ = Φ + − Φ with 1 > p > 0, i.e. each heat � ow being a weighted average between the heat � ows at time t (= 1) and time t + �t (= 2), then the balance becomes (i = l, m, n and j = ±1):

(2) (1)

, , , , , , , ,

(2) (1), , , ,

( ) (1 ) ( )s i j i j l m n s i j s i j i j l m n s i j

l m n l m n

p P P p P P

c Vt

+ + + + + +⎡ ⎤ ⎡ ⎤θ − θ + − θ − θ⎣ ⎦ ⎣ ⎦θ − θ

= ρ ΔΔ

∑ ∑ ∑

For p = 0.5, i.e. when using the arithmetical average for the heat � ows according to Cranck-Nicholson, rearranging the terms gives:

(2)

, , , ,

(1)

, , , ,

2( )

2( )

s i j i j l m n s i j

s i j i j l m n s i j

c VP P

t

c VP P

t

+ + +

+ + +

⎡ ⎤ρ Δ⎛ ⎞θ − θ +⎜ ⎟⎢ ⎥⎝ ⎠Δ⎣ ⎦

⎡ ⎤ρ Δ⎛ ⎞= − θ + θ −⎜ ⎟⎢ ⎥⎝ ⎠Δ⎣ ⎦

∑ ∑

∑ ∑

In this equation, the temperatures at the following time steps are the unknown and the preceding time steps the known quantities. That way, for each time step, we obtain a system of m equations with m unknowns (m equal to the number of control volumes). If the starting and boundary conditions are known, then we can solve the systems. Questions to decide on

1.2 Conduction

1335vch01.indd 47 12.02.2007 11:36:19

48 1 Heat Transfer

beforehand are: the choice of the mesh, � ner in materials with a high thermal diffusivity, and the time step and time resolution, which must be chosen in accordance with mesh density, the time step between the successive boundary condition values and the desired information density. A bad choice of meshing and time step may result in an unstable solution.For p = 0, the system simpli� es to a series of successive forward difference equations, while p = 1 gives a system of backward difference equations.

1.3 Convection

1.3.1 Overview

1.3.1.1 Heat Transfer at a Surface

Up to this point, we assumed that the surface temperatures were known. But in most cases, not these but the air temperatures and sometimes the heat � ow rates from the environment to a surface are the known quantities. For example, the outside air temperature is registered at every meteorological station and measuring indoor air temperatures is not complicated. In fact, logging surface temperatures proves to be more dif� cult.From now on, we will focus on heat transfer across a building element from the environment at one side, to the environment at the other side. To do that, we have to know how heat reaches a surface. The answer is twofold: by convection between the surface and the air and by radiant exchange with all surfaces facing the surface considered. Thus, the inside surface of wall A in Figure 1.24 faces � ve other walls in the room, plus the radiator, furniture, etc. A exchanges radiation with each of these, and at the same time, the air in the room transfers convective heat to A.

Figure 1.24. Heat exchange by radiation and convection in a room.

1335vch01.indd 48 12.02.2007 11:36:19

49

1.3.1.2 Convection

The word convection applies to heat transfer by � uid motion. In the paragraphs that follow, the term is used more restrictively for the heat transferred between a � uid and a surface. In such a case, the convective heat � ow rate and the convective heat � ow are written as:

c fl s( )cq h= θ − θ c c fl s( )h AΦ = θ − θ (1.48)

with �� the temperature in the ‘undisturbed’ � uid, �s the surface temperature and hc the convective surface � lm coef� cient, units: W/(m2 · K). The equation (1.48) is known as Newton’s law. At � rst sight, convection seems very simple. In fact, a linear relationship seems to exist between the heat � ow rate and the driving temperature difference (�� – �s). The equation, however, should be read as a de� nition of the convective surface � lm coef� cient. That coef� cient accommodates the whole complexity induced by convection. In the convective mode, heat and mass transfer go hand in hand. Mass � ow is described by combining the scalar law of mass conservation with the vectorial law of conservation of momentum.

Mass (Scalar)

div ( ) 0ρ =v

Momentum (Navier-Stokes, Vectorial)

2d( )

dP

t

ρ = ρ − + μ ∇vg grad v

where is the density and � the dynamic viscosity of the � uid, P total pressure and g the gravity force gradient. The unknown factors in both equations are the velocity components vx, vy, vz and total pressure P.In turbulent � ow conditions, turbulency equations have to be added. The (k, �) model is often used, where k is the turbulent kinetic energy and � the turbulent energy dissipation. Both require an extra scalar equation. Along with the three scalar equations and one vector equation, whose split into the x-, y-, and z-directions produces three scalar equations, energy conservation has to be added. In case of constant properties and negligible kinetic energy and friction heat, that equation looks like:

2 d

da

t

θ∇ θ = (1.49)

where a is the thermal diffusivity of the � uid. In both the Navier-Stokes and the energy equation, / dd t accounts for the total derivative:

(1) (3)(2)

d x y zd x y z

v v vt x t y t z t t x y z t

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= + + + = + + +∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂��� ������

The terms (1), (2) and (3) describe the enthalpy transport, i.e. the fact that a � uid moving at a temperature ��with a velocity v (components vx, vy, vz)� carries c � units of heat.Energy conservation adds temperature � as the seventh unknown quantity. All information about convection follows from the solution of that system of seven scalar Partial Differential

1.3 Convection

1335vch01.indd 49 12.02.2007 11:36:19

50 1 Heat Transfer

Equations (PDE’s). Once the temperature of the � uid is known, then the heat � ow rate against a surface follows from Fourier’s First Law (at least when the element belonging to the surface is air tight!). Indeed, at the surface, convection becomes conduction across a laminar boundary layer. Consequently, the following applies for the convective surface � lm coef� cient:

sc fl

fl s

(grad )h

θ= −λ

θ − θ (1.50)

where (grad �)s is the temperature gradient at the surface in the � uid and �� is the temperature in the undisturbed � ow. What that means has to be agreed upon for each case.Solving the seven PDE’s is analytically possible for simple problems, such as natural convection by laminar � ow along a semi-in� nite vertical surface or forced convection by laminar � ow along a semi-in� nite horizontal surface. In the last case, the formulas simplify to:

0yxvv

x y

∂∂+ =

∂ ∂

2

fl 2yx x

x y

vv vv v

x y y

∂∂ ∂+ = ν

∂ ∂ ∂

2

fl 2x yv v ax y y

∂θ ∂θ ∂ θ+ =∂ ∂ ∂

giving a velocity and temperature equation which are simular. If moreover thermal diffusivity a� equals kinematic viscosity �� , then the two are identical. So, we may presume that convective heat transfer is a copy of the � ow pattern. Solving the system gives as convective surface � lm coef� cient:

flfl

c,

0.664

x

v x

hx

∞λν

=

where x is the distance along the surface from the free edge of the surface and v� is the � uid velocity outside the boundary layer. This equation shows that the convective surface � lm coef� cient is not a constant but a function of the position along the horizontal surface.Any attempt to calculate analytically convective heat transfer at the walls in a room, at the outside surfaces of a building or any other construction is doomed to fail. However, a numerical calculation with CFD (Computerized Fluid Dynamics) is possible. Yet, even in extremely complex situations, it is a rule that convection depends on all the parameters and properties which determine � ow and heat transfer:

Fluid properties: Flow parameters:Thermal conductivity GeometryDensity Surface roughnessSpeci� c heat capacity Temperature difference (�� – �s)Kinematic viscosity Nature and direction of the � owVolumetric expansion coef� cient Velocity components vx, vy, vz

1.3.2 Convection Typology

Which type of convection will develop, is de� ned by the driving forces and the kind of � ow.

�����

1335vch01.indd 50 12.02.2007 11:36:19

51

1.3.2.1 Driving Forces

When differences in density in a � uid caused by gradients in temperature and concentrations are the driving force, then natural convection develops. In natural convection, the � ow pattern follows the � eld of temperatures and/or concentrations. In buildings, this form of convection governs the indoor situation. If on the contrary the driving force is an imposed pressure differ-ence, then forced convection follows. In such case the � ow pattern is independent of temperature and concentration. This form of convection is seen in HVAC systems. Wind also causes it. Forced and natural convection, however, never occur in a clearly separate way. An important natural component is added to low velocity forced convection. An example is convection by wind around a building. Generally, the wind velocity is too low to eliminate buoyancy � ow. On the other hand, pure natural convection is more an exception than a rule. Even inside, there is always forced air momentum. Both together generate mixed convection.

1.3.2.2 Type of � ow

The � ow can be laminar, turbulent or in transition. A laminar � ow consists of diverging and converging streamlines which never cross. Particle velocity and � ow velocity are equal. In turbulent � ow, the momentum becomes chaotic and creates whirling eddies with particle velocities in every direction. Particle velocity and � ow velocity are no longer identical. In the eddies, a turbulent kinetic energy (k) builds up, after which the whirling motion dies away by turbulent dissipation (�). A description of the exact � ow pattern is impossible, even though a fractional eddy approach could be used. Transient � ow occurs between laminar and turbulent � ows. Small disturbances in the � ow pattern suf� ce to switch from laminar to turbulent, after which the turbulences die away and the � ow becomes laminar again. Intense mixing during turbulent � ow accounts for a larger heat transfer than laminar � ow, where the heat transfer perpendicular to the � ow direction proceeds by conduction only. Transient � ow lies in-between.

To summarize:

Driving forceFlow

Laminar Transient Turbulent

Density differences

Density and pressures

Pressures

1.3.3 Calculating the Convective Surface Film Coef� cient

1.3.3.1 Analytically

See above. The few cases which can be solved analytically show that the convective surface � lm coef� cient differs from spot to spot. If heat transfer is the point of interest, that complication is circumvented by using a surface average:

c cA1

dh h AA

= ∫

1.3 Convection

1335vch01.indd 51 12.02.2007 11:36:20

52 1 Heat Transfer

Hereinafter, the ‘convective surface � lm coef� cient c′h typically stands for that average. For calculating surface temperatures, local values should be used.

1.3.3.2 Numerically

Intensive scienti� c work has been done to understand convection using CFD. Yet, one problem remains. The velocity pro� les near a surface are described by surface functions. These have to be ‘assumed’, which means that the boundary layer is not calculated but de� ned beforehand.

1.3.3.3 Dimensional Analysis

Because of the limited number of analytical solutions and the dif� culties CFD poses, most information on convection has been and is still obtained from experiment, using dimensional analysis. That technique determines which dimensionless ratios between � uid properties and geometrical and kinematic parameters de� ning the movement must have the same value in the experiment as in reality to allow extrapolation of the measured results to real world situations. These dimensionless ratios are obtained directly from the differential equations (ex.: �/a) or follow from the Buckingham �-theorem. That theorem assumes that if a problem depends on n single-valued physical properties with involve p basic dimensions in their units, then n – p dimensionless numbers are required to � nd the solution. The description of forced convection, for example, needs seven single physical properties (L, � , v� , � , �� , c� , hc). These properties induce four basic dimensions: [L (length)] [t (time)] [M (mass)] [� (temperature)]. Consequently, three dimensionless numbers (�1, �2, �3) are required to estimate forced convection. As an expression: �1 = f (�2, �3). The three are found by rewriting the three �-functions, which look in general as fl fl fl fl fl c

a b c d e f gL v c hπ = λ ρ μ , in the form of an equation that only contains the intervening basic dimensions:

2

3 3 2 3[ ]

fb gec da M L L M M L M

Lt L tt L t t

⎡ ⎤⎡ ⎤ ⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤π = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥θ θ θ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦

Because � is dimensionless, the sum of the exponents for each dimension should be zero, or:

for M b + d + e + g = 0

for L A + b + c – 3 d – e + 2 f = 0

for t – 3 b – c – e – 2 f – 3 g = 0

for � – b – f – g = 0

The three dimensionless numbers are now found by calculating the general �-function three times, � rst for g = 1, c = 0, d = 0, than for g = 0, a = 1, f = 0 and � nally for g = 0, e = 1, c = 0:

Solution 1 c1

fl

1, 1, 0, 0 orh L

a b e f= = − = = π =λ

Solution 2 fl fl2

fl

0, 1, 1, 1 orv L

b c d eρ

= = = = − π =μ

Solution 3 fl fl3

fl

0, 1, 0, 1 orc

a b d fμ

= = − = = π =λ

1335vch01.indd 52 12.02.2007 11:36:20

53

For natural convection 4 dimensionless numbers are needed, two of which can be combined into one. The three (forced convection) or four (natural convection) dimensionless numbers thus are:

The Reynolds Number

2Re ( )v L= = πν

(1.51)

with v the velocity, � the kinematical viscosity of the � uid and L the characteristic length (representing the geometry of the problem). For pipes, L equals the hydraulic diameter; for a wall it equals the dimension in the � ow direction; for more complex geometry it is an assumed quantity. The Reynolds number re� ects the ratio between the inertia force and viscous friction. If Re is small, then viscous friction gains and the result is laminar � ow. If Re is large, inertia becomes much more important and turbulent � ow occurs. Hence, Re determines the nature of the � ow: laminar for Re � 2000, turbulent for Re � 20,000 and transient for 2000 < Re < 20,000.

The Nusselt Number

c1

fl

Nu ( )h L

= = πλ

(1.52)

with � thermal conductivity of the � uid. Multiplying the left and right term in equation (1.50) with the characteristic length, gives: c fl s fl s/ (grad ) /[ ) / ]h L Lλ = θ θ − θ , which shows that the Nusselt number represents the ratio between the temperature gradient against the surface and the average temperature gradient along the characteristic length. A large Nu means an important gradient at the surface and a small gradient along the characteristic length, which is the case for high � uid velocities. Physically, Nu underlines that conduction governs the heat transfer close to the surface. Even in turbulent � ow a laminar boundary layer remains, of which thickness becomes smaller with increasing � uid velocity but never goes to zero. Practically, the Nu-number underlines the importance of heat transfer by convection compared to conduction.

The Prandtl Number

3Pr ( )a

ν= = π (1.53)

The Prandtl number combines heat and mass transfer by rating two analogeous quantities, thermal diffusivity a, which determines the ease with which a local temperature change spreads in the � uid, and kinematic viscosity �� which determines the ease with which a local mass velocity spreads in the � uid.

The Grasshof Number

3

22Gr ( in case of natural convection)

g Lβ Δθ= = πν

(1.54)

1.3 Convection

1335vch01.indd 53 12.02.2007 11:36:20

54 1 Heat Transfer

with � the volumetric expansion coef� cient, g acceleration by gravity and �� the representative temperature difference. The number replaces Reynolds in the case of natural convection. Mass velocity (v) indeed is now the result of density differences (� g) caused by temperature differences (��). Velocity in turn in� uences the temperature difference (��), hence, L3 and �2 instead of L and �.

The Rayleigh Number

Ra = Gr Pr (1.55)

The Rayleigh number has no physical meaning. It was introduced because in many formulae for natural convection Gr and Pr appear as a product (Gr Pr).All experimental, numerical and analytical expressions for the convective surface � lm coef� cient can now be written as:

natural convection Nu = c (Ra)n mixed convection Nu = F (Re / Gr1/2, Pr)forced convection Nu = F (Re, Pr)

where the coef� cient c, the exponent n and the function F( ) differ from geometry to geometry, while depending on the nature of the � ow and, in case of natural connection, on the � ow direction. The relations allow for mutation between model and reality. As already said, couples where Nu1 = Nu2, Ra1 = Ra2 or Nu1 = Nu2, Re1 = Re2, Pr1 = Pr2 are identical.

1.3.4 Values for the Convective Surface Film Coef� cient

1.3.4.1 Walls

Natural and Forced Convection (L = Characteristic Length)

The � rst question is what characteristic length to use? For vertical surfaces, it is the height, for square horizontal surfaces, the side and for rectangular horizontal surfaces the average of length and width, see Table 1.1. All properties receive the value they have at a temperature equal to the average between wall surface and undisturbed � uid: fl s( ) / 2θ + θ .

Table 1.1. Natural and forced convection.

Natural Conditions Functions

Vertical surfaces RaL � 109

RaL > 109

NuL = 0.56 RaL1/4

NuL = 0.025 RaL2/5

Horizontal surfaces

Heat upwards 105 < RaL � 2 � 107

2 � 107 < RaL < 3 � 1010

NuL = 0.56 RaL1/4

NuL = 0.138 RaL1/3

Heat downwards 3 � 105 < RaL < 1010 NuL = 0.27 Ra1/4

���

1335vch01.indd 54 12.02.2007 11:36:20

55

Table 1.1. (continued)

Forced Conditions Functions

Laminar � ow Pr > 0.1ReL < 5 � 105 NuL = 0.644 ReL

1/2 Pr1/3

Turbulent � ow Pr > 0.5ReL > 5 � 105 NuL = 0.036 Pr1/3 (ReL

4/5 – 23,200)

Simpli� ed Expressions

In and around buildings, the � uid is air at atmospheric pressure. This allows simplifying the equations for ambient temperatures to:

Natural convection:

c

b

h aL

Δθ⎛ ⎞= ⎜ ⎟⎝ ⎠

where �� is the temperature difference between the undisturbed air and the wall surface. For a, b and L the values shown on Table 1.2 should be applied:

Table 1.2. Natural convection.

Conditions a b L

Vertical surfaces 10–4 < L3 �T � 77 < L3 �T � 103

1.41.3

1/41/3

Height1

Horizontal surfaces

Heat upwards 10–4 < L3 �T � 0.140.14 < L3 �T � 200

1.31.5

1/41/3

eq. side1

Heat downwards 2 � 10–4 < L3 �T � 200 0.6 1/4 1

Forced convection:for v � 5 m/s: natural convection plays a role. Hence the constant 5.6; hc = 5.6 + 3.9 vfor v > 5 m/s: hc = 7.2 v0.78

In both formules, v is the meteorological wind velocity.The formulae show that the surface � lm coef� cient for natural convection is largely determined by the temperature difference between the surface and the air. For forced convection, the air velocity is the main parameter. The increase with air velocity is a direct consequence of the decreasing thickness of the surface boundary layer (see the Nusselt number). In any case, the equations apply to air � owing along freestanding surfaces only. Angles between two surfaces and corners between three surfaces have a disturbing effect. Moreover, if the surfaces and the angles form a room, the overall � ow must satisfy the continuity equation. This makes convection so complex that, for the sake of simplicity, most standards impose a constant convective surface � lm coef� cient (see Table 1.3).

�

�

1.3 Convection

1335vch01.indd 55 12.02.2007 11:36:20

56 1 Heat Transfer

In a room, the reference temperature is the air temperature, 1.7 m above � oor center. Outside it is the air temperature measured at the closest meteorological station. When calculating surface temperatures using local convective surface � lm coef� cients, the reference to be used is the air temperature just outside the boundary layer. In case very correct calculations are needed (large temperature differences, complex geometry or protected surfaces), convection should be calculated using the formulae given in Table 1.1 or the literature.

1.3.4.2 Cavities

The word cavity refers to an air or gas layer with a thickness that is small compared to the other dimensions. In a cavity, heat transfer is a combination of conduction, convection and radiation (Figure 1.25).Convection occurs between the cavity surfaces and the gas in the cavity. Against the warmer surface, the heat � ow rate totals: qc1 = hc1 (�s1 – �c), while against the colder surface, it is: qc2 = hc2 (�c – �s2), where �c is the air or gas temperature in the middle of the cavity. Without cavity ventilation, both convective heat � ow rates are on average equal and the convective heat � ow rate thus becomes:

Figure 1.25. Heat exchange between the two surfaces in a cavity.

Table 1.3. Constant convective surface � lm coef� cients.

Heat loss Surface temperatures

Natural convection (= inside surfaces, European standard)

Vertical surfaces 3.5 2.5

Horizontal surfaces

Heat upwards (�)Heat downwards (�)

5.5 1.2

2.5 1.2

Forced convection (= outside surfaces) 19.0 19.0

1335vch01.indd 56 12.02.2007 11:36:20

57

c1 c2c s1 s2

c1 c2

( )h h

qh h

= θ − θ+

(1.56)

With hc1 = hc2 = hc, equation (1.56) simpli� es to: c s1 s2/ 2 ( )cq h= θ − θ . As conduction and convection cannot be measured separately, the conduction equation is mostly used instead of equation (1.56), whereby we multiply thermal conductivity�� of the gas by the Nusselt number:

sc fl c s( Nu)q q h

d

Δθ+ = λ = ′ Δθ (1.57)

In that equation d is cavity thickness in m and ��s the temperature difference over the cavity in °C.

In� nite Cavity

In horizontal cavities, convection causes Bénard cells to develop. These can be described as local eddies which are circular and symmetric. In vertical cavities, local air rotation occurs. In both cases, Nusselt is given by:

2 8dd d

d

RaNu max 1, 1 (10 Ra 10 )

Ra

rm

n

⎡ ⎤= + ≤ ≤⎢ ⎥+⎢ ⎥⎣ ⎦

(1.58)

with m, n, r following Table 1.4.

Table 1.4. In� nite cavity.

m n r

Horizontal cavity

Heat transfer downwardsHeat transfer upwards

00.07

3,200 1.33

Vertical cavity 0.024 10,100 1.39

Tilted cavity, slope below 45°

Heat transfer upwardsHeat transfer downwards

0.0430.025

4,10013,000

1.361.36

The temperature difference �� in the Raleigh number is the difference in temperature between both cavity surfaces, while cavity thickness d � gures as characteristic length. Ra < 100 means pure conduction, Nu = 1.

Finite Cavity

In a � nite cavity, convection diverges strongly from an in� nite cavity. The Nusselt numbers one gets are as shown in Table 1.5 (with d cavity thickness in m, H cavity height in m, L cavity length in m, Lam a superscript meaning laminar, turb a superscript meaning turbulent, transient a superscript meaning transient � ow):

1.3 Convection

1335vch01.indd 57 12.02.2007 11:36:20

58 1 Heat Transfer

Table 1.5. Finite cavity.

Vertical cavity Nud

Boundaries for the applicability of NudH /d 5 20 40 80 100

lam turb transientd d dmax (Nu , Nu , Nu ) , with

0.273lam dd

RaNu 0.242

d

H

⎛ ⎞= ⎜ ⎟⎝ ⎠

turb 0.33d dNu 0.0.0605 Ra=

0.3330.293transient dd 1.36

d

0.104 RaNu 1

1 (6310 / Ra )

⎡ ⎤⎛ ⎞⎢ ⎥= + ⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦

Rad,max 108 2 � 106 2 � 105 3� 104 1.2 � 104

Horizontal cavity Nud

Heat transfer upwards Rad � 1708 1

Rad > 1708

1

42max 1,1537 dL

⎡ ⎤Δθ⎛ ⎞⎢ ⎥

⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦

Heat transfer downwards 1

Tilted cavity: see literature

1.3.4.3 Pipes

From experimental and semi-experimental work on convection between a pipe and the ambient � uid, the following formulae have been derived:

Natural Convection

(Characteristic length is the outside diameter of the pipe)

Vertical pipe:Rad � 109 1 / 4

L dNu 0,555 Ra=Rad > 109 2 / 5

L dNu 0,021 Ra=

Horizontal pipe:Rad � 109 1 / 4

L dNu 0,530 Ra=

Forced Convection

Red < 500 1 / 2d dNu 0, 43 0, 48 Re= +

Red > 500 L dNu 0, 46 0,00128 Re= +

In these equations, all properties assume a temperature equal to fl s( ) / 2θ + θ .

1335vch01.indd 58 12.02.2007 11:36:21

59

1.4 Radiation

1.4.1 Overview

1.4.1.1 Thermal Radiation

Thermal radiation differs completely from conduction and convection. In fact, radiant heat transfer is caused by electromagnetic waves, which are emitted by every surface warmer than 0 K and whose absorption by other surfaces causes thermal agitation of atoms and electrons in them. The term thermal radiation includes all electromagnetic waves with a wave length between 10–7 and 10–3 m: Ultraviolet (UV), light (L) and infrared (IR).

The wave length (�) is given by c / f, where c is the propagation velocity of the electromagnetic wave in m/s and f is the frequency in Hz. Electromagnetic radiation encloses a very large interval of wave lengths:

� � 10–6 �m10–6 < � � 10–4 �m10–4 < � � 10–2 �m

cosmic radiationgamma raysx-rays

10–2 < � � 0.38 �m0.38 < � � 0.76 �m0.76 < � � 103 �m

UV-radiation (UV)light (L)IR-radiation (IR)

103 < � radio waves

Because of its electromagnetic nature, thermal radiation does not require a medium. Instead, it only develops unhindered in a vacuum where the electromagnetic photons move at a velocity of 299,792.5 km/s. Each surface warmer than absolute zero is an emitter, with the emission depending upon its nature and temperature. Resultant heat transfer is only possible between surfaces at different temperatures.

1.4.1.2 Quantities

Table 1.6 outlines the way thermal radiation is quanti� ed. Per quantity the spectral derivative stands for the derivative to wave length. Monochromatic radiation is charac terized by a single wavelength, colored radiation by several wavelengths.

1.4.1.3 Re� ection, Absorption and Transmission

If a radiant heat � ow rate qRi, emitted by a surface at temperature T, touches another surface then (Figure 1.26):

A part is absorbed: Ra Ri/q qα = (1.59)A part is re� ected: Rr Ri/q qρ = (1.60)A part is transmitted: Rt Ri/q qτ = (by transparent layers) (1.61)

where �, � and are the average absorptivity, re� ectivity and transmissivity at temperature T. Conservation of energy dictates: � + � + = 1. If the three belong to different temperatures, that conservation law does not hold. Put another way, never add absorbed, re� ected and transmitted radiant heat � ow rates which are emitted at different temperatures.

1.4 Radiation

1335vch01.indd 59 12.02.2007 11:36:21

60 1 Heat Transfer

Diffuse re� ection differs from specular re� ection. The latter obeys the laws of optics: incident and re� ected beam in the same plane, re� ection angle r equal to the incident angle i (Figure 1.27). Most surfaces re� ect diffusely, which means that the re� ected radiation is scattered in all directions. Re� ectivity can also be de� ned in relation to the radiation intensity incident on a surface at an angle . For diffuse re� ection the result is then Rr Ri/I Iφ α φρ = , where IRr � is the re� ected intensity in a direction �. For specular re� ection it is, Rr Ri( / )I Iφ φρ = where � is a function of the angle of incidence.

Table 1.6. Quantities.

Quantity De� nition and units Equations

Radiant heat QR The amount of heat, which is emitted or received as electromagnetic waves. QR is a scalar, units: J.

Radiant heat � ow �R The amount of radiant heat per unit of time. �R is a scalar, units: W.

Rd

d

Q

t

Radiant heat � ow rate qR The radiant heat � ow per unit of surface. qR is a scalar (reason: a surface emits radiation in all directions) with units W/m2. The term irradiation, symbol E, is used to characterize the incoming radiant heat � ow rate, the term emittance, symbol M, to characterize the emitted radiant heat � ow rate.

2Rd

d d

Q

A t

Radiation intensity I The radiant energy, which is emitted in a speci� c direction. I is a vector, units W/(m2 · rad). d� is the elementary solid angle in the direction considered.

2R Rd d

ofd d d

q

A

Φω ω

Luminosity L The ratio between the radiant heat � ow rate in a direction and the apparent surface, seen from that direction. L is a vector, units W/(m2 · rad). The luminosity describes how a receiving surface sees an emitting surface.

2Rd

cos( ) d dA

Φφ ω

Figure 1.26. Radiant heat transfer, re� ection, absorption and transmission at a surface.

Figure 1.27. Thermal radiation, specular re� ection.

1335vch01.indd 60 12.02.2007 11:36:21

61

Most building and insulation materials are non-transparent for thermal radiation ( = 0). The incoming radiation is absorbed in a very thin surface layer, for metals 10–6 m thick, for other materials 10–4 m thick. For that reason, the term ‘absorptive surface or body’ and not ‘absorpting material’ is currently used. At the other hand, most gases, most � uids and some solids (glass, synthetics) are selectively transmittent ( � 1). At the same time, they all show selective mass absorption which is described by the extinction coef� cient a (Figure 1.28):

R

R

dd

qa x

q= − (1.62)

For a transmittent layer with thickness d, the transmitted radiant heat � ow rate then becomes: Rt Ri exp( )q q a d= − , where qRi is the incoming radiant heat � ow rate. The absorbed heat � ow

rate is: Ra Ri Rt Ri [1 exp( )]q q q q a d= − = − − . Absorptivity and transmissivity thus become: 1 exp( )a dα = − − , exp ( )a dτ = − .

Specular re� ectivity is given by:

21 i 2 tr

i 1 i 2 t

cos( ) cos( )

cos( ) cos( )

n nI

I n n

⎡ ⎤φ − φρ = = ⎢ ⎥φ + φ⎣ ⎦

where n1 and n2 are the refractive indexes of the media at both sides of the incident surface (n = 1 for air), i is the angle of incidence and t the angle of transmittance into the second medium.For all materials, absorptivity, re� ectivity and transmissivity vary with wavelength and, consequently, with temperature. If the three are de� ned as radiation intensity ratios, then also the angle of incidence intervenes. The wavelength can have an important impact. An example is glass, where the transmissivity for visible light is large and the absorptivity for visible light small, while for UV and IR transmissivity approaches zero and absorptivity 0.9 and more. The remarkable difference in properties causes the greenhouse effect. Solar radiation (short-wave radiation, high radiant temperature), which the glass transmits, is absorbed by all inside surfaces and re-emitted as low temperature radiation (IR radiation, low radiant temperature). That IR is absorbed by the glass at its inside surface, leaving conduction through it as the only means to return part of the short wave gain back to the outside. Most solar heat is thus absorbed by the inside environment and only released slowly. As a consequence, the inside temperature may reach uncomfortably high values. Transparent synthetic materials have analogous characteristics although some are transparent for IR.

Figure 1.28. Thermal radiation, mass absorption.

1.4 Radiation

1335vch01.indd 61 12.02.2007 11:36:21

62 1 Heat Transfer

1.4.1.4 Radiant Surfaces

The black body, a surface which absorbs all incident radiation (� = 1, � = 0, = 0, � f (�, )) is used as reference. Although just as non-existing as ideal gases or pure elastic materials, the study of a black body is enlightening for real surfaces. These are divided in:

Grey bodies: for each wavelength absorptivity is a constant, independent of direction (� = Cte, < 1).Blank bodies: are gray surfaces with absorptivity zero.Colored bodies: absorptivity depends upon wavelength (= temperature) and direction.

Real bodies are by de� nition colored. Therefore also the blank and grey ones are just models used to keep absorption, re� ection and transmission independent of wavelength and direction. In all further discussions, we assume that all bodies are blank or grey. In order not to put too much strain on reality, a difference is made between solar shortwave and ambient long wave radiation. For both, bodies are considered grey, although with different absorptivity and re� ectivity, which is expressed by using an index L for longwave and S for shortwave.

1.4.2 Black Bodies

1.4.2.1 Characteristics

Of all surfaces, black bodies emit the highest radiant energy � ow rate, independent of temperature. Their emissivity e (de� nition: see hereafter) is therefore 1. This starts with the second law of thermodynamics. In a closed system, black bodies at different temperatures must evolve irrevocably towards temperature equilibrium. Because each emits the same amount of radiation as it absorbs at equilibrium, the emissivity of a black body must equal its absorptivity, i.e. 1.With respect to direction, a black body obeys Lambert’s law: luminosity (Lb) constant. Hence, radiation intensity (Figure 1.29, suf� x b for ‘black radiation’) becomes:

b b cos( )I Lφ = φ (1.63)

�

�

�

Figure 1.29. The cosine law.

1335vch01.indd 62 12.02.2007 11:36:22

63

This equation is known as the cosine law. It offers a simple relation between emittance and luminosity. From the de� nitions it follows:

b b cos( ) dM Lω

= φ ω∫

where Ω∫ is the integral over the hemisphere.

The elementary solid angle d� can be calculated if we assume a hemisphere with radius ro above the surface dA with emittance Mb. Then the following applies: 2

od sin ( ) d dω = φ φ ϑr , while, on the hemisphere, the intensity decreases from Ib to 2

b o/φI r , or:

/ 22 2 / 2

b b o b 020 0 o

2 cos( )sin ( ) d d [ cos ( )]

ππ

π φ= φ φ ϑ = −π φ∫ ∫M L r Lr

,

which reduces to (Figure 1.30):

b bM L= π (1.64)

Spectral density of the emittance is given by Planck’s law:

2 5

b2

exp 1

c hM

c h

k T

−

λπ λ=⎛ ⎞

−⎜ ⎟⎝ λ ⎠

(1.65)

where c is the velocity of light in m/s, h is Planck’s constant (6.624 · 10–34 J · s) and k is Boltzmann’s constant (1.38047 · 10–23 J/K). The products 2 � c2 h and c h / k are called the radiation constants for black bodies, symbols C1 (3.7415 · 10–16 W · m2) and C2 (1.4388 · 10–2 m · K). Figure 1.31 gives the spectral density for different values of the absolute temperature. The emittance increases very quickly with temperature (= surface under the curve), while the spectral maxima occur at ever smaller wavelengths. The maxima have as a geometric locus in the [�, MB�]-coordinate system a hyperbole of the 5e order. Their wavelengths satisfy Wien’s law:

�M T = 2898 (�M in �m) (1.66)

For example, at 20 °C the maximum lies in the infrared part of the spectrum with �M = 2898 / 293.16 = 9.9 �m. For the sun, with a radiant temperature of 5800 K, the maximum lies in the middle of the visible light part, �M = 2898 / 5800 = 0.5 �m.

Figure 1.30. Cosine law, proof.

1.4 Radiation

1335vch01.indd 63 12.02.2007 11:36:22

64 1 Heat Transfer

The emittance Mb follows from the integration of (1.65) to the wavelength, with as integration boundaries 0 and �:

5 44 4

b b 2 30

2d

15

kM M T T

c hλ

∞ π= λ = = σ∫ (1.67)

Here � stands for Stefan’s constant, 5.67 · 10–8 W/(m2 · K4). Equation (1.67) is known as the Stefan-Boltzmann law. This law and Wien’s law were known before Planck’s law. For that we had to wait until the theory on quantum mechanics was developed. Most of the time, Stefan-Boltzmann law is written as:

4

b b 100

TM C

⎛ ⎞= ⎜ ⎟⎝ ⎠ (1.68)

Figure 1.31. Spectral density of the emittance.

1335vch01.indd 64 12.02.2007 11:36:22

65

where Cb is the black body constant, 5.67 W/(m2 · K4), and T / 100 is the reduced radiant temperature. The luminosity and the radiation intensity then become:

4b b

b 100

M C TL

⎛ ⎞= = ⎜ ⎟⎝ ⎠π π

42Rb b

b bd

cos( ) cos( )d d 100

C TI L

AφΦ ⎛ ⎞= = φ = φ⎜ ⎟⎝ ⎠ω π

1.4.2.2 Radiation Exchange Between Two Black Bodies: The Angle Factor

If two black bodies, dA1 and dA2 are located as in Figure 1.32 without a medium in-between, then we have as elementary radiant heat � ow between dA1 and dA2:

2 b1R,1 2 b1 1 1 1 1 1d d d cos( ) d d

MI A A→Φ = ω = φ ω

π

where d�1 is the solid angle under which dA1 sees dA2. d�1 equals 22 2d cos( ) /A rφ , or:

2 b1 2R,1 2 1 2 1 2

dd cos( ) cos( ) d

M AA

r→Φ = φ φ

π

Analogously:

2 b2 2R,2 1 1 2 1 2

dd cos( ) cos( ) d

M AA

r→Φ = φ φ

π

The resulting radiant heat � ow between both bodies then becomes:

From body 1 to body 2

2 2 2 b1 b2 1 2 1 2R,12 R,1 2 R,2 1 2

( ) cos( ) cos( ) d dd d d

M M A A

r→ →

− φ φΦ = Φ − Φ =

π

From body 2 to body 1

2 2 2 b2 b1 1 2 1 2R,21 R,2 1 R,1 2 2

( ) cos( ) cos( ) d dd d d

M M A A

r→ →

− φ φΦ = Φ − Φ =

π

Figure 1.32. Calculation of the angle factor.

1.4 Radiation

1335vch01.indd 65 12.02.2007 11:36:23

66 1 Heat Transfer

If both bodies are � nite in shape:

From body 1 to body 2

1 2

1 2 2 1R,12 b1 b2 1 2

1

cos( ) cos( ) d d1( )

A A

A AM M A

A r

⎡ ⎤φ φΦ = − ⎢ ⎥

π⎢ ⎥⎣ ⎦∫ ∫ (1.69)

From body 2 to body 1

2 1

1 2 1 2R,21 b2 b1 2 2

2

cos( ) cos( ) d d1( )

A A

A AM M A

A r

⎡ ⎤φ φΦ = − ⎢ ⎥

π⎢ ⎥⎣ ⎦∫ ∫ (1.70)

The term between brackets in both formulas is called the angle factor, symbol F. If we look from A1 to A2, then the angle factor is written as F12. Vice versa, the symbol F21 is used. Angle factors are pure geometrical quantities indicating which fraction of the radiant heat � ow emitted by a body reaches the other. They depend upon the size of each body, their form, the distance and the angle at which both face each other. Its maximum value is one. In that case all emitted radiation falls on the other body.

1.4.2.3 Properties of Angle Factors

Reciprocity exists in a sense that A1 F12 = A2 F21 (this follows directly from the de� ni-tion).If surface A1 is completely surrounded by surface A2, than the angle factor F12 is one (Figure 1.33). This characteristic can be extended to each surface, surrounded by n other surfaces, which together form a closed volume. In such a case:

12

1n

jj

F=

=∑

�

�

Figure 1.33. Angle factor between a body 1, completely surrounded by body 2.

1335vch01.indd 66 12.02.2007 11:36:23

67

1.4.2.4 Calculating Angle Factors

An analytical calculation of the angle factor is possible for some simple con� gurations:

Two in� nitely sized, parallel surfaces (e.g. the surfaces of a cavity):

F12 = F21 = 1

A point (equals an in� nitesimally small sphere) at a distance D from a rectangle with sides L1 and L2. The angle factor can be calculated as the ratio between the solid angle, at which the point sees the rectangle, and the total solid angle (4 �) (point = A1, rectangle = A2):

2

12 22

1 cos( )d

4 A

F Ar

φ=π ∫

If the point is situated directly under a corner of the rectangle, then the integral becomes (cos( ) /D rφ = and r2 = D2 + x2 + y2):

1 2

12 2 2 2 3 20 0

1d d

4 ( )

L LD

F y xD x y

=π + +∫ ∫

with:

2 21 2

121 2

1 1atan

8 4

⎛ ⎞+= − ⎜ ⎟

π ⎜ ⎟⎝ ⎠

D L LF

L L

as a solution. Each other position of the point is converted to that case by the construction of Figure 1.34. The resulting angle factor is then: F12 = F1a + F1b + F1c + F1d. Radiation between the head and a ceiling is an example of that.

An in� nitely small surface dA1 at an orthogonal distance D from a parallel rectangle with sides L1 and L2 and surface A2. The formula for the angle factor in that case becomes:

1 2

1 2 2 112 2

1 d

cos( ) cos( ) d d1

d A A

A AF

A r

φ φ=

π∫ ∫

�

�

�

Figure 1.34. Angle factor, point at a distance D of a surface A2.

1.4 Radiation

1335vch01.indd 67 12.02.2007 11:36:24

68 1 Heat Transfer

If dA1 is in� nitely small with respect to A2, then the way dA1 sees each in� nitely small surface dA2 on A2 is independent of the position of dA1, or:

2 1

1 2 212 2

seen by d

cos( ) cos( ) d1

A A

AF

r

φ φ=π ∫

Suppose surface dA1 lies under the corner (0, 0, D) of rectangle A2 (Figure 1.35). The angle factor then becomes:

1 21 2

12 20 0

cos( ) cos( )1d d

L L

F x yr

φ φ=π ∫ ∫

As 1 2cos cos /D rφ = φ = and r2 = D2 + x2 + y2, that equation simpli� es to

1 2 2

12 2 2 2 20 0

1 d d

( )

L LD y x

FD x y

=π + +∫ ∫

with:

1 2 2 112 2 2 2 2 2 2 2 2

1 1 2 2

1tan tan

2

L L L LF a a

D L D L D L D L

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= +⎜ ⎟ ⎜ ⎟⎢ ⎥π ⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠⎣ ⎦

as a solution.For any other position of the in� nitesimal small surface dA1 the same method as that for the [point to surface]-con� guration is applied.

All other con� gurations demand a numerical solution. That is the case for an in� nitely small surface dA1, perpendicular to a rectangle with sides L1 and L2 and surface A2. Under no circumstances does dA1 face that part of surface A2 which lays behind the intersection of the plane containing dA1 with A2, leaving a quarter of a globe as reference solid angle. The numerical formula for the angle factor then becomes:

Figure 1.35. Angle factor, in� nitesimal small surface dA1 parallel to a surface A2 at a distance D of that surface.

1335vch01.indd 68 12.02.2007 11:36:24

69

1 2

2 2

12 2 2 2 2/ 2 to / 2 step / 2 to / 2 step2 ( )x x L x x y y L y y

x yD x yF

x y D=Δ −Δ Δ =Δ −Δ Δ

⎡ ⎤+Δ Δ ⎢ ⎥=π + +⎢ ⎥⎣ ⎦

∑ ∑

where D is the orthogonal distance between the surfaces dA1 and A2. Other con� gurations which are quite common in rooms and where the angle factor can only be calculated numerically are two identical surfaces in parallel and two surfaces with a common side, perpendicular to one another. Numerics are easily programmed and calculated, using spread sheet software.The angle factor allows writing the radiant heat � ow and radiant heat � ow rate as:

R,12 1 12 b1 b2

R,21 2 21 b2 b1

( )

( )

A F M M

A F M M

φ = −

φ = −

R,12 12 b1 b2

R,21 21 b2 b1

( )

( )

q F M M

q F M M

= −

= −

(1.71)

For n black bodies radiating to one another, both heat � ow and heat � ow rate per body become:

R1 1 1 b1 bj2[ ( )]

n

n jj

A F M M=

φ = −∑

R1 1 b1 bj2[ ( )]

n

n jj

q F M M=

= −∑

(1.72)

where 1 is the body considered and 2 to n are the n – 1 others.

1.4.3 Grey Bodies

1.4.3.1 Characteristics

The laws for grey bodies are similar to the ones for black bodies. Only radiation exchange changes:

For each wavelength and each direction, grey bodies emit the same constant fraction of radiation a black body produces. The ratio b/M M is called emissivity e. Conservation of energy tells us that absorptivity � must equal emissivity e. Re� ectivity � in turn is given by � = 1 – � = 1 – e. Grey bodies with re� ectivity 1 are blank.

A grey body follows Lambert’s law (L = Cte). Consequently, radiant heat � ow rate obeys the cosine law, while the emittance is given by: M = � L.

Spectral emittance obeys Planck’s law, multiplied by the emissivity e. Total emittance is:

4

b 100

TM e C

⎛ ⎞= ⎜ ⎟⎝ ⎠ (1.73)

Each grey body emits and re� ects radiation. Hence, the radiant heat � ow rates between a number of grey bodies can be described as follows: let M be the emittance of one of them and E the irradiation by all others. The radiosity of the one grey body, i.e. the radiant impression it gives, is then:

bM e M E′ = + ρ (1.74)

The emitted radiant heat � ow rate is now equal to the difference between radiosity and irradiation, or:

Rq M E= ′ − (1.75)

�

�

�

�

1.4 Radiation

1335vch01.indd 69 12.02.2007 11:36:25

70 1 Heat Transfer

Eliminating the unknown irradiation between the equations (1.74) and (1.75) gives:

bR b( )

M e M eq M M M

−′= − = −′ ′

ρ ρ (1.76)

The received radiant heat � ow rate is then:

R b( ) e

q M M= −′ρ

(1.77)

Equations (1.76) and (1.77) can be read as follows: treat a grey body as an equivalent black body with a grey � lter in front, and between the body and the � lter a radiant resistance equal to the ratio between re� ectivity and emissivity (�/e). The black body has an emittance Mb, the grey � lter a radiosity M�.

1.4.3.2 Radiation Exchange Between Grey Bodies

Two Bodies

In the case of two grey bodies, separated by a transparent medium, the resulting radiant � ow is:

Equivalent black body 1 � grey � lter 1: 1R,11 b1 1 1

1

( )e

M M AΦ = − ′ρ

Grey � lter 1 � grey � lter 2: R,12 12 1 2 1( )F M M AΦ = ′ − ′

The second equation is identical to the equation for the radiant heat � ow between two black bodies. Indeed, both the emittance of a black body and the radiosity of a grey body represent diffuse radiation which obeys Lambert’s law.

Grey � lter 2 � equivalent black body 2: 2R,22 2 b2 2

2

( )e

M M AΦ = −′ρ

In the three equations, the radiant heat � ows �R,11, �R,12 and �R,22 are identical (energy conservation). After elimination of the unknown radiosities 1M ′ and 2M ′ , the radiant heat � ow from body 1 to body 2 becomes:

R,12 b1 b2 11 2 1

1 12 2 2

1( )

1M M A

A

e F e A

⎡ ⎤⎢ ⎥⎢ ⎥Φ = −ρ ρ⎢ ⎥+ +⎢ ⎥⎣ ⎦

(1.78)

For the radiant heat � ow from body 2 to body 1 we get:

R,21 b2 b1 22 1 2

2 21 1 1

1( )

1M M A

A

e F e A

⎡ ⎤⎢ ⎥⎢ ⎥Φ = −ρ ρ⎢ ⎥+ +⎢ ⎥⎣ ⎦

(1.79)

1335vch01.indd 70 12.02.2007 11:36:25

71

The term between brackets is called the radiation factor, symbol: FR. If A1 is seen as the emitting body, then the factor is written as FR,12,. If it is A2, then we use FR,21. The radiant heat � ow rate is found by dividing both equations by the surface seen as emitting. In (1.78) this is A1, in (1.79) A2.

Frequent con� gurations of two grey bodies are:

Two in� nitely large, parallel isothermal surfaces. In that case: F12 = F21 = 1, A1 = A2 and (1.78) becomes (� = 1 – e):

R,12R,12 b1 b2

1

1 2

1( )

1 11

q M MA

e e

⎡ ⎤⎢ ⎥Φ⎢ ⎥= = −⎢ ⎥+ −⎢ ⎥⎣ ⎦

(1.80)

In building construction, the term between brackets stands for the radiation fac-tor of an in� nite cavity. If one of the two surfaces is blank (say: surface 1), then

R,12 21 /(1 / 0 1 / 1) 0F e= + − = . If one of the two surfaces is black (i.e. surface 2), then: R,12 1 11 /(1 / 1 /1 1)F e e= + − = .

An isothermal surface 1 surrounded by an isothermal surface 2. Now F12 is 1 and (1.78) becomes:

1 2R,12 b1 b2 1

1 2 12 1 1 2

2

( )e e

M M Ae A

e e eA

Φ = −ρρ + +

If both bodies are almost black (e > 0.9), then the denominator in the radiation factor becomes � 1, and: R,12 1 2 b1 b2 1( )e e M M AΦ = − . If the surrounded surface A1 is more-over very small compared to the surrounding surface A2 (A1/A2 � 0), then the equation further simpli� es to:

R,12 1 b1 b2 1( )e M M AΦ = − (1.81)

In such case, the resulting radiant heat � ow only depends on the emissivity of body 1.

Multiple Surfaces

In case of multiple isothermal grey bodies, each at a different temperature, the radiant heat � ow between body 1 and the n – 1 others can be written as:

Equivalent black body 1 � grey � lter 1: 1R,11 b1 1 1

1

( )e

M M AΦ = − ′ρ

Grey � lter 1 � n – 1 other grey � lters: R,1 1 1 12

( )n

j j jj

F M M A=

Φ = ′ − ′∑

Because of energy conservation, both � ows are identical (�R,11 = �R,1j), or:

1 1b1 1 1 1

2 21 1

1 ( ) n n

j j jj j

M F M F Me e= =

⎛ ⎞ρ ρ= + −′ ′⎜ ⎟⎝ ⎠

∑ ∑ (1.82)

�

�

1.4 Radiation

1335vch01.indd 71 12.02.2007 11:36:25

72 1 Heat Transfer

When body 1 is completely surrounded by the surfaces 2 to n, then (1.82) converts into:

1 1b1 1

21 1

( )n

j jj

MM F M

e e =

′ ρ= − ′∑ (1.83)

For each grey body we obtain an equation (1.82) or (1.83), whereby the radiosities jM ′ are the unknown and the (black body) emittances Mb1 the known quantities. The result is a formula of n equations with n unknown radiosities:

b .[ ] [ ] [ ]j n n n j nM F M= ′ (1.84)

[F]n.n is called the radiation matrix, in the above case, of a system of n isothermal grey bodies. The solution of the system gives the radiosities jM ′ as a function of the n emittances Mbj. The radiant heat � ows are then found by inserting jM ′ in the equations for the heat exchange between the different equivalent black bodies and their grey � lters.

1.4.4 Colored Bodies

For colored bodies, the emissivity, absorptivity and re� ectivity depend on wavelength (which is a function of temperature) and direction. For each of them, Kirchoff’s law (e = �) applies. Lambert’s law, however, is no longer applicable since it requires an emission independent from direction. Per wavelength, the spectral emittance differs from a black body. The average emissivity at a temperature T follows from the ratio between the emittance of the colored body and that of a black body at the same temperature.In order to simplify things, we consider colored bodies as grey, but with temperature dependent emissivity. For the emittance and irradiance at strongly different temperatures, Kirchoff’s law no longer applies. This is the case for ambient radiation and solar radiation. Therefore, �S eL, with �S the short wave absorptivity for sunlight and eL the long wave emissivity for ambient radiation (take polished aluminum: �S = 0.2 to 0.4 and eL = 0.05).

1.4.5 Practical Formulae

As has been shown, thermal radiation can be modeled quite accurately. However, calculating all angle factors is quite cumbersome and the number of equations in system (1.84) may become very large. A more simple approach is therefore welcomed. In a � rst step, each con� guration of n radiant bodies is replaced by a (� ctive) system of two radiant bodies: the surface considered and the remaining n – 1 surfaces which form the environment. In a second step, we suppose the environment to be a black body at a temperature �r, the radiant temperature of the environment facing the surface. This is the temperature the environment as a black body should have, if the surface considered, hereafter called surface 1, is to receive or lose the same radiant heat � ow as in reality. The radiant temperature is found by solving the system of equations (1.84) for radiation between all surfaces in the environment. That way, we get the radiosity 1M ′ of surface 1 as a linear combination of the (black body) emittances of all surfaces included 1:

1 r b1

n

i ii

M a M=

=′ ∑

Inserting that result in equation (1.77) and equating with (1.81) gives for the radiant temperature:

1335vch01.indd 72 12.02.2007 11:36:25

73

4 44 r 1 111

1273,15

n

r i ii

a T e T=

⎛ ⎞θ = − −⎜ ⎟ρ ⎝ ⎠∑

In analogy to convection, the radiant heat � ow and radiant heat � ow rate are now written as:

r r s1 r r r s1 r( ) ( )q h h A= θ − θ Φ = θ − θ (1.86)

In both formulas, hr represents the surface � lm coef� cient for radiation, with units of: W/(m2 · K). �s1 is the temperature of surface 1 while �r is the radiant temperature of the environment as seen by the surface. The surface � lm coef� cient for radiation varies with the kind of two-radiant bodies con� guration used as the reference. If the environment surrounds surface 1, a situation which is typical indoors, then the surface � lm coef� cient follows from equaling (1.86) with (1.81) where-in 4

b b s1( /100)M C T= :

4 4s1 r

r bs1 r

100 100

T T

h e C

⎡ ⎤⎛ ⎞ ⎛ ⎞−⎢ ⎥⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎢ ⎥= ⎢ ⎥θ − θ⎢ ⎥⎢ ⎥⎣ ⎦

(1.87)

In (1.87) the term between parentheses is called the temperature ratio for radiation, symbol FT. FT can be simpli� ed to:

2 2 3s1m r m

T4

5000 100 100 100 100

TT T TF

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + ≈⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

with m s1 r( ) / 2T T T= + .

Calculations show that FT varies little between –10 °C and +50 °C. Hence, the simpli� ed expression almost always suf� ces, which allows considering equation (1.86) as linear. Inserting FT in (1.87) gives a surface � lm coef� cient for radiation of:

r 1 b Th e C F= (1.88)

Other combinations than the body, surrounded by the environment, are possible:

Body and environment act as parallel surfaces. If the environment, i.e. the other surface, is isothermal, temperature �s2, then the detour via the radiation temperature is super� uous and the radiant heat � ow rate may be written directly as:

Tr r s1 s2 r

1 2

5,67( ) with

1 11

Fq h h

e e

= θ − θ =+ −

Only part of the environment participates in the radiant exchange (this happens when some surfaces are at the same temperature as surface 1). The radiant temperature could then only include the surfaces with a temperature different from surface 1 r( )= θ′ . In such case, the surface � lm coef� cient for radiation becomes:

�

�

1.4 Radiation

1335vch01.indd 73 12.02.2007 11:36:26

74 1 Heat Transfer

1 b 12 Tr r s1 r r

1 1 12

( ) withe C F F

q h he F

= θ − θ′ =+ ρ

F12 is thereby the angle factor between surface 1 and the enclosing surfaces at a different temperature. If surface 1 is almost black, the denominator in hr becomes 1, and:

r 1 b 12 T 1 12 T5.67h e C F F e F F= = (1.89)

Take for example a surface in an exterior corner formed by two identical walls. Both walls are at the same temperature. That results in a radiant exchange with only half the space, i.e. the part not shielded by the other wall. Angle factor: 1/2. Surface � lm coef� cient for radiation: r 1 b T 1 T/ 2 2,84h e C F e F= = . Of course, surfaces at the same temperature may also be included in the radiant temperature. The angle factor then remains 1 but the radiant temperature changes.

1.5 Applications

1.5.1 Surface Film Coef� cients and Reference Temperatures

1.5.1.1 Overview

In reality, the three modes of sensible heat transfer, which were treated separately, always occur together. Consider for example a massive outside wall. Heat is transferred through the wall by conduction. Between the inside environment and the inside surface and between the outside environment and the outside surface, heat is transferred by convection and radiation. Both modes are represented be the product of a surface � lm coef� cient (hc, hr) and a temperature difference. But convection at a surface and radiation at that surface are so profoundly intertwined that, where possible, the separate surface � lm coef� cients are combined into one single value for heat transfer inside (hi) and one single value for heat transfer outside (he), Both single values are de� ned with respect to a reference temperature (�ref,i en �ref,e). Of course, convection and radiation can also be kept separate. In that case, each environment will be characterized by an air temperature and a radiant temperature, the last as sensed by the surface considered.

