Answer:
The mechanical efficiency of the pump is 91.8 %
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;
[tex]E = mgh + \frac{m}{2} (v_2^2 - v_1^2) \\\\E = \rho V g h + \frac{\rho V}{2} [(\frac{V_2}{\pi r_2^2} )^2 - (\frac{V_1}{\pi r_1^2})^2]\\\\E = PV + \frac{\rho V^3}{2\pi^2} [\frac{1}{ r_2^4} - \frac{1}{ r_1^4}]\\\\E = (500 *10^3)(0.1) + \frac{(860)(0.1)^3}{2\pi^2} [\frac{1}{ 0.06^4} - \frac{1}{ 0.04^4}]\\\\E = 50000 -13653.51\\\\E = 36346.48 \ W\\\\E = 36.347 \ kW[/tex]
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 %
