At the end of the filling process, the pond has the water
needed for one day's irrigation. That gives
msystem@final =
mtotal
At the beginning
of the filling process, the pond is empty.
msystem@initial = 0
From the mass balance, the mass imported to
the pond can be determined.
min = Δmsystem =
msystem@final - msystem@initial
=
mtotal -
0
=
Vtotalρ
=
(90,000/1,000)(996)
=
89,640 kg
min= 89,640 kg
The flow rate of the pump is 20 L/s, or =
20 L/s. According to the relation between the mass flow rate and volume
flow rate, the mass flow rate can be determined.
= ρ = (20/1,000)(996) = 19.9 kg/s
The total time needed for the
pump to be operated is:
min = t
t = min/ =
(89,640)/(19.9) = 4505 s = 1.25 h |