A pump moves water horizontally at a rate of  0.02 m^3  /s. Upstream of the pump, where the pipe diameter is 90 mm, the pressure is 120 kPa. Downstream of the pump, where the pipe diameter is 30 mm, the pressure is 400 kPa. If the loss in energy across the pump due to fluid friction effects is 170 N.m/kg. Determine the hydraulic efficiency of the pump.

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8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
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Chapter6: Forced Convection Over Exterior Surfaces
Section: Chapter Questions
Problem 6.14P
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Correct this problem with step-bystep please: A pump moves water horizontally at a rate of  0.02 m^3  /s. Upstream of the pump, where the pipe diameter is 90 mm, the pressure is 120 kPa. Downstream of the pump, where the pipe diameter is 30 mm, the pressure is 400 kPa. If the loss in energy across the pump due to fluid friction effects is 170 N.m/kg. Determine the hydraulic efficiency of the pump.

Method of Solution:
Loss of energy =
170
0.02 * 100 * 9.81
H =
= 0.806
A₁
A₁
y1
y2
=
=
=
TU
TU
==*
Q
A₁
Q
A₂
H=
400 * 100
1000 * 9.81
Q = A1y1 = A2y2
0.092 6.3610-³m²
+
0.03² = 0.706 *
0.02
6.36 10 -3
=
0.706
Pg
9
X
0.02
X
P2 V₂
+
2g
28.32²
2 * 9.81
10-3
= 3.144m/s
68.917 +0.806
10-³m²
= 28.32m/s
P₁
Pg
H = 68.917m
68.917
+
120 * 1000
1000 * 9.810
2g
= 98.41
3.14²
2 * 9.81
Transcribed Image Text:Method of Solution: Loss of energy = 170 0.02 * 100 * 9.81 H = = 0.806 A₁ A₁ y1 y2 = = = TU TU ==* Q A₁ Q A₂ H= 400 * 100 1000 * 9.81 Q = A1y1 = A2y2 0.092 6.3610-³m² + 0.03² = 0.706 * 0.02 6.36 10 -3 = 0.706 Pg 9 X 0.02 X P2 V₂ + 2g 28.32² 2 * 9.81 10-3 = 3.144m/s 68.917 +0.806 10-³m² = 28.32m/s P₁ Pg H = 68.917m 68.917 + 120 * 1000 1000 * 9.810 2g = 98.41 3.14² 2 * 9.81
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