Answer to Question #137320 in Mechanical Engineering for Chinny

Question

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.

Answer 1

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 = $400×9.75×20 = 78000 J$

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

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.