Basic Equationsin Integral Form
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Transcript of Basic Equationsin Integral Form
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Chapter 4 Basic Equations in Integral Form for a Control Volume
4-1 Reynolds Transport Theorem (RTT)4-2 Continuity Equation4-3 The Linear Momentum Equation4-4 The Angular Momentum Principle4-5 The First Law of Thermodynamics4-6 The Second Law of Thermodynamics
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
4-1 Reynolds Transport Theorem (RTT) (1)
Conservation of Mass:
4-1
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
System and fixed control volume.
4-1 Reynolds Transport Theorem (RTT) (2)
4-2
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
( ) (inflow if negative) (11-41)net out inCS
B B B b V n dA = = r
42)-(11 dV bBCV
CV =43)-(11 )(
CSCV += dAnVbdV bdtddt
dB :General sys
4-1 Reynolds Transport Theorem (RTT) (3)
4-3
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Interpreting the Scalar Product:
4-1 Reynolds Transport Theorem (RTT) (4)
4-4
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
44)-(11 )( CS = dAnVbdt
dB :flowSteady sys
:flow ldimensiona-One
45)-(11 -in exiteach for out exiteach for CV += 4342143421 iiiieeeesys AVbAVbdV bdtddt
dB
Special Case 1: Steady Flow
Special Case 2: One-Dimensional Flow
46)-(11 -inoutCV += iieesys bmbmdV bdtddt
dB
4-1 Reynolds Transport Theorem (RTT) (5)
4-5
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Extensive and Intensive Properties:
4-1 Reynolds Transport Theorem (RTT) (6)
4-6
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
47)-(11 )(0 CS += dAnVdVdtd :equation Continuity CV
48)-(11 =in
iout
e mm :flowSteady
An Application: The Continuity Equation
49)-(11 or AA 221 AVAVVV :constant stream,Single
22111 ===
4-2 Continuity Equation (1)
4-7
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
4-2 Continuity Equation (2)
Conservation of Mass:9 Incompressible Fluids
9 Steady, Compressible Flow
4-8
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
The amount of mass flowing through a control surface per unit time is called the mass flow rate and is denoted The dot over a symbol is used to indicate time rate of change.Flow rate across the entire cross-sectional area of a pipe or duct is obtained by integration
While this expression for is exact, it is not always convenient for engineering analyses.
c c
n cA A
m m V dA = = &
m&
m&
4-2 Continuity Equation (3)
4-9
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
m&1
c
avg n cc A
V V dAA
= avg cm V A=&
c
n c avg c cA
V V dA V A VA= = =&V&
V&
m V= &&
9Integral in can be replaced with average values of and Vn
9For many flows variation of r is very small: 9Volume flow rate is given by
9Note: many textbooks use Q instead of forvolume flow rate.9Mass and volume flow rates are related by
Average Velocity and Volume Flow Rate:
4-2 Continuity Equation (4)
4-10
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
The conservation of mass principlecan be expressed as
CVin out
dmm mdt
=& &
Conservation of Mass Principle:
4-2 Continuity Equation (5)
4-11
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
in outm m= & &
n n n nin out
V A V A=
9For steady flow, the total amount of mass contained in CV is constant.9Total amount of mass entering must be equal to total amount of mass leaving
9For incompressible flows,in out
m m= & &
4-2 Continuity Equation (6)
SteadyFlow Processes:
4-12
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
50)-(11 )V(mdtd
dtVdm amF
===51)-(11 dVV
dtdF
sys =
52)-(11 )()(
+=CSCV
sys dAnVVdVVdtd
dtVmd
53)-(11 ( )dAnVVdVVdtdF:General
CSCV
+=
4-3 The Linear Momentum Equation (1)
4-13
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
4-3 The Linear Momentum Equation (2)
4-14
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Stress vector acting on the control surface.
