[Pump]-Pump System Analysis and Sizing (2002).pdf

by J. Chaurette, engineer

PUMP SYSTEM ANALYSIS AND SIZING

BY JACQUES CHAURETTE p. eng.

5th Edition February 2003

Published by Fluide Design Inc. www.fluidedesign.com

Copyright 1994

I

TABLE OF CONTENTS

Foreword Introduction Symbols Chapter 1 - An Introduction to pump systems 1.0 Hydrostatic pressure and fluid column height ................................................... 1.1 1.1 The three forms of energy ................................................................................. 1.2 1.2 The relationship between elevation, pressure and velocity in a fluid ................. 1.4 1.3 The difference between pressure and head ...................................................... 1.7 1.4 Fluid systems ................................................................................................... 1.8 1.5 The driving force of the fluid system ................................................................. 1.9 1.6 The components of Total Head ........................................................................ 1.10 1.7 Negative (relative) pressure................................................................................ 1.13 1.8 The siphon effect ................................................................................................ 1.17 1.9 Specific gravity .................................................................................................. 1.22 Chapter 2 - The application of thermodynamics to pump systems

2.0 Energy and thermodynamic properties .............................................................. 2.1 2.1 Closed systems and internal energy .................................................................. 2.2 2.2 Closed systems, internal energy and work ......................................................... 2.3 2.3 Open systems and enthalpy .............................................................................. 2.4 2.4 Open systems, enthalpy, kinetic and potential energy ....................................... 2.5 2.5 Work done by the pump .................................................................................... 2.6 2.6 Fluid and equipment friction loss ........................................................................ 2.6 2.7 The control volume ............................................................................................ 2.7 2.8 The determination of Total Head from the energy balance................................. 2.9 2.9 System or Total Head equation for a single inlet-single outlet system ............... 2.9 Example 2.1-Calculate the Total Head for a typical pumping system................. 2.13 2.10 Method for determining the pressure head at any location .............................. 2.18 Example 2.2 Calculate the pressure head at the inlet of the control valve ...... 2.21 2.11 System or Total Head equation for a single inlet-double outlet system ............ 2.26 2.12 General method for determining Total Head in a system with multiple inlets and outlets .................................................................................. 2.29 2.13 General method for determining Total Head in a system with multiple pumps, inlets and outlets ..................................................................... 2.31 2.14 General method for determining the pressure head anywhere in a system

with multiple pumps, inlets and outlets............................................................. 2.34

II

Chapter 3 - The Components of Total Head

3.0 The Components of Total Head ......................................................................... 3.1 3.l Total Static Head (HTS) ...................................................................................... 3.1 3.2 Suction Static Head (HSS) ................................................................................. 3.2 3.3 Net Positive Suction Head Available (N.P.S.H.A.) .............................................. 3.3 Example 3.1-Calculate the Net Positive Suction Head Available ...................... 3.12 3.4 Pump intake suction submergence..................................................................... 3.13 3.5 Discharge Static Head (HDS) ............................................................................. 3.16 Example 3.2 Calculate the Suction & Discharge Static Head .......................... 3.17 3.6 Velocity Head Difference (HV)........................................................................... 3.18 3.7 Equipment Pressure Head Difference (HEQ) ..................................................... 3.18 3.8 Pipe Friction Head Difference for newtonian fluids (HFP) .................................. 3.20 3.9 Fitting Friction Head Difference for newtonian fluids, K method and 2K method (HFF).................................................................................................... 3.23 3.10 Pipe Friction Head Difference for wood fiber suspensions (HFP) .................... 3.28 Chapter 4 - Pump Selection, Sizing, Interpretation of Performance Curves

4.0 Pump classes .................................................................................................... 4.1 4.1 Coverage chart for centrifugal pumps ................................................................ 4.2 4.2 Performance curve chart ................................................................................... 4.2 4.3 Impeller diameter selection ................................................................................ 4.5 4.4 System curve ..................................................................................................... 4.6 4.5 Operating point ................................................................................................... 4.7 4.6 Safety factor on Total Head or capacity ............................................................. 4.9 4.7 Pump operation to the right or left of Best Efficiency Point (B..E.P.) .................. 4.11 4.8 Pump shut-off head ........................................................................................... 4.14 4.9 Pump power ....................................................................................................... 4.15 4.9 Affinity laws ........................................................................................................ 4.17 Chapter 5 - Field Measurements

5.0 Real live measurements .................................................................................... 5.1 5.1 Total Head ......................................................................................................... 5.1 5.2 Net Positive Suction Head Available (N.P.S.H.A.) .............................................. 5.4 5.3 Shut-off head ..................................................................................................... 5.5 5.4 Equipment head difference................................................................................. 5.7 5.5 Flow measurement ............................................................................................ 5.8 5.6 Calculating flow based on power consumed by the motor.................................. 5.9

III

Glossary Bibliography Appendix A

Useful equations (metric and imperial systems) The definition of viscosity Rheological (viscous behavior) properties of fluids Appendix B

The Newton-Raphson iteration technique applied to the Colebrook equation

Appendix C

The determination of slurry density based on the volume and weight concentration of the solid particles Appendix D

The use of imperial system (FPS) units Appendix E

Power factors and efficiency values for ABB electric motors

IV

Foreword

One of my goals in writing this book was to make sense of the various terminology, equations and miscellaneous tips and tricks that are published in the general literature on centrifugal pump sizing and system head calculations. There is no lack of articles or books on the subject but usually only certain isolated aspects of the topic are treated. It seems that so far nobody has put all the relevant concepts and principles together. I had some difficulty starting this book. For starters, I was concerned about the quality of my writing skills, I think they have improved. Then I wondered if I had anything original to say. That was more difficult. At first, I believed that I could simply write down what I knew on the subject. I asked myself a few tough questions, and quickly discovered major gaps in my knowledge. That was a good starting point. Another source of inspiration came from conversations with colleagues on pump system problems (I was not much help), another gold mine. I found that most books on the subject just do not give the big picture, so here is the big picture.

V

Introduction

The purpose of this book is to describe how pressure can be determined anywhere within a pump system. The inlet and outlet of a pump are two locations where pressure is of special interest. The difference in pressure head (the term pressure head refers to the energy associated with pressure divided by the weight of fluid displaced) between these two points is known as the Total Head. A system equation will be developed based on fundamental principles from which the Total Head of the pump can be calculated, as well as the pressure head anywhere within the system. These principles can be applied to very complex systems. Friction loss due to fluid flow in pipes is the most difficult component of Total head to calculate. The methods used to calculate friction loss for different types of fluids such as water and viscous fluids of the Newtonian type and wood fiber suspensions (or stock) will be explained. The fluids considered in this book belong to the categories of viscous and non-viscous Newtonian fluids. Wood fiber suspensions are a special type of slurry. There is an excellent treatment on this subject by G.G. Duffy in reference 2. For the readers benefit, a condensed version is provided. Slurries, which are an important class of fluids, are not considered. I recommend reference 7, which provides a complete treatment of the subject. However, all the principles for Total Head determination described in this book apply to slurry fluid systems. The only exception is the methods used to calculate pipe friction head. Centrifugal pumps are by far the most common type of pump used in industrial processes. This type of pump is the focus of the book. The challenge in pump sizing lies in determining the Total Head of the system, not the particular pump model, or the materials required for the application. The pump manufacturers are generally more than willing to help with specific recommendations. Information on models, materials, seals, etc., is available from pump manufacturer catalogs. Often when approaching a new subject, our lack of familiarity makes it difficult to formulate meaningful questions. Chapter 1 is a brief introduction to the components of Total Head. I hope it proves as useful to you as it did to me.

VI

Symbols

Variable nomenclature Imperial system (FPS units)

Metric system (SI units)

A area in2 (inch square) mm2 (mm square) CW solids concentration ratio by weight in a

slurry non-dimensional

CV solids concentration ratio by volume in a slurry

non-dimensional

D pipe diameter in (inch) mm (millimeter) F force lbf (pound force) N (Newton) f pipe friction factor non-dimensional g acceleration due to gravity: 32.17 ft/s2 ft/s2 (feet/second

squared) m/s2 (meter/second squared)

E energy Btu (British Thermal Unit) kJ (kiloJoule) E specific energy Btu/lbm kJ/kg En enthalpy variation of the system Btu kJ H head ft (feet) m (meter) HP Total Head ft m HDS discharge static head ft m HEQ equipment head difference ft m HF friction head difference ft m HSS suction static head ft m HTS total static head ft m Hv velocity head difference ft m KE kinetic energy variation of the system Btu kJ L length of pipe ft m m mass lbm (pound mass) kg (kilogram) M mass flow rate tn/h t/h PE potential energy variation of the system Btu kJ p pressure psi (pound per square

inch) kPa (kiloPascal)

P power hp (horsepower) W (watt) Re Reynolds number non-dimensional SG specific gravity; ratio of the fluid density

to the density of water at standard conditions

non-dimensional

T temperature F (degrees Fahrenheit) C (degrees Celsius)

Q heat loss Btu kJ q volumetric flow rate ft3/s m3/s U internal energy variation of the system Btu kJ V volume ft3 m3 v velocity ft/s m/s W work Btu kJ z vertical position ft m

VII

Variable nomenclature Imperial system

(FPS units) Metric system (SI units)

Greek terms delta: the difference between two terms epsilon: pipe roughness ft m nu: kinematic viscosity SSU (Saybolt

Universal Second) cSt (centiStoke)

eta: efficiency non-dimensional mu: dynamic viscosity cP (centiPoise) rho: density lbm/ft3 kg/m3 gamma: specific weight lbf/ft3 N/m3

Note: A dot above the symbol (i.e. &m , &Q , &W ) indicates the rate of change of the variable. A term with multiple subscripts such as HEQ1-2 means the total or sum of all equipment head between points 1 and 2. *FPS: Foot-pound-second system (Imperial) used in the U.S. and anglophone countries. **SI: Systme internationale, the metric system.

