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;
![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](https://tex.z-dn.net/?f=E%20%3D%20mgh%20%2B%20%5Cfrac%7Bm%7D%7B2%7D%20%28v_2%5E2%20-%20v_1%5E2%29%20%5C%5C%5C%5CE%20%3D%20%5Crho%20V%20g%20h%20%2B%20%5Cfrac%7B%5Crho%20V%7D%7B2%7D%20%5B%28%5Cfrac%7BV_2%7D%7B%5Cpi%20r_2%5E2%7D%20%29%5E2%20-%20%28%5Cfrac%7BV_1%7D%7B%5Cpi%20r_1%5E2%7D%29%5E2%5D%5C%5C%5C%5CE%20%3D%20PV%20%2B%20%5Cfrac%7B%5Crho%20V%5E3%7D%7B2%5Cpi%5E2%7D%20%5B%5Cfrac%7B1%7D%7B%20r_2%5E4%7D%20%20-%20%5Cfrac%7B1%7D%7B%20r_1%5E4%7D%5D%5C%5C%5C%5CE%20%3D%20%28500%20%2A10%5E3%29%280.1%29%20%2B%20%5Cfrac%7B%28860%29%280.1%29%5E3%7D%7B2%5Cpi%5E2%7D%20%5B%5Cfrac%7B1%7D%7B%200.06%5E4%7D%20%20-%20%5Cfrac%7B1%7D%7B%200.04%5E4%7D%5D%5C%5C%5C%5CE%20%3D%2050000%20-13653.51%5C%5C%5C%5CE%20%3D%2036346.48%20%5C%20W%5C%5C%5C%5CE%20%3D%2036.347%20%5C%20kW)
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 %