Chapter 20 Pumps And Fans n2z2x

EE THEN II = I: EE=E
PUMPS AND FANS 0 613 130 LPRINT"II,E(l)»Hj11jEE 140 READ HH.QQ,NSYN,Dl,Nil DATA 60.,200.,3600.,72.,225. 150 LPRINTwHHvQQvNSYNtDlfNlsM{HHiQQ;NSYN;Ql;Nl 160 D*(CQQ/Q1UI>)~2*H1UI>/HH)~.25*D1: N*N1 *SQR (HH/H t (II) >*D1/D 170 LPRINT"D,N*";D;N: I*FIX(NSYN/N) 180 NN1*FIX(NSYN/I>: NN2*FIX <1♦1)>: LPRINT"N,NN1.NN2=M;N;NNl;NN2 190 IF (NNl-NK (N-NN2) THEN N*NNt ELSE N*NN2 200 D*0itNl *SQR(HH/Hl < 11) ) /N: 00* (D/Di >A3>Nt0l

>~2: C2«(D/Dl)~3#N/N1 220 FOR 1*1 TO 15: H(I)*C1*H1 ; : NEXT I: LPR I NT HI*60.0 57.5 55.0 52.5 50.0 47.5 45.0 42.5 40.0 37.5 35.0 32.5 30.0 27.5 25.0 Q=200.0 228.0 256.0 280.0 303.0 330.0 345.0 362.0 382.0 396.0 411.0 425.0 438.0 449.0 459.0 E*69.0 75.0 BO.O 83.7 86.0 87.3 88.0 87.4 86.3 84.4 82.0 79.0 75.0 71.0 66.5 I I. E (I) * 7 88 HH,QQ,NSYN,D1,N1* 60 200 3600 72 225 D,N= 51.01563 366.6749 N,NN1,NN2* 366.6/49 400 360 D, N,QQ= 51.96153 360 207.4853 H- BO. O 76.7 73.3 70.0 66.7 63.3 60.0 56.7 53.3 50.0 46.7 43.3 40.0 36.7 33.3 Q* 120.3 137.1 154.0 168.4 182.2 198.5 207.5 217.7 229.7 238.2 247.2 255.6 263.4 270.0 276.0

20.18

Develop the characteristic curve for a homologous pump of the series of Prob. 20.16 for 21-in-diameter discharge and 1600 rpm. Q = QXN/MKD/D,)3 = G,W(1.75/6)3 = 0.176Q,

I

H = H^ND/N^f = W,[(1600)(1.75)/(225)(6)]2 = 4.30«, «1

H

Qi

Q

60 55 50 45 40 35 30 25

258 236 215 194 172 150 129 108

200 256 303 345 382 411 438 459

35.2 45.1 53.3 60.7 67.2 72.3 77.1 80.8

See curve plotted in Fig. 20-2.

20.19

Determine the size and synchronous speed of a pump homologous to the 72-in pump of Prob. 20.16 that will produce 2.5 m3/s at 90-m head when operating at highest efficiency.

I

Q/ND* = QjNtD 3 ,

e, = 2.5/0.3048 3

3

= 88.3cfs

«,= 90/0.3048 = 295.3 ft 3

«,Z>? = (QJQ)ND = (88.3/345)(225)(H) = 12 439 H/N 2 D 2 = HJN\D\ N\D\ = (HJH)N 2 D 2 = (295.3/45 )(225)2(T!)2 = H 959 650

614 0 CHAPTER 20 D x = 1.897 ft, or 22.76 in, and N x = 1822 rpm. Use N = 1800 rpm (one pair of poles). Then, D, = y/HJH(N/N t )D = V295.3/45(^)(6) = 1.92 ft. 20.20

Sketch the head versus discharge curve for a centrifugal pump (Fig. 20-la) having r, = 55 mm, r 2 = 110 mm, ft, = 27.5 mm, b 2 = 22 mm, N = 1300 rpm, and ft = 33°. f H = u\jg - [ ( U 2 Q cot ft)/(2nr 2 b 2 g)] where u 2 = N(2jt/60)r 2 = 1300(2JT/60)(0. 110) = 14.975 m/s, H = (14.9752/9.807) - {[(14.975 cot 33°)0]/[2/r(O.llO)(O.O22)(9.8O7)]} = 22.87 - 173.940. See Fig. 20-3.

20.21

A centrifugal pump (see Fig. 20-la) has an impeller with dimensions r x = 75 mm, r 2 = 150 mm, ft, = 50 mm, b 2 = 30 mm, ft = ft = 30°. For a discharge of 70 L/s and shockless entry to vanes, compute (a) the speed, (b) the head, (c) the torque, and (d) the power. Neglect losses (o') = 90°). V; = ^, = 0/2^6, = (70 xl0“3)/[(2)(^)(0.075)(0.050)] = 2.971 m/s

I

a, = V,/(tan /3,) = 2.971/(tan 30°) = 5.146 m/s (a)

N = (a,/r,)(60/2jr) = [5.146/(0. 075)](60/2JT) = 655.2 rpm

(ft) V r2 = Q/2jzr 2 b 2 = (r t b,/r 2 b 2 )V, = [(75)(50)/(150)(30)](2.971) = 2.476 m/s u 2 /u x = r2/r, = 150/75 = 2 u 2 = 2M, = (2)(5.146) = 10.292 m/s V u2 = u 2 - V r2 cot /32 = 10.292 - 2.476 cot 30° = 6.003 m/s H = u 2 VJg = (10.292)(6.003)/9.807 = 6.30 m (c) 20.22

T = pQ(r 2 V u2 ) = (1000)(70 x 10 3)(0.150)(6.003) = 63.0 N • m

{d) P = T(o = T(ujr { ) = (63.0)[5.146/(0.075)] = 4323 W A centrifugal pump with impeller dimensions r, = 2 in, r 2 = 5 in, ft, = 3 in, ft2 = 1.5 in, jS2 = 60° is to pump 6 cfs at a 64-ft head. Determine (a) /3,, (6) the speed, (c) the pressure rise across the impeller. Neglect losses and assume no shock at the entrance (or, = 90°).

I

V rl = QI2nr x b x = 6/[(2)0r)(n)(^)] = 22.92 ft/s V* = 0/2^6, = (r,b x /r 2 b 2 )V rX = [(2)(3)/(5)(1.5)](22.92) = 18.34 ft/s H = u 2 V u2 /g = u 2 (u x — v r2 cot /?2) ul-u 2 V r2 cotp 2 -gH = 0

U = (K2 cot p 2 + Vv22 cot2 fi2 + AgH)/2= [18.34 cot 60° + V18.342 cot2 60° + (4)(32.2)(64)]/2 = 51.0 ft/s 2

«i = (r x /r 2 )u 2 = (1)(51.0) = 20.4 ft/s (а)

tan /3, = VJu x = 22.92/20.4 = 1.124

(б)

/3, = 48.33°

N = u 2 /r 2 = [51.0/(^)](60/2^r) = 1169 rpm

(C) V, = K,/(sin /3,) = 22.92/(sin 48.33°) = 30.68 ft/s

V 2 = ^/(sin ft) = 18.34/(sin 60°) = 21.18 ft/s

Ap = (p/2)(u 2 — u x — V 2 + V X ) = (1.94/2)(51.0 - 20.4 - 21.182 + 30.682) = 2597 Ib/ft2 or 18.0 lb/in2 2

20.23

2

2

Evaluate r x , r 2 , /3,, ft, ft,, and ft2 of a centrifugal impeller that will take 36 L/s of water from a 100-mm-diameter suction line and increase its energy head by 18 in (TV = 1440 rpm; or, = 90°). Neglect losses.

PUMPS AND FANS £7 615 f Select sample values r, = 50 mm, r 2 -100 mm, f>, = 40 mm, b 2 = 25 mm. Then find /3] and (3 2 . V, = QI2jzr t b x = (36 x 10"3)/[(2)(;r)(0.050)(0.040)] = 2.865 m/s u, = (or, = [(2)(JT)(1440)/60](0.050) = 7.540 m/s u 2 = (r2/r,)u, = (f?)(7.540) = 15.08 m/s Since or, = 90°, tan /3, = Vj/u, = 2.865/7.540 = 0.37997, /3, = 20.8°. Vu2 = tfg/u2 = (18)(9.807)/15.08 = 11.71 m/s Vr2 = V rt {r t bjr 2 b 2 ) = (2.865)[(50)(40)/(100)(25)] = 2.292 m/s tan |S2 = VJ(u 2 - V u2 ) = 2.292/(15.08 - 14.04) = 2.20385 /32 = 65.6° Note; It is best to choose r, = 50 mm (100-mm inlet diameter). Then r2, b 2 , and 6, can be arbitrarily selected within reasonable limits. 20.24

A mercury-water differential manometer, R' = 740 mm, is connected from the 100-mm-diameter suction pipe to the 80-mm-diameter discharge pipe of a pump, as shown in Fig. 20-4. The centerline of the suction pipe is 330 mm below the discharge pipe. For Q = 3.42 m3/min of water, calculate the head developed by the pump. f V, = Q/A, = (3.42/60)/[(jr)(0.100)2/4] = 7.257 m/s V 2 = V,(£>?/D|) = (7.257)(1002/802) = 11.34 m/s The energy equation yields pjy + V,/2g + H p = p 2 /y + V\/2g + h, H p = Ap/y + (V2— V 2 ,)/2g + h. The manometer equation yields Ap/y = R'(s.g. — 1) — h = (0.740)(13.6 — 1) — 0.330 = 8.99 m of water, H p = 8.99 + (11.342 - 7.2572) / [(2)(9.807)] + 0.330 = 13.19 m.

