Scribd
Upload
375 views2 pages

Pump Mechanical Efficiency Calculation

The document provides information to calculate the mechanical efficiency of an oil pump. It states that the pump draws 35kW of power to pump oil at 0.1m3/s with a pressure rise of 400kPa. Given inlet and outlet diameters and densities, the useful head is calculated as 30.4m. This corresponds to useful power of 25.6kW. The shaft power is 31.5kW based on motor efficiency of 90%. The mechanical efficiency is then calculated as 81.3%.

Uploaded by

antima meena
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
375 views2 pages

Pump Mechanical Efficiency Calculation

The document provides information to calculate the mechanical efficiency of an oil pump. It states that the pump draws 35kW of power to pump oil at 0.1m3/s with a pressure rise of 400kPa. Given inlet and outlet diameters and densities, the useful head is calculated as 30.4m. This corresponds to useful power of 25.6kW. The shaft power is 31.5kW based on motor efficiency of 90%. The mechanical efficiency is then calculated as 81.3%.

Uploaded by

antima meena
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 2

Example TET401: Material Engineering & Thermofluid

An oil pump is drawing 35kW of electric power while pumping oil with  = 860kg/m3 at rate of
0.1m3/s as shown in Figure 1. The inlet and outlet diameters are 8cm and 12cm respectively. If the
pressure rise of oil in the pump is measured to be 400kPa and the motor efficiency is 90%,
determine the mechanical efficiency of the pump. Take the kinetic energy correction factor to be
1.05.

Figure 1

Solution

Assumptions:
(i) The flow is steady and incompressible.
(ii) The elevation difference across the pump is negligible.
(iii) All the losses in the pump are accounted for by the pump efficiency and thus hL = 0.
(iv) The kinetic energy correction factors are given to be 1 = 2 =  = 1.05.

The density of oil is given to be  = 860 kg/m3.

We take points 1 and 2 at the inlet and the exit of the pump, respectively. Noting that z1 = z2, the
energy equation for the pump reduces to
P1 V2 P V2 P2  P1  (V 22  V12 )
  1 1  z1  hpump, u  2   2 2  z 2  hturbine, e  h L  hpump, u  
g 2g g 2g g 2g

where
V V 0.1 m 3 /s
V1     19.9 m/s
A1 D12 / 4  (0.08 m) 2 / 4
V V 0.1 m 3 /s
V2     8.84 m/s
A2 D22 / 4  (0.12 m) 2 / 4

1
Example TET401: Material Engineering & Thermofluid

Substituting, the useful pump head and the corresponding useful pumping power are determined to
be
400,000 N/m 2  1 kg  m/s 2  1.05[ (8.84 m/s) 2  (19.9 m/s) 2 ]
hpump, u     47.4  17.0  30.4 m
(860 kg/m 3 )(9.81 m/s 2 )  1N 
 2(9.81 m/s 2 )
 1 kN  1 kW 
W pump,u  Vghpump, u  (860 kg/m 3 )(0.1m 3 /s)(9.81m/s 2 )(30.4 m) 
 1 kN  m/s   25.6 kW
 1000 kg  m/s
2

Then the shaft pumping power and the mechanical efficiency of the pump become
W pump,shaft   motorW electric  (0.90)(35 kW)  31.5 kW

W 25.6 kW
pump   pump, u   0.813  81.3%
Wpump, shaft 31.5 kW

You might also like