Question

A pump increases the pressure in the flow of water from 120 kPa to 400 kPa. The inlet and outlet diameters are 9 cm and 3 cm respectively. If the flow rate is 57 m3 /hr (neglecting losses) what is power delivered by the pump to the water? Neglect elevation changes.

Answer 1

Answer:

4.433 kW

Explanation:

Suppose the flow rate is the same at the inlet and outlet at 57 m3/hr.

Since 1 hour = 60 minutes = 3600 seconds we can calculate the flowrate in m3/s

$\dot{V} = 57 / 3600 = 0.0158 m^3/s$

As we are neglecting loss from friction and elevations changed, the power delivered by the pump is the product of the change in pressure and the flow rate

$P = \Delta P \dot{V} = (400 - 120)*0.0158 = 4.433 kW$