20.25

An air blower is to be designed to produce pressure of 90 mmH 20 when operating at 3200 rpm; r 2 = 1. lr,; f} 2 = /3,; width of impeller is 100 mm; a, = 90°. Find . Assume a temperature of 30 °C.

f Q = 2;rr1i>1 V, = 2nr 2 b 2 V 2 . Since r2 = 1. lr, and 6, = b 2 , V r2 = Vri/1.1 = VJl.l w2= (r2/r,)Mi = 11 «1

h air = WH o(yH o/yair) = (0.090)[(9.77)(1000)/11.4] = 77.13 m of air 2

2

V„2 = ghju 2 = (9.807)(77.13)/1.1«, = 687.6/M, M2 = V„2 + Vr2 cot /8 Since /3, = /32 = /3 and M, = V, cot /3, 1.1M, = 687.6/M, + (V, cot /3)/l.l = 687.6/M, + M,/1.1, M, = 60.0 m/s; r, = M,/ct) = 60.0/[(3200)(2JT)/60] = 0.179 m, or 179 mm. 20.26

In Prob. 20.25, calculate the discharge for j3, = 32°.

f 20.27

What is the cavitation index at a point in flowing water where T = 20 °C, p = 16 kPa, and the velocity is 13 m/s. I

20.28

V, = M, tan j3, = 60.0 tan 32° = 37.5 m/s Q =2jir l b 1 V l = (2)(jt)(0.179)(0.100)(37.5) =4.218 m3/s or 253m3/min

a' = (p -

P V )I( P V

2

I2) = (16 - 2.34)(1000)/[(998)(13)2/2] = 0.162

Two reservoirs A and B are connected with a long pipe which has characteristics such that the head loss through the pipe is expressible as h L = 20Q2, where h, is in feet and Q is the flow rate in hundreds of gpm. The water-surface elevation in reservoir B is 35 ft above that in reservoir A. Two identical pumps are available for use to pump the water from A to B. The characteristic curve of the pump when operating at 1800 rpm is given in the table on page 616. At the optimum point of operation the pump delivers 200 gpm at a head of 75 ft. Determine the specific speed N s of the pump and find the rate of flow under the following conditions: (a) a single pump operating at 1800 rpm; (6) two pumps in series, each operating at 1800 rpm; (c) two pumps in parallel, each operating at 1800 rpm.

f The head-capacity curves for the pumping alternatives are plotted in Fig. 20-5 and so is the h L versus Q curve for the pipe system. In this case h = Az + h L = 35 + 20Q2- The answers are found at the points of

616 0 CHAPTER 20

operation at 1800 rpm head,ft

flow rate, gpm

100 90 80 60 40 20

0 110 180 250 300 340

intersection of the curves. They are as follows: (a) single pump, 156 gpm; (6) two pumps in series, 224 gpm; (c) two pumps in parallel, 170 gpm. If Az had been greater than 100 ft, neither the single pump nor the two pumps in parallel would have delivered any water. If Az had been —20 ft (i.e., with the water-surface elevation in reservoir B 20 ft below that in A), the flows would have been (a) 212 gpm; (&) 258 gpm; and (c) 232 gpm.

20.29

The diameter of the discharge pipe of a pump is 8 in, and that of the intake pipe is 10 in. The pressure gage at discharge reads 32 psi, and the vacuum gage at intake reads 12 inHg. If Q = 4.0 ft3/s of water and the brake horsepower is 49.0, find the efficiency. The intake and the discharge are at the same elevation.

I

V, = Q/A, = 4.0/[(jr)(}§)2/4] = 7.334 ft/s V p = 4.0/[(jr)(£)2/4] = 11.46 ft/s vacuum pressure =—(12/29.9)(14.7) =—5.900 lb/ft2 H p =pjy + V 2 J2g + z d - (pjy + V 2 J2g + z s ) = (32)(144)/62.4 + 11.462/[(2)(32.2)] + 0 - {(—5.900)(144)/62.4 + 7.3342/[(2)(32.2)] + 0} = 88.67 ft P = QyH p = (4.0)(62.4)(88.67)/550 = 40.24 hp »/ = 40.24/49.0 = 0.821 or 82.1%

20.30

A centrifugal pump with an impeller diameter of 2.94 in delivers 260 gpm of water at a head of 820 ft with an efficiency of 62 percent at 21000 rpm. Compute the peripheral velocity u, specific speed N„ and peripheral-velocity factor

I u = (or = [(21000)(2;r)/60][(2.94/2)/12] = 269 ft/s N, = NQ 0S /H° 75 = (21 000)(260)° 7820°75 = 2210 u = <j>y/2gh 20.31

269 = ($) V(2)(32.2)(820)

= 1.171

Select the specific speed of the pump or pumps required to lift 5835 gpm of water 350 ft through 9400 ft of 2.5-ft-diameter pipe (/ = 0.022). The pump rotative speed is to be 1700 rpm. Consider the following cases: single pump, two pumps in series, two pumps in parallel.

PUMPS AND FANS Q 617

f

Q = (5835)(0.002228) = 13 ft3/s V = Q/A = 13/[(jr)(2.5)2/4] = 2.648 ft/s h f = (f)(L/D)(V 2 /2g) = (0.022)(9400/2.5) (2.6482/ [(2)(32.2)]) = 9.0 ft h p = 350 + 9.0 = 359.0 ft N s = NQ 05 / H 015

Single pump Two pumps in series Two pumps in parallel 20.32

gpm per pump

K per pump

Ns

5835 5835 2918

359.0 179.5 359.0

1575 2648 1113

The pump of Fig. 20-6« is used to lift water from one reservoir to another through a 9-in-diameter pipe (/ = 0.022), 692 ft long. The difference Az in surface elevation between the reservoirs fluctuates from 20 ft to 110 ft. Plot delivery rate Q versus pump head h p . Plot also the corresponding efficiencies. The pump is operated at a constant speed of 1450 rpm. Neglect minor losses.

f With Q in gpm, V = 0.002228Q/A = 0.002228Q/[(^)(f2)2/4] = 0.0050(2 h p = Az + (f )(LID)(V 2 /2g) = Az + (0.022)[692/(£)]{(0.0050Q)2/[(2)(32.2)]} = Az + (8.0 x lO ^Q2 Plot the pump characteristic curve and pipe system curves for a variety of Az’s. Ac, ft 20 40 60 80 100 110

gpm

2500 2320 2100 1860 1350 940

efficiency, %

70 76 81 83 77 66

See plotted curves in Fig. 20-66.

Fig. 20-6

618 D CHAPTER 20

Fig. 20-6 (continued)

(b) 20.33

Repeat Prob. 20.32 for the same pump operating at 1250 rpm. Assume efficiency pattern and values remain the same.

f The pump characteristic curve must be transformed using Q « N and h « N 2 . Thus point A in Fig. 20-66 transforms to A', where Q' = (2000)({fi) = 1724 gpm and h' p = (lOOKw,)2 = 74.3 ft, etc.

Az, ft

G< gpm

efficiency, %

20 40 60 80 82.3

2000 1750 1350 550 (shutoff)

73 82 84 58 0

The transformed pump characteristic curve is shown in Fig. 20-66. (The transformed efficiency curve is not shown in the figure.) 20.34

Repeat Prob. 20.32 for a homologous pump whose diameter is 21% smaller. Assume efficiency pattern and values remain the same.

f In this case the pump characteristic curve is transformed using Q « D3 and h <*■ D 2 . Thus point A in Fig. 20-66 transforms as follows: Q = (2000)(0.79)3 = 986 gpm, h = (100)(0.79)2 = 62.4 ft.

