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Finite Control Volume Analysis

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(1)

Chap. 5

Finite Control Volume

Analysis

(2)

5.1 Conservation of Mass – The Continuity Equation

 Law of Conservation of Mass:

 To express the control volume basis, Reynolds transport equation with B=mass and b=1 will result in

Dt 0 Dm

sys

“The amount of mass in a system is constant.”

 

  









 

 

 

 

 

 

1 cv cs

b

cs cv

sys sys

dA n V t dV

dA n V b t bdV

0 t bdV

Dt DB

 

 

(3)

 Conservation of Mass (Continuity Equation):



 

 



 

 



 

surface control

e through th

mass of

flow of

rate net volume

control the

of mass the

of change of

rate time system

the of mass the

of change of

rate time

  



 

1 cv cs

b

sys

dV V n dA

t Dt

DB  

  V n dA 0

t

cv

 dV 

cs

  

  

0 m

m

t

cv

 dV 

out

in

  

or

(4)

0 m

m

t

cv

 dV 

out

in

  

m 



 





 

 





volume control

the leaving

flow mass

of Rate volume

control the

entering

flow mass

of Rate volume

control the

of inside mass

of

on accumulati of

Rate

Here, means mass flowrate(flux, [kg/s]) across the control volume surface.

  V n dA A V Q

m  

cs

   

where is the average value normal to the control surface.

Q is called volume flowrate(flux, [m

3

/s]):

V  

A

dA n V V cs



V

A

Q 

(5)

 When the flow is steady, the continuity equation becomes

(if the flow has only one stream)

 If the steady flow is also incompressible,

(if the flow has only one stream) 0

m m

out

 

in

2 2 2 1

1

1

A V A V

m     

0 Q

Q

out

 

in

2 2 1

1

V A V

A

Q  

(6)

Ex. 5.1

0 m

m

t cvdV outin

 

zero in steady flow

(7)

Ex. 5.2

Severe changes in

pressure and temperature The flow should be

treated as compressible.

(8)

Ex 5.3

0 m

m

t

cv

 dV 

out

in

  

zero

(9)

Ex. 5.4

0 m

m

t cvdV out in

 

 

V n dA 0

t cvdV cs

 

 

V n dA

 

V n dA 0

t cvdV out in

 



0

 

A

dA n V V cs



(10)

Ex. 5.5

0 m

m

t cvdV outin

 

0 m

t  cvdV in

 

0 V

dt A ) Vol ( d

j

j

Neglecting this column of water

0

Incompressible:

0 V dt A

) h A ( d

j j

B  

AB

j j

j

B

A V Q

dt

A dh  

B j

A Q dt

dh 

(11)

5.1.3 Moving, Nondeforming Control Volume

  V n dA 0

t

cv

 dV 

cs

  

  

V 

W 

  



 

b1 Vcv Acs

sys

dV W n dA

0 t Dt

Dm  

  W n dA 0

t

Vcv

 dV 

Acs

  

  

The Reynolds transport equation for a moving control volume

:

where

V 

cv

W 

cv

V 

V

cv

V

W   

(12)

Ex. 5.6

2021 )

971 ( 1050 V

V

W2 2 cv Vcv

V W

absolute velocity

V2 V10

971 )

971 ( 0 V V

W1 1 cv Vcv

1

2

(13)

Ex. 5.7

(14)

Ex. 5.8

0 m

m

t

cv

 dV 

out

in

  

0 )

m m

(

t dV

2 leakage

t ) Vol (

cv

   







0

  ( Q Q ) 0

A t

2 leakage

V 1

p

 

 

 

Q Q0

V

A

1 p

 

2

leakage

1

2 2

1 leak 2

p

A

Q 1 . 0 Q

A Q

V Q 

 

  A ( Q Q ) 0

t

1

 

2

leakage

  

(15)

5.2 Newton’s Second Law

– The Linear Momentum and Moment of Momentum Equations

Derivation of the Linear Momentum Equation If we let b= ,

That is,

V 

 

  



 

 

 

 

 

 

 

cs cv

sys sys

sys sys

dA n V V dV

t V

t F V dV m

t V Dt

DB

 

 

  

F t 

cv

V dV

cs

V V n dA

Linear momentum equation

(16)

    





in

out A n

A n

cv

V dV V V dA V V dA

F  t   

or

  M

cv

M

out

M

in

F  t   

 