4-3 The Linear Momentum Equation (3)
4-15
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
54)-(11 )( dAnVVF :flowSteady CS
=
55)-(11 V -V ie +=
ini
oute
CV
mmdVVdtdF
:flow ldimensiona-One
Special Cases
56)-(11 V -V ie =
ini
oute mmF
:flow ldimensiona-oneSteady,
4-3 The Linear Momentum Equation (4)
4-16
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
57)-(11 )V-V(mF :exit)-oneinlet,-(one
flowldimensiona-oneSteady,
12
=58)-(11 )V-V(mF :coordinate x Along x1,x2,x
=
4-3 The Linear Momentum Equation (5)
4-17
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Choosing a Control Volume:CV is arbitrarily chosen by fluid dynamicist,however, selection of CV can either simplify or complicate analysis.
Clearly define all boundaries. Analysis is often simplified if CS is normal to flow direction.Clearly identify all fluxes crossing the CS.Clearly identify forces and torques of interest acting on the CV and CS.
Fixed, moving, and deforming control volumes.For moving CV, use relative velocity,
For deforming CV, use relative velocity all deforming control surfaces,
4-3 The Linear Momentum Equation (6)
4-18
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Forces Acting on a CV:Forces acting on CV consist of body forces that actthroughout the entire body of the CV (such as gravity, electric, and magnetic forces) and surface forces that act on the control surface (such as pressure and viscous forces, and reaction forces at points of contact).
Body forces act on each volumetricportion dV of the CV.
Surface forces act on each portion dA of the CS.
4-3 The Linear Momentum Equation (7)
4-19
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Body Forces: The most common body force is gravity, which exerts a downward force on every differential element of the CVThe different body force
Typical convention is thatacts in the negative z-direction,
Total body force acting on CV
4-3 The Linear Momentum Equation (8)
4-20
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Surface Forces: Surface forces are not as simple to analyze since they include both normal and tangential componentsDiagonal components xx, yy, zz are called normal stresses and are due to pressure and viscous stressesOff-diagonal components xy, xz, etc., are called shear stresses and are due solely to viscous stressesTotal surface force acting on CS
4-3 The Linear Momentum Equation (9)
4-21
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Body and Surface Forces Surface integrals are cumbersome.Careful selection of CV allowsexpression of total force in terms ofmore readily available quantities likeweight, pressure, and reaction forces.Goal is to choose CV to expose onlythe forces to be determined and aminimum number of other forces.
4-3 The Linear Momentum Equation (10)
4-22
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
4-3 The Linear Momentum Equation (11)
Special Case: Control Volume Moving with Constant Velocity:
4-23
Momentum Equation for Inertial Control Volume with Rectilinear Acceleration
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Angular Momentum:
*Motion of a rigid body can be considered to be thecombination of -the translational motion of its center of mass (Ux, Uy, Uz)-the rotational motion about its center of mass (x, y, z)
*Translational motion can be analyzed with linear momentumequation.
*Rotational motion is analyzed with angular momentum equation.*Together, the body motion can be described as a 6degreeoffreedom (6DOF) system.
4-4 The Angular Momentum Principle (1)
4-24
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Review of Rotational Motion:Angular velocity is the angular distance traveled per unit time, and angular acceleration is the rate of change of angular velocity.
