fluid_mechanics
236 Chapter 5 ■ Finite Control Volume Analysis If the velocity profile at any section where flow crosses the control surface is not uniform, inspection of the energy equation for a control volume, Eq. 5.64, suggests that the integral cs V 2 2 rV nˆ dA The kinetic energy coefficient is used to account for nonuniform flows. will require special attention. The other terms of Eq. 5.64 can be accounted for as already discussed in Sections 5.3.2 and 5.3.3. For one stream of fluid entering and leaving the control volume, we can define the relationship V 2 cs 2 rV nˆ dA m# a a outV 2 out 2 where a is the kinetic energy coefficient and V is the average velocity defined earlier in Eq. 5.7. From the above we can conclude that m # aV 2 V 2 2 A rV nˆ dA 2 for flow through surface area A of the control surface. Thus, a inV 2 in b 2 a A 1V 2 22rV nˆ dA m # V 2 2 (5.86) Parabolic (laminar) Turbulent Uniform = 2 ~ 1.08 = 1 It can be shown that for any velocity profile, a 1, with a 1 only for uniform flow. Some typical velocity profile examples for flow in a conventional pipe are shown in the sketch in the margin. Therefore, for nonuniform velocity profiles, the energy equation on an energy per unit mass basis for the incompressible flow of one stream of fluid through a control volume that is steady in the mean is p out r a outV 2 out gz 2 out p in r a inV 2 in gz 2 in w shaft loss net in On an energy per unit volume basis we have p out ra outV 2 out 2 gz out p in ra inV 2 in 2 and on an energy per unit weight or head basis we have gz in rw shaft r1loss2 net in (5.87) (5.88) p out g a outV 2 out z 2g out p in g a inV 2 in z 2g in w shaft net in h g L (5.89) The following examples illustrate the use of the kinetic energy coefficient. E XAMPLE 5.26 Energy—Effect of Nonuniform Velocity Profile GIVEN The small fan shown in Fig. E5.26 moves air at a mass flowrate of 0.1 kgmin. Upstream of the fan, the pipe diameter is 60 mm, the flow is laminar, the velocity distribution is parabolic, and the kinetic energy coefficient, a 1 , is equal to 2.0. Downstream of the fan, the pipe diameter is 30 mm, the flow is turbulent, the velocity profile is quite uniform, and the kinetic energy coefficient, a 2 , is equal to 1.08. The rise in static pressure across the fan is 0.1 kPa and the fan motor draws 0.14 W. FIND Compare the value of loss calculated: (a) assuming uniform velocity distributions, (b) considering actual velocity distributions.
5.3 First Law of Thermodynamics—The Energy Equation 237 SOLUTION Application of Eq. 5.87 to the contents of the control volume shown in Fig. E5.26 leads to or solving Eq. 1 for loss we get 0 1change in gz is negligible2 To proceed further, we need values of w shaft net in , V 1 , and V 2 . These quantities can be obtained as follows. For shaft work power to fan motor w shaft net in m # or p 2 r a 2V 2 2 gz 2 2 p 1 r a 1V 2 1 gz 2 1 loss w shaft net in For the average velocity at section 112, V 1 , from Eq. 5.11 we obtain V 1 m# rA 1 m # (4) r1pD 2 142 0.479 ms For the average velocity at section 122, V 2 , (a) For the assumed uniform velocity profiles 1a 1 a 2 1.02, Eq. 2 yields loss w shaft net in loss w shaft net in a p 2 p 1 b a 1V 2 1 a 2V 2 2 r 2 2 w shaft 10.14 W2311 N # ms2W4 160 smin2 net in 0.1 kgmin 84.0 N # mkg 10.1 kg min2 11 min60 s2 11000 mmm2 2 11.23 kgm 3 23p160 mm2 2 44 V 2 10.1 kg min2 11 min60 s2 11000 mmm2 2 11.23 kgm 3 23p130 mm2 2 44 1.92 ms a p 2 p 1 b V 2 1 r 2 V 2 2 2 Using Eqs. 3, 4, and 5 and the pressure rise given in the problem statement, Eq. 6 gives loss 84.0 N # m kg 10.1 kPa211000 Pa kPa211 Nm 2 Pa2 1.23 kgm 3 10.479 ms2 2 11.92 ms2 2 231 1kg # m21N # s 2 24 231 1kg # m21N # s 2 24 (1) (2) (3) (5) (6) Turbulent flow or (Ans) (b) For the actual velocity profiles 1a 1 2, a 2 1.082, Eq. 1 gives If we use Eqs. 3, 4, and 5 and the given pressure rise, Eq. 7 yields or Section (2) α 2 = 1.08 D 2 = 30 mm Control volume loss 84.0 N # mkg 81.3 N # mkg 0.115 N # mkg 1.84 N # mkg 0.975 N # mkg loss w shaft net in loss 84 N # mkg 10.1 kPa211000 Pa kPa211 Nm 2 Pa2 1.23 kgm 3 210.479 ms2 2 1.0811.92 m s2 2 231 1kg # m21N # s 2 24 231 1kg # m21 # s 2 24 loss 84.0 N # mkg 81.3 N # mkg 0.230 N # mkg 1.99 N # mkg 0.940 N # mkg (7) (Ans) COMMENT The difference in loss calculated assuming uniform velocity profiles and actual velocity profiles is not large compared to for this fluid flow situation. w shaft net in D 1 = 60 mm Section (1) α 1 = 2.0 F I G U R E E5.26 Laminar flow m • = 0.1 kg/min a p 2 p 1 b a V 1 2 r 1 2 a V 2 2 2 2
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5.3 First Law of Thermodynamics—The Energy Equation 237<br />
SOLUTION<br />
Application of Eq. 5.87 to the contents of the control volume<br />
shown in Fig. E5.26 leads to<br />
or solving Eq. 1 for loss we get<br />
0 1change in gz is negligible2<br />
To proceed further, we need values of w shaft net in , V 1 , and V 2 . These<br />
quantities can be obtained as follows. For shaft work<br />
power to fan motor<br />
w shaft <br />
net in m #<br />
or<br />
p 2<br />
r a 2V 2 2<br />
gz<br />
2 2 p 1<br />
r a 1V 2 1<br />
gz<br />
2 1<br />
loss w shaft<br />
net in<br />
For the average velocity at section 112, V 1 , from Eq. 5.11 we obtain<br />
V 1 m#<br />
rA 1<br />
m #<br />
<br />
(4)<br />
r1pD 2 142<br />
0.479 ms<br />
For the average velocity at section 122, V 2 ,<br />
(a) For the assumed uniform velocity profiles 1a 1 a 2 1.02,<br />
Eq. 2 yields<br />
loss w shaft<br />
net in<br />
loss w shaft<br />
net in<br />
a p 2 p 1<br />
b a 1V 2 1<br />
a 2V 2 2<br />
r 2 2<br />
w shaft 10.14 W2311 N # ms2W4<br />
160 smin2<br />
net in 0.1 kgmin<br />
84.0 N # mkg<br />
10.1 kg min2 11 min60 s2 11000 mmm2 2<br />
11.23 kgm 3 23p160 mm2 2 44<br />
V 2 10.1 kg min2 11 min60 s2 11000 mmm2 2<br />
11.23 kgm 3 23p130 mm2 2 44<br />
1.92 ms<br />
a p 2 p 1<br />
b V 2 1<br />
r 2 V 2 2<br />
2<br />
Using Eqs. 3, 4, and 5 and the pressure rise given in the problem<br />
statement, Eq. 6 gives<br />
loss 84.0 N # m<br />
kg<br />
10.1 kPa211000 Pa kPa211 Nm 2 Pa2<br />
1.23 kgm 3<br />
10.479 ms2 2<br />
11.92 ms2 2<br />
<br />
<br />
231 1kg # m21N # s 2 24 231 1kg # m21N # s 2 24<br />
(1)<br />
(2)<br />
(3)<br />
(5)<br />
(6)<br />
Turbulent<br />
flow<br />
or<br />
(Ans)<br />
(b) For the actual velocity profiles 1a 1 2, a 2 1.082, Eq. 1<br />
gives<br />
If we use Eqs. 3, 4, and 5 and the given pressure rise, Eq. 7<br />
yields<br />
or<br />
Section (2)<br />
α 2 = 1.08<br />
D 2 = 30 mm<br />
Control volume<br />
loss 84.0 N # mkg 81.3 N # mkg<br />
0.115 N # mkg 1.84 N # mkg<br />
0.975 N # mkg<br />
loss w shaft<br />
net in<br />
loss 84 N # mkg 10.1 kPa211000 Pa kPa211 Nm 2 Pa2<br />
1.23 kgm 3<br />
210.479 ms2 2<br />
<br />
1.0811.92 m s2 2<br />
231 1kg # m21N # s 2 24 231 1kg # m21 # s 2 24<br />
loss 84.0 N # mkg 81.3 N # mkg<br />
0.230 N # mkg 1.99 N # mkg<br />
0.940 N # mkg<br />
(7)<br />
(Ans)<br />
COMMENT The difference in loss calculated assuming uniform<br />
velocity profiles and actual velocity profiles is not large<br />
compared to for this <strong>fluid</strong> flow situation.<br />
w shaft net in<br />
D 1 = 60 mm<br />
Section (1)<br />
α 1 = 2.0<br />
F I G U R E E5.26<br />
Laminar flow<br />
m • = 0.1 kg/min<br />
a p 2 p 1<br />
b a V 1 2<br />
r<br />
1<br />
2 a V 2 2<br />
2<br />
2