An oil pump is drawing 44 kw of electric power while pumping oil with rho = 860 kg/m3 at a rate of 0.1 m3 /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.

The mechanical efficiency of the pump is 91.66 %.

Explanation:

We have given that an oil pump is drawing 44 kW of electric power

inlet diameter = 8 cm and outlet diameter = 12cm

The volume flow rate of the pump (V) =  0.1 cubic meter/s

The pressure rise across the pump (ΔP) = 500kPa

The efficiency of the motor() = 90%

The total mechanical energy by neglecting potential energy is given by

Where ΔP =

The initial velocity,

Similarly, final velocity

By substituting values in the equation (1) we get,

Thus the power delivered by pump is

The shaft power of the pump is :

Therefore the mechanical efficiency of the pump is expressed as,

The mechanical efficiency of the pump is 91.66 %

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.

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