Chap. 5
Finite Control Volume
Analysis
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
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
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
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
Ex. 5.1
0 m
m
t cvdV out in
zero in steady flow
Ex. 5.2
Severe changes in
pressure and temperature The flow should be
treated as compressible.
Ex 5.3
0 m
m
t
cv dV
out
in
zero
Ex. 5.4
0 m
m
t cvdV out in
V n dA 0t cvdV cs
V n dA
V n dA 0t cvdV out in
0
A
dA n V V cs
Ex. 5.5
0 m
m
t cvdV out in
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
5.1.3 Moving, Nondeforming Control Volume
V n dA 0
t
cv dV
cs
V
W
b1 Vcv Acssys
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
cvW
cvV
V
cvV
W
Ex. 5.6
2021 )
971 ( 1050 V
V
W2 2 cv Vcv
V W
absolute velocity
V2 V10
971 )
971 ( 0 V V
W1 1 cv Vcv
1
2
Ex. 5.7
Ex. 5.8
0 m
m
t
cv dV
out
in
0 )
m m
(
t dV
2 leakaget ) Vol (
cv
0
( Q Q ) 0
A t
2 leakageV 1
p
Q Q 0
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
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
in
out A n
A n
cv
V dV V V dA V V dA
F t
or
M
cvM
outM
inF 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
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 muin 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 MM
in out
in
n out
z n in
out
M
cvM
outM
inF 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
Ex. 5.10
M cv Mout MinF t
in out
x mu mu
F
2
1 1m 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 10 i
V V1 1
cos i sin k
V
V2 2
Ex. 5.11
M cv Mout MinF 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
Thrust
Turbo-Jet 엔진 : 전투기용
압축기 연소실 배기 노즐
초음속용
소음 과다
연료 과소비
Turbo-fan 엔진 : 여객기용
아음속 용
소음 감소
효율 증가
Turbo-prop 엔진 : 프로펠러기용
Turbo-shaft 엔진 :헬리콥터용
헬기 로터
비행기에 작용하는 힘
Thrust Vectoring
Cobra
Bell 끄루크
Vertical Take-off of Harrier
JSF F-35 Joint Strike Fighter
JSF F-35
Vertical Take-off
Ship
V
max 60 Knots (108km/h)
Waterjet
Waterjet Propulsion for Amphibian Tank
Waterjet Propulsion for Amphibian Tank
PNU ME CFD LAB.
Configuration of Waterjet
Waterjet Propulsion
Front oblique view Side view Near the rotor blade
Streamlines past the rotor of waterjet
Flow Analysis of Waterjet
Front oblique view Side view Near the rotor blade
Streamlines past the stator of waterjet
Flow Analysis of Waterjet
Pressure
Flow Analysis of Waterjet
Streamline
PNU ME CFD LAB.
PNU ME CFD LAB.
Armored Vehicle (Amphibian)
Armored Vehicle (Amphibian)
Armored Vehicle (Amphibian)
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 2 in 1
2 1
1 2
A
m v v A p p
F
out in
1 1 2 2A
m v v p A p A
F
since
in out
cv
y mv mv
t
F Mv
0
P1,abs
Gage pressure
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 2 in 1
in out
cv
x mu mu
t
F Mu
Using
RT AV AV p
m
If Rx has negative value, Rx acts direction
Ex. 5.14
in out
cv
z
m w m w
t
F Mw
12 R 20 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 2 122 2 z
2
1 R w 2 r dr R w
A p 1
p
0 velocityconst across the cross-section
Q: p1-p2=?
2 2 1R 3 w
4
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
Ex. 5.16
in out
cv
x mu mu
t
F Mu
0 R
b 2 H
Fx 1 2 x
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
1f 2
x 2
x h b F u hb u u Hb u
2 R 1
b 2 H
F 1
2
2
1
1f 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 2b 1 2 H
R 1
Galilean transformation
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
cvV
W
V
V
cvwhere
: Relative velocity of the fluid relative to control volume
: Absolute velocity of the fluid : Velocity of the control volume
Vcv
V V
cvW
W
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
xW
ndA
in x xout x
in
n x
out
nA W W A W mW mW
W
In component
F
y
cs W
yW
ndA
out
WnA
Wy
in
WnA
Wy
out mWy
in mWy
F
z
cs W
zW
ndA
out
WnA
Wz
in
WnA
Wz
out mWz
in mWzEx. 5.17
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 cvr
V dV
csr
V V
n dA M t
F
r
Thus,
r V dV m
r V m
r Vt
V r A V V
r A V dV
V t r
M
in cv out
in
n out
cv n sys
If in steady flow,
r V m
r Vm M
is that
V r A V V
r A V M
in out
sys
in
n out
n sys
,
Application of Angular Momentum Eq.
Euler (Pump and Turbine) Equation
r V dV k m
r V k m
r V kt
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 stateTorque:
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
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
Velocity diagram:
V U W
Rotor
velocity Relative velocity viewed from a coordinate
system fixed to the rotor Absolute
velocity
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:
Ex. 5.18
2 2 1 1
shaft
m r V r V
T
0
Ex. 5.19
2 2 1 1
shaft
m U V U V
W
Since V1=radial V1=0
2 2 1 1
shaft
m U V U V
W
0
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,
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
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
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
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
Ex. 5.22
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
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
QW g
m W g
w
h net in
shaft in
net shaft in
net shaft
s
g loss Energy
hL :head loss