PUMP SELECTION AND APPLICATION
245
13.1 to 13.14: Answers to questions in text.
13.17 1112
2 2
2 1
1 1
0.5 0.25 aaaa
N N
h h h h
N N
: ha divided by 4.
13.20 1112
22
12
1 1
0.75 0.5625 aaaa
D D
h h h h
D D
: 44% reduction.
13.22 1
12 3 6
13.24 1
1 3 6
2
CHAPTER THIRTEEN
PUMP SELECTION AND APPLICATION
Chapter 13
246
a
max
R
13.27 From Problem 13.26, 1
a
h = 248 ft: Let 2
a
h = 1.15 1
a
h = 285 ft
13.28
13.30 Throttling valves dissipate energy from fluid that was delivered by pump. When a lower
13.34 a. Rotary or 3500 rpm centrifugal
b. Rotary
13.35 Q = 350 gal/min; H = 550 ft; D = 12 in; N = 3560 rpm
13.36 Q = 2525 gal/min; H = 200 ft; D = 15 in; N = 1780 rpm
Point in Figure 13.34 lies in radial flow centrifugal region.
6 in 7 in 8 in 9 in 10 in
PUMP SELECTION AND APPLICATION
247
13.38 Ns = 3 / 4 0.75
(1750) 5000
(100)
N Q
H
= 3913
13.42 Same method as Problems 13.38 to 13.41.
a. Ns = 1463 radial b. Ns = 260 too low
13.43 to 13.46 See text.
13.48 Elevating reservoir raises pressure at pump inlet and increases NPSHa.
13.52 NPSHa = hsp hs hf hp: Some data from Prob. 7.14.
3 2
atm
14.4lb ft 144in
p
b. Water at 180 F, vp
h = 17.55 ft;
13.53 NPSHa = hsp hs hf hp = 34.48 4.8 2.2 6.78 = 20.70 ft
Chapter 13
248
13.54
NPS
H
a
=
h
sp
+
h
s
h
f
h
p
= 11.47
+
2.6
0.80
1
.
5
5
=
11.72
m
13.55 NPSHa = hsp hs hf hp
Friction 3 in Foot valve Elbow Friction 2 in K1 = 75f3T = 75(0.017) = 1.28
3 =
2 23
3
3 2
300 L/min 1 m /s 1.05 m (1.05)
;
4.768 10 m 60000 L/min 2 2(9.81)
Q gsA
= 0.0560 m
For all problems 13.56 13.65: NPSHa = hsp hs hf hvp
See Section 13.11, Equation 13-14. See Figure 13.37 for vapor pressure head hvp.
13.56 Find NPSHa: Carbon tetrachloride at 150 F; sg = 1.48; patm = 14.55 psia; hs = 3.6 ft; hf = 1.84 f
t
hvp = 16.3 ft
13.57 Find NPSHa: Carbon tetrachloride at 65 C; sg = 1.48; patm = 100.2 kPa; hs = 1.2 m; hf = 0.72 m
PUMP SELECTION AND APPLICATION
249
13.58
Find
NPS
H
a
:
G
a
s
o
l
i
n
e
a
t
4
0
C;
s
g
=
0.6
5
;
p
atm
=
9
9
.
2
k
P
a;
h
s
=
2.7
m
;
h
f
= 1.1
8
m
13.59 Find NPSHa: Gasoline at 110 F; sg = 0.65; patm = 14.28 psia; hs = +4.8 ft; hf = 0.87 ft
13.60 Find NPSHa: Carbon tetrachloride at 150 F; sg = 1.48; patm = 14.55 psia; hs = +3.66 ft; hf = 1.84 f
t
13.61 Find NPSHa: Gasoline at 110 F; sg = 0.65; patm = 14.28 psia; hs = 2.25 ft; hf = 0.87 ft
13.62 Find NPSHa: Carbon tetrachloride at 65 C; sg = 1.48; patm = 100.2 kPa; hs = +1.2 m; hf = 0.72 m
hvp = 4.8 m
hvp = 14.0 m
13.64 Find required pressure above the fluid in a closed, pressurized tank so that NPSHa 4.0 ft.
Propane at 110 F; sg = 0.48; patm = 14.32 psia; hs = +2.50 ft; hf = 0.73 ft
13.65 Find required pressure above the fluid in a closed, pressurized tank so that NPSHa 150.0 m.
Chapter 13
250
13.66