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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
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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
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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.
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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
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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
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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
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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.
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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 ==
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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
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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
=
=
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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
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V1
=?
SOLUTION
IV Th d l f th d i t
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This formula would be applied for each process and we can come out with
a final table that summarizes all formula seeing before.
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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
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1
Lesson Prepared By M. Talla Jerry William