ch5 flow analysis cv

8/8/2019 Ch5 Flow Analysis CV

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Fluid Mechanics - I : Chapter 5 11

Chapter 5

Flow Analysis

using

Control Volumes

Fluid Mechanics - 1


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Introduction

Practical problems in fluid mechanics require analysis of the behavior of the

contents of a finite region in space ( control volume ; CV ) Many important questions can be readily answered with finite control volume

analyses

The bases of this analysis method are some fundamental principles of physics,applied to a CV namely,

Conservation of mass Newtons second law of motion

The first and second1 laws of thermodynamics

The resultant techniques presented in this Chapter are powerful and applicable toa wide variety of fluid mechanical circumstances that require engineeringjudgment

Furthermore, the finite control volume formulas are easy to interpret physicallyand thus are not difficult to use

In fluid mechanics, the control volume or Eulerian view is generally lesscomplicated and, therefore, more convenient to use than the system orLagrangian view


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Continuity Equation : Conservation of Mass

A system is defined as a collection of unchanging contents, so the conservation of

mass principle for a system is simply stated as

Time rate of change of the system mass = 0

or

where Msys is the system mass,

For a system and a fixed, non-deforming control volume that are coincident at aninstant of time, the Reynolds transport theorem (Eq. 4.19) with that B = Mass and b=

1, allows us to state

(5.3)

.

,

0!Dt

DMsys

dvMsys V

dAndvt

dvDt

D

Dt

DM

CSCVsys

sysy

x!| VVVHV


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Continuity Equation : Conservation of Mass

.,

dAbdvbdvbC

CV

y!| VVVV

CVdvb

tV

H

dAnbCS

y VV

Discussion on basic element of above eqn


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Continuity Equation /(Contd.)

In other word, the integral over control surface is represented as

For conservation of mass (continuity eqn ), thus eqn 5.1 becomes

(5.5)

In words, above Eqn states that to conserve mass the time rate of change of the

mass of the contents of the control volume plus the net rate of mass flow through

the control surface must equal zero

Actually, the same result could have been obtained more directly by equating the

rates of mass flow into and out of the control volume to the rates of accumulationand depletion of mass within the control volume (See Section 3.6.2)

It is reassuring, however, to see that the Reynolds transport theorem works for this

simple-to-understand case

!yo

in

o

outCS

mmdAnVV

0!Dt

DMsys

0!yx

dAndvt CSCV

VVV

H


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Continuity Equation /(Contd.)Mass Flow Rate

Mass Flow Rate through a section of control surface having area A

and uniform properties of incompressible flow, is represented as

(5.6)

For compressible flow the density and velocity are not uniform.

For compressible flows, we will normally consider a uniformly

distributed fluid density at each section of flow and allow density

changes to occur only from section to section

(5.8)

AVQm VV !!0

VA

dAnb

VCS

!

y

!

V

V V


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Fixed, Non-D

eforming Control Volume In many applications of fluid mechanics, an appropriate

control volume to use is fixed and non-deforming

Several example problems that involve the continuity

equation for fixed, non-deforming control volumes (Eq. 5.5)illustrates use of such CVs

See Examples 5.1 to 5.5 of text Book


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Ex


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Ex


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x


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where


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Continuity Equation /(Contd.)Summary Observations on Application of Cont. Eqn. :

( to fixed, non-deforming Control Volumes) The vector Dot Product , has following sign convention :

+ve for flow out of CV

- ve for flow into CV

When the flow is steady, the time rate of change of CV is zero, i.e

Thus net mass flow rate through CS is zero, i.e

or

For incompressible the density is constant thus

For Un-Steady Flows , but if cyclic, the term can be taken as zero or

steady flow, based on time averaged basis value

Its value is +ve when mass of CV in increasing and is ve when it decreases

0!yx

dAndvt

CSCV

VVV

HnyV

0!x

CV dvt VH

0 !! o

i

o

out

VV

0!o

in

o

out QQ

!o

in

o

out mm

x CV dvt VH


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Moving, Non-deforming Control Volume

