fluid_mechanics

664 Chapter 12 ■ Turbomachines
(2)
Δz = 10 ft
K L = 1.5
Pump
Diameter of pipe = 6 in.
Total pipe length = 200 ft
Water
K L = 1.0
(1)
K L = 0.5
(a)
100
80
Head
Head, ft
Efficiency, %
60
40
Efficiency
20
0
0 400 800 1200 1600 2000 2400
F I G U R E E12.4a, b
Flowrate, gal/min
(b)
SOLUTION
Application of the energy equation between the two free surfaces,
points (1) and (2) as indicated, gives
p 1
g V 2 1
2g z 1 h a p 2
g V 2 2
2g z 2
f / V 2
(1)
D 2g a K V 2
L
2g
Thus, with p 1 p 2 0, V 1 V 2 0, z z 2 z 1 10 ft,
f 0.02, D 6/12 ft, and 200 ft, Eq. 1 becomes
where the given minor loss coefficients have been used. Since
Eq. 2 can be expressed as
1200 ft2
h a 10 c 0.02
1612 ft2
10.5 1.5 1.02d
2132.2 ft/s 2 2
V Q A Q1ft 3 /s2
1/4216/12 ft2 2
h a 10 4.43 Q 2
V 2
(2)
(3)
where Q is in ft 3 /s, or with Q in gallons per minute
h a 10 2.20 10 5 Q 2
Equation 3 or 4 represents the system equation for this particular
flow system and reveals how much actual head the fluid will need
to gain from the pump to maintain a certain flowrate. Performance
data shown in Fig. E12.4b indicate the actual head the
fluid will gain from this particular pump when it operates at a certain
flowrate. Thus, when Eq. 4 is plotted on the same graph with
performance data, the intersection of the two curves represents
the operating point for the pump and the system. This combination
is shown in Fig. E12.4c with the intersection (as obtained
graphically) occurring at
Q 1600 gal/min
(4)
(Ans)
with the corresponding actual head gained equal to 66.5 ft.
Another concern is whether the pump is operating efficiently
at the operating point. As can be seen from Fig. E12.4c, although
this is not peak efficiency, which is about 86%, it is close
(about 84%). Thus, this pump would be a satisfactory choice,
assuming the 1600 gal/min flowrate is at or near the desired
flowrate.