26389078-finite-cv-analysis.pdf

7/27/2019 26389078-Finite-CV-Analysis.pdf

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CHAPTER 5. Finite Control Volume Analysis

Applications of Reynolds Transport Theorema) Conservation of Fluid Mass (Continuity Equation)

b) Newtons 2nd

law of fluid motion (Fluid dynamics)

c) 1st

and 2nd

laws of Thermodynamics

Note: An assumption through entire chapter,

Flow properties)Uniform over cross-sectional areas (CS)

Application 1. Conservation of Mass (The Continuity Equation)

Let B = mass ) b = 1

Mass of a system: Must be conserved i.e. 0=Dt

DMsys

Consider a system and a fixed, nondeforming control volume as shown

System

Control Volume

Vr

Vr

Vr

t - t t + tt

Dependent to

choice ofB

(Coincident at an instant of time t)


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Then, Reynolds transport theorem:

= syssys

VdDt

D

Dt

DM =

cv

Vdt

+ CS dAnV r

Finally,Fluid Mass in a control volume

0 =+

CSCV dAnVVdt

r : The continuity equation

e.g. For a steady flow,

Mass in CV, 0=

CV Vd

t

& CV

Vd = Constant

Then, CS dAnV r

= 0 Meaning: Mass flow leaving (+) CV

= Mass flow entering () CV

i.e. Across Control surface,

0= inout mm&&

wheredt

dmm =& = A dAnV

r AVQ == ,

for uniform flow over a projected areaA, (Vr

)

= 0, because of mass

conservation of SYS

Time rate of change

ofFluidMass in CVNet flowrate of

mass throu h CS

= +


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Useful analysis tips forMass Conservation

1. nV r

over the CS: Negative () for flowentering the CV

Positive (+) for flowleaving the CV

2. For asteady flow,

CVVd

t = 0

and thus, 0== ininoutoutinout QQmm &&

3. For asteady flow ofincompressible fluid ( = constant)

0== ininoutoutinout VAVAQQ

4. For anon-uniform velocity distribution over the CS,

m& = AV (V : average value)

5. For more than onesteadynon-uniformstream,

=== ininininoutoutoutout mVAVAm &&


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Ex. 1Fixed and Non-deformingControl Volume

Air flows steadily betweentwo sections in a long, straight

portion of 4-in. inside diameterpipe as indicated in figure. Theuniformly distributed temperatureand pressure at each section aregiven.

If the average air velocity(nonuniform velocity distribution) at section (2) is 1000 ft/s, calculatetheaverage air velocity at section (1).

Sol.) Necessary Eq.: The continuity equation,

CVVd

t + CS dAnV

r = 0 (Steady flow & 0=

CV Vdt

)

Then, 0 )1(at)2(at == inoutCS mmdAnV &&r

Nonuniform velocity distribution at (1) and (2)

& Express the Equation using theAverage velocity

VVVAVAmm inout 122111222)1(at)2(at == &&

or 21

2

1 VV

= (Compressible Air&

Constant)

Since we know the pressure and temperature at Sections (1) and (2)

221

122

1

21 V

Tp

TpVV ==

(Using the ideal gas law, RTp = )


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Ex. 2Moving and Non-deformingControl Volume

An airplane moves forward at a speed of 971 km/h as shown. Thefrontal intake area of the jet engine is 0.80 m

2and the entering air

density is 0.736 kg/m3. A stationary observer determines that relative to

the earth, the jet engine exhaust gases move away from the engine with aspeed of 1050 km/h. The engine exhaust area is 0.558 m

2, and the

exhaust gas density is 0.515 kg/m3. Estimate the mass flowrate of fuel

into the engine in kg/h.

In case ofmoving CV,

Dt

DMsys=

CVVd

t + CS dAnW

r

where cvVVWrrr

= : Relative velocity

Then, the continuity equation for a moving, nondeforming CV

CV

Vd

t

+

CS

dAnW r

= 0

Sol) Necessary Equation:

CVVd

t + CS dAnW

r = 0

(Since the air flow relative to moving CV (Engine) is steady,if the air surrounding the engine: Assumed to be stationary.)


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Then, CS dAnW r

= 0 or 0= inout mm &&

a) Outflow of mass of CV: 222)2( AWdAnW =

r

b) Inflow of mass of CV: )1( dAnWr

+ (Fuel supply)

fuelmAW &+= 111

Thus, 222 AW 0111 = fuelmAW &

fuelm& 222 AW= 111 AW

Note: 1W = (Velocity of the air = 0) (Velocity of plane = 971 km/h)

= 971 km/h (From left to right)

2W = (Velocity of the exhaust air = 1050 km/h)

(Velocity of plane = 971 km/h)

= 2021 km/h (From left to right)

C.f. Deforming and moving Control Volume

: Change in volume size & Control surface movement

CVVd

t + CS dAnW

r = 0: Still applicable

a) 0

CV Vdt : Boundary of integration changes

b) CS dAnW r

: Determined with Wr


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Application 2. Newtons Second Law (The Force Equation)

Let B = momentum & b = Vr

Then, Reynolds transport theorem:

sys VdVDtD

r

=

CVVdV

tr

+ CS dAnVV rr

(1)

y Newtons 2nd law oflinear motion ofa system (No rotation)

&

i.e. sys VdVDtD r =

sysF r (2)

Time rate of changeoflinear momentum

of the system

Time rate of changeoflinear momentum

in CV

Net flowrate oflinear momentum

through the CS= +

= External)(

Fdt

Vmd

Object

r

Linear momentum of Fluid(Mass Velocity)

Objectdt

Vmd )(r

sys VdVDtD

r


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Consider a system and a fixed, nondeforming control volume

At the instant of coincidence (Initial instant),

sysFr

=CVcoincidentofContentsF

r

By combining Eqs. (1) and (2),

CVVdV

tr

+ CS dAnVV rr


r

:Linear momentum equation (CV must be in inertial system)


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Control Volume analysis forLinear Momentum Conservation

Step 1. Choose the appropriate CV.