In many applications, Fluid Flow Analysis involves moving bodies, Gas TurbineEngine of flying aircraft and we use relative velocity (W) in our analysis asdiscussed in last chapter; Section 4.4.6

That is, V = W + VCVWhere V, is the fluid velocity seen by a stationary observer andWis relative velocity

For a system and a moving, non-deforming control volume that are coincident atan instant of time, the Reynolds transport theorem (Eqn. 4.23, 5.5) for a movingcontrol volume leads to

(5.15)

From Eqns. 5.1 and 5.15, we can get the control volume expression forconservation of mass (the continuity equation) for a moving, non-deforming control

volume, namely,(5.16)

See Examples 5.6 & 5.7 for application of above equation (5.16) .

dAbdvbtDt

DM

S

sysnW y

x! VVH

0!yx

dbdbS

nW VVH


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Ex. 5.7 & 5.8


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Deforming Control Volume

Occasionally, a deforming control volume can simplify the solution of a problem. A deforming control volume involves changing volume size and control surface movement. Thus,

the Reynolds transport theorem for a moving control volume can be used for this case, and

Eqs. 4.23 and 5.1 lead to

(5.17)

The time rate of change term, (!st Term on RHS) is usually nonzero and must be carefullyevaluated because the extent of the control volume varies with time

The second term on RHS (mass flowrate term), must be determined with the relative velocity, W,the velocity referenced to the control surface.

Since the control volume is deforming, the control surface velocity (VCS) is not necessarilyuniform and identical to the control volume velocity, as was true for moving, non-deforming

control volumes For Deforming CV, the absolute velocity is : V = W + VCS

where VCS is the velocity of the control surface as seen by a fixed observer. The relativevelocity, W, must be ascertained with care wherever fluid crosses the control surface

See examples 5.8 & 5.9 illustrating the use of Eqn 5.17

0!yx

! dAbdvbtDtDM

S

sysnW VV

H


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Ex 5.8


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Fluid Mechanics - I : Chapter 520


Ex 5.9


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21

The Linear Momentum

Newtons second law of motion for a system is

Since momentum is mass times velocity, thus the momentum of a small particle of

mass is

Thus, the momentum of the entire system is and Newtons law can bewritten as

Any reference or coordinate system for which this statement is true is called

inertial

A fixed coordinate system is inertial A coordinate system that moves in a straight line with constant velocity and is

thus without acceleration is also inertial

We proceed to develop the control volume formula for this important law

dvV dvVV

syssys FdVVsys

dvVV


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The Linear Momentum /(Contd.)

When a control volume is coincident with a system at an instant of time, the forces

acting on the system and the forces acting on the contents of the coincident control

volume (see Fig ) are instantaneously identical, that is,

For a system and fixed coincident non-deforming CV

the Reynolds transport theorem for (Velocity and System

Momentum) can be written as

(5.21)

or

For Fixed CV, the above can be written in following form and is called as Linear

Momentum Equation (5.22):

dAndvt

dvt CS Vsys

y! VVVV VVV

!y CVcoinofcontentCSCV FdAndvt .VVV VVH


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The Linear Momentum /(Contd.)

Application of the Linear Momentum Equation

The forces involved in Eq. 5.22 are body and surface forces that act on what is contained inthe control volume

The only body force we consider in this chapter is the one associated with the action ofgravity. We experience this body force as weight.