AN INTRODUCTION TO PUMP SYSTEMS

1.0 HYDROSTATIC PRESSURE AND FLUID COLUMN HEIGHT

Our starting point will be pressure and how it is developed within a pumping system. It is easy to build pressure within a solid, a fluid however requires containment walls. A fluid can only be pressurized if it is in a container. Sometimes the container is very big like an ocean. This does not mean that the container has to be closed. Even if the container has an outlet, it is still possible to build pressure. An experiment with a common syringe will demonstrate this fact. With the syringe full of water, it is quite easy to generate significant pressure within the fluid (this is evidenced by the amount of force applied to the plunger) even while fluid squirts from the tip. Hydrostatic pressure is the pressure associated with a motionless body of water. Pressure within the body of water varies and is directly proportional to the vertical position with respect to the free surface. Divers are well aware of this fact since each foot of vertical descent increases pressure on the eardrums.

Fluid weight is the cause of hydrostatic pressure. In Figure 1-1, a thin slice of fluid is isolated so that the forces surrounding it can be visualized. If we make the slice very thin, the pressure at the top and bottom of the slice will be the same. The slice is compressed top and bottom by force vectors opposing each other. The fluid in the slice also exerts pressure in the horizontal direction against the pipe walls. These forces are balanced by stress within the pipe wall. The pressure at the bottom of the slice will be equal to the weight of fluid above it divided by the area.

Figure 1-1 Pressure vs. hydrostatic head.

1 2 AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD

The weight of a fluid column of height (z) is:

AzVAzVVgF ==== since The pressure (p) is equal to the fluid weight (F) divided by the cross-sectional area (A) at the point where the pressure is calculated :

zA

AzAFp ===

[1-1]

where F : force due to fluid weight V : volume g : acceleration due to gravity (32.17 ft/s2) : fluid density in pound mass per unit volume : fluid density in pound force per unit volume Note that the relationship between pressure (p) and fluid column height (z) is independent of the total volume in the container. The pressure generated by the weight of water at the deep end of a pool (i.e. 10 feet down) is the same as ten feet below the surface of a lake.

1.1 THE THREE FORMS OF ENERGY There are three forms of energy that are related and always occur together in a fluid system. They are potential, kinetic and pressure energies. This section briefly provides the definition for each to give the reader some familiarity with the terms. If we divide energy by the weight of the fluid element we obtain specific energy or head.

Figure 1-2 Fluid container shapes vs. pressure head.

AN INTRODUCTION TO THE COMPONENTS OF TOTAL HEAD 1 3

POTENTIAL SPECIFIC ENERGY KINETIC SPECIFIC ENERGY PRESSURE SPECIFIC ENERGY

Figure 1-3 Potential specific energy provided by the difference in elevation of fluid particles.

Figure 1-4 Kinetic specific energy provided by moving fluid particles.

Figure 1-5 Pressure specific energy provided by the weight of a fluid column.

zenergyspecificPotential =

gvenergyspecificKinetic

2

2

=

penergyspecificPressure =


1.2 THE RELATIONSHIP BETWEEN ELEVATION, PRESSURE AND VELOCITY IN A FLUID There is a relationship between the energies associated with elevation, pressure and velocity of fluid particles. The energy terms are: the elevation energy (z), the pressure energy (p/), and the velocity energy (v2/g). The sum of these 3 types of energy must be constant, since energy cannot be lost. Stated in another way: the energy at point 1 must be equal to the energy at point 2 (see Figure 1-6). The sum of the three forms of energy must be constant: CONSTANTEg

vpz ==++ 22

Or if we wish to describe the relationship between the energy levels of fluid particles in different locations of the system such as point 1 and 2 in Figure 1-6 then:

gvpzg

vpz 22222

2

211

1 ++=++

Relationship between pressure, elevation and velocity A variation in one or two of these terms implies a variation in the third. The total energy at point 1 in a fluid system must be equal to the total energy at point 2 (see Figure 1-6). For example, if we were to increase the velocity at point 1 by reducing the section, keeping all the other terms the same, the pressure p1 will decrease.

Figure 1-6 The relationship between pressure, elevation and velocity.


Relationship between pressure and elevation There are many areas in a system where the velocity is constant. In that case, it is only pressure and elevation that are related. In particular, if the velocity is zero as in a static system, we have the relationship between pressure and fluid column height previously mentioned. The following equation gives the relationship between elevation and pressure when the velocity is constant.

2

21

1pzpz +=+

Figure 1-8 shows a real system with a pressure gauge at the low part of the system (near the pump) and one in the upper part (near the discharge tank). The pressure p1 will be greater than p2 due to the elevation difference.

Figure 1-7 The relationship between pressure and elevation.

Figure 1-8 Pressure variation due to elevation in a real system.


Relationship between pressure and velocity If the elevation is constant, then there is a relationship between pressure and velocity. It is this relationship that helps us calculate the flow rate in a venturi tube. In Figure 1-9, the pressure p2 will be lower than p1 due to the increase in velocity at point 2. The following equation gives the relationship between pressure and velocity when the elevation is constant.

gvp

gvp

222211

22+=+

A venturi tube is used to measure flow rate. The flow rate (q) is proportional to the difference in pressure at points 1 and 2 ( see Figure 1-9).

12 ppKq =

Figure 1-9 The relationship between pressure and velocity.


The 3 forms of energy (elevation, pressure and velocity) are always present in a fluid system. Using a simple container (see Figure 1-10), these types of energy can be clearly demonstrated.

1. Potential energy. The fluid particles at elevation z vs. those at the bottom have potential energy. We know this type of energy is present because we must have spent energy moving the fluid particles up to that level.

2. Pressure energy. The weight of the fluid column produces pressure p at the

bottom of the tank. The pressure energy is transformed to kinetic energy when we open the valve at the bottom of the tank.

3. Kinetic energy. If fluid is allowed to exit the container at the bottom of the tank, it

will exit the tank with a velocity v. Pressure energy has been converted to kinetic energy.

1.3 THE DIFFERENCE BETWEEN PRESSURE AND HEAD

In a moving fluid, the velocity of the particles must be considered. The principle of conservation of energy states that the energy of a fluid particle as it travels through a system must be constant. This can he expressed as:

mgz mgp

mv E CONSTANT+ + = =

12

2

[1-2]

where (E) is the total energy of fluid particles with mass (m) and velocity (v). The total energy consists of the potential energy (mg z), the pressure energy (mg p/), and the kinetic energy (m v2/2g). By dividing all the terms in the above equation by (mg), we obtain equation [1-3] which is known as Bernoullis equation. E now becomes the specific energy of the fluid particle or the energy per unit weight E . All the terms on the left-hand

Figure 1-10 The 3 forms of energy in a fluid system, potential, kinetic and pressure.


side of Bernoullis equation are known as head. Bernoullis equation expresses the relationship between the elevation head (z), the pressure head (p/), and the velocity head (v2/2g). Bernoullis equation:

zp v

gE C O N S T A N T+ + = =

2

2 [1-3]

Bernoullis equation will be expanded later to include the pump Total Head and the friction losses. It is important at this point to make a clear distinction between pressure and head. Head is a generic term for a type of specific energy (i.e. elevation, pressure or velocity head). When we need to calculate pressure at a specific point in a system, we will refer to it as pressure head. Pressure head can be converted to pressure by equation [1-1]. Pressure can be measured anywhere in the system quite easily and can provide valuable information. However, since it is not an energy term and cannot be used for calculations involving head, especially Total head. The pressure measurement must be converted to pressure head (see equation [1-5]) to be useful in these calculations.