1.5.1.2 Inside Environment

Average Surface Film Coef� cient for Heat Transfer hi

Calculation

Suppose surface j is isothermal, with a temperature �si. For the average convective heat � ow rate between the air and the surface we have: ci ci, i, si,[ ] ( )j j j jq h= θ − θ where hci,j is the area-averaged convective surface � lm coef� cient at that surface and �i,j is the average air temperature outside the boundary layer. If not that temperature, but the air temperature in the center of the room at a height of 1 m (�i) is taken as the reference (hereafter called the central air temperature), then the average convective heat � ow rate changes into (suf� x j omitted):

ci ci i si( )q h= θ − θ

1335vch01.indd 74 12.02.2007 11:36:26

75

where hci is the average surface � lm coef� cient for convective heat transfer with the central air temperature �i as reference temperature:

i, sici ci,

i si

jjh hθ − θ⎛ ⎞

= ⎜ ⎟θ − θ⎝ ⎠

For the radiant heat � ow rate between the inside environment and surface j, we have:

ri ri ri si( )q h= θ − θ

where �ri is the radiant temperature seen by surface j. The total heat � ow rate between the inside environment and surface j then becomes:

i ci ri ci i si ri ri si( ) ( )q q q h h= + = θ − θ + θ − θ

or:

ci i ri rii ci ri

ci ri

( ) sih h

q h hh h

⎛ ⎞⎡ ⎤θ + θ= + − θ⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦

(1.90)

The sum hci + hri is called the inside surface � lm coef� cient (for heat transfer). Symbol hi, units W/(m2 · K). The term between the brackets is the reference temperature, as seen by surface j. Symbol �ref,i. That reference temperature equals the weighted average between the central air temperature and the radiant temperature, seen by surface j, with the respective surface � lm coef� cients for convective and radiant heat transfer as weighting factors.

Values

Standard values for the surface � lm coef� cient for convective heat transfer hci are (in W/(m2 · K), EN-standard):

Vertical surfaces � 3.5

Horizontal surfaces

Heat upwards (q �)Heat downwards (q �)

5.51.2

The surface � lm coef� cient for radiation is (surface completely surrounded): hri = 5.67 eL FT, where eL is the long wave emissivity of the surface considered and FT is the temperature ratio for radiation in the interval �si – �ri. Inside, FT scatters around 0.95, or hri � 5.4 eL. Since most � nishing materials have a long wave emissivity of 0.8 � eL � 0.9, we get: 4.3 � hri � 4.95 W/(m2 · K). The inside surface � lm coef� cient is then (W/(m2 · K), EN-standard):

Vertical surfaces 7.7

Horizontal surfaces

Heat upwards (q �)Heat downwards (q �)

106

1.5 Applications

1335vch01.indd 75 12.02.2007 11:36:26

76 1 Heat Transfer

These values are not accurate for cases which deviate substantially from normal, such as the use of re� ective � nishings, the presence of large temperature differences, etc. In such cases, one should return to the theory and de� ne case-relevant inside surface � lm coef� cients. It is still better is to decouple radiation and convection.

Interpretating the Reference Temperature

Calculating the radiant temperature �ri is quite complex. That’s why, provided that the room is beam shaped and all walls are grey bodies with a long wave emissivity � 0.9, the radiant temperature �ri is usually replaced by an area-weighed average temperature of all inside surfaces:

sri

1

1

nk kn

kk

k

A

A=

=

θθ = ∑

∑ (1.91)

Vertical, sloped and horizontal exterior walls, the last with upward heat � ow:If the surface considered behaves as a grey body, the reference temperature becomes:

ref,i i ri0.44 0.56θ = θ + θ . That value is close to a temperature which governs thermal comfort in residential and commercial buildings: the resulting temperature. Symbol: �rs. The resulting temperature is usually calculated as the average between the central air and the radiant temperature, or:

i riref,i rs 2

θ + θθ = θ = (1.92)

For re� ective surfaces, the resulting temperature equals the central air temperature: �rs � �i. In such a case, only convection is left as heat transfer mode.

Horizontal inside partitions and exterior walls, downward heat � ow:For a grey surface, the reference temperature can be expressed as:

�ref,i = 0.2 �i + 0.8 �ri (1.93)

Vertical, sloped and horizontal inside partitions, the last with the heat � owing upwards:The larger impact of the ‘colder’ outside walls on the radiant temperature seen by the inside partitions, makes the above simpli� cations for the inside reference temperature less evident.

Local Surface Film Coef� cient for Heat Transfer

An average surface � lm coef� cient gives information on area-averaged phenomena, such as heat loss and heat gain. When we need to know the surface temperature on a speci� c spot or a speci� c surface part, then local surface � lm coef� cients are needed. A simpli� ed calculation is given here.In general, the following holds (left and right part describe the same heat transfer):

ix ref, six cix ix six rix rix six( ) ( ) ( )ih h hθ − θ = θ − θ + θ − θ (1.94)

with hix the local surface � lm coef� cient linked to a reference temperature �ref,i. �six is the local surface temperature, hcix the local convective surface � lm coef� cient, �ix the local air

�

�

�

1335vch01.indd 76 12.02.2007 11:36:26

77

temperature just outside the boundary layer, hrix the local surface � lm coef� cient for radiation, and �rix the radiant temperature of the environment seen by the spot. If each partial surface can be represented by an equivalent thermal resistance between the inside surface and the environment at the other side (R�), then the following holds:

six ref,ix ref, six

( )( ) j

ihR

θ − θθ − θ =

′ (1.95)

Equation (1.95) is an approximation. In fact, the equivalent thermal resistance depends on the distribution of local surface � lm coef� cients (hix) over the inside surface. �ref,j, the reference temperature at the other side, is the outside temperature (j = e) for outside walls, and the reference temperature in the adjacent room (j = i) for inside partitions. Eliminating �six from (1.94) and (1.95) and solving for the local surface � lm coef� cient (hix) then gives:

cix rix Tix

T1

h h ph

R p

+ −=

+ ′ (1.96)

where:

cix ref, ix rix ref, rixT

ref ref,

( ) ( )i i

i j

h hp

θ − θ + θ − θ=

θ − θ (1.97)

If the reference temperature �ref,i, the relationships between �ref,i and �ix and �ref,i and �rix and the local surface � lm coef� cients hcix and hrix are known, then the combination of (1.96) with (1.97) gives the local surface � lm coef� cient for heat transfer (hix) on condition that the equivalent thermal resistance R� is calculated in advance. For example:

1. Reference temperature inside (�ref,i) The central air temperature in the room, 1.7 m above � oor level.

2. Relationship between �ref,i and �ix

Suppose the central air temperature increases linearly with height. The height gradient becomes smaller as the room is better insulated and the heating elements produce less convection. In a formula:

ix ref,c m

ref, ref,

1 0,2 ( 1, 7)i

i j

p U yθ − θ

= + −θ − θ

(1.98)

with y the height ordinate, �ref,i the reference temperature in the room, �ref,j the reference temperature in the adjacent room or outside, pc the convection factor of the heating element (1 for air heating, 0.9 for convectors, 0.4 to 0.8 for radiators, 0.4 for � oor heating) and Um the average thermal transmittance (see next paragraph) of the room:

i i im

i

U A aU

A= ∑

∑ (1.99)

In this formula ai is a weighing factor for the inside partitions ref, ref, ref, ref,( ( ) /( ))i j i e= θ − θ θ − θ with �ref,j equal to the central air temperature in the adjacent room). Equation (1.98) followed from measurements in a test room, where the heating system and the insulation of the exterior walls could be varied.

1.5 Applications

1335vch01.indd 77 12.02.2007 11:36:26

78 1 Heat Transfer

3. Relationship between �ref,i and �rix

We replace the speci� c radiant temperature, as seen by the surface, by a uniform radiant temperature �ri, which is proportional to the reference temperature with a gradient depending on the local convective surface � lm coef� cient, the convection factor of the heating element and the average thermal transmittance of the room:

ri ref, cix

c mref, ref,cix

( 0, 4)

0,6

i

i j

hp U

h

θ − θ= −θ − θ +

(1.100)

For the interpretation of the different parameters, see above. Equation (1.100) followed from computer simulations of the radiant exchanges in rooms of different shape.

4. Local convective surface � lm coef� cient Take hcix equal to 2.5 W/(m2 · K).

5. Local surface � lm coef� cient for radiation hrix?

Corner between three exterior walls or between two exterior walls and an inside partition

5.5 eL Surfaces at more than 0.5 m from the corner line and the corner point

3.4 eL Surfaces within the 0.5 m strip around the corner line but at more than 0.5 m from the corner point

2.2 eL Surfaces within the 0.5 m strip around the corner point

Corner between an exterior wall and an inside partition or between two exterior walls

5.5 eL Surfaces at more than 0.5 m from the corner line

3.4 eL Surfaces within the 0.5 m strip around the corner line

Outside wall or inside partition wall

5.5 eL –

6. Remark Furniture close to an outside wall is equivalent to a surface � lm coef� cient of 2 W/(m2 · K).

1.5.1.3 Outside Environment

Calculation

The heat balance at a surface is more complicated now. In fact, not two but four heat � ow rates intervene:

1. Convection between the outside environment and the surface

The average convective heat � ow rate is given by: ce ce, e, se,( )j j jq h= θ − θ with hce,j the average surface � lm coef� cient for convection and �e,j the average temperature of the air outside the boundary layer. If the outside temperature measured at the nearest weather station is used as the reference (the air temperature in an open � eld under a thermometer hut, 1.7 m above grade) then the average heat � ow rate becomes:

1335vch01.indd 78 12.02.2007 11:36:26

79

ce ce e se( )q h= θ − θ

where hce is the average convective surface � lm coef� cient for the reference temperature:

se exce cex

se e

h hθ − θ

=θ − θ

2. and 3. Longwave radiation between the surface, the terrestrial outside environment (2) and the sky (3)

The surface, the terrestrial outside environment and the sky form a system of three radiant bodies, the sky black and the others grey. Radiation balances:

From the surface (s) to the outside environment (e) and the sky (sk)

Ls Lsbs se ssk s se e ssk bsk

Ls Ls

1 ( ) ( )M F F M F M F Me e

⎡ ⎤ρ ρ= + + − +′ ′⎢ ⎥⎣ ⎦

From the outside environment (e) to the surface (s) and the sky (sk)

Le Lebe es esk e es s esk bsk

Le Le

1 ( ) ( )M F F M F M F Me e

⎡ ⎤ρ ρ= + + − +′ ′⎢ ⎥⎣ ⎦

In both equations, eLs and �Ls are the longwave emissivity and re� ectivity of the surface and eLe and �Le the average longwave emissivity and re� ectivity of the terrestrial outside environment. Fse is the angle factor between the surface and the terrestrial outside environment, Fssk the angle factor between the surface and the sky, Fes the angle factor between the terrestrial outside environment and the surface and Fesk the angle factor between the terrestrial outside environment and the sky. Mbs, Mbe and Mbsk are the black body emittances of the surface, the terrestrial outside environment and the sky, while sM ′ and eM ′ are the radiosities of the surface and the terrestrial outside environment. Now, Fse + Fssk equals 1 (together, the terrestrial outside environment and the sky surround the surface completely), while Fes equals 0 (the surface is in� nitely small compared to the terrestrial outside environment,) and Fesk equals 1 (all radiation from the terrestrial outside environment goes to the sky). The two balances then simplify to:

Ls Lebs s se e ssk bsk be e bsk

Ls Ls Le Le

1 1( )M M F M F M M M M

e e e e

ρ ρ= − + = −′ ′ ′

Solving the equations for sM ′ and inserting the result in rse rssk Ls bs s Ls( ) /q q e M M+ = − ρ′ gives (Fse + Fssk = 1, or eLs (Fse + Fssk) Mbs = eLs Mbs):

serse rssk Ls se bs Le be Ls ssk bs Le bsk

ssk

( ) 1F

q q e F M e M e F M MF

⎡ ⎤⎛ ⎞+ = − + − ρ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

This ‘exact’ result is usually replaced by an approximation, presuming that the terrestrial outside environment is a black body at outside temperature. Using the experimentally proven fact that during clear nights the sky temperature is some 21 °C lower than the air temperature, the radiant heat � ow rate between the surface and the terrestrial outside environment plus the sky becomes:

rse rssk Ls b se Tse ssk Tssk e se ssk Tssk[( ) ( ) 21 (1 0.87 )]q q e C F F F F F F c+ = + θ − θ − −

1.5 Applications

1335vch01.indd 79 12.02.2007 11:36:27

80 1 Heat Transfer

with FTse the temperature ratio for radiation between the surface and the terrestrial outside environment, FTws the temperature ratio for radiation between the surface and the sky and c the cloud factor, zero for a clear sky and one for a completely overcast sky.

4. Solar radiation

Solar radiation is shortwave. It combines three components: direct, diffuse and re� ected. A square meter of surface absorbs the product S EST as radiant heat, with EST total solar radiation and S the shortwaved absorptivity of the surface. Even though solar radiation is present only during the day, its impact is large.The resulting ‘simpli� ed’ heat balance then becomes:

e ce e se Ls se Tse ssk Tssk e se

Ls ssk Tssk c K ST

( ) 5,67 ( ) ( )

120 (1 )

q h e F F F F

e F F f E

= θ − θ + + θ − θ− − + α

If the term 5.67 eLs (Fse FTse + Fssk FTssk) is called the surface � lm coef� cient for radiation (hre), then the equation reduces to:

K ST Ls ssk Tssk ce ce re e se

ce re

120 1 )( )

E e F F fq h h

h h

⎧ ⎫⎡ ⎤α − −⎪ ⎪= + θ + − θ⎨ ⎬⎢ ⎥+⎪ ⎪⎣ ⎦⎩ ⎭ (1.101)

With the term above the brace, which has degrees centigrade as a dimension, � guring as the reference sol-air temperature (symbol *

eθ ), and the sum hce + hre called the outside surface � lm coef� cient for heat transfer he, with units of W/(m2 · K), (1.101) may be simply written as:

*e e e se( )q h= θ − θ (1.102)

Values

As standard value for the convective surface � lm coef� cient, the number 19 W/(m2 · K) has been forwarded. If the temperature ratios for radiation FTse and FTssk are given the same value FT, then, knowing that Fse + Fssk = 1, we get as surface � lm coef� cient for radiation (hre):

re L T5.67h e F= (1.103)

Since the temperature factor for radiation FT is between 0.8 and 0.9, the most probable interval for this surface � lm coef� cient is: 4.4 eL � hre � 5.1 eL W/(m2 · K). Provided the outside surface is grey with longwave emissivity (eL) 0.9, then a value 4 W/(m2 · K) is very likely. Thus, the standard outside surface � lm coef� cient for heat transfer is 19 plus 4, or:

he = 23 W/(m2 · K) (1.104)

The European Standard (EN) uses 25 W/(m2 · K) instead. The deduction of the outside surface � lm coef� cient, as given, is based on considerable assumptions. If a more accurate analysis is needed, one should return to the complete heat balance.

Interpreting the Reference Temperature

The sol-air temperature *e( )θ does not exist. In fact, it is a ‘� ctive’ air temperature which

ensures that the heat exchange with the outside surface remains the same as the one we have in reality by solar radiation, longwave radiation and convection, provided that the convective surface � lm coef� cient equals he. The sol-air temperature depends on the radiation properties of the surface, its inclination and orientation, the average climate data, the speci� c value of the outside surface � lm coef� cient he, etc. Its value differs according to applications.

1335vch01.indd 80 12.02.2007 11:36:27

81

1.5.2 Steady-state, One-dimension: Flat Walls

1.5.2.1 Thermal Transmittance and Interface Temperatures

The use of the surface � lm coef� cient hi and he simpli� es the calculation of the steady-state heat � ow rate, environment to environment, across a � at wall.

Outside Walls

Consider the wall of Figure 1.36. Inside, the reference temperature is �rs. Outside, the reference temperature is *

eθ . Assume the inside environment is heated while it is cold outside (this is the case during the heating season). From the inside environment to the inside surface the heat � ow rate is: 1 i rs si( )q h= θ − θ with �si the inside surface temperature. As heat � ow rate through the wall we have: 2 si se T( ) /q R= θ − θ with �se the outside surface temperature and RT the thermal resistance of the wall T i( )R R= ∑ . From the outside surface to the outside environment, the heat � ow rate is: *

3 e se e( )q h= θ − θ . In steady-state the three heat � ow rates have to be equal. If their common value is q, then by rearrangement and addition, we get:

rs sii

T si se

*se e

e

*T rs e

i e

1 1

= θ − θ

= θ − θ

= θ − θ

⎛ ⎞+ + = θ − θ⎜ ⎟⎝ ⎠

q

h

q R

q

h

q Rh h

The sum can also be written as *rs e( )q U= θ − θ , where

e i

11 1

UR

h h

=+ +

(1.105)

1.5 Applications

Figure 1.36. Thermal transmittance or U-value.

1335vch01.indd 81 12.02.2007 11:36:27

82 1 Heat Transfer

U is called the thermal transmittance of the � at wall. Units: W/(m2 · K). The smaller its value, the less heat � ows on average through the wall. Hence, the U-value is a measure for the insulation quality. The inverse 1/U is called the thermal resistance environment to environment, symbol Ra, units m2 · K/W. The thermal transmittance shows that calculations from environment to environment, i.e. taking radiation and convection at the inside and outside surfaces into account, means the addition of two extra thermal resistances: a surface resistance 1/hi at the inside, marked Ri, standard value 0.13 m2 · K/W (vertical surface), 0.1 m2 · K/W (horizontal surface, heat � ow upwards) and 0.17 m2 · K/W (horizontal surface, heat � ow downwards), and a surface resistance 1/he at the outside, marked Re, standard value 0.04 m2 · K/W.

Inside Walls

For inside walls, we have hi as surface � lm coef� cient at both sides. Thus, the thermal transmittance becomes:

i

12

UR

h

=+

(1.106)

Temperatures

With the surface resistances Ri and Re, radiation and convection are treated as ‘equivalent conduction’. Both can be seen as the thermal resistances of a � ctive surface layer, 1 m thick, with thermal conductivity hi or he. That way the reference temperatures become ‘surface temperatures’ and the heat transfer becomes steady-state conduction from � ctive surface to � ctive surface. Hence, for an outside wall, the temperature course in the [R, �]-plane is a straight line through the points *

e[0, ]θ and [Ra, �rs]. For inside walls, the straight line goes through the points [0, �rs1] and [Ra, �rs2]. However, when constructing the temperature line, the correct layer sequence must be respected, included the surface resistances! The slope of the line represents the heat � ow rate through the wall (q) (Figure 1.37).

Figure 1.37. Interface temperatures.

1335vch01.indd 82 12.02.2007 11:36:27

83

As temperature at the inside surface we have:

Outside wall

i*

*rs esi rs i rs rs e

a i

( )hU

RR h

θ − θθ = θ − = θ − θ − θ

Inside wall

irs1 rs2si rs1 i rs1 rs1 rs2

a i

( )hU

RR h

θ − θθ = θ − = θ − θ − θ (1.107)

The suf� x ihU underlines the necessity to calculate the U-value with the denominator surface

� lm coef� cient. The temperatures at the interfaces are: xx rs i si( )q R Rθ = θ − + .

Average Thermal Transmittance of Walls in Parallel

Consider a wall with surface AT, composed of n parallel parts with surfaces Ai. Each part has a different section (think of a facade made of transparent and non-transparent elements, windows with posts and glass, etc.). If lateral conduction between the parts is negligible, then the partial heat � ow through each part is: i i iU AΦ = Δθ (Figure 1.38). The total heat � ow (T) through the wall equals the sum of the partial � ows:

T i i i1 1

( )n n

i iU A

= =Φ = Φ = Δθ∑ ∑ (1.109)

This sum can be rewritten as:

T m i1

n

iU A

=Φ = Δθ ∑ (1.110)

where Um is the average thermal transmittance of the n parts in parallel:

Figure 1.38. Thermal transmittance of a wall, composed of n parallel parts.

1.5 Applications

1335vch01.indd 83 12.02.2007 11:36:28

84 1 Heat Transfer

i i i i1 1

mT

i1

( ) ( )n n

i in

i

A U A UU

AA

= =

=

= =∑ ∑

∑ (1.111)

The average U-value of n parts in parallel is thus given by the surface weighted average of the U-values of the composing parts. This result is typically used for the envelope of a building, where the average U-value is set equal to the surface weighted average of the U-values of the separate � oor-, façade-, window- and roof elements.

The Electrical Analogy

Equation (1.111) can be seen as the electrical conductivity of a circuit of n parallel electrical conductivities. Conversion to resistances gives:

T i Tam

1 iam ai

1 ai

orn

ni

i

A A AR

AR RR

=

=

= =∑∑

As long as lateral conduction in the contacts between parts is negligible, the electrical analogy can be used to solve quite complex problems. Take as an example a cavity wall with the � ll perforated by the ties. Let At be the tie section, Rt the thermal resistance of the ties, Ais the total surface of the wall and Ris the thermal resistance of the insulation (Figure 1.39).For the thermal resistance (R) of the insulation, included the ties, the following applies:

is

is t t

is t

AR

A A A

R R

= −+

Figure 1.39. Electrical analogy, cavity wall, the � ll perforated by cavity ties.

1335vch01.indd 84 12.02.2007 11:36:28

85

If R1 + Ri is the thermal resistance between the insulation layer and the inside and R2 + Re the thermal resistance between the insulation layer and the outside, then the total thermal resistance equals the sum of three series connected resistances Ri + R1, R and R2 + Re:

iT i 1 2 e i 1 2 e

i t t

is t

( ) ( ) ( ) ( )A

R R R R R R R R R RA A A

R R

= + + + + = + + + +−+

1.5.2.2 Thermal Resistance of a Non-ventilated In� nite Cavity

Heat transfer in a non ventilated in� nite cavity proceeds by conduction, convection and radiation:

g b TT 1 2

1 2

Nu( )

1 11

C Fq

de e

⎛ ⎞⎜ ⎟λ

= + θ − θ⎜ ⎟⎜ ⎟+ −⎜ ⎟⎝ ⎠

In the formula, �g is the thermal conductivity of the cavity gas and e1 and e2 the emissivities of the cavity surfaces. As for a wall, the heat resistance is de� ned as the ratio between the temperature difference across and the heat � ow rate through the cavity or:

1

g b Tc

1 2

Nu

1 11

C FR

de e

−⎛ ⎞⎜ ⎟λ

= +⎜ ⎟⎜ ⎟+ −⎜ ⎟⎝ ⎠

(1.112)

Figure 1.40 gives the thermal resistance Rc for an in� nite air cavity as a function of cavity width and temperature difference across it. Average temperature: 10 °C. One cavity surface is grey, the other grey or re� ecting. Note the prevalence of radiation in the heat transfer (thermal resistance increases quickly with decreasing emissivity of one of the surfaces), and the absence of any resistance gain for widths above 20 mm (radiation is independent of cavity width while increasing convection with larger widths compensates for decreasing conduction). At low emissivity, even the temperature difference over the cavity gains in� uence.For � rst order calculations, the following thermal resistances from Table 1.7 apply for non-ventilated cavities:

Table 1.7. Thermal Resistances.

Rc (m2 K/W)

Both surfaces greyRc (m

2 K/W)One surface re� ecting

Vertical cavity 0.17 0.35

Horizontal cavity

Heat � ow upwardsHeat � ow downwards

0.140.20

0.280.40

1.5 Applications

1335vch01.indd 85 12.02.2007 11:36:28

86 1 Heat Transfer

1.5.2.3 Solar Transmittance

De� nition

The solar heat � ow rate through transparent and opaque parts can be written as:

S STq g E= (1.113)

where EST is the total solar radiation incident on the outside surface and qS is the transmitted heat � ow rate, both in W/m2. The factor g is called the solar transmittance. That factor includes the direct and indirect solar gains. The direct gains are given by Sd S STq E= τ , where �S is the total shortwave transmissivity of the part. If opaque, transmissivity is zero. Opaque parts lack direct gains. For a transparent part the shortwave transmissivity differs from zero, giving

Figure 1.40. Thermal resistance of an in� nite cavity.

1335vch01.indd 86 12.02.2007 11:36:29

87

direct gains. The indirect gains occur because opaque and transparent parts absorb a fraction of the incoming sunlight, warm up, conduct the absorbed radiation as heat to the inside and dissipate it there by convection and longwave radiation. For simple transparent parts such as single and double glass, the indirect gains are quite easily estimated.

Single Glass

Let S be the shortwave absorptivity of the glass. The indirect gains are part of the heat � ow by convection and radiation at the inside surface:

Si i si rs( )q h= θ − θ (1.114)

with �si the unknown inside surface temperature. If we equate �si with the glass temperature �x and suppose the glass equally warm over its thickness, then: �si = �x = �se, with �se the outside surface temperature. The thermal balance for 1 m2 glass then becomes: the sum of the absorbed solar radiation plus the heat � ow rate between outside and outside surface plus the heat � ow rate between inside and inside surface zero (S EST + he (�e – �x) + hi (�rs – �x) = 0 with �e the outside and �rs the inside temperature). For the glass temperature that gives:

S ST e e i rsx si

i e i e

E h h

h h h h

α θ + θθ = θ = +

+ +

The second term on the right represents the glass temperature when no solar radiation is absorbed, while the � rst term quanti� es the temperature increase by absorption. Only that part is the result of incident solar radiation. Combination with (1.114) gives as heat � ow at the inside surface:

i S ST i e e rsSi

i e i e

( )h E h hq

h h h h

α θ − θ= +

+ + (1.115)

For the solar transmittance only the indirect solar gains, i.e. the � rst term in the right part of (1.115), count, or:

i S STSi

i e

h Eq

h h

α=

+

The solar transmittance of single glass then becomes:

Sd Si

ST e i1 /S

Sq q

gE h h

+ α= = τ +

+ (1.116)

The solar gains not only depend on the shortwave transmissivity of the glass, but also on its shortwave absorptivity and the ease with which the absorbed heat is dissipated to the inside (as the ratio e i/h h decreases, the g-value increases).

Double Glass

Let �S1 be the shortwave transmissivity, �S1 the shortwave re� ectivity and S1 the shortwave absorptivity for the one pane and �S2, �S2 and S1 the same for the other pane. Solar radiation transmitted now includes some re� ected radiation from the cavity (Figure 1.41). This turns shortwave transmissivity into a sum of a geometric series with ratio �S1 �S2. Direct gains then become:

1.5 Applications

1335vch01.indd 87 12.02.2007 11:36:29

88 1 Heat Transfer

S1 S2Sd ST

S1 S21q E

τ τ=

− ρ ρ (1.117)

In general, the denominator 1 – �S1 �S2 stays close to 1, giving as a simple calculation rule: total shortwave transmissivity of double glass is given by the product of the transmissivities of each of the panes. The indirect gains look like: qSi = hi �x2 (�e = �rs = 0 °C, both panes isothermal), with �x2 the temperature of the inside pane. That temperature ensues from the heat balance of both panes (1 = outside, 2 = inside):

Pane 1:

S1 S2 S1 S2 x2 x1S1 ST e x1

S1 S2 c

10

1E h

R

− ρ ρ + τ ρ θ − θα − θ + =

− ρ ρ

Pane 2:

S1 x1 x2S2 ST i x2

S1 S2 c

01

E hR

τ θ − θα + − θ =

− ρ ρ

If we call S1 S2 S1 S21

S1 S2

1

1f

− ρ ρ + τ ρ− ρ ρ

and S12

S1 S21f

τ− ρ ρ

, then that system produces as a

solution:

S1 1S2 2 e

c cx2 ST

i e 2c c c

1

1 1 1

ff h

R RE

h hR R R

⎛ ⎞α+ α +⎜ ⎟⎝ ⎠

θ =⎛ ⎞ ⎛ ⎞

+ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Entering that result in qSi = hi �x2 gives the indirect gain. Solar transmissivity of double glass then becomes:

Figure 1.41. Solar transmittance of double glass.

1335vch01.indd 88 12.02.2007 11:36:29

89

S1 1S2 2 e

c cS1 S2i

S1 S2i e 2

c c c

1

1 1 1 1

ff h

R Rg h

h hR R R

⎛ ⎞α+ α +⎜ ⎟⎝ ⎠τ τ

= +− ρ ρ ⎛ ⎞ ⎛ ⎞

+ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(1.118)

With increasing complexity of the transparent element, the calculation gets more complicated. However, the result for double glass shows how to in� uence solar gains. In all cases, the value is reduced by limiting the direct transmission. At the same time, lowering the inside surface � lm coef� cient compresses the indirect gains. The same is obtained by decreasing the shortwave absorptivity of the glass.

Multiple-layer Glass

A same approach applies as for double glass. Total shortwave transmissivity, governing the direct gains, is approximated by:

KT Ki1

n

i=τ = τ∏ (1.119)

For the indirect gains (qSi = hi �xn) the system of heat balances extends with one equation per additional glass pane.

1.5.3 Steady State, Cylindrical Coordinates: Pipes

Introducing surface � lm coef� cients is quite simple. At the exterior surface of a pipe, the heat � ow is given by:

1 1 2 s,2 ref,22 ( )n nR h+ +Φ = π θ − θ

with h2 the surface � lm coef� cient in the environment outside the pipe, �ref, 2 the reference temperature there and Rn + 1 the outside radius of the pipe or of the layer around it. At, the interior surface, the � ow becomes:

1 1 1 ref,1 s,12 ( )R hΦ = π θ − θ

where h1 is the surface � lm coef� cient in the pipe, �ref, 1 the temperature of the � uid there and R1 the inside radius of the pipe. Through the pipe wall, we have:

s,1 s,21, 1

1

1

ln ( / )

2

n ni i

i i

R R+

+

=

θ − θΦ =

⎡ ⎤⎢ ⎥π λ⎣ ⎦

∑

where indicates that the pipe wall may consist of several layers. In steady state, the heat � ows n + 1, 1 and 1, n + 1 must be equal. Reshuf� ing and adding the three equations gives:

1s,2 ref,2

1 22n

nR h+

+

Φ= θ − θ

π

11, 1 s,1 s,2

1

ln ( / )

2

ni i

ni i

R R++

=

⎡ ⎤Φ = θ − θ⎢ ⎥π λ⎣ ⎦

∑

1.5 Applications

1335vch01.indd 89 12.02.2007 11:36:29

90 1 Heat Transfer

11 1

ref,1 s,1

2 R hΦ

= πθ − θ

Sum: 11, 1 ref,1 ref,2

11 2 1 1

ln( / )1 1

2 2 2

ni i

nin i

R R

R h R h+

+=+

⎧ ⎫⎡ ⎤⎪ ⎪Φ + + = θ − θ⎨ ⎬⎢ ⎥π π λ π⎪ ⎪⎣ ⎦⎩ ⎭∑

That sum may be written as 1, 1 pipe ref,1 ref,2( )n U+Φ = θ − θ with Upipe the thermal transmittance per meter run of pipe, a quantity equal to:

pipe1

11 2 1 1

1

ln ( / )1 1

2 2 2

ni i

in i

UR R

R h R h+

=+

=⎡ ⎤

+ +⎢ ⎥π π λ π⎣ ⎦∑

(1.120)

The Upipe-value � gures as equivalent for the thermal transmittance of a � at wall. As with a � at wall, we may restrict the heat losses or gains of pipes transporting � uids by adding insulation. But there is a difference. The gain by thicker insulation diminishes faster for pipes than for � at walls.

1.5.4 Steady-state, Two and Three Dimensions: Thermal Bridges

1.5.4.1 Calculation by the Control Volume Method (CVM)

‘Environment to environment’ is implemented in the control volume method (CVM) by adding surface layers with the surface � lm resistances as thermal resistance. That way the reference temperature at either side becomes a ‘� ctive’ surface temperature. Heat transfer through these surface layers occurs perpendicularly to the actual surface. If the temperatures on the actual surface are of interest, then local surface � lm resistances are used. The standard surface � lm resistances apply for the heat losses.As heat balance in a point on the surface, we have 4 adjacent points on the surface, 1 in the material and 1 at the surface � lm resistance (surface parallel to the [y, z]-plane) (Figure 1.42):

Figure 1.42. CVM-method, control volume against the inside surface.

1335vch01.indd 90 12.02.2007 11:36:30

91

Example

Heat � ow through the surface layer (from point (i, m, n) where the temperature equals the environmental temperature �1 to the central point (s, l, m, n) on the actual surface):

, , 2, , , , , , , ,( )Φ = θ − θi m n

s l m n i i m n s l m nh a

Heat � ow from the 4 adjacent points on the actual surface to the central point (s, l, m, n):

, , ,, , 1, 1 , , 1, , , ,( )

2s l m ns l m n s l m n s l m n

a+ +Φ = λ θ − θ

, , ,, , 1, 1 , , 1, , , ,( )

2s l m ns l m n s l m n s l m n

a− −Φ = λ θ − θ

, , ,, , , 1 1 , , , 1 , , ,( )

2s l m ns l m n s l m n s l m n

a+ +Φ = λ θ − θ

, , ,, , , 1 1 , , , 1 , , ,( )

2s l m ns l m n s l m n s l m n

a− −Φ = λ θ − θ

Heat � ow from the adjacent point in the material to the central point:

, , ,1, , 1 1, , , , ,( )− −Φ = λ θ − θs l m n

l m n l m n s l m n a

Sum zero:

, , 1, , , 1, , , , 1 , , , 1, , 1

1 1, , 1 , , ,

2( 3 ) 0

+ − + −

−

θ + θ + θ + θθ + λ

+ λ θ − + λ θ =

s l m n s l m n s l m n s l m ni i m n

l m n i s l m n

a h

a h

In this equation a hi �i, m, n is the known term. For points in corners, etc., analogous equations have to be constructed.

1.5.4.2 Thermal Bridges in Practice

The term thermal bridge refers to spots on the envelope where heat transfer develops in two or three dimensions (the word envelope indicates the entirety of � oors, facades, windows and roofs, which separate the inside from the outside, the inside from grade and the inside from all adjoining, unheated volumes). The name may be taken literally: not only do these spots experience larger heat losses and gains than the adjacent elements (unless glass), but also the inside surface temperatures are lower there than on the adjacent elements during the heating season. We distinguish two types of thermal bridges:

Geometric thermal bridgesA consequence of the three dimensional character of a building: angles and corners, inner and outer reveals around windows, etc. (Figure 1.43).

Structural thermal bridgesThe consequence of structural decisions. Examples: steel or concrete girders and columns that penetrate the envelope, discontinuities in the thermal insulation (Figure 1.44). Structural thermal bridges could be there for reasons of structural integrity. Take a balcony. The cantilever moment cannot be balanced without continuity with the � oor inside.

�

�

1.5 Applications

1335vch01.indd 91 12.02.2007 11:36:30

92 1 Heat Transfer

Rules to follow are:

Neutralize geometric thermal bridges by assuring continuity of the thermal insulation.Avoid structural thermal bridges by paying attention to continuity of the insulation layer. It should be possible to go around the building drawings with a tracer in the insulation, without crossing any element that forms an easy path for heat � ow between the inside and the outside.If complete avoidance of thermal bridging is impossible, reduce the negative impact of structural thermal bridges on the heat loss or gain and the inside surface temperature as much as can be managed.

In practice, using CVM or FEM to evaluate thermal bridges during a design is neither practicable nor economical. Therefore an alternative approach has been developed. As far as additional heat loss is concerned, thermal bridges are quanti� ed by a linear or local thermal transmittance. With respect to the lowest inside surface temperature, they are characterized by a temperature ratio.

Linear and Local Thermal Transmittances

The linear thermal transmittance, symbol �, units: W/(m · K), gives the extra heat transfer per Kelvin temperature difference between the inside and the outside for each meter of linear, that means of two dimensional thermal bridge. The local thermal transmittance, symbol �, units W/K gives the extra heat transfer per Kelvin temperature difference between the inside and the outside for a local, three-dimensional thermal bridge. In both cases, the calculation demands a well-de� ned one-dimensional reference situation (Figure 1.45). When � xing that reference, one should ignore all hidden envelope details which cause thermal bridging, i.e. the real situation must be changed in a composition of the visible plane elements with dots where hidden thermal bridge effects exist (for this, the facade drawings are the best to use). When calculating the one-dimensional heat losses or gains for the reference, the exterior dimensions are used. After that, the correct structural drawings form the basis for the two and three-dimensional heat transfer and inside surface temperature calculation.

�

�

�

Figure 1.43. Geometric thermal bridges.

Figure 1.44. Structural thermal bridge.

1335vch01.indd 92 12.02.2007 11:36:30

93

When nD is the two- or three-dimensional heat � ow and o the heat � ow for the one-dimensional reference, then the linear (�) and the local thermal transmittance (�) follows from:

2D o 3D o

L

Φ − Φ Φ − Φψ = χ =

Δθ Δθ (1.121)

with L the length of the linear thermal bridge in the one-dimensional reference. In many cases, local thermal bridges emerge where linear ones cross each other. In such case, two reference situations must be judged: (1) a one-dimensional case per linear thermal bridge (allows for the calculation of the linear thermal transmittances) and (2) a case which includes all linear thermal bridges (allows for the calculation of the point thermal transmittance). Once the linear and local thermal transmittances are known, then the average whole wall thermal transmittance of a � at part containing thermal bridges is calculated as:

1 1o

( )n m

i i ji j

L

U UA

= =ψ + χ

= +∑ ∑

(1.122)

with Uo the thermal transmittance of the base part, A its surface, n the number of linear thermal bridges, Li their length and m the number of local thermal bridges.

Figure 1.45. Using linear thermal transmittances, from reality to calculation model (only � at parts; thermal bridges replaced by lines perpendicular to the section).

1.5 Applications

1335vch01.indd 93 12.02.2007 11:36:31

94 1 Heat Transfer

Temperature Factor fhi

The temperature factor is given by:

i

s,min ref,e

ref,i ref,ehf

θ − θ=θ − θ

(1.123)

where �ref,i is the reference temperature inside, �ref,e the reference temperature outside and �s,min the lowest inside surface temperature on the thermal bridge. The suf� x hi indicates that the local surface � lm coef� cient for the coldest point on the thermal bridge has to be used in the calculation. The temperature factor is a dimensionless surface temperature, which is found directly by a CVM- or FEM-calculation when taking 1 K as the temperature difference.The severity of a thermal bridge grows, the higher its linear (�) or local thermal transmittance (�) and the lower its temperature factor

i( )hf are. ‘More severe’ means that the inside surface

collects more dirt, suffers from a higher risk on fungal growth, becomes a preferred spot for surface condensation, shows an increased sensitivity to crack formation, while taking a disproportionately high share of the overall heat � ow.Over the past years, several hard copy based thermal bridge catalogues have been published which contain linear thermal transmittances (�), local thermal transmittances (�) and temperature factors

i( )hf for typical details where thermal bridging is quite a common problem

(reveals, lintels, dormer windows, balconies, etc.). In these catalogues, various combinations of materials and various layer thicknesses are considered. Interactive CD-ROM’s with thermal bridge data and failure examples are also on the market.

1.5.5 Transient, Periodic: Flat Walls

The periodic response of a � at wall, environment to environment, is simpli� ed the same way as a thermal bridge, i.e. by considering the surface � lm resistances as the thermal resistance of two surface-connected � ctive layers. In that, the two layers are assumed to be 1 m thick, to have as thermal conductivity the values hi or he and to possess a volumtric speci� c heat capacity (� c) of zero. The latter turns both into pure conductors. This way, the reference temperatures become ‘� ctive’ surface temperatures. For conduction-active layers with a volumetric speci� c heat capacity zero, the following applies:

20n

i n c

T

π ρ λω = = cosh ( ) 1n Rω =

sinh ( ) 0n n Rω ω =

sinh ( ) sinh ( )0lim

0n n

nn n

R RR

→∞

⎛ ⎞ω ω= = =⎜ ⎟ω ω⎝ ⎠

Turning the complex surface � lm matrixes into:

ii

11

0 1

hW

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

ee

11

W

0 1

h⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

1335vch01.indd 94 12.02.2007 11:36:31

95

Transposed in real numbers as:

i

ii

11 0 0

10 1 0

0 0 1 0

0 0 0 1

h

Wh

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

e

ee

11 0 0

10 1 0

0 0 1 0

0 0 0 1

h

Wh

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(1.124)

The wall matrix, environment to environment, thus becomes:

Outside wall: Wna = Wi Wn1 Wn2 Wn3 … Wnn We

Inside wall: Wna = Wi Wn1 Wn2 Wn3 … Wnn Wi

For a monolayer wall, this product reduces to:

Outside wall: Wna = Wi Wn We

Inside wall: Wna = Wi Wn Wi

Apparently, each monolayer wall becomes three-layered instead of monolayered. The model of course simpli� es reality. On the one hand, the radiant part in the surface � lm resistance involves all surfaces, seen by the one considered, into the periodic response. On the other hand, even convection introduces inertia, linked to the volumetric heat capacity of the air, the limited air velocity, the interaction with other wall surfaces and with furniture. But taking all that into account is too complicated to be treated here.

1.5.6 Heat Balances

The usage of surface � lm coef� cients for heat transfer, the U-value, etc., simplify reality. Too much simplicity often prevents one from � nding the correct answer to a problem. In cases these simple concepts do not work, a common technique, already applied in previous paragraphs, is to use heat balances:

First, the interfaces where temperature and heat � ow rate act as unknown are selected. These surfaces form the so-called calculation points. Their number de� nes the size of the system of equations which ensues from the heat balances.Then, in each calculation point, conservation of energy is applied: the sum of all heat � ows or heat � ow rates from the environment or the neighbour points to the calculation point zero. That way, each point supplies an equation, in which the temperatures in the neighbour points and the calculation point are the variables. The dif� culty resides in � nding all heat � ows between the calculation points and then between these points and the larger environment. These have to be written as a function of the known and unknown temperatures.Finally, the system is solved. The solution gives the requested temperatures and/or heat � ow rates and heat � ows.

�

�

�

�

�

�

�

1.5 Applications

1335vch01.indd 95 12.02.2007 11:36:31

96 1 Heat Transfer

1.6 Problems

(1) Find the thermal transmittance of an outside wall with the following section (from inside to outside):

Layer Thickness (d)cm

Thermal conductivity (�)W/(m · K)

Thermal resistance (R)m2 · K/W

Render

Inside leaf

Cavity � ll

Air cavity

Brick veneer

1

14

8

4

9

0.3

0.5

0.04

0.9

0.17

Answer

Before calculating the thermal transmittance, all quantities have to be expressed in the correct basic SI-units. Meter applies, not cm.

�

�

i e

1

1 / 1 /

1

1 / 8 0.01 / 0.3 0.14 / 0.5 0.08 / 0.04 0.17 0.09 / 0.9 1 / 25

j

Uh R h

=+ +

=+ + + + + +

∑

or

U = 0.36 W/(m2 · K)

Never give more than a double signi� cant.

(2) Find the thermal transmittance of a � at roof with the following section (from inside to outside):

Layer Thickness (d)cm

Thermal conductivity (�)W/(m · K)

Render

Concrete � oor

Screed

Vapor barrier

Thermal insulation

Membrane

1

14

10

1

12

1

0.3

2.5

0.6

0.2

0.028

0.2

(3) Find the clear wall thermal transmittance (i.e. without considering the studs) of a timber framed outside wall with the following section (from inside to outside):

�

1335vch01.indd 96 12.02.2007 11:36:31

97

Layer Thickness (d)cm

Thermal conductivity (�)W/(m · K)

Thermal resistance (R)m2 · K/W

Gypsum board

Air space

AFVR

Thermal insulation

Outside sheathing

Air cavity

Brick veneer

1.2

2

0.02

20

2

2

9

0.2

0.2

0.04

0.14

0.9

0.17

0.17

(4) Calculate the outside equivalent temperature for a solar irradiation of 750 W/m2. The out-side temperature is 30 °C and the outside surface � lm coef� cient for heat transfer 12 W/(m2 · K). The surface considered has a shortwave absorptivity 0.9. The long wave emissivity to the clear sky reaches 100 W/m2 while the longwave emissivity of the surface is 0.8 (� at roof).Repeat the calculation for a daily mean temperature of 24 °C, a daily mean solar irradiation of 169 W/m2 and a daily mean long wave emissivity to the clear sky of 50 W/m2 (= warm summer day, south oriented wall). Do it again for a daily mean temperature of –15 °C, a daily mean solar irradiation of 109 W/m2 and a daily mean long wave emissivity to the clear sky of 50 W/m2 (= cold winter day, south oriented wall). The respective shortwave absorptivity and longwave emissivity of the wall are 0.5, and 0.8. The outside surface � lm coef� cient for heat transfer reaches 16 W/(m2 · K). Cloudiness in winter is 0.8.

Answer

The equivalent temperature for the extreme situation is given by:

* * S S L Le e

e

0.9 750 0.8 10030 79.6 °C

12

− ⋅ − ⋅θ = θ + = + =a E e q

h

which is rather high.

The mean equivalent temperature for a south oriented wall during a hot summerday is:

* * S S L Le e

e

0.5 169 0.8 5024 31 °C

16

− ⋅ − ⋅θ = θ + = + =a E e q

h

During a cold winterday, we have:

* * S S L Le e

e

0.5 109 0.8 0.8 5015 10.9 °C

16

− ⋅ − ⋅ ⋅θ = θ + = − + = −a E e q

h

(5) Return to the cavity wall of problem (1). Calculate the highest and lowest daily mean temperatures in all interfaces, knowing that the equivalent temperature has the values calculated in problem (4). The inside temperature is 21 °C in winter and 25 °C in summer. Draw a � gure of the result.

1.6 Problems

1335vch01.indd 97 12.02.2007 11:36:31

98 1 Heat Transfer

Answer

The outside surface � lm coef� cient now is 16 W/(m2 · K). The temperatures are given by:

*j i i e

a

( )

j

iR

Rθ = θ − θ − θ

∑

As a table and a � gure:

Interface � Rjm2 · K/W

Temperature °C

Winter Summer

Insidesi1234seoutside (air)

00.130.160.442.442.612.712.77

21 19.6 19.2 16.0 –7.1 –9.0–10.2–15.0

2525.325.325.930.330.630.824.0

Note that the thermal insulation absorbs the largest temperature difference. It is as if the wall is split in two parts, one that leans towards the inside and one that leans towards the outside. The last one experiences the greatest temperature movement.

(6) Do problem (5) again, now for the timber framed wall of problem (3). The in- and outside reference temperatures and the in- and outside surface � lm coef� cients are the same as in problem (5). Draw a � gure of the result.

1335vch01.indd 98 12.02.2007 11:36:31

99

(7) Take the � at roof of problem (2). Calculate the highest and lowest daily mean temperatures in all interfaces, knowing that the daily mean equivalent temperature in summer may reach 40 °C for a daily mean outside air temperature of 24 °C, while in winter the value may drop down to –19.5 °C for a daily mean outside air temperature of –15 °C. The outside surface � lm coef� cient on such windless cold and warm days is 12 W/(m2 · K). Inside, the temperature is 21 °C in winter and 25 °C in summer. The inside surface � lm coef� cient equals the standard value. Draw a � gure of the result.

(8) A building systems manufacturer introduces a new sandwich panel on the market with a design (from inside to outside) of:

Layer Thickness (d)cm

Thermal conductivity (�)W/(m · K)

Thermal resistance (R)m2 · K/W

AluminumVIP (vacuum insul.)Air cavityGlass panel

0.2221

2300.006

Assume �0.15

The panel is used as cladding in a curtain wall system. Suppose the outside temperature is 35° C and the inside temperature 24 °C. Solar irradiation on the glass panel reaches 500 W/m2. No longwave radiation has to be considered. The outside heat transfer coef� cient is 15 W/(m2 · K). Inside, we have the standard value. Radiant properties of the glass: aS = 0.06, rS = 0.19, S = 0.75. The VIP has a shortwave absoptivity 1. What temperature will be noted in the glass panel? How large is the heat � ow rate that enters the building through the panel?

Answer

The problem is solved by writing two heat balances: one for the glass and the other for the cavity side of the VIP:

Glass (temperature �1)

2 1e e 1 S S

cav

( ) 0θ − θ

θ − θ + + =h a ER

VIP (temperature �2)

1 2 i 2S S

cav VIP alu i

01 /

θ − θ θ − θ+ τ + =

+ +E

R R R h

or

1 2

1 2

(15 1 / 0.15) / 0.15 0.05 500 15 35

/ 0.15 (1 / 0.15 1 /(0.02 / 0.005 0.0002 / 230 1 / 8)

0.75 500 1 /(0.02 / 0.005 0.0002 / 230 1 / 8) 24

− + θ + θ = − ⋅ − ⋅⎧⎪θ − θ + + + =⎨⎪− ⋅ − + + ⋅⎩

Solving that system of two equations gives: �1 = 60.2 °C, �2 = 113.2 °C. The heat � ow rate to the inside thus becomes: 21.6 W/m2. The high temperatures show the panel acts as a solar collector. The heat � ux to the inside equals the value a U = 1.97 W/(m2 · K)-panel

1.6 Problems

1335vch01.indd 99 12.02.2007 11:36:31

100 1 Heat Transfer

should transfer without solar irradiation. The U-value of the manufactured panel is only 0.23 W/(m2 · K).What solutions could be applied to lower the temperatures in, and heat � ow rate through the panel?

(9) Solve problem (8) in the case a heat absorbing glass, aS = 0.3, rS = 0.19, S = 0.51, is used and the shortwave absorptivity of the VIP is lowered to 0.5 (shortwave re� ectivity: 0.5)?

(10) The roof of a chalet in the mountains is covered with 40 cm of snow (� = 0.07 W/(m · K), aS = 0.15). The temperature outside is –15 °C, inside we have 22 °C. Solar irradiation reaches 600 W/m2. Surface � lm coef� cients: 15 W/(m2 · K) outside, 10 W/(m2 · K) inside.What insulation thickness does the roof need to avoid the snow from melting in the contact interface with the roof surface? What heat � ow rate through the roof will be noted inside? The thermal resistance surface to surface of the roof without insulation is 0.5 m2 · K/W.

(11) An intensely ventilated attic receives an insulated ceiling, composed of metallic girders, h.o.h. 60 cm, with a 120 thick thermal insulation in between. The metal pro� le used is the one given in the � gure. Suppose the insulation has a thermal conductivity 0 W/(m · K). For the metal, the value is � W/(m · K). The surface � lm coef� cients are 25 W/(m2 · K) at the attic side and 6 W/(m2 · K) at the inside. The attic temperature is –10 °C, the inside temperature 20 °C.Does it make any difference in heat loss if the pro� le is mounted with the broader � ange to the inside (1) or vice versa (2)? What is the metal temperature in both cases? What could be the U-value of the ceiling?

Answer

The heat balance for the metallic pro� le in case (1) is:

x x0.2 6 (20 ) 0.05 25 ( 10 ) 0⋅ ⋅ − θ + ⋅ ⋅ − − θ =

The heat balance for the metallic pro� le in case (2) is:

x x0.05 6 (20 ) 0.2 25 ( 10 ) 0⋅ ⋅ − θ + ⋅ ⋅ − − θ =

This leads to the following results:YesCase (1): 4.7 °C; case (2): –8.3 °CCase (1): U = 1.02 W/(m2 · K)Case (2): U = 0.47 W/(m2 · K) (1/0.6 pro� les per meter run)

�

�

�

1335vch01.indd 100 12.02.2007 11:36:32

101

(12) Solve problem (11) for a metallic pro� le with � anges of 100 mm each.

(13) A reinforced concrete square column with side 0.4 m is positioned between two glass panels in a way the glass lines up with the column’s inside surface. The glass may be considered as a surface with thickness zero. The inside temperature is 21 °C, the outside temperature 0 °C. Surface � lm coef� cients: inside: 8 W/(m2 · K), outside 25 W/(m2 · K). What is the temperature � eld in, and the heat loss through the column?

Answer

Consider the column to be a � at wall. U-value:

213.1 W/(m K)

1 / 25 0.4 / 2.5 1 / 8= ⋅

+ +

Temperature on the inside surface: 21 3.1 (21 0) / 8 12.9 °C− ⋅ − = . Heat loss: 3.1 0.4 21 25.8 W/m⋅ ⋅ = . U-value and inside surface temperature are close to the values for double glazing. Temperature in the middle of the column: 7.8 °C. Temperature factor: 0.65.

Let us do it better and apply a simple CVM-grid: center of the column as the only calculation point. Heat balance:

x x0.2 (21 ) 0.2 (0 )3 0

1 / 8 0.2 / 2.5 1 / 25 0.2 / 2.5

− θ − θ+ =

+ +

giving as central column temperature 3.4 °C and as temperature factor 0.49. That value is 24.6% lower than calculated above. Heat loss: 34.3 W/m, i.e. 33.7% more then found with the � at wall assumption.

�

�

1.6 Problems

1335vch01.indd 101 12.02.2007 11:36:32

102 1 Heat Transfer

A further upgrade is found by re� ning the grid: not 1 central point but 9 equidistant points, of which 8 lie along the perimeter. Heat balances (we may consider only half the column which gives us six points to balance):

Point 1,1

1,1 1,2 1,1 2,1 1,12.5 0.1 2.5 0.1

8 0.1 (21 ) ( ) ( ) 00.2 0.2

⋅ ⋅⋅ ⋅ − θ + θ − θ + θ − θ =

Point 2,1

2,1 1,1 2,1 2,2 2,1

2,1

2.5 0.1 2.5 0.18 0.1 (21 ) ( ) ( )

0.2 0.225 0.1 (0 ) 0

⋅ ⋅⋅ ⋅ − θ + θ − θ + θ − θ

+ ⋅ ⋅ − θ =

Point 1,2

1,1 1,2 1,3 1,2 2,2 1,22.5 0.1 2.5 0.1 2.5 0.2

( ) ( ) ( ) 00.2 0.2 0.2

⋅ ⋅ ⋅θ − θ + θ − θ + θ − θ =

Point 2,2

2,1 2,2 1,2 2,2 2,3 2,2

2,2

2.5 0.1 2.5 0.2 2.5 0.1( ) ( ) ( )

0.2 0.2 0.225 0.2 (0 ) 0

⋅ ⋅ ⋅θ − θ + θ − θ + θ − θ

+ ⋅ ⋅ − θ =

Point 1,3

1,2 1,3 1,3 2,3 1,32.5 0.1 2.5 0.1

( ) 25 0.1 (0 ) ( ) 00.2 0.2

⋅ ⋅θ − θ + ⋅ ⋅ − θ + θ − θ =

Point 2,3

2,3 2,2 2,3 1,3 2,32.5 0.1 2.5 0.1

2 25 0.1 (0 ) ( ) ( ) 00.2 0.2

⋅ ⋅⋅ ⋅ ⋅ − θ + θ − θ + θ − θ =

Solving the system gives as temperatures in the column:

0.4 0.9 0.4

1.4 2.9 1.4

4.9 8.1 4.9

�

1335vch01.indd 102 12.02.2007 11:36:32

103

Lowest temperature factor: 0.25 that means as bad as single glazing.Heat loss: = 2 · 0.1 · 8 · (21 – 4.9) + 0.2 · 8 · (21 – 8.1) = 46.4 W/m, i.e. 80% more than with the � at wall assumption.

Further re� nement comes from considering more control volumes. The correct answer in terms of temperatures is given by the � gure below:

(14) Suppose aerated concrete is chosen as envelope material. The seller promises a very good transient thermal behavior, which is formulated in terms of a high equivalent thermal resistance. Is this true? Material properties:

Situation DensityKg/m3

Thermal conductivity (�)W/(m · K)

Speci� c heat capacityJ/(kg · K)

Just applied (humid) 450 0.30 2700

After some years (airdry) 450 0.13 1120

Thickness: variable (10, 20 and 30 cm). Surface � lm coef� cients: 25 W/(m2 · K) outside, 8 W/(m2 · K) inside.

Answer

The best way to evaluate that promise is by calculating the harmonic properties of the wall. A high dynamic thermal resistance underlies the statement. A low admittance, however, means the high equivalent thermal resistance will not suf� ce to stabilize the inside climate in case of important solar and internal gains.

�

1.6 Problems

1335vch01.indd 103 12.02.2007 11:36:32

104 1 Heat Transfer

To show how the harmonic calculation works, we calculate the temperature damping, the dynamic thermal resistance and the admittance for the d = 10 cm, � = 0.3 W/(m · K) case (i.e. the initially wet aerated concrete).