The surface forces are basically exerted on the contents of the control volume by material just

outside the control volume in contact with material just inside the control volume. For example,a wall in contact with fluid can exert a reaction surface force on the fluid it bounds

Similarly, fluid just outside the control volume can push on fluid just inside the control volumeat a common interface, usually an opening in the control surface through which fluid flowoccurs

An immersed object can resist fluid motion with surface forces

The linear momentum equation for an inertial control volume is a vector equation (Eq. 5.22) In engineering applications, components of this vector equation resolved along orthogonal

coordinates, for example, x, y, and z (rectangular coordinate system) or r, and x (cylindrical

coordinate system), will normally be used

A simple example involving steady, incompressible flow is considered first,(see Ex 510 & 5.11)

!y

xCVcoi

ofco

te

tCSCVFddv

t .VVV VV

H


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Example 5.10, 5.11


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Several important generalities about the application of the linear momentum equation (Eqn.5.22) can be concluded as :

1. When the flow is uniformly distributed over a section of the control surface The integraloperations are simplified. Thus, one-dimensional flows are easier to work with than flowsinvolving non-uniform velocity distributions.

2. Linear momentum is directional; it can have components in as many as three orthogonalcoordinate directions. Furthermore, along any one coordinate, the linear momentum of afluid particle can be in the positive or negative direction and thus be considered as a positiveor a negative quantity.

3. The flow of positive or negative linear momentum into a control volume involves anegative product. Momentum flow out of the control volume involves a positive product. Thecorrect algebraic sign to assign to momentum flow will depend on the sense of the velocity (in positive coordinate direction, in negative coordinate direction) and the product (for flow outof the control volume, for flow into the control volume).

4. The time rate of change of the linear momentum of the contents of a non-deformingcontrol volume is zero for steady flow. The momentum problems considered in this text all

involve steady flow. 5. If the control surface is selected so that it is perpendicular to the flow where fluid enters or

leaves the control volume, the surface force exerted at these locations by fluid outside thecontrol volume on fluid inside will be due to pressure


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6. The forces due to atmospheric pressure acting on the control surface may needconsideration as indicated by Eq. 13 for the reaction force between the nozzle and the w2

w1 1i.e., 00t cv Vr dV2 fluid. When calculating the anchoring force, the forces due toatmospheric pressure on the control surface cancel each other 1for example, aftercombining Eqs. 12 and 13 the atmospheric pressure forces are no longer involved2 andgage pressures may be used.

7. The external forces have an algebraic sign, positive if the force is in the assigned positivecoordinate direction and negative otherwise.

8. Only external forces acting on the contents of the control volume are considered in the

linear momentum equation (Eq. 5.22). If the fluid alone is included in a control volume,reaction forces between the fluid and the surface or surfaces in contact with the fluid [wettedsurface (s)] will need to be in Eq. 5.22. If the fluid and the wetted surface or surfaces arewithin the control volume, the reaction forces between fluid and wetted surface1s2 do notappear in the linear momentum equation 1Eq. 5.222 because they are internal, not externalforces. The anchoring force that holds the wetted surface1s2 in place is an external force,however, and must therefore be in Eq. 5.22.

9. The force required to anchor an object will generally exist in response to surface pressureand or shear forces acting on the control surface, to a change in linear momentum flowthrough the control volume containing the object, and to the weight of the object and the fluidcontained in the control volume. In Example 5.11 the nozzle anchoring force was requiredmainly because of pressure forces and partly because of a change in linear momentum flowassociated with accelerating the fluid in the nozzle. The weight of the water and the nozzlecontained in the control volume influenced the size of the anchoring force only slightly. FA,


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Examples 5.12 to 5.17


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Moment-of-Momentum Equation Using Newtons second law of motion, we studied a useful relationship between

forces and linear momentum flow. Now, by forming the moment of the linear momentum and the resultant force

associated with each particle of fluid with respect to a point in an inertial coordinatesystem, we will develop a moment-of-momentum equation that relates torques andangular momentum flowfor the contents of a control volume

When torques are important, the moment-of-momentum equation is often more

convenient to use than the linear momentum equation. Application of Newtons second law of motion to a particle of fluid yields

(5.30)

If we form the moment of each side of Eq. 5.30 with respect to the origin of aninertial coordinate system, we obtain

Where r is the position vector from origin of inertial coordinate system to the fluid

The above can be rearranged as

(5.35)

particlessys

FdvDt

DHV !

