Agriculture Reference
In-Depth Information
(a) Pump capacity, Q = V / t = 48,000/864,000 = 0.05556 m 3 /s (Ans.)
(b) Considering pump efficiency, capacity of the motor (output rated capac-
ity),
P m =
( Q
×
9.81
×
H )
/
E p =
(0.05556
×
9.81
×
25)
/
0.85
=
16.03 kW
(c) Rated capacity of the motor = P m / E m = 16.03/0.8 = 20.04 kW (Ans.)
Example 12.5
Maize crop of 20 ha is to be irrigated from a submergible pump. The maximum
permissible interval between two irrigations at peak period is 12 days and the depth
of each irrigation is 60 mm. If the maximum allowable operating period of the pump
is 10 h/day, determine the pump capacity to meet the water demand of the farm.
Solution
Area of the field, A = 20 ha = 200,000 m 2
Depth of irri. req., d =60mm=0.06m
Max. interval period
12 days
Daily max. pumping period
=
10 h
We get, total volume of water to be discharged in 12 days, V
=
=
A
×
d
0.06 m 3
=
200, 000
×
= 12,000 m 3
Total pumping period in 12 days, t
=
(12
×
10)
×
60
×
60
=
332, 000 s
Pumping rate or pump capacity required, Q
=
V
/
t
=
12, 000
/
432, 000
0.027778m 3
s(Ans.)
Or, 0.98154 ft 3 /s (or cfs or cusec)
[Since 1 ft 3 /s
=
/
0.0283 m 3 /s]
=
Example 12.6
In a residential area having population of 2,000 and expected population growth rate
of 5%, the average daily water demand per capita is 100 l/day. Projecting for a time
period of 30 years, determine the required capacity of the pump to satisfy the water
demand of that area. Assume that the pump can be operated 8 h/day at its maximum.
If the pump is installed at 25 m below the ground surface, the velocity head of the
flowing water is 1.5 m/s, friction loss in the discharge pipe and within pump casing
is 10% of the discharge head, determine the optimum size of the motor to operate
the pump.
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