 

where denotes the momentum flow.M

 

 

 

 

 

 

 

 

 

 

 

CV entering

flow

momentum of

Rate

CV leaving

flow

momentum of

Rate CV

of momentum of

n accumultio of

Rate volume

control the

on acting

forces

external of

Sum

(17)

 Looking into the momentum flux terms:

 

V m V

m M

M

V A V V

A V dA

n V V M

M

in out

in out

in

n out

CS n in

out

 

 





 

 



 

x-component: M M

V A

u

V A

u mu mu

in out

in

n out

x n in

out

y-component:

z-component:

Writing in components,

 V A  v  V A  v m v m v M

M

in out

in

n out

y n in

out

V A

w

V A

w mw mw M

M

in out

in

n out

z n in

out

(18)

  M

cv

M

out

M

in

F  t   

 

 

 

 

in out

cv

y

m v m v

t

F Mv  

x-component:

 

in out

cv

x

m u m u

t

F Mu  

y-component:

z-component:

Therefore, the linear momentum equation becomes

 

in out

cv

z

m w m w

t

F Mw  

(19)

Ex. 5.10

 

M cv Mout Min

F t  

in out

x mu mu

F  

 

2

 

1 1

m 2

x

AV V cos AV V

F

2

 



0

p1=patm=p2 (Bernoulli equation across the straignt streamiles)

No net pressure force exerted on cv.

(Height difference was neglected.)

From the incompressible continuity equation, V1=V2

AV

 

V cos 1

Fx   1 1 

in out

z mw mw

F  

 AV V sin

F

z 1 1

0 i

V V1 1

 

cos i sin k

V

V2 2  

(20)

Ex. 5.11

 

M cv Mout Min

F t   

 

 

 

in out

cv

z mw mw

t

F Mw  

in in out

out

w 2

2 1 1 A z

w m w

m

A p A p F F

 

P2=0 p1

out in

w 2

2 1 1

A p A p A m w w

F

2 1

w 2

2 1 1

A p A p A m w w

F       

k w w

; k w

wout 2 in 1

2 1 2

1 1

1 A

w Q

; A

w  Q 

0

since

where

(21)

Thrust

(22)

Turbo-Jet 엔진 : 전투기용

압축기 연소실 배기 노즐

초음속용

소음 과다

연료 과소비

(23)

Turbo-fan 엔진 : 여객기용

아음속 용

소음 감소

효율 증가

(24)

Turbo-prop 엔진 : 프로펠러기용

(25)

Turbo-shaft 엔진 :헬리콥터용

헬기 로터

(26)

비행기에 작용하는 힘

(27)

Thrust Vectoring

(28)

Cobra

Bell 끄루크

(29)

Vertical Take-off of Harrier

(30)

JSF F-35 Joint Strike Fighter

JSF F-35

Vertical Take-off

(31)

Ship

V

max

 60 Knots (108km/h)

Waterjet

(32)

Waterjet Propulsion for Amphibian Tank

Waterjet Propulsion for Amphibian Tank

(33)

PNU ME CFD LAB.

Configuration of Waterjet

Waterjet Propulsion

(34)

Front oblique view Side view Near the rotor blade

Streamlines past the rotor of waterjet

Flow Analysis of Waterjet

(35)

Front oblique view Side view Near the rotor blade

Streamlines past the stator of waterjet

Flow Analysis of Waterjet

(36)

Pressure

Flow Analysis of Waterjet

Streamline

PNU ME CFD LAB.

(37)

PNU ME CFD LAB.

Armored Vehicle (Amphibian)

(38)

Armored Vehicle (Amphibian)

(39)

Armored Vehicle (Amphibian)

(40)

Ex. 5.12

in in out

out

2 2 1

1 A

y

v m v

m

A p A

p F

F

j v v

; j v

vout 2in 1

2 1

 

1 2

A

m v v A p p

F      

out in

1 1 2 2

A

m v v p A p A

F

  

since

 

in out

cv

y mv mv

t

F Mv

0

P1,abs

Gage pressure

(41)

EX. 5.13

in in out

out

2 2 1 1 x

x

u m u

m

A p A p R

F

 

in in out

out 2

2 1

1

x

p A p A m u m u

R

  