V

particleFrvDtD

r HHxx

!V

? A parti lFrvrDt

DHHV xx !V


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Moment-of-Momentum Equation /(Contd.)

For a system (collection of fluid particles) we need to use sum of both sides of Eqn.

5.35, i.e. (5.39)

where

or

Further, for the system and the contents of the coincident control volume that isfixed and non-deforming, the Reynolds transport theorem (Eq. 4.19) when appliedto LHS of above Eqn, leads to

Thus Eqn 5.39 becomes: ; (5.42)

or

? A ! syssysrvr

DtD FHVV

!! CVsysparticle rrr FFF

? A An. dvrvrt

vrt

csCVsys

HVHVHV xx

! VVV

CV

csCV

dvrvr

t !

x

xFxrAn.xx HVHV VV


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Application of Moment-of-Moment Equation

An important category of fluid mechanical problems that is readily solved with thehelp of the moment-of-momentum equation (Eq. 5.42) involves machines that rotate

or tend to rotate around a single axis

Examples of these machines include rotary lawn sprinklers, ceiling fans, lawn mower

blades, wind turbines, turbochargers, and gas turbine engines. As a class, these

devices are often called turbo-machines Eqn. 5.42 can be simplified in several ways :

We assume that flows considered are one-dimensional (uniform distributions

of average velocity at any section)

Steady or steady-in-the-mean cyclical flows i.e.

We work only with the component of Eq. 5.42 resolved along the axis of rotation

Using above simplifications, we will see how it looks like and is applied to various

situations

0! vrtC

HVVx

C

csC

dvrvrt

!xx

FxrAn.xx HVHV VV


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Each term on both sides of eqn can be written as follows by applying

the said simplifications :-

For steady Flow :

For rating body/system :

For 1-D flow : ; as

Thus above equation becomes:

(5.50)

where the sign for is taken as per Right-hand-Rule

Now note that shaft Power is related to shaft Toque as;

Thus above equation becomes

CV

cCV

dvrvr

t !

x

xFxrAn.xx HVHV VV

o

mrVAn.x UHV ! drcs

V

shaftC

T! Fxr 0x !

x

x vrtCV

HVV

UHV rVrdvcs

!! xV&mAn.o

outin

VrmVrmTout

o

outin

o

inshaft UU s

s

!

xVr[shaft

oshaftoTW !

outi

VrmVrmW out

o

outi !

o

i !shaft

o

UU [[ s

s

!


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Example 5.18


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Now relating angular velocity ([) with equivalent Linear Velocity (U)as U = r[, we can write last equation as

We also know that from conservation of mass, thus the

specific shaft power can be written as

Sign of the identity is taken as per Right Hand Rule

Example 5.19

outin

UmUmWout

o

outin

o

inshaft

o

UU s

s

!

outin

rm

rmW out

o

outin

o

inshaft

o

UU s

s

!

o

out

o

in mm !

outin

UUwoutinshaft

o

UU ss!


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Example 5.19


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The Energy Equation

First Law of Thermodynamics

In words, it can be written as :

In symbolic form :

where e=u+V2 / 2 + g z , where uis internal energy per unit mass

Now Reynolds Transport Theorem (replacing b by e) we have

Putting the above values in RTT, we get(5.59)

Work transfer rate is also called Power and is considered +ve when work is doneon the system / CV by the surroundings

sys

o

innet

o

innetsysWQdv

t ! Ve

!

o

out

o

i "

o

out

o

i "sys

WWQQvDt

DVe

dAdvt

dvt C

#

CV$ y$y

x! Veee VVHV

CV

%

& ' t

%

netCSCV inin

WdAndvt

!y

x Vee VVH


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The Energy Equation /(Contd.) Also note that work is transferred to and from system/CV through Rotating Shaft,

