A water pump delivers 6 hp of shaft power when operating. The pressure differential between the outlet and the inlet of the pump is measured to be 1.2 psi when the flow rate is 10 ft3/s, and the velocity changes to 5 ft/s from 2 ft/s as the water passes through the pump. Determine the mechanical efficiency of this pump assuming the water density to be 62.4 lbm/ft3 (written in decimal form with 3 significant figures).

Answer: Pump efficiency = 0.585 = 58.5%

Explanation: Pump efficiency = (power gained by the fluid) / (power supplied by the shaft)

Power gained by the fluid = Q (ΔP) + m (((v2) ^2) - ((v1) ^2)) / 2

Where Q = volumetric flow rate = 10 ft3/s = 0.283 m3/s

ΔP = 1.2 psia = 8273.709 Pa

m = mass flow rate = density * volumetric flow rate

Density = 62.4 lbm/ft3 = 999.52 kg/m3

m = 999.52 * 0.283 = 282.86 kg/s

v2 = 5ft/s = 1.524 m/s

v1 = 2ft/s = 0.61 m/s

Q (ΔP) = 0.283 * 8273.709 = 2341.46 W

Power from change in kinetic energy = m (((v2) ^2) - ((v1) ^2)) / 2 = 275.92 W

Power gained by the fluid = 2341.46 + 275.92 = 2617.38 W

Power supplied by shaft = 6hp = 6 * 746 = 4476 W

Efficiency = 2617.38/4476 = 0.58475 = 0.585 to 3s. f

QED!