Step 2. Draw a free-body diagram.

i.e. Find all forces acting on the chosen CV

Step 3. Apply the force equation for eachx, y, z components.

=

CVxVdu

tF + CS dAnVu

r :x-component

=

CVyVdv

tF + CS dAnVv

r :x-component

=

CVxVdw

t

F + CS dAnVw r

:x-component

where kwjviuV ++=r

Step 4. Check the steadiness of flow.

i.e. In case of asteady flow, 0=

CVVdV

tr

Step 5. Use the boundary conditions to determine the velocity on CS

(inlets and outlets, etc.)


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Ex. 1 (Fixed and non-deformingcontrol volume) As shown in Figure,a horizontal jet of water exits a nozzle with a uniform speed of 1V =

10 ft/s, strikes a vane, and is turned through an angle . Determinethe anchoring force needed to hold the vane stationary. Neglectgravity and viscous effects

AxF and AzF :x andz components of the anchoring force

Then linear momentum equations,

x - comp.:

CV

Vdut

+ CS dAnVu r

= Axx FF =r

z - comp. :

CV

Vdwt

+ CS dAnVw r

= Azz FF =r

(i) Boundary conditions,

At section (1), 1Vu = , 0=w & 1 VnV =r

At section (2), cos2Vu = , sin2Vw = & 2 VnV =r

x

z


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(ii) In addition, Bernoulli eq. between sections (1) & (2)

22

2212

112

1

2

1zVpzVp ++=++

where 021 == pp : Atmospheric pressure

21 zz = : Neglect the gravity effect (Special case)

Then, 21 VV =

Inserting all values to the linear momentum equations,

211111)2()1(cos)( AVVAVVdAnVudAnVu +=+ rr AxF=

21111)2()1(sin)()0( AVVAVdAnVudAnVu +=+

rr AzF=

(iii) From the continuity equation,

2211 VAVA = 21 AA = , since 21 VV =

Finally,

AxF = 12

112

112

1 )1(coscos AVAVAV =+

AzF = 12

1sin AV

Note. a) If 0= (Flat vane), 0== AzAx FF (No anchoring force)

b) If o90= (Jet:), AzAx FF = : Negative

c) If o180= (Vertical vane), AxF : Negative & AzF = 0


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Ex. 2. (Inertially moving, nondeforming control volume) A vane onwheels moves with constant velocity 0V

rwhen a stream of water

having a nozzle exit velocity of 1Vr

is turned by the vane as indicated

in figure. Determine the magnitude and direction of the force, Fr

,

exerted by the stream of water on the vane surface. The speed of the

water jet leaving the nozzle is 100 ft/s, and the vane is moving to the

right with a constant speed of 20 ft/s.


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Case of an inertial, moving, nondeforming control volume- Coincident at an initial time

Reynolds transport theorem (Chap. 4),

sys VdVDtD

r

=

CVVdV

tr

+ CS dAnWV rr

where cvVVWrrr

= : Relative velocity

Then, linear momentum equation;

CVcoincidentofContentsFr

=

CVVdV

tr

+ CS dAnWV rr

= ( ) +

CVCV VdVW

t

rr+ ( ) +CS CV dAnWVW

rrr

(since Inertial CV) Constant CVVr

) Steady flow)

In addition,

( ) +CS CV dAnWVW rrr

= CS dAnWW rr

+ CSCV dAnWV rr

Finally, for an inertial, moving, nondeforming control volume

CS dAnWW rr


r

since the continuity equation for a steady flow,

CVVd

t + CS dAnW

r = 0


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Back to the problem,

xR ( zR ): Reaction of vane (Exerting on the CV) alongx andz

( Exerting force by jet, RFrr

= )

ww : Fluid Weight within CV

Then linear momentum equations,

x - comp.: CS x dAnWW r

= xR

z - comp.: CS z dAnWW r

= wz wR

(i) Boundary conditions,

At section (1), 1WWx = , 0=zW & 1 WnW =r

At section (2) cos2WWx = , sin2WW= & 2 WnV =r

(ii) For simplicity, neglect viscous effect and elevation (gravity) effect,

then Bernoulli eq. between sections (1) & (2)

21 WW=

where 011 VVW=

(See Ex. 1)

Inserting all values to the linear momentum equations,

211111)2()1(cos)( AWWAWWdAnWWdAnWW xx +=+

rr= xR

wzzz wRAWWdAnWWdAnWW =+=+ 211)2()1( sin0 rr

(iii) From the continuity equation, 21 AA = (See Ex. 1)

=xR )cos1(12

1 AW

=zR wwAW + sin12

1

26389078-finite-cv-analysis.pdf

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