Water-Lifting Devices – Pumps - Practices of Irrigation and On-farm Water Management - page 476

Solution

Given,
Present population, P = 2,000 nos
Growth rate, r = 5% = 0.05
Projection time period in yr, t =30
Daily per capita water 100 l

We get, population after 30 yrs = P (1 + r ) t = 2,000 (1 + 0.05) 30

=

8, 643.9
Assuming no change in per capita demand, daily total demand ( V ) will be
=

=

864.39 m 3

× = =

8, 643.9 100 864, 390 l

= = × =

Daily pump operating period, t 8h 8 3600 28,800 s

0.030 m 3 /s ( Ans. )

Pump discharge rate req., Q = V / t =

Given, delivery head, H d =25m
Velocity of flowing water, v =1.5m/s
Thus, velocity head, Hv v 2 (0.5) 2 = = / (2 × 9.81) = 0.115 m

Friction loss (or friction head), Hf 2.5 m

Thus, total head, H = H d +Hv+Hf=25+0.115+2.5=27.625m
Power required to deliver the water, P

= 25 × 10 / 100 =

= Q × 9.81 × H

= 0.030 × 9.81 × 27.625 = 8.13 kW

Considering efficiency of the pump 80%

Capacity of the motor, P m = P / E p = 8.13/0.80 = 10.16 kW (Ans.)

=

Example 12.7

Four pumps are connected in series, each one pumping 30 GPM at 25 PSI. What is
the total output in flow volume and pressure?

Solution

Tot a l flow 30 GPM,

Total pressure = 25 + 25 + 25 + 25 = 100 PSI

=

Example 12.8

Four pumps are connected in parallel, each one pumping 30 GPM at 25 PSI. What
is the total output in flow volume and pressure?

Total pressure = 25 PSI,

Tot a l flow = 25 + 25 + 25 + 25 = 100 GPM

Questions

(1) What is a pump? Describe its purpose.
(2) What are the principles of water pumping?
(3) Classify the pumps under different perspectives.