CHAPTER 5. Finite Control Volume Analysis
 Applications of Reynolds Transport Theorem
a) Conservation of Fluid Mass (Continuity Equation)
Dependent to
choice of B
b) Newton’s 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
DM sys
i.e.
Dt
0
Consider a system and a fixed, nondeforming control volume as shown
(Coincident at an instant of time t)
System

V

V

V
Control Volume
t - t
t
t + t
Then, Reynolds transport theorem:
DM sys
Dt


D
dV +
=

d
V


cv
sys
t
Dt
= 0, because of mass
conservation of SYS
=

CS V  nˆdA
Time rate of change
of Fluid Mass in CV
+
Net flowrate of
mass through CS
Finally, Fluid Mass in a control volume




d
V


V

CS  nˆdA  0 : The continuity equation
t CV
e.g. For a steady flow,

 d V  0 
t CV
Mass in CV,
Then,

CS V  nˆdA = 0
CV dV
= Constant
Meaning: Mass flow leaving (+) CV
= Mass flow entering () CV
i.e. Across Control surface,
 m out   m in  0

dm
where m 
= A V  nˆdA  Q  AV ,
dt

for uniform flow over a projected area A, ( V )
 Useful analysis tips for Mass Conservation

1.V  nˆ over the CS:
Negative () for flow entering the CV
Positive (+) for flow leaving the CV
2. For a steady flow,
and thus,

dV = 0
t CV
 m out   m in  out Qout  in Qin  0
3. For a steady flow of incompressible fluid (  = constant)
 Qout   Qin   AoutVout   AinVin  0
4. For a non-uniform velocity distribution over the CS,
m = AV (V : average value)
5. For more than one steady non-uniform stream,
 m out   out AoutVout   in AinVin   m in
Ex. 1 Fixed and Non-deforming Control Volume
Air flows steadily between
two sections in a long, straight
portion of 4-in. inside diameter
pipe as indicated in figure. The
uniformly distributed temperature
and pressure at each section are
given.
If the average air velocity
(nonuniform velocity distribution) at section (2) is 1000 ft/s, calculate
the average air velocity at section (1).
Sol.) Necessary Eq.: The continuity equation,



ˆ

d
V

V

n
dA
+
=
0
(Steady
flow

 d V  0 )
CS
t CV
t CV
Then,

CS V  nˆdA  m out at (2)  m in at (1)  0
Nonuniform velocity distribution at (1) and (2)

Express the Equation using the Average velocity
m out at ( 2)  m in at (1)   2 A2V2  1 A1V1   2V2  1V
or
V1 
2
V
1 2
(Compressible Air

  Constant)
Since we know the pressure and temperature at Sections (1) and (2)
 V1 
2
pT
V2  2 1 V2
1
p1T2
(Using the ideal gas law, p  RT )
Ex. 2 Moving and Non-deforming Control Volume
An airplane moves forward at a speed of 971 km/h as shown. The
frontal intake area of the jet engine is 0.80 m2 and 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 a
speed of 1050 km/h. The engine exhaust area is 0.558 m2, and the
exhaust gas density is 0.515 kg/m3. Estimate the mass flowrate of fuel
into the engine in kg/h.
 In case of moving CV,



d
V
+

W
CS  nˆdA
t CV
Dt
  
where W  V  Vcv : Relative velocity
DM sys
=
Then, the continuity equation for a moving, nondeforming CV


  dV +  W  nˆdA = 0
CS
t CV



d
V

W
+
CS  nˆdA = 0
t CV
(Since the air flow relative to moving CV (Engine) is steady,
if the air surrounding the engine: Assumed to be stationary.)
Sol) Necessary Equation:

Then, CS W  nˆdA = 0 or
 m out   m in  0
a) Outflow of mass of CV:
b) Inflow of mass of CV:

(2) W  nˆdA  2W2 A2

(1) W  nˆdA + (Fuel supply)
 1W1 A1  m fuel
 fuel  0
Thus,  2W2 A2  1W1 A1  m
 m fuel  2W2 A2  1W1 A1
Note: W1 = (Velocity of the air = 0) – (Velocity of plane =  971 km/h)
= 971 km/h (From left to right)
W2 = (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

dV +
t CV
CS

W  nˆdA = 0: Still applicable
a)

 d V  0
t CV
b)
CS

W  nˆdA
: Boundary of integration changes

: Determined with W
Application 2. Newton’s Second Law (The Force Equation)
Let
B = momentum


b=V
Then, Reynolds transport theorem:

D
VdV
Dt sys
Time rate of change
of linear momentum
of the system
=
=


VdV
t CV
CS
+
Time rate of change
of linear momentum
in CV
+
 
VV  nˆdA
(1)
Net flowrate of
linear momentum
through the CS
 Newton’s 2nd law of linear motion of a system (No rotation)

d (mV )
  FExternal
dt Object
i.e.