1 2

 

out in

x A p p m u u

R     

1 2

 

2 1

x

A p p m v v

R     

j v u

; j v

uout 2in 1

 

in out

cv

x mu mu

t

F Mu

Using

RT AV AV p

m 

 

 

If Rx has negative value, Rx acts  direction

(42)

Ex. 5.14

 

in out

cv

z

m w m w

t

F Mw  

 

    

   

12 R 2

0 2 2

1 1

CS 2 2

in in out

CS out out

in in out

out

2 2 1 1 z z

w R dr

r 2 w

w Aw dA

j j w j w

w m dA n

w w

w m w

m

A p A p R F





   





0R 2 12

2 2 z

2

1 R w 2 r dr R w

A p 1

p

0 velocityconst across the cross-section

Q: p1-p2=?

2 2 1R 3 w

4

(43)

Ex. 5.15

 

in out

cv

x mu mu

t

F Mu

0

in in out

out

2 2 1 1 th x

u m u

m

A p A p F

F

 

2 1

2 2 1

1

in in out

out 2

2 1

1 th

u u

m A

p A

p

u m u

m A

p A

p F

1 1 1 1 1

1

1 A V

RT V p

A

m 

 



where

(44)

Ex. 5.16

 

 

in out

cv

x mu mu

t

F Mu  

0 R

b 2 H

Fx  1  2x

b 2 H

Rx 1 2

(a) when the gate is closed

0 0 0

no flow

(b) when the gate is open

 

in out

cv

x mu mu

t

F Mu

2

2

1

1

f 2

x 2

x h b F u hb u u Hb u

2 R 1

b 2 H

F  1         

2

2

1

1

f 2

2

x h b F u hb u u Hb u

2 b 1 2 H

R  1       

If H>>h, u1<<u2 x 2 h2b Ff

u2hb

u2 2

b 1 2 H

R 1

(45)

Galilean transformation

(46)

Moving Control Volumes

If a control volume is moving with a constant velocity of , V 

cv

  



 

cs

cv A

V

sys

b dV b W n dA

t Dt

DB  

V

cv

V

W   

V 

V 

cv

where

: Relative velocity of the fluid relative to control volume

: Absolute velocity of the fluid : Velocity of the control volume

Vcv

V  V 

cv

W 

W 

(47)

For an inertial (no accelaration),

moving, nonderofming control volume

 For steady linear constant translating, nondeforming control volume

  



 

cs

cv A

V

sys

b dV b W n dA

t Dt

DB  

steady and b=

W 

  

F

cs

W W n dA

  

F

x

cs

W

x

W

n

dA

in x x

out x

in

n x

out

nA W W A W mW mW

W

  

In component

  

F

y

cs

W

y

W

n

dA

out

WnA

Wy

in

WnA

Wy

out mWy

in mWy

  

F

z

cs

W

z

W

n

dA

out

WnA

Wz

in

WnA

Wz

out mWz

in mWz

(48)

Ex. 5.17

(49)

5.2.3 Moment-of-Momentum (Angular Momentum) Equation

 obtained by setting b  r V 

       

      



 



 

 

 

 

 

 

cs cv

sys sys sys sys

sys

dA n V V r dV

V t r

M F

r V

r t m

dV V t r

Dt DB

 

 

 

 

 

 

         

sys

sys cv

r

V dV

cs

r

V V

n dA M t

F

r        

Thus,

     

 

r V dV m

 

r V m

 

r V

t

V r A V V

r A V dV

V t r

M

in cv out

in

n out

cv n sys

 

 

 

 

 

 

 

 

 

 

 

 

 

(50)

 If in steady flow,

     

 

r V m

 

r V

m M

is that

V r A V V

r A V M

in out

sys

in

n out

n sys

 

 

 

 

 

,

(51)

Application of Angular Momentum Eq.

(52)

Euler (Pump and Turbine) Equation

     

 

r V dV k m

 

r V k m

 

r V k

t

k V r A V k

V r A V k

dV V t r

M

in cv out

in

n out

cv n z

 

 

 

 

 

 

 

 

 



 

  

 



 

  

 

 

 

   

2 2

1

1 1

2

in out

shaft

V r m V

r m

k V r

m k

V r

m T

  

 

 

 

 

for steady state

Torque:

(53)

   

2 2 1 1

m m m

1 1 1 2

2 2 shaft

V r V

r m

V r m V

r m T

2 1

for steady state

Power transferred to the fluid by the rotor :

2 2 1 1

shaft

shaft

T m r V r V

W    

  

Since U=r,

2 2 1 1

shaft

m U V U V

W   

(54)

Euler (Pump and Turbine) Equation

2 2 1 1

shaft

m U V U V

W   

For a pump, , so is positive.