1.4 FLUID SYSTEMS

The pump is the heart of a fluid system. It is impossible (or at the very least impractical) to move fluid from one place to another without the energy provided by a pump. Figure 1-11 shows a simple but typical system. All the fluid in the suction tank will eventually be transferred to the discharge tank. The purpose of the system is to displace fluid and sometimes to alter it by filtering, heating or other with the appropriate equipment. The box with the term EQ symbolizes equipment such as: control valves, filters, etc. Any device (or equipment) introduced into the line will have the effect of reducing the pressure head in the line. This requires more head, or energy, from the pump.

Figure 1-11 A typical pumping system.


How is a system designed?

A. The flow rate is determined based on the process and production requirements. B. The location and size of the suction and discharge tanks is established. C. The location, capacity and size of the equipment to be installed in the line is

determined. D. The pump location is fixed. E. The line sizes are determined and the auxiliary equipment such as manual valves

are sized and located. F. The Total Head of the pump is determined as well as the size, model, type, and

power requirement. Where does a system begin and end? We can imagine a boundary (also called control volume) which envelops and determines the extent of a system. A complete system contains fluid that is continuous from inlet to outlet. There can be no gaps or empty spaces between parts of the fluid. In Figure 1-11, the system inlet is at point 1 and the outlet is at point 2. Point 1 is located on the liquid surface of the suction tank. Normally there would be a pipe inserted into the suction tank providing fluid to maintain the elevation of point 1. This fill pipe is not considered part of the system. The outlet of the system, or point 2, is located on the liquid surface of the discharge tank. Again, there is normally a pipe, which controls the level of point 2. This discharge pipe is not part of the system. The reasons for this will be carefully explained when we review control volumes in section 2.7.

1.5 THE DRIVING FORCE OF THE FLUID SYSTEM

The pump supplies the energy to move the fluid through a system at a certain flow rate. The energy is transferred to the fluid by a rotating a disk with curved vanes called an impeller (see Figure 1-12). This movement drives the liquid in a circular path and imparts centrifugal force to it. The pump pushes and pressurizes the fluid up against the casing (or volute). This ability of the pump to pressurize fluids at a certain flow rate provides us with the means to design systems that will meet the process goals (for example, flow and pressure at the desired locations). The Total Head is the difference in head between the inlet and outlet of the pump that produces the flow rate required of the system.


The Total Head of the pump provides the energy necessary to overcome the friction loss due to the movement of the fluid through pipes and equipment. It also provides the energy to compensate for the difference in height, velocity and pressure between the inlet and the outlet.

1.6 THE COMPONENTS OF TOTAL HEAD

Friction Fluids in movement generate friction. There is a difference in pressure (pF1 - pF2 ) required to move a fluid element A (see Figure 1-13) towards the outlet. This difference in pressure is known as the friction pressure loss (p) (see Figure 1-13). When this term is converted to head, it is then known as friction head loss. Reference tables and charts for friction head loss are widely available (see reference 1 and 8). Viscosity is an important factor and higher viscosity fluids generate higher friction. More about viscosity in chapter 3.

Figure 1-12 Principal components of a centrifugal pump.

Figure 1-13 The difference in pressure due to fluid friction.


Equipment Any equipment in the line will create a reduction in pressure head. A filter is a common example of a device producing a pressure drop (see Figure 1-14). Other examples are control valves, heat exchangers, etc. Equipment introduced into an existing system will reduce the flow rate unless the pump is modified to provide more energy (for example by installing a bigger impeller). Velocity The kinetic energy of the fluid increases when it leaves the system at a higher velocity than when it enters and this requires extra energy. The energy required for the velocity increase is typically small and is often neglected. However, certain systems are specifically designed to produce high output velocity. This is done by using nozzles (see Figure 1-15 C) and therefore require a substantial amount of energy that the pump must supply. Figure 1-15 illustrates three systems with progressively higher output velocities. CASE A. DISCHARGE PIPE END SUBMERGED. The output velocity v2 is small and the

velocity head negligible. CASE B. DISCHARGE PIPE END NOT SUBMERGED. The output velocity v2 is small and

the velocity head is non-negligible. CASE C. NOZZLES ARE INSTALLED ON THE DISCHARGE PIPE END. The output velocity v2

is high and the velocity head is high.

Figure 1-14 The effect of equipment in a system.

Figure 1-15 Various systems with varying output velocities.


Elevation It takes energy to pump fluids from a lower level to a higher one. There is often a significant difference in elevation between the inlet of a system (point 1), and the outlet (point 2, see Figure 1-16). Typically, the elevation difference within a system is the largest contributor to the Total Head of the pump.

Figure 1-16 The difference in elevation between suction and discharge tank.

Pressurized Tanks The discharge and suction tank may be positively or negatively pressurized (with respect to the local atmospheric pressure). If the discharge tank is positively pressurized then the pump must provide additional energy to overcome this additional pressure. When the discharge tank is negatively pressurized, there is less resistance to the fluid entering the discharge tank and this reduces the energy required of the pump. The effect on the pump is exactly opposite when the suction tank is positively or negatively pressurized.

Figure 1-17 Pressurized suction and discharge tanks.


In many applications the tanks are not pressurized and the level of pressure in these tanks is zero or the same as the local atmospheric pressure. However, there is a misuse of terminology that tends to muddy the waters occasionally. Have you ever heard someone say: Boy, that water is really coming out, there must be allot of pressure at the end of that hose. In reality, there is no pressure at the outlet of the hose since the fluid comes out into the atmosphere. What feels like pressure, is the mass of water particles hitting at high velocity. The kinetic energy is converted to pressure energy, which produces a force on the hand. The pressure is zero but the water jet has a significant amount of kinetic energy.

1.7 NEGATIVE (RELATIVE) PRESSURE

Figure 1-19 illustrates how easy it is to produce negative relative pressure. The fluid is stationary and the elevation of point 1 is identical to that of points 3, 6 and 9. The pressure at point 2 is higher than the pressure at point 1 due to the depth of point 2 and the pressure caused by the fluid at that depth. The pressure at point 2 is positive and decreases to zero as we reach the level of point 3 (same level as point 1) inside the tube. From 3 to 4, the pressure decreases and becomes negative. The pressure at point 5 is the same as at point 4 since we are on the same level. Pressure then increases from negative to zero at point 6 (same level as point 1). From point 6 to 7 there is a further increase, the pressure remains constant from 7 to 8. The pressure decreases to zero from point 8 to 9 since point 9 is at the same level as point 1.

Figure 1-18 Ouch.

Figure 1-19 The pressure distribution in a static system.


We create relative negative pressure everyday with a straw. Find some flexible tubing and try the experiment shown on Figure 1-20 and Figure 1-24. Try the following simple experiment. Get a small container and a short length of clear plastic tube. Our goal will be to put some water on a shelf so to speak.

1. Suction is applied to the tube and the liquid is lifted up to point 4.

2. Bend the tube as you apply suction to get the fluid past point 5. At this point a siphon (see section 1.8) is established.

3. The tube is bent at points 7 and 8 and the liquid level establishes itself at point 9,

which is the same level as point 1. The liquid in the tube remains stable and suspended at the level of point 4 and 5. Liquid has been raised from a lower elevation at point 1 to a higher one at point 4, like putting a book on a shelf. If the tube was punctured at point 4 or 5, what would happen? Air would enter the tube and the liquid would drop to its lowest level.

We have managed to create negative relative pressure at point 4, which is easily maintained without further intervention.

Figure 1-20 Creating negative pressure.


Use a longer tube this time, fill it with water and put your finger on the end of the tube sealing it off. The lower end of the tube is open (see Figure 1-21a). It is possible to suspend a column of water 34 feet high by sealing the top. This creates a volume of zero pressure at the top end, between the finger and the fluid surface. At the bottom end, which is open, atmospheric pressure is pushing on the fluid in an upward direction. The weight of the liquid column is balanced by the force generated by atmospheric pressure.

Figure 1-21a Water suspended from an open tube 34 feet high.

Figure 1-21b Difference in pressure in a water column suspended from an open tube 34 feet high.


A common unit in North America for measuring pressure is the psig (or pound per square inch gauge). Zero psig corresponds to the level of pressure in the local atmosphere. The g stands for gauge, meaning dial gauge. The equivalent to 0 psig in pressure head is 0 feet of fluid. These units are relative to the local atmospheric pressure or atmospheric pressure head. Pressure in certain parts of the system can drop below the local atmospheric pressure, and become negative. An area of negative pressure is under a vacuum. A breach of the containment wall, in an area under vacuum, will cause air to be drawn into the system. This is what would happen if the tube were punctured at point 4 in Figure 1-20. The units often used to express low pressure are psia (pounds per square inch absolute). The levels of pressure head are expressed in feet of fluid absolute or in Hg (inch of Mercury). These units are absolute, and are therefore not relative to any other pressure. A perfect vacuum corresponds to 0 psia. Figure 1-22 shows graphically the relationship between absolute and relative pressure. The atmospheric pressure at sea level is 14.7 psia. Not all plants are at sea level, for example, Johannesburg is 5,200 feet above sea level and the local atmospheric pressure is 12 psia. This effect has to be considered when calculating the available N.P.S.H. (Net Positive Suction Head ) at the pump suction (see Chapter 3).