Thermal diffusivity

72.469 10 m/sac

−λ= = ⋅ρ

Xn-value

7

3.14180.1 1.2135

2.46910 3600 24n

nX d

a T −π= = =

⋅ ⋅

Functions Gn1(Xn) tot Gn6 (Xn)

1

2

3

4

5

6

0.640429

1.437199

0.92792

0.48581

1.43083

2.73295

n

n

n

n

n

n

G

G

G

G

G

G

−

Layer matrices

Surface matrix (h = 8 W/(m2 · K)):

1 0 0.125 0

0 1 0 0.125

0 0 1 0

0 0 0 1

|1|

Layer matrix:

0.640429 1.437199 0.309307 0.161937

1.437199 0.640429 0.161937 0.309307

4.292490 8.198851 0.640429 1.437199

8.198851 4.292490 1.437199 0.640429

− −−− − −

|2|

Surface matrix (h = 25 W/(m2 · K)):

1 0 0.04 0

0 1 0 0.04

0 0 1 0

0 0 0 1

|3|

�

�

�

�

1335vch01.indd 104 12.02.2007 11:36:33

105

Matrix multiplication

0.640429 1.437199 0.389360 0.341587

1.437199 0.640429 0.341587 0.3893602 1

4.292490 8.198851 0.103868 2.462055

8.198851 4.292490 2.462055 0.103868

− −× =

−− − −

|4|

0.468730 1.765153 0.3993515 0.4400690

1.765153 0.468730 0.4400690 0.39935153 4

4.292490 8.198851 0.1038680 2.4620550

8.198851 4.292490 2.4620550 0.1038680

− −× =

−− − −

Harmonic propertiesTemperature damping:

2 20.46873 1.765133 1.83Dθ = + =

1.765133 12tan 5 h

0.46873aθ

⎛ ⎞φ = =⎜ ⎟⎝ ⎠ π

Dynamic thermal resistance:

2 2 2q 0.3993515 0.38936 0.59 m K/WD = + = ⋅

q0.38936 12

tan 3.2 h0.3993515

a⎛ ⎞φ = =⎜ ⎟⎝ ⎠ π

Admittance:

2

q

3.09 W/(m K)D

AdDθ= = ⋅

Ad q 1.8 hθφ = φ − φ =

The other cases may be calculated by the reader. Use a spreadsheet program. The results:

Case 1d = 10 cm� = 0.3W/(m · K)

Case 2d = 20 cm� = 0.3W/(m · K)

Case 3d = 30 cm� = 0.3W/(m · K)

Case 4d = 10 cm� = 0.13 W/(m · K)

Case 5d = 20 cm� = 0.13 W/(m · K)

Case 6d = 30 cm� = 0.13 W/(m · K)

D� 1.83 6.56 23.1 1.63 5.72 19.1

���(h) 5 h 00 9 h 48 14 h 36 4 h 33 9 h 19 13 h 57 Dq (m

2 · K/W) 0.59 1.93 6.56 1.02 3.16 10.4�q (h) 3 h 18 7 h 54 12 h 42 2 h 24 6 h 53 11 h 28 Ad (W/(m2 · K) 3.09 3.39 3.53 1.60 1.81 1.84�Ad (h) 1 h 48 1 h 52 1 h 57 2 h 09 2 h 26 2 h 29

�

�

1.6 Problems

1335vch01.indd 105 12.02.2007 11:36:33

106 1 Heat Transfer

The table shows that aerated concrete is not the wonder promised. To get suf� cient temperature damping (D� > 15), a thickness beyond 20 cm is needed. For the dynamic thermal resistance Dq to pass 4 m2 · K/W, the same holds. The � nal admittance is quite low, especially once the concrete gets rid of its initial moisture. The material therefore will not function as an effective heat storage medium.

(15) Assume a living room, surface 4 � 6.5 m, ceiling height 2.5 m. The room has two exterior walls, one 4 � 2.5 m large and the other 6.5 � 2.5 m large which are completely glazed with a gas-� lled, low-e double glazing, U-value 1.3 W/(m2 · K). The two inside walls and the ceiling have a thermal resistance between the environment at the other side and the surface in the living room of 0.505 m2 · K/W. The room is equipped with � oor heating, covered by a screed with a thermal resistance of 0.1 m2 · K/W. Walls, � oor and ceiling are grey bodies with emissivity 0.9. The glass is a grey body with emissivity 0.92. As outside air ventilation rate in the room we have 1 h–1. Assume a surface � lm coef� cient for convection equal to hc,i = 3.5 W/(m2 · K)

Calculate the glass, wall and ceiling temperatures knowing that the inside air temperature.is 21 °C and the outside temperature is –8 °C.

Answer

An approximation of the correct solution is found by considering the room as a six grey radiant surfaces system: the two windows (surface temperatures Ts1 and Ts2) the two inside walls (surface temperatures Ts3 and Ts4), the ceiling (surface temperature Ts5) and the � oor (surface temperature Ts6). The � oor heating’s temperature is T� .

We have to write seven heat balances, one convective heat balance for the room and 6 wall surface balances.

Room Balance

6

v c j sj1

(21 ) 0j

Q h A=

+ − θ =∑ , or, with v a a e i( )Q c V= ρ θ − θ

and

�i = 21 °C, �e = –8 °C, V = 65 m3

A1 = A3 = 4 � 2.5 = 10 m2

A2 = A4 = 6.5 � 2.5 = 16.25 m2

A5 = A6 = 6.5 � 4 = 26 m2

ca = 1008 J/(kg · K)

�a and hc = 3.5 W/(m2 · K):

s1 s2 s3 s4 s5 s6633.36 35 56.875 35 56.875 91 91 7680.75 0T T T T T T− + + + + + + − =

1335vch01.indd 106 12.02.2007 11:36:33

107

Surface Balances

RadiationPer surface, we have:

LR b

L

( )e

q M M= −′ρ

Linearization of the black body emittance for a temperature interval 10 to 25 °C gives:

Mb = 307.75 + 5.57 �s, r2 = 0.999

The radiosity M in turn is given by:

6bj j

j ji j2j j i

MM F M

e e =

ρ′ = − ′∑

with Fji the angle factors between each surface and the � ve other ones (values from references [1.5] and [1.36]):

Surface 1 Surface 2 Surface 3 Surface 4 Surface 5 Surface 6

Surface 1 – 0.187 0.070 0.187 0.278 0.278

Surface 2 0.115 – 0.115 0.210 0.280 0.280

Surface 3 0.070 0.187 – 0.187 0.278 0.278

Surface 4 0.115 0.210 0.115 – 0.280 0.280

Surface 5 0.107 0.175 0.107 0.175 – 0.436

Surface 6 0.107 0.175 0.107 0.175 0.436 –

Black body emittance of each surface:

s1 b1 s1 s2 s3 s4 s5 s61 0.08

(0.187 0.07 0.187 0.278 0.278 )0.92 0.92

M M M M M M M= − + + + +′ ′ ′ ′ ′ ′

s2 b2 s2 s1 s3 s4 s5 s61 0.08

(0.115 0.115 0.21 0.28 0.28 )0.92 0.92

M M M M M M M= − + + + +′ ′ ′ ′ ′

s3 b3 s3 s1 s2 s4 s5 s61 0.1

(0.07 0.187 0.187 0.278 0.278 )0.9 0.9

M M M M M M M= − + + + +′ ′ ′ ′ ′ ′

s4 b4 s4 s1 2 s3 s5 s61 0.1

(0.115 0.210 0.115 0.28 0.28 )0.9 0.9 sM M M M M M M= − + + + +′ ′ ′ ′ ′ ′

s5 b5 s5 s1 s2 s4 s4 s61 0.1

(0.107 0.175 0.107 0.175 0.436 )0.9 0.9

M M M M M M M= − + + + +′ ′ ′ ′ ′ ′

s6 b6 s6 s1 s2 s4 s4 s51 0.1

(0.107 0.175 0.107 0.175 0.436 )0.9 0.9

M M M M M M M= − + + + +′ ′ ′ ′ ′ ′

�

1.6 Problems

1335vch01.indd 107 12.02.2007 11:36:33

108 1 Heat Transfer

Inverting the matrix of that system of six equations allows us to write the radiosities of the six surfaces as a function of the black body emittances:

Matrix:

1.0870 0.0163 0.0061 0.0163 0.0242 0.0242

0.0100 1.0870 0.0100 0.0183 0.0243 0.0243

0.0078 0.0208 1.1111 0.0208 0.0309 0.0309

0.0128 0.0233 0.0128 1.1111 *0.0311 0.0311

0.0119 0.0194 0.0119 0.0194 1.1111 0

− − − − −− − − − −− − − − −− − − −− − − − − .0485

0.0119 0.0194 0.0119 0.0194 0.0485 1.1111− − − − −

Inverted:

1

2

3

4

5

6

0.9208 0.0150 0.0058 0.0146 0.0219 0.0219

0.0092 0.9214 0.0090 0.0162 0.0221 0.0221

0.0074 0.0187 0.9010 0.0182 0.0273 0.0273

0.0115 0.0207 0.0112 0.9017 0.0275 0.0275

0.0108 0.0176 0.0105 0.0172 0.9032 0.0408

0

H

H

H

H

H

H

′′′=

′′′

s1

s2

s3

s4

s5

s6

307.75 5.57

307.75 5.57

307.75 5.57

307.75 5.57

307.75 5.57

307.75 5.57.0108 0.0176 0.0105 0.0172 0.0408 0.9032

+ θ+ θ+ θ

×+ θ+ θ+ θ

Introducing that result in the qR-equation eliminates the constant 307.75.

Radiation, convection and conduction

The overall heat balance per surface is:

R C cond 0q q q+ + =

or

s1 s2 s3 s4 s5 s6 fl

s1 s2 s3 s4 s5 s6 fl

s1 s2 s3 s4 s5

9.9437 0.9599 0.3726 0.9352 1.4023 1.4023 0 62.5295

0.5907 9.9041 0.5771 1.0391 1.4129 1.4129 0 62.5295

0.3726 0.9378 10.444 0.9136 1.3699 1.369

− θ + θ + θ + θ + θ + θ + θ =θ − θ + θ + θ + θ + θ + θ =θ + θ − θ + θ + θ + s6 fl

s1 s2 s3 s4 s5 s6 fl

s1 s2 s3 s4 s5 s6 fl

s1 s2 s3 s4

9 0 115.084

0.5755 1.0391 0.5622 10.410 1.3765 1.3765 0 115.084

0.5393 0.8831 0.5269 0.8603 10.335 2.0454 0 115.084

0.5393 0.8831 0.5269 0.8603

θ + θ =θ + θ + θ − θ + θ + θ + θ =θ + θ + θ + θ − θ + θ + θ =θ + θ + θ + θ s5 s6 fl2.0454 18.355 10 73.5+ θ − θ + θ = −

In these equations, the diagonal terms consist of:

�s1, �s2 1 0.92

3.5 0.9208(1 /1.3 1 / 25) 0.08

⎡ ⎤⎛ ⎞− + + ⎜ ⎟⎢ ⎥⎝ ⎠−⎣ ⎦

�s3, �s4, �s5 1 0.9

3.5 (0.901, 0.9017, 0.9032)0.505 0.1

⎡ ⎤⎛ ⎞− + + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

�

1335vch01.indd 108 12.02.2007 11:36:33

109

�s6 1 0.9

3.5 0.90320.1 0.1

⎡ ⎤⎛ ⎞− + + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Adding the room balance and solving the system of seven equations gives as temperatures:

Window 4.0 � 2.5 m2 �s1 = 18.1 °C

Window 6.5 � 2.5 m2 �s2 = 18.2 °C

Wall 4.0 � 2.5 m2 �s3 = 21.9 °C

Wall 6.5 � 2.5 m2 �s4 = 21.8 °C

Ceiling �s5 = 22.3 °C

Floor �s6 = 28.7 °C

Floor heating �� = 35.2 °C

(16) Repeat exercise (15) for single and double glass instead of argon � lled, low-e double glass.

1.7 References

[1.1] De Grave, A. (1957). Bouwfysica 1. Uitgeverij SIC, Brussel (in Dutch).

[1.2] Cammerer, J. S. (1962). Wärme und Kälteschutz in der Industrie. Springer-Verlag, Berlin, Heidelberg, New York (in German).

[1.3] Welty, Wicks, Wilson (1969). Fundamentals of Momentum, Heat and Mass Transfer. John Wiley & Sons, New York.

[1.4] Haferland, F. (1970). Das Wärmetechnische Verhalten mehrschichtiger Außenwände. Bauverlag, Wiesbaden (in German).

[1.5] Rietschel, Raiss (1970). Heiz- und Klimatechnik, 15. Au� ., Springer-Verlag, Berlin, Heidelberg, New York (in German).

[1.6] TU-Delft, Faculteit Civiele Techniek, Vakgroep Utiliteitsbouw-Bouwfysica (1975–1985). Bouwfysica, naar de colleges van Prof A. C. Verhoeven (in Dutch).

[1.7] CSTC (1975). Règles de calcul des charactéristiques thermiques utiles des parois de construction de base des bâtiments et du coef� cient G des logement et autres locaux d’habitation. DTU (in French).

[1.8] Kreith, F. (1976). Principles of Heat Transfer. Harper & Row Publishers, New York.

[1.9] Feynman, R., Leighton, R., Sands, M. (1977). Lectures on Physics, Vol. 1. Addison-Wesley Publishing Company, Reading, Massachusetts.

[1.10] Hens. H. (1978, 1981). Bouwfysica, Warmte en Vocht, Theoretische grondslagen, 1e en 2e uitgave. ACCO, Leuven (in Dutch).

[1.11] NEN 1068 (1981). Thermische isolatie van gebouwen. NNI (in Dutch).

[1.12] El Sherbiny, S., Raithby, G., Hollands, K. Heat Transfer by Natural Convection Across Vertical and Inclined Air Layers. Journal of Heat transfer, No. 104, pp. 96–102.

[1.13] DIN 4701 (1983). Regeln für die Berechnung des Wärmebedarfs von Gebäuden. DIN (in German).

1.7 References

1335vch01.indd 109 12.02.2007 11:36:34

110 1 Heat Transfer

[1.14] Standaert, P. (1984). Twee- en driedimensionale warmteoverdracht: numerieke methoden, experimentele studie en bouwfysische toepassingen. Doktoraal Proefschrift, KU-Leuven (in Dutch).

[1.15] Tavernier, E. (1985). De theoretische grondslagen van het warmtetransport. Kursus ‘Thermische isolatie en vochtproblemen in gebouwen’. TI-KVIV (in Dutch).

[1.16] Carslaw, H. S., Jaeger, J. C. (1986). Conduction of Heat in Solids. Oxford Science Publica-tions.

[1.17] Mainka, G. W., Paschen, H. (1986). Wärmebruckenkatalog. B. G. Teubner Verlag, Stuttgart (in German).

[1.18] NBN B62-003 (1987). Berekening van de warmtedoorgangscoëf� ciënt van wanden. BIN (in Dutch).

[1.19] Lecompte, J. (1989). De invloed van natuurlijke convectie op de thermische kwaliteit van geïsoleerde spouwconstructies. Doktoraal Proefschrift, KU-Leuven (in Dutch).

[1.20] Lutz, Jenisch, Klopfer, Freymuth, Krampf (1989). Lehrbuch der Bauphysik. B. G. Teubner Verlag, Stuttgart (in German).

[1.21] Hauser, G., Stiegel, H. (1990). Wärmebrücken Atlas für den Mauerwerksbau. Bauverlag, Wiesbaden, Berlin (in German).

[1.22] Taveirne, W. (1990). Eenhedenstelsels en groothedenvergelijkingen: overgang naar het SI. Pudoc, Wageningen (in Dutch).

[1.23] IEA-Annex 14 (1990). Condensation and Energy: Guidelines and Practice. ACCO, Leuven.

[1.24] Hens, H. (1992, 1997, 2000, 2003). Bouwfysica 1, Warmte en Massatransport, 3e, 4e, 5e en 6e uitgave. ACCO, Leuven (in Dutch).

[1.25] PREN 31077 (1993). Windows, Doors and Shutters, Thermal Transmittance, Calculation Method. CEN.

[1.26] Dragan, C., Goss, W. (1995). Two-dimensional forced convection perpendicular to the outdoor fenestration surface-FEM solution. ASHRAE Transactions, Vol. 101, Pt 1.

[1.27] de Wit, M. H. (1995). Warmte en vocht in constructies. Diktaat TU-Eindhoven (in Dutch).

[1.28] Murakami, S, Mochida, A, Ooka, R., Kato, S. (1996). Numerical Prediction of Flow Around a Building with Various Turbulence Models: Comparison of k – �, EVM, ASM, DSM and LES with Wind Tunnel tests. ASHRAE Transactions, Vol. 102, Pt 1.

[1.29] Physibel, C. V. (1996). Kobra Koudebrugatlas (edited in several languages).

[1.30] Blomberg, T. (1996). Heat Conduction in Two and in Three Dimensions, Computer Modelling of Building Physics Applications. Report TVBH-1008, Lund.

[1.31] Roots, P. (1997). Heat transfer through a well insulated external wooden frame wall. Report TVBH-1009, Lund.

[1.32] Vogel, H. (1997). Gerthsen Physik. Springer-Verlag, Berlin, Heidelberg (in German).

[1.33] Hagentoft, C. E. (2001). Introduction in Building Physics. Studentlitteratur, Lund.

[1.34] ASHRAE (2001). Handbook of Fundamentals. SI Edition, Tullie Circle, Atlanta, GA.

[1.35] Saelens, D. (2002). Energy Performance Assessment of Single Storey Multiple-Skin Facades. Doctoral Thesis, KU-Leuven.

[1.36] Van der Marcke, B., Iakob (2004). Inventarisation, analysis and optimalisation of thermal bridges. CD-ROM.

[1.37] ASHRAE (2005). Handbook of Fundamentals. SI Edition, Tullie Circle, Atlanta, GA.

1335vch01.indd 110 12.02.2007 11:36:34

2 Mass Transfer

2.1 In General

2.1.1 Quantities and De� nitions

The term ‘mass transfer’ points to the transfer of air, water vapour, water, dissolved solids, gases and liquids in and through materials and building constructions. As examples, we have the air� ow in a room, the transport of water vapour through a roof, the movement of water and salts in bricks, the diffusion of blowing agents out of insulation materials, the absorption of CO2 by fresh lime plaster, etc. Mass � ow can only develop in open-porous materials, i.e., in materials that have accessible pores with an equivalent diameter larger than the diameter of the molecules that try to pass through them. In building materials without pores, in materials with smaller pores than the said diameter or in materials with only closed pores, mass transfer does not occur.Air and moisture are of the utmost importance for the physical integrity of buildings. Air carries heat (enthalpy) and water vapour. When the open pores in a material are not � lled with water, they contain humid air. Water can enter a pore only when the humid air is pushed out. Air transfer has positive and negative effects. On the one hand, the passage of dry air increases the drying potential of a construction and discharges water vapour before it may condense. On the other hand, air out� ow affects the thermal and moisture performance, while buoyancy � ow around the thermal insulation increases heat loss and gain. Cavity ventilation in turn facilitates condensation by clear sky radiant cooling. At the same time, comfort, health and indoor air quality require correctly ventilated buildings with a continuous � ow pattern between fresh air intakes and air exhausts.Of all the burdens on a building, moisture is the most destructive one. Consequently, a correct moisture tolerance is a challenge for each designer and builder. The word ‘moisture’ indicates that water in porous materials is present in its two or three phases, with different substances dissolved in the liquid phase. In other words, ‘moisture’ includes:

For temperatures below 0 °C:Ice, water, water vapour and diverse substances dissolved in the liquid phase (such as salts).

For temperatures above 0 °C:Water, water vapour and diverse substances dissolved in the liquid phase (such as salts).

Water vapour consists of separate water molecules with a diameter close to 0.28 � 10–9 m (0.28 nm), while water is composed of clusters of molecules which as clusters have a much larger diameter. As a consequence, pores that are permeable for water vapour may not be accessible for water. Thus, some materials are waterproof but not water vapour proof. Ice is crystalline. If water transforms into ice, its volume expands by 10%, which is why ice formation can be quite destructive.

The amount of humid air, moisture or another � uid a material may contain, depends on:

Density � [kg/m3]Mass per volume-unit of material. A porous solid substance has a density which is smaller than the speci� c density. For liquids and gases, the density equals the speci� c density.

�

�

�

1335vch02.indd 111 12.02.2007 16:55:03

Building Physics - Heat, Air and Moisture: Fundamentals and Engineering Methods with Examples and Exercises by Hugo Hens

Copyright © 2007 Ernst & Sohn Verlag für Architektur und technische Wissenschaften GmbH und Co. KG, Berlin

112 2 Mass Transfer

Total porosity � [% m3/m3]Volume of pores per volume-unit of material.

Open porosity �o [% m3/m3]Volume of open pores per volume-unit material. What fraction of the porous system is ‘open’ depends on the � uid migrating through the material. In general, open porosity is smaller than total porosity (�o � �).

The following relationship exists between porosity and density (pores � lled with humid air, speci� c density of the material �s):

s

s a s

1ρ − ρ ρΨ = ≈ −ρ − ρ ρ

(2.1)

with �a the density of air.

The air in a material is indicated as:

Air content wa [kg/m3]Air mass per volume-unit of material.

Air ratio Xa [% kg/kg]Air mass per mass-unit of dry material.

Volumetric air ratio �a [% m3/m3]Volume of air per volume-unit of material.

Air saturation degree Sa [%]Ratio between the current air content and the maximum possible air content. May also be de� ned as the fraction of pores � lled with air against those accessible for air.

Moisture presence is quanti� ed in an analogous way:

Moisture content w [kg/m3]Mass of moisture per volume-unit of material.

Moisture ratio X [% kg/kg]Mass of moisture per mass-unit of dry material.

Volumetric moisture ratio � [% m3/m3]Volume of moisture per volume-unit of material.

Moisture saturation degree S [%]Relationship between the moisture content present and the maximum moisture content possible. May also be de� ned as the fraction of open pores � lled with moisture against those accessible for moisture.

The four de� nitions can be extended to any kind of � uid. The � rst three average the presence of a � uid over the material, although in reality no air, moisture, dissolved salt, etc., are contained in the matrix. The real ‘content’ in the pores is equal to � S with � the density of the � uid and S the saturation degree for the � uid considered.

�

�

�

�

�

�

�

�

�

�

1335vch02.indd 112 12.02.2007 16:55:04

113

The following relationships hold between air content and air ratio and, air content and volumetric air ratio (� is the density of the porous material):

aa 100

wX =

ρ

aa

a

100w

Ψ =ρ (2.2)

Identical formulas link moisture content to moisture ratio and, moisture content to volumetric moisture ratio:

100w

X =ρ

1001000 10

w wΨ = = (2.3)

At the same time, air and moisture content are coupled by:

a a o 1000

ww

⎛ ⎞= ρ Ψ −⎜ ⎟⎝ ⎠ (2.4)

The saturation degrees also form a twin (on condition that no other substances � ll the pores):

a 1S S+ = (2.5)

Equations (2.4) and (2.5) indicate that no air will be present where water sits and vice versa. This is not the case for water vapour, which mixes with air.In the case of air, the choice of what quantity to use (wa, Xa of �a) is not bound by rules. This is different for moisture. The equation (2.3) shows that the same moisture reality induces important numerical differences among the three quantities. For materials with a density above 1000 kg/m3 the moisture ratio gives the smallest value. For materials with a density below 1000 kg/m3 that honour goes to the volumetric moisture ratio. While ‘moisture’ is co-notated negatively in practice, there is a temptation to use the lowest and apparently best scoring number for each material, without indication of units. Therefore, the following rules are set: (1) moisture content w (kg/m3) is used for stony materials; (2) moisture ratio X (% kg/kg) is used for wood-based materials; (3) volumetric moisture ratio � (% m3/m3) is used for highly porous materials, and (4) never forget to mention the units.

2.1.2 Saturation Degree Scale

If all open pores in a porous material are � lled with water, then moisture content is maximal and air content zero, meaning that the moisture saturation degree is 100% and the air saturation degree is zero. Conversely, if the pores do not contain any water, then the material is dry and air content is maximal, or, the air saturation degree is 100%. The real air and moisture saturation degrees lie between these two extremes and � zero, whereby the two evolve in opposite direction. If the moisture saturation degree increases, then the air saturation degree decreases and vice versa. Consequently, for each material, a saturation degree scale for air and moisture can be constructed. That scale contains some important areas and values, which have a speci� c role to play in the material response (see Figure 2.1 and Table 2.1).

2.1 In General

1335vch02.indd 113 12.02.2007 16:55:04

114 2 Mass Transfer

Table 2.1. The air and moisture saturation degree scale.

Saturation degree Meaning

Moisture Air

S = 0 Sa = 1 Dry materialOpen pores contain only air. S = 0 / Sa = 1 is physically not possible, but S � 0.

0 � S � SH – Hygroscopic intervalRelative humidity of the air in the pores determines the moisture saturation degree. SH represents the moisture saturation degree at a relative humidity of 98%.

Scr – Critical moisture saturation degreeBelow the critical moisture saturation degree, no water transport in the material is possible; above, it is. Air transport is hardly in� uenced by the critical moisture saturation degree.

Sc Sa,cr Capillary moisture saturation degree, critical air saturation degreeAt the capillary moisture saturation degree, air � ow through a porous material becomes impossible. That is why the value coincides with the critical air saturation ratio. Below that value, we have free � ow of air; above, not. When moistening and drying are in balance, water absorption can hardly exceed the capillary moisture saturation degree.

Sc < S � 1Scrf

This interval of the moisture saturation degrees is reached when a porous material contacts liquid water for a long period of time without drying. Dissolution of air left in the pores pushes the moisture saturation ratio beyond capillary. The critical moisture saturation degree for frost Scrf lies anywhere here. That value determines the saturation degree beyond which frost damage becomes inevitable. For materials with high tensile strength its value is close to 0.9. The closer the critical moisture saturation degree for frost to the capillary moisture saturation degree, the poorer the frost resistance of a material.

S = 1 Sa = 0 Moisture saturationMoisture saturation is only possible after all air has been sucked from the material. This requires water absorption under vacuum. An alternative is boiling the material.

In general, the following inequality holds: SH < Scr < Sc < Scrf. For a non capillary-porous material the critical saturation degree, capillary saturation degree and the critical saturation degree for frost lose their meaning.

1335vch02.indd 114 12.02.2007 16:55:04

115

2.1.3 Air and Moisture Transfer

Air and moisture transfer in a porous material could be accurately described if all related physical laws were known and if the pore system could be quanti� ed as a hydraulic network. Neither of the two is the case. The transport laws can be approximated and a global knowledge of the pore system is built up from measuring the pore distribution with mercury porosity, from material images under the electronic microscope and from X-ray tomography.However, this is not suf� cient to obtain an accurate three-dimensional topology of the network. An additional complication resides in the change of state from water into water vapour and vice versa, from water into ice and vice versa and from ice into water vapour and vice versa, all the more because water vapour is blending with the air in the pores. Hence, only a phenomenal treatment is possible, using potentials and transport coef� cients, which are all related to the pores in the material, wherein transport occurs, and the nature of the � uid in movement.For air transport, the air pressure gradients are the driving force. These are caused by external forces, by water migrating through the material, by wind and by differences in temperature and air composition. Moisture transport is a combination of water vapour and liquid displacement, included the movement of dissolved substances. The mechanisms responsible for that are: equivalent diffusion, air transport, capillary suction, gravity and external pressures.

Figure 2.1. Air and moisture saturation degree scale.

2.1 In General

1335vch02.indd 115 12.02.2007 16:55:04

116 2 Mass Transfer

Mechanism Driving force

Water vapour

Equivalent diffusion Gradient in partial water vapour pressure in the air that � lls the pores and is present at both sides of a construction part.

Air transport Gradient in total air pressure. Water vapour migrates together with the air moving in and through a construction part.

Water

Capillarity Gradient in capillary suction. Capillary suction is connected to pore width. The larger a pore, the smaller the suction but the more intense the � ow.

Gravity Difference in weight between different moisture columns. Gravity is mainly activated in pores that show limited capillary suction because of their width.

Pressure Gradient in external total pressure. Air as well as water columns are respon-sible for gradients in external pressure. Generally, air pressure gradients are small compared to the water pressure gradients. At any rate they may cause water � ow in very large pores and in cracks.

What combinations occur in which material depends on the nature of the material (Table 2.2).

Table 2.2. Vapour and moisture transfer depending on material.

Material Vapour Water

Eq. diffusion Air � ow Capillarity Gravity Pressure

Non capillary, non hygroscopica) � � – � �

Non capillary, hygroscopicb) � � – – × –

Capillary, non hygroscopicc) × – × – –

Capillary, hygroscopicd) × – × – –

a) Materials with very large pores only (ex.: coarse concrete).b) Materials with macro-pores, i.e. pores too large to cause suction, and micro-pores,

i.e. sorption-active pores with too high frictional resistance to generate capillary transport. Air � ow is possible through a macro-porous material. In a micro-porous material it is not.

c) Materials with pores small enough to cause suction but too large to generate sorption of any importance.

d) Materials with pores narrow enough to cause suction and sorption.

1335vch02.indd 116 12.02.2007 16:55:04

117

2.1.4 Moisture Sources

The moisture sources � x the starting and boundary conditions that govern moisture response in buildings. At any moment, humid air � lls part of the pores in the materials and layers that form the fabric. Humid air is also the gas mixture which contacts the bounding surfaces, except the ones below grade and those under water. Water vapour in the air inevitably entrains hygroscopic moisture in open porous materials. A material is called dry as long as the moisture content does not pass the hygroscopic equilibrium for the relative humidity present. The reasons why more moisture in excess of that equilibrium may invade the open pores are:

Water as a Source

1. By accident: leaking water pipes, leaking drain pipes, leaking sewage pipes, a leaking joint sealer around a shower, etc.

2. Rising damp: all moisture absorbed from the bottom upwards into a wall.3. Rain: the most important source of moisture in a wet climate.4. External pressure (water, air): attacks walls and � oors below the water table, is a problem

in water tanks, swimming pools, etc.5. Construction moisture: refers to all moisture in a construction at the time of decommission-

ing. There are many causes that entrain moisture in the building fabric during construction: precipitation, chemical reactions, etc.

Water Vapour as a Source

6. Relative humidity: in open-porous materials, hygroscopic moisture is entrained spontane-ously in equilibrium with relative humidity.

7. Surface condensation: indicates condensation on an inside or outside surface. On an inside surface it is seen as quite harmful. On an outside surface it is not, although condensation there competes with wind driven rain in the amounts of water deposited.

8. Interstitial condensation: concerns condensation in a construction. Generally, the pheno-menon remains unnoticed for a long time. Panic starts when consequences such as rotting, dripping moisture or corrosion become visible.

Of these eight, the triumvirate of ‘contact with water vapour’ is the most feared, although this is not always justi� ed in cool, rainy climates. Also, its understanding seems more scienti� c than the contact with water sources. So, professional literature deals extensively with water vapour, even though often not critically and far from exhaustively. Craftsmen instead are much more concerned about rain penetration and rising damp.

2.1.5 Air, Moisture and Durability

The presence of air in a material or a construction part does not bring about many dif� culties. The effects become more negative if the air is migrating in and through the part. In fact, moving air entrains water vapour into the part where it may condense, causing an increased moisture presence. At the same time, air displacement adds enthalpy transport to heat conduction.Neither should a high quantity of moisture by de� nition be harmful. Nobody cares about a brick veneer of an insulated cavity wall becoming wet by wind-driven rain. Things change

2.1 In General

1335vch02.indd 117 12.02.2007 16:55:04

118 2 Mass Transfer

when moisture stains appear inside. Also, nobody minds about surface condensation on single glass. However, once the condensate affects the wood of the window frame, then it turns into a problem. In any case, moisture is de� nitely one of the major causes of early degradation and damages. Studies show that up to 70% of all construction problems are directly or indirectly related to moisture. This is a consequence of the properties of water: a strong polar molecule, therefore an excellent solvent, an ef� cient chemical catalyst and a prerequisite for biological activity. As invisible damage we have:

Increase in thermal conductivity. For insulation materials which can absorb water or become wet by interstitial condensation, this is a problem.Larger heat � ow as a result of latent heat transfer.Decrease of strength and rigidity. The phenomenon is seen with all wet porous materials. Limited for stony materials, it is very pronounced for wood-based products that contain resins.Swelling when the moisture content increases and shrinkage when it decreases. This phenomenon, observed in all materials, is very distinct for wood-based materials.

Visible damages are:

For wood-based materialsSwelling and hydrolysis of the binder resin in particle boards, strawboards, plywood and OSB-boards. As a result, all lose strength and stiffness, sometimes to the extent that they � nally crack when loaded at a fraction of the design load.Damage by mold, mildew, fungus and bacteria. Fungus especially is to be feared. It digests cellulose and produces water as a residual product, which disintegrates completely a wood-based material.

For stony materialsMoss and algae growth. This is mainly an aesthetical problem, although the roots of mosses can degrade masonry joints, while the organic acid they produce may cause metal corrosion.Salt attack. Hydration and crystallization have to be feared. Salt attack may completely pulverize a stony material.Chemical reactions which degrade the material. Examples are carbonation in concrete, the alkali silicate reaction, etc.Frost damage. Again, the result is a pulverization of the stony material.

For metalsCorrosion. In many cases, corrosion goes together with volume expansion. This is the reason why reinforced concrete suffers from ‘concrete decay’, a consequence of corrosion of the reinforcement. Metal oxides are also porous and hygroscopic, and swell when they get wet.

For synthetic materialsHydration. The synthetic material becomes weak and loses its cohesion over the time.Foams sometimes experience an irreversible deformation as a result of the combined actions of temperature and vapour pressure � uctuations in the pores. Water which entered the pores by interstitial condensation is responsible for the latter.

�

�

�

�

�

�

�

�

�

�

�

��

1335vch02.indd 118 12.02.2007 16:55:04

119

2.1.6 Linkages between Mass- and Energy Transfer

Mass transfer causes energy displacement. Indeed, movement means:Mechanical energyIncludes kinetic energy coupled to mass velocity and potential energy as a result of gravity, suction, pressure, etc. Most mass transfer phenomena in porous materials generates mass velocities which are too low for the kinetic energy to play any role. Instead differences in potential energy are generally the cause of displacement.

Enthalpy transferEach gas, each liquid, each solid, which has a temperature �, contains a quantity of heat per mass-unit, given by:

o( )Q c= θ − θ (2.6)

where c is speci� c heat (J/(kg · K)); � temperature and �o a reference temperature, generally 0 °C. A mass � ow G is responsible for a sensible heat � ow, equal to:

o( )G cΦ = θ − θ (2.7)

If phase changes occur in the moving mass, then a latent heat � ow, equal to the heat of transformation (symbol l or lo at reference temperature, units J/kg), is also carried along:

o o[ ( ) ]G c lΦ = θ − θ + (2.8)

In building physics most processes are isobaric. In such case c is the speci� c heat at constant pressure. The quantities c (� – �o) and c (� – �o) + lo are then named the enthalpy of the migrating mass (symbol: h, units: J/kg). It is precisely the enthalpy transfer that damages the thermal quality of a building part subjected to air and moisture intrusion.

2.1.7 Conservation of Mass

As stated, this chapter deals with air, vapour and moisture transfer. This of course does not mean that other masses such as radon, CO2, SO2, etc., cannot � ow through an open-porous material nor that salts and other soluble substances may not use moving liquid water as a carrier to migrate into the porous system. First some de� nitions:

Quantity of mass M [kg]Quantity, which indicates how much mass is present or migrates. The quantity of mass is a scalar.

Mass � ow G [kg/s]The quantity of mass which migrates per unit of time. Just as the quantity of mass, it is a scalar.

Mass � ow rate g [kg/(m2 · s)]The quantity of mass, which � ows per unit of time through a unit surface. The mass � ow rate is a vector having the same direction as the surface. Components: gx, gy, gz in a Cartesian co-ordinate system, gR, g�, g in a polar coordinate system.

The kind of substance which migrates is given by the suf� x: a for air; l for dry air; v for water vapour, m for moisture, w for water; the chemical formula for gases

2CO(ex.: )g , the name for

�

�

�

2.1 In General

1335vch02.indd 119 12.02.2007 16:55:04

120 2 Mass Transfer

other liquids. If water vapour, water and moisture are the only mass components dealt with, then the suf� xes v, m and w are often omitted.Solving a mass transfer problem means calculating the values for the potentials (Po) that cause the transfer and quantifying the mass � ow rates (g). Thus, two unknowns intervene: Po(x,y,z,t) and g(x,y,z,t), the one a scalar (Po), the other a vector (g). In order to calculate both, a scalar and a vector equation are needed. The scalar equation follows from the mass conservation axiom: mass � ow exchanged between a system (an elementary material volume, a layer, a space) and its environment (the rest of the material, other layers, other spaces) plus the production (source) or removal (sink) of that mass per unit of time in the system equal to the change in quantity of this mass per unit of time in the system:

xxdiv ( )

wS

t

∂± =

∂xg (2.9)

where Sx is the source or sink, in kg/(m3 · s). The � ows can be either diffusive or convective. In the � rst case, they depend on the gradient of the driving potential. In the second case, they depend on the driving potential itself. The relationship between � ow and driving potential delivers the vector equations needed.

2.2 Air Transfer

2.2.1 In General

Dry air, suf� x l, is a mixture of 21 % m3/m3 oxygen (O2), 78 % m3/m3 nitrogen (N2) and traces of other gases (CO2, SO2, Ar, Xe). Hereafter, it is assumed that dry air is behaving as an ideal gas, equation: pl V = ml Rl T where pl is the (partial) dry air pressure in Pa, T the temperature in K, ml the mass of (dry) air in the volume V in kg, and Rl the gas constant for dry air, equal to 287.055 J/(kg · K). In reality, dry air is not an ideal gas, and the following ‘exact’ equation should be used:

aa aaal2

l l l

1( / ) ( / )

= + +B Cp V

n R T V n V n

with nl the number of moles of dry air in volume V (m3); R the general gas constant (8314.41 J/(mol · K)); and Baa and Caaa the viral coef� cients, given by:

24

aa 2 3

0.668772 10 2.10141 92.47460.349568 10B

T T T

−− ⋅= ⋅ − − + [m3/mol]

2aaa 2

0.190905 63.24670.125975 10C

T T−= ⋅ − + [m6/mol2]

At ambient temperature, these viral coef� cients are almost zero. For the dry air concentration the ideal gas law gives:

l ll

l

m p

V R Tρ = = (2.10)

1335vch02.indd 120 12.02.2007 16:55:04

121

For moist air, a mixture of dry air and water vapour, suf� x a, the ideal gas equation and concentration become: a a aP V m R T= , a a a/ ( )P R Tρ = where Pa is the air pressure in Pa, equal to the sum of the partial dry air pressure and the partial water vapour pressure (Pa = pl + pv, large P’s indicate total pressures, small p’s partial pressures).With respect to dry air, the presence of water vapour modi� es the air mass ma, the gas constant Ra and the concentration �a. However, for temperatures under 50 °C, the effect is so small that the following holds: Ra � Rl, Pa � pl and �a � �l or: a a l l/ ( ) / ( )P R T p R T≈ .

Air transfer through materials, construction elements and buildings occurs when:Wind, stack forces and fans create air pressure differences over the various elements, between different rooms and between a building and its environment.The construction elements are air permeable. Air permeability may be a desired characte-ristic, the case when ventilation grids are used. However, it is typically an unwanted consequence of using air permeable materials, of � ssures and cracks in and between parts and of the several meters of joints in layers made of planks, small elements and boards.

2.2.2 Air Pressure Differences

2.2.2.1 Wind

Wind pressure is given by:

22a

w p p0,62

vP C C v

ρ= ≈ (2.11)

The equation follows from Bernoulli’s law, applied to a horizontal wind, velocity v, colliding with an in� nite obstacle and obtaining a velocity zero (conversion from kinetic to potential energy; Figure 2.2). Cp is a correction term, called the pressure factor, which takes care of the difference between an in� nite and any � nite obstacle. Finite dimensions de� ect the air � ow to the upper and side edges where it forms turbulent vortexes. At the same time, a lee zone develops behind the obstacle. Consequently the pressure factor Cp links the real pressure at a speci� c point on the surface to the reference velocity. Its value is function of:

The reference velocity. In principle this concerns the velocity measured in an open � eld at a height of 10 m. Of course, another reference can be used depending on the situation. For buildings, pressures on the facades are often correlated to the wind velocity above the ridge. A change in reference alters the pressure factors:Wind directionLocation of the building (open, wooded, urban area, close to high-rise buildings)The geometry of the buildingThe façade spot considered

Cp positive means overpressure, Cp negative means depressurization. Overpressure is found at the weather side, depressurization at the lee side and along all surfaces more or less parallel to the wind direction (Figure 2.2). Wind pressure differences are extremely variable over time. They may also change sign.

�

�

�

����

2.2 Air Transfer

1335vch02.indd 121 12.02.2007 16:55:05

122 2 Mass Transfer

2.2.2.2 Stack Effects

Stack or buoyancy in gases and liquids has two causes: (1) differences in temperature and (2) differences in composition. Con� ned to atmospheric air, the decrease of air pressure at a height h is given by: a ad dP = − ρ g h where g is the acceleration of gravity. Height zero stands for sea level. Inserting the ideal gas law gives:

a

a a

d dP

P R T= − g h

(2.12)

With temperature T constant, the solution becomes:

a a0a

expP PR T

⎛ ⎞= −⎜ ⎟⎝ ⎠

g h (2.13)

This result is known as the barometric equation. Air pressure apparently depends on the height. However, temperature differences (T variable) and different composition (Ra variable) will induce air pressure gradients at the same height. They produce ‘stack’. For the pressure difference between two points, one at height h1, the other at height h2, we have:

oa a,h 2 1 a( ) / ( )P P h h R TΔ = −g where the suf� x ho denotes the average height over the interval h2 – h1. The ‘stack potential’ in a point at height h is now given by the pressure difference between that point in an air mass with variable temperature and composition, and the point at identical height in an air mass at constant temperature and composition:

o oStack a,h a,ha a o a m a oo

d 1 1

( ) ( ) ( ) ( ) ( )

h

h

P P PR h T h R T R T R T=

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦∫

h hg g h

where (Ra T)m is the harmonic average of the product Ra(h) T(h) at height h:

a

ao

( )d

( ) ( )

m h

h

R T

R h T h=

⎡ ⎤=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

∫

h

h

Figure 2.2. Bernouill’s law and the wind pressure � eld around a building.

1335vch02.indd 122 12.02.2007 16:55:05

123

Resulting stack between two points at the same height h, but in a column of air at different temperature and composition, then becomes:

oT,1 2 Stack2 Stack1 a,ha m2 a m1

1 1

( ) ( )p P P P

R T R T−⎡ ⎤

= − = −⎢ ⎥⎣ ⎦

g h

If between height zero and height h, the temperatures T1 and T2 are different but constant and if the composition of both air volumes is the same, then the equation simpli� es to:

oa,h a1 1 a2 2 0 a1 1 a2 2T,1 2

a1 1 a2 2 am12 m12

a 1 2

( ) ( )

( - )

aP R T R T R R

pR T R T R T−

− ρ θ − θ≈ ≈

≈ ρ β θ θ

g h g h

g h

(2.14)

where ��= 1 / Tm12) is the compressibility of air. If, on the contrary, temperature is constant but the composition variable, then stack becomes:

T,1 2 a aa2 a1

1 1p R

R R−⎛ ⎞

≈ ρ β −⎜ ⎟⎝ ⎠g z (2.15)

In case the air volumes are connected by leaks, a neutral plane, where stack is zero, develops somewhere between the lowest and the highest leak. Without other causes of pressure difference, the air in that plane is horizontally in equilibrium. Above that plane air goes from warm to cold, or, in case of different vapour concentration, from higher to lower concentration. Below, it goes from cold to warm or from lower to higher concentration. The location of the neutral plane depends on the distribution and the size of the leaks.Unless for high-rise buildings, air pressure differences by thermal stack are small. Pressure differences by water vapour concentration linked stack may even be neglected.

Example

If during winter the inside temperature is 20 °C and the outside temperature 0 °C, then over the height of a room (z = 2.5 m) thermal stack reaches (�a0 = 1.2 kg/m3, g = 9.81 m/s2, = 1/273.16 K–1, �m12 – �0 = 20 °C, z = 2.5 m):

t1

1.2 9.81 20 2.5 2.15 Pa273.16

p = ⋅ ⋅ ⋅ ⋅ =

For a high-rise building of 250 m instead, thermal stack between the lowest and the highest � oor equals 215 Pa, which makes it as important as the pressure difference between the windward and leeside for a wind velocity of 62 km/h.

Stack is much more stable over time than wind pressure differences.

2.2.2.3 Fans

Fans are part of any air heating, air conditioning or forced ventilation system. The pressure differences they create are typically slightly to quite a lot higher than those caused by stack and wind. Fan induced pressure differences are also very stable over time.

2.2 Air Transfer

1335vch02.indd 123 12.02.2007 16:55:05

124 2 Mass Transfer

2.2.3 Air Permeances

We have to differentiate between:

1. Open porous materials, such as no-� nes concrete, mineral � bre and wood-wool cement.

2. Air permeable layers (layers made of small elements with joints in between), leaks, cracks, cavities and intentional or unintentional gaps. All these are found in great numbers in building constructions:Air permeable layersAll covering, panelling and � nishing made of scaly or plate elements (tiles, slates, corrugated sheets, lathed � nishes, wooden laths) and all layers composed of board-like or strip formed materials (plywood, OSB, insulation layers, underlay). Even masonry is an air permeable layer (bricks, joints and cracks between bricks and joints!).Joints and cracksJoints can be found between construction parts and between boards and other elements. Cracks form spontaneously when tensile strength is superseded. LeaksEach nail hole, electricity boxes in walls, light spots in ceilings, etc. form leaks. Cavities, voidsAre found in a variety of facade and roof solutions. Openings in the envelopeOccur intentionally (ventilation grids) or unintentionally (damage, bad joint � ller, loose connection, etc.).

In an open porous material, air displacement follows Poiseuille’s law of proportionality between the air � ow rate and the driving force, in the case being the change of air pressure per unit length. The proportionality factor is called the air conductivity or air permeability ka, units s, of the material:

a ak P= −ag grad (2.16)

The minus sign in the equation indicates that air � ows from points at higher to points at lower air pressure. The permeability ka is a scalar for isotropic materials and a tensor with three main directions for anisotropic materials. In that case its value is different for each main direction. Permeability increases with the open porosity and the number of macro pores in a material.For air permeable layers, leaks, cracks, joints, cavities and openings, the � ow equation becomes:

Air permeable layers (per m2) a a ag K P= − Δ , in s/m

Joints, cavities (per m) a a aG K Pψ= − Δ , in kg/(m · s) and aKψ in s (2.17)

Leaks, openings (per unit) a a aG K Pχ= − Δ , in kg/s and aKχ in m · s

where Ka, aKψ and aK χ are the air permeances and Pa the air pressure differential over the air open layer, the joint, cavity, leak or opening. Because the � ow is not necessarily laminar, the air permeance depends on the pressure differential. This dependence can be expressed through an exponential relationship: 1

a a( )x bK a P −= Δ , where a is the air permeance coef� cient for a pressure differential of 1 Pa, and b is the air permeance exponent. For laminar � ow, b is 1, for turbulent � ows it is 0.5. For transition � ows, b lays between 0.5 and 1. Normally, the air

�

�

�

�

�

1335vch02.indd 124 12.02.2007 16:55:05

125

permeance is only quanti� able through experiment. For joints, leaks, cavities and openings with a known geometry the value can be calculated using hydraulics. Pressure losses to take into account are:

Frictional losses:

2a

aH H

0.422

vL LP f f

d d

ρΔ = ≈ 2

ag (2.18)

where f is the friction factor, dH the hydraulic diameter of the section with length L, and v the average � ow velocity in that section (Figure 2.3).

Local losses near bends, widening, narrowing, entrances and exits (Figure 2.3):

22a

a a0.422

vP g

ρΔ = ξ ≈ ξ (2.19)

where � is the factor of local loss.

Values for the friction factor are given in Table 2.3. Table 2.4 shows some factors of local loss. The calculation of air permeance then stems from a direct application of the continuity equation, which states that in each section, the same air � ow has to be found (= de� nition of mass conservation, Figure 2.3). In other words, along the � ow line, the air � ow Ga is constant.

�

�

Table 2.3. Friction factor ‘f ’.

Reynolds numbera) Flow Friction Factor f

Re � 2500 Laminar 96 / Re

2500 � Re � 3500 Critical ,Re 25000.038 (3500 Re) (Re 2500)

1000Tf =− + −

Re > 3500 Turbulent fT b)

Re � 3500 Stable turbulent fT = C te, single function of roughness

a) HRev d

=ν

where v is the average � ow velocity, � the kinematic viscosity and dH the hydraulic

diameter, given by:

– for a circular section: the diameter of the circle

– for a rectangular section: 2 a b

a b+, where a and b are the sides of the rectangular

– for a cavity: 2 b, where b is the width of the cavity

Re can also be written as: Re � 56,000 ga dH where ga is the air � ow rate

b)

210

4.793 log 0.2Re

2 log 0.2698ReTf

−⎡ ⎤⎛ ⎞⎛ ⎞+ ε⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎢ ⎥= − + ε⎜ ⎟⎢ ⎥⎜ ⎟

⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

where � is relative roughness

(see Figure 2.4).

2.2 Air Transfer

1335vch02.indd 125 12.02.2007 16:55:06

126 2 Mass Transfer

Table 2.4. Factor of local loss �.

Local resistance �

Entering an opening 0.5

Leaving an opening 1.0

Widening Re � 1000 –0.036 + 9.6 � 10–5 Re + �� = A0 / A1 1000 < Re � 3000 1.28 � 10–5 Re1.223 + �A0 small section Re > 3000 0,.21 Re0.012 + �A1 large section � � 0.5 � = 0.78 – 1.56 �

� > 0.5 � = 0.48 – 0.96 �

Narrowing Re � 1000 0.98 Re–0.03 + A

� = A0 / A1 1000 < Re � 3000 10.59 Re–0.37 + A

A0 large section Re > 3000 0.57 Re–0.01 + A

A1 small section A = 0.0373 �2 – 0.067 �

Leak 2.85

Angle or curve 0.86Re0 g 0 1

gH 0 0

where 0.885 and( )

−ξ ⎛ ⎞ξ = ⎜ ⎟⎝ ⎠

k k b

d b

b0: width of the admission channel � 3000 � Re < 40,000 Re � 40,000

b1: width of channel after the curve kRe0 k0 / (dH)0 kRe0 k0 / (dH)0

0: refers to the admission channel 0 45 f0 1 1.1 1

f0: friction factor in admission channel 0–0.001 45 f0 1 1.0 1 + 0.5 � 103 ��: relative roughness > 0.001 45 f0 1 1.0 1.5

Figure 2.3. Friction losses, local losses and continuity.

1335vch02.indd 126 12.02.2007 16:55:06

127

Further on, total pressure loss – the sum of the frictional losses and all local losses – must equal the driving pressure. Both relationships allow for the calculation of air � ow Ga as function of pressure difference Pa.

2.2.4 Air Transfer in Open-porous Materials

2.2.4.1 Conservation of Mass

For air, we have (neither source nor sink):

aadiv

wg

t

∂= −

∂ (2.20)

Air content wa depends on total open porosity �o, saturation degree for air Sa, air pressure Pa and temperature T: a 0 a a 0 a a a/( )w S S P R T= Ψ ρ = Ψ . While the gas constant Ra and the open porosity �0 are quite invariable, saturation degree Sa, air pressure Pa and temperature T may vary continuously. Therefore, the differential of air content with respect to time becomes:

a 0 a a a aa

aa

w P S S P TS P

t R T t t T t

∂ Ψ ∂ ∂ ∂⎛ ⎞= + −⎜ ⎟⎝ ⎠∂ ∂ ∂ ∂

If the saturation degree is constant, then the equation simpli� es to:

a 0 a a a

a

w S P P T

t R T t T t

∂ Ψ ∂ ∂⎛ ⎞= −⎜ ⎟⎝ ⎠∂ ∂ ∂

In isothermal conditions, 0T

t

∂ =∂

(that condition of course excludes thermal stack):

0 a a

a

divS P

R T t

⎛ ⎞Ψ ∂= − ⎜ ⎟ ∂⎝ ⎠ag (2.21)

2.2 Air Transfer

� = [d1 + d2 + d3 + d4) / 4] / dH

Figure 2.4. De� nition of relative roughness for a circular tube.

1335vch02.indd 127 12.02.2007 16:55:06

128 2 Mass Transfer

The ratio 0 a a/( )S R TΨ is called the isothermal volumetric speci� c air content, symbol ca, units kg/(m3 · Pa). If the open porosity and saturation degree for air are lower than 1, replacing Ra by its value, 287 J/(kg · K), gives: 0 a a/( ) 0.00348 /S R T TΨ < , which is a very small number.

2.2.4.2 Flow Equation

The second relationship between air � ow rate and air pressure is given by equation (2.16) with the air pressure gradient including wind, stack and fan operation. Isothermal air transfer at any rate excludes thermal stack.

2.2.4.3 Air Pressures

Inserting equation (2.16) in the isothermal mass balance (2.21) gives:

aa a adiv ( )

Pk P c

t

∂=

∂grad

In case air permeance ka is constant, we may also write:

2 a aa

a

c PP

k t

∂∇ =

∂ (2.22)

In equation (2.22) the ratio ka / ca stands for the isothermal air diffusivity of the material; symbol Da, units m2/s. That value is generally so high that each adjustment in air pressure happens within a few seconds. Only for pressures that � uctuate over extremely short periods of time, such as wind gusts, will the air pressure response show damping and time shift. With respect to periods of some minutes or longer, damping turns to one and the time shift to zero.

Example

Assume mineral wool has a density of 40 kg/m3.

Properties:

Thermal Air

� = 0.036 W/(m · K) ka = 8 � 10–5 s

� c = 33,600 J/(m3 � K) ca = 1.17 � 10–5 kg/(m3 · Pa)

Giving as diffusivities:

Thermal Air

1.1 � 10–6 m2/s 6.8 m2/s

Consequently, the isothermal air diffusivity is 6,355,000 times greater than the thermal diffusivity.

Thus, when combining fast air transfer with much slower heat and moisture transfer, the isothermal equation (2.22) can be limited to steady-state:

2a 0P∇ = a ak P= −ag grad

1335vch02.indd 128 12.02.2007 16:55:06

129

Both equations form a copy of steady-state heat conduction. For a non-isothermal condition, the same conclusion holds: steady-state suf� ces. Yet, the isobaric volumetric speci� c air capacity is larger than the isothermal one (� 3525 / T2), resulting in a little more inertia. At any rate, for periods longer than some minutes, air pressure damping remains one. Because thermal stack now interferes, both equations are written as:

2ao a( ) 0P∇ − ρ =g z a ao a( )k P= − − ρag grad g z

Solution is only possible in combination with the thermal balance.