For a turbine, , so is negative.

1 1 2

2V U V

U W shaft

1 1 2

2V U V

U  W shaft

(55)

Velocity diagram:

V  U  W 

Rotor

velocity Relative velocity viewed from a coordinate

system fixed to the rotor Absolute

velocity

(56)

2 2 1 1

shaft

m U V U V

W   

2 2 2 1 1 1

shaft

m U V cos U V cos

W      

   

2 2

2 2

2

1 1 1

1 1 1

1 1

n 1

b r 2 sin

V

b r 2 sin

V b

r 2 V m

Mass flowrate:

(57)

Ex. 5.18

2 2 1 1

shaft

m r V r V

T  

0

(58)
(59)
(60)

Ex. 5.19

2 2 1 1

shaft

m U V U V

W   

Since V1=radial V1=0

(61)

2 2 1 1

shaft

m U V U V

W   

0

(62)

5.3 First Law of Thermodynamics – The Energy Equation

 setting b=e (total energy per unit mass)

  





 

cs cv

sys

sys

e dV e V n dA

dV t t e

Dt

DB  

e=internal energy + kinetic energy + potential energy = u

V

2

/ 2

gz

.

Reynolds transport eq:

in net in

net

W

Q   

 

in net in

cs net

cv

e dV e V n dA Q W

t

 

 

 

  

Hence,

(63)

 If considering the flow work by pressure in ,

 If the flow is steady and are uniform,

in

W 

net

 

in net shaft in

cs net

2

cv

gz V n dA Q W

2 V u p

dV t e

 



  

 

  

 

 

 

gz and 2 ,

, V , p u

2

 

in net shaft in

cs net 2

W Q

dA n V

2 gz V

u p 





in net shaft in

net in

2

out

2

(enthalpy) h

W Q

m 2 gz

V u p

m 2 gz

V

u p    

 

 

  

 









 

(64)

in net shaft in

net in

2

out

2

W Q

m 2 gz

h V m

2 gz

h V    

 

 

 

  

 

 

   

in net shaft in

net in

2

out

2

(enthalpy) h

W Q

m 2 gz

V u p

m 2 gz

V

u p    

 

 

  

 

 

 

 

 

 

 

(65)

Ex. 5.20

in net shaft in

net in

2

out

2

(enthalpy) h

W Q

m 2 gz

V u p

m 2 gz

V

u p





(66)

Ex. 5.21

ve ' W

, Q

in net in

net   

when heat and work are added into the system

ve ' W

, Q

in net in

net   

in net shaft in

net in

2

out

2

W Q

m 2 gz

h V m

2 gz

h V









when heat and work are extracted from the system

(67)

Ex. 5.22

(68)

5.3.3 Comparison of the Energy Equation with the Bernoulli Equation

in net in

2

out 2

Q 2 gz

V u p

2 gz V u p

m    







 

  

 

 

 

  

 

Energy equation:

m Q u

u 2 gz

V gz p

2 V

p innet

in out

quation Bernoullie

in 2

out 2

 

 

 

 

  

 



 

  

If the flow is incompressible and inviscid (i.e., frictionless), there is no changes in internal energy and no heat transfer occurs.

Assumptions for Bernoulli eq: zero -Steady state

-Inviscid flow

-Incompressible flow -Along the streamline

(69)

 Energy equation:

loss energy

w 2 gz

V gz p

2 V p

in net shaft

equation

Bernoulli

in 2

out

2

  

 

  

 

 

 

  

        

 

m W w net in

shaft

in net shaft

m / Q u

u loss energy

in net in

out

L s

head unit in the equation

Bernoulli

in 2

out 2

h h

g z 2 V z p

g 2 V

p   

 

  

 

 

 

  

          

 

Q

W g

m W g

w

h net in

shaft in

net shaft in

net shaft

s

g loss Energy

hL :head loss

(70)

Ex. 5.24

(71)

Ex. 5.25

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