Figure 1-22 Absolute vs. relative pressure scales.


The relationship between an absolute (psia) and a relative (psig) pressure measurement is:

)()()( psiappsigppsiap A+= [1-4] where pA is the local barometric or atmospheric pressure in psia (i.e. pA = 14.7 psia at sea level).

1.8 THE SIPHON EFFECT

At first glance, a fluid moving vertically upwards without assistance creates a surprising effect. Figure 1-23 compares the movement of a rope with that of a ball. Both objects are solid, however the rope can emulate the behavior of a fluid where a ball cannot. A ball moves toward an incline and encounters a rise before it gets to a sharp drop; can it get over the hump without any intervention? No, not if it has a low velocity. Imagine the ball stretched into the shape of a rope, lying on a smooth surface, and draped across the hump. Even when starting from rest, the rope will slide down and drag the overhung part along with it. A fluid in a tube will behave in the same way as the rope. A volume of fluid is formless. A fluid will take on the shape of its container, whatever shape it may be. Its natural tendency is to lie flat. Gravity (or potential energy) is responsible for this behavior. Gravity is also responsible for the siphon effect.

Figure 1-23 A siphon as a rope.


Figure 1-24 shows an experiment that requires a few feet of flexible tube. Fill the tube with water and keep the bottom end sealed in some manner. Pinch the end of the tube at point 1, turn it around l80 to form a U shape as in position C. Release the end of the tube at point 1. What happens? No fluid leaves the tube. Why? If the water were to come out on the short side then a void would be created in the tube on the level of point 2. A void would produce a volume under low pressure. Low relative pressure acts as a pulling force on the short fluid column. Actually, it is not low pressure that pulls the fluid up but atmospheric pressure that pushes the short column of fluid upward. Therefore, fluid coming out at point 1 would create a vacuum at point 2 and a vacuum stops the movement so that there can be no movement out at point 1. What this means is that the pressure at point 1 (atmospheric pressure) is sufficient to prevent the short column of fluid from moving downward. Why, because if it did not, a void would have to be created at point 2. The effect of creating a void at point 2 would reduce the pressure to 0 psia. This low pressure can suspend a column of water 34 feet high. This is clearly not necessary for the short column of fluid to the left of 2. Something less than a void needs to be created to suspend the short column of fluid. In Figure 1-24 E, a reduced pressure is present at point 2 which suspends the short column. There is no void required to suspend the short column. The following experiment will show how easy it is to create low pressure. The effect produced is surprising and shows up an unusual property of fluids, the ability of fluids to be suspended in mid air without apparent means of support.

Figure 1-24 Water suspended in an open tube.


Experiment No.1 The experiment consists of a vertical tube, full of water, and closed at the bottom end. We take the top end and turn it around vertically downward as shown in Figure 1-25. What happens to the fluid in the tube? Is it in equilibrium, is it suspended, or will some portion of it drop out of the tube? If the fluid is suspended, there must be a balance of forces that holds it in place, preventing it from falling out of the short end of the tube. The pressure p0 within the fluid produces a force F0 at the section of point 0 (see Figure 1-25). Similarly, the atmospheric pressure pA produces a force FA at the end of the tube. W is the weight of that portion of fluid between point 0 and A.

Figure 1-25 Balance of forces between points 0 and A.

The balance of forces is:

thenApFalsoWFFA =+= 0 AzApApA += 0

Therefore zppA += 0 For the fluid to be in equilibrium, p0 must be smaller than pA. This means that p0 will be negative with respect to the atmospheric pressure.


There are two principles in the following argument that should be made clear:

In a static system, the fluid particles on the same level are at the same pressure.

Fluids are incompressible.

We know that for the fluid to be stable there must be a balance of forces. How does this balance of forces come about? Lets see if we can duplicate this experiment without bending any tubes. Experiment No.2 Lets start with a vertical tube full of water with one end closed as shown in Figure 1-26. If we apply a source of vacuum, the fluid will remain exactly where it is since it is incompressible. In other words, if the pressure is lowered the fluid does not expand and alternatively, if the pressure is raised it does not contract.

Attach a tee to the top end of the tube, then a short leg to the tee and connect one branch to a reservoir. Reattach the vacuum source to the tee as in Figure 1-27. The fluid is pulled up the short leg causing it to be in balance and therefore suspended. If we shut off the vacuum source and remove the reservoir, we have exactly the exact same system we started out with in Figure 1-25.

Figure 1-26 The level in the tube does not change when we apply a vacuum.


We managed to create a similar system to the original one (Figure 1-25) without bending the tube. We know that the pressure in the tube drops as we raise a fluid vertically from the outlet. It drops just sufficiently to suspend the column of fluid. This was made apparent as we applied a vacuum to lift the fluid. The bending of the tube in Figure 1-25 disguises this effect, giving us a suspended column of fluid without a clear indication of the mechanism. Experiment no. 2 shows the mechanism by which the fluid is suspended.

A typical siphon situation is shown in Figure 1-29. The graph shows how the pressure head varies within the system. In the next chapter, we will develop the method required to calculate the pressure head at any location in the system.

Figure 1-27 Lifting fluid with a source of vacuum.

Figure 1-28 The vacuum source is disconnected and the fluid is suspended.

Figure 1-29 The pressure variation within a simple siphon system.


1.9 SPECIFIC GRAVITY

We often need to calculate the pressure head that corresponds to the pressure. Pressure can be converted to pressure head or fluid column height for any fluid. However, not all fluids have the same density. Water for example has a density of 62.34 pounds per cubic foot whereas gasoline has a density of 46.75 pounds per cubic foot. Specific gravity is the ratio of the fluid density to water density at standard conditions. By definition water has a specific gravity (SG) of 1. To convert pressure to pressure head, the specific gravity SG of the fluid must be known. The specific gravity of a fluid is:

S G FW

=

where F is the fluid density and W is water density at standard conditions. Since

p zg z

gFF

c= =

and therefore

F WW

cSG p SG

g zg

= =

where F is the fluid density in terms of weight per unit volume and F is the density in terms of mass per unit volume. The constant gc is required to provide a relationship between mass in lbm and force in lbf (see Appendix D). The quantity Wg/gc ( W = 62.34 lbm/ft3 for water at 60 F) is:

=

=

ftinlbf

inft

slbfftlbmsftftlbm

gg

c

W222

23

31.21

)(144)(1

)/(17.32)/(17.32)/(34.62

After simplification, the relationship between the fluid column height and the pressure at the bottom of the column is:

)(31.21)( fluidofftzSGpsip = [1-5]

Figure 1-30 Pressure vs. elevation in a fluid column.


Here he goes again with the siphon. OK, maybe not everyone has the same interest in siphons as I do. The explanation for the behavior of a siphon is quite simple: a siphon is like a rope: if you pull on one end the other will follow. Why is it important to know how a siphon works? Siphons by their very nature produce an area of low pressure. A feed pipe with top entry to a tank behaves like a siphon. Any area in a piping system that is higher than the discharge point will likely be under low pressure.

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

2.0 ENERGY AND THERMODYNAMIC PROPERTIES

This chapter requires some introduction to thermodynamic properties and states. No need to panic, we will use only what we need to know to calculate Total Head. Several measurable quantities are used to define the state of a substance: temperature (T), pressure (p), velocity (v), and elevation (z). A specific type of energy corresponds to each of these quantities. Kinetic Energy The formula for kinetic energy is:

KE mv=

12

2

The above equation states that the kinetic energy of a body is equal to one half the mass times the velocity squared. Potential Energy Another form of energy is potential energy:

PE mg z= which means that the potential energy of a body is equal to the weight (mg) times the vertical distance (z) above a surface upon which the object would come to rest. These two types of energies interact. For example, an object at the top of an incline has a certain amount of potential energy. After it's release, potential energy is gradually converted to kinetic energy. When it reaches the bottom, all the potential energy has converted to kinetic energy.

Figure 2-1 Transformation of potential energy to kinetic energy.

2 2 THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS

The principle of conservation of energy states that energy can neither be created or lost. Consequently, if one form of energy decreases then another form must increase. This allows us to make an energy balance that describes the energy variation of the object in Figure 2-1.

KE PE+ = 0 [2-1] In other words, equation [2-1] reads: the sum of the potential and kinetic energy variation between points 1 and 2 is equal to zero. Energy levels are associated with positions within a system, position 1 is the top of the slope and position 2 the bottom.

