Water-Lifting Devices – Pumps - Practices of Irrigation and On-farm Water Management - page 474

Example 12.3

A submersible pump lifts 70,500 l of water/h against a total head of 25 m. Determine
the power requirement to lift the water in (i) kilowatt, and (ii) horse power.

Solution

Given,
Q = 70,500 l/h = 0.0195833 m 3 /s
H =25m

(i) We know, power, P = Q × 9.81 × H = 0.0195833 × 9.81 × 25

= 4.833 kW(Ans.)

(ii) We know, 1 kW = 1.341 hp

Thus, P = 4.803 × 1.341 hp = 6.441 hp (Ans.)

Example 12.4

In a wheat growing area, the cultivable land is 80 ha and wheat will be cultivated to
all of the lands. The permissible interval between two irrigations at peak period is
15 days and the depth of irrigation required for that particular soil & agro-climatic
region at peak period is 6.0 cm. If the total head for pumping is 25 m, pump effi-
ciency is 85%, motor efficiency is 80%, and the maximum allowable operating
period of the pump is 16 h/day, determine:

(a) The pump capacity required for that command area,
(b) Capacity of the motor

Solution

Given,
A = 80 ha = 800,000 m 2
Irri. interval

15 days
Depth of irri., d =6cm=0.06m
Daily pump operating period

=

= 16 h

Total head, H =25m
Pump efficiency, E p = 85% = 0.85
Motor efficiency, E m = 80%
We get,

total volume of water required for irrigation in 15 days,

(800, 000 m 2

V = × 0.06 m)

= 48,000 m 3
Total pumping period in 15 days, t = 15 × 16 × 3, 600 = 864, 000 s