Az, ft

G> gpm

efficiency, %

20 40 60 76.8

1400 1200 800 (shutoff)

60 76 81 0

The transformed pump characteristic curve is shown in Fig. 20-7. 20.35

Repeat Prob. 20.32 if the pump diameter and rotation speed are reduced by 21% and 13.8%, respectively. Assume efficiency pattern and values remain the same.

f In this case the pump characteristic curve is transformed using Q « ND 2 and h <*- N 2 D 2 . Thus point A in Fig. 20-66 for the solution of Prob. 20.32 transforms as follows: Q = (2000)(0.862)(0.79)3 = 850 gpm, h = (100)(0.862)2(0.79)2 = 46.4 ft. Transforming the pump characteristic curve according to these rules and following

PUMPS AND FANS D 619

Fig. 20-7

Flowrate

the procedure as outlined in Prob. 20.34 gives

Az, ft

gpm

20 40 52.6

20.36

1080 700 Shutoff

efficiency, % 69 83 0

A pump with a critical value of a c of 0.10 is to pump against a head of 450 ft. The barometric pressure is 14.4 psia, and the vapor pressure of the water is 0.6 psia. Assume the friction losses in the intake piping are 6 ft. Find the maximum allowable height of the pump relative to the water surface at intake. f

(z,)max = (PoWy -Pvly- o c h-h L pjy = (14.4)(144)/62.4 = 33.2 ft pjy = (0.6)(144)/62.4 = 1.385 ft (zx)max = 33.2 - 1.385 - (0.10)(450) - 6 = -19.2 ft (submerged)

20.37

A pump draws water at 212 °F from a boiler with a friction loss of 3 ft in the intake pipe. Barometric pressure is 29 in of mercury, a c for the pump is 0.10, and the total pumping head is 300 ft. Locate the pump intake relative to the water surface in the boiler.

f

y Pv/y - o c h -h L p 0

(2s)max = (Po)abs/

= (i)(847.3/144)

=

14.2 psia p„/y = (14.2)(144)/59.8 = 34.2 ft

pjy = (14.7)(144)/59.8 = 35.4 ft (zs)max = 34.2 - 35.4 - (0.10)(300) - 3 = -34.2 ft Thus the water surface must be 34.2 ft above the pump suction. 20.38

Water is to be pumped at a head of 142 ft, the water temperature being 100 °F and the barometric pressure being 14.3 psia. At intake, the pressure is a vacuum of 15 inHg and the velocity is 13 fps. What are the values of NPSH and a? f NPSH = ( p X J y + V V l g - p J y ( p X u s = p J y + P s l y = (14.3)(144)/62.0 + (—{i)(847.3/62.0) = 16.13 ft

V 2 J2g = 132/[(2)(32.2)] = 2.62 ft p j y = 135/62.0 = 2.18 ft NPSH = 16.13 + 2.62 - 2.18 = 16.57 ft a = 16.57/142 = 0.117 20.39

A pump discharges 7700 gpm of water at 140 °F at a head of 250 ft, the barometric pressure being 13.8 psia. Determine the reading on a pressure gage at the suction flange when cavitation is just starting. Assume the suction pipe diameter equals 2.1 ft and neglect the effects of prerotation. Take a c = 0.085. f a = [(p,)abs/y + V 2 J2g -p v /y]/h. Let p = gage pressure at suction flange. (p s ) abs = p Mm + P = 13.8 + p V S

= QIA S = (7700)(0.002228)/[(JT)(2.

l)2/4] = 4.953 ft/s

pjy = 416/61.4 = 6.775 ft 0.085 = {(13.8 + p)(144)/61.4 + 4.953 2/[(2)(32.2)] - 6.775J/250 p — —2.013 psia = +2.013 psia vacuum

620 0 CHAPTER 20 20.40

If the maximum efficiency of the pump of Prob. 20.32 is 81%, approximately what would be the efficiency of the pump of Prob. 20.34?

I (1 - 7j,)/(l - rj2) = (D2/D,)1/5 20.41

(1 - 0.81)/(1 - rj2) = (0.79/1.0)1'5 JJ2 = 0.801 or 80.1%

Figure 20-8 shows the dimensions and angles of the diff vanes of a centrifugal pump. The vane ages are 0.75 in wide perpendicular to the plane of the figure. If the impeller delivers water at the rate of 3 cfs under ideal and frictionless conditions, what is the rise in pressure through the diff?

I

V 2 J2g = V\l2g + Ap/y V = Q/A A, = (2jr)(^)(sin 15°)(0.75/12) = 0.05505 ft 2 V, = 3/0.05505 = 54.50 ft/s

A 2 = (2jr)(9/12)(sin 28°)(0.75/12) = 0.1382 ft2

V 2 = 3/0.1382 = 21.71 ft/s 54.502/[(2)(32.2)] = 21.712/[(2)(32.2)] + Aply Ap/y = 38.80 ft

20.42

Ap = (62.4)(38.80)/144 = 16.8 lb/in 2

Two pumps whose characteristics are given in Prob. 20.28 are to be used in parallel. They must develop a head h = 35 + 20Q 2 as in Prob. 20.28. One pump is to be operated at 1800 rpm. The speed of the other pump is to be gradually reduced until it no longer delivers water. At approximately what speed will this happen?

I From Prob. 20.28, with a single pump operating at 1800 rpm, the flow rate is 156 gpm and the head is 83.8 ft. If we add a pump in parallel to this pump with a shutoff head of 83.8 ft, the second pump will convey no flow. Thus, h / h ' = {nln'f, 83.8/100 = (n/1800)2, n = 1648rpm. 20.43

A pump is installed to deliver water from a reservoir of surface elevation zero to another of elevation 290 ft. The 10-in-diameter suction pipe (/ = 0.022) is 90 ft long, and the 8-in-diameter discharge pipe (/ = 0.028) is 4900 ft long. The pump characteristic at 1100 rpm is defined by h p = 365 — 20Q2 where h p , the pump head, is in feet and Q is in cubic feet per second. Compute the rate at which this pump will deliver water under these conditions assuming the setting is low enough to avoid cavitation.

I h„

= 365 - 20Q2. Also, since h f = ( f ) ( L / D ) ( y 2 / 2 g ) , h p = 290 + (0.022)[90/({§)] Vf0/2g + (0.028)(4900)(£)Fi/2g V = Q/A

V m = e/[(jr)G§)2/4] = 1.833G

V g = QI[(n){hflA] = 2.865Q

h p = 290 + (0.022)[90/G§)]{(1.833e)7[(2)(32.2)]}

+ (0.028)[4900/(£)]{(2.865G)7[(2)(32.2)]} = 290 + 26.35G 2 Equating the two expressions for h p gives 365 - 20Q 2 = 290 + 26.35Q2, Q = 1.27 ft’/s. 20.44

Repeat Prob. 20.43 for the case of two such pumps in parallel.

I From Prob. 20.43, h p per pump = 290 + 26.35Q2 = 365 - 20(Q/2)2, or Q = 1.55 fC/s.

PUMPS AND FANS 0 621 20.45

A centrifugal pump with a 14-in-diameter impeller is rated at 690 gpm against a head of 91 ft when rotating at 1800 rpm. What would be the rating of a pump of identical geometry with a 9-in impeller? Assume pump efficiencies and rotative speeds are identical.

I 20.46

h*D 2 h = (91)(^>)2 = 51.2 ft Q«D3 Q = (690)(^)3 = 291 gpm

The following are data for a centrifugal water pump: r, = 4 in, r 2 = 7 in, /3 = 30°, b\ = b = 1.75 in, estimate (a) design-point discharge; (6) water horsepower; (c) head.

= 20°, speed = 1440 rpm. If

2

f (a) The angular velocity is co = 2jr(Jgr) = 150.8 rad/s. Thus the tip speeds are u, = cur, = 150.8(^) = 50.3 ft/s, and u 2 = cur = 150.8(r2) = 88.0 ft/s. From the inlet-velocity diagram, Fig. 20-9a, with a , = 90° for design point, we compute V„, = w, tan 30° = 29.Oft/s, whence the discharge is Q = 2Jirtb, F„, = 2;r(i|)(1.75/12)(29.0) = 8.86 ft3/s = 3980 gal/min. (6) The outlet radial velocity follows from Q: V n2 — Ql(2itr 2 b 2 ) = 8.86/[2jr(^)(1.75/12)| = 16.6 ft/s. This enables us to construct the outlet-velocity diagram as in Fig. 20-9b, given /32 = 20°. The tangential component is V,2 = u 2 — V„2cot jS2 = 88.0 — 16.6 cot 20° = 42.4 ft/s, <x 2 = tan- '(16.6/42.4) = 21.4°. The power is then computed with V tt = 0 at the design point: P w = pQu 2 V l2 — [(1.94)(8.8)(88.0)(42.4)]/550 = 117 hp. (c) The head is estimated as H = PJpgQ = (117)(550)/[(62.4)(8.86)] = 116 ft. 2

(a)

20.47

(b)

Fig. 20-9

The 32-in pump of Fig. 20-10 is to pump 24 000 gal/min of water at 1170 rpm from a reservoir whose surface is at 14.7 lb/in2 abs. If head loss from reservoir to pump inlet is 6 ft, where should the pump inlet be placed to avoid cavitation for water at ( a ) 60 °F, p v = 0.26 lb/in2 abs, s.g. = 1.0 and (b ) 200 °F, p v = 11.52 lb/in2 abs, s.g. = 0.9635?

f For either case read from Fig. 20-10 at 24 000 gal/min that the required NPSH is 40 ft.

