FLOW IN CLOSED CONDUITS 9.233

0 251

A p ipe of mean diameter 5 ft and length 6000 ft delivers water to a facility 1300 ft below the water surface at i ntake. Assume / = 0.025. When the pipe delivers 300 cfs, what is the horsepower delivered? I
2 f

u = e M = 3 00/[(^)(?Í ) /4] = 15.28 ft/s
2 2

2

h = (f)(L/d)(v /2g) = 0 .025[6000 /(fi )]{15.28 /[(2 )(32 .2)]} = 108.76 ft h = (K)(v /2g) = 0.50{15.28 /[(2)(32.2)]} = 1.81 ft
m 2 2

2

h = h + h = 108.76 + 1.81 = 110.57 ft
L f m

P = Q y(Az - h ) = (300)(62.4)(1300 - 110.57) = 2.227 x 10 ft • lb/s = (2.227 x 1 0 )/550 = 40 491 hp
L

7

7

9234

Find the kilowatt loss in 500 m of 50-cm-diameter pipe for which e = 0.05 mm when dye at 45 °C (s.g. = 0.86, v = 4.4 x 10" f t /s) flows at 0.22 m /s. Neglect minor losses.
6 2 3

f
R

hf = (f)(L/d)(v /2g)
6

2

u = f 2M =
5

0.22/[(^)(m)74]

= 1.120m/s
5 1

N = dv/v = (0.50)(1.120)/(4.4 x 10~ ) = 1.27 x 10

e/d = (5 x 10" )/(5 x 10" ) = 0.0001

F rom Fig. A - 5 , / = 0.018. h, = 0.018[500/0.50]{1.1207[(2)(9.807)]} = 1.151 m, P = Qyh, = 0.22[(0.86)(9.79)](1.151) = 2.13 kW. 9.235 L inseed oil, of kinematic viscosity 0.0005 ft /s and weight density 59.8 lb/ft , is pumped through a 3-in pipe ( e = 0.001 i n), ( a) A t what maximum velocity would the flow still be laminar? ( 6) W hat would then be the loss in energy head per 1000 ft of pipe? I (a) A ssume laminar flow exists for N < 2000. N = dv/v, 2000 = ( ^)(u )/0 .0005, v = 4.00 f t/s.
R R 2 3

( 6)

/ = 64/N = £5 = 0.032
R f

h = (J)(L/d)(v /2g) = 0.032[1000/(A)]{4.00 /[(2)(32.2)]} = 31.80 ft
f

2

2

p = yh = (59.8)(31.80)/144 = 13.2 psi per 1000 ft 9.236 R epeat Prob. 9.235 if the velocity is three times the maximum velocity for laminar flow. I
R

v = (3)(4.00) = 12.00 f t/s N = dv/v = (^)(12.00)/0.0005 = 6000

h = (f)(L/d)(v /2g)
f

2

e/d = 0.001/2 = 0.00050

F rom Fig. A - 5 , / = 0.036. h = 0.036[1000/(á)]{12.007[(2)(32.2)]} = 322.0 ft
f

p = yh = (59.8)(322.0)/144 = 134 psi per 1000 ft
f

9237

W ater flows upward at 3 m/s through a vertical 150-mm-diameter pipe standing in a body of water with its lower e nd 1.0 m b elow the surface. Considering all losses and with / = 0.022, find the pressure at a point 3 m above t he surface of the water. ' PJY + f f/2g + z, = p /y + v\/2g +z + h . L et point 1 be at the water surface and point 2 be 3 m above the water surface. h = h + h , h = (J)(L/d)(v /2g) = 0.022[(3 + 1.0)/0.150]{3 /[(2)(9.807)]} = 0.269 m, h = ( K)(u /2g). F or entrance loss, assume K = 0.8. h = 0.8{3 /[(2)(9.807)]} = 0.367 m, h = 0.269 + 0.367 = 0.636 m, 0 + 0 + 0 = p /9 .79 + 3 /[(2)(9.807)] + 3 + 0.636, p = - 40 .1 k Pa.
2 2 L 2 2 L f m f m 2 2 m L 2 2 2

9.238

W ork Prob. 9.237 if the flow is downward. f p / y + u|/2g + z = p , / y + u 72g + z, + /i .
2 2 i

h =h +h
L f

m

hf = 0.269 m For exit loss, assume K = 1.0.

(from Prob. 9.237)

h = (K)(v /2g)
m

2

h = 1.0{3 /[(2)(9.807)]} = 0.459 m
m 2 2

2

h = 0.269 + 0.459 = 0.728 m
L

p /9 .79 + 3 /[(2)(9 .807)]+ 3 = 0 + 0 + 0 + 0.728

p =-26.7kPa
2