5.119 Water is to be pumped from the large tank shown in Fig. P5.119 with an exit velocity of 6 m/s. It was determined that the original pump (pump 1) that supplies 1 kW of power to the water did not produce the desired velocity. Hence, it is proposed that an additional pump (pump 2) be installed as indicated to increase the flowrate to the desired value. How much power must pump 2 add to the water? The head loss for this flow is h, = 250 Q², where h, is in m when Q is in m/s. Nozzle area = 0.01 m2 Pipe area = 0.02 m² 6 m/s

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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5.119 Water is to be pumped from the large tank shown in Fig.
P5.119 with an exit velocity of 6 m/s. It was determined that the
original pump (pump 1) that supplies 1 kW of power to the water
did not produce the desired velocity. Hence, it is proposed that an
additional pump (pump 2) be installed as indicated to increase the
flowrate to the desired value. How much power must pump 2 add
to the water? The head loss for this flow is h, = 250 Q², where h, is
in m when Q is in m³/s.
Nozzle area = 0.01 m²
Pipe area = 0.02 m²
%3!
V = 6 m/s
Pump
L#2 _
Pump
#1
2 m
Figure P5.119
Transcribed Image Text:5.119 Water is to be pumped from the large tank shown in Fig. P5.119 with an exit velocity of 6 m/s. It was determined that the original pump (pump 1) that supplies 1 kW of power to the water did not produce the desired velocity. Hence, it is proposed that an additional pump (pump 2) be installed as indicated to increase the flowrate to the desired value. How much power must pump 2 add to the water? The head loss for this flow is h, = 250 Q², where h, is in m when Q is in m³/s. Nozzle area = 0.01 m² Pipe area = 0.02 m² %3! V = 6 m/s Pump L#2 _ Pump #1 2 m Figure P5.119
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