2.2.4.4 One Dimension: Flat Walls

The results are analogical as for steady-state heat conduction. Thus, in a single-layered wall, the air pressures form a straight line (Figure 2.5) while the constant air � ow rate is given by:

a aa a

a/

P Pg k

d d k

Δ Δ= = (2.23)

The ratio d / ka is called the (speci� c) air resistance of the wall; symbol W, units m/s. The inverse gives the air permeance; symbol Ka, units s/m.For a constant air � ow rate through a composite wall, we have:

aa

1 a,

ni

i i

Pg

d

k=

Δ=∑

(2.24)

where � di / ka,i is the total air resistance of the composite wall; symbol WT, units m/s. The inverse 1 / WT is called the total air permeance KaT. di / ka,i represents the air resistance of layer i, symbol Wi. The inverse 1 / Wi = ka,i / di is called the air permeance Ka,i of layer i.

Figure 2.5. Air � ow through a single-layered � at wall.

2.2 Air Transfer

1335vch02.indd 129 12.02.2007 16:55:07

130 2 Mass Transfer

Air pressure on a random interface in the wall is given by:

1aj a1 a2 a1 a1 a 1

T

( )

j

ii

j

WP P P P P g W

W== + − = +∑

(2.25)

That result is also found graphically, after transposing the wall section into the [W, Pa] plane, where the composite wall behaves as if it was single-layered, and transferring the intersections with the interfaces there to the wall section in the [d, Pa] plane (Figure 2.6). As air pressure difference �Pa,j over any layer j in the wall we have:

a, a2 a1T

( )jj

WP P P

WΔ = −

showing that the largest air pressure difference in a composite wall is found over the most air-tight layer (Figure 2.7). Deliberately building-in such a layer diminishes air permeance

Figure 2.6. Air� ow through a composite wall. The graph at the left gives the air pressure distribution in a [Wa, Pa]-plane, the graph at the right in the [da, Pa]-plane.

Figure 2.7. Effect of an airtight layer on the air pressure distribution in a wall. It attracts the total pressure difference.

1335vch02.indd 130 12.02.2007 16:55:07

131

of a wall substantially, which is why it is called an ‘air barrier’ or ‘air retarder’. The purpose of an air barrier is to limit the air � ow through a construction part. Of course, the air barrier must be able to withstand wind load.

2.2.4.5 Two- and Three-dimensions

For isotropic open porous materials the calculation is a copy of two- and three-dimensional heat conduction. Of course, mass conservation substitutes energy conservation. Sum of air � ows from the adjacent to the central point zero: � Ga, i + j = 0. The algorithm then becomes:

a, a, a, , , a,, , , ,1 1

( ) 0i j i j l m n i ji l m n i l m nj j

K P P K+ + += ==± =±

− =′ ′∑ ∑

with a,i jK +′ the air permeance between each of the adjacent and the central point. Value: ka A / a, where A is the surface of the control volume seen by the adjacent point and a is the distance along the mesh between adjacent and central point (Figure 2.8). For p control volumes, the result is a system of p equations with p unknowns:

a a a, , , a, , ,[ ] [ ] [ ]pp p i j k i j k pK P K P⋅ =′ ′ (2.26)

where a[ ]p pK ⋅′ is a p rows p columns permeance matrix, [Pa]p is a column matrix of the p unknown air pressures and a, , , a, , ,[ ]i j k i j k pK P′ is a column matrix containing the p known air pressures and/or air � ows. Once the air pressures are known, then the � ows follow from:

a, , a, , a, a,( )i i j i i j i j iG K P P+ + += −′ (2.27)

For anisotropic materials the same algorithm applies, except that per mesh line the direction-related aK ′-value has to be taken. Besides, it is implicitly understood that the mesh lines coincide with the main directions of the air permeability tensor. Under non-isothermal conditions, thermal stack intervenes and the thermal and air balances have to be solved simultaneously. Most of the time this requires iteration. CVM has been translated into software packages, which allow the calculation of air transfer through the most complicated details.

Figure 2.8. Conservation of mass: sum of air� ows from adjacent to central point zero.

2.2 Air Transfer

1335vch02.indd 131 12.02.2007 16:55:08

132 2 Mass Transfer

2.2.5 Air Flow Through Permeable Layers, Apertures, Joints, Leaks and Cavities

2.2.5.1 Flow Equations

We refer to paragraph 2.1.3. For operable window and door frames, the air permeance exponent b in the � ow equation is typically set at 2/3:

2 / 3a aG a L P= Δ

with L the casement length in m and a the air permeance per meter run of operable frame, units kg/(m · s · Pa2/3).

2.2.5.2 Conservation of Mass, Equivalent Hydraulic Circuit

Most constructions combine air-open layers with open porous materials, joints and cavities. In such a case, expressing the conservation law as a partial differential equation is not possible. One approach consists of transforming the construction into an equivalent hydraulic circuit, i.e., a combination of well chosen points, connected by air permeances (Figure 2.9). In each point the sum of the air� ows coming from the adjacent points must be zero, or: � Ga = 0. As each of the air � ows is given by a a a

xG K P= Δ , insertion in the conservation law results in:

a, a, a, , , a,, , , ,1 1

( ) 0x xi j i j l m n i j

i l m n i l m nj j

K P P K+ + += ==± =±

− =∑ ∑ (2.28)

With p points, a system of p non-linear equations ensues. The p unknown air pressures are found by solving that system for the known boundary conditions. Insertion of these air pressures into the � ow equations gives the air transfer between points. The non-linearity makes iteration necessary, starting with the assumption of p air pressures values. Then, the air permeances are calculated with these values and the system solved with these permeances. The p new air pressures allow recalculating the air permances and solving the system again. Iteration that way is continued until the standard deviation between the preceding and present p air pressures falls below a suitable value (for example < 1/1000).

Figure 2.9. Equivalent hydraulic circuit.

1335vch02.indd 132 12.02.2007 16:55:08

133

As soon as � at building parts contain air permeable layers, leaks, joints and cavities, one-dimensional air transfer becomes the exception and the hydraulic circuit method should also be applied there. In case we have one-dimensional � ow by coincidence, then the part can be represented by a series circuit of linear and non-linear air permeances. As the same air � ow rate ga must migrate through each layer, the following holds:

Layer 1

11 1 1

ba a a 1 ag K P a P= Δ = Δ

or

1

1

1

ba

a1

gP

a

⎛ ⎞= Δ⎜ ⎟⎝ ⎠

Layer 2

22 2 2

ba a a 2 ag K P a P= Δ = Δ

or

2

2

1

ba

a2

gP

a

⎛ ⎞= Δ⎜ ⎟⎝ ⎠

…

Layer n

ba a a a

nn n nng K P a P= Δ = Δ

or

1

ba

an

nn

gP

a

⎛ ⎞= Δ⎜ ⎟⎝ ⎠

Sum:

11

ba

a a1

b

i

ii

gg P

a

−⎡ ⎤⎢ ⎥

= Δ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

∑ (2.29)

where �Pa is the air pressure difference over the � at part. Equation (2.29) requires an iterative solution. After the air � ow rate ga is determined, the air pressure distribution in the � at building part is calculated using the layer equations.

2.2.6 Combined Heat- and Air Transfer

2.2.6.1 Open-porous Materials

Air displacement through an open-porous material adds enthalpy � ow, given by q = H = ca ga �, to heat conduction, with ca speci� c heat capacity, for air 1008 J/(kg · K), and ga air � ow rate. The reference temperature for enthalpy is typically 0 °C. If we assume that air progresses so slowly through the porous system that the material matrix and the air in the pores keep the same temperature at each spot, then energy conservation for an elementary volume dV becomes:

a a a oa

( )div ( )

c c Sc

t

∂ ρ + ρ Ψ θ+ = − ± Φ′

∂θaq g

with �a ca the volumetric speci� c heat capacity of air, Sa the saturation degree for air and �o the open porosity of the material. In most cases the product �a ca Sa �o is much smaller than the volumetric speci� c heat capacity �c of the material, which allows simplifying the balance to:

adiv ( )c

ct

∂ρ θ+ = − ± Φ′∂

θaq g

2.2 Air Transfer

1335vch02.indd 133 12.02.2007 16:55:08

134 2 Mass Transfer

Introducing Fourier’s conduction law and expanding the operator term (div ga = 0) gives (� and � c constant):

2ac c

t

∂θ−λ ∇ θ + ∇θ = −ρ ± Φ′∂ag (2.30)

Steady-state

In ‘steady-state’, boundary conditions do not depend on time. Heat sources and sinks never change and the material properties remain constant. In such cases equation (2.30) becomes:

2acλ ∇ θ − ∇θ = ±Φ′ag (2.31)

One-dimension: Flat Walls

If the air and heat � ow rate develops perpendicular to the wall surface, then equation (2.31) becomes (neither heat source nor sink):

2a

2

d d0

dd

c

xx

θ θ− =λ

ag (2.32)

i.e., a second order differential equation with a solution of:

a a1 2 exp

c g xC C

⎛ ⎞θ = + ⎜ ⎟⎝ ⎠λ (2.33)

wherein C1 and C2 are the integration constants. They ensue from the boundary conditions. The heat � ow rate by conduction, the enthalpy � ow rate (also called convective heat � ow rate) and the total heat � ow rate are then given by:

Conduction

a aa a 2

dexp

d

c g xq c g C

x

θ ⎛ ⎞= −λ = − ⎜ ⎟⎝ ⎠λ

Enthalpy (convection)

a ac a a 1 2 exp

c g xH q c g C C

⎡ ⎤⎛ ⎞= = +⎢ ⎜ ⎟⎥⎝ ⎠λ⎣ ⎦

Total

T c a a 1q q H q q c g C= + = + =

While the conduction and enthalpy � ow rate change exponentially over the wall, the total heat � ow rate remains constant. In other words, the wall functions as a heat exchanger between the conduction and convection mode, heating or cooling the migrating air.

1335vch02.indd 134 12.02.2007 16:55:09

135

Single-layered Walls

Let d be the thickness of the wall. We may rewrite (2.33) as:

a a1 2 exp

c g d xC C

d⎛ ⎞θ = + ⎜ ⎟⎝ ⎠λ

(2.34)

The dimensionless ratio ca ga d / � is called the Péclet number, symbol Pe. The Péclet number links the convective to the conductive heat � ow rate. The more air is � owing through 1 m2 of the wall and the higher its thermal resistance, the higher the number. As boundary conditions we may have:

Type 1: The air � ow through the wall and the temperatures on both wall surfaces are known: x = 0: � = �s1; x = d: � = �s2

Type 2: The air � ow through the wall and the heat � ow rates by convection and radiation on both wall surfaces are known: x = 0, q = h1 (�1 – �s1) + ca ga �; x = d, q = h2 (�2 – �s2) + ca ga ��

Type 1

In this case the integration constants follow from: �s1 = C1 + C2, �s2 = C1 + C2 exp(Pe), or:

s2 s11

exp (Pe)

1 exp(Pe)C

θ − θ=

−

s1 s22 1 exp(Pe)

Cθ − θ

=−

Temperatures in the wall, conductive heat � ow rate, enthalpy � ow rate and total heat � ow rate become:

s1 1 s2 s1( ) ( ) F xθ = θ + θ − θ

s2 s12 ( )q F x

d

θ − θ= λ

c Peq

d

λ θ=

s2 2 s1 2T

(0) ( )F F dq

d

θ − θ= λ

(2.35)

with:

1

1 exp Pe( )

1 exp(Pe)

x

dF x

⎛ ⎞− ⎜ ⎟⎝ ⎠=

− 2

Pe exp Pe( )

1 exp(Pe)

x

dF x

⎛ ⎞⎜ ⎟⎝ ⎠

=−

Both F1(x) and F2(x) depend on the wall characteristics and the air � ow rate through the wall. If the air � ows from d to 0, then the � ow rate gets a negative sign. In the opposite case, it has a positive sign. The air � ow is responsible for the exponential change in temperature, convex when the air goes from warm to cold, concave when � ow develops from cold to warm (Figure 2.10). The higher the Péclet number, i.e., the more air passes through, the more concave or convex is the exponential. Compared to pure conduction (qo), the change in total heat � ow rate as a percentage is (�s1 = 0 °C (x = 0), �s2 = 1 °C (x = d)): 100 (qT – q0) / q0 = 100 [abs (F2(0)) – 1]. For air � owing from warm to cold abs (F2(0)) is larger than 1. In such case, enthalpy transfer forces total heat transfer to increase. For air � owing from cold to warm, abs (F2(0)) becomes smaller than 1 and enthalpy transfer diminishes total heat transfer.

2.2 Air Transfer

1335vch02.indd 135 12.02.2007 16:55:09

136 2 Mass Transfer

Type 2

Boundary conditions of type 2 are converted into type 1 by adding a � ctitious surface � lm layer to each side of the wall, one with a thermal resistance 1 / h1 in x = 0 and one with a thermal resistance 1 / h2 in x = d. The two � ctitious layers are 1 m thick and have a thermal conductivity h1 or h2. As a consequence, the reference temperatures �1 and �2 turn into � ctitious surface temperatures while the wall changes from single-layered to composite. Such a composite wall behaves as single-layered in the [�, R]-plane (��for temperature, R for thermal resistance). This not only applies for pure conduction but also for the combination of enthalpy transfer and conduction. Writing equation (2.32) with � and R as co-ordinates, gives:

2

a a2

d d0

ddc g

RR

θ θ− = (2.36)

Solution:

1 2 a aexp( )C C c g Rθ = +

Boundary conditions:

10 :R = θ = θ

n 21 2

1 1:= = + + θ = θR R R

h h

Integration constants:

2 1 a a n1

a a n

exp ( )

1 exp( )

θ − θ=

−c g R

Cc g R

1 22

a a n1 exp( )

θ − θ=

−C

c g R

Temperatures in the wall, heat � ow rate by conduction, enthalpy � ow rate and total heat � ow rate become:

1 2 1 1( ) ( )F Rθ = θ + θ − θ

2 2 1( ) ( )q F R= θ − θ c a aq c g= θ T 2 2 1 2 n(0) ( )= θ − θq F F R (2.37)

Figure 2.10. Single-layered wall, combined conduction and enthalpy � ow. Upper temperature curve for warm air mitigating to the cold side. Lower temperature curve for cold air mitigating to the warm side.

1335vch02.indd 136 12.02.2007 16:55:09

137

with:

a a1

a a n

1 exp( )( )

1 exp( )

−=

−c g R

F Rc g R

a a a a2

a a n

exp( )( )

1 exp( )=

−c g c g R

F Rc g R

(2.38)

In these equations, R is the thermal resistance between the wall surface at R = 0 and the chosen interface in the wall. Again, the functions F1 and F2 depend on the wall characteristics and the air � ow rate. For buildings with a leaky envelope, in� ltration occurs (i.e. air � ows from outside to inside) at the windward side while out� ow develops (i.e. air � ows from inside to outside) at the leeward side. Yet, comparison of the sum of the total heat � ow at the windward and leeward sides (= T1 + T2) with the total heat � ow one should have in the case conduction ( G = | (�R1 – �R2) (A1 / R1 + A2 / R2) |) and in� ltration ( V = | ca Ga (�R1 – �R2) | (with Ga the in� ltration � ow) occur separately shows that T1 + T2 is always smaller than G + V. In other words, the passage of air through the envelope results in energy conservation. Indeed, part of the higher conduction losses at the windward side are used to warm up the in-� owing air, while the warm, out-� owing air diminishes the conduction losses at the leeward side. However this advantage is hardly usable because air passage creates too many unwanted de� cits.

Composite Walls

The solution for a single-layered wall with boundary conditions of type 2 also applies for a composite wall.

Boundary conditions(outside wall, R = 0 is outside, R = Rn inside)

e0 :R = θ = θ

n i1 2

1 1:= = + + θ = θR R R

h h

Temperatures in the interfaces(surface 1 sees the outside environment, interface 1 is closest to surface 1, etc.)

Surface 1 s1 e i e 11

1( ) F

h

⎛ ⎞θ = θ + θ − θ ⎜ ⎟⎝ ⎠

Interface 1 1 e i e 1 11

1( ) F R

h

⎛ ⎞θ = θ + θ − θ +⎜ ⎟⎝ ⎠

Interface 2 2 e i e 1 1 21

1( ) F R R

h

⎛ ⎞θ = θ + θ − θ + +⎜ ⎟⎝ ⎠

etc.

Heat � ow rates(�e (R = 0) <��i (R = Rn))

2 i e( ) ( )q F R= θ − θ c a aq c g= θ T i 2 e 2 n(0) ( )= θ − θq F F R

with:

2.2 Air Transfer

1335vch02.indd 137 12.02.2007 16:55:10

138 2 Mass Transfer

a a1

a a n

1 exp( )( )

1 exp( )

−=

−c g R

F Rc g R

a a a a2

a a n

exp( )( )

1 exp( )=

−c g c g R

F Rc g R

In the [�, R]-plane, the temperature curve is similar to the one in a single-layered wall (Figure 2.11). In out� ow mode (air migrates from the inside, where it is warmer, to the outside, where it is colder), the curve turns convex, in the in� ow mode, concave. With inside insulation, out� ow increases the wall temperature at the cold side of the insulating layer signi� cantly.

Two- and Three Dimensions

In two and three dimensions analytical solutions for equation (2.31) hardly exist. When using CVM or FEM, � rst the air � ows are calculated and then the temperatures and heat � ow rates. If thermal stack intervenes, iteration is needed between the air and the heat balances. The heat � ows between adjacent points and a central point may be written in two ways:

By using an integral numerical technique. In such a case, the heat � ow between two points becomes:

Air � ow from 2 to 1 21 21 2 1 a a 2 2 21 a a 1 21( ) ( )P c G P c G PΦ = ′ θ − θ + θ = θ ′ + − θ ′

Air � ow from 1 to 2 21 21 2 1 a a 2 2 21 1 21 a a( ) ( )P c G P P c GΦ = θ − θ − θ = θ − θ +′ ′ ′ (2.39)

where 21P′ is the surface related thermal permeance.

By using the analytical solution for � at walls. For a series connection of surface related thermal resistances:

21 1 2 2 2 21(0) ( )F F RΦ = θ − θ ′

where 21R′ is the surface related thermal resistance between both points. In case of a parallel connection of series paths between the two points, the equation is used to calculate the � ow per series path. The total heat � ow then is:

2

21 211

i

i=Φ = Φ∑

�

�

Figure 2.11. Composite � at wall, temperature curve for warm air � owing to the cold side.

1335vch02.indd 138 12.02.2007 16:55:10

139

Transient Regime

Equation (2.30) is the starting point. For a constant air � ow rate and periodically varying temperatures in a � at wall, the complex calculation method, dealt with in heat conduction, is applicable. For each harmonic:

2( ) ( ) exp

π⎛ ⎞θ = α ⎜ ⎟⎝ ⎠i n t

R RT

where �(R) is the complex temperature. In the absence of a heat source or sink, the complex temperature ensues from:

2

a a2

d d 20

dd

i n cc g

R TR

α α π ρ λ− − α =

In this equation, i is the imaginary unit and T the period. The solution is:

1 1 2 2( ) exp ( ) exp ( )R C r R C r Rα = + (2.40)

where r1 and r2 are the roots of the quadratic equation 2a a

20

n i cr c g r

T

π ρ λ− − = :

21 a a a a

1 8( )

2

n i cr c g c g

T

⎡ ⎤π ρ λ= + − +⎢ ⎥⎣ ⎦

22 a a a a

1 8( )

2

n i cr c g c g

T

⎡ ⎤π ρ λ= − − +⎢ ⎥⎣ ⎦

Equation (2.40) can thus be written as:

a a 1 2 1 21 1 1

( ) exp ( ) sinh ( ) cosh2 2 2

R c g R C C a R C C a R⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞α = − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

with 2a a

8( )

n i ca c g

T

π ρ λ= − +

For an air � ow rate (ga) zero, the solution for conduction is obtained. The integration constants C1 and C2 follow from the boundary conditions (example: on surface R = 0 the complex temperature �s and the complex heat � ow rate sα′ are known). It is up to the reader the make these and other calculations. The most important result she or he will � nd is that in� ow decreases the damping and time shift of a wall, while out� ow increases both.For two and three dimensional details, only CVM or FEM can help in � nding a solution. For CVM, the energy balance per control volume becomes:

m c V Vt

ΔθΦ ≈ ρ Δ ± Φ Δ′Δ∑

where �V is the control volume, � the heat source or sink and m the heat � ow by conduction and enthalpy transfer, during time interval �t, from each of the adjacent points to the central point.

2.2 Air Transfer

1335vch02.indd 139 12.02.2007 16:55:10

140 2 Mass Transfer

2.2.6.2 Air Permeable Layers, Joints, Leaks and Cavities

In General

In the case of air permeable layers, leaks, joints and cavities, steady-state and transient air and heat transfer problems can only be solved by combining the equivalent hydraulic network with CVM for the heat balances, where the heat � ows are given by equation (2.39).

Ventilated Cavities

Consider the cavity of Figure 2.12, with a width of dcav m and a length or height of L m (ordinate z). The air velocity in the cavity is v m/s. As convective surface � lm coef� cient at both bounding surfaces, we have hc (W/(m2 · K)). The thermal resistance between the environment 1 and bounding surface 1 is R1 m

2 · K/W, between environment 2 and bounding surface 2 R2 m

2 · K/W. The heat balance for each of the surfaces is:

Cavity surface 1:

1 s1c cav s1 r s2 s1

1

( ) ( ) 0h hR

θ − θ+ θ − θ + θ − θ =

Cavity surface 2:

2 s2c cav s2 r s1 s2

2

( ) ( ) 0h hR

θ − θ+ θ − θ + θ − θ =

with hr the surface � lm coef� cient for radiation.

For the cavity, the following heat balance applies (the cavity stretches out along the z-axis):

c s1 cav c s2 cav a a cav cav[ ( ) ( )] d dθ − θ + θ − θ = ρ θh h z c d v

Figure 2.12. Ventilated cavities, a combination of heat conduction through both leafs with enthalpy � ow in the cavity.

1335vch02.indd 140 12.02.2007 16:55:11

141

Unknowns in the three equations are the temperatures �s1, �s2 and �cav, i.e, the temperature in the cavity and on both bounding surfaces. The solution is found by � rst calculating the surface temperatures �s1 and �s2 from the surface balances as a function of cavity temperature �cav, and then by introducing both in the cavity heat balance. To do so, we � rst de� ne the constants D, A1, A2, B1, B2, C1, C2:

2c r c r r

1 2

1 1D h h h h h

R R

⎛ ⎞ ⎛ ⎞= + + + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

r2

11

1ch h

RA

D R

+ +=

r2

1

hA

D R=

r1

2

hB

D R=

c r1

22

1h h

RB

D R

+ +=

c c r r c2

1

1h h h h h

RC

D

⎛ ⎞+ + +⎜ ⎟⎝ ⎠

=

c c r r c1

2

1h h h h h

RC

D

⎛ ⎞+ + +⎜ ⎟⎝ ⎠

=

The surface temperatures then become:

s1 1 1 1 2 1 cavA B Cθ = θ + θ + θ s2 2 1 2 2 2 cavA B Cθ = θ + θ + θ (2.41)

For the cavity balance, we get:

a a cav1 2 1 1 2 2cav cav

1 2 c 1 2

1 1

( ) ( )d d

2 (2 )

⎡ ⎤⎢ ⎥

ρ+ θ + + θ⎢ ⎥− θ = θ⎢ ⎥− − − −⎢ ⎥⎢ ⎥⎣ ⎦������������� �������

c d vA A B Bz

C C h C C

a b

with a solution of 1 cav

1

lnaz

b C

− θ⎛ ⎞= − ⎜ ⎟⎝ ⎠.

The integration constant C follows from the boundary conditions. For z in� nite, the ratio z / b1 also then becomes in� nite, turning the solution into:

1 cav,ln ∞− θ⎛ ⎞= −∞⎜ ⎟⎝ ⎠

a

C

which equals 1 cav, 0a ∞− θ = or 1 cav,a ∞= θ . Consequently, a1 represents the equilibrium temperature in an in� nitely long cavity. That value does not differ from the temperature one gets in a non-ventilated cavity. Indeed, z / b1 becomes also in� nite when the air velocity goes to zero (v = 0)! For z = 0, �cav equals �cav,0, giving:

cav,0cav, cav,0

cav,ln 0 or CC ∞

∞θ − θ⎛ ⎞= = θ − θ⎜ ⎟⎝ ⎠

2.2 Air Transfer

1335vch02.indd 141 12.02.2007 16:55:11

142 2 Mass Transfer

The cavity temperature thus equals:

cav cav,01

cav, cav,( ) expz

b∞ ∞⎛ ⎞

θ = θ − θ − θ −⎜ ⎟⎝ ⎠

Or, the temperature in a ventilated cavity changes exponentially, from the in� ow value to a value which approaches the one without ventilation. The term b1 has the dimensions of a length. The number is named the ‘ventilation length’. The higher the air velocity in the cavity, the more this length increases. It decreases for a larger heat transfer to the cavity. The temperatures on both cavity surfaces (�s1, �s2) are found by introducing �cav into the equations (2.41):

s1 1 1 1 2 1 cav,01

cav, cav,( ) expz

A B Cb∞ ∞

⎡ ⎤⎛ ⎞θ = θ + θ + θ − θ − θ −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

s2 2 1 2 2 2 cav,01

cav, cav,( ) expz

A B Cb∞ ∞

⎡ ⎤⎛ ⎞θ = θ + θ + θ − θ − θ −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

What is the in� uence of cavity ventilation on the heat loss through an outside wall? Let ‘2’ be the inside leaf. For the heat � ow through that leaf, the following applies:

2 s22 0

2 1 2 2 2 cav,02 10

2 s2 2 1 sp sp 0

2 2 1

cav, cav,

1( ) d

1(1 ) ( ) exp d

( ) ( )exp 1

L

L

zR

zA B C z

R b

L C b L

R R b

∞ ∞

∞ ∞

Φ = θ − θ

⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪= − θ + − θ − θ − θ − θ −⎨ ⎬⎢ ⎥⎜ ⎟⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭θ − θ θ − θ ⎡ ⎤⎛ ⎞

= − − −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

∫

∫

The average U-value thus is:

2 1 sp sp 0o

1 2 2 1 2 1

( )1 exp

( ) ( )

C b LU U

L L R b∞θ − θ ⎡ ⎤⎛ ⎞Φ= = + − −⎢ ⎥⎜ ⎟θ − θ θ − θ ⎝ ⎠⎣ ⎦

Since the temperature ratio sp sp 0 1 2( ) /( )∞θ − θ θ − θ can be approximated by: 1 a 1 o/R R R U= , this formula simpli� es to:

2 1 1o

2 1

1 1 expC b R L

U UL R b

⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪= + − −⎨ ⎬⎢ ⎥⎜ ⎟⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

Hence, the effect of ventilation depends on the thermal resistances R2 and R1 of the inside leaf and outside veneer. If R2 is large and R1 small, i.e., if a partial insulation � ll is pressed against the inside leaf, then ventilation hardly has an impact. If in turn R2 is small and R1 large, i.e. if the insulation touches the veneer, then the effect becomes signi� cant. Improper workmanship when installing the insulation may be responsible for that.At the air intake, the heat � ow rate is 2 s 2,0 2( ) / Rθ − θ (air intake at z = 0).

1335vch02.indd 142 12.02.2007 16:55:11

143

2.3 Vapour Transfer

2.3.1 Water Vapour in the Air

2.3.1.1 Overview

Air contains water vapour. Proof of that are damp windows during winter. Those who wear glasses also experience it when they enter a heated, crowded room coming from the cold outdoors. For convenience, moist air is considered as a mixture of two ideal gases: dry air and water vapour, even though the term ‘ideal’ is less accurate for water vapour than for dry air. The gas equation of dry air has been presented in the previous paragraph. For water vapour we have: p V = mv Rv T, where p is the partial water vapour pressure (Pa), T the ‘water vapour’ temperature (K), mv the water vapour mass in the volume V (kg) and Rv the speci� c gas constant for water vapour, equal to 461.52 J/(kg · K). The suf� x v in Rv is typically omitted. More accurate is the equation:

vv vvv2

v v v v

1( / ) ( / )

= + +B Cp V

n R T V n V n

where nv are the moles of water vapour in volume V (m3), Rv is the general gas constant (8314.41 J/(mol · K)) and Bvv and Cvvv are the viral coef� cients for water vapour.As vapour concentration, we have:

vv

v

ρ = =m p

V R T (2.42)

Normally the gases in dry air do not react with water vapour. As a result, the Dalton law applies for moist air: total air pressure in a volume V equal to the sum of the partial pressures:

a lP p p= + (2.43)

For building constructions, Pa is the atmospheric pressure, at sea level � 101.3 kPa (1 Atm). Equation (2.13) linked air pressure to the altitude. Thanks to its limited solubility in water, the presence of dry air does not in� uence the balance between water, water vapour and ice (Raoult’s and Henry’s law). Hence, in air, the diagram of state for water is still applicable: with the triple point at 0 °C, etc. The kinetics at any rate change. In air, evaporation and condensation evolve slower than in vacuum.

2.3.1.2 Quantities

The presence of water vapour in the air is described by the following quantities:

Partial water vapour pressure p [Pa]Together with temperature and total air pressure a fundamental variable of state

Water vapour concentration �v [kg/m3]Mass of water vapour per volume-unit of air or any other gas. As a derived state variable, the vapour concentration is a function of partial water vapour pressure and temperature (see equation (2.42))

Water vapour ratio x [kg/kg]Mass of water vapour per mass-unit of dry air. Also the vapour ratio is a derived variable of state.

�

�

�

2.3 Vapour Transfer

1335vch02.indd 143 12.02.2007 16:55:11

144 2 Mass Transfer

Following relations exist between water vapour ratio x and partial water vapour pressure p:

v a

l v a a

0.62

( )

R p px

R P p P p

ρ= = =

ρ − − (2.44)

Further on, the indications ‘partial’ and ‘water’ are omitted

2.3.1.3 Maximum Vapour Pressure and Relative Humidity

As stated, the presence of air hardly changes the equilibrium between water, water vapour and ice. Therefore, for each air temperature, the equilibrium vapour pressure between liquid, solid and vapour � gures as the maximum attainable vapour pressure, also called the vapour saturation pressure. Related vapour concentration and vapour ratio are also saturation values. All increase with temperature. At 100 °C, the saturation pressure equals the atmospheric pressure. Saturation values have the suf� x ‘sat’: psat, �v,sat, xsat. Figure 2.13 and Table 2.5 to 2.7 give the vapour saturation pressure as a function of temperature. For temperatures between –30 °C and 41 °C, the table lists the vapour saturation pressure in steps of 0.1 °C. For temperatures between 45 and 95 °C, 5 °C is used as a step. Vapour saturation pressure follows a more or less exponential curve. The most accurate approximation for temperatures between 0 and 50 °C is:

csat c,sat exp 2.3026 1

Tp p

T

⎡ ⎤⎛ ⎞= κ −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦ (2.45)

with Tc the critical temperature of water (i.e. the temperature above which water only exists as vapour) and pc,sat related saturation pressure. (Tc = 647.4 K, pc,sat = 217.5 � 105 Pa). The parameter � depends on temperature in K:

2 3

4.39553 6.2442 9.953 5.1511000 1000 1000

T T T⎛ ⎞ ⎛ ⎞ ⎛ ⎞κ = − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Figure 2.13. Water vapour saturation pressure psat(�) as function of temperature, line of 50% relative humidity.

1335vch02.indd 144 12.02.2007 16:55:12

145

Table 2.5. The water vapour saturation pressure in Pa, temperatures between 0 and –30 °C.

� (°C) –0.0 –0.1 –0.2 –0.3 –0.4 –05 –0.6 –0.7 –0.8 –0.9

–0 611 606 601 596 591 586 581 576 572 567

–1 562 558 553 548 544 539 535 530 526 522

–2 517 513 509 504 500 496 492 488 484 479

–3 475 471 467 464 460 456 452 448 444 441

–4 437 433 430 426 422 419 415 412 408 405

–5 401 398 394 391 388 384 381 378 375 371

–6 368 365 362 359 356 353 350 347 344 341

–7 338 335 332 329 326 323 321 318 315 312

–8 310 307 304 302 299 296 294 291 289 286

–9 284 281 279 276 274 271 269 267 264 262

–10 260 257 255 253 251 248 246 244 242 240

–11 237 235 233 231 229 227 225 223 221 219

–12 217 215 213 211 209 207 206 204 202 200

–13 198 196 195 193 191 189 188 186 184 183

–14 181 179 178 176 174 173 171 170 168 167

–15 165 164 162 160 159 158 156 155 153 152

–16 150 149 148 146 145 143 142 141 139 138

–17 137 136 134 133 132 131 129 128 127 126

–18 125 123 122 121 120 119 118 116 115 114

–19 113 112 111 110 109 108 107 106 105 104

–20 103 102 101 100 99 98 97 96 95 94

–21 96 92 91 90 90 89 88 87 86 85

–22 84 84 83 82 81 80 80 79 78 77

–23 76 76 75 74 73 73 72 71 70 70

–24 69 68 68 67 66 66 65 64 64 63

–25 62 62 61 60 60 59 59 58 57 57

–26 56 56 55 55 54 53 53 52 52 51

–27 51 50 50 49 49 48 48 47 47 46

–28 46 45 45 44 44 43 43 42 42 41

–29 41 41 40 40 39 39 38 38 38 37

–30 37 36 36 36 35 35 35 34 34 33

Table 2.6. The saturation pressure in water vapour in Pa, temperatures between 45 and 95 °C in steps of 5 °C.

� (°C) p� (Pa) � (°C) p� (Pa) � (°C) p� (Pa) � (°C) p� (Pa)

45 9,582 60 19,917 75 38,550 90 70,108

50 12,335 65 25,007 80 47,356 95 84,524

55 15,741 70 31,156 85 57,800

2.3 Vapour Transfer

1335vch02.indd 145 12.02.2007 16:55:12

146 2 Mass Transfer

Table 2.7. The water vapour saturation pressure in Pa, temperatures between 0 and 4 °C.

� (°C) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0 611 615 620 624 629 634 638 643 647 652 1 657 662 666 671 676 681 686 691 696 701 2 706 711 716 721 726 731 736 742 747 752 3 758 763 768 774 779 785 790 796 802 807 4 813 819 824 830 836 842 848 854 860 866 5 872 878 884 890 896 903 909 915 922 928 6 935 941 948 954 961 967 974 981 987 994 7 1001 1008 1015 1022 1029 1036 1043 1050 1057 1065 8 1072 1079 1087 1094 1101 1109 1117 1124 1132 1139 9 1147 1155 1163 1171 1178 1186 1194 1203 1211 121910 1227 1235 1243 1252 1260 1269 1277 1286 1294 130311 1312 1320 1329 1338 1347 1356 1365 1374 1383 139212 1401 1411 1420 1429 1439 1448 1458 1467 1477 148713 1497 1506 1516 1526 1536 1546 1556 1566 1577 158714 1597 1608 1618 1629 1639 1650 1661 1671 1682 169315 1704 1715 1726 1737 1748 1760 1771 1782 1794 180516 1817 1829 1840 1852 1864 1876 1888 1900 1912 192417 1936 1949 1961 1973 1986 1999 2011 2024 2037 205018 2063 2076 2089 2102 2115 2128 2142 2155 2169 218219 2196 2210 2224 2237 2251 2265 2280 2294 2308 232220 2337 2351 2366 2381 2395 2410 2425 2440 2455 247021 2486 2501 2516 2532 2547 2563 2579 2595 2611 262722 2643 2659 2675 2691 2708 2724 2741 2758 2774 279123 2808 2825 2842 2859 2877 2894 2912 2929 2947 296524 2983 3001 3019 3037 3055 3073 3092 3110 3129 314825 3166 3185 3204 3224 3243 3262 3281 3301 3321 334026 3360 3380 3400 3420 3440 3461 3481 3502 3522 354327 3564 3585 3606 3627 3649 3670 3692 3713 3735 375728 3779 3801 3823 3845 3868 3890 3913 3935 3958 398129 4004 4028 4051 4074 4098 4122 4145 4169 4193 421830 4242 4266 4291 4315 4340 4365 4390 4415 4440 446631 4491 4517 4543 4569 4595 4621 4647 4673 4700 472732 4753 4780 4807 4835 4862 4889 4917 4945 4973 500133 5029 5057 5085 5114 5143 5171 5200 5229 5259 528834 5318 5347 5377 5407 5437 5467 5498 5528 5559 559035 5621 5652 5683 5715 5746 5778 5810 5842 5874 590736 5939 5972 6004 6037 6071 6104 6137 6171 6205 623937 6273 6307 6341 6376 6410 6445 6480 6516 6551 658738 6622 6658 6694 6730 6767 6803 6840 6877 6914 695139 6989 7026 7064 7102 7140 7178 7217 7255 7294 733340 7372 7412 7451 7491 7531 7571 7611 7652 7692 7733

1335vch02.indd 146 12.02.2007 16:55:12

147

Less accurate approximations are:

–10 � � � 50 °C sat exp (65.8094 7066.27 / 5.976 ln( ))p T T= − −

–30 � � � 0 °C 3 6 2 6 3sat 611 exp(82.9 10 288.1 10 4.403 10 )p − − −= ⋅ θ − ⋅ θ + ⋅ θ

0 � ��� 40 °C 3 6 2 6 3sat 611 exp(72.5 10 288.1 10 0.79 10 )p − − −= ⋅ θ − ⋅ θ + ⋅ θ

� � 0 °C sat22.44

611 exp272.44

pθ⎛ ⎞= ⎜ ⎟⎝ ⎠+ θ

� > 0 °C sat17.08

611 exp234.18

pθ⎛ ⎞= ⎜ ⎟⎝ ⎠+ θ

0 � � � 80 °C sat exp (23.5771 4042.9 ( 37.58))p T= − −

The approximations 4 and 5 are of interest because they allow, with the vapour saturation pressure known, for calculation of the related temperature.Normally, moist air contains less water vapour then the saturation value. The relationship between the actual vapour concentration �v and the saturation concentration �v,sat at the same temperature, is called relative humidity (RV, ):

v

v,sat

100ρ

φ =ρ

(2.46)

Units: – or %. Implementing (2.42) gives:

v

v,sat sat

100 100p

p

ρφ = =

ρ (2.47)

or, relative humidity also equals the ratio between vapour pressure and vapour saturation pressure at the same temperature. For example, in Figure 2.13 relative humidity in Q [� = 20 °C, p = 1168.5 Pa] is 50%. All points with relative humidity 50% lie on a curve through Q, similar in shape as the saturation curve, at a 0.5 ratio. Equation (2.46) and (2.47) show that relative humidity cannot exceed 100%. Indeed, for each temperature, the related vapour saturation pressure psat is the highest attainable vapour pressure and vapour saturation concentration �v,sat, the highest attainable vapour concentration. If an attempt is made to go beyond that, the pixel representing the moist air will stop at the saturation curve. At the same time condensation starts. This means that water vapour precipitates as water while releasing the heat of evaporation (lb). Evaporation of water, in turn, requires supply of that heat. For water at 0 °C, the heat of evaporation (lbo) equals 2.5 � 106 J/kg.Among relative moisture, air pressure and air temperature, a simple relationship exists. Suppose an air mass at a relative humidity 1, an air pressure Pa1 and a temperature �1. Assume the air pressure changes from Pa1 to Pa2 and temperature from �1 to �2, while the quantity of water vapour in the air volume remains the same. The relative humidity 2 in the new situation then becomes:

sat1 a22 1

sat2 a1

p P

p Pφ = φ (2.48)

2.3 Vapour Transfer

1335vch02.indd 147 12.02.2007 16:55:12

148 2 Mass Transfer

The proof is as follows. Because the quantity of dry air and water vapour in the air volume does not change, the ratio v l v/ ( )m m m+ stays constant.

In the situation [Pa1, �1] we have:

1 1

v 1 a 1 sat1 a1

a1 1l v a1 a1

(4)a 1

p V

m p R p RR TP Vm m P R P RR T

φ= = =

+ �����

In the situation [Pa2, �2] that ratio becomes:

2 2

v 2 a 2 sat2 a2

a2 2l v a2 a2

(4)a 2

p V

m p R p RR TP Vm m P R P RR T

φ= = =

+ �����

In both formulae, the moist air gas constant Ra is v l l a( ) /m R m R m+ . Equating step (4) in both proves (2.48). Under atmospheric conditions, all changes of state are isobaric (Pa1 = Pa2), converting equation (2.48) into:

sat12 1

sat2

p

pφ = φ (2.49)

That equation shows what happens with relative humidity when temperature changes. Because the vapour saturation pressure psat increases with temperature, cooling will achieve an increase, and warming-up a decrease in relative humidity. For example, raising the temperature of the cold ventilation air in winter to � 20 °C, means a lower relative humidity inside than outside. Cooling down the ventilation air during the summer will give a higher relative humidity inside than outside.For a correct de� nition of any mixture of two gases, three variables of state have to be known. For moist air, the obvious choices are the total air pressure, the vapour ratio and the temperature. In most building physics applications, the total air pressure deviates very little from the atmospheric. Thus, two variables remain. Put another way, the relative humidity does not suf� ce to describe the state of moist air. The only thing known at that moment is that the point representing the air in a diagram of state lies on a curve of equal relative humidity. In order to locate the point and to describe the dry air/vapour mixture, we also have to know the temperature, or the vapour pressure, or the saturation pressure or other factors.

2.3.1.4 Changes of State in Humid Air

From point Q in Figure 2.13, there are in� nite possibilities to reach saturation. Two are quite speci� c:

If, without changing air temperature, vapour is added, then Q moves parallel to the vapour pressure axis to Q � on the saturation curve. Continuing to inject water vapour will block Q in Q �, meaning that the vapour pressure remains equal to the vapour saturation pressure psat,Q � and that all extra water vapour condenses. This change of state is called isothermal.

�

1335vch02.indd 148 12.02.2007 16:55:12

149

Instead of adding water vapour, the air temperature may be lowered. Now, Q moves parallel to the temperature axis to reach Q on the saturation curve. At this point, the temperature touches a value �d. Any further decrease in temperature will give condensation with Q descending along the saturation curve. The related difference in saturation pressure (psat,Q – psat,Q�) drives that change of state from vapour to liquid, without telling anything about the kinetics. This change of state is called isobaric with the temperature �d representing the dew point of the air. Figure 2.13 shows that all points on the same parallel as Q have an identical dew point. Thus, just as for relative humidity, the dew point alone does not suf� ce to describe moist air. A second parameter is needed. Mathematically, the dew point is found as follows: the relative humidity for each point on Q Q is Q psat,Q / 100. When the temperature equals the dew point, relative humidity reaches 100%, meaning that psat,Q = Q psat,Q / 100 = psat,Q. In other words, the dew point is the temperature for which the actual vapour pressure becomes the saturation pressure (psat(�d) = p). Once the air temperature falls below the dew point, condensation occurs.

Each real change of state consists of a combination of isothermal and isobaric, i.e., both humidi� cation or drying and temperature change. The isobaric change of state explains in part phenomena such as soiling, mold growth and surface condensation (see hereafter).

2.3.1.5 Enthalpy of Moist Air

The enthalpy of moist air, containing x kg of water vapour per kg of (dry) air, is given by (reference temperature 0 °C):

a v bo( )h c x c l= θ + θ + (2.50)

where ca and cv are the isobaric speci� c heat of dry air (� 1008 J/(kg · K)) and water vapour (� 1860 J/(kg · K)), and lbo is the heat of evaporation at 0 °C.

2.3.1.6 Characterizing Moist Air

Measuring the presence of water vapour in the air can be done directly or indirectly.

Directly

Mirror dew point meter: measures the temperature and the dew point.Psychrometer: measures the dry (�) and the wet air temperature (�w). The wet temperature is the temperature reached by air at dry temperature � and relative humidity when cooled by adiabatic saturation with water vapour. All other variables of state can be deduced once the dry and the wet temperature are known. For example vapour pressure:

sat ( ) 66.71 ( )n np p= θ − θ − θ (�2.51)

Indirectly

Hair hygrometer: determines the relative humidity from the hygric expansion or contraction of a hair bundle. Capacitive meter: determines the relative humidity from the change in dielectric constant of a hygroscopic polymer.

An instrument widely used for the analysis of the changes of state of moist air is Mollier’s diagram, of which different versions exist (ASHRAE, VDI).

�

�

�

�

�

2.3 Vapour Transfer

1335vch02.indd 149 12.02.2007 16:55:13

150 2 Mass Transfer

2.3.1.7 Applications

Typical problems taken care of by building physics are the water vapour balance in a building, mold growth on inside envelope surfaces, dust mite infection and surface condensation.

Water Vapour Balance in a Building

In rooms without full air conditioning, vapour pressure and relative humidity are determined by the heat, air and vapour balance. In the framework of this book, we limit the analysis to the simple case of a room at constant temperature, without sorption-active surfaces, where the ventilation air comes from outside or from adjacent rooms. In case of ideal mixing of room and ventilation air, then air temperature �i, vapour pressure pi and relative humidity í in the middle of the room, ±1.8 m above the � oor, suf� ce to describe the overall air condition.

Balances

First, the air � ows are calculated. For each room the sum of all air exchanged between the room, the outside and adjacent rooms must be zero. This gives an equation with the air pressures in the room and adjacent rooms (de� ned by wind, stack and fans in service) as unknowns. For n rooms the result is a system with at least n equations with n unknown air pressures. In case the room has leaks at different heights or large openings, more equations than the number of rooms are needed. Anyhow, once the air pressures Pa,i are known, calculation of all air � ows is straight forward, using as a formula a ai T,i aj T,j( )bG a P p P p= − − + . Of course, the boundary conditions have to be known.The air � ows play an important role in the vapour equilibrium per room. According to conservation of mass, water vapour which enters a room with the supply air (for outside air, the vapour � ow is xve Ga with xve the vapour ratio outside in kg/kg, and Ga the air � ow from outside in kg/s) plus the water vapour produced or removed by air drying in that room (GvP) plus the water vapour added to the air by surface drying or withdrawn from it by surface condensation (Gvc/d), should equal the water vapour removed with the exhaust air (xvi �Gai,j) or withdrawn by diffusion plus the water vapour stored in the room air [d(xvi Ml) / dt] (Figure 2.14).In a formula (one room, supply air comes from outside, incoming and outgoing air � ow equal; diffusion negligible):

vi lve a vP vc/d vi a

d ( )

d

x Mx G G G x G

t+ + = + (2.52)

The air mass Ml in the room numbers �a V with V the room volume in m3. The production term GvP includes vapour from people, animals and plants, release linked to activities (cooking,

Figure 2.14. Vapour balance in a one-zone building. Storage in the air is not shown in the � gure.

1335vch02.indd 150 12.02.2007 16:55:13

151

washing, cleaning, etc.), evaporation from exposed water surfaces (an indoor swimming pool) etc. Sometimes it also contains the water vapour released by wet construction parts when drying (rising damp, wetting by rain, built in moisture). If neither surface condensation nor surface drying occurs (that case will be discussed in the section on water vapour surface � lm coef� cients), then from the ideal gas law follows:

ve eve a ae

ae e

( ) ( )p

x G n V n VR T

ρ= ρ =

ρ

vi ivi a ai

ai i

( ) ( )p

x G n V n VR T

ρ= ρ =

ρ

viai

aivi ai i

i

dd( ) d

d d d

Vx V pV

t t R T t

⎛ ⎞ρρ⎜ ⎟ρ⎝ ⎠ρ

= =

In these equations, n is the ventilation rate, i.e. the number of room volume exchanges per time unit, pe vapour pressure outside, pi vapour pressure inside, Ti inside temperature in K and R gas constant of water vapour. The supply air undergoes two changes of state: a warm up from Te to Ti and humidi� cation or dehumidi� cation from xve to xvi. The � rst change allows to write:

eve a

i

( )p

x G n VR T

=

All these transformations convert the equilibrium equation (2.52) in a differential equation of � rst order with the inside vapour pressure as dependent variable:

i ie vP i

d1

d

R T pp G p

n V n t+ = + (2.53)

Solutions

In steady-state (GvP = Cte, n = Cte, pe = Cte, dpi / dt = 0) – also called average regime – the inside vapour pressure becomes:

ii e vP

R Tp p G

n V= + (2.54)

For relative humidity, the result is:

ei ii vP

sat,i sat,i sat,i

100pp R T

Gp p p n V

φ = = + (2.55)

When water vapour is produced inside, the average inside vapour pressure can never drop below the value outside. At the same time, the difference between both increases proportionally to vapour production, and inversely proportional to ventilation rate. This means that the additional bene� t of ventilation slows down quickly when the rate continues to increase (Figure 2.15).

2.3 Vapour Transfer

1335vch02.indd 151 12.02.2007 16:55:13

152 2 Mass Transfer

In a transient regime, various cases demands consideration:

1. Suppose vapour release inside increases suddenly, while vapour pressure outside and ventilation rate remain constant (pe = C te, n = C te). Initial conditions then are: t < 0: GvP = GvP1, t 0: GvP = GvP2. Solution:

i i i o i( ) exp( )p p p p n t∞ ∞= + − − (2.56)

where pio is the vapour pressure inside for t = 0 and pi� the steady state vapour pressure for a vapour release GvP2. The solution shows that after a sudden increase or decrease in vapour release, vapour pressure inside lags behind, especially when the ventilation is low (the time constant of the exponential is 1/n). Or, less ventilation seems all of a sudden positive. This advantage of course is completely annulled because of the higher end value reached (pi�) (Figure 2.16).

Figure 2.15. Effect of better ventilation on vapour pressure excess inside (building with a volume of 400 m3, daily vapour release (GvP) 14 kg, 0 °C outside, p = 580 Pa outside, 20 °C inside).

Figure 2.16. Inside vapour pressure, sudden vapour release Inertia effects as a consequence of air buffering (building with a volume of 400 m3, daily vapour release (GvP) 14 kg, 0 °C outside, p = 580 Pa outside, 20 °C inside).

1335vch02.indd 152 12.02.2007 16:55:13

153

2. Suppose the vapour pressure outside changes suddenly without any variation in ventilation rate and vapour release inside (n = Cte, GvP = Cte). Initial conditions then are: t < 0 pe = pe1, t 0: pe = pe2. The solution does not differ from (2.56). Now, pio is the vapour pressure inside at the moment t = 0, while pi� is the steady state vapour pressure inside which will occur for a value pe2 outside. Lagging behind remains.

3. Suppose the ventilation rate remains constant (n = Cte) but vapour release inside and vapour pressure outside change periodically. Solution:

ii i

1 2Re exp

⎛ ⎞ ⎡ ⎤π⎛ ⎞= + + ⎜ ⎟⎢ ⎥⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎣ ⎦vP

eG

pZ

R T i tp p

n V T (2.57)

with ip the vapour pressure inside for the average vapour release inside and average vapour pressure outside, pe the complex vapour pressure outside, GvP the complex vapour release inside and Z the complex damping (the term complex is used here to pinpoint the analogy with complex temperatures). These are given by:

12 22

1Zn T

⎡ ⎤⎛ ⎞π⎢ ⎥= + ⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

2arg ( )

⎛ ⎞π= ⎜ ⎟⎝ ⎠Z b g t g

n T

eeˆ exp( )pp i= ϕep vPevP

ˆ exp( )GG i= ϕvPG

where Z, ep and vPG are the amplitudes of the damping factor, the vapour pressure outside and the vapour release inside.

epϕ is the time lag vapour pressure outside experiences,

vPeGϕ the time lag vapour release inside experiences and arg(Z) the time shift the complex damping factor imposes. Compared to the average response (Z = 1, arg(Z) = 0), periodical changes in vapour release inside and vapour pressure outside give a dampened and time shifted vapour pressure inside. Both damping and time shift increases when ventilation rate decreases.

4. If all parameters are variable, than a � nite difference calculus gives the solution.

Dust Mites, Mold and Surface Condensation

All three are connected to relative humidity on surfaces.The number of dust mites on a surface may increase explosively when the monthly average relative humidity in the direct environment passes 52 + 1.2 (�s – 15) (for Dermatophagoides farinae). Mold growth becomes unavoidable if the monthly average relative humidity on a surface passes 80%. In such case, the monthly mean vapour pressure in the direct environment of the surface (ps) equals 0.8 times the monthly mean saturation pressure at that surface (psat,s). Surface condensation occurs when the relative humidity on a surface touches 100%, i.e. every time the temperature of the surface drops below the dew point of the ambient air.Things get nasty when all this happens on inside surfaces. In case of ideal air mixing, mold will develop when the monthly mean inside vapour pressure (pi) passes 0.8 times the monthly mean vapour saturation pressure (psat,s) on a surface (Figure 2.17). For surface condensation to happen, vapour pressure inside must equal the vapour saturation pressure on that surface (Figure 2.18). As explained, vapour saturation pressure on a surface depends on surface temperature. On any envelope part, surface temperatures are given by:

2.3 Vapour Transfer

1335vch02.indd 153 12.02.2007 16:55:14

154 2 Mass Transfer

Figure 2.17. Concrete roof edge forms a thermal bridge. The consequence inside is mold in the corner.

Figure 2.18. Surface condensation on the ceiling of a water closet. Main reasons: � at roof not insulated, insuf� cient ventilation (only air inlet, no air outlet).

1335vch02.indd 154 12.02.2007 16:55:14

155

is e i e( )hfθ = θ + θ − θ (2.58)

with fhi the temperature factor on the spot considered. Clearly, surface temperature and vapour saturation pressure decrease and mite, mold and surface condensation risk increases, when:

1. It is cooler outside (�e lower).2. One heats less (�i lower).3. Temperature factor of an envelope part is lower. That occurs when a part is hardly insulated,

when notable thermal bridges exist and when the surface � lm coef� cient for heat transfer hi drops.

The outside temperature is a fact that a single human cannot change. In moderate and cold climates instead, humans intervene with the inside temperature through their heating habits. These depend among others on the extent of the heating bill. Temperature factors in turn are de� ned by the thermal quality of the building fabric (Good thermal insulation? Thermal bridges solved?), the ratio between the envelope surface and the total wall surface in a room (de� nes the radiant part in the surface � lm coef� cient hi), and cupboards or other furniture standing against outside walls (lowers the surface � lm coef� cient hi substantially).At the same time, equation (2.54) shows that vapour pressure inside (pi) changes with vapour pressure outside, vapour release inside and ventilation � ow. The higher the vapour pressure inside, the greater the likelihood is to get mite overpopulation, mold problems and long lasting surface condensation. Likely boundary conditions are:

4. A higher vapour pressure outside. That is the case in moderate and cold climates during summer. Of course, the outside temperature is also higher in summer.

5. More water vapour released inside (GvP / V higher).6. On average, less ventilation (n lower).

Again, a single human has no impact on the outside vapour pressure. Water vapour release, instead, is mainly determined by the amount of building users, their usage habits and the function the building has (dwellings, of� ces, swimming pool, etc.). Also the average ventilation rate depends to some extend on the way the users air, but mainly on the means and devices foreseen in the building fabric and building services to ventilate correctly. Classifying the six factors according to their links with building design, gives:

Factors 1 and 4: outside climate. Much cannot be done about this.Factor 5: Mainly de� ned by living habits and function. Much cannot be done about this. Factors 2 and 6: Partly de� ned by habitation habits, partly design and construction de� ned. An energy wasting design almost automatically leads to lower inside temperatures, simply because comfortable heating may be too expensive. If, analogously, no attention is paid to a correct ventilation design, then ventilating suf� ciently becomes a stochastic reality. Factor 3: Mainly design de� ned. Insuf� cient, careless thermal insulation without attention for thermal bridge avoidance and air tightness.