(a) For this case p g = 62.4 lb/ft3. NPSH ≤ [(p a — p v ) / p g ] - Z, — h f i or 40 ft ≤ [(14.7 — 0.26)(144)/62.4] -

Z, — 6.0 or Z, ≤ 27.3 - 40 = -12.7 ft. The pump must be placed at least 12.7 ft below the reservoir surface to avoid cavitation.

(From Ingersoll-Rand Corporation, Cameron Pump Division)

Fig. 20-10

622 D CHAPTER 20 (b) For this case pg = 62.4(0.9635) = 60.1 lb/ft3. 40 ft ≤ [(14.7 -11.52)(144)/60.1] -Z,-6.0 or Z, ≤ 1.6 40 = 38.4 ft. The pump must now be placed at least 38.4ft below the reservoir surface. These are unusually stringent conditions because a large, high-discharge pump requires a large NPSH. 20.48

A pump from the family of Fig. 20-11 has D = 21 in and n = 1500 rpm. Estimate ( a ) discharge, (6) head, (c) pressure rise, and (d) brake horsepower of this pump for water at 60 °F and best efficiency. f ( a ) In BG units take D = % = 1.75 ft and n = ^ = 25 r/s. At 60 °F, p of water is 1.94 slugs/ft3. The BEP parameters are known from Fig. 20-11. The BEP discharge is thus Q* = C Q .nD 3 = 0.115(25)(1.75)3 = 15.4 ft3/s = 6920 gal/min. (6) Similarly, the BEP head is H* = C H .n 2 D 2 !g = 5.0(25)2(1.75)z/32.2 = 297 ft water. (c) Since we are not given elevation or velocity-head changes across the pump, we neglect them and estimate Ap = pgH = 1.94(32.2)(297) = 18 600 lb/ft2 = 129 lb/in2. (d) Finally, the BEP power is P* = C„.pn 3 D 5 = 0.65(1.94)(25)3(1.75)5/550 = 588 hp.

20.49

We want to build a pump from the family of Fig. 20-11, which delivers 3000 gal/min water at 1200 rpm at best efficiency. Estimate (a) the impeller diameter, (b) the maximum discharge, (c) the shutoff head, and (d) the NPSH at best efficiency.

f (a) 3000 gal/min = 6.68 ft3/s and 1200 rpm = 20 r/s. At BEP we have Q* = Ce.nD3 = 6.68 ft3/s = (0.115)(20)£>3 or D = [6.68/0.115(20)]1/3 = 1.43 ft = 17.1 in. (b) The max Q is related to Q* by a ratio of capacity coefficients: Q max = Q*C Q . max /C Q. = 3000(0.23)/0.115 = 6000 gal/min. (c) From Fig. 20-11 we estimated the shutoff-head coefficient to be 6.0. Thus H(0) ~ C H (0)n 2 D 2 /g = 6.0(20)2(1.43)2/32.2 = 152 ft. (d) Finally, the NPSH at BEP is approximately NPSH* = C HS .n 2 D 2 /g = 0.37(20)2(1.43)2/32.2 = 9.4 ft. Since this is a small pump, it will be less efficient than the pumps in Fig. 20-11, probably about 82 percent maximum. 20.50

We want to use a centrifugal pump from the family of Fig. 20-11 to deliver 100 000 gal/min of water at 60 °F with a head of 25 ft. What should be (a) the pump size and speed and (b) brake horsepower assuming operation at best efficiency?

f (a) Enter the known head and discharge into the BEP parameters: H* = 25 ft = C„.n 2 D 2 /g = 5.0n2D2/32.2, Q* = 100 000 gal/min = 222.8 ft3/s = C Q .nD 3 = 0.115n£>3. The two unknowns are n and D. Solve simultaneously for D = 12.4 ft, n = 1.03 r/s = 62 rpm.

PUMPS AND FANS D 623 (ft) The most efficient horsepower is, then, bhp* = C p .pn 3 D s = 0.65(1.94)(1.03)3(12.4)5/550 = 734 hp. The solution to this problem is mathematically correct but results in a grotesque pump: an impeller more than 12 ft in diameter, rotating so slowly one can visualize oxen walking in a circle turning the shaft. 20.51

We want to use the 32-in pump of Fig. 20-10 at 1170 rpm to pump water at 60 °F from one reservoir to another 120 ft higher through 1500 ft of 16-in-ID pipe with friction factor / = 0.030. (a) What will the operating point and efficiency be? (ft) To what speed should the pump be changed to operate the BEP?

f (a) For reservoirs the initial and final velocities are zero; thus the system head is H s = z 2 — z x + (V 2 /2g)(fL/D) = 120 + (V2/2g)[0.030(1500)/ff]. From continuity in the pipe, V = Q/A = Ql[\n(If)2], and so we substitute for V above to get H s = 120 + 0.269Q2 Ginft3/s

(1)

Since Fig. 20-10 uses thousands of gallons per minute for the abscissa, we convert Q in Eq. (1) to this Umt

'

H s = 120+ 1.335<22 Q in 103 gal/min

(2)

We can plot Eq. (2) on Fig. 20-10 and see where it intersects the 32-in pump-head curve, as in Fig. 20-12. A graphical solution gives approximately H ~ 430 ft, Q ~ 15 000 gal/min. The efficiency is about 82 percent, slightly off design. An analytic solution is possible if we fit the pump-head curve to a parabola, which is very accurate: tfpump ~ 490 - 0.26Q 2 Q in 103 gal/min 2

(3) 2

Equations (2) and (3) must match at the operating point: 490 - 0.26Q = 120 + 1.335Q or Q 2 = (490 - 120)/(0.26 -I-1.335) = 232, Q = 15.2 x 103 gal/min = 15 200 gal/min; H = 490 0.26(15.2)2 = 430 ft. (ft) To move the operating point to BEP, we change n, which changes both Q « n and H « n 2 . From Fig. 20-10, at BEP, H* ~ 386 ft; thus for any n, H* = 386(n/1170)2. Also read Q* = 20 x 103 gal/min; thus for any n, Q* = 20(n/1170). Match H* to the system characteristics, Eq. (2), H* = 386(n/1170)2 ~ 120 + 1.335[20(n/1170)]2, which gives n2< 0. Thus it is impossible to operate at maximum efficiency with this particular system and pump. H

Fig. 20-12

20.52

Investigate extending Prob. 20.51 by using two 32-in pumps in parallel to deliver more flow. Is this efficient?

f Since the pumps are identical, each delivers \Q at the same 1170 rpm speed. The system curve is the same, and the balance-of-head relation becomes H = 490 - 0.26(jQ)2 = 120 + 1.335Q2 or <22 = (490 - 120)/(1.335 + 0.065), Q = 16 300 gal/min. This is only 7 percent more than a single pump. Each pump delivers \Q = 8150 gal/min, for which the efficiency is only 60 percent. The total brake horsepower required is 3200, whereas a single pump used only 2000 bhp. 20.53

Suppose the elevation change in Prob. 20.51 is raised from 120 to 500 ft, greater than a single 32-in pump can supply. Investigate using 32-in pumps in series at 1170 rpm.

f Since the pumps are identical, the total head is twice as much and the constant 120 in the system-head curve

is replaced by 500. The balance of heads becomes H = 2(490 — 0.26Q 2 ) = 500 + 1.335g2 or Q 2 = (980 — 500)/(1.335 + 0.52), Q = 16.1 x 103 gal/min. The operating head is 500+ 1.335(16.1)2 = 846 ft, or 97 percent more than a single pump in Prob. 20.50. Each pump is operating at 16.1 x 10 3 gal/min, which from Fig. 20-10 is 83 percent efficient, a pretty good match to the system. To pump at this operating point requires 4100 bhp, or about 2050 bhp for each pump.

624

a CHAPTER 20

20.54

A piston positive-displacement pump (PDP) has a 6-in diameter and a 2.5-in stroke. Its crankshaft rotates at 300 rpm. Calculate its output at 100 percent volumetric efficiency. ^

20.55

2

P = QyE p = (12.27/60)(62.4)(24) = 306.3 ft-lb/s Prequlred = (306.3/550)/0.72 = 0.77 hp

A pump delivers 1.6 m3/min of water at 20 °C against a pressure rise of 287 kPa. Kinetic- and potential-energy changes are negligible. If the driving motor supplies 10 kW, what is the overall efficiency?

I 20.57

= [(^XTI) /4](T!) = 0.04091 ft3 Q = (0.04091)(300) = 12.27 ft3/min

If the PDP of Prob. 20.54 delivers water against a total head of 24 ft, what horsepower is required at 72% efficiency?

I 20.56

^displaced

p = QyE p = ApQ = (287)(1.6/60) = 7.65 kN- m/s = 7.65 kW r)= 7.65/10 = 0.765

A 26-hp pump delivers 475 gpm of gasoline (y = 42.5 lb/ft3) at 20°C with 78% efficiency. What head and pressure rise result across the pump? #

Required

(26)(550) = [(475)(0.002228)](42.5)(£„)/0.78

= Gy£p/»?