4-4 The Angular Momentum Principle (2)
4-25
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Review of Angular Momentum:Moment of a force:
Moment of momentum:
For a system:
Therefore, the angular momentum equation can be
written as:
To derive angular momentum for a CV, use RTT with
and
4-4 The Angular Momentum Principle (3)
4-26
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
4-4 The Angular Momentum Principle (4)
Basic Law, and Transport Theorem:
4-27
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
4-4 The Angular Momentum Principle (5)
4-28
General form
Approximate form using average properties at inlets and outlets
Steady flow
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
4-5 The First Law of Thermodynamics (1)
Basic Law, and Transport Theorem:
4-29
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
General Energy Equation:
, ,sys
net in net in
dEQ W
dt+ =& &
The energy content of a closed system can be changed by two mechanisms: heat transfer Q and work transfer W.Conservation of energy for a closed system can be expressed in rate form as
Net rate of heat transfer to the system:
Net power input to the system:,net in in outQ Q Q= & & &
,net in in outW W W= & & &
4-5 The First Law of Thermodynamics (2)
4-30
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
2
) V dA
V2
systemcv cs
shaft normal shear other
dEQ W e dV edt t
e u gz
W W W W W
= = + = + += + + +
vv& &
& & & & &
b=e
4-5 The First Law of Thermodynamics (3)
4-31
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
pressure boundary pistondsW W PA PAVdt
= = =& &
( )pressure nW PdAV PdA V n = = r r&
Where does expression for pressure work come from?When piston moves down ds under the influence of F=PA, the work done on the system is Wboundary=PAds.If we divide both sides by dt, we have
For generalized control volumes:
Note sign conventions: is outward pointing normal
Negative sign ensures that work done is positive when is done on the system.
nr
4-5 The First Law of Thermodynamics (4)
4-32
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Work Involves9 Shaft Work9 Work by Shear Stresses at
the Control Surface9 Other Work
4-5 The First Law of Thermodynamics (5)
Recall that P is the flow work,which is the work associated with pushing a fluid into or out of a CV per unit mass.
4-33
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
As with the mass equation, practical analysis is often facilitated as averages across inlets and exits
Since e=u+ke+pe = u+V2/2+gz
( ), , ,
C
net in shaft net inout inCV
cA
d P PQ W edV m e m edt
m V n dA
+ = + + + =
& & &r r
2 2
, , , 2 2net in shaft net in out inCV
d P V P VQ W edV m u gz m u gzdt
+ = + + + + + + + & & &
4-5 The First Law of Thermodynamics (6)
4-34
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
For steady flow, time rate of change of the energy content of the CV is zero.This equation states: the net rate of energy transfer to a CV by heat and work transfers during steady flow is equal to the difference between the rates of outgoingand incoming energy flows with mass.
2 2
, , , 2 2net in shaft net in out inV VQ W m h gz m h gz
+ = + + + + & &
4-5 The First Law of Thermodynamics (7)
Energy Analysis of Steady Flows:
4-35
-
Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
For single-stream devices, mass flow rate is constant.
( )( )
2 22 1
, , , 2 1 2 1
2 21 1 2 2
, , 1 2 2 1 ,1 2
2 21 1 2 2
1 2 ,1 2
2
2 2
2 2
net in shaft net in
shaft net in net in
pump turbine mech loss
V Vq w h h g z z
P V P Vw gz gz u u q
P V P Vgz w gz w e
+ = + +
+ + + = + + +
+ + + = + + + +
4-5 The First Law of Thermodynamics (8)
Energy Analysis of Steady Flows:
4-36
-
Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
Divide by g to get each term in units of length
Magnitude of each term is now expressed as an equivalent column height of fluid, i.e., Head
2 21 1 2 2
1 21 22 2
pump turbine LP V P Vz h z h h
g g g g + + + = + + + +
Energy Analysis of Steady Flows:
4-5 The First Law of Thermodynamics (9)
4-37
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Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
If we neglect piping losses, and have a system without pumps or turbines
This is the Bernoulli equation3 terms correspond to: Static, dynamic, and hydrostatic head (or pressure).
2 21 1 2 2
1 21 22 2P V P Vz z
g g g g + + = + +
4-5 The First Law of Thermodynamics (10)
Limitations on the use of the Bernoulli Equation
Steady flow: d/dt = 0Frictionless flowNo shaft work: wpump=wturbine=0Incompressible flow: = constantNo heat transfer: qnet,in=0Applied along a streamline (except for irrotational flow)
4-38
-
Chapter 4: Basic Equations in Integral Form for a Control Volume
Fluid Mechanics Y.C. Shih Spring 2009
4-6 The Second Law of Thermodynamics
4-39