�

�

�

�

2.3 Vapour Transfer

1335vch02.indd 155 12.02.2007 16:55:14

156 2 Mass Transfer

2.3.2 Water Vapour in Open-porous Materials

2.3.2.1 Overview

Under atmospheric circumstances all open pores in a dry open-porous material are � lled with air. Thus, it seems logical to apply the theory for humid air on the air volume �0 V, where �0 is the open porosity and V the volume taken by the material. The amount of water vapour in V m3 of material and the moisture content in equilibrium with the relative humidity in the air would then be:

0m v 0 462

p VG V

T

Ψ= ρ Ψ =

0mH 462

pGw

V T

Ψ= =

Both relationships underline that the equilibrium moisture content (wH) in a porous material should vary linearly with relative humidity, while its temperature-dependence should re� ect the vapour saturation curve and become steeper with increasing temperature.

Example

Concrete has an open porosity of approximately 15%. At 20 °C and 65% relative humidity, 1 m3 of concrete should show a moisture content equal to:

0mH 3

0.15 2340 kg0.0017

462 462 293.16 m

pGw

V T

Ψ ⋅= = = =⋅

At 50 °C and 65% of relative humidity, the moisture content should be:

0mH 3

0,15 12 335 kg0.0081

462 462 293.16 m

pGw

V T

Ψ ⋅= = = =⋅

The reality, however, is totally different. At 20 °C and 65% relative humidity, the equilibrium moisture content in 1 m3 of concrete reaches 40 to 50 kg/m3! Furthermore, moisture content does not vary linearly but S-shaped with relative humidity, while a higher temperature results in a very limited drop in moisture content instead of an increase. It seems as if the behaviour of the air and water vapour mixture changes drastically in the pores of a porous material, in the case being concrete.

2.3.2.2 Sorption Isotherm and Speci� c Moisture Ratio

As a function of relative humidity, each material is characterized by a series of equilibrium moisture contents, which lie on a curve of the form shown in Figure 2.19. This S-curve, with a hysteresis between humidi� cation from 0 to 100% and drying from 100 to 0% is called the sorption isotherm of the material. Each point on and in-between the two lines represents a possible equilibrium for the relative humidity abscissa value. What the equilibrium moisture content will precisely be depends on the moisture history of the material. At any rate, the equilibrium at 98% is called the maximum hygroscopic moisture content.Each material has its own sorption isotherm. Nevertheless, all materials showing hardly any sorption, except at a very high relative humidity, are called non hygroscopic. Examples are bricks, synthetics, most insulation materials, etc. On the contrary, materials showing already considerable sorption at low relative humidity are called hygroscopic. Examples are all cement

1335vch02.indd 156 12.02.2007 16:55:15

157

bonded and timber-based materials. In literature, mathematical expressions are found for the sorption isotherm. Although no one is universal, it seems that, provided the roots of the denominator lie outside the interval [0, 1], the following relationship can be used for a relative humidity lower than 95% (0.95):

H 2H H H

wa b c

φ=φ + φ +

(0 � � 0.95) (2.58)

with aH, bH and cH material-speci� c constants, which differ between sorption and desorption.Another expression, applicable above 20% relative humidity, is:

1

H cln

1c

w wb

−φ⎡ ⎤= −⎢ ⎥⎣ ⎦

(0.2 � � 1) (2.59)

where wc is the capillary moisture content and b and c are material-speci� c constants.When the sorption isotherm is known, then the speci� c moisture ratio can be calculated. That property represents the quantity of moisture a material absorbs or emits for a unit change in relative humidity on a scale from 0 to 1. As a formula it is:

Hd

d

Xφξ =

φ (2.60)

with XH moisture ratio. Units: kg/(kg � RV).

Figure 2.19. Sorption isotherm. The two mechanisms that bind water molecules are:(1) adsorption of water molecules at the pore walls at low relative humidity; (2) capillary condensation on the water menisci formed by the adsorbed layers at higher relative humidity.

2.3 Vapour Transfer

1335vch02.indd 157 12.02.2007 16:55:15

158 2 Mass Transfer

The speci� c moisture ratio is comparable with the speci� c heat. It gives the moisture storage per kg of dry material for a unit change in driving force, in the case being the relative humidity (hence the suf� x ). The equivalent of the volumetric speci� c heat capacity is found by multiplying � by density:

Hd

d

wφρ ξ =

φ (2.61)

where wH is the hygroscopic moisture content.� � is called the speci� c moisture content; units kg/(m3 � RV). Both the speci� c moisture ratio and the speci� c moisture content depend on relative humidity. Using the equations (2.58) and (2.59):

2H HH H2

d

d

w cw aφ

⎛ ⎞ρ ξ = = −⎜ ⎟φ ⎝ ⎠φ

1H Hd ln

1d

w w

c b b

−

φφ⎛ ⎞ρ ξ = = −⎜ ⎟⎝ ⎠φ φ

2.3.2.3 The Physics Behind

Hygroscopic moisture is built up by water molecules, which diffuse into the pores and stick to the pore walls. Two mechanisms are responsible for this:

At low relative humidity (0 to 40%): Molecular adsorption against the pore-walls. Explains why the hygroscopic curve is convex there (see Figure 2.19).At high relative humidity (40 to 100%): Capillary condensation on the water menisci formed in the pores by the adsorbed water molecule layers. Makes the curve concave there (see Figure 2.19).

Molecular adsorption occurs in two steps. First we have monolayer adsorption of water molecules against the pore walls. Once that � rst layer is formed at a relative humidity of � 20%, multi-layer adsorption starts. Monolayer adsorption is described by the Langmuir equation:

7w PH P

w

2.62 101 1

M A C Cw A

A N C C−φ φ= = ⋅

+ φ + φ (2.62)

where Mw is the mass of 1 mol of water (0.018016 kg), Aw the surface taken by 1 water molecule (11.4 � 10–20 m2), N the Avogadro number (6.023 � 1023 molecules/mol), AP the speci� c pore surface in the material (m2/m3), the relative humidity on a scale from 0 to 1, and C the energy exchange between a water molecule and the pore wall (J/kg). That heat exchange is given by: a bexp[( ) / ]C k l l R T= − , with k adsorption constant, la heat of adsorption (J/kg), lb heat of evaporation (J/kg) and R the gas constant of water vapour. For multi-layer adsorption, the BET equation applies:

17

H P 1

1 ( 1)2.62 10

1 1 ( 1)

n n

n

C n nw A

C C

+−

+

⎡ ⎤φ − + φ + φ= ⋅ ⎢ ⎥− φ + − φ − φ⎣ ⎦ (2.63)

where n is the number of water molecule layers. Both equations show that hygroscopic moisture content at low relative humidity increases when both the energy exchange and the speci� c pore surface increase. For a given open porosity �o, the latter demands a large number of very small pores.

�

�

1335vch02.indd 158 12.02.2007 16:55:15

159

Example

Total open porosity of sand-lime stone is quite the same as for medium-weight brick (� 33%). However, its average pore diameter is only 0.1 �m, while in brick it is 8 �m. Consequently, sand-lime stone has a � 6000 times larger speci� c pore surface than brick. Hence, it will be more hygroscopic at low relative humidity, which measurements con� rm.

A higher temperature in turn tempers low relative humidity sorption somewhat (heat exchange C lower).Capillary condensation starts in the smallest pores, when the adsorbed water layers touch each other. At that moment surface tension rearranges the molecules into the more stable form of a water-� lled capillary with at each end a meniscus. Yet, vapour saturation pressure above a water surface not only depends on temperature but also on the meniscus formed. Indeed, escaping from a concave meniscus is more dif� cult for a molecule than escaping from a � at surface. The reverse is true for a convex meniscus. Therefore, the vapour phase above a concave meniscus contains less and above a convex meniscus more molecules than the vapour phase above a � at water surface. Conversely, condensation on a concave meniscus happens for fewer molecules in the vapour phase than condensation on a � at water surface. Water deposition will thus start at a relative humidity below 100%, while on a convex meniscus a relative humidity above 100% is needed. This explains why talking about capillary condensation. In equilibrium, the phenomenon is described by Thompson’s law:

Wsat sat

w 1 2

cos 1 1expp p

R T r r

⎡ ⎤⎛ ⎞σ ϑ= − +′ ⎢ ⎥⎜ ⎟ρ ⎝ ⎠⎣ ⎦

(2.64)

with satp′ the saturation pressure is above the curved meniscus, with psat the saturation pressure above a � at water surface, �w the density of water, �w the surface tension of water, � the contact angle between the meniscus and the pore wall and r1 and r2 the two curvature radii of the meniscus (see capillarity). If we assume each pore to be represented by an equivalent circular one with diameter deq, then equation (2.64) simpli� es to sat sat w w eqexp[ 4 cos / ( )]p p R T d′ = − σ ϑ ρ .

Figure 2.20. Thompson’s law for a temperature of 20 °C.

2.3 Vapour Transfer

1335vch02.indd 159 12.02.2007 16:55:15

160 2 Mass Transfer

Equating the ratio sat sat/p p′ to relative humidity () gives (see Figure 2.20):

w

w eq

4 cosln( )

R T d

σ ϑφ = −

ρ (2.65)

The wider the pores, the higher the relative humidity needed for capillary condensation. Below a relative humidity of 20%, the equivalent diameter deq drops below 10–9 m, i.e., below the sphere of in� uence of a water molecule. In such small pores, molecules no longer behave statistically and the condensation theory no longer applies. At 100% relative humidity the equivalent pore diameter equals in� nity. All open pores in a material should then be � lled with water, which means saturation. Hence, air in the pores does not allow this. The moisture content will remain at a value which prohibits air from escaping the porous system. We name that value the ‘capillary moisture content’. However, when a 100% relative humidity situation continues for years, dissolution of the entrapped air into the liquid phase will slowly increase the moisture content, up to saturation. Temperature also intervenes. The lower it is, the lower the relative humidity at which capillary condensation starts.The term –4 �w cos (�) / deq in Thompson’s law equals capillary suction pc (or s) in a pore with diameter deq. Thus, equation (2.65) may also be written as:

cln ( )p

R Tφ =

ρ (2.66)

which shows that an unequivocal relationship exists between two apparently different quantities: capillary suction and relative humidity.In a pore with equivalent diameter deq, capillary condensation starts when the relative humidity in the environment equals sat sat100 /p p′ . That process should be understood as follows: for each relative humidity in all pores with an equivalent diameter smaller than or equal to deq(), condensation occurs, while in all pores with an equivalent diameter larger than 2 deq() multi-layer adsorption continues and in all pores with an equivalent diameter deq() < deq < 2 deq() condensation on the adsorbed layers develops. The very strong increase in hygroscopic moisture content beyond a relative humidity of 90% is caused by the large share of these wide pores in the total pore volume.Lastly, hysteresis between sorption and desorption still demands an explanation. Many causes intervene. Among others, water uptake and release are so slow that testing may stop too early, resulting in pseudo equilibriums being measured. As a consequence, too low moisture contents are noted in sorption and too high in desorption. Also the interaction between the liquid and the air in the pores is different during sorption and desorption. Indeed, testing in a vacuum gives a much smaller hysteresis. Further-on, sorption gives other menisci then desorption, mainly because the contact angle ϑ at the pore wall changes a little between both modes.

2.3.2.4 Impact of Salts

With salts in the pores sorption is augmented. This is a direct consequence of the vapour saturation depression caused by a saturated salt solution. Each solution in fact is characterized by a salt-speci� c equilibrium relative humidity:

Saturated salt solution Equilibrium-RV (%)

MgCl2 33NaCl 75KaCl 86…

1335vch02.indd 160 12.02.2007 16:55:16

161

Thus, with NaCl, a sudden increase in hygroscopic moisture content is seen at 75%. As a result, masonry which has been in contact with sea water for a long period of time hardly dries in a climate where the average relative humidity is above 75%. Increased sorption may also be used to demonstrate the presence of salts in a porous material.

2.3.2.5 Consequences

Hygroscopic moisture is an important element in the hygric response of materials and construction parts. In porous materials, relative humidity is the actual moisture potential with the sorption isotherm as a link with moisture content. Sorption also creates hygric inertia with the speci� c moisture content playing the same role in humidi� cation and dehumidi� cation that the volumetric speci� c heat capacity does in thermal processes. A favourable effect of hygric inertia is its inhibitory impact on interstitial condensation. What a steady-state calculation takes for condensation often is no more than an increase in hygroscopic moisture content.

Hygroscopic moisture also has undesired effects. Indeed, material dimensions depend on it and, as a result, on relative humidity. Decreasing relative humidity causes shrinkage, and increasing relative humidity causes expansion. Wood-based materials are extremely sensitive, but also cement bonded and even burnt materials react that way. For thin layers, hygroscopic expansion and shrinkage causes most of the cracking. Above a relative humidity of 75 to 80% the amount of water adsorbed on surfaces suf� ces to stimulate biological activity, such as mold growth.

2.3.3 Vapour Transfer in the Air

Suppose the dry air and water vapour concentration in moist air are �l (= ml / V) and �v (= mv / V) respectively. Moist air concentration is then:

l va l v

m m

V

+ρ = = ρ + ρ

Convection is the main driver behind moist air movement. Let va be the moist air velocity. When moving, the dry air and water vapour component may have their own velocity, different from the moist air value. With vl the dry air velocity and vv the water vapour velocity, the relationship between the three becomes:

l v l v

l v a

ρ + ρ ρ + ρ= =

ρ + ρ ρl v l v

av v v v

v

with gl = �l vl the dry air � ow rate, gv = �v vv the water vapour � ow rate and ga = �a va the moist air � ow rate, equal to the sum of both the dry air and water vapour � ow rate (ga = gl + gv).

The difference between the three velocities is caused by a combination of two displacement modes:

The bulk � ow, provoked by convection

Dry air gl1 = �l va

Water vapour gv1 = �v va

�

2.3 Vapour Transfer

1335vch02.indd 161 12.02.2007 16:55:16

162 2 Mass Transfer

A diffusive � ow coupled to concentration equalization by Brown’s molecular movement in mixtures of liquids or gases. Diffusive � ows steer many processes such as carbonation of lime plaster, mortar and concrete (CO2/moist air), the lime to gypsum transformation (SO2/moist air) and chlorine corrosion in concrete (Cl2/moist air). Diffusion in moist air starts whenever differences in vapour to dry air ratio exist.

In a right-angled coordinate system which moves along at the same velocity va as the moist air, Fick’s empirical law for diffusion holds:

Water vapour:

va vl

a

D⎛ ⎞ρ

= −ρ ⎜ ⎟ρ⎝ ⎠v2g grad

Dry air:

la vl

a

D⎛ ⎞ρ

= −ρ ⎜ ⎟ρ⎝ ⎠l2g grad

In both formulas, Dvl is the binary diffusion coef� cient between water vapour and dry air; units m2/s, given by Schirmer’s equation:

1.81

vla

2.26173

273.15

TD

P⎛ ⎞= ⎜ ⎟⎝ ⎠

(2.67)

For a moist air pressure of 1 atmosphere (1 Bar, 101,300 Pa) that relation converts into: Dv1 = 8.69 � 10–10 T1.81. Total water vapour � ow rate gv thus becomes:

vv a vl

a

Dρ

= + = ρ − ρρv v1 v2 ag g g v grad

In that sum of two displacement modes – convection and diffusion – convective � ow is usually by far the largest. A diffusive water vapour � ow in one direction now generates an equally large diffusive dry air � ow in the opposite direction. So, if diffusion is the only mode, neither a resulting moist air transfer nor a change in moist air concentration is noted. At atmospheric conditions and temperatures below 50 °C, moist air pressure is much larger than vapour pressure, while the sum p / R + pl / Rl is close to constant. That simpli� es the equation for the total water vapour � ow rate to:

vl l

la l

l

pD pp p R

ppR T T R RR R

⎛ ⎞= − +⎜ ⎟ρ ⎝ ⎠ +

v ag g grad

or

vl

a

Dp p

R T R T≈ −ρ

av

gg grad (2.68)

Equation (2.68) describes vapour transfer in moist air under atmospheric conditions, with vapour pressure p as a driving force. Vapour � ow rate thus depends on the driving force and its gradient, a fact also seen in combined heat conduction and enthalpy � ow.

�

1335vch02.indd 162 12.02.2007 16:55:16

163

Two speci� c cases are worth considering:

1. Convection and vapour diffusion push water vapour in the same direction, while the dry air � ow rate remains zero. In such case the moist air � ow rate ga consists of water vapour only, converting (2.68) into:

vl vl a

v a

a

1

1

D D pp p

R T R T p p= − ≈ −ρ −−

ρ

vg grad grad

(� if Ra � Rl). Evaporation through the boundary layer just above water is a typical example of that (Figure 2.21).

Figure 2.21. Evaporation from a water surface (ga = 0, gv > 0).

2. Pure diffusion which demands a moist air � ow rate ga zero. Vapour � ow rate then becomes:

vlDp

R T≈ −vg grad (2.69)

while for dry air, we get:

vll

l

Dp

R T≈ −vg grad

To keep the analogy with heat and air transport, the ratio Dvl / (R T) is called the vapour permeability of air. Symbol �a, units s. The vapour � ow rate in general and for the two speci� c cases thus becomes:

aa a a

a a

Pp p p p

R T P p= − δ = −δ = −δ

ρ −a

v v vg

g grad g grad g grad

The last expression resembles Fourier’s law for heat conduction. We just have to replace vapour permeability �a by thermal conductivity � and vapour pressure p by temperature �. Vapour permeability, however, is an explicit function of total air pressure and temperature while thermal conductivity could be taken as a constant.

2.3 Vapour Transfer

1335vch02.indd 163 12.02.2007 16:55:16

164 2 Mass Transfer

2.3.4 Vapour Transfer in Materials and Construction Parts

Moisture transport only develops in open-porous materials. At low moisture content, displacement happens to a large extent, if not only, in the vapour phase. For non-capillary materials vapour transport is even the only mode until saturation. In other words, talking about vapour � ow in materials is physically correct.

2.3.4.1 Flow Equation

At � rst sight, vapour � ow in an open-porous material should not differ from vapour � ow in the air. However, displacement is only possible in the pores accessible for dry air and water vapour molecules. At the same time, convection plays a much smaller role than in air, while hygric buffering is of great importance in hygroscopic materials. As in air, the (much smaller) convective vapour � ow rate can be written as:

a

pR T

=ρ

av1

gg (2.70)

while vapour diffusion (� < 50 °C, Pa � p) looks like:

p= −δv2g grad (2.71)

with � the vapour permeability of (the porous system in) the material (units s). Hence, the part per unit surface of material occupied by open pores varies from smaller to much smaller than the unit surface. Thus, vapour � ow rate through a unit surface of material must be smaller than through a unit surface in air. Therefore: � is smaller than �a, the vapour permeability in air. Instead of using a vapour permeability �, Krischer introduced a vapour resistance factor � as material characteristic. The vapour resistance factor indicates how many times the vapour permeability of a material is smaller than that in stagnant air, at the same conditions of temperature and total pressure. In a formula:

aδμ =δ

(2.72)

Units: none. The lowest value of the vapour resistance factor is one, the highest in� nite (for a material with only closed pores or no pores at all). What value we will have, depends on the pore system. The following play a role:

1. Open pore area Ao per unit surface of material. An increasing open pore area lowers while a decreasing open pore area heightens the vapour resistance factor (Figure 2.22). Thus: � � 1 / Ao. For an open pore area of zero, i.e. if no open pores are present, the vapour resistance factor touches in� nity. In such material, air and vapour � ow are impossible, or, neither diffusion nor convection has to be considered.

Figure 2.22. �-value increases when the open-pore area per unit-surface of material decreases.

1335vch02.indd 164 12.02.2007 16:55:16

165

2. Path length lo. Is the average distance the diffusing molecules have to travel in the pores compared to the thickness of the layer (d0). The higher the ratio lo / do above one, the larger the diffusion resistance factor. The closer the ratio is to one, the smaller the diffusion resistance factor (Figure 2.23). Hence: � � lo / do.

Figure 2.23. �-value increases with path length lo.

3. Tortuosity or ‘deviousness’ of the pore system (Figure 2.24).

Figure 2.24. �-value increases with tortuosity.

In general, the vapour resistance factor equals the ratio between tortuosity �l and total open porosity �o:

l

o

Ψμ =

Ψ

For hygroscopic materials, relative humidity in the pores also plays a role. A hygroscopic material in fact does not experience pure Fickian diffusion. In pores, which have a diameter hardly larger than the free path length of a water molecule, friction diffusion, also called Knudsen diffusion, occurs. Moreover, at high relative humidity, water transfer in the adsorbed water layers and the pores � lled by capillary condensation adds to diffusion. Further on, when hygroscopic moisture content increases (i.e. with increasing relative humidity), diffusion through the pores turns into diffusion from water islet to water islet (Figure 2.25), which shortens the diffusion length. The three effects together result in a decrease of the vapour resistance factor with increasing relative humidity. The � rst two and changes in pore structure also create linkages to temperature.

2.3 Vapour Transfer

1335vch02.indd 165 12.02.2007 16:55:16

166 2 Mass Transfer

Implementing the vapour resistance factor into the diffusion equation gives:

a pδ

= −μv2g grad

Yet, water vapour permeability of air (�a) is a very small number (� 1.87 � 10–10 at 20 °C). Since small quantities give the impression that the phenomena they re� ect are unimportant, the inverse number, called the diffusion constant, is often used (symbol N, units s–1):

1p

N= −

μv2g grad (2.73)

For the value of the diffusion constant, see Figure 2.26.

Figure 2.26. The diffusion constant as a function of temperature and air pressure (upper curve: Pa = 110,000 Pa, middle curve: Pa = 101,300 Pa, lower curve: Pa = 90,000 Pa).

2.3.4.2 Conservation of Mass

Mass conservation for water vapour is given as (with vapour content converted into vapour pressures):

Figure 2.25. Diffusion length shortened because of capillary condensation.

1335vch02.indd 166 12.02.2007 16:55:16

167

ocdiv

pG

t R T

Ψ⎛ ⎞∂± ′ = − ⎜ ⎟∂ ⎝ ⎠vg (2.74)

Whenever vapour condensates, evaporates or sublimates, then the source term cG′ differs from zero. At right, open porosity �� decreases when water vapour condenses and vice versa, if water present evaporates, it increases. With a relative humidity permanently below 100%, the source term becomes zero. However, for hygroscopic materials at a relative humidity below 100%, changes in hygroscopic moisture include both sources and sinks and changing vapour content. If we assume isothermal conditions and if the relative humidity stays below a value M < 100%, which marks the end of the vapour transfer and surface � ow only interval, then the mass equilibrium simpli� es to:

Hdivw

t tφ∂ ∂φ= − = −ρ ξ∂ ∂vg (2.75)

At a relative humidity beyond M (the exact value of M is unknown. Often 95 to 98% is postulated, even though measurements may show a lower value), water � ow develops. That makes equation (2.75) no longer applicable.

2.3.4.3 Vapour Transfer by ‘Equivalent’ Diffusion

The term ‘equivalent’ is used to indicate that diffusion in porous materials combines molecular and friction diffusion with surface � ow and water transfer in pores randomly liquid � lled by capillary condensation. Because that complex � ow situation is dominated by the vapour phase, it is described using the Fickian diffusion law. Whenever convection is negligible, ‘equivalent’ diffusion provides a usable model to evaluate ‘vapour’ transfer through construction parts. For convection to be extremely small, the part should not contain cavities. This restriction is commonly omitted, because it limits the application of the diffusion model to massive constructions. At least one layer should have a very low air permeance, i.e., should consist of a material without open porosity or with micro-pores only. Unless the micro-pores represent a high open porosity, such materials have at the same time a high vapour resistance factor. Also the air pressure differences should remain extremely small (this assumption is purely hypothetical, because the different temperature at both sides of a vertical or inclined porous building part automatically activate air pressure gradients).Hence, equivalent diffusion is in no way the rule. The further text, therefore, has less a practical than a theoretical value. This is even truer because the diffusion approach assumes each building part to be dry at the start, while in reality, each building construction begins its service life quite humid by excess moisture, moisture generated by chemical reactions taking place, wind driven rain, melting snow, etc. That sets the starting relative humidity beyond the value M which marks the start of liquid � ow in the pores.

Steady-state

In steady-state, the derivative H /w t∂ ∂ turns zero. That converts equation (2.75) into div gv = 0. Introducing the diffusion equation then gives:

1div 0p

N

⎛ ⎞=⎜ ⎟⎝ μ ⎠

grad

2.3 Vapour Transfer

1335vch02.indd 167 12.02.2007 16:55:17

168 2 Mass Transfer

One-dimension: Flat Walls

In � at walls the equation simpli� es to:

1 d0

d d

d p

x N x

⎛ ⎞=⎜ ⎟⎝ μ ⎠

Let us assume that the vapour resistance factor � is constant. The diffusion constant N anyhow remains an explicit function of temperature and total air pressure. Under atmospheric conditions, when air pressure is a constant, temperature may still vary. This obliges us to differentiate between isothermal and non-isothermal isobaric situations. When isothermal, the diffusion constant N takes a � xed value.

Single-layered Wall, Isothermal Conditions

A constant N allows one to reduce the equation to 2 2d / d 0p x = , with a solution of: 1 2p C x C= + . If the vapour pressures on both wall surfaces (ps1, ps2), the thickness of

the wall (d) and the vapour resistance factor of the material (�) are known, then from the boundary conditions x = 0: p = ps1, x = d: p = ps2, ps1 < ps2 (Figure 2.27) we get: C2 = ps1,

1 s2 s1( ) /C p p d= − . The vapour pressures in the wall thus become:

s2 s1s1

p pp x p

d

−= + (2.76)

As vapour � ow rate gv, we get:

s2 s1v

1 d

d

p ppg

N x N d

−= − = −

μ μ

or, in absolute value:

s2 s1v

p pg

N d

−=

μ (2.77)

Figure 2.27. Diffusion under isothermal conditions in a single-layered wall.

1335vch02.indd 168 12.02.2007 16:55:17

169

Thus, the steady state vapour pressure in a single-layered wall in an isothermal regime changes linearly, while the vapour � ow rate is constant, proportional to the vapour pressure difference and inversely proportional to the wall thickness and the vapour resistance factor. We call the product � N d the ‘speci� c’ diffusion resistance of the wall; symbol Z, units m/s. The larger that diffusion resistance is, the smaller is the vapour � ow rate. A large diffusion resistance means a large wall thickness or the use of a material with high vapour resistance factor i.e., one that acts as vapour retarder! The diffusion constant N in (2.77) is � 5.4 � 109 s–1. That very large number allows concluding that diffusion only generates very small vapour � ow rates. Displacement of ‘signi� cant’ quantities of water vapour by that mechanism requires a lot of time. In other words, diffusion is a very slow process.

Composite Wall, Isothermal Conditions

We assume the vapour pressure on both wall surfaces (ps1, ps2), the thickness of all layers (di) and their vapour resistance factors (�i) to be known. Contrary to heat conduction, the likelihood to have a resistance Zci in each interface i between successive layers is high now. Its size depends on how the layers contact each other (glued, cast, etc.) As a result, each interface i generates two unknown vapour pressures: pi1 and pi2 (Figure 2.28). For each layer and for the contact resistances, the following holds:

Layer 1:

11 s1v

1 1

p pg

N d

−=

μ

Contact resistance 1:

12 11v

c1

p pg

Z

−=

Layer 2:

21 12v

2 2

p pg

N d

−=

μ

Contact resistance 2:

22 21v

c2

p pg

Z

−=

…

Contact resistance n – 1:

1,2 1,1v

, 1

n n

c n

p pg

Z− −

−

−=

Layer n:

s2 1,2v

n

n n

p pg

N d−−

=μ

2.3 Vapour Transfer

1335vch02.indd 169 12.02.2007 16:55:17

170 2 Mass Transfer

In these equations, p11, p12, …, pn – 1,1, pn – 1,2 are the unknown interface vapour pressures and gv is the unknown but constant vapour � ow rate. Rearranging and adding gives:

s2 s1v

i i ci( )

p pg

N d Z

−=

μ +∑ ∑ (2.78)

The denominator in equation (2.78) is called the ‘speci� c’ diffusion resistance of the composite wall; symbol ZT, units m/s. �i N di is the diffusion resistance and �i di the diffusion thickness of layer i, units m. The higher the diffusion resistance of the wall, the smaller the vapour � ow rate. A large diffusion resistance ZT may be achieved by inserting a layer with high diffusion thickness. If a thin foil with high �-value is concerned, the name vapour retarder or vapour barrier is used (the latter when the diffusion resistance factor lies close to in� nity). Large contact resistances are obtained by gluing layers together over the whole interface. The vapour pressures follow from the series of layer equations for the vapour � ow rate. The algorithm is:

x s1 v s1x

x s2 v s2x

p p g Z

p p g Z

= += −

(2.79)

In the [Z, p]-plane, the vapour pressures form a straight line through the points (0, ps1) and (Z, ps2) with the vapour � ow rate gv as slope. The vapour pressure curve in the [x, p]-plane is obtained by transposing the intersections with the interfaces and contact resistances between the layers from the [Z, p] to the [x, p]-plane and connecting the successive intersections with line segments (Figure 2.28). The highest difference in vapour pressure occurs over the most vapour-tight layer or the highest contact resistance, which in such a case acts as a vapour retarder. In the [Z, p]-plane, a composite wall behaves as a single-layered wall.

Figure 2.28. Diffusion under isothermal conditions in a composite wall. Remark the contact resistances between the different layers.

1335vch02.indd 170 12.02.2007 16:55:17

171

One problem is that contact resistances Zci are usually unknown quantities. Therefore, in practice, they are generally ignored. This is only correct for a free contact with an extremely thin air gap left between the contacting layers.

Single-layered Wall, Non-isothermal Conditions

In non-isothermal steady-state conditions, temperature is a linear function of the ordinate x. So, the diffusion constant N, which is a function of temperature, will also depend on x.

Vapour Flow Rate

The vapour � ow rate is obtained by dividing the wall into in� nitesimally thin layers, each with a thickness dx and a temperature 1 2 1( ) ( ) /x x dθ = θ + θ − θ . One so has (Figure 2.29):

v1 d

[ ( )] d

pg

N x x= −

μ θ

The rate must be constant in steady-state, or:

s2

s1

v 1 2 10

( ) d dpd

p

xg N x p

d⎡ ⎤μ θ + θ − θ = −⎢ ⎥⎣ ⎦∫ ∫ (2.80)

Yet, the diffusion constant N is given by:

6 0.81a

vl

(5.25 10 )R T

N P TD

−= = ⋅ (2.67)

Implementing that function into equation (2.80) and solving the integrals results in:

s2 s1v 6

a5.25 10 ( )

p pg

d P F T

−= −

μ ⋅ (2.81)

with 0,19 0,19s2 s1 s2 s1( ) ( ) / [0.19 ( )]F T T T T T= − − . Numerically that expression hardly differs

from the integral of a straight line between �s1 and �s2, divided by the integration-interval

�

Figure 2.29. Diffusion under non-isothermal steady state conditions in a single-layered wall.

2.3 Vapour Transfer

1335vch02.indd 171 12.02.2007 16:55:18

172 2 Mass Transfer

(�s2 – �s1) (Figure 2.30) As temperature changes linearly with x (� = aT x + bT), F(T) also matches the integral of a straight line aN,x + bN between 0 and x, divided by x, giving the average value of �s1 and �s2 between 0 and x. The vapour � ow rate through the single-layered wall thus becomes (�m being the average temperature in the wall):

s2 s1v

m( )

p pg

d N

−=

μ θ (2.82)

That equation is similar to the isothermal one, however with the diffusion constant for the average temperature in the wall.

Vapour Pressures

Vapour pressures ensues from:

s1 v0

( ) dx

p p g N x x= + μ ∫

with the integral equal to:

0.196 0.19

a s1 s2 s1 s1

s2 s1

(5.25 10 ) ( )

0.9 ( )

xP d T T T T

d

T T

⎧ ⎫⎪ ⎪⎡ ⎤⋅ + − −⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭−

an expression which closely matches the product N(�m) x, or:

s2 s1s1

p pp x p

d

−= + (2.83)

As in an isothermal regime, the vapour pressure line is almost straight (almost, because the product N(�m) x is only a close approximation of the correct function).

�

Figure 2.30. F(T) as a function of the average temperature in °C. In fact a straight line (�2 = 0 °C).

1335vch02.indd 172 12.02.2007 16:55:18

173

Composite Wall, Non-isothermal Conditions

Again, we assume the vapour pressure on both wall surfaces (ps1, ps2), the boundary tempera-tures *

e i( , )θ θ , the thickness (di), the vapour resistance factor (�i) and thermal conductivities (�i) of all layers known.

Vapour Flow Rate

For each layer in the wall, the diffusion constant to be used is the one for the average temperature in that layer. For the contact resistances, it is the contact temperature:

s2 s1v

i i mi ci i,i 1[ ( )] ( )

p pg

d N Z +

−=

μ θ + θ∑ ∑ (2.84)

As for isothermal conditions, i i mi ci i,i 1[ ( )] ( )d N Z +μ θ + θ∑ ∑ represents the diffusion resistance ZT of the composite wall, while i i mi( )d Nμ θ is the diffusion resistance Zi of layer i.

Vapour Pressures

The vapour pressure curve hardly differs from the isothermal one, provided that the diffusion constant of each layer is the one for the layer’s average temperature and the diffusion constant for each contact resistance is the one for the contact temperature. As an algorithm:

x s1 v s1x

x s2 v s2x

p p g Z

p p g Z

= += −

(2.85)

In the [Z, p]-plane, vapour pressures lie on a straight line through the points (0, ps1) and (Z, ps2) with the vapour � ow rate gv as slope. The curve in the [x, p]-plane is found by transposing the successive intersections with the interfaces and contact resistances from the [Z, p] plane to the [x, p]-plane and connecting them there with line segments. Again, the contact resistances Zci are mostly unknown, which is why they are typically set at zero.

Interstitial Condensation

‘Interstitial’ indicates that a fraction of the water vapour which migrates through a wall condenses somewhere in it. For that to happen, non-isothermal conditions and a vapour pressure line intersecting the saturation pressure are needed, which is found by transposing the temperatures in the wall into the related saturation pressures. In general, when comparing the vapour with the saturation pressures, two possibilities exist:

1. No intersection. Under no condition will interstitial condensation occur and the vapour pressure line found is the correct one.

2. Intersection (Figure 2.31). In such a case, interstitial condensation is a fact demanding further analysis. For that purpose the wall is redrawn in the [Z, p]-plane, the temperature curve calculated, related saturation pressures traced and the vapour pressures at both surfaces connected with a straight line that intersects the saturation line (Figure 2.32). This creates a physically impossible situation. In fact, between the crossing points, vapour pressures above saturation pressure appear! The correct solution follows from conservation of mass. As the construction part is initially dry, the mass balance in the condensation zone is:

�

�

2.3 Vapour Transfer

1335vch02.indd 173 12.02.2007 16:55:18

174 2 Mass Transfer

satc

dd 1

d d

pG

x N x

⎛ ⎞= ′⎜ ⎟⎝ μ ⎠

(2.86)

with cG′ the condensation rate per m3 and psat the local saturation pressure. Outside the condensation zones, that balance changes into:

d 1 d0

d d

p

x N x

⎛ ⎞=⎜ ⎟⎝ μ ⎠

(2.87)

In a [Z, p]-plane, both equations transform into:

within the condensation zones

2sat c2

d

d

p G

NZ

′= ±

μ

outside the condensation zones

2

2

d0

d

p

Z=

In the [Z, p]-plane outside the condensation zone, the vapour pressure curve thus remains a straight line. Within the condensation zone, the liquid deposited (in kg/(m3 · s)) is proportional to the second derivative of the saturation line. Suppose now the correct vapour pressures consisted of those found in case of no condensation outside the condensation zones and

Figure 2.31. Diffusion under non-isothermal steady state conditions, composite wall. Vapour pressure passes saturation pressure! Means interstitial condensation.

1335vch02.indd 174 12.02.2007 16:55:19

175

Figure 2.32. Successive steps in the tangent line method:(1) Transpose the wall into the [Z, p]-plane.(2/3) Trace vapour pressure (p) and saturation pressure line (psat) in the wall. If they do not intersect, no interstitial condensation will develop. If they intersect, interstitial condensation is a fact (the case here).(4) Correct solution in case of intersection = vapour pressure line replaced by the tangents;(5) Searches the diffusion resistance to be added at the warm side of the thermal insulation to avoid interstitial condensation.

2.3 Vapour Transfer

1335vch02.indd 175 12.02.2007 16:55:19

176 2 Mass Transfer

the saturation values inside the condensation zones. In the contacts, whether approached along the vapour pressure or the saturation line, the vapour � ow rate, which in both cases is given by the slope of the lines, must be identical. In that case, this is only possible if the contacts accommodate a vapour source, which con� icts with the assumption of an initially dry construction. Thus, the vapour pressure straight segments must change direction. As the derivatives of both these lines and the saturation curve represent vapour � ow rates in the [Z, p]-plane, the following should hold:

satd d

d d cc

p p

Z Z ←→

⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ (2.88)

meaning that in each contact, the gradient of the saturation pressure line must equal the slope of the vapour pressure lines, which is only possible if these become tangent to the saturation line (Figure 2.32).Mathematically, the points of tangency on the saturation line (Zc, psat,c) are found from (equa-tions written for the inward tangent line, starting at the wall’s surface ZT, called surface 2):

1. Vapour and saturation pressure are equal, or: sat,c2 s2 v2 T c2( )p p g Z Z= − − with gv2 the unknown inward vapour � ow rate.

2. Vapour pressure line coming from the non-condensation zone and saturation pressure line with the same gradient:

sat satv2

c2 c2

d d

d d

p p qg

Z N

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎢ ⎥ θ λ μ⎣ ⎦ ⎣ ⎦

with q the heat � ow rate through the wall and � and � thermal conductivity, respectively vapour resistance factor of the layer which contains the points of tangency. For the ingoing as well as for the outgoing tangent, these points coincide most of the time with an interface between two layers, which means singular points on the vapour saturation line. Moreover, in many cases the two points coincide, restricting the number of condensation interfaces to one.

The tangent line solution is named after its developer, the Glaser method. It provides an answer to the following four questions:

Vapour pressure line in case of interstitial condensation?Outside the condensation zones the vapour pressure line coincides with the tangent lines. Inside the condensation zones, it coincides with the vapour saturation line.

Where are the condensation zones?Condensate is deposited in all zones where the vapour pressure and saturation line coincide, i.e., between the points of tangency. When the ingoing and outgoing vapour pressure tangent touches the saturation curve in the same interface, the condensate is deposited in that interface only.

What is the condensation rate?The condensation rate gc is given by the difference in gradient between the in- and outgoing vapour pressure tangent lines. If the diffusion resistance between wall surface 1 (Z = 0) and the point of tangency for the outgoing line is Zc1, if the diffusion resistance between the same wall surface 1 and the point of tangency for the ingoing line is Zc2 and if the saturation pressures in the points of tangency are psat,c1 and psat,c2, then the condensation rate (gc) becomes:

1335vch02.indd 176 12.02.2007 16:55:19

177

s2 sat,c2 sat,c1 s1c

T c2 c1

− −= −

−p p p p

gZ Z Z

(2.89)

In the [x, p]-plane, the distribution of water deposited in a condensation zone is given by:

22 2 2c sat sat

c 2 2 2

d d d1 1 d d

d dd d d

g p p pG N

x N N xZ x

θ′ ⎛ ⎞= = μ = =′ ⎜ ⎟⎝ ⎠μ μ θ (2.90)

The amount of condensate and the moisture behaviour of the materials within the condensation zones determine the acceptability of the phenomenon.

How can interstitial condensation be avoided?The three measures that guarantee avoidance are easily found by analyzing the [Z, p]-picture of the wall:

1. Lower the vapour pressure at the warm side to the extent that the connecting line with the vapour pressure at the cold side no longer intersects the saturation pressure: ps2 � p2,max. A better ventilation of the building and/or a lower vapour release inside are the actions to be undertaken (Figure 2.33).

2. Increase the diffusion resistance at the warm side to the extent that the vapour pressure line does no longer intersect saturation pressure. This option demands the addition of a vapour resistance �Z (= Zvapour retarder) to the existing resistance ZT – Zc2. The size of �Z is seen on the Z-axis between the diffusion resistance of the wall and the intersection of the outgoing vapour pressure tangent with the horizontal through the vapour pressure ps2 (Figure 2.32).

s2 s1vapor retarder c1 T

sat,c1 s1

−Δ = ≥ −

−p p

Z Z Z Zp p

(2.91)

The measure implies the inclusion of a vapour retarding foil at the warm side of the insulation layer.

Figure 2.33. Tangent line method (Glaser’s method). Highest admissible vapour pressure at the warm side to avoid interstitial condensation.

2.3 Vapour Transfer

1335vch02.indd 177 12.02.2007 16:55:19

178 2 Mass Transfer

3. Lower the diffusion resistance at the cold side of the condensation zone (Zc1) to the extent that vapour pressure and saturation pressure no longer intersect. That value i

c1Z sits on the Z-axis between the point of contact of the outgoing vapour pressure tangent and the intersection of the ingoing vapour pressure tangent and the horizontal through the vapour pressure at the cold side (ps1) (Figure 2.34).

Analytically:

sat,c1 s1c1 exterior clad. T c1

s2 sat,c1

( )−

′ = ≤ −−

p pZ Z Z Z

p p (2.92)

The measure implies the usage of a vapour permeable cladding or � nish.Of course, evaluating diffusion as a cause of interstitial condensation may be done beforehand with the wall drawn on the thickness axis without tangent construction:

1. ‘No interstitial condensation’ means ‘no intersection between vapour and saturation pressure’. For that purpose, what is needed is a convex saturation and a concave vapour pressure line from warm to cold (Figure 2.35a). Because saturation pressures and temperatures are identical in shape, this implies a convex temperature and concave vapour pressure line. Hence, the rule of the thumb is simple. Be sure that thermal conductivity and vapour resistance factor decrease from warm to cold, i.e. locate the best insulating but more vapour permeable layers at the cold side of a construction part and the least insulating vapour retarding layers at the warm side of a construction part.

2. ‘Interstitial condensation hardly avoidable’ translates into ‘most probably intersection between vapour and saturation pressure’. For that purpose, all that is needed is a concave saturation and a convex vapour pressure line from warm to cold (Figure 2.35b). Because the saturation pressure and the temperature curve are identical in shape, this implies a concave temperature and convex vapour pressure line. Hence, the rule of the thumb again is simple. Condensation may be a problem if thermal conductivity and vapour resistance

Figure 2.34. Tangent line method (Glaser’s method). Minimum diffusion resistance of an outside cladding to avoid interstitial condensation.

1335vch02.indd 178 12.02.2007 16:55:20

179

factors increase from warm to cold, i.e. locate the best insulating but more vapour permeable layer at the warm and the least insulating but vapour retarding layer at the cold side of the construction part. It of course still depends on the boundary conditions if interstitial condensation will occur or not.

3. ‘Interstitial condensation unlikely’ is obtained when vapour and saturation pressure line are both concave or convex from warm to cold. If either or no interstitial condensation will occur depends on the boundary conditions and the thermal conductivity and vapour resistance factor sequence in the wall (Figure 2.35c).

However, the simple measures analyzed, let alone the rules of the thumb, do not guarantee moisture tolerance of a facade or roof design. In fact, many vapour permeable layers are unsuitable as outside � nish (not rain-tight, too capillary, not strong or rigid enough). Also, the so-called warm and cold side may change with the seasons. In a moderate climate, for example, the outside surface forms the cold side during winter. In summer instead, cold may shift to the inside surface. In warm and humid climates, the outside surface is warm and the inside surface cold. Diffusion thus generates nice rules with regard to wall construction (applying a vapour retarder, using a vapour permeable outside lining, etc.) which under various circumstances may be quite inadequate.

Single-layered Wall with Variable Vapour Resistance Factor, Isothermal Conditions

Such a wall behaves as a composite wall of n layers, each with an in� nitesimal thickness dx.

Figure 2.35. Interstitial condensation, simple evaluation of the risk:a) � convex and p concave, risk zero;b) ��concave and p convex, large risk;c) both � and p convex or concave, risk unclear.

2.3 Vapour Transfer

(a) (b) (c)

1335vch02.indd 179 12.02.2007 16:55:20

180 2 Mass Transfer

Two- and Three-dimensions

Steady State, Isothermal Regime

The CVM-approach, which should be applied here, does not differ from heat conduction. Conservation of mass – the sum of the vapour � ows from the adjacent volumes to the central volume zero – leads to n equations of the form:

v, ,, ,

1, 1, 1

0i ji l m nj l m n

G== ± ± ±

=∑ (2.93)

Each vapour � ow Gv,i,j equals the difference in vapour pressure between the adjacent and central volume, multiplied by the vapour permeance d( )P′ connecting both. For two adjacent volumes in the same material d 1 2( ) /( )P a a N b′ = + μ with a1 and a2 half the mesh width at both sides of the grid line through both points and b the distance in between along the grid line.

Steady-state, Non-isothermal Regime

In that case, applying CVM is more complicated. First, we have to calculate the temperature � eld and convert it into saturation pressures. Then, using the same mesh, the vapour pressures are solved. If they do not pass saturation pressure in any of the control volumes, then interstitial condensation is impossible. If, on the contrary, vapour pressure is somewhere larger than saturation, then interstitial condensation is a fact. To � nd the condensation zone, the amounts condensing and the correct vapour pressures, the following applies:

In all volumes where vapour pressure goes beyond saturation, we assume equality and introduce the condensation rate as the new variable c( )G V′ Δ . In these volumes, mass equilibrium changes to:

v, , c, ,

1, 1, 1

i ji l m nj l m n

G G V== ± ± ±

= Δ′∑ (2.94)

giving a new system of n equations with n unknown (p or cG′). Solving results in a � rst distribution of vapour pressures and condensation rates.

If in some volumes the condensation rate cG′ turns negative because the accumulated deposit drops below zero, then cG′ is set at zero and vapour pressure is reintroduced as unknown. Solving that system gives a new distribution of vapour pressures and condensation rates. We have to repeat that step until part of the volumes show a positive condensation rate or accumulated deposit and the others a vapour pressure below saturation. At that point the condensation zone, the amounts condensing and the correct vapour pressures are known.

Transient

In a transient regime, the equations (2.74) and (2.75) are the ones to be used:

Non-hygroscopic materials:

0cdiv

pG

t R T

Ψ⎛ ⎞∂± =′ ⎜ ⎟∂ ⎝ ⎠vg (2.74)

�

�

1335vch02.indd 180 12.02.2007 16:55:20

181

Hygroscopic materials:

Hdivw

t tφ∂ ∂φ= − = −ρ ξ∂ ∂vg (2.75)

both with 1

pN

= −μvg grad

Hygroscopic Materials

Below a relative humidity �M the diffusion equation reshuf� es to:

satsat sat

d1 1( )

d

pp p

N N⎛ ⎞= − φ = − φ + φ θ⎜ ⎟⎝ ⎠μ μ θvg grad grad grad (2.95)

which is an expression with temperature and relative humidity as potentials. As for tempera-ture, provided there are no contact resistances, the relative humidity in all interfaces is continuous and unequivocal, making it a real potential. The step from relative humidity and temperature to moisture content goes via the sorption curve. If sat / ( )p Nμ is called D�, and

sat/ [ ] d / dN p Dθφ μ θ , then the balance equation (2.75) becomes:

( )div ( )D D

tφ

φ θ∂ ρ ξ φ

φ + θ =∂

grad grad (2.96)

The transport coef� cients D� and D� and the speci� c moisture content � ��, are functions of relative humidity and temperature. Therefore equation (2.96) has to be solved together with the heat balance, using FEM or CVM.Under isothermal conditions the balance simpli� es to:

( )div ( )D

tφ

φ∂ ρ ξ φ

φ =∂

grad

In theory, isothermal moisture transfer is only possible in a material with an in� nite thermal conductivity. Otherwise, the conversion of latent to sensitive heat irrevocably results in temperature differences. If we assume a constant transport coef� cient D� and constant speci� c moisture content � ��, then the balance can be written as:

2

D tφ

φ

ρ ξ ∂φ∇ φ =∂

(2.97)

Constant speci� c moisture content � �� reduces the sorption curve to a straight line between 0% and �m% relative humidity (see Figure 2.36). That line ensues from a least square analysis on the real curve in the relative humidity interval 30 to 86%. However it is annoying that a constant transport coef� cient is only possible in non-hygroscopic materials! Indeed, the vapour resistance factor of hygroscopic materials decreases and their transfer coef� cient D� increases with relative humidity. Therefore, equation (2.97) is only applicable if the differences in relative humidity remain small. Anyhow, even then, the transport coef� cient D��(via psat and N) as well as the speci� c moisture content � �� depend on temperature.

2.3 Vapour Transfer

1335vch02.indd 181 12.02.2007 16:55:20

182 2 Mass Transfer

Equation (2.97) resembles Fourier’s second law of heat conduction. Solving is done the same way. For a periodic change in relative humidity, the wall characteristics per harmonic and per temperature for example become:

a wall matrix for diffusion Wnd;

a moisture damping nDφ with phase shift nφφ , for single-layered walls given by:

n vncoshd

DDφφ

⎛ ⎞ω= ⎜ ⎟

⎝ ⎠

n vnarg coshd

Dφφ

⎛ ⎞ωφ = ⎜ ⎟

⎝ ⎠

a dynamic moisture resistance ngD with phase shift n

gφ , for single-layered walls given by:

vn

ng

n

sinhd

DD

φ

⎛ ⎞ω⎜ ⎟⎝ ⎠

=ω

vn

ng

n

sinh

arg

d

Dφ

ω⎛ ⎞⎜ ⎟⎜ ⎟φ =

ω⎜ ⎟⎜ ⎟⎝ ⎠

a hygric admittance nvAd with phase shift n

Advφ , for single-layered walls given by:

nnv n

g

AdD

D

φ=

n n nAdv gφφ = φ − φ

In these equations, �vn is the complex hygroscopic pulsation:

vn

2 D i n

Tφ φρ ξ π

ω =

For composite walls, the same product rules as for heat conduction govern layer and wall matrixes.

�

�

�

�

Figure 2.36. Schematic hygroscopic curve. At a relative humidity 0%, moisture content jumps from w = 0 to w = b. Then moisture content increases linearly between 0% and �m%. Beyond �m% (set at 95%), moisture content follows a second line, up to capillary.

1335vch02.indd 182 12.02.2007 16:55:21

183

The change in relative humidity along a semi-in� nite solid caused by a sudden increase in relative humidity ��s at the surface can be written as:

s0H

1 erf4

sx

a t

⎡ ⎤φ = φ + Δφ −⎢ ⎥

⎢ ⎥⎣ ⎦

where aH is moisture diffusivity in m2/s, equal to /Dφ φρ ξ . For the vapour � ow rate at the surface, we have:

v 0(grad )x

Dg D

tφ φ

φ =ρ ξ

= − φ = Δφπ

with Dφ φρ ξ the water vapour sorption coef� cient or vapour effusivity; units kg/(m2 · s1/2).

For two- and three dimensional problems, the analysis demands the use of FEM or CVM.

Non-hygroscopic Materials

Introducing the diffusion equation into (2.74) gives:

0 satc

1div grad

pp G

N t R T

Ψ⎛ ⎞ ⎛ ⎞∂± ′ =⎜ ⎟ ⎜ ⎟⎝ μ ⎠ ∂ ⎝ ⎠

That differential equation can only be solved in conjunction with the heat balance and the equation of state psat(�). If the vapour pressure equals saturation pressure, then the condensation/evaporation term cG′ differs from zero:

0 satc sat

1div grad

pG p

N t R T

Ψ⎛ ⎞ ⎛ ⎞∂± = − +′ ⎜ ⎟ ⎜ ⎟⎝ μ ⎠ ∂ ⎝ ⎠

In an isothermal, dry construction, the partial differential equation simpli� es to:

2 0 N pp

R T t

Ψ μ⎛ ⎞ ∂∇ = − ⎜ ⎟⎝ ⎠ ∂ (2.98)

Equations (2.97) and (2.98) are, as equations, identical. They generate analogous characteristics, now called vapour diffusivity 0/( )R T NΨ μ and the vapour effusivity 0 / ( )R T NΨ μ .In a periodic regime, the properties that de� ne the response of a � at wall are vapour pressure damping, dynamic diffusion resistance and vapour admittance.

2.3.4.4 Vapour Transfer by (Equivalent) Diffusion and Convection

As stated, pure diffusion is the exception rather than the rule. Combined convection and diffusion, also called advection, delivers already a better picture of reality. In fact, many construction parts are more or less air permeable. Wind, stack and fans in operation then suf� ce to activate air in and out� ow, wind washing, inside air ingress and air looping. The mass balance is:

0cdiv

pG

t R T

Ψ⎛ ⎞∂± ′ = − ⎜ ⎟∂ ⎝ ⎠vg

with = +v v1 v2g g g (2.99)

2.3 Vapour Transfer

1335vch02.indd 183 12.02.2007 16:55:21

184 2 Mass Transfer

gv1 is the convective and gv2 the diffusive vapour � ow rate. Introducing the � ow equations gives:

0c

a

1div

pp p G

N R T t R T

⎛ ⎞ Ψ⎛ ⎞∂− ± =′ ⎜ ⎟⎜ ⎟μ ρ ∂ ⎝ ⎠⎝ ⎠ag

grad (2.100)

Solving equation (2.100) is not straight forward. Indeed, a solution with CVM or FEM is only possible in combination with air and heat balances. Further discussion is therefore limited to � at walls in steady-state. Also constructions with ventilated cavities are dealt with.

One-dimension: Flat Walls

In steady-state, the derivative with time in equation (2.100) evaporates. All contact resistances are set at zero. To get a one-dimensional situation, air and vapour � ows must develop perpendicular to the wall. In that case, equation (2.100) simpli� es to:

a

a

d 1 d0

d d

gpp

x N x R T

⎛ ⎞− =⎜ ⎟μ ρ⎝ ⎠

(2.101)

Let the vapour resistance factor () be a constant, i.e., no layer in the wall is hygroscopic. Normally, the diffusion constant N is a function of temperature. However, as for diffusion, N may be set as a constant ( 5.4 � 109 s–1).

Isothermal Regime

Equation (2.101) is transposed to the [Z, p]-plane:

2 2a

P2 2a

d d d d0

d dd d

gp p p pa

R T Z ZZ Z− = − =ρ

with

a aP

a a

0.62g ga

R T P= ≈

ρ

The result is a differential equation of 2nd order in Z, which holds for single-layered as well as for composite walls. Solution:

1 2 pexp( )p C C a Z= + (2.102)

The integration constants C1 and C2 ensue from the boundary conditions. For the diffusive vapour � ow rate, the convective vapour � ow rate and total vapour � ow rate, the result is:

Diffusion v P 2 Pd

exp( )d

pg a C a Z

Z= − = −

Convection vc P P 1 2 P[ exp( )]g a p a C C a Z= = +

Total vT v vc P 1g g g a C= + =

While the diffusion and convection � ow rates vary along the wall, total vapour � ow rate remains constant. When the vapour pressures on the wall surfaces are known (Z = 0: p = ps1; Z = ZT = � ( N d): p = ps2), then solving the system: ps1 = C1 + C2, ps2 = C1 + C2 exp (aP ZT) gives as integration constants:

1335vch02.indd 184 12.02.2007 16:55:22

185

s2 s1 P T1

P T

exp( )

1 exp( )

p p a ZC

a Z

−=

−

s2 s12

P T1 exp( )

p pC

a Z

−=

−

Vapour pressure and vapour � ow rate in the wall so become:

s1 s2 s1 v1( ) ( ) p p p p F Z= + −

v v2 s2 s1( ) ( ) g F Z p p= − vc Pg a p= vT s2 v2 s1 v2 T(0) ( )g p F p F Z= −

with:

Pv1

P T

1 exp( )( )

1 exp( )

a ZF Z

a Z

−=

−

P Pv2

P T

exp( )( )

1 exp( )

a a ZF Z

a Z=

− (2.103)

Combined diffusion and convection changes vapour pressure in the [Z, p]-plane from a straight line through the points [0, ps1] and [ZT, ps2] to an exponential through these points. When the air � ows from high to low vapour pressure, then the curve is convex. In the opposite case, it is concave (Figure 2.37). The higher the coef� cient aP, i.e., the larger the air � ow, the more concave or convex the curves. The vapour pressure line in the [x, p]-plane is obtained by transferring the successive crossing points between the exponential and the interfaces from the [Z, p]-plane and linking with exponential segments.At � rst sight, the results are similar to those found for combined heat and air � ow. But there is an important difference:

Figure 2.37. Advection under isothermal steady state conditions. At the left vapour pressure in a single layered wall in the [x, p]-plane. At the right vapour pressure in a composite wall in the [Z, p]-plane.

