Question

An oil pump is drawing 44 kW of electric power while pumping oil with rho = 860 kg/m^3 at a rate of 0.07 m^3/s. The inlet and outlet diameters of the pipe are 8 cm and 12 cm, respectively. If the pressure rise of oil in the pump is measured to be 500 kPa and the motor efficiency is 90 percent, determine the mechanical efficiency of the pump. (Round the final answer to one decimal place.)

Answer 1

Answer:

mechanical efficiency  = 54%

Explanation:

given data

power We = 44 kW

density = 860 kg/m³

rate of flow = 0.07 m³/s

inlet diameter d1 = 8 cm

outlet diameter d2 = 12 cm

pressure = 500 kPa

efficiency η = 90%

to find out

mechanical efficiency

solution

we get here shaft power is here

shaft power = efficiency × electric power

shaft power = 0.9 × 44 = 39.6

and

now we get DEmech that is

DE(mech)= m g h + $\frac{m}{2}$ (v² - u ²) .....................1

that is we can write as

DEmech = ρVgh +$\frac{\rho V}{2} [(\frac{V}{\pi r2^2})^2- (\frac{V}{\pi r2^2})^2 ]$

= VDP + $\frac{\rho V^3}{2\pi ^2}$ × $(\frac{1}{\pi r2^2} - \frac{1}{\pi r1^2})^4$

Plugging in values we now have   DEmech that is

DEmech = (0.07m^3/s × 500kPa) + $\frac{860*0.1^3}{2\pi ^2} *( \frac{1}{0.06^4} -\frac{1}{0.04^4} )$  

solve we get

DEmech = 21342 W

DEmech = 21.3 kW

mechanical efficiency is

mechanical efficiency = $\frac{21.3}{39.6}$

mechanical efficiency = 0.54

mechanical efficiency  = 54%