PE PE PE and KE KE KE= = 1 2 1 2

mg z z m v v( ) ( )1 2 12

221

20 + =

v v g z z2

212

1 22 = ( ) [2-2] Equation [2-1] expresses the principle of conservation of energy for this system. This leads to equation [2-2] that describes how velocity changes with respect to height. The principle of conservation of energy makes it possible to account for all forms of energy in a system. The energies present in the system are in continuous change and the term delta () is used in equation [2-1] to indicate a change or variation in energy. Thermodynamic properties Thermodynamic properties are the different types of energies associated with a body (for example, potential, kinetic, internal or external). It is characteristic of a thermodynamic property is that its value is independent of the method or path taken to get from one value to another or from an initial state 1 to another state 2. The potential energy (PE) and the kinetic energy (KE) are thermodynamic properties. We require other energy quantities (work and heat) to account for real world conditions. These quantities are dependent on the path or method used to get from state 1 to state 2 (more about paths later).

2.1 CLOSED SYSTEMS AND INTERNAL ENERGY

The system shown in Figure 2-2 is a fluid contained in a sealed tube with its inlet connected to its outlet forming a closed system.

THE APPLICATION OF THERMODYNAMICS TO PUMP SYSTEMS 2 3

Internal Energy All fluids have internal energy (U). If we apply a heat source to the system, the temperature, pressure and internal energy of the fluid will increase. Internal energy is the energy present at the molecular level of the substance.

Closed Systems In a closed system no mass enters or leaves the boundary. What happens if we put a fluid within a closed environment such as a sealed container and apply heat? Pressure and temperature in the fluid will increase. The temperature rise indicates that the internal energy of the fluid has increased. The heat source increases the internal energy of the fluid from U1 to U2 . The energy quantities present in this system are the internal energy (U) and the heat loss (Q). Therefore, the energy balance is:

Q Q Q U U UC E= = = 1 2

QC is the quantity of heat absorbed by the fluid from the source and QE is the heat loss of the fluid to the environment. The internal energy has changed from its level at instant 1 to another level at instant 2 (after heat is applied). The internal energy (U) is a thermodynamic property.

2.2 CLOSED SYSTEMS, INTERNAL ENERGY AND WORK

Another way to increase the internal energy of a fluid is to do work on it by means of a pump. In this way, without applying heat, the work done on the fluid through the pump raises the internal energy of the fluid from U1 to U2. This causes the temperature of the fluid to increase, producing a heat loss QE to the environment.

Figure 2-2 The relationship between internal energy and heat gain within a closed system.

Figure 2-3 The relationship between internal energy, work and heat loss within a closed system.


In this situation, the energy balance is: Q W U U UE = = 1 2

The sign convention is positive energy for heat leaving the system and negative energy for heat or work entering the system.

2.3 OPEN SYSTEMS AND ENTHALPY

An open system is one where mass is allowed to enter and exit. The mass entering the system displaces an equal amount of mass that exits. A clearly defined boundary, called the control volume, envelops the system and intersects the inlet and outlet. The control volume allows us to apply the principle of conservation of energy. Each unit of mass (m1) and volume (V1) that enters the system is subject to a pressure (p1) at the inlet. The same is true of the mass leaving the system (m2) of volume (V2) and subjected to a pressure (p2). The principle of conservation of mass requires that m1 = m2. In this book, we are dealing only with incompressible fluids where V1 = V2. The pressure at the inlet produces a certain quantity of work that helps push the fluid through the system. The pressure at the outlet produces work opposing that movement. The difference between these two work components is the work associated with pressure at the inlet and outlet of the system. This difference is the energy term (pV) and is the only difference between the systems shown in Figure 2-3 vs. the system shown in Figure 2-4.

Figure 2-4 The relationship between internal energy, heat loss, work, and a pressurized environment for an open system.


The energy balance is: EnpVUWQE == )( U and (pV) are terms that always occur together in an open system. For that reason their sum has been given the name enthalpy (En). Enthalpy is also a thermodynamic property.

2.4 OPEN SYSTEMS, ENTHALPY, KINETIC AND POTENTIAL ENERGY

In closed systems, the fluid particles move from one end of the container to another but they do not leave the container. It is impossible to have a net effect on the velocity and therefore on the kinetic energy of the system. In open systems, fluid particles move from the inlet to the outlet and leave the system. If the velocity varies between inlet and outlet, the kinetic energy varies. The pump is the energy source that compensates for the variation. The same is true for potential energy in a closed system. A fluid particle that goes from the bottom of a container to the top will eventually return to the bottom. There can be no net effect on the potential energy of this system. In an open system, fluid particles can leave the system at a different elevation than they enter producing a variation in potential energy. The pump also compensates for this energy variation. Potential Energy The difference in vertical position at which the fluid enters and leaves the system causes the potential energy of a fluid particle to increase or decrease. The potential energy increases when the fluid leaves at a higher elevation than it enters. The pump provides energy to compensate for the increase in potential energy. If the fluid leaves the system at a lower elevation than it enters, the potential energy decreases and it is possible to convert this energy to useful work. This is what happens in a hydroelectric dam. A channel behind the dam wall brings water to a turbine located beneath the dam structure. The flow of water under pressure turns a turbine that is coupled to a generator producing electricity. The difference in elevation or potential energy of the fluid is the source of energy for the generator. Kinetic Energy There is often a velocity difference between the inlet and outlet of a system. Usually the velocity is higher at the outlet versus the inlet; this produces an increase in kinetic energy. The pump compensates for this difference in kinetic energy levels. The complete energy balance for an open system is:

PEKEEnWQE ++= [2-3]


Equation [2-3] has the thermodynamic properties on the right-hand side. The heat loss (QE) and the work done on the system (W), which are not thermodynamic properties, are on the left-hand side of the equation. To determine QE and W, it is necessary to know how the heat is produced (i.e. path that the fluid must take between inlet and outlet) and how the work is done.

2.5 WORK DONE BY THE PUMP

The role of a pump is to provide sufficient pressure to move fluid trough the system at the required flow rate. This energy compensates for the energy losses due to friction, elevation, velocity and pressure differences between the inlet and outlet of the system. This book does not deal with the method by which a pump can pressurize a fluid within its casing. The reader is referred to reference 15, a very detailed treatment of this subject. For our purpose, the pump is a black box whose function is to increase the fluid pressure at a given flow rate.

2.6 FLUID AND EQUIPMENT FRICTION LOSS

The total heat loss (QE) is the sum of fluid friction loss and equipment friction loss.

EQFE QQQ +=

Fluid Friction Loss (QF) How is it possible to go from a nebulous concept such as fluid friction to the pressure drop due to friction? Friction occurs within the fluid and between the fluid and the walls.. A detailed method of fluid friction calculation is presented in Chapter 3. Friction in fluids produces heat (QF ) which dissipates to the environment. The temperature increase is negligible. If we can estimate the friction force, we know that we must supply an equal and opposite force to overcome it.

Figure 2-5 The work provided by the pump.

Figure 2-6 Heat loss due to fluid friction within pipes.


The energy corresponding to the heat loss (QF) must be supplied by the pump. The net force (FF) required to balance the friction force is F1-F2 These forces are the result of the action of pressures p1 and p2 (see Figure 2-6). The difference between p1 and p2 is the pressure differential or pressure drop due to friction for a given length of pipe. Many publications (reference 1 and 8 for example) provide pressure head drop values in the form of tables. The pressure head drop can also be calculated by the Colebrook equation, which we will discuss in chapter 3.

Equipment Friction Loss (QEQ) Many different types of equipment are present in typical industrial systems. Control valves, filters, etc., are examples of equipment that have an effect on the fluid. Because of the great variety of different equipment that can be installed, it is not possible to analyze everyone of them. However we can quantify the effect of the equipment by determining the pressure drop across the equipment much as we did with the pressure drop across the pump. Again it is not necessary to know exactly what happens inside the equipment to be able estimate this effect. Pressure drop (difference between inlet and outlet pressure) is the outward manifestation of the effect of equipment. This pressure drop produces heat (QEQ) which is then lost through the fluid to the environment. The heat loss is calculated in the same manner as the pump work (W).

Q p

mEQ EQ=

The product supplier normally makes available the amount of pressure drop for the equipment at various flow rates in the form of charts or tables.

2.7 THE CONTROL VOLUME

Up to now we have been using control volumes without defining their purpose. A control volume is an imaginary boundary that surrounds a system and intersects all inlets and outlets. This makes it possible to apply the principle of conservation of mass. The boundary of the system is determined by appropriately locating the inlet and the outlet. The term appropriate refers to locating the boundary at points where the conditions (pressure, elevation and velocity) are known. The control volume encompasses all internal and external the energy sources affecting the system. This makes it possible to apply the principle of conservation of energy. How is the control volume positioned in real systems? Figure 2-7 shows a typical industrial process. Suppose that we locate the inlet (point 1) on the inlet pipe as shown in Figure 2-7. Is this reasonable? Where exactly should we locate it, will any position due? Clearly no. Point l could be located anywhere on the feed line to the suction tank. The problem with locating point 1 on the inlet line is precisely that the location is indeterminate.