2.3 Vapour Transfer

1335vch02.indd 185 12.02.2007 16:55:22

186 2 Mass Transfer

Example

Take a single-layered wall made of mineral � bre, 10 cm thick, thermal conductivity 0.036 W/(m · K), air permeability 8 � 10–5 s, vapour resistance factor 1.2. For an air pressure difference of 1 Pa, the air � ow rate is 0.0008 kg/s (= 2.4 m3/h). The exponent for x = d then becomes:

thermal : 2.2 hygroscopic: 3.2

or, even for a well insulated, extremely vapour-permeable wall, vapour pressure is more in� uenced by air � ow than temperature. Thus, the vapour pressure line will be more convex or concave than the temperature line. For less insulating, air permeable walls, that difference becomes even more pronounced.

The increase in total vapour � ow rate compared to diffusion (gv0) becomes: 100 (gvT – gv0) / gv0 = 100 [ZT abs (Fv2 (ZT)) – 1]. If the air migrates from high to low vapour pressure, then abs (Fv2 (ZT)) passes 1 and total vapour � ow rate increases. When the air takes the opposite direction, then even a limited in� ow may invert the resulting vapour � ow: no longer from high to low but from low to high vapour pressure.

Non-isothermal Conditions

Both saturation (psat) and vapour pressure lines have to be considered now, the � rst as a transposition of the exponential temperature curve. If the two do not intersect, interstitial condensate is not deposited and the isothermally calculated vapour pressure line remains unchanged. If, however, both intersect, condensation is a fact. In case the rate is small enough to neglect the heat of evaporation (heat source) only the vapour pressure line in the wall changes, not the temperatures and saturation pressures. Indeed, at the spot where interstitial condensation begins, vapour pressure must equal saturation pressure for identical in and outgoing vapour � ows. The same holds at the spot where condensation stops. So, at the limits of a condensation zone, we have:

satsat P sat P

d d

d d(1) (2)

p pp p a p a p

Z Z= − = −

(2.104)

The saturation pressure psat and diffusion resistance Z are unknown in both equations. Combining the two simpli� es the second to:

satdd

d d

pp

Z Z⎡ ⎤⎡ ⎤ = ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

(2.105)

Hence, where a condensation zone starts and ends, vapour and saturation pressure have a common value and a common tangent. In practice, the solution of the system (2.104) to (2.105) is as follows:

1. First, the saturation and the vapour pressure curves are calculated.

2. Then, all interfaces where the vapour pressure exceeds the saturation pressure are tested for condensation, starting with the interface where the excess is highest. Let this be interface j. The vapour pressure line from surface s1 (= inner surface) to interface j then becomes (the diffusion resistance is calculated with surface s1 as the origin):

1335vch02.indd 186 12.02.2007 16:55:22

187

s1 sat, s1 v1 s1( ) ( )jp p p p F Z= + −

with

P s1v1 s1

P s1

1 exp[ ]( )

1 exp[ ( )]

z

j

a ZF Z

a Z

−=

−

with s1jZ the diffusion resistance between surface s1 and interface j, and s1 zZ the diffusion

resistance from s1 to a plane Z between s1 and j. If this new vapour pressure line does not intersect the saturation pressure between j and s1, then it stands for the correct ingoing exponential during condensation. For the outgoing vapour pressure, an analogous reasoning applies, except that the boundary conditions are now saturation pressure psat,j in interface j and the vapour pressure at surface s2:

sat, s2 sat, v1( ) ( )j jp p p p F Z= + −

with

Pv1 s2

P

1 exp( )( )

1 exp( )

zj

j

a ZF Z

a Z

−=

−

with s2jZ the diffusion resistance between interface j and surface s2, and z

jZ the diffusion resistance from interface j to Z, now between j and s2. Without intersection with the saturation pressure, this new vapour pressure line represents the outgoing exponential during condensation.

3. In case the new ingoing vapour pressure exponential intersects saturation pressure, a step by step procedure allows � nding the correct solution. If the intersection zone contains no interfaces, then a point l on the saturation curve between s1 and j is looked for, where:

P P s1 satsat, s1

P s1

exp[ ] d( )

d1 exp[ ]

l

l ll

a a Z pp p

Za Z

⎡ ⎤− = ⎢ ⎥− ⎣ ⎦

If present, condensation starts there. With interfaces, these have to be tested first on condensation. The same reasoning holds between j and s2, giving the additional condensation interfaces or the endpoint k of the condensation zone:

s2P P sat

s2 sat, s2P

exp ( ) d( )

d1 exp( )k

kkk

a a Z pp p

Za Z

⎡ ⎤− = ⎢ ⎥− ⎣ ⎦

In case k and 1 do not coincide or no continuous condensation zone stretches between both, then additional vapour pressure tangent exponentials exist. In such case, the wall contains two or more condensation zones or interfaces.

With air � owing from low to high temperature and low to high vapour pressure the more concave vapour pressure line never intersects the less concave saturation curve, making interstitial condensation impossible in such a case. However, when the air � ows from high to low temperature and high to low vapour pressure, then a high enough vapour pressure at the warm side unavoidably ends in interstitial condensation (see Figure 2.38). At the same time, the air � ow changes reaction times, from very slow for diffusion only, to almost instantaneously.The above methodology, which is called the advection method, gives an answer to the following questions:

1. Vapour pressure line in case of interstitial condensation

See above.

2.3 Vapour Transfer

1335vch02.indd 187 12.02.2007 16:55:22

188 2 Mass Transfer

2. Condensation zones

Air ingress may increase the number of condensation zones in a wall compared to diffusion.

3. Condensation � ow rate

In case of a single condensation interface, the rate is given by (s1 is outside and s2 inside):

sat,1 sat,1s2 s2s1 s1 s1 s1

sat,1 sat,1s1 s1

s2 sat,1c s2 v2 sat,1 v2 s1 s1 sat,1

sat,1sat,1 v2 s1 v2 s1

[ (0) ( ) ]

(0) ( )

− −= − −

− +

Z Z Z Z

Z Z

g p F p F Z Z p

p F p F Z

(2.106)

withsat,1s2

s1 s1 Pv2 s2 sat,1

P s1 s1

(0)1 exp[ ( )]

− =− −

Z Z aF

a Z Z sat,1s2

s1 s1s2 sat,1

s2 sat,1 P P s1 s1v2 s1 s1 s2 sat,1

P s1 s1

exp[ ( )]( )

1 exp[ ( )]

− −− =

− −Z Z a a Z Z

F Z Za Z Z

sat,1s1 P

v2 sat,1P sl

(0)1 exp[ ]

=−

Z aF

a Z

sat,1s1

sat,1sat,1 P P sl

v2 s1 sat,1P sl

exp[ ]( )

1 exp[ ]=

−Z a a Z

F Za Z

Figure 2.38. Advection under non-isothermal steady state conditions. Vapour pressures in the com po-site wall (thin line) intersects vapour saturation pressure. The wall thus suffers from interstitial con den-sa tion. The correct vapour pressure line shows condensate to be deposited at the backside of the veneer.

1335vch02.indd 188 12.02.2007 16:55:23

189

where psat.l is the saturation pressure in the condensation interface. Firstly, compared to diffusion, the amounts of condensate increase signi� cantly with increasing air out� ow. At the same time, the temperature in the condensation interface climbs to higher values which gradually slows down the rise in condensate, until a maximum is reached whereafter the deposit drops quickly to zero once the temperature in the condensation interface nears the dew point at the warm side (Figure 2.39).

4. How to avoid interstitial condensation

Combined diffusion and convection provides two options:The wall is air-tight with a correct combination of diffusion resistances in its composing layers. With respect to energy ef� ciency, this is an excellent choice.The wall is so air permeable that the air out� ow heats the condensation interface beyond the dew point at the warm side. That option, however, does not serve energy ef� ciency!

Other possibilities are:In moderate climates, combining good ventilation with low vapour release indoors may keep vapour pressure in the building below the saturation value noted in the coldest condensation interface inside the envelope.Maintaining the building in an under-pressurized state in cold and moderate climates will limit air migration to in� ow only and exclude interstitial condensation. In warm, humid climates, the building must be over-pressurized to gain the same bene� t.

In diffusion mode, the Glaser diagram allowed � xing the vapour pressure at the warm side which excludes interstitial condensation. Also, the vapour retarder needed at the warm side or the vapour permeability of the outside cladding for avoiding interstitial condensation could be deduced. Hence, in advection mode the only justi� cation is for the maximum allowable vapour pressure at the warm side. The rules with respect to vapour retarders and vapour permeable claddings have no effect anymore. Thus, the most important measure is: make the envelope as air-tight as possible.

�

�

�

�

Figure 2.39. Interstitial condensation by advection in a composite wall, deposit in kg/week as function of the air out� ow rate.

2.3 Vapour Transfer

1335vch02.indd 189 12.02.2007 16:55:23

190 2 Mass Transfer

2.3.5 Surface Film Coef� cients for Diffusion

Until now, we assumed the vapour pressures at the bounding surfaces (ps) of a construction part to be known. Yet, measuring the vapour pressure on a surface is an impossible task. Logging vapour pressures in the air through temperature and relative humidity measurements, instead is not a problem. The preference therefore goes to using the vapour pressures in the air as a boundary condition and to describe vapour transfer through a construction part from environment to environment. The question then is: does water vapour, and in general, each gas, experience resistance when migrating from the air to a surface and vice versa, just as heat does? The answer is yes. A laminar air layer, called the boundary layer, in which diffusion is the only transport mode possible, sticks to each surface. Up to this layer, vapour transfer occurs mainly by convection. Vapour migration through the boundary layer is described by an equation of the form:

v s( )g p p= β − (2.107)

where p is vapour pressure outside the boundary layer (the reference in practice is the vapour pressure in the centre of the � oor at a height of 1.8 m), and ps the vapour pressure at the surface. � is called the surface � lm coef� cient for diffusion, units s/m. The reverse 1/� is the surface resistance for diffusion; symbol Z, units m/s. The suf� x i will be used to indicate inside (�i), the suf� x e to indicate outside (�e).The surface � lm coef� cient for diffusion is calculated the same way as the surface � lm coef� cient for heat transfer by convection. The basic equations are the mass balance for air, momentum equilibrium as expressed by the Navier-Stokes equations, the turbulence equations and conservation of water vapour. Solving this system of 7 partial differential equations can be done analytically, numerically or by means of similarity. The last option generates a set of dimensionless numbers, which are used to quantify the surface � lm coef� cient for diffusion:

Forced convection Mixed convection Free convection

Reynolds number:

Rev L=ν

Re / Gr1/2

Grashoff’s number:3

a2

Grg LΔρ

=ν

Sherwood’s number:

a

ShLβ=

δ

replaces Nusselt’s number

Idem

Schmidt’s number:

a

ScR T

ν=δ

replaces Prandl’s number

Idem

1335vch02.indd 190 12.02.2007 16:55:23

191

Like the Nusselt number Sherwood stands for the relationship between total vapour transfer and the vapour transfer by diffusion. Schmidt’s number shows the � eld of � ow velocities to be similar in shape to that of vapour pressures. Between the numbers identical relations as those found for convective heat transfer exist. Thus, forced laminar � ow along a horizontal � at wall results in a surface � lm coef� cient for diffusion given by: 1/2 1 / 3

L LSh 0.664 Re Sc= . The similarity with heat convection leads to the following relationship between both surface � lm coef� cients (Lewis equation):

0.67a pc

a a p a

R T ch

R T c

δ⎛ ⎞β = ⎜ ⎟ρ λ⎝ ⎠

(2.108)

with �a the density, Ra the gas constant, cp the speci� c heat at constant pressure and a the thermal conductivity of air. Under atmospheric circumstances, i.e., for most building applications, equation (2.108) may be simpli� ed to:

ca

1h

Nβ =

λ (2.109)

Equation (2.109) ensues from the boundary layer theory, assuming it is equally thick for heat and vapour transfer. That condition is met when the Schmidt and Prandl numbers have the same value. Let d be the common thickness (Figure 2.40).

For heat, we have:

s aa c s

c

( ) orq h dd h

θ − θ λ= λ = θ − θ =

Vapour gives:

sv s

1( ) or

p pg p p d

N d N

−= = β − =

βEquating both results in equation (2.109). For air, with a diffusion constant N 5.4 � 109 s–1 and a thermal conductivity a 0.025 W/(m · K), equation (2.109) reduces to:

9c7.7 10 h−β ≈ ⋅ (2.110)

Figure 2.40. Surface � lm coef� cient for diffusion, the boundary layer theory.

2.3 Vapour Transfer

1335vch02.indd 191 12.02.2007 16:55:24

192 2 Mass Transfer

Also see Table 2.8.

For hand and spreadsheet calculations, the following two values are typically used:

Inside: � = 18.5 � 10–9 s/m; Z = 54 � 106 m/s

Outside: � = 140 � 10–9 s/m; Z = 7.2 � 106 m/s

As for the thermal resistance, the diffusion resistance from environment to environment through a � at wall becomes:

Outside wall: a i T eZ Z Z Z= + +

Inside wall: a i T2Z Z Z= +

giving as steady-state diffusion � ow rate:

Outside wall: i ev

i T e

p pg

Z Z Z

−=

+ +

Inside wall: i1 i2v

i T2

p pg

Z Z

−=

+

Vapour pressure on both bounding surfaces looks like:

Inside surface: si i v ip p g Z= −

Outside surface: se e v ep p g Z= +

Nevertheless, there is a big difference between the surface � lm coef� cients for heat transfer (Ri and Re) and the ones for diffusion (Zi and Ze).

Table 2.8. The inside and outside surface � lm coef� cient for diffusion.

Surface � lm coef� cient for diffusion �i(pa = 1 Atm, 0 � �i � 20 °C)

Surface � lm coef� cient for diffusion �e(pa = 1 Atm, –20 � �i � 30 °C)

�i – �siK

�� 10–9 s/m

vam/s

�e� 10–9 s/m

2 4 6 8101214161820

28.630.031.432.834.236.037.438.840.241.6

< 15

5 to 1025

� 110212280849

�i = [27 + 0.73 (�i – �si)] � 10–9

r2 = 1�e = 49.9 � 10–9 va

0.875

r2 = 0.998

1335vch02.indd 192 12.02.2007 16:55:24

193

Example

Assume a moderately insulated, vapour permeable wall, made of air-dry cellular concrete, � = 480 kg/m3, d = 0.3 m, � = 0.15 W/(m · K), � = 5. Thermal resistance and diffusion resistance are:

R = 0.3 / 0.15 = 2 m2 · K/W Z = 0.3 · 5.5.4 · 109 = 8.1 � 109 m/s

Thermal resistance and diffusion resistance from environment to environment:

Ra = 2 + 0.168 = 2.168 m2 · K/W Za = 8.1 � 109 + 61.2 � 106 = 8.16 � 109 m/s

In terms of percentage, the surface resistance counts for 7.7% of the thermal resistance, but only for 0.7% of the diffusion resistance. In other words the share of the surface resistance in the total resistance is much smaller for vapour than for heat.

The conclusion is clear: including the surface resistances when studying diffusion through a construction gives such small differences that vapour pressure on a surface hardly differs from vapour pressure in the adjacent air! Thus, replacing surface vapour pressure by vapour pressure in the air and setting the surface resistances at zero has hardly any effect on the result of a Glaser or a convection + diffusion-calculation, also called an advection calculation.Things change when the vapour � ow rate at a surface is the looked for quantity. Then, the surface � lm coef� cient for diffusion plays a decisive role. This is the case for drying, surface condensation and sorption. With ps the vapour pressure at the surface and p the vapour pressure in the air, vapour � ow rate to and from a surface becomes:

v s( )g p p= β − (2.111)

2.3.6 Some Applications

2.3.6.1 Diffusion Resistance of a Cavity

If the air or the gas in a cavity has the same temperature, pressure and composition everywhere, then water vapour moves by diffusion only and the diffusion resistance of that cavity (�a = 1) becomes:

Zc = N d (2.112)

If, instead, gas pressure, temperature and composition vary along the cavity, then convection will short-circuit diffusion, reducing the diffusion resistance of the cavity to the sum of the surface resistances at the bounding surfaces:

1 2c 1 2

1 2

Z Z Zβ + β

= + =β β

(2.113)

This value is typically so small compared to the diffusion resistance of all other layers in a part with cavity, that it may be set at zero. Of course, in cases where the vapour � ow rates at the surfaces are the looked for quantities, this cannot be done. As an example of this, we have ventilation of a cavity behind a wet veneer wall.

2.3 Vapour Transfer

1335vch02.indd 193 12.02.2007 16:55:24

194 2 Mass Transfer

2.3.6.2 Cavity Ventilation

A calculation of the effect of ventilating a cavity (think of a partially � lled cavity wall with open head joints below and at the top of the veneer) can be done with an approximate model, combining diffusion through the wall with air � ow along the cavity (Figure 2.41). Assume the cavity is dcav m wide and L m long. The air velocity in the cavity is v m/s, while the surface � lm coef� cient for diffusion at both surfaces is � (s/m). The diffusion resistance from the environment (vapour pressure p1) to the cavity of the leaf at one side is Z1 m/s, and the diffusion resistance from the environment (vapour pressure p2) to the cavity of the leaf at the other side is Z2 m/s.At both bounding surfaces, the vapour balances are (vapour balances resemble heat balances: the surface functioning as a calculation point, all masses � owing from the environment to the calculation points):

Cavity surface 1: 1 s1cav s1

1

( ) 0p p

p pZ

−+ β − =

Cavity surface 2: 2 s2cav s2

2

( ) 0p p

p pZ

−+ β − =

with a solution of:

1cav

1s1

1

1

pp

Zp

Z

+ β=

+ β

2cav

2s2

2

1

pp

Zp

Z

+ β=

+ β

(2.114)

where pcav is the vapour pressure in the cavity. The � rst term in both balances describes vapour transfer through both leafs from the environment to the related cavity surface. The second term concerns the vapour transfer between the air in the cavity and the cavity surface.

Figure 2.41. Cavity ventilation, a combination of one-dimensional diffusion through the inside leaf and the veneer with air � ow along the cavity.

1335vch02.indd 194 12.02.2007 16:55:24

195

Vapour balance in the cavity:

cavs1 cav s2 cav cav

cav

[ ( ) ( )] d dd v

p p p p z pR T

β − + β − =

or, the vapour � ow rates coming from the cavity surfaces change the vapour content in the passing air� ow. If the cavity temperature Tcav is replaced by its asymptote Tcav,� and the surface vapour pressures ps1 and ps2 by the equations (2.114), then the cavity balance becomes:

1 2

1 2 2 1 cavcav cav

cav, 1 11 2 2 1 1 2

1 1

d d1 1 1 1 1 1

p p

Z Z Z Z d vp z p

R TZ Z Z Z Z Z

∞ − −

⎡ ⎤⎛ ⎞ ⎛ ⎞+ β + + β⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥− =⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ β + + β⎢ ⎥ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ + β + β⎝ ⎠⎣ ⎦

If vapour pressure on surface 1 is known, which is the case for a wet veneer, then that balance simpli� es to:

21

2 2 cavcav cav

cav, 122

1

d d2 1

pp

Z Z d vp z p

R TZ Z −∞

⎡ ⎤⎛ ⎞+ β +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥− =⎢ ⎥ ⎛ ⎞+ β⎢ ⎥ β +⎜ ⎟⎢ ⎥ + β⎝ ⎠⎣ ⎦

Setting

21

2 2

2

1

2

pp

Z Za

Z

⎛ ⎞+ β +⎜ ⎟⎝ ⎠

=+ β

and

cav

cav, 12

1

d vb

R TZ −∞

=⎛ ⎞β +⎜ ⎟+ β⎝ ⎠

,

the solution becomes:

cavlna pz

b C

−= −

with C the integration constant. The cavity height or length in� nite gives z / b = �, or: cav,ln [( ) / ]a p C∞− = −∞ which transposes to cav, 0a p ∞− = and 1 cav,a p ∞= . So, the

number a represents the asymptote value of the vapour pressure in case of an in� nite cavity. It also represents the vapour pressure in a non-ventilated cavity. Indeed, for v = 0, z / b again becomes �. For z = 0, pcav equals pcav,0, giving: cav, cav,0ln [( ) / ] 0p p C∞ − = , or

cav, cav,0C p p∞= − . Hence, the solution can be written as:

cav , , cav,01

( ) expcav cavz

p p p pb∞ ∞

⎛ ⎞= − − −⎜ ⎟⎝ ⎠

(2.115)

In a ventilated cavity, the vapour pressure changes exponentially, from the in� ow value to the value without ventilation. In the equation, b has the dimensions of a length and is called the ventilation-in� uence length. The smaller the air velocity and the higher the vapour transfer to the cavity, the smaller that length.

2.3 Vapour Transfer

1335vch02.indd 195 12.02.2007 16:55:24

196 2 Mass Transfer

One of the questions posed with outside walls is: how much water vapour cavity may ventilation carry away? The approximate answer is:

cav,L cav,0 cavcav cav, cav,0

cav, cav, 1

( ) 1 expv

p p b v LG b v p p

R T R T b∞∞ ∞

⎡ ⎤− ⎛ ⎞= = − − −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

(2.116)

Drying becomes minimal, when the ventilation � ow is small and the vapour pressure of the incoming air is hardly different from the vapour pressure in a non-ventilated cavity.

2.3.6.3 Water Vapour Balance in a Room in Case of Surface Condensation and Drying

In case vapour buffering does not intervene, we learned to write the water vapour balance in a room as:

ve a vP vc / vi a( )

dvi l

dd x M

x G G G x Gt

+ + = +

In the discussion on mold and dust mites, the term Gvc / d, indicating surface condensation and surface drying was overlooked. If present, the balance can be written as follows (for condensation or drying restricted to one surface):

i ie vP i i sat,A i

d1[ ( )]

d

R T pp G A p p p

n V n t+ − β − = + (2.117)

Solutions

In a steady-state (GvP = Cte, n = Cte, pe = Cte, dpi / dt = 0), we get:

i ie vP i sat,A

i vP i sat,A ei e

i ii i

[ ( )]

1 1

R T R Tp G A p

R T G A p pn V n Vp p

R T R TA n V An V n V

+ + β + β −= = +

⎛ ⎞+ β + β⎜ ⎟⎝ ⎠

or, with ii1

R TC A

n V= + β :

ii io vP i sat,A e[ ( 1) ( )]

R Tp p G C A p p

n V C= − − + β − (2.118)

Here, pio is the vapour pressure inside without surface condensation or surface drying.For the same vapour production and ventilation rate, vapour pressure inside is clearly lower with than without surface condensation. The difference increases with the size of the condensation surface and a lower value of the surface’s saturation pressure. Some use that conclusion to argue that replacing well insulating glass panels by single glass may cure mold problems. This is a regrettable conclusion. Surface condensation only generates a limited drop in vapour pressure, compelling to the use of large surfaces of single glass, while drying gives the opposite effect; a vapour pressure increase. The large surfaces of single glass needed also jeopardize thermal comfort and increase the heating bill.

1335vch02.indd 196 12.02.2007 16:55:25

197

The transient solutions remain the same as those without condensation or drying, provided that equation (2.118) is used for the asymptote pi� and that the ventilation rate in the exponential is replaced by the product of ventilation rate and constant C.Surface condensation and drying go hand in hand with latent heat release and latent heat absorption. Per m2:

i sat,A( ) bq p p l= β −

with lb the latent heat in J/kg.

2.4 Moisture Transfer

2.4.1 Overview

In the previous section, we assumed moisture content in a material to be low enough to only support vapour transfer. This is true below a relative humidity threshold m. Sorption beyond that value, interstitial condensation, rain absorption, built in moisture and rising damp all push the moisture content into the “above hygroscopic” zone. Once there, aside from vapour transport, unsaturated water � ow develops with capillary suction, gravity, external pressure differences and internal pressure differences as driving forces (caused by the presence of water, air and mechanical stress). The term moisture transfer covers both vapour and unsaturated water � ow, and includes all dissolved substances (such as salts).Air and vapour transfer were analysed in a phenomenological way, introducing macroscopic material properties such as the vapour resistance factor, the air permeability and others. These are meaningful beyond the scale of a representative material volume, a term referring to the smallest volume which has the same properties as the material. For example, a pore is not a representative volume, simply because it does not contain any material but is a microspace � lled with moist air, water and dissolved substances. The size of a representative volume depends on the type of material. For a � ne-porous, homogeneous material, it may be microscopically small. For highly heterogeneous materials, such as concrete, even a few cm3 may not be representative.Moisture � ow analysis starts at the microscale. Then the results are upgraded to a representative volume and larger. Therefore moisture transfer in a pore is studied � rst, after which we move to porous materials.

2.4.2 Moisture Transfer in a Pore

2.4.2.1 Capillarity

Although the phenomena described in this section are seen whenever a gas x, a liquid y and a material z contact each other, we restrict the discussion to the trio of air/water/pore wall. In a pore � lled with water and air, capillary action originates from the cohesion between the water molecules (i.e. the attraction between molecules of the same substance), the impairment of that cohesion in the contact between water and air and the adhesion between the water and air molecules and the pore-wall.

2.4 Moisture Transfer

1335vch02.indd 197 12.02.2007 16:55:25

198 2 Mass Transfer

Surface Tension

Cohesion causes surface tension. The following simple model, shown in Figure 2.42, helps to understand that. Take a water molecule deep in the � uid. All surrounding molecules attract it. Of course, for those farther away that attraction weakens to such extent that, once beyond a distance of 1 nm (= 10–9 m), not much attraction is left. In other words, each water molecule can be represented by a sphere with a radius of 1 nm. In case that sphere lies deep enough under the contact surface with air, no resulting attraction is felt. If the sphere intersects the contact surface and the attraction between air and water is smaller than the cohesion in the water, then a resulting force oriented towards the water occurs. This force attaches the water surface, whose thickness is only a little larger than the sphere of a water molecule, to the underlying liquid. If we want to separate them, that force has to be broken. For that purpose work is required, or, the potential energy of the thin surface layer, called the meniscus, is larger than of the liquid below. The energy difference per surface-unit is named the surface energy, symbol w, units J/m2 (the suf� x w indicates water). Given that 1 J/m2 equals 1 N/m, the term surface tension is also used. The surface tension for water in contact with air decreases with temperature: w = (75.9 – 0.17 �) � 10–3.

Figure 2.42. How surface tension comes into being.

Suction

If a water meniscus touches a pore wall, then, provided the adhesion to the wall is larger than the attraction between water and air, the meniscus creeps up against the wall with an angle of �. In such case, that angle, called the contact angle, lies between 0 and 90° (Figure 2.43). Creeping up occurs in practically all stony and wood-based materials. Therefore they are called hydrophilic. If instead, adhesion to the wall is smaller than the attraction between water and air, then the meniscus is pushed away. This produces a contact angle between 90 and 180° (Figure 2.43). That occurs in most synthetics and all water repellent surfaces. They react hydrophobically.When a circular pore in a hydrophilic material contacts water, then a raised meniscus develops all around at the pore wall. For a suf� ciently small diameter, those raised menisci touch each other, forming a concave water surface. For ever smaller diameters, that concave surface pulls the water into the pore, creating capillarity. The capillary force per unit surface exerted by the surface, which is called suction pc or s with Pa as units, is (Figure 2.44):

wc

4 cosp

d

σ ϑ= − (2.119)

For a crack with width b suction becomes (Figure 2.44):

wc

2 cosp

b

σ ϑ= − (2.120)

1335vch02.indd 198 12.02.2007 16:55:25

199

The formulas show that suction is inversely proportional to the pore radius or crack width. The nature of the gas and liquid intervenes via the surface tension (here water and air) while the material plays a role through the contact angle �. Suction rapidly becomes important. For example, in an ideal hydrophilic pore with diameter 1 �m (� = 0), 300,000 Pa is attained, which is the pressure exercised by a water column of 30 m. As a comparison, wind pressure on a wall under stormy conditions hardly passes 1000 Pa. At any rate this 1000 Pa may push water through a 0.15 mm wide crack. Figure 2.45 gives the suction in an ideal hydrophilic pore that contacts water: zero at the water surface, equal to pc (–4 w / d) at the concave water surface or meniscus, in between given by pc x / L with L the wet length along the pore and x the distance from the water surface. As the air above the meniscus experiences the same pressure as the air above the � at water surface, suction also equals the difference between the water pressure right below and the air pressure just above the meniscus: pc = Pw – Pa.Capillary repulsion (or negative suction) by hydrophobic materials is also inversely proportional to the pore diameter or crack width. A water repellent surface with micro-cracks more than 0.15 mm may thus be short-circuited by wind. Just as suction, repulsion equals the difference between the pressure in the water right under and the pressure in the air just above the meniscus.

Figure 2.43 (a) Water is pulled up against a pore wall, giving a contact angle between 0° and 90°. (b) Water is pushed away from the pore wall, giving a contact angle between 90° and 180°.

Figure 2.44. Capillary suction in a circular pore and a crack.

2.4 Moisture Transfer

1335vch02.indd 199 12.02.2007 16:55:25

200 2 Mass Transfer

2.4.2.2 Water Transfer

Poiseuille’s Law

Fluids move under the in� uence of pressure differences. In a capillary active pore, suction induces such differences. The nature of the movement or � ow is determined by the Reynolds number: laminar, transitional or turbulent. Water at 10 °C has a kinematic viscosity � of 1.25 � 10–6 m/s2, giving a Re-number of 800,000 v dH with dH the hydraulic diameter (the diameter for a circular pore, twice the width for a crack). The � ow remains laminar as long as Re stays below 2000, which means as long as the velocity stands below 0.0025 / dH. This relationship shows that turbulent � ow in pores is only possible at supersonic water velocities (2500 m/s voor d = 1 �m!!).

Thus, the � ow must be laminar and the water � ow rate should be given by:

2w H

w ww

grad32

dg P

ρ= −

η (2.121)

with �w the dynamic viscosity of water, units N � s/m2, value:

Temperature �wN � s/m2

0 0.001820

20 0.001025

100 0.000288

For a circular pore, the proof is as follows. Laminar � ow without surface slip has a water velocity zero at the pore wall. For each concentric � ow cylinder with length dy, the equilibrium between the pressure difference and the viscous friction along the perimeter is (Figure 2.46):

2w w

dd 2 d

d

vr P r y

rπ = π η , giving

w

w

d1d ( d )

2 d

Pv r r

y=

η

Figure 2.45. Suction in an ideally hydrophilic pore (cos � = 1).

1335vch02.indd 200 12.02.2007 16:55:26

201

Integration:

0w

w

d1d d

2 d

R

v r

Pv r r

y=

η∫ ∫

or

2 2 w

w

d1( ) ( )

4 d

Pv r R r

y= − −

η

The water � ow through the pore thus becomes:

4w

w w ww0

d( ) 2 d

8 d

R PRG v r r r

y

π= ρ π = −ρη∫

giving as an average rate:

2w w

w w2w

4 d

32 d

G Pdg

yd= − = −ρ

ηπ

For a pore with circular section, the diameter d equals the hydraulic diameter dH. That closes the proof.Equation (2.121) may be extended to any � uid. It suf� ces to replace the density �w and the dynamic viscosity �w of water by the ones of the � uid. 2

H32 / dη is called the speci� c � ow resistance of a pore, symbol Wm, units kg/(m3 · s). The property differs among � uids by viscosity. The speci� c � ow resistance for air, for example, is seventy times smaller than the speci� c � ow resistance for water (�a = 1.74 � 10–5 N � s/m2 versus �w = 1.25 � 10–3 N � s/m2). The property can be made � uid-independent by rewriting equation (2.121) as:

2w wH

wd d1

32 d d

P Pdg

y W y

⎛ ⎞ρ ρ= − = − ⎜ ⎟⎝ ⎠η η ′

with W� the generalized speci� c � ow resistance of a pore.Reshuf� ing equation (2.121) in terms of water pressure differences gives:

w w ww m2 2

w

32 d 8 dd

y g yP v

d r

⎛ ⎞η η= − = −⎜ ⎟ρ⎝ ⎠

with vm the average water velocity in the pore (= dy / dt).

Figure 2.46. Poiseuille’s law, an equilibrium between pressure difference and viscous friction.

2.4 Moisture Transfer

1335vch02.indd 201 12.02.2007 16:55:26

202 2 Mass Transfer

Isothermal Water Transfer in a Pore That Contacts Water

Let us take a pore with circular section, radius r, and length L. The pore has a slope � to the vertical and contains air. One side contacts water. Application of Newton’s law on the sucked water column gives (Figure 2.47):

22

i w 2

d( )

d

lF r l

t= ρ π∑ (2.122)

with �w � r2 l the mass of the water column, l the ordinate of the meniscus in the pore and d2l / dt2 the acceleration. The most important intervening forces Fi are:

Gravity:2

g cosF g r l= −ρ π α

Capillary suction:

2 wc w

2 cos2 cosF r r

r

σ ϑ= π = π σ ϑ

Water friction:

2fw w2

8 d d8

d dw l l l

F r lt tr

η ⎛ ⎞ ⎛ ⎞= − π = − π η⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Air friction above the meniscus:

fa ad

8 ( )d

lF L l

t⎛ ⎞≈ − π η − ⎜ ⎟⎝ ⎠

The sign � indicates that Poiseuille’s equation is an approximation for air. Indeed, air is a compressible � uid. As a consequence, when suction starts, a peak in air pressure will be noted.

Figure 2.47. Suction by a pore from a � at water surface.

1335vch02.indd 202 12.02.2007 16:55:26

203

That in� uences the initial water � ow. In any case, introducing the four forces into the dynamic equilibrium (2.122) gives as an (approximate) equation of motion:

2w a w

2 2w ww

8 ( ) 2 cosd d1 cos 0

dd

L ll lg

l t r lt r

⎡ ⎤ ⎛ ⎞η η − σ ϑ+ + − − α =⎢ ⎥ ⎜ ⎟η ρ⎝ ⎠ρ ⎣ ⎦

(2.123)

i.e. a differential equation of second order with variable coef� cients. That equation cannot be solved analytically. However, the second derivative hardly plays a role. In fact, because of the importance of friction (� 1 / r2!) the water meniscus moves so slowly that shortly after the start acceleration becomes negligible. So, for t > 0 + dt, the equation of motion simpli� es to:

2w w

mw

w a

2 cosd 1cos

( )d 8

rlv g

L lt r ll

⎡ ⎤⎢ ⎥ ⎡ ⎤ρ σ ϑ

= = − α⎢ ⎥ ⎢ ⎥− ρ⎣ ⎦⎢ ⎥η + η⎣ ⎦

(2.124)

The term with the two dynamic viscosities �w (water) and �a, (air) acts as a variable viscosity, in� nite large at the beginning and approaching the viscosity of water as suction progresses. The equation shows the existence of a maximum suction length lmax. Indeed, velocity goes to zero for max w w2 cos /( cos )l g r= σ ϑ ρ α . That length imposes a maximum suction height

max max cos( )h l= α (Figure 2.48).

The general solution of equation (2.124) is:

w a w a max a2

w a maxw

8 ( ) ( )ln 1

cos

l L lt l

lg r

⎧ ⎫⎡ ⎤ ⎛ ⎞η − η η − η + η⎪ ⎪= − − −⎨ ⎬⎢ ⎥ ⎜ ⎟η − η ⎝ ⎠ρ α ⎪ ⎪⎣ ⎦⎩ ⎭ (2.125)

with time as a function of the wet length! Figure 2.49 shows that relationship for a vertical pore, the model used to simulate rising damp in a capillary wall. Rain absorption by a capillary façade material instead is simpli� ed to water suction by a horizontal pore (Figure 2.50).

Figure 2.48. The maximum suction height.

2.4 Moisture Transfer

1335vch02.indd 203 12.02.2007 16:55:27

204 2 Mass Transfer

Horizontal Pore

In a horizontal pore with length L and radius r, the cosine of the slope � is zero. That turns equation (2.124) into:

w

w a a

cosd d

d d 4 [( ) ]

rl x

t t x L

σ ϑ= =

η − η + η

with a quadratic equation as solution: 2w a a w2 ( ) 4 cos 0x L x r tη − η + η − σ ϑ = .

Positive root:

2a a w

w a w a w a

cos

2 ( )

L L rx t

⎛ ⎞η η σ ϑ= − + +⎜ ⎟η − η η − η η − η⎝ ⎠

The wet thickness x not only depends on time t, but also on the radius and the length of the pore. The longer and narrower it is, the slower the water is absorbed. In other words, the velocity at which a porous facade material will absorb rainwater will decrease for a larger layer thickness and a � ner pore structure. If friction in the air column is neglected, the solution becomes:

Figure 2.49. Relationship between wet height and time for a 15 m long vertical pore (time and height on a logarithmic scale). Radius 1 �m, hmax = 14.78 m / radius 10 �m, hmax = 1.478 m / radius 100 �m, hmax = 0.1478).

Figure 2.50. Rain absorption and rising damp.

1335vch02.indd 204 12.02.2007 16:55:27

205

w

w

cos

2

σ ϑ=

������

rx t (2.126)

For short but wide pores, neglecting the air is of little in� uence. For � ner pores, the error increases substantially, especially at the start of the suction process as shown by Figure 2.51. The expression above the brace in equation (2.126) is called the capillary water penetration coef� cient; symbol B, units m/s1/2. Multiplying both parts by the water density gives:

w wm B t A t= ρ = wherein A = �w B and mw is the absorbed water in kg/m2.

The coef� cient A is called the capillary water sorption coef� cient, units kg/(m2 · s1/2). The smaller A is, the slower a horizontal pore is sucking and the longer it takes before the water reaches the other side.Both the correct and simpli� ed t -relationships show that horizontal suction never halts. Whatever the length of a pore is, the other end will be reached. Once there, the water meniscus straightens. This turns suction to zero and brings the system in equilibrium. Hence, an out� ow of water by capillary suction is impossible. Thus, rain absorption by capillary porous materials may result in wet stains inside but never in a water � ow along the inner surface. Such � ow can only be caused by outside pressures or gravitational forces. If that happens with rain, paths must be present where gravity causes in� ltration.The model of a horizontal pore with invariable section is easily expanded to pores with variable section. As an illustration, take plaster on any substrate. The two-layer-system can be modeled as two serial connected pores that contact each other. The � rst pore – the plaster – has a radius r1, a contact angle �1 and a length l1; the second pore a radius r2, a length l2 and a contact angle �2 (Figure 2.52). First, we neglect air friction. As long as the meniscus remains in the pore [r1, �1, l1], the following applies: 1m A t= . Water absorption is consequently slower for a smaller capillary water sorption coef� cient, i.e. as the pore becomes � ner and/or the contact angle larger. When the meniscus reaches the second, thinner pore, then that pore

Figure 2.51. Procentual deviation from the t -law if air friction in the capillary is taken into account (pore length: 30 cm, lowest curve for a radius of 1 �m, upper curve for a radius of 0.001 �m).

2.4 Moisture Transfer

1335vch02.indd 205 12.02.2007 16:55:27

206 2 Mass Transfer

takes over suction while the � rst pore starts functioning as hydraulic resistance. If the second pore is wider, then the meniscus straightens out in the � rst pore until the curve equals that of a meniscus in the second pore. The latter then takes over, while the � rst pore again becomes a hydraulic resistance only. With the water in the second pore, total friction is:

2w 1 2

fw w21

8 d8

d

l r xF x

tr

⎛ ⎞π η= + π η⎜ ⎟⎝ ⎠

(2.127)

where x is the wet length in the second pore. Solving the force equilibrium differential equation Ff = Fc for x gives:

22 2 w1 2

2w1

cos20

2

rl rx x t

r

⎛ ⎞ ⎛ ⎞σ ϑ+ − =⎜ ⎟ ⎜ ⎟η⎝ ⎠⎝ ⎠

with a positive root of:

22 221 2 1 222 2

1 1

l r l rx B t

r r

⎛ ⎞= − + +⎜ ⎟⎝ ⎠

Clearly, water absorption by the second pore (x) slows down when the ratio 2 21 2 1/l r r increases,

i.e. when the � rst pore is longer and � ner and the second wider. Hence, a coarse plaster cannot increase the rain resistance of a wall made of � ne porous material. Only a water repellent plaster (� � 90°, B very small) may move its wetting time beyond the maximum duration of a wind driven rain event.When air friction is not neglected, then the solution of the quadratic equation giving the position of the meniscus in the � rst pore is:

2a 1 a 2 1 a 1 a 2 1 w

2 2w a w a w aw a 2 w a 2

(1) (1)(2) (2)

cos

2 ( )( ) ( )

d d r d d r rx t

r r

η η η η σ ϑ⎡ ⎤= − + + +⎢ ⎥η − η η − η η − ηη − η η − η⎢ ⎥⎢ ⎥⎣ ⎦

����� ������������ �������

For the radius r2 < r1 and the layer thickness d2 � d1 the term (2) is many times larger than the term (1) and the movement of the meniscus in the � rst, wide pore is a lot slower than it was

Figure 2.52. A serial connected two pores model in case of water contact (for example: stucco on whatever substrate).

1335vch02.indd 206 12.02.2007 16:55:28

207

without air friction. Or, a coarse plaster on a � ne porous substrate will absorb water slower than the same plaster without substrate. But be careful with that conclusion. The pore model used is one-dimensional. The pore system in a porous material is three dimensional. Also surface humidi� cation never occurs uniformly. Even the out� ow of air is not one-dimensional.

Vertical Pore

For a vertical pore, the cosine of the slope � becomes 1 and equation (2.124) converts into:

2w w

mw

w a

2 cosd 1

d 8

rzv g

L zt r zz

⎛ ⎞ρ σ ϑ= = −⎜ ⎟− ρ⎝ ⎠η + η

Velocity goes to zero for w w2 cos / ( )r z gσ ϑ ρ = , i.e. for a maximum suction height, equal to:

wmax

w

2 cosz h

g r

σ ϑ= =

ρ (2.128)

Maximum height increases with smaller pore diameter and/or smaller contact angle. Thus, the � ner the pores and the more hydrophilic a material, the higher sucked water may rise. Instead, for a material with a contact angle of 90°, the height is zero. The same holds for very wide pores and materials without pores. Rising damp in capillary walls, in other words, can be prevented in three ways: (1) by including a coarse porous layer (r large) above grade, (2) by injecting a water repellent substance above grade which moves the contact angle beyond 90° (cos� between 0 and –1), (3) by inserting a water barrier with zero porosity above grade. The suction exerted by the coarse porous layer or the water repellent zone should stay below �w g h, with h the distance between grade or the water table and the upper side of the layer (Figure 2.53).

Figure 2.53. Combating rising damp, some possibilities.

2.4 Moisture Transfer

1335vch02.indd 207 12.02.2007 16:55:28

208 2 Mass Transfer

Isothermal Water Transfer in a Pore After Interrupting the Water Contact

Constant Contact Angle, Constant Section

Consider a pore with circular section in contact with water. After a time, the contact is interrupted. The result is a water island in the pore with a concave meniscus at both sides (Figure 2.54). In a horizontal pore, both meniscuses exercise the same traction. That excludes the water island from moving and water from being transferred. By inclining the pore, gravity makes the lower meniscus less and the upper more concave. Flowing starts when the water island is heavy enough to straighten out the lower meniscus, i.e. when its length equals the length of maximum rise lmax (lmax = hmax / cos �). Hence, the wider a pore, the easier gravitational � ow develops. The same holds for external pressures.

Constant Contact Angle, Variable Section

The pore has a variable section (Figure 2.54). In such case, the meniscus with the smaller diameter d1,(the other is d2) exerts the highest traction. As a result, the water island moves in that direction (i.e. the direction with the largest suction, (4 �w cos � / d1 > 4 �w cos � / d2). The movement lasts until the island arrives at a location in the pore where both meniscuses mirror each other. There, the difference in suction goes to zero and the water � ow stops. The water � ow rate increases as the water island between both meniscuses becomes shorter and the difference in suction larger, i.e. when the suction gradient increases. The � ow rate can thus be written as:

2w

c ww32

dp k s

⎛ ⎞ρ= − = −⎜ ⎟η⎝ ⎠wg grad grad (2.129)

with kw the water permeability of the pore, units s. A water island located partly in a narrow and partly in a wider pore reacts as in a pore with variable section, with the narrow pore draining the wider one. Gravity and external pressures can hinder as well as favor such motion.

Figure 2.54. Water transport in a pore after the water contact is interrupted. Isothermal conditions, constant contact angle, at the left constant section, at the right variable section.

Variable Contact Angle, Constant Section

In such case, the water island moves towards the smaller contact angle, i.e. in the direction of the largest suction, as shown by equation (2.129).

Variable Contact Angle, Variable Section

The island moves towards the largest suction. This convenes not necessarily with the smallest pore diameter, but with the largest ratio cos � / d.

1335vch02.indd 208 12.02.2007 16:55:28

209

Non Isothermal Water Transfer in a Pore After Interrupting the Water Contact

Constant Contact Angle, Constant Section

Suction increases with surface tension. As surface tension of water grows with decreasing temperature, temperature differences will force a water island to move to the colder side. That effect can be explained by writing suction as:

wo

wo

s sσ

=σ

(2.130)

where so is suction at a reference temperature, for example 20 °C (�w = �w,o – 0.17 � 10–3 �). The suction gradient can now be expressed as:

3w oo

wo wo

0.17 10s

s s −σ= + ⋅ θ

σ σgrad grad grad (2.131)

Since the suction gradient at reference temperature goes to zero in a pore with constant section, the � ow equation becomes:

23 o w

,wwo w

0.17 1032

s dK−

θ

⎡ ⎤⎛ ⎞ρ= − ⋅ θ = − θ⎢ ⎥⎜ ⎟σ η⎝ ⎠⎢ ⎥⎣ ⎦

wg grad grad (2.132)

K�,w is called the thermal water permeability; units kg/(m · s · K). In the further text, equation (2.131) is not used. Instead, capillary suction s in equation (2.129) is handled as function of temperature.

Variable Contact Angle, Variable Section

Equation (2.129) holds.

Remark

As discussed earlier, water � ow in a pore is driven by three forces: gradients in suction, gravity and pressure or tension. Suction is temperature dependent. In narrow pores, gravity and pressure gradients can be neglected. Not so in wider pores. There, the total driving force becomes: grad s + grad (�w g z) + grad p. Suction (s) further stands for the difference in water and air pressure over a meniscus. When air is expelled from a pore, air pressure at the meniscuses rises somewhat. But air ingress as a consequence of moving water islands lowers air pressure a little.

2.4.2.3 Vapour Transfer

Water transfer is caused by moving water islands. Vapour � ow instead develops in the air inclusions between the islands. Or, both occur in series. The required succession of islands and air inclusions occurs when the contact with a water surface is interrupted repeatedly. The two reasons for vapour transfer are embedded in Thompson’s law: gradients in pore diameter and temperature.

2.4 Moisture Transfer

1335vch02.indd 209 12.02.2007 16:55:29

210 2 Mass Transfer

Isothermal Vapour Transfer

Constant Contact Angle, Constant Section

As we have the same vapour saturation pressure everywhere in the pore (p�sat = psat(�) exp [s / (�w R T]), vapour transfer in the air inclusions is impossible (Figure 2.55). However, vapour transfer to the environment may go on. For this to happen, the only condition is that the relative humidity in the environment is different from sat sat100 ( / )p p′ .

Constant Contact Angle, Variable Section

The variable section induces suction differences between the meniscuses of the sequential water islands (Figure 2.55). The result is differences in vapor saturation pressure and a vapour displacement in between. The equation is:

sat va sat a a

w

pp s s

s

∂ ρ′= −δ = −δ = −δ′

∂ ρvg grad grad grad (2.133)

Thompson’s law states that for a meniscus in equilibrium vapour pressure and suction are interchangeable. In principle, transfer unbalances the equilibrium. However, for small water � ow rates, the unbalance is too small to exclude interchangeability.

Variable Contact Angle, Constant Section/Variable Contact Angle, Variable Section

Again, we have a difference in suction between the meniscuses of the sequential water islands with as a result, vapour transfer in the air inclusions in between. See equation (2.133).

Non Isothermal Vapour Transfer

Constant Contact Angle, Constant Section

Differences in temperature cause differences in water vapour saturation pressure satp′ between the meniscuses of the sequential water islands. Vapour � ow rate:

sat sata sat a

w

(1)

d

d

p p s sp

R T T

⎡ ⎤∂⎛ ⎞= −δ = −δ φ + − θ′ ⎜ ⎟⎢ ⎥⎝ ⎠θ ρ ∂θ⎢ ⎥⎢ ⎥⎣ ⎦

vg grad grad���������

Only for very small pore diameters, the term (1) in the formula is of the same order of magnitude as the term dpsat / d�, or, the equation may usually be simpli� ed to:

sata

d

d

p= −δ φ θ

θvg grad (2.134)

Figure 2.55. Vapour transfer in the air inclusions between the water isles in, isothermal conditions, constant contact angle, at the left constant section, at the right variable section.

1335vch02.indd 210 12.02.2007 16:55:29

211

Variable Contact Angle, Variable Section

Now, both causes join forces, turning the general equation for the vapour � ow rate in the air inclusions, the � rst island and the environment and the last island and the environment into:

v sata

w

d

d

ps

⎡ ⎤ρ= −δ + φ θ⎢ ⎥ρ θ⎣ ⎦

vg grad grad (2.135)

Of course, that equation can also be written as:

sata a sat

d

d

pp p

⎛ ⎞= −δ = −δ φ + φ θ⎜ ⎟⎝ ⎠θvg grad grad grad (2.136)

2.4.2.4 Moisture Transfer

Moisture transfer in a pore consists of a serial transfer of liquid and water vapour: liquid in the water islands and vapour in between. As a consequence, the islands with the smallest meniscuses and the lowest temperature increase, while those with the larger meniscuses and the higher temperature decrease. Moisture transfer comes to a halt once the islands are rearranged in such way that suction gradients no longer exist.

2.4.3 Moisture Transfer in Materials and Construction Parts

2.4.3.1 Transfer Equations

Capillary Porous Materials, Moisture Content Between Zero and Capillary

For capillary porous materials, we distinguish three moisture content intervals between dry and saturated: (1) from dry to hygroscopic at a relative humidity m, (2) from m to capillary, (3) from capillary to saturation. This section deals with intervals 1 and 2.Each capillary porous material consists of a matrix and a labyrinth of pores. These pores are neither circular nor straight. They change continuously their section, deviate, come together, split, etc. In each ‘representative material volume’, the labyrinth as a whole is present. Moisture transfer in such capillary porous material has the same driving forces as in a pore. However, it is neither possible to differenciate between vapour and water nor between series and parallel. If we were to make a cut in a representative volume, then for some pores the cut would cross a water island, while for others it would cross an air inclusion. Hence, in each cut, a parallel � ow of liquid and water vapour is seen, activated by the same driving forces: suction and temperature. By exchanging the water permeability of a pore for the water permeability of the material in the water transfer equation (2.129), and the vapour permeability of air for the vapour permeability of the material in the vapour transfer equation (2.135), moisture transfer becomes:

w

v sat

w

m

d

d

( )

k s

ps

T

k s Kθ

= −ρ

+ = −δ − δ φ θρ

= − + θ

w

v

m

g grad

g grad grad

g grad grad

(2.137)

2.4 Moisture Transfer

1335vch02.indd 211 12.02.2007 16:55:29

212 2 Mass Transfer

with

v satm w

w

d

d

pk k Kθ

⎛ ⎞ρ= + δ = δ φ⎜ ⎟ρ θ⎝ ⎠

km is called the moisture permeability, units s, and K� the thermal moisture diffusion coef� cient, units kg/(m · s · K) of a porous material. Unsaturated water transfer plays a leading role in the moisture permeability, while vapour transfer dominates the thermal moisture diffusion coef� cient. For most materials, both properties are tensors. As temperature, suction s is a real potential. In case of an ideal contact between materials it is unequivocally de� ned in each interface.

All Materials

Besides hygroscopic capillary materials (group 1) and non-hygroscopic capillary materials (group 2), we have a group non-hygroscopic non-capillary materials (group 3) and a group hygroscopic non-capillary materials (group 4). Group 3 includes all materials with only wide pores. In these, capillary transfer hardly exists, while gravity and pressure only play a role at higher moisture contents. Moisture transfer is thus limited to vapour diffusion and vapour convection. Group 4 consists of materials with very � ne pores or a combination of very � ne and very wide pores. For such materials, the � ow resistance in the very � ne pores is so high that capillarity, gravity and external pressures hardly intervene. Also vapour transfer by convection is highly improbable. Only diffusion remains. In other words, in group 3 as well as in group 4, the use of capillary suction as a potential has no physical meaning. Relative humidity on the contrary remains a useful quantity. Hence, with relative humidity as a potential, we get an expression which is more universal than equation (2.137).

Without vapour convection

,m( )k Kφ θ= − φ + θmg grad grad (2.138)

With vapour convection (material group (1) to (3))

6,m a sat( ) 6.21 10k K g p−

φ θ= − φ + θ + ⋅ φmg grad grad (2.139)

For the transfer coef� cients, the following applies:

Material k,m K�

Group 1 and 2, capillary porous md

d

sk

φ see (2.136)

Group 3, non-hygroscopic, non-capillary psatsatd

d

pδ φ

θ

Group 4, hygroscopic, non-capillary psatsatd

d

pδ φ

θ

1335vch02.indd 212 12.02.2007 16:55:29

213

In capillary porous materials, moisture transfer below a relative humidity m encompasses (equivalent) diffusion only. While equation (2.138) remains applicable there, equation (2.137) is no longer once below the relative humidity zone for capillary condensation.

Moisture Content Above Capillary

In the above capillary interval, equation (2.137) applies for drying. In all other cases except the external water pressure head case, water transfer is activated by vapour diffusion between water islands and the dissolution of air inclusions in the pore water. An external water pressures head activates saturated water � ow, which can be described by an equation of the form: gm = –kw grad Pw, with Pw water pressure and kw the water permeability at saturation (some-times called the Darcy coef� cient). Saturated water � ow does not show inertia (�2Pw = 0). That limits the solutions to steady state. Thus, the saturated water � ow rate through a single-layered wall with thickness d (water � ow rate gw, kg/(m2 · s)) is given by:

w w ww w

w w/

P P Pg k

d d k W

Δ Δ Δ= = =

where Ww is the water resistance of the wall, units m/s. For a composite wall, we have:

Flow rate Water pressures

w ww

wTw,

1

n

ii

P Pg

WW

=

Δ Δ= =

∑ w, w1 w w,1 w1 w2( )xxP P g W P P= + <

i.e. analogous equations as for steady-state heat conduction, diffusion and air � ow. To make a wall waterproof, a layer of in� nite water resistance must be inserted. As an alternative, the water front can be prevented from reaching the surface that should stay dry. Therefore, water movement from the other side must remain smaller than the drying � ow rate at that surface.