Normal and tangential Stresses / Pressure, where W = -P

Eqn 5.59 is valid for inviscid flow as it derivation did not account for fluid viscosity orWorkDone by system to overcome viscous effects / stresses

For viscous flow the eqn can be modified as :

or (5.64)

The above eqn is also valid for compressible flows

For a flow having uniform flow properties across any cross section area, theintegrant (2nd term of LHS) is

For a single stream of flow (with on exit and one entry point to CV) and it reduces to

dAnPWQdAndvt CSCV

o

net

o

netCSCV inin

y

!yx

VVee VVHdAnPWQdAnPgzVudv

t CSCV

(

net

(

netCSCV inin

2

2

y

!y

x VVe VVVH

!y

inflow

o

outflow

o

CS m

P

gz

V

um

P

gz

V

udAn

P

gz

V

u VVVV 222

222

V

o

in

in

o

out

out) S

Pgzum

PgzudAn

Pgzu

!y

VV

VV 22

2

222

V


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The Energy Equation /(Contd.) Thus for a steady 1-D mean flow in channel, the energy equation (5.65) can be

written as

(5.67)

Using definition of enthalpy[ h = u + pv ], it can also be written as

(5.69)

This equation is often used for solving compressible flow problems Examples 5.20 and 5.21 illustrates the use of above equations

Thus for a steady 1-D mean flow with no work, Eqn 5.67 can be written as

Comparing and re-arranging it in the form of Continuity Eqn, we get :

where

See Examples 5.20 to 5.25

CV

o

net

o

net

inout

inoutinout

inout

o

inin

PPzzgVVuu

!

VV2

22

CV

o

net

o

netinoutinout

inout

o

ininWQzzg

VVhhm

!

2

22

o

net

inout

inoutinout

inout

o

in

PPzzg

VVuu !

VV2

22

lossesP

gzV

uP

gzV

u

inout

!

VV 22

22

innetinoutquuosses !


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The Energy Equation /(Contd.)Application of Energy Eqn to Non-Uniform Flow

For this type of flow, the only part in above equation which require attention is. The other terms would remain same and can be simplified as done

earlier for particular applications

For one stream entering and leaving CV, we define the following expression for above integrant as

where E is Kinetic Energy Coefficientand V is average velocityas defined earlier in eqn. 5.7

At any cross section area A, ite the above eqn as

Thus

E= 0for Uniform flows and is always > 1 for non-uniform flows

Examples 5.26, 5.27 & 5.28

dAnPWQdAnPVudvt CSCV

onet

onet

CSCV inin

0

y

!y

x VVe VVVH

dV

CS

2

2

y V

!y

22

2

222

ininoutouto

1S

mdAnEE

V V

2

2

22 VmdAn

V o

AEV !y V

2

22

2

Vm

AV

2

A

y

! VV

E


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Irreversible Flow

Second Law of Thermodynamics


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Irreversible Flow /(Contd.)

nss ,

dt

sD

dt

Da

VV

!|

dt

sD

sDt

D

!a

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

!t

s

t

s

s

s

t

sstt

s

st

a

s

s

xx

s

xsx

xx

xx

!s

sss

a

s


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Irreversible Flow /(Contd.)

xx

xx

!s

ss

s

a

x

xs

s

Rs

s

!xx

Rs

sor

s

s

R

s 1!!

H

HHH

0psH

xxs

s

Rs

s

s

s

s

!!

x

x

p H

H

H 0

lim

nR

Vs

s

VV

2

x

x!a


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Irreversible Flow/(Contd.)


44/46




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Relation to material derivatives

Moving Control Volumes

z

w

yv

xttDt

D

xx

xx

xx

xx

!xx

| .V

dbdvbtDt

DB

CSCV

sysy

x! WVVH


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Assignments / Self Study

By now you must have solved and understood

Examples 5.1 to 5.29

Complete Solving at least 60 (out of 136)

problems from Chapter 5 of text book

ch5 flow analysis cv

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