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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