Moisture Permeability (km)

For capillary porous materials, moisture permeability is a function of suction or relative humidity. The larger the suction is (i.e. the � ner a pore and the smaller the contact angle) or the lower relative humidity , the lower is the moisture permeability. This dependence was studied for a single pore. For the water part it was:

22 2w w w

ww w

cos

32 2

dk

s

ρ ρ σ ϑ⎡ ⎤= = ⎢ ⎥η η ⎣ ⎦

Moisture permeability also depends on temperature. In principle, the property can be determined experimentally, as the product of moisture diffusivity Dw (see hereinafter) and speci� c moisture content (the derivative of the suction characteristic w(s), which stands for the relationship between moisture content and suction; see Figure 2.56. However, this approach usually leads to hardly usable results. It is better to predict the moisture permeability using a pore model of the material.

2.4 Moisture Transfer

1335vch02.indd 213 12.02.2007 16:55:29

214 2 Mass Transfer

2.4.3.2 Conservation of Mass

In General

According to conservation of mass, the following applies for each representative material volume Vrep:

rep rep(div ) d dw

G V Vt

∂⎛ ⎞± = −′ ⎜ ⎟⎝ ⎠∂mg

where w is the average moisture content, G� the moisture source or moisture sink and div the average divergence in the volume Vrep.

No vapour convection:hygroscopic, capillary materials,non-hygroscopic, capillary materials,hygroscopic, non-capillary materials with � ne pores only.

If the representative volume Vrep is assumed to be in� nitely small, then combining conservation of mass with the general transfer equation gives:

,m wdiv ( )w

k K Gt tφ θ φ

∂ ∂φφ + θ ± = = ρ ξ′∂ ∂

grad grad (2.140)

With vapour convection:non-hygroscopic, non-capillary materials,hygroscopic, non-capillary materials with wide as well as � ne pores.

�

�

�

�

�

Figure 2.56. Suction characteristic of calcium-silicate stone (round robin, measured results for 6 laboratories).

1335vch02.indd 214 12.02.2007 16:55:30

215

Here the result is:

6,m a sat wdiv [( ) 6.21 10 ]k K g p G

t−

φ θ φ∂φφ + θ + ⋅ φ ± = ρ ξ′∂

grad grad (2.141)

In both cases, the product �w � is the speci� c moisture content, i.e. the derivative of the suction characteristic for the relative humidity . That quantity is found by taking the derivative for s (Figure 2.56) and multiplying it by the derivative of s for : � = �s s / . The result should equal the derivative for �of the full hygroscopic curve between 0 and 100%. Non-hygroscopic, non-capillary materials have as speci� c moisture content:

o sat

w

( / )p T

R

Ψ ∂ φρ ∂φ

For hygroscopic materials, the capillary as well as the hygroscopic moisture embody the source and sink term. That way, interstitial condensation disappears as a separate phenomenon.

2.4.3.3 Starting, Boundary and Contact Conditions

Starting Conditions

In most cases, the starting condition is a construction with moisture content far beyond hygroscopic (the so-called initial moisture).

Boundary Conditions

Contact with moist air:As vapour � ow rate at a surface, we have: gv = � (p – ps) where p is vapour pressure in the air, ps vapour pressure at the surface and � the surface � lm coef� cient for water vapour diffusion. Vapour pressure written as psat,s � with psat,s the saturation pressure at the surface, introduces ‘relative humidity’ as a potential. ‘Contact with moist air’ acts as boundary condition when surface condensation, drying, sorption, desorption and interstitial condensation are considered.

Known � ow rate:Typical for precipitation without water run off.

Contact with water:Typical for surface condensation and precipitation, once water runs off.

Contact Conditions

Suction contact:Ideal suction contact is characterized by continuity of water and vapour � ow. This seldom happens. Most of the time suction contacts are non-ideal, i.e. introduce a contact resistance, a typical example being a mortar/brick interface.

Diffusion contact:Means continuity of vapour � ow only. Such condition applies when layers are separated by small air gaps. Also non-capillary materials only experience diffusion contacts.

�

�

�

�

�

2.4 Moisture Transfer

1335vch02.indd 215 12.02.2007 16:55:30

216 2 Mass Transfer

Contact with water:As a result of interstitial condensation and rain penetration, water can � ll gaps between layers. Gravity and water pressure may force run off, while capillary forces may immobilize the water layer.

2.4.3.4 Remark

Equations (2.140) and (2.141) are correct as long as a capillary-porous material does not contact liquid water. If that is the case, water always moves from the wider to the narrower pores. If it does contact liquid water, the wider pores may suck water from the narrower ones (see isothermal water transfer in a serial two-pore system that contacts water). That difference in moisture movement is accounted for by valuing moisture permeability and speci� c moisture content between dry and capillary and not dry and saturated. For non-capillary materials, water contact then means a 100% relative humidity in the contact plane, while for capillary materials, it means suction zero, 100% relative humidity and moisture content equal to capillary.

2.4.4 Simpli� ed Moisture Transfer Model

2.4.4.1 Assumptions

The transport properties as well as the buffering property in the equations (2.140) and (2.141) are suction, relative humidity and temperature dependent. This excludes any analytical solution. Applying CVM or FEM of course is possible. This requires a software tool, such as WUFI, MATCH and HygIRC, which are commercially available. However, a disadvantage of both CVM and FEM and the software tools mentioned is that they are black box-like and hardly help in understanding the physics involved. That opens room for simpli� ed models, which are easier to read and to understand. For that purpose, we start with introducing moisture content as an ‘improper’ driving force. ‘Improper’ means that its value changes in the interfaces between materials. Water transfer is then written as:

wD w= −wg grad (2.142)

with Dw the water diffusivity of the material; units m2/s. Between Dw (Dm) and kw (km) the following relationship exists:

ww w

w s

d

d

ksD k

w= =

ρ ξ (2.143)

Dw is a function of moisture content. Often an a priori equation is used. For example:

2

wc c

expA w

D a bw w

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

where A is the capillary water sorption coef� cient, wc the capillary moisture content and a and b material constants.In the simpli� ed model, these relations are reduced to a step function, zero for moisture contents under the critical value (wcr) (the moisture content at a relative humidity m) and constant above (Figure 2.57). Below critical moisture transfer concerns vapour only (gw = – grad p). Beyond critical, unsaturated water � ow dominates, with vapour transfer only intervening when the situation is non-isothermal:

�

1335vch02.indd 216 12.02.2007 16:55:31

217

The model learns that a sharp moisture front exists between wet and air-dry in a partially above-critical moist material (Figure 2.58). If we assume a one-dimensional, hygrically steady-state, isothermal regime, then each elementary material layer must see the same moisture � ow rate gm passing. Thus, between sub- (1) and above-critical (2) gm1 should equal gm2. As

Figure 2.57. Moisture diffusivity as function of moisture content, simpli� ed model.

Figure 2.58. Moisture front between above-critical wet and airdry (i.e. sub-critical).

2.4 Moisture Transfer

Summary w < wcr w � wcr

Isothermal gv = – grad p gw = –Dw grad w

Non-isothermal gv = – grad p satw

d

d

pD w= − − δ θ

θwg grad grad

1335vch02.indd 217 12.02.2007 16:55:31

218 2 Mass Transfer

gm1 = – (grad p)1 = – dp / dw (grad w)1 = –Dw1 (grad w)1 and gm1 = gm2 = –Dw2 (grad w)2, with Dw1 (� 0) much smaller than Dw2, (grad w)1 must be much larger than (grad w)2. Therefore, the moisture content at the sub-critical side of the contact has to increase vertically, while it should be very � at in the above-critical zone. The result is the named moisture front between ‘dry’ and ‘wet’. Simple observations con� rm this. An ink stain on blotting paper spreads out for a while and then stops from expanding. At that moment, we have the critical ink content in the stain and very little ink outside the stain. In the out� ow zone of the waves at the beach, the contact between dry and wet sand is clearly drawn. After rain, we easily observe the boundary between dry and wet on a brick façade, etc.Even though the transfer equations seem simular, the fact that a moisture front is formed creates a clear difference between unsaturated water � ow at one side and saturated water � ow, vapour diffusion, air transfer and heat conduction at the other side. Solving the simpli� ed model analytically for example is only possible when the form of the moisture pro� le is assumed beforehand.

2.4.4.2 Applications

Capillary Suction

Capillary suction is one of the most important wetting mechanisms for materials and construction parts. Think of rising damp and wind driven rain.

Hydrophilic, Homogeneous Material

If a hydrophilic sample is weighed regularly while sucking water, then, just as for a pore, a linear relation between the quantities sucked per m2 of contact surface and the square root of time ( )t is found (Figure 2.59):

cm A t= (2.144)

where A is the water sorption coef� cient, units kg/(m2 · s1/2). If the moving moisture front is monitored, then the following relationship applies between location and time:

h B t= (2.145)

where B is the water penetration coef� cient, units m/s1/2. As soon as the moisture front rea-ches the other side of the sample, the capillary line switches to a second line [ , ]t m with a much smaller slope. Here, moisture absorption is the result of diffusion and solution of air inclusions in the pore water. That second line is called the line of secondary water sorption. The related slope gives the coef� cient of secondary water sorption A�. The intersection with the capillary line � xes the capillary moisture content wc.Between the triplet A, B and wc the following relationship is assumed:

cA B w≈ (2.146)For that relation to be correct, the capillary moisture content should extend over the whole height of the sample. The water penetration coef� cient B and the capillary moisture content wc are net material properties. The larger that B is, the more capillary is a material. The larger wc, the more moisture it buffers. The capillary water sorption coef� cient A instead is a composite property, comparable with the contact coef� cient in the case of heat transfer. The number underlines how easily a material absorbs water, just as the contact coef� cient shows how easily a material absorbs heat.

1335vch02.indd 218 12.02.2007 16:55:31

219

The linear relationship c[ , ]t m does not account for air transfer. As has been shown for one pore, air egress induces some anomaly, especially when high samples of � ne porous material or samples sealed at the top are tested. In the � rst case the curve bends towards the time axis to form a kind of concave parabola. In the second case, water sorption retards as time progresses while, at the water side, air bubbles out of the sample.

Hydrophilic, Non-homogeneous Material

In non-homogeneous materials, the suction curve deviates from the linear in t .

Hydrophobic Materials

Capillary moisture uptake by hydrophobic materials, if at all, advances very slowly without any linear linkage to t . Calculating the water sorption coef� cient is generally impossible, except as a slope of tangents to the suction curve.

Simulation with the Simpli� ed Transfer Model

Does the simpli� ed model predict the linear relationship between water uptake and t ? The answer is as follows. We assume that the capillary � ow is one-dimensional. Application of mass conservation then gives:

2

w 2

w wD

tx

∂ ∂=∂∂

(2.147)

i.e., an equation, analogue to heat conduction in plane layers without source or sink. Capillary water sorption can be compared to a sudden temperature increase at the surface of a semi in� nite medium. The temperature step is replaced by a moisture content step from w = wH to w = wc. However, the existence of a critical moisture content makes the solution less uniquivocal. Besides the starting and boundary condition t � 0, 0 � x � �: w = wH; t � 0, x = 0: w = wc, two additional conditions are needed for the moisture front (xfr): (1) during suction, moisture content at the front is always critical, (2) the related water � ow rate supplies the moisture needed to move the front over a distance dxfr per unit of time, or:

2.4 Moisture Transfer

Figure 2.59. Capillary suction: wm A t= .

1335vch02.indd 219 12.02.2007 16:55:31

220 2 Mass Transfer

fr fr

frfr cr w kr

d0, : (1) , (2)

dx x x x

xwt x x w w D w

x t= =

∂ ⎛ ⎞⎛ ⎞> = = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∂

Mathematically, x = xfr is a singular point. This excludes an analytical solution. Anyhow, an approximate solution can be given for two extreme cases.

Critical Moisture Content Close to Zero

In case hygroscopic (wH) and critical moisture content (wkr) are close to zero, then, for a step �w = wc at the surface, moisture content in an in� nitely long sample at a distance x from the suction surface becomes:

w

2c

2

2exp( ) d

xq

D t

w w q q∞

=

⎛ ⎞⎜ ⎟

= −⎜ ⎟π⎜ ⎟

⎜ ⎟⎝ ⎠

∫ (2.148)

The term above the brace is the inverse error function ‘erfc’ (see Figure 2.60). In the wet zone, the very low critical moisture content is found at the moisture front. At the water contact, moisture content is capillary. For the quantity of water absorbed between t = 0 and t = t, the

following holds: c c w2 /m w D t= π . Or, between the capillary water uptake and the square

root of time a linear relationship is found. The water sorption coef� cient in turn is given by

c w2 /A w D= π , which convenes with a constant moisture diffusivity:

2

wc4

AD

w

⎛ ⎞π= ⎜ ⎟⎝ ⎠ (2.149)

�

Figure 2.60. Capillary suction, moisture content in the material when critical is laying far below capillary (wcr � wc).

1335vch02.indd 220 12.02.2007 16:55:32

221

The position of the moisture front is given by equation (2.148) for the moisture content equal

to critical. The water penetration coef� cient then becomes: 1cr c w2 erf ( / )B c w w D−= .

Materials which follow the rule that the critical moisture content is so much lower than capillary are rare.

Critical and Capillary Moisture Content Are Hardly Different

Let us assume the moisture content pro� le is linear between the sucking surface and the moisture front. Mass balance (Figure 2.61):

cr c c crwd d

2

w w w wx D t

x

+ −=

giving for x:

c crw

c cr

2w w

x D tw w

−=

+

i.e. again the experimentally proven linear relationship between the position of the moisture front and square root of time. The water penetration coef� cient now is:

c crw

c cr

2w w

B Dw w

−=

+ (2.150)

�

Figure 2.61. Capillary suction, moisture content in the material when critical and capillary are hardly different (wcr � wc).

2.4 Moisture Transfer

1335vch02.indd 221 12.02.2007 16:55:32

222 2 Mass Transfer

Absorbed water between t = 0 and t = t (kg/m2):

c crc w c cr c cr( ) ( )

2

w wm x D w w w w t

+= = + −

Water sorption coef� cient: 2 2w c cr( )A D w w= − , giving as constant moisture diffusivity:

2

w 2 2c cr( )

AD

w w=

− (2.151)

Moisture Diffusivity Function of Moisture Content

In reality Dw is a function of moisture content. Rewriting equation (2.147) gives:

ww w

Dx x t

∂ ∂ ∂⎛ ⎞ =⎜ ⎟⎝ ⎠∂ ∂ ∂ (2.152)

In case Dw(w) is determined unequivocally, the material volume is close to semi-in� nite, moisture content is w0 at time zero and w(x = 0) remains constant for t � 0, then the Bolzmann transformation /x tλ = can be applied. That converts the partial differential equation (2.152) into a differential equation of second order:

wd d d

( )2 d d d

w wD w

λ ⎡ ⎤− = ⎢ ⎥λ λ λ⎣ ⎦ (2.153)

with moisture content depending on the Bolzmann variable � only. In a [�, w] Cartesian coordinate system, all time related curves melt together into one single curve, with a surface between that curve and the �-axis of:

mte

0

( ) dw Cλ

λ λ =∫ (2.154)

The moisture absorbed between t = 0 and t = t is

m

w0

( ) ( ) dx

t

m t w x x⎡ ⎤

= ⎢ ⎥⎢ ⎥⎣ ⎦∫

with m mx t= λ . As the integration for � is independent of time, combination with equation (2.154) gives the following for the water sorption coef� cient:

mte

0

( ) dA C wλ

= = λ λ∫ (2.155)

This proves the relationship cm A t= holds in general, provided that air � ow has no in� uence and the function Dw(w) is unequivocal. When assuming a pro� le w(�), equation (2.155) allows calculating moisture diffusivity as a function of moisture content, correctly scaled to the water sorption coef� cient.

�

1335vch02.indd 222 12.02.2007 16:55:33

223

Wind Driven Rain

In General

The absorption of wind driven rain depends on two successive boundary conditions. As long as the moisture content at the wetted surface is lower than capillary (wc), the sucked water � ow rate gws remains constant, equal to the wind driven rain intensity. Once the surface is capillary wet, a water � lm forms and capillary suction occurs (Figure 2.62).

Simulation with the Simpli� ed Transfer Model

Critical Moisture Content Much Lower Than Capillary

As long as the moisture content at the surface (ws) lies below capillary, the boundary conditions are: t � 0, 0 � x � �: w = 0; t � 0, x = 0: –Dw (dw / dx)x = 0 = gws where gws is the wind driven rain intensity in kg/(m2 · s). The solution of the balance equation (2.147) then looks like:

2ws w

w w w

2exp erf

4 2 2

g D t x x xw c

D D t D t

⎡ ⎤⎛ ⎞⎛ ⎞⎢ ⎥= − − ⎜ ⎟⎜ ⎟π ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦

At the surface (x = 0) the moisture content is:

ws ws

w

2 g D tw

D=

π (2.156)

Once capillary, a water � lm forms and water uptake changes into horizontal suction from a water layer. The moment of � lm formation is given by:

2 2 2w c

f 2 2ws ws4 16

D w At

g g

π π= = (2.157)

Figure 2.62. Rain absorption by a surface before a water � lm is formed (wcr � wc (ws < wc until ws = wc)).

2.4 Moisture Transfer

1335vch02.indd 223 12.02.2007 16:55:33

224 2 Mass Transfer

In other words, the smaller the water sorption coef� cient of the surface material and the higher the wind driven rain intensity, the quicker rain runs off, resulting in an extra water load on lower facade parts. That difference is clearly seen between brick and stuccoed walls. Most bricks have a high water sorption coef� cient (0.2 � A � 0.8 kg/(m2 · s1/2)), making run off rather exceptional (only during an extremely heavy or long lasting wind driven rain event) Water repellent stucco instead has a low water sorption coef� cient (A < 0.005 kg/(m2 · s1/2), giving an almost instant run off, with water spreading over the lower facade parts. In such case, joints and the way they are sealed becomes more critical.Until run off starts, the quantity of water absorbed is mw = gws tf. Just before � lm formation starts, as water � ow rate at the surface and thus as wind driven rain intensity we have (see equation (2.157)):

w wws c c

ff f

10.89 0.79

4

D D Ag w w

tt t

π= = =

If the solution for capillary sorption applies to run off, then, after � lm formation, water sorption becomes:

w wws2 c c ws

ff

1 20.56

2

D D Ag w w g

tt t= = = =

π π

or, as the � lm forms, the facade material is supposed to suddenly absorb less water. Physically, such discontinuity is impossible. Therefore, after � lm formation a transition regime occurs between ‘constant water � ow rate’ and ‘suction from a water surface by an uniformally wet material’. Mathematically, transition is a result from the condition at the moment of � lm formation: no dry or uniformally wet material but a material with a moisture pro� le.

Critical and Capillary Moisture Content Hardly Different

We assume a moisture content in the facade material changing linearly from the outside surface down to the critical value at a depth x. As long as the moisture content at the surface (ws) lies below capillary (wc), the surface-related water � ow rate in the material looks like:

s crws w

w wg D

x

−=

From the mass balance gws dt = ws dx the following equation applies for the moisture front:

2w cr ws

2ws w cr

21 1

D w g tx

g D w

⎛ ⎞= + −⎜ ⎟⎜ ⎟⎝ ⎠

while the moisture content (ws) at the surface equals wcr + gws x / Dw. Once that surface is capillary wet (ws = wc), a water � lm forms. At that moment, the moisture front sits at xfr = [Dw (wc - wcr)] / gws or, with 2 2 2

w c cr/ ( )D A w w= − , xfr = A2 / [wc + wcr) gws]. The moment of � lm formation so becomes: 2 2

f ws/ (2 )t A g= .

1335vch02.indd 224 12.02.2007 16:55:33

225

Drying

Experiment

Drying occurs when the relative humidity in the environment drops below the relative humidity at a surface. Experimentally, drying is studied by sealing � ve of the six sides of a water saturated sample, hanging it in a climate chamber at constant temperature and constant relative humidity and weighing it regularly. If the drying � ow rate (= weight decrease per time unit) is depicted as a function of the remaining average moisture content in the sample, then the curve of Figure 2.63 is obtained.Two drying stages appear: the � rst at higher average moisture content when the drying � ow rate is quasi constant and the second at lower average moisture content when the drying � ow rate decreases rapidly. A transition zone exists between both.

Simulation with the Simpli� ed Transfer Model

Does the simpli� ed transfer model explain the experimental curve? For this, assume an above critical moist sample. The lateral sides and the rear side are vapour-sealed, resulting in a one-dimensional drying through the top side. The sample hangs in a climate chamber at constant temperature (�) and constant relative humidity (). In the sample, the relative humidity is 100%. As vapour pressure on the drying surface we have the saturation pressure at surface temperature (psat,s(�s)). The drying � ow rate to the environment is:

vd sat, sat( )sg p p= β − φ (2.158)

with the surface � lm coef� cient for diffusion, the relative humidity in the chamber and psat the saturation pressure at the chamber temperature. Since drying means evaporation, the drying � ow rate carries along the heat of evaporation. As a result, the temperature and the

Figure 2.63. Isothermal drying, the two stages.

2.4 Moisture Transfer

1335vch02.indd 225 12.02.2007 16:55:33

226 2 Mass Transfer

saturation pressure at the drying surface will drop below that in the chamber. For the purpose of simplicity, we assume that the sample has an in� nite thermal conductivity and that no heat transfer by radiation occurs. In such a case, the saturation pressure on the drying surface equals the value in the chamber and equation (2.158) may be rewritten as:

vd sat (1 )g p= β − φ (2.159)

Apparently, drying velocity only depends on the climate in the chamber: air temperature, relative humidity and air velocity via the surface � lm coef� cient, whatever the material is. The higher the temperature, the lower the relative humidity and the higher the air velocity, the faster the sample dries. Many people apply this principle without knowing it. The laundry is hung outside when the weather is sunny (= high temperature) and dry (= low relative humidity). If it is windy, (= higher), then the laundry dries even faster.Drying decreases the moisture content at the drying surface. As a consequence, a moisture content gradient, initiating moisture � ow, develops in the material towards the drying surface. At any time at the drying surface, the following applies: x = 0, w 0 sat(d / d ) (1 )xD w x p=− = β − φ. The situation – capillary moisture � ow from the material to the drying surface, evaporation there and decreasing moisture content in the material –, called the � rst drying stage, continues until the moisture content at the drying surface turns critical. At that moment, moisture diffusivity (Dw)x = 0 drops to zero there and moisture transfer becomes vapour only. As a result, the moisture front detaches from the surface and shifts into the material (see Figure 2.63), giving as drying � ow rate:

satvd

(1 )

1 /

pg

N x

− φ=

β + μ (2.160)

where � is the vapour resistance factor of the material and x the distance between the surface and the moisture front in the material. As x increases, the drying rate decreases: the second drying stage. This lasts until the sample has regained hygroscopic equilibrium with the environment. Shortly after the start of the second stage, the sample will appear super� cially dry. Too often, this is taken as a proof that the material is dry, although it may still contain quite some moisture. This misinterpretation occurs regularly when screeds and wall � nishes are judged on dryness. Once the surface looks dry, craftsmen start painting the wall, installing a vapour retarding � oor covering, etc. Afterwards, problems arise.The � rst drying stage can be described by equation (2.147) with starting and boundary conditions of (x = 0: dry surface, x = d: vapour-tight surface): t � 0, 0 � x � �: w = w0; t � 0, x = 0: dw / dx = –gvd / Dw, x = d: dw / dx = 0. Solution:

vd0

w

2 2 2 2

w2 2 2 2

3 6 2 2 ( 1) ( )exp cos

6

= − −

⎧ ⎫⎡ ⎤⎛ ⎞− + − π π −⎛ ⎞⎪ ⎪+ −⎢ ⎥⎜ ⎟⎜ ⎟ ⎝ ⎠⎨ ⎬π ⎝ ⎠⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭

∑

⋅

�����������������������

vd

n

g t g dw w

d D

x d x d n t n d xD

dd n d

The term above the brace describes how the moisture gradient moves from the drying surface to the vapour-tight rear surface. Once there, the term becomes negligible and the sample is left with a parabolic moisture pro� le:

2 2vd vd

0 2w

3 6 2

6

g t g d x d x dw w

d D d

⎛ ⎞− += − − ⎜ ⎟⎝ ⎠ (2.161)

1335vch02.indd 226 12.02.2007 16:55:33

227

The moisture content is lowest at the drying surface (x = 0):

vd vd0

w3

g t g dw w

d D= − − (2.162)

As an average moisture content at any moment, we have:

vd0

0

1g t

w ww d

⎛ ⎞= −⎜ ⎟⎝ ⎠

(2.163)

The second drying stage starts when the drying surface becomes critically moist:

0 crsecond

vd w3

w w dt d

g D

⎡ ⎤−= −⎢ ⎥

⎣ ⎦ (2.164)

The moisture left in the sample � xes the transition moisture content (wtr):

vdtr cr

w3

g dw w

D= + (2.165)

Equation (2.165) shows that the difference between critical and transition moisture content increases at higher drying rates. Hence, faster drying during the � rst stage does not guarantee a material will be dry sooner. The second stage simply starts at higher average moisture content. For highly capillary materials (Dw large), the difference between transition and critical moisture content is smaller at equal drying � ow rate than for less capillary materials (Dw small). A small diffusion resistance factor, large moisture diffusivity and low critical moisture content constitute favourable conditions. Such materials dry easily. Bricks are an example of this. Instead, less capillary materials with critical moisture content near capillary show a short � rst and a very long second drying stage. They dry slowly. Concrete is an example of that.The second drying stage can be approached as follows. Assume the transition and critical moisture content (wtr – wcr) are close (a hypothesis which simulates reality best when the drying � ow rate during the � rst stage is low). Then, drying causes nothing more than the moisture front to withdraw. Mass equilibrium (x is the moisture front, x = 0 the drying surface) consequently says:

satcr H

(1 )d ( ) d

1p

t w w xN x

− φ= −

+ μβ

(2.166)

with wH the hygroscopic moisture content at the prevailing climate conditions. In steady-state and with the vapour resistance factor (�) constant, x, the position of the moisture front then becomes:

2sat

cr H

2 (1 )11 1

⎛ ⎞− φ μ β= + −⎜ ⎟⎜ ⎟μ β −⎝ ⎠

p N tx

N w w (2.167)

If the vapour resistance � N x is much larger than the surface resistance 1 / , then the equation simpli� es to:

sat

cr H

2 (1 )

( )

px t

N w w

− φ=

μ − (2.168)

2.4 Moisture Transfer

1335vch02.indd 227 12.02.2007 16:55:34

228 2 Mass Transfer

So, also drying results in a square root relation between the position of the moisture front and time. In a [ , ]x t axis ssytem, the second drying stage is represented by a straight line. Equation (2.168) gives drying quantities when multiplying the left and right term by the difference wcr – wH:

sat cr Hd

2 (1 ) ( )p w wm t A t

N

− φ −= = ′

μ (2.169)

When the critical and the hygroscopic moisture content are known, then equation (2.169) allows calculating the vapour diffusion factor (�) from a drying test. It also shows that a higher value slows down drying.Despite all simpli� cations the model is quite usable at low drying rates and easy heat supply. For high drying rates, the isothermal character gets lost. The heat of evaporation absorbed lowers the temperature and the � ow rate at the drying front. Moreover, the drying front temperature varies. Indeed, as the front shifts deeper into the material and the drying � ow rate decreases, less heat of evaporation is absorbed and a move back to a relationship dm A t= ′ occurs. Hence, the real d[ , ]m t curve will not be straight, but bends towards the t -axis at the beginning to change into a straight line afterwards.

Drying of a Wet Layer in a Composite Wall

The drying model is easily extended to composite walls, which have a wet interface between two non-capillary layers (interface j) or which contain one or more moist layers (moisture content beyond critical, boundaries k, l). In such a case, the vapour � ow rate between the interface j or the boundaries k, l and the environment at both sides (thermal and hygric steady-state, environment 1 (k) and 2 (l)) looks like (Figures 2.64 and 2.65):

Figure 2.64. Drying of a moist interface in a construction part.Total drying � ow rate equal to the sum of the drying � ow rates to both environments.

1335vch02.indd 228 12.02.2007 16:55:34

229

Wet interface

sat, 1 sat, 2 2 sat, sat, 1md

1 2 2 1

j j j jp p p p p p p pg

Z Z Z Z

− − − −⎛ ⎞= − + = −⎜ ⎟⎝ ⎠

(2.171)

Wet zone

sat, 1 sat, 2 2 sat, sat, 1md

1 2 2 1

k l l kp p p p p p p pg

Z Z Z Z

− − − −⎛ ⎞= − + = −⎜ ⎟⎝ ⎠

(2.170)

where psat,j is the vapour saturation pressure in the interface j, while psat,k and psat,l are the vapour saturation pressures at the boundaries k and l of the wet zone. In that zone, relative humidity is 100% and vapour pressure equals saturation pressure. Z1 and Z2 represent the diffusion resistances between j and the two environments alongside the wall, or, the diffusion resistance between the boundary k and the environment 1 and between the boundary 1 and the environment 2. p1 is vapour pressure in environment 1, while p2 is vapour pressure in environment 2.In the [Z, p]-plane, the two terms in both equations represent the slope of the line connecting the saturation pressure in the wet interface j or the boundaries k and l to the points [0, p1] and [ZT, p2]. At the cold side of the wet interface or zone (assume this is the environment 1) that line may intersect the saturation pressure curve. Part of the vapour that diffuses from j or k to the cold side will then condense in the wall. We call this drying condensation. Where that condensation takes place, how much moisture is deposited and how the correct vapour pressure line looks, is given by the tangents to the saturation curve, drawn with the saturation pressures psat,j or psat,l and the vapour pressure p1 as starting points (Figure 2.65). Drying condensation may become a problem in any part which contains initial moisture, in parts where one or more layers absorb rain or in parts which suffered from interstitial condensation of vapour, produced inside. If diffusion is supplemented by vapour convection, drying generally occurs faster, but drying condensation can take on more serious forms.

Figure 2.65. Drying of a wet zone in a construction part (w > wkr).The drying process induces interstitial condensation in a separate interface closer to the cold side.

2.4 Moisture Transfer

1335vch02.indd 229 12.02.2007 16:55:34

230 2 Mass Transfer

Interstitial Condensation by Diffusion of Vapour, Released Inside

Vapour transfer by diffusion, or worse, advection, in single-layered (only for advection) and composite walls may induce interstitial condensation of vapour, released inside. What limit state is reached when annual accumulation occurs, i.e. does condensation go on for ever? Take a wall made of a capillary (Dw > 0), vapour permeable material, diffusion resistance Z2, with low thermal conductivity, � nished at the cold side with a vapour retarding clad, diffusion resistance Z1 (Figure 2.66). Vapour transfer through convection does not occur. We take yearly average climate conditions. Temperature and vapour pressure at the warm side (�2, p2, Z = Z1 + Z2) causes interstitial condensation at the backside of the clad (Z = Z1). In other words, in the interface Z1 the condensation � ow rate (gwc) is:

1 12 sat, sat, 1wc v2 v1

2 1

z zp p p pg g g

Z Z

− −= − = −

with p1 vapour pressure at the cold side, p2 vapour pressure at the warm side and psat,Z1 the

vapour saturation pressure in Z1. The capillary material will now absorb the condensate deposited at the interface Z1, initiating a water � ow rate, given by:

1 12 sat, sat, 1crw

2 1

Z Zp p p pw wD

x Z Z

− −−= −

As a result, a wet zone develops in the capillary layer, extending slowly to the warm side. Inside that zone, moisture content is slightly above critical, while at the interface between moist and dry (dry means hygroscopically humid), it is critical (wcr). In the wet zone, vapour pressure equals saturation pressure. The moving front forces the ingoing vapour pressure line in the [Z, p]-plane to shift and rotate from the ‘vapour pressure inside/saturation pressure in the interface Z1 contact point to the ‘vapour pressure inside/saturation pressure at the front (psat,f)’ contact point (Figure 2.66).As a result, the ingoing vapour � ow rate gv2 decreases. Once the moisture front arrives at the point where the ingoing vapour � ow rate gv2 equals the outgoing one (gv1), condensation stops and the limit state is reached. At any rate, in the moist zone, a capillary water � ow rate from

Figure 2.66. Annually accumulating condensate. Limit state in case a capillary, vapour permeable material is � nished with a non-capillary, vapour retarding clad at the cold side.

1335vch02.indd 230 12.02.2007 16:55:34

231

cold to warm persists, with an identical vapour � ow rate from warm to cold in equilibrium. The ingoing vapour � ow rate left equals:

1 1sat, sat, cr2 sat,v2 v1 w

2 w w w w

f Z Zf p p w wp pg g D

Z N d N d d∞− −−

= = = −− μ μ

(2.172)

with 1Zw the moisture content in the original condensation interface Z1, psat,f the vapour

saturation pressure at the moisture front, � the vapour resistance factor of the dry capillary material, �w the vapour resistance factor of the critically wet material and dw the thickness of the wet zone. Solving equation (2.172) for

1Zw gives:

1

1

sat, sat,cr v1 w

w w

1 f ZZ

p pw w g d

D N

−⎛ ⎞= + −⎜ ⎟μ⎝ ⎠

(2.173)

In general, the added term

1sat, sat,v1 w

w w

1 f Zp pg d

D N

−⎛ ⎞−⎜ ⎟μ⎝ ⎠

is much smaller than the critical moisture content (wcr), so that 1 crZw w≈ . Hence, there is a

limit state of a critically moist zone of thickness dw in the capillary material, extending from the interface with the vapour retarding clad up to the warm side. Its thickness can be approximated on a Glaser diagram. It suf� ces to draw a parallel with the outgoing tangent line through the vapour pressure inside. The thickness dw is then given by the distance between the intersection with the saturation line and the backside of the vapour retarding clad (Z = Z1). That thickness increases with higher vapour pressure at the warm side.The time it takes to reach the limit state can be calculated from:

1sat, 12 sat,cr w

2 1

d dZf p pp p

w x tZ N x Z

−−⎛ ⎞= −⎜ ⎟− μ⎝ ⎠

(2.174)

where x is the thickness of the wet zone. To solve that equation, the relationship between the saturation pressure psat,f and the temperature at the moisture front and therefore between psat,f and x has to be known. An approximate solution is found by assuming

1sat, sat,f Zp a Z p= + ,

1

tesat,Zp C= , Z N x= μ . The result is:

1 2 3cr

ln (1 )N

t a Z a a Zw

μ = − − (2.174)

with:

1v1

1a

a g=

−

v12

2 v2 v1( )

a ga

Z g g

−=

−

23

v1( )

Za

a g=

−

1sat, 1v1

1

Zp pg

Z

−=

12 sat,v2

2

Zp pg

Z

−=

The diffusion resistance (Zw) and thus the thickness of the wet zone (xw) are thereby known as an implicit function of time. For the condensation deposit in kg/m2 after t days, we have:

c cr86, 400 ( ) / ( )m w Z t N= μ .

2.4 Moisture Transfer

1335vch02.indd 231 12.02.2007 16:55:35

232 2 Mass Transfer

Analogously, the limit state for other layer combinations can be found, see Table 2.9. Yet, the assessment of whether or not interstitial condensation is acceptable, will not only depend on the nature of the material in or against which condensation occurs and the quantity deposited, but also on how the limit state looks. From that point of view case (4) is very negative. In practice, we have it each time a capillary layer is sandwiched between the thermal insulation at the warm side and a vapour retarding � nish at the cold side. A typical example in cold climates is a stony wall with vapour-tight ouside clad, which is insulated at the inside. Another example is the deck above the air space in a ventilated low slope roof. An annoying example in warm, humid climates is a wall with a vinyl paper � nish on a gypsum board inside lining, with the thermal insulation in the fabric and a vapour permeable outside clad.

Table 2.9. Limit state in case of annual accumulation of condensate.

Layer Limit State

Cold side Warm side

Case (2) Vapour tight, non-capillary

Vapour permeable, non-capillary

The layer at the warm side is saturated slowly from the interface between both layers inward, with water accumulated over a thick-ness d. Typical example: synthetic insulation covered with roo� ng felt. The thickness of the saturated layer is approximately equal to the distance between the initial condensation interface and the intersection of a tangent parallel to the outgoing vapour pressure through the vapour pressure inside with the saturation curve.

Case (3)Vapour permeable, capillary

Vapour permeable, non-capillary

The layer at the cold side gets critically wet over a thickness d from the condensation interface outward. The thickness of the saturated layer is approximately equal to the distance between the initial condensation interface and the intersection with the satura tion curve of a tangent parallel to the ingoing vapour pressure through the vapour pressure outside.

Case (4)Vapour perme-able, capillary layer, � nished at its cold side with a vapour tight lining

Vapour permeable, non-capillary

Here, interstitial condensation starts in (1) the interface between the vapour retarding � nish and the capillary layer, (2) the interface be-tween the capillary layer and the non-capillary layer at its warm side, (3) the capillary layer itself. At the limit state, the capillary layer will turn more than capillary wet, while at its warm side, in the interface with the non-capillary layer, water will drip or run off permanently.

Case (5)Vapour permeable, capillary

Vapour permeable, capillary

Both layers turn critically moist by back-sucking the condensate from the interface in between. At the limit state, the thickness of the critically moist layer in both is such that the ingoing and outgoing vapour pressure lines in the [Z, p]-plane have the same slope. This thickness cannot be determined graphically. In a � rst approximation, we may assume:

2 21 2

1 22 2 2 2kr,1 kr,21 2 1 2

1 1,

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

A Ad d

w wA A A A

1335vch02.indd 232 12.02.2007 16:55:35

233

2.5 Problems

(1) Outside, the monthly average temperature is 0 °C for a dewpoint of –2 °C. Inside, we have 22 °C. The lowest temperature factor on the inside surface of the envelope is 0.4. Indoors, the daily average vapour release reaches 20 kg. What outside air � ow in m3/h do we need to keep the mold risk zero (which means a relative humidity below 80%).

Answer

Vapour pressure outside:

d,ee

d,e

22.44611 exp 518 Pa

272.44p

⎛ ⎞θ= =⎜ ⎟+ θ⎝ ⎠

(�d,e is the outside dewpoint)

Lowest inside surface temperature:

�s = 0 + 22 � 0.4 = 8.8 °C

Allowable vapour pressure inside:

si

s

17.080.8 611 exp 907 Pa

234.18p

⎛ ⎞θ= ⋅ =⎜ ⎟+ θ⎝ ⎠

Ventilation � ow needed:

v i v,P 3a

i e

462 295.15 20 / 24292 m /h

907 518

R T GV

p p

⋅ ⋅= = =− −

�

If surface condensation was the risk to avoid, then 184 m3/h should suf� ce, i.e., a ventilation � ow which is only 63% of the one needed to exclude mold from developing.

(2) Assume the ventilation � ow calculated in exercise (1) should be delivered by fan extraction. The designer, however, forgot to provide the necessary air inlets. As a result, air leakage through the envelope has to do the job. The envelope’s air permeance equals 0.083 �Pa

–0.37, expressed in kg/(s · Pa). What pressure difference between indoors and outdoors is needed to generate the necessary ventilation � ow?

Answer

The relation between air � ow and air pressure difference is given by:

0.63a a a a0.083G K P P= Δ = Δ

A volumetric � ow of 292 m3/h means a weight related � ow of 345.3 kg/h or 0.095922 kg/s. Pressure difference:

1 / 0.67a (0.095922 / 0.083) 1.24 PaPΔ = =

That result allows concluding that the envelope considered is highly air permeable. In fact, for a pressure difference of 50 Pa, we get a volumetric air � ow of 3468.5 m3/h, which is much too high.

2.5 Problems

1335vch02.indd 233 12.02.2007 16:55:35

234 2 Mass Transfer

(3) Repeat problem (2) in case the airtightness of the envelope is 10 times higher. What is your conclusion?

(4) Outside, the temperature touches 34 °C for a relative humidity of 80%. The air conditioning succeeds in providing 24 °C inside, at a relative humidity of 60%. The building has a volume of 600 m3. The ventilation rate the HVAC induces is 0.6 h–1. The daily mean vapour release equals 15 kg per day. How much vapour must be removed by the system (in g/h)? Could we in� uence that quantity by selecting an appropriate glazing system?

Answer

Let us start from the formula describing the inside vapour pressure excess:

ii,e

R T Xp

n VΔ =

with X the vapour � ow to be removed from the outside air.

In that case, the vapour pressure excess reaches

17.08 24 17.08 34611 0.6 exp 0.8 exp 2468 Pa

234.18 24 234.18 34

⎡ ⎤⋅ ⋅⎛ ⎞ ⎛ ⎞⋅ − ⋅ = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠+ +⎣ ⎦

which gives us a total removal of 6471 g/h vapour from the outside air, with 625 g/h added from the vapour released inside!The glazing system chosen has no impact on that number. In fact, as the outside temperature surpasses the inside temperature, its inside surface temperature will be higher than the temperature inside. Condensation on the inside glass surface is prohibited that way.

(5) Take a dwelling where the bathroom is directly accessible from the parent’s bedroom. The bathroom has a volume of 15 m3, while the daily vapour release there reaches 2 kg per day. In the bedroom, with a volume of 50 m3, an additional 1 kg/day is produced. What average vapour pressure will be measured in the bedroom when the ventilation air which enters the bathroom at a rate of 1.7 h–1 (related to the air volume in the bathroom) passes into the sleeping room and mixes there with an additional outside air � ow of 30 m3/h. Outside, we have a temperature of –10 °C and a relative humidity of 90%. The temperature in the bathroom is 24 °C, in the bedroom 18 °C.

Answer

First we calculate the vapour � ow which enters the bedroom from the bathroom. Vapour pressure in the bathroom:

v,Pi,bath e i

bathroom( )

Gp p R T

n V

⎡ ⎤= + ⎢ ⎥

⎣ ⎦

with d,ee

22.44 100.9 611 exp 234 Pa

272.44 10p

⋅ − θ⎛ ⎞= ⋅ =⎜ ⎟−⎝ ⎠

1335vch02.indd 234 12.02.2007 16:55:35

235

So, pi becomes:

i,bath2 / 24

234 462 297.15 682.5 Pa1.7 15⎛ ⎞= + ⋅ =⎜ ⎟⎝ ⎠⋅

p

and the vapour � ow from the bathroom to the bedroom reaches 0.127 kg/h. In the bedroom, we the get as vapour pressure:

i,bed e0.127 1 / 24

462 291.15 642 Pa1.7 15 30

p p+⎛ ⎞= + ⋅ =⎜ ⎟⎝ ⎠⋅ +

Thus, between bathroom and bedroom there is some diminution in vapour pressure. The main reason is the additional ventilation � ow in the bedroom.

(6) Repeat problem (5) for the case that the only ventilation in the bedroom comes from the air� ow leaving the bathroom. Conclusions?

(7) Given a cavity wall with section:

Layer Thicknessm

�-valueW/(m · K)

Air permeabilitym3/(m2 · s · Pa)

Inside leaf (no � nes blocks) 0.14 1.00 33 � 10–4 �Pa–0.42

Cavity � ll (mineral � ber) 0.10 0.04 0.00081

Air layer 0.01 0.067 �Brick veneer 0.09 1.00 0.32 � 10–4 �Pa

–0.19

Calculate the air � ow rate through the wall, knowing that wind induces an average pressure difference between both sides of 6 Pa (higher outdoors than indoors) and assuming that the � ow develops normally to the wall’s surface.

2.5 Problems

1335vch02.indd 235 12.02.2007 16:55:35

236 2 Mass Transfer

Answer

The air � ow rate is given by:

a 1 11 1

0.58 0.81a a

1 14 40.58 0.81

6

1

0.00081(33 10 ) (0.32 10 )

g

g g− −

− −

=

+ +

⋅ ⋅

That formula has to be solved by iteration, starting from an assumed air � ow rate in the right hand term. As a starting value we use the rate in case the wall consisted of a � ll only: 6 � 0.00081 = 0.00486 m3/s. Succesive iterations give as a � nal result: ga = 0.000133 m3/s. The � gure on the left below shows a good approximation is found after quite a limited number of iteration steps. Thus, 0.48 m3 of air per hour, passses through each square meter of cavity wall!

The air pressures over the successive layers are found by introducing the air � ow rate into the three � ow equations (ga = Ka �Pa). The interface air pressures are then given by:

5

a a,o a1i

P P P=

= − Δ∑

see � gure on the right. Apparently the veneer wall has the largest pressure difference, which means that excessive wind loads may threaten its stability.At the same time, the large pressure difference may help rain to in� ltrate the joints and run off at the backside of the veneer. In other words, such a situation should be avoided! A much better solution is to have the pressure difference over the load bearing part of the wall, in this case the inside leaf. This can be realized by pargetting the inside leaf with a gypsum or a lime plaster.

1335vch02.indd 236 12.02.2007 16:55:36

237

New section:

Layer Thicknessm

�-valueW/(m · K)

Air permeabilitym3/(m2 · s · Pa)

Render 0.01 0.30.1 � 10–4 �Pa

–0.23

Inside leaf (no � nes blocks) 0.14 1.00Cavity � ll (mineral � ber) 0.10 0.04 0.00081Air layer 0.01 0.067 �Brick veneer 0.09 1.00 0.32 � 10–4 �Pa

–0.19

Rendering the inside leaf with a gypsum plaster decreases the in� ltration rate by 75%, down to 0.12 m3/(m2 � h). Also the interface air pressures change substantially as the � gure on the right shows. In fact, the render faces the largest difference now, while the veneer wall is left

2.5 Problems

1335vch02.indd 237 12.02.2007 16:55:36

238 2 Mass Transfer

with some 20% of the total. This creates a much better situation than the original one: less wind loading on the veneer and less pressure difference which may force run off to cross the head joints.

(8) We return to the original cavity wall of problem (7). The temperature outdoors is –10 °C, and indoors is 21 °C. Calculate the temperatures in the wall and evaluate the heat loss by conduction at the inside surface in case the wall faces a wind speed of 4 m/s, resulting in a pressure difference of 6 Pa over the wall (higher outside than inside).

Answer

A temperature difference in combination with wind results in combined heat and air � ow through the wall. In that case, the temperatures are given by:

a ae i e 1 1

a a T

1 exp( )( ) ( ) with ( )

1 exp( )

c g RF R F R

c g R

−θ = θ + θ − θ =

−

The original wall shows an in� ltration rate ga of 0.000133 m3/s or 0.00016 kg/(m2 · s), giving a unit enthalpy � ow of 0.1602 W/(m2 · K) (ca = 1008 J/(kg · K)).

Layer Thick nessm

�-valueW/(m·K)

Rm2 · K/W

� Rm2 · K/W

F1(R)–

Temp.°C

21.0

Surface � lm resistance (Ri) 0.125 0.125 0.0514 19.4

Inside leaf (no � nes blocks) 0.14 1.00 0.14 0.265 0.1077 17.7Cavity � ll (mineral � ber) 0.10 0.04 2.5 2.765 0.9260 –7.7Air layer 0.01 0.067 0.15 2.915 0.9654 –8.9Brick veneer 0.09 1.00 0.09 3.005 0.9885 –9.6Surface � lm resistance (Re) 0.045 3.05 1 –10.0

As a � gure:

1335vch02.indd 238 12.02.2007 16:55:37

239

The thick line gives the temperatures in case ex� ltration intervenes. The thin line accounts for conduction only. Clearly, in� ltration cools down the wall somewhat. Heat � ow rate by conduction at the inside surface is:

a a a a T2 i e 2

a a T

exp( )( ) with

1 exp( )

c g c g Rq F F

c g R= θ − θ =

−

RT = 3.05 m2 · K/W and ca ga = 0.1602 W/(m2 · K) give: 12.9 W/m2. Without in� ltration, the result should have been 10.2 W/m2. In� ltration apparently increases conduction at the inside surface. Simultaneously, an enthalpy � ow of 3.36 W/m2 enters the enclosure

(9) Take the cavity wall pargetted at the inside of problem (7). Calculate the temperatures and heat � ow rate by conduction at the inside surface for the same temperatures indoors and outdoors as in problem (8). What are the conclusions?

(10) Assume the cavity wall of problem (7) is 2.7 m high. The outdoor temperature is –10 °C and the indoor temperature is 21 °C. Since the wall is air permeable, thermal stack develops. What happens if the window-less envelope of a one story high enclosure is built with that type of wall?

Answer

In the case considered, the neutral plane for stack is formed at mid-height, resulting in a triangular pressure pro� le. Just above the � oor a maximum pressure difference of 1.8 Pa forces outside air to in� ltrate. That in� ltration turns into ex� ltration at mid-height, with a maximum activated by a 1.8 Pa pressure difference, just below the ceiling.The air � ow rate at different heights is calculated the same way as done in problem (7). For the air � ow rate, see the � gure below at the left.The air � ow rate pro� le stack generates in the wall is close to linear. The total in- and ex� ltrating air � ow equals 0.13 m3/h per running meter of wall. Not really much. As a consequence, the heat � ow by conduction on the inner surface will vary along the height: highest just above the � oor, pure conduction at mid-height and lowest under the ceiling, see the � gure at the right. Calculating those heat � ow rates does not differ from the way it was done in problem (8).

2.5 Problems

1335vch02.indd 239 12.02.2007 16:55:37

240 2 Mass Transfer

(11) Given a timber framed wall with section:

Layer Thicknessm

Densitykg/m3

�-valueW/(m · K)

�-value–

Air permeabilitym3/(m2 · s · Pa)

Gypsum board lining 0.012 – 0.1 12 3.1 � 10–5 �Pa–0.19

Insulation (mineral � ber) 0.15 20 0.04 1.2 0.00081

OSB-sheathing 0.01 400 0.13 15 3.5 � 10–4 �Pa–0.31

Building paper 0.0005 – 0.2 200 4.2 � 10–4 �Pa–0.4

Cavity 0.03 – 0.18 0 �

Brick veneer 0.09 1600 0.9 5 0.32 � 10–4 �Pa–0.19

Will the design suffer from unacceptable interstitial condensation?

If so, what vapour retarder should be added where?

Boundary conditions to be considered (north-east orientation):

J F M A M J J A S O N D

Outdoors

Air temp. (°C) 2.3 3.3 5.9 8.8 13.5 16.1 16.8 15.7 13.5 8.8 5.5 2.9

Eq. temp. (°C) 3.1 4.2 7.2 10.6 16.1 19.1 19.9 18.6 16.1 10.6 6.7 3.8

Vapour press. (Pa) 619 675 825 997 1277 1427 1470 1406 1277 997 804 653

Indoors

Air temp. (°C) 20.0 20.3 21 21.8 23.1 23.8 24 23.7 23.1 21.8 20.9 20.2

Vapour press. (Pa) 1224 1253 1330 1418 1561 1638 1660 1627 1561 1418 1319 1242

Surface resistances for heat transport: inside 0.13 m2 · K/W, outside 0.04 m2 · K/W. Calcula-tions, if any, have to be done with the equivalent temperature. That value in fact accounts for solar gain and the exponential relationship between temperature and water vapour saturation pressure

Answer

Suppose diffusion is the only mechanism moving water vapour in and out of the wall. To evaluate if interstitial condensation occurs, we apply a Glaser analysis using the climate data for the coldest month, in that case January.

1335vch02.indd 240 12.02.2007 16:55:38

241

Step 1: Temperatures in all interfaces

Step 2: Water vapour saturation pressure in all interfaces

Step 3: Water vapour partial pressure

2.5 Problems

1335vch02.indd 241 12.02.2007 16:55:38

242 2 Mass Transfer

Step 4: Water vapour saturation and water vapour pressure intersect. Interstitial condensation is thus a fact. To � nd the condensation interface, calculate the correct water vapour pressures and quantify the deposit, we redraw the wall in the [vapour diffusion resistance/vapour pressure]-plane and construct the tangents to the saturation line. For the result, see the � gure. Condensation is deposited at the backside of the OSB-sheating wetting it slowly.

Amounts for January:

29 9 9

1224 842 842 6193600 24 31 0.41 kg/m

1.72 10 5.62 10 1.72 10

− −⎛ ⎞− ⋅ ⋅ ⋅ =⎜ ⎟⎝ ⎠⋅ ⋅ − ⋅

As the sheathing is 1 cm thick and has a density of 400 kg/m3, an increase in moisture ratio by weight of 13% may be expected! Looking to the condensation/drying cycle over a whole year, the result is:

J F M A M J J A S O N D

Deposit 0.41 0.33 0.16 –0.16 –0.95 –1.48 –1.72 –1.43 –0.92 –0.17 0.18 0.38Acc. 0.98 1.31 1.46 1.30 0.35 0 0 0 0 0 0.18 0.56

As a � gure:

1335vch02.indd 242 12.02.2007 16:55:38

243

Condensation accumulates during winter. The increase in moisture ratio by weigth reaches 36.5% which is unacceptable for a timber-based material. However drying the next springtime goes quite fast. In summer, nothing happens, at least, as long as wind driven rain is not considered.As the amounts are unacceptable, the question is raised what vapour retarder should be needed to return to acceptability. Typically, one accepts that the condensation deposit may increase the average moisture ratio by weight with 3%, giving a maximum winter accumulation of 0.12 kg/m2 in the case being. An iterative calculation shows that a vapour retarding layer at the inside or between the gypsum lining and the insulation with a vapour permeance below 3.3 � 10–10 s/m suf� ces to reach that objective. That requirement is far from severe. Expressed in terms of an equivalent diffusion thickness, we only need 0.56 m. A vapour-retarding paint should suf� ce!The question of course is if such result re� ects reality? In fact, three important elements are not considered: the in� uence of air in- and out� ow and coupled vapour transport, the effect of hygric inertia and the consequences of wind driven rain sucked by the capillary veneer.

Correction 1

Assume the difference in monthly mean inside and outside air temperature to be the main reason for air out� ow through the upper part of the 2.7 m heigh wall. For reasons of simplicity, the neutral plane is � xed at mid height. Air pressure difference becomes:

J F M A M J J A S O N D

0.41 0.33 0.16 –0.16 –0.95 –1.48 –1.72 –1.43 –0.92 –0.17 0.18 0.38

Applying combined heat, air and vapour transport gives as deposit per month and accumulated condensation deposit:

J F M A M J J A S O N D

Deposit 0.54 0.43 0.25 –0.10 –0.92 –1.47 –1.71 –1.42 –0.89 –0.10 0.28 0.50Acc. 1.32 1.75 2.00 1.90 0.98 0 0 0 0 0 0.28 0.78

As a � gure:

2.5 Problems

1335vch02.indd 243 12.02.2007 16:55:39

244 2 Mass Transfer

Although the continuous gypsum lining makes the wall quite airtight, air ex� ltration remains strong enough to increase the accumulated condensate behind the OSB-sheathing with some 21%, resulting in a change in moisture ratio by weigth of 44%, a number which is far too high. The � rst thing to do is to improve the airtighness of the wall. That measure should be combined with an improved vapour resistance.