£„=248 ft

Ap = yE p = (42.5)(248) = 10 540 lb/ft2 or 73.2 lb/in2 20.58

A sump pump turning at 1600 rpm provides a discharge of 9 gpm against a head of 18 ft. Compute the minimum power input. / JV, = NQ° VH° = (1600)(9)° 5/180 75 = 549. From Fig. A-41, r, max = 0.32. Pmin = QyH/ Vmax = [(9)(0.002228)](62.4)(18)/0.32 = 70.4 ft-lb/s, or 0.128 hp. 75

20.59

A model pump delivering water at 180 °F at 900 gpm and 2500 rpm begins to cavitate when the inlet pressure and velocity are 13 psia and 22 fps. Find the required NPSH of a prototype which is 4 times larger and runs at 1100 rpm.

I

NPSHmode, = [(13)(144) - 1086]/60.6 + 222/[(2)(32.2)] = 20.49 ft

NPSH = (p, -p v )/y + Vf/2g

NPSHproto = (NPSH model )(N p /N m ) 2 (Dp/D m ) 2 = (20.49)(M$)2(4)2 = 63.5 ft. 20.60

Determine the specific speed of the pump whose operating characteristics are shown in Fig. 20-13. If this pump were operated at 1200 rpm, what head and discharge would be developed at rated capacity, and what power would be required? f N,= NQ 0 5 IH° 15 . At its rated capacity of 700 gpm, this pump develops 120 ft of head when operating at 1450 rpm. Thus N s = (1450)(700)° 5/120°75 = 1058. If the efficiency remains constant with the speed change, at 1200 rpm, Q = (jfi)(700) = 579 gpm, H = (jfi)2(120) = 82.2 ft, P = (|ti)3(26) = 14.7 hp. 20 0 18 0 16 0 14 0 12 0 10 0 0 01

200

400

600

Capacity, gpm

800

1000

Fig. 20-13

X

20.61

80

A mixed-flow centrifugal pump (N s = 400, a c = 0.30) is to develop 50 ft of head. Find the maximum permissible suction lift on the pump if the pump60is at sea level and the water is at 80 °F. 40 20

PUMPS AND FANS 0 625

I Neglecting head loss between reservoir and pump, z„ ~ (patm/y - p„/y) - NPSH, NPSH = o c h = (0.30)(50) = 15.0 ft, z, = [(14.7)(144)/62.4 - 73.5/62.4] - 15.0 = 17.7 ft. 20.62

A centrifugal pump (efficiency 79%) driven by an electric motor (efficiency 88%) lifts water through a total height of 135 ft from reservoir to discharge. Pumping is through 977 ft of 3-in-diameter pipe at the rate of 290 gpm. If / = 0.025 and electricity costs $7 per megawatt-hour, find the cost of pumping one million gallons. f

Q = (290) (0.002228) = 0.6461 ft3/s V = Q/A = 0.6461/[(x)(^)2/4] = 13.16 ft/s hf = (f)(L/D)(V 2 /2g) = (0.025)[977/(n)] {13.162/[(2)(32.2)]} = 262.7 ft E„ = 135 + 262.7 = 397.7 ft P„2o = QyE„ = (0.6461)(62.4)(397.7) = 16 034 ft-lb/s = 21.74 kW P mo,o, = 21.74/[(0.79)(0.88)] = 31.27 kW = 0.03127 MW Cos, 03127

20.63

$12.59

A model of a mixed-flow water pump has been tested in the laboratory to give performance curves as is shown in Fig. 20-14. What would be the total head AH delivered by a prototype pump with an impeller size of 1.2 m operating at a speed of 1750 rpm and delivering 1.300 m3/s flow? What is its mechanical efficiency? Consider that dynamic similarity can be achieved between model flow and prototype flow.

I We first compute the flow coefficient for the prototype flow: (Q/ND\ = 1.300/{[(1750)(2JT)/60](1.23)} = 0.00411. Examining Fig. 20-14 we see that the efficiency for this operating point is about 75 percent for the prototype. Also note that the head coefficient is about 0.17. Therefore we have for the head AH of the prototype (g AH)/N 2 D 2 = 0.17, AH = (0. 17/9.81)[1750(2JT/60)]2(1.22) = 838 m. Head coefficient

Efficiency

Fig. 20-14 20.64 A centrifugal blower compresses 1.4 lbm/s of air at a temperature of 60 °F and a pressure of 13.00 psia to a pressure of 14.50 psia. The impeller runs at 3500 rpm. The exit area of the blower is 0.14 ft 2 and the inlet area is 0.4 ft2. How much power is needed to run the blower at 80% efficiency? I

( d W p / d t ) t h e o = {[(VyLo - V]]I2 + c,7i[(p2/pl)<*-,>'* - l]}m

Evaluate the two velocities: p, = p,/?T, p, = (13)(144)/[(53.3)(g)(520)] = 0.002098 slug/ft 3

p,/p2 = (pjp 2 ) vt

p2 = (0.002098)(14.5/13)m 4 = 0.002268 slug/ft3 m = (1.4/g) = p 2 (V 2 ) t hco A 2

1.4 lg = (0.002268)(V2)theo(0.14) (V2)theo = 136.93 ft/s

Also rh = 1.4/g = pi V,A, = (0.002098)(V,)(0.4), V, = 51.81 ft/s. Consequently, ( d W p / d t \ h e o = {[(136.932 - 51.812)/2] + (0.24)(778)(g)(520)[(14.5/13)° 4,14 - l]}(1.4/g)/550 = 8.47 hp (id W p / d t )act = 8.47/0.80 = 10.59 hp

626

a CHAPTER 20

20.65

In Prob. 20.64, if the impeller has an outside radius of 9 in and is 1.6 in wide, what should the blade angle )S 2 be at exit? I

(dW p /dt) (8.47)(550)

=

theo

= (
[(3500)(27r)/(60)](;’

)(1.4/g){[(3500)(2^)/60](^)

+

(V r ) 2

cot

&} (1

) Writing the continuity equation as m = p 2 (V r ) 2 (2jir 2 )(b) and taking p2 from Prob. 20.64, we find 1.4/g = (0.002268)(K)2(2jr)(ft)(1.6/12)

(V r ) 2 = 30.51 ft/s

Now Eq. (i) is solved to give )S2 = 14.86°. 20.66

An air compressor has a mass flow of 0.7 kg/s, taking the fluid from a pressure of 100 kPa at the inlet to 101.67 kPa at the outlet. The inlet temperature is 15 °C. The exit angle fi 2 is 160 °. The inlet area is 8000 mm2 and the exit area is 7000 mm2. Determine the speed of the compressor, if the diameter of the rotor is 175 mm, the width of the impeller is 80 mm, and the compressor efficiency is rj = 0.80. Base your calculation on actual, not theoretical, torque.

f

dW p /dt = {[(V2)LO - Vi]/2 + c p n(p 2lp s r

t)lk

- \Wmln

From the continuity equation, p2(Vr2)theoA2 = 0.7 kg/s, V2 = 0.7/[(1.2244)(7000 x 10-6)] = 81.7 m/s; hence, dW p /dt = {[(81.72 - 71.52)/2] + (1003)(288)[(101.67/100)O4/14 - l]}(0.7)/0.80 = 1883 N • m/s Now, dW p ldt = T s m = r 2m[(U ,)2 + (V r ) 2 cot /32]
=

(0.175/2)(0.7)[(w)(0.175/2)

+

(V r ) 2

cot

160°]« (1

) By the continuity equation, (1.2244)(Vrr2)[(^:)(0.175)](0.08) = 0.7, or (V r ) 2 = 13.00 m/s. Then (i) gives: (o 2 - 408.2« - 3.513 x 105 = 0, « = [408.2 ± V408.22 + (4)(3.518 x 105)(1)]/(2)(1) = 831 rad/s = 7935 rpm.

20.67

A performance chart is shown in Fig. 20-15 for a 3000-rpm blower from a bulletin of the Buffalo Forge Co. For a 21-in wheel having a flow of 3000 ft3/min, what is the efficiency and the required torque? The exit diameter is 8 in and the entrance diameter is 16 in. At the entrance p = 14.7 lb/in2 and T = 70 °F. The capacity is given at l(Ap) = 24.4apply inH2to 0 =the (24.4)(62.4)/12 = 126.9 psf curves apply to the right s.a.ic the exit pressure. Note: The upper curves left ordinate and the lower ordinate. p 2 = (14.7)(144) + 126.9 = 2244 psf w = p 2Q = [(0.002378)(§f$),/*](aiQ) = 0.1228 slug/s f From the chart: v = Q/A 2 = (^a)/[x(^)2/4] = 143.2 ft/s

r/ip=2i

2

Now get V,. p, ViAi = m = 0.1228 slug/s (0.002378)(Vt)(;r)Gi)2/4 = 0.1228 V, = 36.98 ft/s dW p /dm = [(F2/2) - (V?/2)] + c P U (p 2 / p - 1] dW p /dt = (dW p /dm)m = (0.1228){[(143.22/2) - (36.982/2)] + (0.24)(778)(g)(460 + 70)[(§|f)(1 4“1)/l 4 - l]}/550 = 11.49 hp e = [(21 - 11.49)/21](100) = 45.3% T = (11.49)(550)/[3000(2JT/60)] =20.12 ft-lb

20.68 A centrifugal pump in a horizontal pipe drives 213 gal/min of water from an entrance pressure p, = 13 psia. Pump data are: inside radius of impeller is 3 in; outside radius of impeller is 0.8 ft; pump outputs 25 hp running at 1650 rpm; pump efficiency is 80%; width b of the vanes is 2 in. Determine the exit pressure p 2 . Take p = 1.938 slug/ft3 for water.

