Ch5_CV_&_Continuity Presentation Guide Fluids

1
Control Volume Approaches &
Control Volume Approaches &
Continuity Principle
Continuity Principle
Chapter 5
CE319F: Elementary Mechanics of Fluids

2
Rate of Flow (Flow Rate)
• Volume rate of flow
– Constant velocity over
cross-section
– Variable velocity
• Mass flow rate
VA
Q 


A
VdA
Q
Q
VdA
VdA
m
A
A


 






3
Flow Rate
(velocity vector not normal to area)
• Only x-direction component
of velocity (u) contributes to
flow through cross-section
A
V
V


 

 




Q
or
dA
Q
or
dA
V
udA
VdA
Q
A
A A
A

cos
If velocity constant over area
Note: In all cases with dot product, only normal component of velocity multiplied by area

4
Mean Velocity
Vmean = Q/A
• Does not tell us anything about distribution of velocity
• Turbulent flow in pipe: Vmean may be close for much of
cross-section velocity

5
Example: Textbook Problem 5.4
A pipe whose diameter is 8 cm transports air with a temp. of 20o
C and
pressure of 200 kPa abs. The air velocity is 20 m/s.
Find: The mass flow rate?

6
• Find:
)
1
(
)
(
R
r
V
r
v o 

o
V
V

7
• Find: m
V
Q 
,
,

8
Approaches to Solving Fluids Problems
• Experimental Analysis
• Differential Analysis
• Control Volume Analysis
– most valuable tool available

9
Systems
• Laws of Mechanics
– Written for systems
– System = arbitrary quantity
of mass of fixed identity
– Fixed quantity of mass, m
0

dt
dm
dt
m
d )
( V
F



dt
dW
dt
dQ
dt
dE


• Conservation of Mass
– Mass is conserved and
does not change
• Momentum
– If surroundings
exert force on
system, mass
will accelerate
• Energy
– If heat is added to
system or work is
done by system,
energy will change

10
Control Volumes
)
(extensive
energy
momentum,
mass,

 



CV
CV
d
b
bdm
B 
)
(intensive
mass
unit
per
of
amount B
m
B
b



11
CV Inflow & Outflow
Area vector always points
outward from CV
A
V




Q
 







CS
in
out A
V
A
V
Q
Q
A
V
A
V
A
V






1
1
2
2
1
1
2
2

12
CV Inflow & Outflow


 





CS
CS
in
out
in
out
net m
b
b
m
b
m
b
B
B
B 






 A
V

B
m
b 
 

13
Reynolds Transport Theorem

 





































CS
CV
sys
net
CV
in
out
t
t
CV
t
t
CV
t
t
CV
in
out
t
t
CV
t
t
CV
t
t
t
sys
b
d
b
dt
d
dt
dB
B
dt
dB
t
B
B
t
B
B
t
B
B
B
B
t
B
B
dt
dB
A
V





0
,
,
0
,
,
0
,
0
lim
lim
lim
lim

14
Continuity Equation
• Reynolds Transport Theorem
)
(extensive
sys
M
B 
)
(intensive
1



dm
dM
dm
dB
b
sys
 

 

CS
CV
sys
b
d
b
dt
d
dt
dB
A
V




 

 

CS
CV
d
dt
d
A
V




0  

CS
A
V



0
Unsteady Case Steady Case

15
• Continuity Eq.
• Steady flow
• Incompressible fluid
1-D Flow in a Conduit
 

 

CS
CV
d
dt
d
A
V




0
 

CS
A
V



0
2
1
2
2
1
1
2
2
1
1
0
Q
Q
A
V
A
V
A
V
A
V






16

17
CS
VA=2VB
VB
h

18
CS
Find: At 22 s, will water surface be rising or falling?

19

20
Water forced out of cylinder by piston as shown. Piston driven at 5 ft/s. What
is speed of efflux of water from the nozzle if d = 2 in. and D = 4 in.?
Also, determine the force F required to drive the piston.

22
Example – Venturi Tube
Given: Water 20o
C, V1=2 m/s, p1=50 kPa, D=6 cm,
d=3 cm
Find: p2 and p3
D D
d
1
2
3
Nozzle: velocity
increases, pressure
decreases
Diffuser: velocity
decreases, pressure
increases

24
When gage A reads 120 kPa gage, cavitation just starts to occur
in the venturi meter. If D = 40 cm and d = 10 cm, what is the
water discharge in the system for a condition of incipient
cavitation? The atmospheric pressure is 100 kPa gage. The
water temperature is 10 o
C. Neglect gravitational effects.

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