basic equationsin integral form

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



4-1 Reynolds Transport Theorem (RTT) (1)

Conservation of Mass:

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System and fixed control volume.


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( ) (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


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Interpreting the Scalar Product:


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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


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Extensive and Intensive Properties:


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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)

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Conservation of Mass:9 Incompressible Fluids

9 Steady, Compressible Flow

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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&


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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:


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The conservation of mass principlecan be expressed as

CVin out

dmm mdt

=& &

Conservation of Mass Principle:


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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= & &


SteadyFlow Processes:

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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




4-14



Stress vector acting on the control surface.


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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,


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57)-(11 )V-V(mF :exit)-oneinlet,-(one

flowldimensiona-oneSteady,

12

=58)-(11 )V-V(mF :coordinate x Along x1,x2,x

=


4-17



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,


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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.


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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


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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


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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.


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Special Case: Control Volume Moving with Constant Velocity:

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Momentum Equation for Inertial Control Volume with Rectilinear Acceleration



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



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.


4-25



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


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Basic Law, and Transport Theorem:

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4-28

General form

Approximate form using average properties at inlets and outlets

Steady flow



4-5 The First Law of Thermodynamics (1)

Basic Law, and Transport Theorem:

4-29



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= & & &


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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


4-31



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


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Work Involves9 Shaft Work9 Work by Shear Stresses at

the Control Surface9 Other Work


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



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

+ = + + + + + + + & & &


4-34



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

+ = + + + + & &


Energy Analysis of Steady Flows:

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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

+ = + +

+ + + = + + +

+ + + = + + + +



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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 + + + = + + + +



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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 + + = + +


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)

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4-6 The Second Law of Thermodynamics

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basic equationsin integral form

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