However, consider Figure 2-8. The pressure head at the inlet of the pump is proportional to the elevation of the suction tank fluid surface. We locate the inlet of the system (point 1) on the suction tank fluid surface since any change in the elevation of point 1 will affect the pump. The inlet pipe is the means by which we supply the pump with enough fluid to operate. As far as this system is concerned, the inlet feed pipe is irrelevant. This is because the pressure head at the inlet of the pump (or any point within the system) is dependent on the level of the suction fluid tank surface. The other reason that requires us to locate point 1 on the liquid surface of the suction tank is that the control volume must contain all the energy sources that affect the system and if the suction and discharge tanks are open to atmosphere than this pressure will be present at points 1 and 2 (H1 = H2 = 0 = atmospheric pressure).

.

Figure 2-7 Incorrect positioning of the control volume in a pumping system.

Figure 2-8 Proper positioning of the intersections of the control volume with a pumping system.


This leads to a representation of a generalized system that we will use from now on. A general system must take into account the possibility that the suction and discharge tanks are pressurized. The terms H1 and H2 in Figure 2-8 represent respectively the suction and discharge pressure head at the fluid surfaces of these tanks. A general system must take into account the possibility that the system contains no tanks and that points 1 and 2 are connections to pipe lines of other systems. The only difference between a tank and a pipe is that a tank can contain a much greater volume of fluid. Therefore, whether the system has tanks or not is immaterial. However the conditions (i.e. pressure, velocity and elevation) of the inlet and outlet of the system must be known. It will always be possible to disconnect our system from another if the pressure head, velocity and elevation (H , v and z) at the connection point is known (more about disconnecting interconnected systems later in this chapter). Therefore, Figure 2-8 can represent all possible situations.

2.8 THE DETERMINATION OF TOTAL HEAD FROM THE ENERGY BALANCE

We now have all the background information necessary to determine the work required of the pump and therefore the Total Head. First, we calculate the energy lost in friction due to fluid movement through pipes (QF) and equipment (QEQ.). Next, calculate the energy variations due to conditions at the inlet and outlet.

1. Potential energy variation (PE ) is associated with the difference in height of a fluid element between the outlet and inlet of the system.

2. Kinetic energy variation (KE) is associated with the variation in velocity of a

fluid element between the outlet and inlet of the system. 3. Enthalpy variation (En) consists of two components En = U + (pV). The

first is the internal energy (U) of the fluid. In most common fluid transfer situations, there is no significant temperature change and therefore no significant internal energy change (U = U1 - U2 = 0). The other component of enthalpy (pV) is the energy associated with the difference in pressure at the inlet and outlet of the system.

The difference between the inlet and outlet energies added to the total friction loss (QE) gives the resultant work W that must be supplied by the pump.

2.9 SYSTEM OR TOTAL HEAD EQUATION FOR A SINGLE INLET - SINGLE OUTLET

SYSTEM

The simplest case we can consider is a system with one inlet and one outlet. The first law of thermodynamics or the principle of conservation of energy for open systems states:

EPEKnEWQE &&&&& ++= [2-4]


Equation [2-4] expresses the rate of variation of the energy terms instead of the energy variation as in equation [2-3]. This allows us to balance mass flow rates instead of mass between the inlet and outlet of the system. In Figure 2-8, the box with the letters EQ represents the effect of all the equipment present in the line between points 1 and 2. H1 and H2 are the pressure heads present at the suction tank and the discharge tank fluid surface respectively. z1 and v1 are respectively the elevation and the velocity of a fluid particle at the inlet of the system, and z2 and v2 are the same variables for a fluid particle at the outlet.

ENERGY RATE v F V pm

p= = = &&

The following equations are all based on the above which is the work or energy rate required to move a body of mass m at a velocity v. &V and &m are respectively the volumetric and mass flow rate. p is the pressure differential and is the fluid density. The energy rate of heat loss ( & )QE The rate of heat transfer ( & )QE is the heat generated in the system by fluid friction in the pipes and through the equipment:

( )& &Q m p pE F EQ= + 1 2 1 2 [2-5]

where pF1-2 is the pressure drop associated with fluid friction for a fluid particle traveling between points 1 and 2. pEQ1-2 is the sum of all pressure drops produced by all of the equipment between the same points.

The rate of mechanical work ( & )W Similarly, the rate of mechanical energy introduced into the system, such as supplied by a pump is:

& &W

mpP=

[2-6]

where pP is the difference in pressure between the discharge and the suction of the pump (points P and S in Figure 2-8). The rate of enthalpy variation ( & )E The rate of enthalpy variation is composed of the rate of internal energy variation ( &U ), and the difference in pressure energy between the inlet or outlet of the system ( &)pV . &U is normally zero or negligible in most fluid transfer situations, therefore:

& & ( &) & & ( & & ) & & & &E U pV U U p V p V p V p V U U= + = + = =1 2 1 1 2 2 1 1 2 2 1 2 0since


also for incompressible fluids & &V V1 2= , therefore )()( 2121 pp

mppVnE ==&&& [2-7]

where p1 is the pressure at the suction tank liquid surface and p2 is the pressure at the discharge tank liquid surface. The rate of kinetic energy variation ( & )KE Kinetic energy is the energy associated with the velocity (v) of a body of mass (m). The rate of kinetic energy variation of the system is:

& & ( )KE

gm v v

c= 12 1

222

[2-8]

where v1 is the velocity of a particle at the surface of the suction tank or the system inlet velocity. v2 is the velocity of a particle at the surface of the discharge tank or the system outlet velocity. The constant (gc) is required to make the units consistent in the FPS system.

The rate of potential energy variation ( & )PE Potential energy is the energy associated with the vertical position z2 or z1 of a mass (m) subject to the influence of a gravity field. The change of potential energy of fluid particles in the system is:

& & ( )PE m

ggz z

c= 1 2 [2-9]

where z1 and z2 are respectively the elevation of a particle on the fluid surface of the suction and discharge tank. Elevations such as z1 and z2 are often taken with respect to a DATUM or reference plane.

By substituting equations [2-5] to [2-9] in equation [2-4] and dividing by &mggc

, we obtain:

( ) ( )( ) ( )

p pgg

pgg

p pgg

gv v z zF EQ

c

P

c c

1 2 1 2 1 212

22

1 2

12

+ =

+ +

[2-10]

Equation [2-10] is Bernoullis equation with the pump pressure increase (pP) and fluid (pF1-2) and equipment (pEQ1-2) friction terms added. Pressure can be expressed in terms of fluid column height or pressure head as demonstrated in chapter 1.

p

gHgc

=

[2-11]


All pressure terms are replaced by their corresponding fluid column heights with the use

of equation [2-11], for example pgHgc

11=

,

p

g HgF

F

c1 2

1 2

=

, etc.). The constant gc

cancels out. The Total Head is:

)()(

21)()( 1122

21

222121 HzHzvvg

HHfluidftH EQFP +++++= [2-12]

The unit of Total Head (HP) is feet of fluid. The pump manufacturers always express the Total Head (HP) in feet of water, is a correction required if the fluid is other than water? Do we need to convert feet of fluid to feet of water? The terms in equations [2-10] and [2-12] are in energy per pound of fluid, or lbf-lbf/lbf; which is the same as feet. Since head is really energy per unit weight, the density of the fluid becomes irrelevant (or in other words 1 pound of water is the same as 1 pound of mercury). However, we will see later that the motor power required to move the fluid at a certain rate does required that the density of the fluid be considered (see Chapter 4). The pump manufacturers test the performance of their pumps with water. The capacity of a centrifugal pump is negatively affected by the fluids viscosity; therefore the three major performance parameters of a pump (total head, flow rate and efficiency) will have to be corrected for fluids with a viscosity higher than water ( see reference 1 and the web site www.fluidedesign.com).


EXAMPLE 2.1 CALCULATE THE TOTAL HEAD FOR A TYPICAL PUMPING SYSTEM Our first example will take us through the ins and outs of a Total Head calculation for a simple pumping system. The fluid is water at 60 F.

The equation for Total Head is:

)()(

21

112221

2224412441 HzHzvvg

HHHHHHH EQEQPSEQFFPSFP +++++++++=

Point 4 is not essential to solve this problem but required for comparison purposes in example 2.2 where a branch is added at point 4. Friction Head Difference Pipe & Fittings

Since the pipe diameter changes between points 1 and 2, it is necessary to determine the friction occurring between points 1 and S and points P and 2.

POINT 1 TO S Pipe friction loss between points 1 and S.

H H HF F S FP1 2 1 2 = + The friction loss term HF1-S is made up of fluid friction and fittings friction.

Figure 2-9 An example of a calculation for Total Head of a typical pumping system.


Pipe friction = Fluid friction (pipe) + Fluid friction (fittings)

H H H and H H HF S FP S FF S FP FPP FFP1 1 1 2 2 2 = + = = where HFP1-S is the fluid friction between points 1 and S and HFF1-S is the friction produced by fittings (for example elbows, isolation valves, etc.) between the same two points. From the tables (reference 1 or 8). for a 8" dia. Pipe and a flow rate of 500 USGPM: HFP1-S/L = 0.42 ft/100 ft of pipe.

