Answer to Question #137320 in Mechanical Engineering for Chinny

Question #137320
There are 400 kg/min of water being handled by a pump. The lift is from a 20-m deep well and the delivery velocity is 15 m/s. Find (a) the change of potential energy in kJ/kg, (b) the kinetic energy in kJ/kg, and (c) the required power of the pumping unit in kW. Assume g = 9.75 m/s2.
Expert's answer

In one minute: 400 kg of water is lifted 20 m and accelerated to a speed of 15 m/s (presumably from rest).

a. Change in PE = mgh = "400\u00d79.75\u00d720 = 78000 J"

b. Change in KE = "\\frac{1}{2}mv^2 = \\frac{400}{2} \u00d7 15^2 = 45000 J"

c. Total energy imparted to water per minute = 123 kJ

Thus required power = 123 / 60 = 2.05 kW

This assumes that we neglect all resistances.

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