A pump moves water from one reservoir to another as sketched. The elevation diff

Question

A pump moves water from one reservoir to another as sketched. The elevation difference between the two reservoir surfaces is hR = 30.5 m. The reservoirs are very large, and thus the elevations of the water at their surfaces changes by a negligible amount. The pipe diameter is D = 15.2 cm. The desired volume flow rate is 0.156 m3/s, and the total irreversible head loss through the pipes and valves, etc. is proportional to the square of the average flow speed V through the pipe. i.e.. hL = 41.3 V2/(2g). [Note: Later on in the course (Chapter 8). you will learn how to estimate the irreversible head loss: for now, it is given to you.] If the pump is 82.5% efficient, calculate the power (in kW and in lip) required to rotate the pump shaft.

Explanation / Answer

v=Q/A= 0.156*4/D2 = 8.59 m/s

hl=41.3v2/2g

=155.736 m

total head h= hl+hR = 186.23 m

power = hgQ*100/82.5 = 345.11 kW

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