40

30

20

10

1000

500

1500

2000

2500

3000

3500

4000

4500

3

Capacity ft of air per minute

Fig. 20-15 (Courtesy Buffalo Forge Company. Reprinted with permission.)

f First, find the blade angle from T s =

(theoretical power)/<w = r 2 rh[(U,) 2

+ (V r ) 2 cot j32]:

(25/0. 80)(550)/[(1650)(2JT)/60]} = (0.8)[(213)(231)/(172S)(60)](1.938)([(1650)(2jr)/60](0.8) + {[(213)(231)/(1728)(60)]/[(2JT)(0.8)(A)]} cot /S2) f} 2 = 169.4° Next compute the exit velocity: (F2)rel sin (jr - ft) = (V r ) 2 (V2)rel = {[(213)(231 )/(1728)(60)]/[2^r(0.8)(ft)]}/[sin (180° - 169.4°)] = 3.079 ft/s and, from the law of cosines, V 2 = (U,)l + (V rcl )l - 2(t/,)(Vrei)2 cos (Jt - ft) 2

= {[(1650)(2nr)/60](0.8)}2 + 3.0792 - (2){[(1650)(2jr)/60](0.8)}(3.079 cos 10.6°) = 18 280 ft 2/s2 Since the entrance velocity is V, = [(213)(231)/(1728)(60)]/[(2?r)(ft)(ft)] = 1.813 ft/s, we can now write 18 280 — 1.813‘ Ap ' theoretical power = (^)(550) = ^ = (W)(Mi)(l-938) + 2 1.938.

j

Solving, Ap = 18 507 psf = 129 psi and p 2 = 13 + 129 = 142 psi. 20.69

A centrifugal pump moves 9.0 L/s of water from a pressure of 90 kPa to a pressure of 520 kPa. The section inlet has a diameter of 50 mm and the discharge has a diameter of 40 mm. What power is required if there is an efficiency of 65 percent and the discharge pipe is 0.3 m above the inlet pipe?

Horsepowe r

Static pressure, inches of water

PUMPS AND FANS D 627

628 0 CHAPTER 20

f (dW p /dt) t hco = m({[(Ff),he„ - V?]/2} + (p 2 - Pl )/p +

g(z 2 - z,)) m = (0.0090)(1000) = 9 kg/s

K, = 2M, = 0. 0090/[(JT)(0.050)2/4] = 4.584 m/s V2 = GM2 = 0.0090/[(^)(0.040)2/4] = 7.162 m/s t/Wp/df = (9){[(7.1622 - 4.5842)/2] + [(520 - 90)/1000] x 103 + 9.81(0.3)} = 4033 W P = 4033/0.65 = 6.20 kW 20.70

Shown in Fig. 20-16 are performance curves as found in a Buffalo Forge Co. bulletin for multistage centrifugal pumps. The number designations 2 x 3, 4 x 4, etc., give the exit and the inlet diameters, respectively. For a 2x3 (four-stage) pump operating at 25 hp with a flow Q of 80 gal/min of water, compute the efficiency. What is the total required torque on the four rotors?

I A// (output) = 540 ft A//(input) = [(hp)(550)/Qy] = (25)(550)/{[(80)(231)/(1728)(60)]62.4} = 1236 ft

Total head, ft

V = tsk(lOO) = 44 percent Torque = (25)(550)/[(3500)(2JT)/60] = 37.5 ft-lb

25 30

40

50 60 70 80 90100

150

200

250 300

400 500 600

800 1000 1250

U.S. gallons per minute

Fig. 20-16 (Courtesy Buffalo Forge Company. Reprinted with permission.)

20.71

Using Fig. 20-16, select a centrifugal pump to move 120 gal/min of water from tank A to tank B in Fig. 20-17. The inlet and outlet pipes are steel. Neglect head loss for the short inlet piping. Water is at 60 °F.

I Use the first law for pump and 2-in outlet pipe. (1 /g)(dW p /dm) + (V?/2g) + (Pi/y) + 0 = (Vf/2g) + (p 2/y) + 390 + [{h L )lg]

(1)

PUMPS AND FANS D 629 For p, (gage pressure) apply Bernoulli’s equation between free surface of tank A and pump inlet: (V%/2g) + ( P B J Y ) + 1 = (V?/2g) + ( P J Y ) 2

V, = [(120)(231)/(1728)(60)]/[(?r)(3 )/(4)(144)] = 5.447 ft/s

(2) 2

p,/y = l- (5.447 /2g) = 0.539 ft

V 2 = [(120)(231)/(1728)(60)]/[(^)(22)/(4)(144)] = 12.25 ft/s N R = (12.25)(£)/(1.217 x 10“5) = 1.398 x 105 e/D = 0.00015/(n) = 0.0009

/ = 0.021 h L =/(L/d)(V2/2) = (0.021)(390/|)(12.252/2) = 3687 ft

Now substitute in Eq. (1): (Ug)(dW p /dm) + (5.4472/2g) + 0.539 = (12.252/2g) + [(y)(20)/y] + 390 + (3687/g) = 527 ft Use 2 X 3(four-stage) 30-hp pump.

----------- -

2 Oft B 2

D = 2 in

Fig. 20-17

20.72

A pump impeller, 12 in in diameter, discharges 5.25 cfs when running at 1200 rpm. The blade angle /3 2 is 160° and the exit area A 2 is 0.25 ft2. Assuming losses of 2.8(u2/2g) and 0.38(Vf/2g), compute the efficiency of the pump (exit area A 2 is measured normal to v 2 ). f The absolute and relative velocities at exit must be calculated first. Velocities u 2 and v 2 are u 2 = r 2 w = (fcX^XW) = 62.8 ft/s,- v 2 = Q/A 2 = 5.25/0.25 = 21.0ft/s. From the vector diagram shown in Fig. 20-18, the value of the absolute velocity at exit is V 2 = 43.7 ft/s. Head furnished by impeller, H' = («|/2g) - (vj/lg) + (Vj/2g) = [(62.8)2/2g] - [(21.0)2/2g] + [(43.7)2/2g] = 84.0 ft. Head delivered to water, H = H' - losses = 84.0 - (2.8[(21.0)2/2g] + 0.38[(43.7)2/2g]} = 53.6 ft. Efficiency e = H/H' = 53.6/84.0 = 63.8 percent. The value of H' might have been calculated by means of the commonly used expression H' = {u 2 /g)(u 2 + v 2 cos /32) = (62.8/g)[62.8 + 21.0(—0.940)] = 84.0 ft.

Fig. 20-18

20.73

A 6-in pump delivers 1300 gpm against a head of 90 ft when rotating at 1750 rpm. The head-discharge and efficiency curves are shown in Fig. 20-19. For a geometrically similar 8-in pump running at 1450 rpm and delivering 1800 gpm, determine (a) the probable head developed by the 8-in pump. (6) Assuming a similar efficiency curve for the 8-in pump, what power would be required to drive the pump at the 1800-gpm rate? f (a) The homologous pumps will have identical characteristics at corresponding flows. Choose several rates of flow for the 6-in pump and read off the corresponding heads. Calculate the values of Q and H so that

630 0 CHAPTER 20 a curve for the 8-in pump can be plotted. One such calculation is detailed below and a table of values established by similar determinations. Using the given 1300 gpm and the 90-ft head, we obtain from the speed relation H x = (D 8 /D 6 )\N S /N 6 ) 2 H 6 = (l)2(^i)2H6 = 1.22He = 1.22(90) = 109.8 ft. From the flow relation, Q/D 3 N = constant, we obtain Q s = (D s /D 6 ) 3 (NJN 6 )Q 6 = (l)3(jli)Q6 = 1.964Q6 = 1.964(1300) = 2550 gpm. Additional values, which have been plotted as the dashed line in Fig. 20-19, are as follows:

For 6-in pump at 1750 rpm

For 8-in pump at 1450 rpm

Q, gpm

H, ft

Efficiency, %

Q, gpm

H,tt

Efficiency, %

0 500 800 1000 1300 1600

124 119 112 104 90 66

0 54 64 68 70 67

0 980 1570 1960 2550 3140

151.6 145.5 134.5 127.0 110.0 80.6

0 54 64 68 70 67

From the head-discharge curve, for Q = 1800 gpm the head is 130 ft. (6) The efficiency of the 8-in pump would probably be somewhat higher than that of the 6-in pump at comparable rates of flow. For this case, the assumption is that the efficiency curves are the same at comparable rates of flow. The table above lists the values for the flows indicated. The figure gives the efficiency curve for the 8-in pump and, for the 1800 gpm flow, the value is 67 percent. Then P = Y QH/550e = 62.4[1800/(60)(7.48)](130)/[550(0.67)] = 88.3 horsepower required. 180

160

140

120

C 100

Fig. 20-19

20.74

In order to predict the behavior of a small oil pump, tests are to be made on a model using air. The oil pump is to be driven by a jg-hp motor at 1800 rpm and a ?-hp motor is available to drive the air pump at 600 rpm. Using the s.g. of oil at 0.912 and the density of air constant at 0.00238 slug/ft 3, what size model should be built? f Using the power relation, we obtain prototype [P/(pD 5 N 3 )] = model[P/(pZ)sA3)]. Then ^/[0.912(62.4/32.2)Dp(1800)3] = i/[0.00238D5m(600)3] and DJD P = f. The model should be 10 times as large as the oil pump.

