Chapter 3 x Integral Relations for a Control Volume
3.143 The insulated tank in Fig. P3.143 is
to be filled from a high-pressure air supply.
Initial conditions in the tank are T 20qC
and p 200 kPa. When the valve is opened,
the initial mass flow rate into the tank is
0.013 kg/s. Assuming an ideal gas, estimate
the initial rate of temperature rise of the air in
the tank.
267
Fig. P3.143
Solution: For a CV surrounding the tank, with unsteady flow, the energy equation is
d
dt
³
§
· p V2
in ¨ uˆ gz ¸ Q
Wshaft 0, neglect V 2 /2 and gz
e U dX m
2
U
©
¹
d
dT
dU
in c p Tin UX c v
c v TX
Rewrite as
( UX c v T) | m
dt
dt
dt
where U and T are the instantaneous conditions inside the tank. The CV mass flow gives
d
dt
³ U dX m
in
0, or: X
dU
dt
in
m
Combine these two to eliminate X(dU/dt) and use the given data for air:
dT
dt
p c v )T
m(c
UX c v
tank
qC
(0.013)(1005 718)(293)
| 3.2
s
ª 200000 º
3
« 287(293) » (0.2 m )(718)
¬
¼
3.144 The pump in Fig. P3.144 creates
a 20qC water jet oriented to travel a maximum horizontal distance. System friction
head losses are 6.5 m. The jet may be
approximated by the trajectory of frictionless particles. What power must be delivered by the pump?
Fig. P3.144
Solution: For maximum travel, the jet must exit at T
V2 sin T
2g 'z max
or: V2
Ans.
45°, and the exit velocity must be
[2(9.81)(25)]1/2
m
| 31.32
sin 45q
s
The steady flow energy equation for the piping system may then be evaluated:
p1 /J V12 /2g z1
p2 /J V 22 /2g z 2 h f h p ,
268
Solutions Manual x Fluid Mechanics, Fifth Edition
or: 0 0 15 0 (31.32)2 /[2(9.81)] 2 6.5 h p , solve for h p | 43.5 m
J Qh p
Then Ppump
ªS
º
(9790) « (0.05)2 (31.32) » (43.5) | 26200 W
¬4
¼
3.145 The large turbine in Fig. P3.145
diverts the river flow under a dam as
shown. System friction losses are hf
3.5V 2 /(2g), where V is the average velocity
in the supply pipe. For what river flow rate
in m3/s will the power extracted be 25 MW?
Which of the two possible solutions has a
better “conversion efficiency”?
Ans.
Fig. P3.145
Solution: The flow rate is the unknown, with the turbine power known:
p1
J
V12
z1
2g
where h f
V22
z 2 h f h turb , or: 0 0 50 0 0 10 h f h turb
J 2g
Q
2
3.5Vpipe
/(2g) and h p Pp /(J Q) and Vpipe
(S /4)D2pipe
p2
Introduce the given numerical data (e.g. Dpipe
4 m, Ppump
Q3 35410Q 2.261E6 0, with roots Q
25E6 W) and solve:
76.5, 137.9, and 214.4 m 3 /s
The negative Q is nonsense. The large Q ( 137.9) gives large friction loss, hf | 21.5 m.
The smaller Q ( 76.5) gives hf | 6.6 m, about right. Select Qriver | 76.5 m3/s. Ans.
3.146 Kerosene at 20qC flows through the
pump in Fig. P3.146 at 2.3 ft3/s. Head
losses between 1 and 2 are 8 ft, and the
pump delivers 8 hp to the flow. What
should the mercury-manometer reading h ft
be?
Solution: First establish the two velocities:
V1
2.3 ft 3 /s
(S /4)(3/12 ft)2
ft
1
ft
46.9 ; V2
V1 11.7
s
4
s
Q
A1
Fig. P3.146