D
VdV
Dt sys

=
Linear momentum of Fluid
(Mass × Velocity)

d ( mV )
dt Object

 F sys


D
VdV
Dt sys
(2)
Consider a system and a fixed, nondeforming control volume
At the instant of coincidence (Initial instant),

F
 sys =

F
 Contents of coincident
CV
By combining Eqs. (1) and (2),


V
dV +
t CV
CS
 
VV  nˆdA =

F
 Contents of coincident
CV
: Linear momentum equation (CV must be in inertial system)
 Control Volume analysis for Linear 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 each x, y, z components.
 Fx 

 udV +
t CV

u

V
 nˆdA
CS
: x-component
 Fy 

 vdV +
t CV

v

V
 nˆdA
CS
: x-component
 Fx 

 wdV +
t CV

w

V
 nˆdA : x-component
CS

where V  uiˆ  vˆj  wkˆ
Step 4. Check the steadiness of flow.
i.e. In case of a steady flow,


V
 dV  0
t CV
Step 5. Use the boundary conditions to determine the velocity on CS
(inlets and outlets, etc.)
Ex. 1 (Fixed and non-deforming control volume) As shown in Figure,
a horizontal jet of water exits a nozzle with a uniform speed of V1 =
10 ft/s, strikes a vane, and is turned through an angle  Determine
the anchoring force needed to hold the vane stationary. Neglect
gravity and viscous effects
z
x
FAx and FAz : x and z components of the anchoring force
Then linear momentum equations,



ˆ
u

d
V
F
u

V

n
dA
x - comp.:
+
=
 x  FAx

CS
t CV



ˆ
w

d
V
w

V

n
dA
F
z - comp. :
+
=
 z  FAz

CS
t CV
(i) Boundary conditions,
At section (1), u  V1 , w  0
&
At section (2), u  V2 cos  , w  V2 sin  &

V  nˆ  V1

V  nˆ  V2
(ii) In addition, Bernoulli eq. between sections (1) & (2)
1
1
p1  V12  z1  p2  V2 2  z2
2
2
where p1  p2  0 : Atmospheric pressure
: Neglect the gravity effect (Special case)
z1  z2
Then,
V1  V2
Inserting all values to the linear momentum equations,


(1) uV  nˆdA  (2) uV  nˆdA  V1 (V1) A1  V1 cos V1 A2  FAx


ˆ
u

V

n
dA

u

V
 nˆdA  (0)  (V1 ) A1  V1 sin V1 A2  FAz
(1)
(2)
(iii) From the continuity equation,
A1V1  A2V2

A1  A2 ,
since V1  V2
Finally,
 FAx =  V12 A1  cos V12 A1  (cos   1) V12 A1
FAz = sin V12 A1
Note. a) If   0 (Flat vane),
b) If   90 (Jet:),
FAx  FAz  0 (No anchoring force)
FAx   FAz : Negative
c) If   180 (Vertical vane), FAx : Negative & FAz = 0
Ex. 2. (Inertially moving, nondeforming control
volume) A vane on

wheels moves with constant velocity
V0 when a stream of water

having a nozzle exit velocity of V1 is turned by the vane as indicated

in figure. Determine the magnitude and direction of the force, F ,
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.
 Case of an “inertial”, “moving”, nondeforming control volume
- Coincident at an initial time
Reynolds transport theorem (Chap. 4),


 
D

VdV =  VdV +  VW  nˆdA
CS
Dt sys
t CV
  
where W  V  Vcv : Relative velocity
Then, linear momentum equation;

 


V

d
V
F
=
+
V
 Contents of coincident CV t CV
CS W  nˆdA
 
 


=  W  VCV dV +  W  VCV W  nˆdA
CS
t CV

(since Inertial CV  Constant VCV  Steady flow)
In addition,


 



CS W  VCV W  nˆdA = CS WW  nˆdA + VCV CS W  nˆdA
since the continuity equation for a steady flow,


dV +  W  nˆdA = 0
CS
t CV
Finally, for an inertial, moving, nondeforming control volume

CS
 
WW  nˆdA =

F
 Contents of coincident
CV
Back to the problem,
R x ( Rz ): Reaction of vane (Exerting on the CV) along x and z


( Exerting force by jet, F   R )
ww
: Fluid Weight within CV
Then linear momentum equations,

x - comp.:
W

W
CS x   nˆdA =  Rx
z - comp.:
CS Wz W  nˆdA = Rz  ww
(i) Boundary conditions,
At section (1),
At section (2)

Wx  W1 , Wz  0
& W  nˆ  W1

Wx  W2 cos  , W  W2 sin  & V  nˆ  W2
(ii) For simplicity, neglect viscous effect and elevation (gravity) effect,
then Bernoulli eq. between sections (1) & (2)
W1  W2
where W1  V1  V0
(See Ex. 1)
Inserting all values to the linear momentum equations,


ˆ
W

W

n
dA

W

W
(1) x
(2) x  nˆdA  W1 (W1 ) A1  W1 cos W1 A2 =  Rx


ˆ
W

W

n
dA

W

W
(1) z
(2) z  nˆdA  0  W1 sin W1 A2  Rz  ww
(iii) From the continuity equation,
 Rx  W12 A1 (1  cos  )
Rz  W12 A1 sin  ww
A1  A2
(See Ex. 1)