H

HL

LftFF S

FP S1

1

1000 42

8100

0 03= = =. .

Fittings Friction Loss between points 1 and S In this portion of the line there is (1) 8" butterfly valve, from the tables (reference 1 or 8) K = 0.25

H

Kvg

ftFF SS

1

2 2

20 25 3192 3217

0 04 = =

=

. ..

.

The pipe and fittings friction loss between points 1 and S is: HF1-S = HFP1-S + HFF1-S = 0.03 + 0.04 = 0.07 ft POINT P TO 4. Pipe friction loss between points P and 4. From the tables (reference 1 or 8), for a 4" dia. Pipe at a flow rate of 500 USGPM: HFPP-4 /L= 13.1 ft/100 ft of pipe.

H

HL

LftFPP

FP P

= = =44

100131

18100

2 3. .


Fittings Friction Loss between points P and 4 In this portion of the line there is: 4 dia. elbow, from the tables (reference 1 or 8) K = 0.25; (1) 4 isolation (butterfly) valve; from the tables (reference 1 or 8) KVAL = 0.25

HK K v

gftFFP

ELB VAL =

+ =

+

=442 2

20 25 0 25 12 76

2 32 1713

( ) ( . . ) ..

.

The pipe and fittings friction between points P and 4 is: HFP-4 = HFPP-4 + HFFP-4 = 2.3 + 1.3 = 3.6 ft POINT 4 TO 2 Pipe Friction Loss between points 4 and 2. From the tables (reference 1 or 8), for a 4"dia. pipe at a flow rate of @ 500 USGPM: HFPP-2 /L= 13.1 ft/100 ft of pipe.

H

HL

LftFP

FP4 2

4 2

100131

114100

14 9= = =. .

Fittings Friction Loss between points 4 and 2 In this portion of the line there is (2) 4"dia. elbows, from the tables (reference 1 or 8) KELB = 0.25.

H

K vg

ftFFELB

4 222 22

22 0 25 12 76

2 32 171 3 =

=

=. .

..

The pipe and fittings friction between points 4 and 2 is: HF4-2 = HFP4-2 + HFF4-2 = 14.9 + 1.3 = 16.2 ft


Equipment Pressure Head Difference POINT 1 TO S There is no equipment in this portion of the line. HEQ1-S = 0 POINT P TO 4 The filter between point P and 4 is considered a piece of equipment. The filter manufacturer's technical data states that this filter will produce a 5 psi pressure drop at a flow rate of 500 USGPM. For water 5 psi is equivalent to 11.5 ft of head (5 X 2.31, see equation [1-5]). HEQP-4 = HEQFIL = 11.5 ft POINT 4 TO 2 The control valve between point P and 4 is considered a piece of equipment. The control valve will have a head drop of 10 ft of fluid. The justification for this will be given later in Chapter 3. HEQ4-2 = HEQCV = 10 ft Velocity Head Difference

The velocity head difference between points l and 2 is zero, since v1 and v2 are very small.

H

gv vv = =

12

022

12( )

Static Head Both tanks are not pressurized therefore H1 = H2= 0. The elevation difference z2 - z1 = 137 - 108 = 29 ft


Total Head Once again the equation for Total Head is:

)()(21

112221

2224412441 HzHzvvg

HHHHHHH EQEQPSEQFFPSFP +++++++++=

By substituting the values previously calculated into the above equation: HP= 0.07 + 3.6 + 16.2 + 0 + 11.5 + 10+ 0 + 137 + 0 - (108 + 0 ) = 70.4 ft The result of the calculation indicates that the pump will require a Total Head of 70.4 ft to deliver 500 USGPM. How can we guarantee that the pump will deliver 500 USGPM at 70 ft of Total Head? The answer is in Chapter 4, section 4.5.


2.10 METHOD FOR DETERMINING THE PRESSURE HEAD AT ANY LOCATION

The system configuration in Figure 2-10 illustrates how drastically the pressure can vary in a simple pumping system. Why do we need to calculate the pressure at a specific location in the system? One reason is the pressure head just before the control valve is a parameter required to size the valve. It is calculated by the method described below. In addition, the system may have other equipment where the inlet pressure must be known to ensure proper operation.

A. The pressure head at any location on the discharge side of the pump. First, calculate the Total Head (HP) for the complete system using equation [2-12]. Then draw a control volume (see Figure 2-11) to intersect point X and point 1. Point X can be located anywhere between P and 2. The system equation [2-12] is then applied with point X as the outlet instead of point 2 and solved for the unknown variable Hx instead of HP.

Figure 2-10 The pressure head variation within a typical pumping system.


From equation [2-12], the Total Head for the complete system is:

)()(21

112221

222121 HzHzvvg

HHH EQFP +++++=

Equation [2-12] is applied with the following changes: all terms with subscript 2 are replaced with the subscript X, H2=HX, HF1-2 = HF1-X , HEQ1-2 = HEQ1-X, v2 = vX, z2 = zX. The unknown term Hx is isolated on one side of the equation:

XXXEQXFPX zHzvvgHHHH ++++= 11

22111 )(2

1)( [2-13]

The unknown pressure head Hx can also be determined by using that part of the system defined by control volume 2 (see Figure 2-11) by using the same reasoning as above, in equation [2-12], HP = 0 and all terms with subscripts 1 replaced with X. Therefore, H1=HX, HF1-2 = HFX-2 , HEQ1-2 = HEQX-2, v1= vX, z1 = zX

XXEQXFXX zHzvvgHHH ++++= 22

22222 )(2

1)( [2-14]

Figure 2-11 Using control volumes to determine the pressure head anywhere after the pump discharge


We now have two methods for determining the pressure head at any location. We can use one equation to verify the results of the other. The calculation for HX is often quicker using equation [2-14]. Note that in equation [2-14] the value of HP is not required to calculate Hx. B. The pressure head at any location on the suction side of the pump

We can determine the pressure head anywhere on the suction side of the pump by the same method. In this case, HP = 0 since there is no pump within the control volume (see Figure 2-12). HP = 0 and subscript 2 is replaced with subscript X in equation [2-12]. We obtain:

The velocity vX must be the same as would occur at point X in the complete system. In chapter 3, we will use equation [2-15] to calculate the available Net Positive Suction Head of the pump.

Figure 2-12 Using control volumes to determine the pressure head anywhere before the pump inlet.

XXXEQXFX zHzvvg

HHH ++++= 1122

111 )(21)( [2-15]


EXAMPLE 2.2 CALCULATE THE PRESSURE HEAD AT THE INLET OF A CONTROL VALVE

It is often required to know the pressure head just ahead of a piece of equipment such as a control valve. We will apply the methods in the previous section to calculate the pressure head just ahead of the control valve in example 2.1. CALCULATE THE PRESSURE HEAD AT POINT 7 USING EQUATION [2-13] Applying equation [2-13] where point X is now point 7 (X = 7) and HP = 70.4 from the calculation in example 2.1.

71127

2177117 )(2

1)()(4.70 zHzvvg

HHHHH EQPFPSEQSF +++++=

Friction Head Difference - Pipe and Fittings POINT 1 TO S From example 2.1: HF1-S = 0.07 ft

Figure 2-13 Example of the calculation of the pressure head at the inlet of a control valve.


POINT P TO 7 HFP-7 = HFPP-7 + HFFP-7 Pipe friction loss between point P and 7. From the tables (reference 1 or 8), for a 4"dia. at a flow rate of 500 USGPM: HFPP-7 /L= 13.1 ft/100 ft of pipe.

HHL

LftFPP

FP P

= = =77

100131

125100

16 4. .

Fittings Friction Loss between points P and 7 In this portion of the line there is: (3) 4 dia. elbow, KELB = 0.25; (1) 4 dia. isolation (butterfly) valve, KVAL = 0.25.

HK K v

gftFFP

ELB VAL =

+ =

+

=772 23

23 0 25 0 25 12 6

2 32 172 5

( ) ( . . ) ..

.

The pipe and fittings friction between points P and 7 is: HFP-7 = HFPP-7 + HFFP-7 = 16.4 + 2.5 = 18.9 ft. Equipment Pressure Head Difference POINT 1 TO S HEQ1-S = 0 POINT P TO 7 The filter between point P and 4 has a pressure head drop of: HEQP-7 = HEQFIL = 11.5


Velocity Head Difference The velocity head difference between points 1 and 7 is:

H

gv v

gftv = = =

12

12

0 12 76 2 512

72 2( ) ( . ) .

