Problem on mechanical efficiency of the pump

The oil pump is drawing 25 kW of electric power while pumping oil with ρ = 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are of 8 cm and 12 cm, respectively. When the pressure rise of oil in the pump is measured to be 250 kPa and the motor efficiency is 90%, then find out the mechanical efficiency of the pump. Taking kinetic energy correction factor to be 1.05.

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Inlet Dia, Di = 8 cm = 8 x 10-2 m
Outlet Dia, Do = 12cm = 12 x10-2m

Density of oil,  δ = 80Kg /m3

Flow rate Q = 0.1 m3/s

Pressure rise = 250KPa = 250 x10-3 Pa

Power supplied to the pump = 25Kw = 25 x 10-3 w

Motor efficiency = .90

Kinetic energy correction factor, α= 1.05

Inlet area Ai= Π/4 x D12=-Π/4 x (8 x 10-2)2 = 0.0804 m2
Outlet area A0= Π/4 x D02 = Π/4 x (12 x10-2)2= 0.1809 m2

Average evolution 
Vi = Q/Ai = 0.1/ 0.804 = 1.1235 m/s
V0 = Q/A0 = 0.1/ 0.1809 = 0.5526 m/s

A note of kinetic energy correction factor

K. E correction factor, α = (K. E /See based on actual velocity) / (K. E / See based on average velocity)

The factor α is used when the flow is viscous.

Applying Bernoulli’s equation at the inlet (i) i outlet (0) of the pump.

Pi/ δg + α1 Vi2/ 2g +zi + HP= P0 /δg +α2 Vo2/2g + Z0 + Hf .

Given  αi= α2= α= 1.05     (Z0 –Zi is considered negligible)
HP = head added by the pump
Hf = head loss due to friction

H= HP – Hf = P0–Pi / δg + α ( V02-V12)/ 2g
    = 250 x 103 / 1000 x 9.81 + 1.05 / 2 x 9.81 (0.55262  - 1.2435)
    = 25.42 m

Power of the pump PP= δg QH
            = 1000 x9.81x 0.1 25.42
            = 24934.85 w
            = 24.934Kw

Mechanical efficiency of the pump:

Case (1)  ηm = power output/power input = 24.934/ 25 = 99%
Case (2)  if the  motor is to get 25Kw  considering its efficiency  the supply should be of 25/ 0.9 KW

ηm = 24.934/ (25/0.9) = 89.67%

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