An oil pump is drawing 35 kW of electric power while pumping oil with p= 860 kg/m³at 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 400 kPa andthe motor efficiency is 90 percent, determine the mechanical efficiency of the pump.Hint: Oil flow is steady and incompressible, and the elevation change is negligible.12 cmPump8cmOilAP-400 LP01m%

Answered: n oil pump is drawing 35 kW of electric…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Question

An oil pump is drawing 35 kW of electric power while pumping oil with p= 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 400 kPa and
the motor efficiency is 90 percent, determine the mechanical efficiency of the pump.
Hint: Oil flow is steady and incompressible, and the elevation change is negligible.
12 cm
Pump
8cm
Oil
AP-400 LP
01m%

Expert Solution

Step 1

Given:

The power of motor is, $P_{m o t o r} = 35 k W$ .

Density of oil is, $ρ = 860 k g / m^{3}$ .

The flow rate is, $Q = 0.1 m^{3} / s$ .

The inlet diameter is, $d_{1} = 8 c m = 0.08 m$ .

The outlet diameter is, $d_{2} = 12 c m = 0.12 m$ .

The efficiency of the motor is, $η_{m o t o r} = 0.90$ .

The pressure rise of oil is, $∆ P = P_{2} - P_{1} = 400 k P a$ .

The area of pump at inlet can be calculated as,

$A_{1} = \frac{π}{4} \times d_{1}^{2}$

Substitute the values in the above equation,

$A_{1} = \frac{π}{4} \times {(0.08)}^{2} A_{1} = 0.00503 m^{2}$

Similarly area at outlet,

$A_{2} = \frac{π}{4} \times d_{2}^{2}$

Substitute the values in the above equation,

$A_{2} = \frac{π}{4} \times {(0.12)}^{2} A_{2} = 0.01131 m^{2}$

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