ME:5160 (58:160) Intermediate Mechanics of Fluids

Fall 2022 – HW10 Solution

P6.80 The head-versus-flowrate characteristics of a centrifugal pump are shown in Fig.

P6.80. If this pump drives water at 20C through 120 m of 30-cm-diameter cast-iron pipe,
what will be the resulting flow rate, in m3/s?

Solution: For water, take  = 998 kg/m3 and  = 0.001 kg/ms. For cast iron, take   0.26
mm, hence /d = 0.26/300  0.000867. The head loss must match the pump head:

Fig. P6.80

L V 2 8fLQ 2
h =f = = h pump  80 − 20Q 2 , with Q in m 3 /s
f d 2g  2 gd 5
8f(120)Q 2 80
Evaluate h f = = 80 − 20Q 2 , or: Q 2 
 2 (9.81)(0.3)5 20 + 4080f
1/2
 80  m 3
4 Q
Guess f  0.02, Q =    0.887 , Re =  3.76E6
 20 + 4080(0.02)  s d
 m3
= 0.000867, f  0.0191, Re  3.83E6, converges to Q  0.905 Ans.
d better better s
P6.97 A heat exchanger consists of multiple parallel-plate passages, as shown in Fig. P6.97.
The available pressure drop is 2 kPa, and the fluid is water at 20C. If the desired total flow
rate is 900 m3/h, estimate the appropriate number of passages. The plate walls are
hydraulically smooth.

Fig. P6.97

Solution: For water,  = 998 kg/m3 and  = 0.001 kg/ms. Unlike Prob. 6.88, here we
expect turbulent flow. If there are N passages, then b = 50 cm for all N and the passage
thickness is H = 0.5 m/N. The hydraulic diameter is Dh = 2H. The velocity in each passage
is related to the pressure drop by Eq. (6.58):

L  2  VDh 
p = f V where f = fsmooth = fcn 
Dh 2   
2.0 m 998 kg/m3 2
For the given data, 2000 Pa = f V
2(0.5 m/N ) 2

Select N, find H and V and Qtotal = AV = b2V and compare to the desired flow of 900 m3/h.
For example, guess N = 20, calculate f = 0.0173 and Qtotal = 2165 m3/h. The converged
result is
Qtotal = 908 m3 /h, f = 0.028,
Re Dh = 14400, H = 7.14 mm, N = 70 passages Ans.
*P6.102 A 70 percent efficient pump delivers water at 20 C from one reservoir to
another 20 ft higher, as in Fig. P6.102. The piping system consists of 60 ft of galvanized-iron
2-in pipe, a reentrant entrance, two screwed 90 long-radius elbows, a screwed-open gate
valve, and a sharp exit. What is the input power required in horsepower with and without a 6
well-designed conical expansion added to the exit? The flow rate is 0.4 ft3/s.

Fig. P6.102

Solution: For water at 20C, take  = 1.94 slug/ft3 and  = 2.09E−5 slug/fts. For galvanized
iron,   0.0005 ft, whence /d = 0.0005/(2/12 ft)  0.003. Without the 6 cone, the minor
losses are:

K reentrant  1.0; K elbows  2(0.41); K gate valve  0.16; K sharp exit  1.0

Q 0.4 ft Vd 1.94(18.3)(2/12)

Evaluate V = = = 18.3 ; Re = =  284000
A  (2/12) /4
2
s  2.09E−5

At this Re and roughness ratio, we find from the Moody chart that f  0.0266. Then

V2  L  (18.3)2   60  
(a) h pump = z +  f +  K  = 20 + 0.0266   + 1.0 + 0.82 + 0.16 + 1.0 
2g  d  2(32.2)   2/12  

gQh p (62.4)(0.4)(85.6)
or h pump  85.6 ft, Power = =
 0.70
= 3052  550  5.55 hp Ans. (a)
(b) If we replace the sharp exit by a 6 conical diffuser, from Fig. 6.23, Kexit  0.3. Then

(18.3)2   60  
h p = 20 + 0.0266   + 1.0 + .82 + .16 + 0.3 = 81.95 ft
2(32.2)  2/12  

then Power = (62.4)(0.4)(81.95)/0.7  550  5.31 hp (4% less) Ans. (b)

P6.113 The parallel galvanized-iron pipe system of Fig. P6.113 delivers water at 20C
with a total flow rate of 0.036 m3/s. If the pump is wide open and not running, with a loss
coefficient K = 1.5, determine (a) the flow rate in each pipe and (b) the overall pressure drop.

Fig. P6.11

Solution: For water at 20C, take  = 998 kg/m3 and  = 0.001 kg/ms. For galvanized
iron,  = 0.15 mm. Assume turbulent flow, with p the same for each leg:
L1 V12 V2  L 
hf1 = f1 = hf2 + hm2 = 2  f2 2 + 1.5 ,
d1 2g 2g  d 2 
and Q1 + Q2 = ( /4)d1 V1 + ( /4)d2 V2 = Qtotal = 0.036 m3 /s
2 2

When the friction factors are correctly found from the Moody chart, these two equations
may be solved for the two velocities (or flow rates). Begin by guessing f  0.020:
2
 60  V1 V22   55  
(0.02)   =  (0.02)   + 1.5 , solve for V1  1.10V2
 0.05  2(9.81) 2(9.81)   0.04  
 2  2 m m
then (0.05) (1.10V2 ) + (0.04) V2 = 0.036. Solve V2  10.54 , V1  11.59
4 4 s s
Correct Re1  578000, f1  0.0264, Re 2  421000, f2  0.0282, repeat.
The 2nd iteration converges: f1  0.0264, V1 = 11.69 m/s, f2  0.0282, V2 = 10.37 m/s,
Q1 = A1V1 = 0.023 m3/s, Q2 = A2V2 = 0.013 m3/s. Ans. (a)
The pressure drop is the same in either leg:
L V2  L  V22
p = f1 1 1 =  f2 2 + 1.5  2.16E6 Pa Ans. (b)
d1 2  d2  2
C6.3 The water slide in the figure is to be installed in a swimming pool. The manufacturer
recommends a continuous water flow of 1.39E−3 m3/s (about 22 gal/min) down the slide
to ensure that customers do not burn their bottoms. An 80%-efficient pump under the slide,
submerged 1 m below the water surface, feeds a 5-m-long, 4-cm-diameter hose, of
roughness 0.008 cm, to the slide. The hose discharges the water at the top of the slide, 4 m
above the water surface, as a free jet. Ignore minor losses and assume  = 1.06. Find the
brake horsepower needed to drive the pump.

FigC6.3

Solution: For water take  = 998 kg/m3 and  = 0.001 kg/ms. Write the steady-flow energy
equation from the water surface (1) to the outlet (2) at the top of the slide:

pa 1V12 pa  2V22 1.39E−3 m

+ + z1 = + + z2 + h f − hpump , where V2 = = 1.106
g 2 g g 2g  (0.02) 2
s

V22  L
Solve for hpump = (z2 − z1 )+ 2 + f 
2g  d

Work out Red = Vd/ = (998)(1.106)(0.04)/0.001 = 44200, /d = 0.008/4 = 0.002,

whence fMoody = 0.0268. Use these numbers to evaluate the pump head above:

(1.106)2   5.0  
hpump = (5.0 − 1.0) + 1.06 + 0.0268  0.04   = 4.27 m,
2(9.81)  

gQhpump 998(9.81)(1.39E−3)(4.27)
whence BHPrequired = = = 73 watts Ans.
 0.8