Question

An oil pump is drawing 44kW while pumping oil with a density of 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are 8cm and 12cm, respectively. If the pressure increases by 500kPa going through the pump and the motoreffi ciency is 90%, determine the mechanical efficiency of the pump

Answer 1

The mechanical efficiency of the pump is 91.8 %

Step-by-step explanation:

Given;

input power, p = 44 kw

density of oil, ρ = 860 kg/m³

motor efficiency, η = 90 %

inlet diameter, d₁ = 8 cm

outlet diameter, d₂ = 12 cm

volume flow rate, V = 0.1 m³/s

pressure rise, P = 500kPa

output power = motor efficiency x input power

output power = 0.9 x 44 = 39.6 kW

Thus, the mechanical input power = 39.6 kW

The mechanical output power is given by change in mechanical energy;

`E = mgh + (m)/(2) (v_2^2 - v_1^2) \\\\E = \rho V g h + (\rho V)/(2) [((V_2)/(\pi r_2^2) )^2 - ((V_1)/(\pi r_1^2))^2]\\\\E = PV + (\rho V^3)/(2\pi^2) [(1)/( r_2^4) - (1)/( r_1^4)]\\\\E = (500 *10^3)(0.1) + ((860)(0.1)^3)/(2\pi^2) [(1)/( 0.06^4) - (1)/( 0.04^4)]\\\\E = 50000 -13653.51\\\\E = 36346.48 \ W\\\\E = 36.347 \ kW`

The mechanical efficiency is given by

η = mechanical output power / mechanical input power

η = 36.347 / 39.6

η = 0.918

η = 91.8 %

Therefore, the mechanical efficiency of the pump is 91.8 %