PUMPS AND FANS D 631 20.75

What is the power ratio of a pump and its \ scale model if the ratio of the heads is 4 to 1? f For geometrically similar pumps, P/(D 2 H 312 ) for pump = P/(D 2 H 3 ' 2 ) for model. Then P p /[(5D) 2 (4H) 3 ' 2 ] = PJ(D 2 H 3 ' 2 ) and P p = 25(4)3,2Pm = 200Pm.

20.76

Determine the thrust coefficient of a propeller which is 4 in in diameter, revolves at 1800 rpm, and develops a thrust of 2.50 lb in fresh water.

I

Thrust coefficient = F/(pN 2 D 4 ) = 2.50/[1.94(1®s)2(s)4] = 0.116. The coefficient is dimensionless when Fis in pounds, Win revolutions/per second, and D in feet. 20.77

The power and thrust coefficients of an 8-ft-diameter propeller moving forward at 100 ft/s at a rotational speed of 2400 rpm are 0.068 and 0.095, respectively, (a) Determine the power requirement and thrust in air (p = 0.00237 slug/ft3). (b ) If the advance-diameter ratio for maximum efficiency is 0.70, what is the air speed for the maximum efficiency?

I (a)

Power P = C P pN 3 D 5 in ft-lb/s = [0.068(0.00237)(2$a)3(8)5]/550 = 615 horsepower. Thrust F = C F pN 2 D 4 in lb = 0.095(0.00237)(2$e)2(8)4 = 1476 lb.

(b) 20.78

Since V/ND = 0.70, V = 0.70(^2)(8) = 224 ft/s.

An airplane flies at 180 mph in still air, y = 0.0750 lb/ft3. The propeller is 5.5 ft in diameter and the velocity of the air through the propeller is 320 ft/s. Determine ( a ) the slipstream velocity, (b ) the thrust, (c) the horsepower input, (d) the horsepower output, (e) the efficiency, and (/) the pressure difference across the propeller.

I (a) (6)

V = h(V, + V4)

320 = |[180(H) + V4]

V4 = 376 ft/s (relative to fuselage)

Thrust F = (y/g)G(V4 - V t ) = (0.0750/32.2)[>(5.5)2(320)](376 - 264) = 1983 lb

(c)

Power input P = FV/550 = 1983(320)/550 = 1154 hp

{d)

Power output P 0 = FVJ550 = 1983(264)/550 = 952 hp

(e) Efficiency e = = 82.5 percent or e = 2VJ(V 4 + Vi) = 2(264)/(376 + 264) = 82.5 percent. (/) Pressure difference = (thrust F)/[area (\nD 2 )\ = 1983/[?x(5.5)2] = 83.5 psf or pressure difference = (0.0750/32.2) {[(376)2 — (264)2]/2) = 83.5 psf. 20.79

Water is being pumped from the lower reservoir to the upper reservoir, as shown in Fig. 20-20. Find the pump horsepower required, assuming 75 percent pump efficiency. Neglect minor losses. f pjy + V\/2g + Zj + E p = pzly + Vi/2g + z 2 + h L , From Fig. A-13, with Q = 23 cfs and D = 24 in, hi = 0.0076 ft/ft. h L = h f = (0.0076)(3600) = 27.36 ft 0 + 0 + 97.5 + E p = 0 + 0 + 132.0 + 27.36 E p = 61.86 ft P = QyE p /550 = (23)(62.4),(61.86)/550 = 161 hp

Fig. 20-20

632

a CHAPTER 20

20.80

Water is being pumped from a reservoir to the top of a hill, where it is discharged, as shown in Fig. 20-21. The pump, which is 70 percent efficient, is rated at 150 kW. Find the flow rate at which water is being discharged from the pipe. Neglect minor losses.

I p j y + V?/2g + Zi + E p = p 2 /y + Vl/2 g + z2 + hL

E p = P/Qy = (15O)(O.70)/[(0)(9.79)] = 10.725/0

V = Q/A F2 = 0/[(*)(O.5OO)2/4] = 5.0930 h L = h f = (f)(L/D)(V 2 /2g) Assume/ = 0.018. h L = (O.O18)[975/O.5OO]{(5.O930)7[(2)(9.8O7)]} = 46.42Q 2, 0 + 0 + 111.0 + 10.725/0 = 0 + (5.O930)2/[(2)(9.8O7)] + 150.2 + 46.42Q2, 0 = 0.254 m3/s (by trial and error). This solution was based on the assumed value of /of 0.018. However, the value of/is dependent on the Reynolds number and relative roughness. Therefore a new value of /should be determined based on the computed value of Q of 0.254 m’/s. e/D = 0.00030/0.500 = 0.00060 V = Q/A = 0.254/[(jr)(0.500)74] = 1.294 m/s N„ = DV/v = (0.500)(1.294)/(1.02 x 10~6) = 6.34 X 105 From Fig. A-5, / = 0.018. Hence, the computed flow rate of 0.254 m3/s is correct.

(D

Fig. 20-21

20.81

Oil with a specific gravity of 0.87 is being pumped from a lower reservoir to an elevated tank as shown in Fig. 20-22. The pump in the system is 78 percent efficient and is rated at 185 kW. Determine the flow rate of the oil in the pipe if the total head loss from point 1 to point 2 is 12 m of oil. I

pJy + v\l2g + z l + E p =p 2 ly + v\l2g + z 2 + h L pjy-p 2 /y = v\/2g = 0 Zi = 150m P = Q Y E P (0.78X185) = (0)[(O.87)(9.79)](Ep) E P = 16.94/0 vl/2g = (QIA) 2 /2g = {0/[(*)( 0.160)2/4] }2/[(2)(9.807)] = 126.1202

z 2 = 200 m h L = 12m

0 + 0+150 + 16.94/0 = 0+126.1202 + 200 +12

126.1202 - 16.94/0 + 62 = 0

By trial and error solution, 0 = 0.244 m3/s.

Elev. 200 m

Fig. 20-22

20.82

For the conditions given in Prob. 20-80, find the flow rate at which water is being discharged from the pipe using the computer program written for Prob. 9.295.

PUMPS AND FANS 0 633

I INPUT 11

M

11 ?M i«iiiti)t4iii»miin»»mn4UMiNiff»Mnu)4UMiriiii«4i«u«44iii4r«4iHtiuui«uHPu»Mt(«UMaMUwii}innnMnHminii

1 SAMPLE ANALYSIS 0F INCOMPRESSIBLE FLOW 500.0

975.0

HATER

1.0 .00000102 .00030 (ROUGHNESS = 0.300 Mi)

111.0

150.2 9.79

105.0

SAMPLE ANALYSIS OF INCOMPRESSIBLE FLOW I OUTPUT GIVEN DATA FOR A CIRCULAR CLOSED CONDUIT CARRYING INCOMPRESSIBLE FLOW PRESSURE AT POINT 1 =

0.0 KPA

PRESSURE AT POINT 2 =

0.0 KPA

ELEVATION AT POINT 1 =

111.0 M

ELEVATION AT POINT 2 =

150.2 M

ACTUAL ENERGY ADDED BETWEEN POINTS 1 AND 2 =105.0 KW ACTUAL ENERGY REMOVED BETWEEN POINTS 1 AND 2 = 0.0 KW MINOR HEAD LOSSES BETWEEN POINTS 1 AND 2 = 0.0 M DIAMETER OF CONDUIT =

500.0 MM

LENGTH OF CONDUIT =

975.0 M

FLUID FLOWING IS WATER CONDUIT MATERIAL IS (ROUGHNESS = 0.300 MM)

THE FLOW RATE WILL BE 0.254 CU M/S VELOCITY AT POINT 1 = 0.00 M/S VELOCITY AT POINT 2 = 1.29 M/S

20.83

A pump draws water from a reservoir and then discharges it to an elevated tank, as shown in Fig. 20-23. The pipe’s ends are squared-cornered (i.e., sharp-edged). There are two 90° bends as shown. The pipe roughness (e) is 0.0084 ft and C is 120. If the rate of flow is 8.0 ft3/s and the efficiency of the pump is 75 percent, determine the required horsepower of the pump. Consider both friction and minor losses.

