Water-Lifting Devices – Pumps - Practices of Irrigation and On-farm Water Management - page 473

12.13 Sample Workout Problems on Pump

Example 12.1

Determine the net positive suction head available at the pump inlet from the
following data:

suction head = 5m

friction loss 1m

vapor pressure of the liquid at water temperature

=

= 0.5 m

barometric (or atmospheric) pressure at pump level = 10 m

Solution

We know, net positive suction head available,
NPSHa = BP − SH − FL − VP

Given,

Barometric pressure at pump level, BP = 10 m

Suction head, SH = 5m

Friction loss, FL 1m

Vapor pressure of the liquid, VP

=

= 0.5 m

Putting the values,
NPSHa = 10 − 5 − 1 − 0.5 = 3.5 m (Ans.)

Example 12.2

A centrifugal pump has been installed to a depth of 35 m. The pump is discharging
2.5 cusec (ft 3 /s) water. Determine the capacity of the motor to operate the pump.
Assume motor efficiency of 82%.

Solution

Given,
Q =2.5ft 3 /s = 0.0708 m 3 /s [1 m 3 =35.3ft 3 ]
H d =35m
E m = 82% = 0.82
Taking discharge velocity of water V 2

= 2.0 m/s, velocity head = / 2g

(2.0) 2

= / (2 × 9.81)

= 0.2038 m
Assuming friction loss = 5% of discharge head = 5 × 35 / 100 = 1.75m

Total head, H discharge head + friction head + velocity head

= 25 + 1.75 + 0.204
=36.95m
We get, power of the motor, P m =

=

Q × 9.81 × H / E m = 0.0708 × 9.81 × 36.95

= 31.31 kW (Ans.)