Static Head The suction tank is not pressurized, therefore H1 = 0. The elevation difference z1 - z7 = 108 - 140 = -32 ft The pressure head at point 7 Again the equation for the pressure head at point 7 is:

71127

2177117 )(2

1)()(4.70 zHzvvg

HHHHH EQPFPSEQSF +++++=

By substituting the values calculated into the above equation we obtain: H7 = 70.4 - 0.07-(18.9 + 11.5) - 2.5 + 108 + 0 -140 = 5.4 ft CALCULATE THE PRESSURE HEAD AT POINT 7 USING EQUATION [2-14] This is a good opportunity to verify our calculation for H7 using equation [2-14]. Substituting 7 for X, the equation is:

72227

2227277 )(2

1)( zHzvvg

HHH EQF ++++=

Friction Head - Pipe and Fittings POINT 7 TO 2 HF7-2 = HFP7-2 + HFF7-2


Pipe friction loss between point 7 and 2. From the tables (reference 1 or 8), for a 4" dia. at a flow rate of 500 USGPM: HFP7-2 /L= 13.1 ft/100 ft of pipe.

H

HL

LftFP

FP7 2

7 2

100131

7100

0 92= = =. .

Fittings Friction Loss between points 7 and 2 In this portion of the line, there is no fittings. Therefore HFF7-2 = 0 The pipe and fittings friction between points 7 and 2 is: HF7-2 = HFP7-2 + HFF7-2 = 0.92 + 0 = 0.92 ft. Equipment Pressure ahead Difference POINT 7 TO 2 As previously stated, we allow a pressure head drop of 10 ft for the control valve. HEQ7-2 = 10 ft Velocity Head Difference

The velocity head difference between points 7 and 2 is:

Hgv v

gf tv = = =

12

12

0 1 2 7 6 2 522

72 2( ) ( . ) .

Static Head

The discharge tank is not pressurized, therefore H2 = 0. The elevation difference z2 - z7 = 137 - 140 = -3 ft


The pressure head at point 7 Again the equation for the pressure head at point 7 is:

72227

2227277 )(2

1)( zHzvvg

HHH EQF ++++=

By substituting the values calculated into the above equation: H7 = 0.9 + 10 - 2.5 + 137 + 0 - 140 = 5.3 ft - the same result as previously.


2.11 SYSTEM OR TOTAL HEAD EQUATION FOR A SINGLE INLET - DOUBLE OUTLET SYSTEM

Industrial systems often have more than one outlet (see Figure 2-14). Equation [2-4] is used as the basis for the analysis of the system using the principle of conservation of mass flow rate for a steady state system. EPEKnEWQE &&&&& ++= & & &m m m1 2 3= + The same logic used to deduce equation [2-12] can be used for a system with multiple inlets and outlets. &QE is the sum of all the friction heat losses and equipment losses. &W is the work supplied by the pump. nE& is the sum of the inlet minus the sum of the outlet enthalpies. &KE is the sum of the inlet minus the sum of the outlet kinetic energies. &PE is the sum of the inlet minus the sum of the outlet potential energies. The energy balance is:

( ) ( ) ( )& & & &m p p m p p m p p m pF EQ F EQ F EQ1 1 4 1 4 2 4 2 4 2 3 4 3 4 3 1 + + + + + = & & & & & & & & &mp

mp

mp

mgv

gmv

mv

m ggz

gg

mz

mz

c c c c

11

22

33

112 2

22 3

32 1

12

23

3

1

+

+ +

+ +

Figure 2-14 The use of control volumes to determine the Total Head for a single inlet double outlet system.


replacing all pressure terms by their corresponding fluid column height (for example

pgHgc

11=

,

p

g HgF

F

c1 4

1 4

=

, etc.), and all mass flow terms by their corresponding

volumetric flow rate q, 11 qm =& etc., we obtain:

+++++= )()()( 3434324242414111 EQFEQFEQFP HHqHHqHHqHq

)())((21)( 332211

233

222

211332211 zqzqzqvqvqvqg

HqHqHq +++++++

and since q1 = q2 + q3, the Total Head is:

++++++++= ))(()(21)()( 1122

21

2224244141 HzHzvvg

HHHHH EQFEQFP

))(()(21)()( 2233

22

2324243434

1

3 HzHzvvg

HHHHqq

EQFEQF ++++++ [2-16]

For equation [2-16] to be true, the friction loss in the major branches has to be adjusted by closing the manual valves to balance the flows. The term EQ2 represents equipment on branch 2 such as a valve (see Figure 2-14). Closing the valve, reduces the flow at branch 2 but increase it at branch 3. The conditions in one branch affects the other because of the common connection point 4. The correct flow is obtained at point 3 by adjusting components in branch 2 (for example HF4-3, HEQ4-3 and H4). One would think that the main branch 2, the branch with the highest head and flow requirement, should determine the pump Total Head. So, why is it that the head requirements for branch 3 appear in equation [2-16]? All of the terms associated with branch 3 can be made to disappear after simplification. Determine the value of H4 using equation [2-14] and a control volume that surrounds branch 3:

H H H

gv v z H zF EQ4 4 3 4 3 32

4 32

3 3 4

12

= + + + + ( ) ( ) ( )

or

( ) ( ) ( ) H H Hgv v z H zF EQ4 3 4 3 4 32

4 32

3 3 4

12

+ = + [2-17]

We can also determine the value of H4 with equation [2-14] and a control volume surrounding branch 2:


H H H

gv v z H zF EQ4 4 2 4 2 22

4 22

2 2 412

= + + + + ( ) ( ) ( )

or

( ) ( ) ( ) H H Hgv v z H zF EQ4 2 4 2 4 22

4 22

2 2 4

12

+ = +

[2-18]

v4-3 and v4-2 are respectively the velocity at point 4 for branch 3 and branch 2, substituting equations [2-17] and [2-18] into equation [2-16] we obtain:

H H H H H

gv v z H z HP F EQ F EQ= + + + + + + + ( ) ( ) ( ) ( ( ))1 4 1 4 4 2 4 2 22

12

2 2 1 1

12

))(21( 224

234

1

3 + vvgq

q

[2-19]

Point 4 is the point at which the flow begins to divide and the fluid particles move towards the direction of branch 2 or branch 3. All fluid particles have the same velocity at point 4 although their trajectory is about to change.

When we replace equation [2-20] into [2-19] then the last term in [2-19] disappears and we obtain equation [2-21] which is the equation for a single branch system. This means that the branch has no impact on the Total Head as long as branch 2 represents the path requiring the highest energy.

H H Hgv v z H z HP F EQ= + + + + + 1 2 1 2 22

12

2 2 1 112( ) ( ) [2-21]

In many cases the path requiring the highest energy is obvious. What happens if the flow in branch 3 is equal to the flow in branch 2? Alternatively, if the flow is much smaller in branch 3, but the elevation of point 3 is higher than point 2?

Figure 2-15 Location of H4.

v v v4 4 3 4 2= = [2-20]


These situations make judging which branch will produce the higher head difficult to do at a glance. To resolve this, calculate the Total Head for the fluid traveling through either branch and compare the results. The Total Head for the pump as calculated considering the flow through branch 3 is:

H H Hgv v z H z HP F EQ= + + + + + 1 3 1 3 32

12

3 3 1 112( ) ( ) [2-22]

For the system to work as intended, the highest Total Head as calculated from equations [2-21] or [2-22] will be used for sizing the pump.

2.12 GENERAL METHOD FOR DETERMINING TOTAL HEAD IN A SYSTEM WITH MULTIPLE INLETS AND OUTLETS

The energy rate balance for any system is: EPEKnEWQE &&&&& ++= Figure 2-16 represents a system with (n) inlets and (m) outlets.

In the system shown in Figure 2-16 there are 1 to p branches. These subscripts are used to identify the flow rates (for example q1, q2, ..qp) There are li to ni inlets, which are used to identity the inlet properties (for example v1i, H1i, z1i, v2i, H2i, z2i, ... vni, Hni, zni). Similarly, there are 1o to mo outlets that identify the outlet properties.

Figure 2-16 A general system with a single pump and multiple inlets and outlets.


There is one path, which for a given flow rate, will require the highest Total Head. Often, this path is obvious, but several paths may have to be checked to find the most critical.

The following equations are all based on the general principle for energy rates in a fluid medium:

HqpmPVFvRATEENERGY ===

&&

The friction term &QE is:

(...)()(1

222111 EQtFtt

pt

tPBRpEQFBREQFBR HHqHqHHqHHq +=+++++

=

=

where the subscripts 1 to p stand for branch no.1, 2, etc,. up to branch no. p. The work rate term &W is:

PP Hq

Figure 2-17 The use of the control volume with a generalized system representation.


The enthalpy rare term nE& is:

))...(...11

220112211 =

=

=

=

=++++++mt

ttototiti

nt

tmomooooniniiiii HqHqHqHqHqHqHqHq

The kinetic energy rate term &KE is:

( )

=++++++

=

=

=

=

)21)...(...(

21

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[Pump]-Pump System Analysis and Sizing (2002).pdf