Basic Ref Lessons by m Talla Class 7

8/3/2019 Basic Ref Lessons by m Talla Class 7

1/15

CHAPTER ONE

INTRODUCTION TO FLUIDMECHANICS

1.1 BRIEF HISTORY

There are two main approaches of presenting an introduction offluid mechanics materials. The first approach introduces the fluidkinematic and then the basic governing equations, to be followed bystability, turbulence, boundary layer and internal and external flow. The

second approach deals with the Integral Analysis to be followed withDifferential Analysis, and continue with Empirical Analysis.

The need to have some understanding of fluid mechanics startedwith the need to obtain water supply. For example, people realized thatwells have to be dug and crude pumping devices need to be constructed.Later, a large population created a need to solve waste (sewage) andsome basic understanding was created. At some point, people realized

that water can be used to movethings and provide power.

The first progress in fluid mechanics was made by Leonardo DaVinci (1452-1519) who built the first chambered canal lock near Milan. Healso made several attempts to study the flight (birds) and developedsome concepts on the origin of the forces. After his initial work, theknowledge of fluid mechanics (hydraulic) increasingly gained speed by thecontributions of Galileo,Torricelli, Euler, Newton, Bernoulli family, and

DAl b t


2/15

DAlembert

CHAPTER 2 FLUIDS STATICS

2.1 NOTION OF PRESSURE

a) Definition

Pressure is the (compression) stress at a point in a static fluid. Next

to velocity, the pressure p is the most dynamic variable in fluidmechanics. Differences or gradients in pressure often drive a fluid flow,especially in ducts.

S

FP= in Pascal (Pa)

gdzP =

where differencealtitudedz

atmospherep

=

= 1

kg/m3xN/kgxm=N/m2=Pa

b) The Hydrostatic principle

If the fluid is at rest or at constant elocit the pressure distribution


3/15

A floating body displaces its own weight in the fluid in which itfloats.

Archimedes theorem states that the force balance is at displacedweight liquid (of the same volume) should be the same as the container,the air. Thus,

gvF = Where F is the Archimedes force

APPLICATION EXERCISE (extrait dans Mcaniques des Fluides BTS Industriels page 29)

The figure below is the representation of a simplified central heatinginstallation, in which water flow in closed circuit.

The part B is located at a height Hb, above the part A; The part C islocated at a height Hc above the part A.

A manomether place at A show the pressure PA.

Datas:

Hc= 3m; HB= 5m; Pa= 5.105 Pa ; = 1000 kg /m3 ;g=10m/s2

We suppose that the eating process is stopped and that water doesnot circulate.

1) What is the expression of the pressure PB, in the part B? CalculatePB.


4/15

HB= zB-zA;

The application of the fundamental principle of hydrostatic give:

BAABAB hgPzzgPP ..).(. ==

Numerical Application

PB= 5.105 5.104 = 4.5x105 Pa

2) The line Oz is oriented up, therefore

Hc= zA-zC

The application of the fundamental principle of hydrostatic give:

cAAcAc hgPzzgPP ..).(. +==

Numerical Application

PB= 5.105 + 3.104 = 5.3x105 Pa


5/15

CHAPTER 2

Review of Thermodynamics

I- INTRODUCTION

In this chapter, a review of several definitions of commonthermodynamics terms is presented. This introduction is provided to bringthe student back to current place with the material.

II- Basic Definitions

The following basic definitions are common to thermodynamics andwill be used in this lesson.

a) Work

In mechanics, the work was defined as:

F = PS

W F dl PS dl S dl dV


6/15

W=

a) Isobaric process

It occur at a constant pressure

)()(211

1

1212

VVPVVPPdVWPdVdW ====

b) Isochoric Process

or isometric/isovolumetric process. It Occur at a constantVolume

===1

112 0PdVWPdVdW Because dV=0 (Constant

Volume)

) I th l


7/15

cv and cp respectively specific volume heat and specific pressure heat inKj/Kgok, and the specific heats ratio

Then dTrPdVdTrWrcv111

12

=

=

=

or dTrdVVrT

VrTP

1==

Dividing the last expression by T, thenT

dTrdV

V

r

1=

By intergrading the last term we come out with:

1

1

2

1

2

2

1

1

2

)1(

=

=

= =

P

P

T

T

V

V

P

P

cstePVcsteTV

( )

( )

1

1

112212

1212

1212

=

=

=

VPVPW

TTr

W

TTcW V

e) Application Exercise

1 kg of air describe the below cycle, compose of two isothermalsand two isochoric

processes.


8/15

SOLUTION

1-

Process 1 -2 Isothermal then T1=T2

Process 2-3 Isochoric then V1=V2

Process 3-4 Isothermal then T3=T4

Process 1- 4 Isochoric the V1=V4

PV= mrT here m=1kg then P1V1=rT1

kgmP

rT/86,0

1010

300287V

3

4

1

1

1 =

==

kgm /17,05

86,0V 32 ==


9/15

2- (W+Q)Cycle=0

Process 1-2 isothermal then

JV

VmrTW 63,139579

17,0

86,0ln3002871ln

2

121 ===

Point Pressure Volume Temperature1

2

3

4


10/15

2) Heat Expression

a) Isothermal process

In an isothermal process, the variation of internal energy is zero

dE=0 =dQ+dW=0

dQ=-dw

2

112

2

112

ln

ln

V

VmrTQ

V

VmrTW

=

=


11/15

Cv=Mcvv

p

c

c=

d) adiabatic Process

dQ=0

e) Polytrophic Process

With n than

APPLICATION EXERCISE

1-1kg of azoth occupy a volume v= 0.6m

3

at a temperature T=

120

o

c. What is his pressure? R of azote = 297j/kg

o

c

2-Determine the work and Heat exchanged during the following

122 VPVP


12/15

V1

=?

SOLUTION

IV Th d l f th d i t


13/15

This formula would be applied for each process and we can come out with

a final table that summarizes all formula seeing before.


14/15

1

2

PROCESS EQUATION WORK ( W ) HEAT ( Q )

Internal ENERGY

(U)ENTHALPY (H) ENTROPY (S)

ISOTHERMA

L

ISOBARIC

ISOCHORIC

ADIABATIC

POLYTROPI

C

Lesson Prepared By M. Talla Jerry William


15/15

1

Lesson Prepared By M. Talla Jerry William

Documents

Basic Ref Lessons by m Talla Class 7