fluid_mechanics

5.3 First Law of Thermodynamics — The Energy Equation 231
SOLUTION
We use the control volume for each vent sketched in Fig. E5.23a.
What is sought is the flowrate, Q A 2 V 2 , where A 2 is the vent exit
cross-sectional area, and V 2 is the uniformly distributed exit velocity.
For both vents, application of Eq. 5.79 leads to
0 (no elevation change)
p 2
V 2 2
2 gz 2 p 1
V 2 1
2 gz 1 1 loss 2
0 1V 1 02
where 1 loss 2 is the loss between sections (1) and (2). Solving Eq.
1 for V 2 we get
(1)
Control
volume
Section (1) for
both vents is
in the room and
involves V 1 = 0
p 1 = 1.0 kPa
D 2 = 120 mm
D 2 = 120 mm
V 2
Section (2)
V 2
Section (2)
V 2 2 ca p 1 p 2
b B 1 loss 2 d
(2)
Control
volume
Since
1loss 2 K L V 2 2
2
(3)
F I G U R E E5.23a
where K L is the loss coefficient (K L 0.5 and 0.05 for the two vent
configurations involved), we can combine Eqs. 2 and 3 to get
Solving Eq. 4 for V 2 we obtain
Therefore, for flowrate, Q, we obtain
For the rounded entrance cylindrical vent, Eq. 6 gives
or
Q A 2 V 2 pD 2 2 p 1 p 2
4 B r311 K L 224
p1120 mm22
Q
411000 mmm2 2
V 2 2 ca p 1 p 2
b K V 2 2
B
L
2 d
11.0 kPa211000 PakPa2311Nm 2 21Pa24
B 11.23 kgm 3 2311 0.05224311Ns 2 21kgm24
For the cylindrical vent, Eq. 6 gives us
p1120 mm22
Q
411000 mmm2 2
V 2 B
p 1 p 2
r311 K L 224
Q 0.445 m 3 /s
11.0 kPa211000 PakPa2311Nm 2 21Pa24
B 11.23 kgm 3 2311 0.5224311Ns 2 21kgm24
(4)
(5)
(6)
(Ans)
or
Q 0.372 m 3 /s
(Ans)
COMMENT By repeating the calculations for various values
of the loss coefficient, K L , the results shown in Fig. E5.23b are
obtained. Note that the rounded entrance vent allows the passage
of more air than does the cylindrical vent because the loss associated
with the rounded entrance vent is less than that for the
cylindrical one. For this flow the pressure drop, p 1 p 2 , has two
purposes: (1) overcome the loss associated with the flow, and (2)
produce the kinetic energy at the exit. Even if there were no loss
(i.e., K L 0), a pressure drop would be needed to accelerate the
fluid through the vent.
Q, m 3 /s
0.5
0.4
0.3
0.2
0.1
(0.05, 0.445 m 3 /s)
0
0 0.1 0.2 0.3 0.4 0.5
K L
F I G U R E E5.23b
(0.5, 0.372 m 3 /s)
An important group of fluid mechanics problems involves one-dimensional, incompressible,
steady-in-the-mean flow with friction and shaft work. Included in this category are constant density
flows through pumps, blowers, fans, and turbines. For this kind of flow, Eq. 5.67 becomes
m # c ǔ out ǔ in p out
r p in
r V 2 out V 2 in
g1z
2
out z in 2d Q # net W # shaft
in net in
(5.80)