Correction 2

Calculating the impact of sorption/desorption as correctly as feasible is only possible with an appropriate software tool. A � rst guess can be made by assuming a constant vapour resistance factor and constant speci� c moisture content for the gypsum board and the OSB-sheathing. A solution is then found by modeling the timber framed wall as a system of two resistance coupled capacitances, whereby the hypothesis is made that the cavity between the building paper and the brick veneer is enough ventilated to get the outside vapour pressure there. Equations (forward differences):

Capacitance 1: Gypsum board

i sat,1 1 sat,2 2 sat,1 1

gyps gyps i gyps gyps insul insul OSB OSB

1 1gyps gyps gyps

/ 2 1 / / 2 / 2

t t t t t t t

t t t

p p p p

N d N d N d N d

dt

+Δ

− φ φ − φ+

μ + β μ + μ + μ

φ − φ= ρ ξ

ΔCapacitance 2: OSB

sat,1 1 sat,2 2

gyps gyps insul insul OSB OSB

, ,2 2

OSB OSB bp bp cavity

2 2OSB OSB OSB

/ 2 / 2

/ 2 1 /

t t t t

t t t tsat e e sat

t t t

p p

N d N d N d

p p

N d N d

dt

+Δ

φ − φμ + μ + μ

φ − φ+

μ + μ + β

φ − φ= ρ ξ

Δ

Solving these equations demands knowledge of the boundary conditions. We refer to the data given above. Each month is modelled as a step function with the temperatures and saturation pressures invariable. If, in such case, we reach the steady state vapour pressure solution before the month ends, then Glaser re� ects the vapour addition in winter quite well. Time interval used: one day.Additional material properties:

Layer � dkg/m2

Gypsum board lining 0.46

Insulation (mineral � ber) 0

OSB-sheathing 0.932.83 beyond 90% RH

Building paper 0

1335vch02.indd 244 12.02.2007 16:55:39

245

Starting conditions: 80% relative humidity in the OSB, 50% relative humidity in the gypsum board. The solution shows that, if calculation starts on January 1st, hygroscopic moisture uptake extends until the 29th of that month, absorbing 0.43 kg/m2. Then, condensation at the backside of the OSB starts, depositing 0.048 kg/m2 during the two days left. That deposit causes droplets to be formed, ending in run off down to the sill plate. In February and March all moisture deposits as condensate that runs off, in total 0.485 kg/m2. In April, drying starts.

Correction 3

The solutions mentioned are all valid as long as the veneer wall is not wetted by wind driven rain. Suppose we look to a south-west oriented wall, the rain direction in Western Europe. In cool, humid climates, drying of a brick veneer, wetted by successive rain periods, proceeds so slowly between November and March that 100% relative humidity in the veneer is noted during that period. The building paper with an equivalent diffusion thickness of 0.1 m has been replaced by a high permeance one, equivalent diffusion thickness 0.01 m.

Boundary conditions are:

J F M A M J J A S O N D

Outdoors

Air temp.(°C) 2.3 3.3 5.9 8.8 13.5 16.1 16.8 15.7 13.5 8.8 5.5 2.9

Eq. temp.(°C) 5.5 7.0 9.2 12.3 17.3 20.0 20.8 19.7 17.3 12.3 8.8 6.1

Vapour press. (Pa) 619 675 825 997 1277 1427 1470 1406 1277 997 804 653

Indoors

Air temp.(°C) 20.0 20.3 21 21.8 23.1 23.8 24 23.7 23.1 21.8 20.9 20.2

Vapour press. (Pa) 1224 1253 1330 1418 1561 1638 1660 1627 1561 1418 1319 1242

2.5 Problems

1335vch02.indd 245 12.02.2007 16:55:39

246 2 Mass Transfer

In winter, things hardly change compared to no rain. Glaser gives some ondensation at the backside of the OSB, with an accumulated maximum of 0.47 kg/m2. In reality, that amount means hygroscopic wetting of the OSB, followed by a little deposit of condensate. When performing the Glaser calculation, the brick veneer should be kept on water vapour saturation pressure from November till March.

In summer, however, when the veneer has been wetted accidently by rain and is then heated by the sun, reversed vapour diffusion develops. As a result, condensation deposits on the wrap. On a sunny day, the amount reaches 0.16 kg/m2, which is close to the vapour � ow adsorbed by the OSB during the whole month of December! Without wrap, that condensate wets the OSB! The � gure underlines that the wet veneer situation leads to very high vapour saturation pressures during sunny summer weather.

1335vch02.indd 246 12.02.2007 16:55:40

247

(12) Given a pitched roof with section:

Layer Thicknessm

Densitykg/m3

�-valueW/(m · K)

�-value–

Air permeabilitym3/(m2 · s · Pa)

Lathed ceiling 0.01 450 0.14 85 2 � 10–4 �Pa–0.32

Insulation (mineral � ber) 0.15 20 0.04 1.2 0.00081

Underlay 0.0032 – 0.2 45 4.2 � 10–4 �Pa–0.4

Cavity 0.06 – 0.18 0 �

Tiles 0.012 1800 1 20 1.6 � 10–2 �Pa–0.5

Will the design suffer from unacceptable interstitial condensation? If so, what vapour retarder should be added where? Boundary conditions to be considered (slope: 30°, looking SW):

J F M A M J J A S O N D

Outdoors

Air temp. (°C) 2.3 3.3 5.9 8.8 13.5 16.1 16.8 15.7 13.5 8.8 5.5 2.9

Eq. temp. (°C) 3.9 5.2 8.7 12.6 19.1 22.5 23.5 22.1 19.1 12.6 8.2 4.7

Vapour press. (Pa) 619 675 825 997 1277 1427 1470 1406 1277 997 804 653

Indoors

Air temp. (°C) 20.0 20.3 21 21.8 23.1 23.8 24 23.7 23.1 21.8 20.9 20.2

Vapour press. (Pa) 1224 1253 1330 1418 1561 1638 1660 1627 1561 1418 1319 1242

2.5 Problems

1335vch02.indd 247 12.02.2007 16:55:40

248 2 Mass Transfer

Surface resistance for heat transport: inside 0.13 m2 · K/W, outside 0.04 m2 � K/W. Calcula-tions, if any, have to be done with the equivalent temperature. That monthly mean value in fact accounts for solar gain and the exponential relationship between temperature and water vapour saturation pressure

Answer

It is up to the reader to solve the problem and to formulate comments.

(13) Assume a timber framed wall has been designed for a cold climate. The air and vapour barrier consists of the gypsum board lining, � nished with a vinyl paper. The wall is used in a hot and humid climate now, where cooling and air drying is needed to keep the building livable. What problems could be expected?

Layer properties:

Layer Thicknessm

Densitykg/m3

�-valueW/(m · K)

�-value–

�d-valuem

Vinyl paper – – – 1

Gypsum board lining 0.012 – 0.1 12

Insulation (mineral � ber) 0.15 20 0.04 1.2

OSB-sheathing 0.01 400 0.13 15

Building paper 0.0005 – 0.2 200

Cavity 0.03 – 0.18 0

Brick veneer 0.09 1600 0.9 5

Boundary conditions (North orientation):

J F M A M J J A S O N D

Outdoors

Air temp. (°C) 22.5 23.2 25.3 28.1 30.9 32.9 33.6 32.9 30.9 28.1 25.3 23.2

Vapour press. (Pa) 1911 2001 2246 2581 2916 3160 3250 3160 2916 2581 2246 2001

Indoors

Air temp. (°C) 20.0 20.3 21 21.8 23.1 23.8 24 23.7 23.1 21.8 20.9 20.2

Vapour press. (Pa) 1224 1253 1330 1418 1561 1638 1660 1627 1561 1418 1319 1242

Surface resistances for heat transport: inside 0.13 m2 · K/W, outside 0.04 m2 · K/W.

Answer

It is up to the reader to solve the problem and to make comments.

1335vch02.indd 248 12.02.2007 16:55:41

249

(14) Given a masonry cavity wall.

Layer properties:

Layer Thicknessm

Densitykg/m3

�-valueW/(m · K)

�-value–

�d-valuem

Gypsum plaster 0.01 980 0.3 7

Inside leaf (hollow blocks) 0.14 1400 0.5 5

Insulation (mineral � ber) 0.12 30 0.04 1.2

Cavity 0.03 400 0.18 0

Brick veneer 0.09 1600 0.9 5

The cavity is ventilated. For that purpose, two head joints per meter run are left unpointed at the bottom and the top of the 3 m high wall. Section of the head joints: 1 � 6.5 cm2, depth: 9 cm.Suppose that in some of the walls, the insulation is correctly mounted. In others, it sits centrally in the cavity with an air space left of 1.5 cm at both sides. In a few, the insulation touches the brick veneer. The tray at the bottom guarantees an open contact between the cavities and air layers left and the open head joints, while at the top, the insulation stops just below the open head joints. This of course results in some thermal bridging which is not considered. To what extend is the U-value in the three cases affected by the ventilation � ow when on average a 4 Pa air pressure difference exists between the head joints at the top (highest air pressure) and the bottom (lowest air pressure). Surface � lm coef� cients: he = 25 W/(m2 · K), hi = 8 W/(m2 · K). Indoor temperature: 20 °C, outdoor temperature: –10 °C.

Answer

We solve the problem for the case of the insulation in the centre of the cavity. The two other cases are left for the reader to solve.

Air� ow

The air� ow is � xed by the two head joints per m3. Their hydraulic resistance in fact outweighs the hydraulic resistance of the two 1.5 cm thick air layers at both sides of the insulation. Hydraulic equation:

2 2a head jointa a a

a aa head joint a H H a head joint

22a aa a

96 22 1.5 0.42

2 2 2

0.32 0.08 757, 400 12.30.00065 0.00065

AG GLP

A G d d A

G GG G

⎡ ⎤⎛ ⎞ ⎛ ⎞ν ρρ⎢ ⎥Δ = + ρ ≈⎜ ⎟ ⎜ ⎟⎢ ⎥ρ ρ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

For �Pa = 4 Pa, the air� ow from top to bottom is 0.0023 kg/s. As the air layers at both sides of the insulation are of equal thickness, that � ow splits in two identical parts, one washing the air layer between the insulation and the brick veneer, the other washing the air layer between the insulation and the inside leaf.

2.5 Problems

1335vch02.indd 249 12.02.2007 16:55:41

250 2 Mass Transfer

Temperatures and Equivalent U-Value

As the insulation layer has a thermal resistance of 3 m2 · K/W, we may set the temperature in the air layer between the insulation and the brick veneer equal to the outside temperature. That way the problem does not differ from calculating the average U-value of a wall with a ventilated cavity between two leafs.

Thermal resistancesR1 = 3.070 m2 · K/W (the insulation and half of the air layer)R2 = 0.438 m2 · K/W (the plastered inside leaf, included the inside surface resistance)

ConstantsSee ‘Cavity ventilation’

D = (3.33 + 4.45 + 1 / 3.07) � (3.33 + 4.45 + 1 / 0.438) – 4.452 = 61.9 In D, 3.33 W/(m2 · K) is the convective and 4.45 W/(m2 · K) the radiation surface � lm

coef� cient in the cavity. As for a 1.5 cm wide cavity, the Nusselt number may be taken as 1, and the convective surface � lm coef� cient becomes:

0.025 / 0.015 � 2 = 3.33 W/(m2 · K).A1 = 0.0531A2 = 0.0235B1 = 0.1643B2 = 0.2993C1 = 0.7826C2 = 0.6773

Cavity surface temperatures

�s1 = 2.8 + 0.7826 �cav�s1 = 5.7 + 0.6773 �cav

1335vch02.indd 250 12.02.2007 16:55:41

251

Temperatures

See � gure. It is interesting to see how radiation impacts the cavity surface temperatures. In fact, just below the outside air inlet, the surface at the warm side as well as the one at the cold side are both warmer than the air in the cavity.

Equivalent U-Value

Should be given by

2 1 1o

2 1

1 1 expC b R L

U UL R b

⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪= + − −⎨ ⎬⎢ ⎥⎜ ⎟⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

with a1

c 1 2

1008

(2 )

Gb

h C C=

− −

Result: U = 0.55 W/(m2 · K), while the value without the 1.5 cm thick air layer behind the thermal insulation should have been 0.27 W/(m2 · K)! Hence, an increase over 100%, which of course is unacceptable.It is up to the reader to repeat the calculations for the other two cases (insulation correctly mounted, insulation touching the brick veneer).

2.6 References

[2.1] Langmuir, I. (1918). The adsorption of gases on plane surfaces of glass, mica and platinum. Journal of the American Chemical Society, No. 40.

[2.2] Brunauer, Emmett, Teller (1938). Adsorption of Gases in Multimolecular Layers. Journal of the American Chemical Society.

[2.3] Cranck, J. (1956). The mathematics of Diffusion. Clarendon Press, Oxford.

[2.4] De Grave, A. (1957). Bouwfysica 1. Uitgeverij SIC, Brussel (in Dutch).

[2.5] Glaser, H. (1958). Wärmeleitung und Feuchtigkeitsdurchgang durch Kühlraumisolierungen. Kältetechnik, Nr. 3 (in German).

[2.6] Glaser, H. (1958). Temperatur- und Dampfdruckverlauf in einer homogene Wand bei Feuchtig-keitsausscheidung. Kältetechnik, Nr. 6 (in German).

[2.7] Glaser, H. (1958). Vereinfachte Berechnung de Dampfdiffusion durch geschichtete Wände bei Ausscheiden von Wasser und Eis. Kältetechnik, Nr. 11 and 12 (in German).

[2.8] Glaser, H. (1959). Gra� sches Verfahren zur Untersuchung von Diffusionsvorgängen. Kälte-technik, Nr. 10 (in German).

[2.9] Devries, D. A. (1962). Een mathematisch-fysische behandeling van het transport van warmte en vocht in poreuze media. De Ingenieur, No. 28 (in Dutch).

[2.11] Cammerer, J. S. (1962). Wärme- und Kälteschutz in der Industrie. Springer-Verlag, Berlin, Heidelberg, New York (in German).

[2.12] Krischer, O. (1963). Die wissenschaftlichen Grundlagen der Trocknungstechnik, 2. Au� age. Springer-Verlag, Berlin (in German).

2.6 References

1335vch02.indd 251 12.02.2007 16:55:41

252 2 Mass Transfer

[2.13] Luikov, A. (1966). Heat and Mass Transfer in Capillar porous Bodies. Pergamon Press, Oxford.

[2.14] Harmathy, T. Z. (1967). Moisture sorption of building materials. Technical Paper No. 242, NRC, Division of Building Research, Ottawa.

[2.15] Vos, B. H., Coelman, E. J. W. (1967). Condensation in structures. Report Nr. BI-67-33/23 TNO-IBBC, Rijswijk.

[2.16] Vos, B. H., Tammes, E. (1969). Moisture and Moisture Transfer in Porous Materials. Report Nr. B1-69-96/03.1.001, TNO-IBBC, Rijswijk.

[2.17] Welty, Wicks, Wilson (1969). Fundamentals of Momentum, Heat and Mass Transfer. John Wiley & Sons, New York.

[2.18] Bomberg, H. (1971). Water� ow through porous materials, Part 1: Methods of water transport measurements. Report 19, Lund Institute of Technology, Section of Building Technology.

[2.19] Bomberg, H. (1971). Water� ow through porous materials, Part 2: Relative suction model. Report 20, Lund Institute of Technology, Section of Building Technology.

[2.20] Van der Kooy, J. (1971). Moisture transport in cellular concrete roofs. Uitgeverij Waltman, Delft.

[2.21] Gösele, Schüle (1973). Schall, Wärme, Feuchtigkeit, 3. Au� age. Bauverlag, Wiesbaden, Berlin (in German).

[2.22] Klopfer, H. (1974). Wassertransport durch Diffusion in Feststoffen, 1. Au� age. Bauverlag, Wiesbaden, Berlin (in German).

[2.23] Hens, H. (1975). Theoretische en experimentele studie van het hygrothermisch gedrag van bouw- en isolatiematerialen bij inwendige condensation en droging, met toepassing op de platte daken. doctoraal proefschrift, KU Leuven.

[2.24] Kreith, F. (1976). Principles of Heat Transfer. Harper & Row Publishers, New York

[2.25] Feynman, R., Leighton, R., Sands, M. (1977). Lectures on Physics, Vol. 1. Addison-Wesley Publishing Company, Reading, Massachusetts.

[2.26] TI-KVIV (1976, 1979, 1980, 1981, 1983, 1985) Kursus Thermische Isolatie en Vochtproblemen in Gebouwen (in Dutch).

[2.27] TU-Delft, Faculteit Civiele Techniek, Vakgroep Utiliteitsbouw-Bouwfysica (1975–1985). Bouwfysica, naar de colleges van Prof. A. C. Verhoeven (in Dutch).

[2.28] Hens, H. (1978, 1981) Bouwfysica, Warmte en Vocht, Theoretische grondslagen, 1. en 2. uitgave. ACCO, Leuven (in Dutch).

[2.29] Nevander, E. L., Elmarsson, B. (1981). Fukt-handbok. Svensk Byggtjänst, Stockholm (in Swedish).

[2.30] Kießl, K. (1983). Kapillarer und dampfförmiger Feuchtetransport in mehrschichtigen Bauteilen. Dissertation, Essen (in German).

[2.31] Kohonen, R., Ojanen, S. (1985). Coupled convection and conduction in two dimensional building structures. 4th conference on numerical methods in thermal problems, Swansea.

[2.32] Pato, Sectie Bouwkunde (1986). Syllabus van de leergang: Leidt energiebesparing tot vocht-problemen. Delft, 30 Sept. – 1 Oct. (in Dutch).

[2.33] Kohonen, R., Ojanen, T. (1987). Coupled diffusion and convection heat and mass transfer in building structures. Building Physics Symposium, Lund.

[2.34] Lutz, Jenisch, Klopfer, Freymuth, Krampf (1989). Lehrbuch der Bauphysik. B. G. Teubner Verlag, Stuttgart (in German).

1335vch02.indd 252 12.02.2007 16:55:41

253

[2.35] Taveirne, W. (1990). Eenhedenstelsels en groothedenvergelijkingen: overgang naar het SI. Pudoc, Wageningen (in Dutch).

[2.36] IEA-Annex 14 (1990). Condensation and Energy: Guidelines and Practice. ACCO, Leuven.

[2.37] Pedersen, C. R. (1990). Combined heat and moisture transfer in building constructions. Ph. D. Thesis, Technical University of Denmark.

[2.38] Chaddock, J. B., Todorovic, B. (1991). Heat and Mass Transfer in Buiding Materials and Structures. Hemisphere Publishing Corporation, New York.

[2.39] Hens, H. (1992, 1997, 2000). Bouwfysica 1, Warmte en Massatransport, 3., 4. en 5. uitgave. ACCO, Leuven (in Dutch).

[2.40] Garrecht, H. (1992). Porenstrukturmodelle für den Feuchtehaushalt von Baustoffen mit und ohne Salzbefrachtung und rechnerische Anwendung auf Mauerwerk. Doktors Abhandlung, Universität Karlsruhe (in German).

[2.41] Künzel, H. M. (1994). Verfahren zur ein- und zweidimensionalen Berechnung des gekoppelten Wärme- und Feuchtetransports in Bauteilen mit einfachen Kennwerten. Doktors Abhandlung, Universität Stuttgart (in German).

[2.42] Trechsel, H. R. (Ed.) (1994). Moisture Control in Buildings. ASTM Manual Series, MNL 18.

[2.43] Krus, M. (1995). Feuchtetransport- und Speicherkoef� zienten poröser mineralischer Baustoffe. Theoretische Grundlagen und Neue Meßtechniken. Doktors Abhandlung, Universität Stuttgart (in German).

[2.44] de Witt, M. (1995). Warmte en vocht in constructies. diktaat TUE, Eindhoven (in Dutch).

[2.45] Hens, H., (1996). Modelling, Vol. 1 of the Final Report Task 1, IEA-Annex 24. ACCO, Leuven.

[2.46] Time, B. (1998). Hygroscopic Moisture Transport in Wood. Doctors thesis, NUST Trond-heim.

[2.47] Arfvidsson, J. (1998). Moisture Transport in Porous Media, Modelling Based on Kirchhoffs Potentials. Doctoral Dissertation, Lund University.

[2.48] Time, B. (1998). Hygroscopic Moisture Transport in Wood. Doctors thesis, NUST Trond-heim.

[2.49] Arfvidsson, J. (1998). Moisture Transport in Porous Media, Modelling Based on Kirchhoffs Potentials. Doctoral Dissertation, Lund University.

[2.50] Brocken, H. (1998). Moisture Transport in Brick masonry: the Grey Area between Bricks. Doctoraal proefschrift, TUE, Eindhoven.

[2.51] Janssens, A. (1998). Reliable Control of Interstitial Condensation in Lightweight Roof Systems. Doctoraal proefschrift, KU Leuven.

[2.52] Häupl, P., Roloff, J. (Ed.) (1999). Proceedings of the 10th Bauklimatisches Symposium, Vol. 1 and 2, TU Dresden, Institut für Bauklimatik.

[2.53] Roels, S. (2000). Modelling Unsaturated Moisture Transport in Heterogeneous Limestone. Doctoraal proefschrift, KU Leuven.

[2.54] Sedlbauer, K. (2001). Vorhersage von Schimmelpilzbildung auf und in Bauteilen. Doktors Abhandlung, Universität Stuttgart (in German).

[2.55] Hagentoft, C. E. (2001). Introduction to Building Physics. Studentlitteratur, Lund.

[2.56] Holm, A. (2001). Ermittlung der Genauigkeit von instationären hygrothermischen Bauteil-berechnungen mittels eines stochastischen Konzeptes. Doktors Abhandlung, Universität Stuttgart (in German).

2.6 References

1335vch02.indd 253 12.02.2007 16:55:41

254 2 Mass Transfer

[2.57] Janssen, H. (2002). The in� uence of soil moisture transfer on building heat loss via the ground. Doctoraal proefschrift, KU Leuven.

[2.58] Häupl, P., Roloff, J. (Ed.) (2002). Proceedings of the 11th Bauklimatisches Symposium, Vol. 1 and 2. TU Dresden, Institut für Bauklimatik.

[2.59] Blocken, B., Hens, H., Carmeliet J. (2002). Methods for the Quanti� cation of Driving Rain on Buildings. ASHRAE Transactions, Vol. 108, Part 2, pp. 338–350.

[2.60] Hall, C., D’Hoff, W. (2002). Water transport in brick, stone and concrete. Spon Press, London and New York.

[2.61] Van Mook, J. R. (2003). Driving Rain on Building Envelopes. TU/e Bouwstenen 69.

[2.62] Carmeliet, J., Hens, H., Vermeir, G. (Ed.) (2003). Research in Building Physics. Balkema publishers, Lisse, Abingdon, Exton, Tokyo.

[2.63] Blocken, B. (2004). Wind-driven rain on buildings. Doctoraal proefschrift, KU Leuven.

[2.64] Kalagasidis, A. (2004). HAM-tools, An Integrated Simulation Tool for Heat, Air and Moisture Transfer Analysis in Building Physics. Ph. D. Thesis, Chalmers University of Technology.

[2.65] ASHRAE (2005). Handbook of Fundamentals. Atlanta.

[2.66] Rose, W. B. (2005). Water in Buildings. John Wiley and Sons, Inc., New York.

1335vch02.indd 254 12.02.2007 16:55:41

3 Combined Heat, Air and Moisture Transfer

3.1 Overview

With the exception of some paragraphs, notably ‘combined air and heat transfer’, ‘interstitial condensation’ and ‘drying’, heat and mass have been kept separate in the two preceding parts. In reality, they are intertwined:

Mass transfer means enthalpy and, thus, heat transfer. This was explained in Chapter 2.Temperature differences induce mass displacement. We refer to natural convection in air, water vapour diffusion and pore water movement by changes in surface tension.The thermal characteristics (volumetric speci� c heat capacity, thermal conductivity) are a function of moisture content.Moisture transfer characteristics are a function of temperature.

These interrelationships are the reason why a separate analysis oversimpli� es reality. Much information and understanding is lost that way. Therefore, in this last chapter, we combine heat, air and moisture transfer (HAM).

3.2 Assumptions

An exact mathematical treatment of combined heat, air and moisture is not possible. Precipitation of soluble substances, crystallization and hydration of salts and diffusion of solid substances is such that the material matrix itself is variable, that the pores modify in form and equivalent diameter and that the speci� c pore surface changes, etc. Such effects are excluded in advance. Thus, a � rst assumption is that the material matrix never alters. Consequently, material characteristics, such as the dry density, the pore distribution, the speci� c pore surface, etc. are considered invariable.Furthermore, it is assumed that each porous material is a continuum, composed of an in� nite number of representative volumes (Vrep), each possessing all characteristics of the material. In other words, each representative volume contains the complete pore distribution, has the same air permeability as the material, etc. The transition to a continuous medium assumes that these volumes are in� nitely small. Of course, this is not the case. That is why a continuum approach should be considered as a calculation, using the average values of all quantities at the level of the representative volumes.

3.3 Solution

A problem of combined heat, air and moisture is solved when the temperature, the air pressure, the moisture potential and the heat, air and moisture � ow rates at each point of the ‘solution space’ – in building physics, the material, the construction part, the building – are known. While the three potentials are scalars, the � ow rates are vectors. Consequently, we need a system of three scalar and three vector equations per representative volume, completed with

�

�

�

�

1335vch03.indd 255 12.02.2007 11:37:25

Building Physics - Heat, Air and Moisture: Fundamentals and Engineering Methods with Examples and Exercises by Hugo Hens

Copyright © 2007 Ernst & Sohn Verlag für Architektur und technische Wissenschaften GmbH und Co. KG, Berlin

256 3 Combined Heat, Air and Moisture Transfer

the necessary equations of state. These describe the thermodynamic equilibriums between the separate potentials and between the potentials and certain material characteristics. Even then, a system can only be solved if its geometry is perfectly known and the initial, boundary and contact conditions are given.

3.4 Conservation Laws

Combined heat, air and moisture are governed by the conservation laws of classical physics:

conservation of massconservation of energyconservation of momentum

As far as mass � ow in porous materials is concerned, conservation of momentum is substituted by the empirical � ow equations. For convection in cavities and air volumes and for mass movement through highly permeable materials, reference to the Navier-Stokes equations, combined with a turbulence model, may be necessary

3.4.1 Conservation of Mass

The pores of a humid material contain at least three and often four of the following mass components:

� � 0 air, liquid water, water vapour, ice (salts, and other substances …)� � 0 air, liquid water, water vapour (salts, and other substances …)

In the framework of this book, salts and other substances are not taken into account. For liquid water, water vapour, ice and air, the quantities per unit volume of material are determined by using the concepts of water content, vapour content, ice content and air content:

� < 0 � � 0

Water ll

mw

V= l

lm

wV

=

Water vapour vv

mw

V= v

vm

wV

=

Ice ii

mw

V=

Air aa

mw

V= a

am

wV

=

Moisture content (w) bundles liquid water, water vapour and ice content. Most of the time, water vapour content can be neglected and the following applies: w = wl + wi. For the air content, we may write (not in case other gasses � ll the pores): wa = �of �a where �of is the pore volume which does not contain liquid water or ice and �a is the air density. For water

���

1335vch03.indd 256 12.02.2007 11:37:25

257

vapour content, we have: wv = xv wa with xv the vapour ratio in the pore air in kg/kg. Applying mass conservation to each of the components gives:

Liquid water

l1div

wG

t

∂± = −′

∂wg (3.1)

where gw is the water � ow rate and 1G ′ a water source (condensation, local water supply) or sink (evaporation).

Water vapour

v2div

wG

t

∂± = −′

∂wg (3.2)

where gv is the vapour � ow rate and 2G′ a vapour source (evaporation) or sink (condensation). For hygroscopic materials, the hygroscopic moisture content wH replaces the water vapour content and the buffering term becomes: H /w t−∂ ∂ (see Chapter 2). Above freezing, the sum of the condensation related sources and sinks in equations (3.1) and (3.2) is zero. Condensation, in fact, is a source in the water balance (= generation of water) but a sink in the vapour balance (= removal of water vapour). For evaporation, the reverse is true.

Ice

i3div

wG

t

∂± = −′

∂ig

where gi is the ice � ow rate and 3G′ an ice source caused by freezing of water or condensation and freezing of water vapour, or a sink as a consequence of ice melting and ice sublimation. In porous materials, ice formation starts at 0 °C in the largest pores. As the temperature goes down, the water in the smaller pores also freezes. As ice is a solid substance, gi may be set at zero, in which case the equation becomes:

i3

wG

t

∂± = −′

∂ (3.3)

Summing up the condensation related sources and sinks in equations (3.1), (3.2) and (3.3) gives: 1 2 3 0G G G+ + =′ ′ ′ . That allows us to unite the conservation laws for water, water vapour and ice into one single equation (with w the moisture content in the material):

w vdiv ( )w

g gt

∂+ = −

∂ (3.4)

Air

a aofdiv

∂ ∂ρ= − = −Ψ

∂ ∂agw

t t

where ga is the air � ow rate. Normally in air equilibrium neither sources nor sinks intervene. Further, we may assume that, compared to changes in moisture and heat content, changes in air content occur so fast that air inertia is more or less zero (see air transfer). The equilibrium thus simpli� es to:

div 0=ag (3.5)

3.4 Conservation Laws

1335vch03.indd 257 12.02.2007 11:37:25

258 3 Combined Heat, Air and Moisture Transfer

3.4.2 Conservation of Energy

For heat conduction, the conservation law was limited to media free of mass � ows. Here, we extend the law to porous media which experience air, liquid water and water vapour migration. Mass � ow induces enthalpy � ow:

j o,j( )h h= −j jq g (3.6)

with hj the speci� c enthalpy for mass component j and ho,j the reference value, normally for 0 °C. Units: J/kg. Energy conservation thus looks like (reference value ho,j zero, kinetic energy and frictional heat too small to play a role):

3 4

j 0 01 1

div ( ) ( )j jj j

h c w ht= =

⎡ ⎤ ⎡ ⎤∂+ ± Φ = − ρ θ +′⎢ ⎥ ⎢ ⎥∂⎣ ⎦ ⎣ ⎦∑ ∑jq g (3.7)

with �0 c0 the volumetric speci� c heat capacity of the matrix. The term �� represents heat dissipation by processes others than changes of state. These are included in (wj hj):

Water: hl = cl (T – T0) = cl �

Water vapour: hv = cv (T – T0) + lb0 = cv � + lb0

Ice: hi = ci (T – T0) – lm0 = ci � – lm0

Dry air: ha = ca (T – T0) = ca �

with T0 = 273.15 K. lb0 is the heat of evaporation and lm0 the melting heat, both at 0 °C (respectively 2,500,000 J/kg and 330,000 J/kg). If we suppose the material matrix and all mass components to be at the same temperature in each representative volume Vrep (this hypothesis is correct if the pores are small and � ow velocities small. For air � owing through apertures, cavities, cracks, � ssures etc. it does not apply), then changes in enthalpy can be written as: j jd dh c= θ and j jh c= θgrad grad . Equation (3.7) thus converts into (vector calculation):

3 4

j j 0 01 1

4j

j1

div ( div ) ( )j jj j

j

h h c c wt

wh

t

= =

=

⎧ ⎫⎡ ⎤∂ ⎪ ⎪− = + + ρ + θ⎨ ⎬⎢ ⎥∂ ⎪ ⎪⎣ ⎦⎩ ⎭∂⎛ ⎞

+ ± Φ′⎜ ⎟∂⎝ ⎠

∑ ∑

∑

j jq g g grad

or

3 3 4

j j1 1 1

(2)

4

0 01

(1)

div ( ) ( div )

( )

jj

j j j

j jj

wc h h

t

c c wt

= = =

=

∂⎛ ⎞− − θ = + ⎜ ⎟∂⎝ ⎠

⎧ ⎫⎡ ⎤∂ ⎪ ⎪+ ρ + θ ± Φ′⎨ ⎬⎢ ⎥∂ ⎪ ⎪⎣ ⎦⎩ ⎭

∑ ∑ ∑

∑

j jq g grad g

�������������

�����������

(3.8)

1335vch03.indd 258 12.02.2007 11:37:26

259

Term (1) includes the volumetric speci� c heat capacity of the matrix and of three or four mass components. Water vapour and air can now be neglected, turning the term into:

4

0 0 j j 0 0 1 1 i i 01( ) ( )

jc c w c c w c w c

=

⎡ ⎤ρ + = ρ + + = ρ ′⎢ ⎥⎣ ⎦

∑

or

1 1 i i0

0

c w c wc c

+= +′

ρ (3.9)

Term (2) can be rewritten as (remember that for ice gi = 0):

jj div

⎡ ⎤∂⎛ ⎞+⎢ ⎥⎜ ⎟∂⎝ ⎠⎢ ⎥

⎢ ⎥⎣ ⎦

∑ jg

�������

wh

t

The term above the brace stands for mass conservation per component without source or sink. Therefore, that part simpli� es to:

3

j j1jh G

=′∑ (3.10)

Sums 1 to 3 indicate that changes of state are excluded for mass component number four, i.e. dry air. For mass component number three, ice, the source term ( 3G′) is zero above freezing, or: 1 2G G= −′ ′ . Equation (3.10) then reduces to:

3

j j 1 1 b0 v 1 b1

( ) ( )j

h G G c l c G l=

= θ − − θ = − θ′ ′ ′∑ ,

with b b0 v l( ) ( )l l c cθ = + − θ .

Below freezing, 1 2 3 0G G G+ + =′ ′ ′ and 3

j j 1 b 3 m1

( ) )j

h G G l G l=

= − θ − (θ′ ′ ′∑ .

Inserting these relationships into equation (3.8) gives as � nal result:

Above freezing:

�3

j 1 b 01 (5)(1) (3)

(4)(2)

div ( grad ) ( )=

∂− − θ = − + ρ θ ± Φ′ ′ ′∂∑ jq g��� ��� ������������j

c G l ct

(3.11)

Below freezing:

�3

j 1 b 3 m 01 (5)(1) (3)

(4)(2)

div ( grad ) ( )=

∂− − θ = − − + ρ θ ± Φ′ ′ ′ ′∂∑ jq g��� ������� ������������j

c G l G l ct

(3.12)

3.4 Conservation Laws

1335vch03.indd 259 12.02.2007 11:37:26

260 3 Combined Heat, Air and Moisture Transfer

The terms (1) to (5) represent:

(1) Heat transfer by equivalent conduction. The term equivalent indicates that, at a micro scale, we look at more than pure conduction through the matrix. Indeed, also conduction and convection in the pore gas, radiation between the pore walls, conduction in all absorbed water layers and latent heat transfer by local evaporation/condensation loops can be attributed to heat mitigation. All that turns the material characteristic ‘(equivalent) thermal conductivity’ into a function of temperature, moisture content, thickness, etc.

(2) Sensible enthalpy transfer as a result of mass � ow. That is especially true for air movement. Liquid water and water vapour � ow rates are usually too small to play any role of importance.

(3) Latent enthalpy transfer. Inter-porous changes of state between liquid water, water vapour and ice are the drivers here. Evaporation of liquid water and condensation of water vapour are dominant in that respect.

(4) Stored heat. The material matrix and water content play the leading role in that buffering process.

(5) Heat sources and sinks as a consequence of chemical and electrical processes.

3.5 Flow Equations

Contrary to the conservation axioms, the � ow equations are all mathematical translations of empirical observations (see Chapters 1 and 2).

3.5.1 Heat

Heat conduction is described by Fourier’s law: q = – grad �, with (equivalent) thermal conductivity (W/(m · K)). For anisotropic materials this is a tensor. Cavities and air layers demand a speci� c treatment (see heat transfer).

3.5.2 Mass, Air

3.5.2.1 Open Porous Materials

Here, Poiseuille’s law holds: ga = –ka grad Pa, with ka air permeability (s) and Pa air pressure (Pa), included buoyancy (see Chapter 2, Section 2.2 “Air Transfer”). Again, the air permeability is a tensor for anisotropic materials, while its value drops to zero for capillary wet materials. One may write:

aa a

1 Bk

W= =

ν ν′

where �a is the kinematical viscosity of air and B the penetration coef� cient of the material (m2/s2). B is a pure pore system characteristic, not in� uenced by the kind of � uid migrating through it.

1335vch03.indd 260 12.02.2007 11:37:26

261

3.5.2.2 Air Permeable Layers, Apertures, Joints, Cracks, Leaks and Cavities

In this case a permeance law is used:

Air permeable layers (per m2) a a ag K P= − Δ , Ga in kg/(m2 · s) and Ka in s/m

Joints, cracks, cavities (per m) a a aG K Pψ= − Δ , Ga in kg/(m · s) and aKψ in s

Leaks, voids, apertures (each) a a aG K Pχ= − Δ , Ga in kg/s and aK χ in m · s

where axK is the air permeance and �Pa the difference in air pressure along the � ow direction,

including again buoyancy a( )g z−ρ (in Pa). The air permeance typically is a function of the air pressure difference: 1

a ax bK a P −= Δ .

Apertures, joints, cracks, voids and leaks are part of the geometry. In most cases, however, the speci� c location of the last three is hardly known, which makes air and liquid water movement simulations troublesome, with results many times far from reality.

3.5.3 Mass, Moisture

3.5.3.1 Water Vapour

Diffusion

Water vapour diffusion is described by Fick’s law: gv = – grad p, where is the vapour permeability, units: seconds (s), a tensor for anisotropic materials. Its value re� ects the magnitude and tortuosity of the pore system. At the same time, the property is a function of relative humidity and temperature in the pores. That a diffusion law is used does not mean diffusive vapour transfer in a porous system is purely Fickian. Friction diffusion, liquid � ow in the adsorbed water layers and liquid � ow in the separate water islands in the pores also intervene.

Convection

Convective vapour transfer is given by:

6v

a a

0.622 0.6226.21 10

px

P p P−= = ≈ ≈ ⋅

−a a

v a ag g

g g g

with ga the air � ow rate (kg/(m2 · s)) and Pa air pressure (Pa). The sign � means that those simpli� ed expressions are only applicable for temperatures below 50 °C and very small air pressure differences compared to atmospheric.

3.5.3.2 Water

For unsaturated water � ow, a law similar to Darcy’s for saturated water � ow is used: gw = –kw grad s, with kw water permeability, a tensor for anisotropic materials, units: seconds, and s suction. Water permeability normally stays very small up to a certain moisture content and increases steeply beyond that point.

3.5 Flow Equations

1335vch03.indd 261 12.02.2007 11:37:26

262 3 Combined Heat, Air and Moisture Transfer

3.6 Equations of State

3.6.1 Enthalpy/Temperature and Water Vapour Saturation Pressure/Temperature

See above.

3.6.2 Relative Humidity/Moisture Content

This relationship is known as the sorption curve. Its derivative represents the relative humidity related speci� c moisture content.

3.6.3 Suction/Moisture Content

That relationship is called the moisture characteristic. In case of contact with a water surface, the characteristic has to be known from a moisture content zero up to capillary. The derivative, called the suction related speci� c moisture content, supersedes the relative humidity related speci� c moisture content beyond the relative humidity threshold �m, when capillary condensation has created continuous � ow paths in a porous material.

3.7 Starting, Boundary and Contact Conditions

3.7.1 Starting Conditions

The starting conditions de� ne the heat, air and moisture situation in a construction at time zero. Starting conditions may be assumed, are the result of measurements, and even become super� uous for steady state or harmonic � uctuations over time. For example, interstitial condensation according to the Glaser and the advection method rely on a very simple starting condition: a dry construction.

3.7.2 Boundary Conditions

The term boundary conditions applies to the temperatures and/or heat � ow rates, vapour pressures and/or vapour � ow rates, suction (moisture content) and/or liquid water � ow rates and air pressures and/or air � ow rates that act on the bounding surfaces of the construction during the time covered by the calculation. Boundary conditions are steady-state or transient, and they may be constant or variable along the bounding surfaces, etc. They may even change in nature during the time slot considered. An example of that is wind driven rain at the time of � lm formation: from a known water � ow rate at the surface to contact with water once moisture content at the surface is capillary. No heat, air and moisture problem can be solved without knowledge of the boundary conditions. However, constructing an exact copy of reality is not feasible. This is why in many cases schematized boundary conditions are used.

1335vch03.indd 262 12.02.2007 11:37:26

263

3.7.3 Contact Conditions

‘Contact conditions’ is the general term to indicate the bounding conditions in the interfaces between layers and materials. For heat, air and moisture transfer, the driving potential is supposed to be unequivocally known in each interface while the � ow rates in materials and layers at both sides are assumed to be equal. The following observations apply (see also Chapter 2):

Heat

For materials with a high thermal conductivity, such as metals, contact resistances have to be taken into account.

Moisture

Water vapourVapour pressure and relative humidity are unequivocal in free contacts between layers and materials. In all other cases, contact resistances intervene.

Liquid waterIn ideal contacting interfaces, liquid water and vapour � ow are continuous. Instead, continuity of water transfer is impossible in free contacting interfaces. In such case only water vapour will experience continuity, while rain, interstitial condensation and pressure � ow can cause a water � lm to develop. That � lm is either absorbed or may be retained in the contact zone by capillary suction, may � ow down by gravity or drips off. Real contacts � nally add resistance to the system, as is the case for glued surfaces and physical-chemical interaction.

Air

A free contact may produce air � ows parallel to the interfaces.

3.8 Two Examples of Simpli� ed Models

3.8.1 Heat, Air and Moisture Transfer in Non-Hygroscopic, Non-Capillary Materials

Insertion of the air � ow equation into mass conservation for air gives:

a adiv ( ) 0k P =grad

The overall air pressure Pa includes the actual air pressure aP ′ as well as buoyancy (–�a g z). When air permeability is a constant, which is mostly not the case, the equation simpli� es to:

2a a( ) 0P∇ − ρ =′ g z (3.13)

Thermal buoyancy (–�a g z) does not intervene in isothermal conditions. In non-isothermal conditions, the term differs from zero in all directions except the horizontal.

�

�

3.8 Two Examples of Simpli� ed Models

1335vch03.indd 263 12.02.2007 11:37:26

264 3 Combined Heat, Air and Moisture Transfer

Inserting the water vapour � ow equation into mass conservation yields:

v2

a

0.622div

wp p G

P t

⎛ ⎞ ∂δ − ± =′⎜ ⎟ ∂⎝ ⎠

aggrad

In non-hygroscopic materials, the storage term represents the increase and decrease of water vapour concentration in the pore air:

v 0w p

t t R T

∂ Ψ⎛ ⎞∂= ⎜ ⎟∂ ∂ ⎝ ⎠

The condensation and drying term 2G′ remains zero as long as relative humidity stays below 100%. That is controlled using the equation of state for the water vapour saturation pressure (psat(�)). If so, the equation simpli� es to:

o

a

0.622div

pp p

P R t T

⎛ ⎞ Ψ ∂ ⎛ ⎞δ − = ⎜ ⎟⎜ ⎟ ⎝ ⎠∂⎝ ⎠ag

grad

At 100% relative humidity, we get:

o satsat sat 2

a

0.622 1div

pp p G

P R t T

⎛ ⎞ Ψ∂ ⎛ ⎞δ − ± =′ ⎜ ⎟⎜ ⎟ ⎝ ⎠∂⎝ ⎠ag

grad

Compared to condensation and drying 2( )G′ , the right-hand storage term is negligible. So:

sat sat 2a

0.622div ( )p p G

Pδ − = ± ′ag

grad grad (3.14)

Energy conservation gives:

3

j 2 b 01

div ( ) ( ) ( )j

c G l ct=

∂λ θ − θ = − + ρ θ ± Φ′ ′ ′

∂∑ jgrad g grad

In the enthalpy transfer only air plays a role of importance. Without condensation or drying, no heat of transformation intervenes. In the speci� c heat term, the material dominates. All that simpli� es the expression to:

a 0 0div ( ) c ct

∂θλ θ − θ = ρ

∂agrad g grad (3.15)

With condensation and drying, equation (3.15) turns into:

a b sat b sata

0

0.622div ( ) div ( )c l p l p

P

ct

λ θ − θ + δ −

∂θ= ρ ′

∂

aa

ggrad g grad grad grad

or:

sat satb a b 0

a

d d0.622div

d d

p pl c l c

P t

⎛ ⎞⎡ ⎤ ∂θ⎛ ⎞λ+ δ θ − + θ = ρ ′⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠θ θ ∂⎝ ⎠⎣ ⎦agrad g grad (3.16)

1335vch03.indd 264 12.02.2007 11:37:26

265

The combined heat, air moisture transfer model for non-hygroscopic, non-capillary materials is ready for use now. With the equation of state psat(�), we look to a system of 4 equations with 4 unknowns: temperature �, water vapour pressure p, condensation or evaporation rate

2G′ and air pressure aP ′ . The � ow rates ensue from the � ow equations. Solving the system demands an iterative process. One has to assume a temperature distribution, calculate the air � ows, recalculate the temperatures with these air � ow rates, recheck the air � ows, etc. Without air � ow the model is much simpler:

Neither condensation nor drying

0div ( )p

pt R T

Ψ⎛ ⎞∂δ = ⎜ ⎟∂ ⎝ ⎠

grad

0 0div ( ) ct

∂θλ θ = ρ

∂grad

Both equations are connected when the material properties depend on temperature. At any rate, a solution assuming constant properties may already give valuable information.

Condensation or drying

satc

ddiv

d

pG

⎛ ⎞δ θ = ± ′⎜ ⎟⎝ ⎠θgrad

satb 0 0

ddiv

d

pl c

t

⎡ ⎤ ∂θ⎛ ⎞λ + δ θ = ρ⎢ ⎥⎜ ⎟⎝ ⎠θ ∂⎣ ⎦grad

Temperature is the only driving potential to be calculated. The condensation or drying � ows per volume unit 2( )G′ are also unknown.

3.8.2 Heat, Air and Moisture Transfer in Hygroscopic Materials at Low Moisture Content

Low moisture content means that the material is hygroscopically loaded. In that case, only ‘equivalent’ vapour transfer is left. Mass conservation gives:

H

a

0.622div

wp p

P t

⎛ ⎞ ∂δ − =⎜ ⎟ ∂⎝ ⎠

aggrad

Assuming air transfer to be negligible in � ne-porous materials, the equation simpli� es to: Hdiv ( ) /p w tδ = ∂ ∂grad . Introducing temperature and relative humidity as driving forces

transforms that simpler equation into:

satsat

ddiv

d

pp

tφ∂φ⎛ ⎞δ φ + δ φ θ = ρ ξ⎜ ⎟⎝ ⎠θ ∂

grad grad (3.17)

Conservation of energy yields:

3

j 1 b 01

div ( ) ( ) ( )j

c G l ct=

∂λ θ − θ = − + ρ θ ± Φ′ ′ ′

∂∑ jgrad g grad

The source or sink term �� (other than evaporation/condensation) is zero in most cases. In the enthalpy term only water vapour intervenes. For diffusion only, the � ow rates are too small to take enthalpy displacement into account. Phase transformations are restricted to sorption and desorption of hygroscopic moisture. All that allows writing mass � ow 1G′ as:

3.8 Two Examples of Simpli� ed Models

1335vch03.indd 265 12.02.2007 11:37:27

266 3 Combined Heat, Air and Moisture Transfer

sat1 sat

ddiv

d

pG p

tφ∂φ ⎛ ⎞= ρ ξ = δ φ + δ φ θ′ ⎜ ⎟⎝ ⎠∂ θ

grad grad

Insertion in the energy balance gives:

satb sat b 0

ddiv

d

pl p l c

t

⎡ ⎤ ∂θ⎛ ⎞δ φ + λ + δ φ θ = ρ ′⎢ ⎥⎜ ⎟⎝ ⎠θ ∂⎣ ⎦grad grad (3.18)

That � xes combined heat and moisture for hygroscopic materials at low moisture content with relative humidity (�) and temperature (�) as driving forces. The system of two partial differential equations with variable coef� cients can only be solved by using CVM or FEM.

3.9 References

[3.1] Welty, Wicks, Wilson (1969). Fundamentals of Momentum, Heat and Mass Transfer. John Wiley & Sons, New York.

[3.2] Hens, H. (1978, 1981). Bouwfysica, Warmte en Vocht, Theoretische grondslagen, 1e en 2e uitgave. ACCO, Leuven (in Dutch).

[3.3] Kohonen, R., Ojanen, S. (1985). Coupled convection and conduction in two dimensional building structures. 4th conference on numerical methods in thermal problems, Swansea.

[3.4] Kohonen, R., Ojanen, T. (1987). Coupled diffusion and convection heat and mass transfer in building structures. Building Physics Symposium, Lund.

[3.5] Taveirne, W. (1990). Eenhedenstelsels en groothedenvergelijkingen: overgang naar het SI. Pudoc, Wageningen (in Dutch).

[3.6] Pedersen, C. R. (1990). Combined heat and moisture transfer in building constructions. Ph. D. Thesis, Technical University of Denmark.

[3.7] Chaddock, J. B., Todorovic, B. (1991). Heat and Mass Transfer in Building Materials and Structures. Hemisphere Publishing Corporation, New York.

[3.8] Duforestel, T. (1992). Bases métrologiques et modèles pour la simulation du comportement hygrothermique des composants et ouvrages du bâtiment. Thèse de Doctorat, Ecole Nationale des Ponts et des Chausses, Paris (in French).

[3.9] Garrecht, H. (1992). Porenstrukturmodelle für den Feuchtehaushalt von Baustoffen mit und ohne Salzbefrachtung und rechnerische Anwendung auf Mauerwerk. Dissertation, Universität Karlsruhe (in German).

[3.10] Künzel, H. M. (1994). Verfahren zur ein- und zweidimensionalen Berechnung des gekoppelten Wärme- und Feuchtetransports in Bauteilen mit einfachen Kennwerten. Dissertation, Universität Stuttgart (in German).

[3.11] Trechsel, H. R. (Ed.) (1994). Moisture Control in Buildings. ASTM Manual Series: MNL 18, PCN 28-018094-10, Philadelphia, PA, 19103.

[3.12] Krus, M. (1995). Feuchtetransport- und Speicherkoef� zienten poröser mineralischer Baustoffe. Theoretische Grundlagen und Neue Meßtechniken. Dissertation, Universität Stuttgart (in German).

[3.13] Hens, H. (1996). Modelling. Final report task 1, Annex 24, Vol 1. ACCO, Leuven.

[3.14] Time, B. (1998). Hygroscopic Moisture Transport in Wood. Doctors thesis, NUST Trond heim.

1335vch03.indd 266 12.02.2007 11:37:27

267

[3.15] Arfvidsson, J. (1998). Moisture Transport in Porous Media, Modelling Based on Kirchhoffs Potentials. Doctoral Dissertation, Lund University.

[3.16] Brocken, H. (1998) Moisture Transport in Brick masonry: the Grey Area between Bricks. Doctoraal proefschrift, TUE, Eindhoven.

[3.17] Janssens, A. (1998). Reliable Control of Interstitial Condensation in Lightweight Roof Systems. Doctoraal proefschrift, KU Leuven.

[3.18] Roels, S. (2000). Modelling Unsaturated Moisture Transport in Heterogeneous Limestone. Doctoraal proefschrift, KU Leuven.

[3.19] Holm, A. (2001). Ermittlung der Genauigkeit von instationären hygrothermischen Bayteil-berech nungen mittels eines stochastischen Konzeptes. Dissertation, Universität Stuttgart (in German).

[3.20] Janssen, H. (2002). The in� uence of soil moisture transfer on building heat loss via the ground. Doctoraal proefschrift, KU Leuven.

[3.21] Häupl, P., Roloff, J. (Ed.) (2002). Proceedings of the 11th Bauklimatisches Symposium, Vol. 1 and 2. TU Dresden, Institut für Bauklimatik.

[3.22] Blocken, B., Hens, H., Carmeliet, J. (2002). Methods for the Quanti� cation of Driving Rain on Buildings. ASHRAE Transactions, Vol. 108, Part 2, pp. 338–350.

[3.23] Van Mook, J. R. (2003). Driving Rain on Building Envelopes. TU/e Bouwstenen 69.

[3.24] Carmeliet, J., Hens, H., Vermeir, G. (Ed.) (2003). Research in Building Physics. Balkema Publishers, Lisse, Abingdon, Exton (PA), Tokyo.

[3.25] Xiaochuan, Q. (2003). Moisture transport across interfaces between building materials. Ph. D. Thesis, Concordia University, Montreal.

[3.26] Blocken, B. (2004). Wind-driven Rain on Buildings. Doctoraal proefschrift, KU Leuven.

3.9 References

1335vch03.indd 267 12.02.2007 11:37:27

4 Postscript

A sound knowledge of heat and mass transfer forms the basis for an application-oriented evaluation of the overall thermal performance and moisture tolerance of building parts and buildings. It also allows understanding the consequences for design, production and construction. Nevertheless, that knowledge represents only a small fraction of what is needed for a global performance analysis when designing and constructing buildings, retro� tting existing ones and developing pre-fabricated building parts. In fact, from a building physics point of view, a building and its parts must not only ful� l heat, air and moisture related performance criteria, but also acoustic and lighting related issues. Moreover, architects, builders and developers must consider structural performances, durability, safety, � re, maintenance, costs, sustainability and other factors.Some examples of the global insight the study of heat, air and moisture transfer delivers are:

Heat transfer not only includes conduction but also convection and radiation. Convection combines conduction with gas or liquid movement. Because of its electro-magnetic nature, radiation and the laws governing it is fundamentally different from convection and conduction. Nonetheless, radiation de� nes to a large extent heat exchange in buildings, between buildings and the environment, in air and gas layers and in porous materials. The surface � lm coef� cients used to handle convection and radiation at a surface are a mathematical expediency to facilitate calculations. They transpose convection and radiation into equivalent conduction through a � ctitious surface layer.Thermal insulation is very effective in diminishing heat losses and heat gains. A pre-condition for ef� ciency, however, is air-tightness. When an air permeable insulation layer is mounted in a construction that is not air tight, it loses its ef� ciency and may induce severe condensation problems at the cold side. This illustrates the negative impact air in� ow, out� ow, looping and washing have on all heat related performances and moisture tolerances.Moisture transport is not restricted to vapour. In open-porous materials at higher moisture contents, unsaturated water � ow becomes much more important. Without a good knowledge of the laws that govern water movement, we can neither understand nor cure phenomena such as rain penetration, rising damp, initial moisture and drying.When evaluating vapour flow related moisture damages, not equivalent diffusion, but advection should be considered � rst. Air is much more able to move vapour than diffusion.

Application of this knowledge to the design and construction of building parts and whole buildings is the subject of a discipline called performance analysis. Performance analysis is the only way to come to a science and engineering based approach to building design and construction, as opposed to the ‘construction as an art’ concept. That is undoubtedly the way of the future in modern, industrialized construction.The material discussed in the book has given birth to several software packages, which can be of great help for designers and constructers. They do help, but a sound knowledge of heat, air moisture transport in its overall complexity is today still based on the triumvirate of simulation/testing/practice. Most software tools in fact do not consider air in� ow, air out� ow,

�

�

�

�

1335vch04.indd 269 12.02.2007 11:37:35

Building Physics - Heat, Air and Moisture: Fundamentals and Engineering Methods with Examples and Exercises by Hugo Hens

Copyright © 2007 Ernst & Sohn Verlag für Architektur und technische Wissenschaften GmbH und Co. KG, Berlin

270 4 Postscript

air looping and air washing with all the nasty consequences they may have. They cannot cope with gravity and pressure induced liquid � ow through cracks, leaks and voids, for example as a consequence of rain run-off. Typically, the real geometry with its roughness, unknown cracks, invisible voids, etc., is of a much larger complexity than the virtual geometry, inputted in the one but even in the two and three-dimensional models. Also a universal linkage between heat, air, moisture transport and the many moisture related durability issues is still lacking.

In short, there is a lot to do.

1335vch04.indd 270 12.02.2007 11:37:35