I

Pi/y + v \/2 g + z x + E„ =p 2 / y + v i /2 g + Z2 + h L Pil y = p 2 ly = v\l2g = v\l2g = 0 z x = 10 ft z 2 = 120 ft h L = h f +h m

For the 12-in pipe with (2 = 8 cfs, (h l ) 12 = 0.032 ft/ft. For the 10-in pipe with Q = 8 cfs, (h,)10 = 0.074 ft/ft. h f = (0.032)(40) + (0.074)(200) = 16.08 ft h m = h c + 2h„ + h e h c = K c v\J2g K c = 0.45 v t2 = Q/A = 8. 0/[(JT)(1.0)2/4] = 10.19 ft/s h c = (0.45)(10.19)2/[(2)(32.2)] = 0.73 ft h„ = K„v 2 J2g vw = Q/A = 8. 0/[(TT)(0. 833)74] = 14.68 ft/s With e/D = 0.0084/Oi) = 0.101 and R/D = f§ = 2.0, K„ = 0.6. h b = (0.6)(14.68)7[(2)(32.2)] = 2.01 ft h e = K e v\ 0 /2g K t = 1.0 h e = (1.0)(14.68)7[(2)(32.2)] = 3.35 ft h m = 0.73 + (2)(2.01) + 3.35 = 8.10 ft h L = h f +h m = 16.08 + 8.10 = 24.18 ft

0 + 0+ 10 + £„ = 0 + 0 + 120 + 24.18 E p = 134.18ft

P = QyE p = (8.0)(62.4)(134.18) = 66 983 ft-lb/s Required horsepower = 66 983/[(550)(0.75)] = 162 hp

634 0 CHAPTER 20

20.84

Oil with a specific gravity of 0.86 is being pumped from a reservoir as shown in Fig. 20-24. The pressures at points 1 and 2 are —4.0 psi and 43.0 psi, respectively. The rate of flow in the pipe is 0.50 ft 3/s. The pump is rated at 8 hp. Determine the efficiency of the pump. Neglect energy losses in the system.

I pjy + v\/2g + z 1 + E p = p 2 /y + V22/2g + z 2 + h L pjy = (-4)(144)/[(0.86)(62.4)] = -10.73 ft v\l2g = v\l2g z 1 = h L = 0 p 2 /y = (43)(144)/[(0.86)(62.4)] = 115.38 ft *2 = 3 ft -10.73 + E P = 115.38 + 3 E p = 129.11 ft p = QyE p = (0.50) [ (0.86) (62.4)] (129.11) = 3464 ft-lb/s = W = 6.30 hp Efficiency of pump = (6.30/8)(100) = 78.8 percent

(D p 2 = 43 psi

Fig. 20-24 20.85

A pump circulates water at the rate of 1900 gpm in a closed circuit holding 9000 gal. The total pump head is 290 ft and the pump efficiency is 0.89. Assuming the bearing friction to be negligible and the system to be adiabatic, find the rate of temperature rise in the water. [For water, c = 25 000 ft-lb/(slug-F°).]

f By the first law mc(AT/At) = QyH/r), or AT QyH [(1900)(0.002228](62.4)(290) = At met] [(9000/7.48)(1.94)](25 000)(0.89)

'

or5.3P/h. 20.86

The pump in Fig. 20-25 delivers water at 3 ft3/s to a machine at section 2, which is 20 ft higher than the reservoir surface. The losses between 1 and 2 are given by h f = KV\!2g, where K = 7.5 is a dimensionless loss coefficient. Find the power required to drive the pump if it is 80 percent efficient.

PUMPS AND FANS 0 635 f The flow is steady except for the slow decrease in reservoir depth, which we neglect, taking V ~ 0. We can compute V2 from the given flow rate and diameter: V 2 = Q/A 2 = 3/[(JT/4)(^)2] = 61.1 ft/s. Because of the solid walls and one-dimensional ports, the viscous work is zero. (pjpg) + (V?/2g) + z x = (pjpg) + (V\j2g) + z 2 + h, + h f . With Vi = Zj = 0 and h f = KV\/2g, we can solve for h* = [(Pi - Pi)lpg] - Zz - (1 + K)(Vl/2g)

(1)

The pressures must be in pounds per square foot for consistent units, with pg — 62.4 lb/ft3 for water. Introducing numerical values, we have h s = [(14.7 - 10)(144)/62.4] - 20 - [(1 + 7.5)(61.1)72(32.2)] = -502 ft

(2)

The pump head is negative, indicating work done on the fluid. The power delivered is P = mw s = pQgh s = (1.94)(3)(32.2)(-502) = -94100 ft-lb/s, hp = 94100/550 = 171 hp. The input power thus required to drive the 80 percent efficient pump is Pinput = P/efficiency = 171/0.8 = 214 hp.

Fig. 20-25

20.87

For the pump of Fig. 20-26, total friction head loss is 6 m. If the pump delivers 40 kW of power to the water, determine the exit velocity V e and the flow rate Q.

I P./y + V i / 2 g + Zl + E p =p 2 /Y + V l / 2 g + z 2 + hL

0 + 0 + 0 +£„ = 0+ V7[(2)(9.807)] + 3 + 6

E p = 0.05098V3 + 9 Also, P = QyE p , Q =AV = [(jr)(0.050)74](V) = 0.001964V, (40)(1000) = (0.001964V)[(9.79)(1000)](£„), E p = 2080/V. Equating expressions for E p , 0.05098V3 + 9 = 2080/V, V = 32.7 m/s (by trial and error); Q = (0.001964)(32.7) = 0.0642 m3/s, or 64.2 L/s.

D = 3in 130 ft/s

Fig. 20-26 20.88

Fig. 20-27

A fireboat pump draws sea water (s.g. = 1.025) from a 9-in submerged pipe and discharges it as 130 fps through

a 3-in nozzle, as in Fig. 20-27. Total head loss is 9 ft. If the pump is 72 percent efficient, how much horsepower is required to drive it?

I

P = QyEph Q=AV = [(jr)(f2)2/4](130) = 6.381 ft3/s pjy + V\l2g + z, + E p = pjy + V3/2g + z 2 + h L 0 + 0 + 0 + E p = 0 + 1303/[(2)(32.2)] + 10 + 9 E„ = 281.4 ft

P = {(6.381)[(1.025)(62.4)](281.4)/550} /0.72 = 290 hp

636 D CHAPTER 20

20.89

The pump in Fig. 20-28 discharges water at 30 L/s. Neglecting losses and elevation changes, what power is delivered to the water by the pump?

I pj y + V 2 J2g + z t + E p = p 2 /y + vi/2g + z 2 + h L V, = Q/A , = (30 x 10“3)/[(jr)(0.10)2/4] = 3.820 m/s V 2 = 0.03/[(JT)(0.04)74] = 23.87 m/s 125/9.79 + 3.8207[(2)(9.807)] + 0 + E p = 409/9.79 + 23.877[(2)(9.807)] +0 + 0 E p = 57.31 m P = QyE p = (30 x 10-3)(9.79)(57.31) = 16.8 kW 1<3R H>.,

Fig. 20-28 20.90

What is the flow for the system shown in Fig. 20-29a? The pump has the characteristics shown in Fig. 20-29b. What is the power required?

Assume Q, = 80 cfs. Then V, = 80/[(JT/4)(2.00)2] = 25.5 f t / s , (N R ) i = [(25.5)(2.00)/(0.1217 x 10 4)] = 4.19 X 106, f , — 0.0118. Solve for H, in Eq. (4): H, = 76.1 ft. The point (1) for //, and Qt is above Q curve. We take as a second estimate Q 2 = 70, V2 = 70/[(jr/4)(2)2] = 22.28 ft/s, (N R ) 2 = (22.28)(2)/(0.1217 x 10“4) = 3.66 x 106, f2 = 0.0118, H 2 = 49.3 ft. New point is just below Q line. Third estimate can now be easily made: Q 3 = 71 cfs, V 3 = 71/[(rr/4)(22)] = 22.60 ft/s, (N R ) 3 = (22.60)(2)/(0.1217 x 10 4) = 3.71 x 106, /3 = 0.0118, H 3 = 51.8 ft. We are close enough to the intersection so that we can say: Q = 71 cfs. The power needed for this operation is then Power = g(51.8)(1.938)(71)/[(0.73)(550)] = 572 hp.

PUMPS AND FANS

3,ft3/s

0 637

Fig. 20-29

(b)

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