Agriculture Reference
In-Depth Information
12.8 Power Requirement in Pumping
Power requirement in pumping can be expressed as follows:
mgh
t
( Q T × ρ
)
×
g
×
H
Q T
t
P
=
=
=
× ρ ×
g
×
H
t
i.e.,
P
=
Q
× ρ ×
g
×
H
(12.8)
where
P
=
power requirement, in watt (W)
m
mass of fluid delivered, in kg
Q T =
=
total discharge for the time t ,inm 3
density of fluid, in kg/m 3 (
1,000 kg/m 3 for normal water)
ρ =
9.81 m/s 2 )
g
=
acceleration due to gravity (
H
=
total head of water, in m
t
=
pumping period, in seconds
Q T
t =
discharge rate, m 3 /s
Q
=
Let the efficiency of the electric motor to be used is E m , then the motor size or
capacity would be
1
E m
P m =
( Q
× ρ ×
g
×
H )
(12.9)
where P m =
motor capacity, in W
If the density of water,
1,000 kg/m 3 , g
ρ =
=
9.81 m/s, then the above equation
can be written as
1
E m
P m =
( Q
×
9.81
×
H )
×
[kW]
Units of other elements will be same as mentioned earlier.
1 kilowatt (kW)
1,000 W
Kilowatt can be converted into horse power by the relation:
=
1kW
=
1.341 hp
1,000 kg/m 3 , g
9.81 m/s, 1 m 3
Using
ρ =
=
=
1,000 l, and the relation of 1 kW
=
1.341 hp; energy of discharging water, or “water horse power” can be expressed
as
WHP
=
(discharge in l/s
×
H )/76
where H
=
total head, in m
 
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