A pump lifts 1600 litres of water per minute against a total head of 21 metres. Compute the
water horse power. If the pump has an efficiency of 75%. What size of motor is required to
operate the pump? If a direct drive electric motor having an efficiency of 85% is used to
operate the pump, compute the cost of electrical energy in a month of 30 days. The pump is
operated for 8 hours daily for 30 days. The cost of electrical energy is Rs. 3.5 per unit
Answers
A pump has the characteristics given in Fig. 8-5. What discharge and head will be
produced at maximum efficiency if the pump size is 50 cm and the angular speed is 30 rps?
What power is required when pumping water under these conditions?
At maximum efficiency, from Fig. 8-5,
CQ = 0.64, CP = 0.60, CH = 0.75
Q = CQnD
3 + 0.64 · 30 s
−1 · (0.5 m)
3 = 2.40 m
3 /s
∆h = CHn
2D 2 g = 0.75 · (30 s −1 )
2 · (0.5 m)2
9.81 m/s
2 = 17.2 m.
P = CPρD
5 n
3 = 0.60 · 1000 kg/m
3 · (0.5 m)5 · (30 s −1 )
3 = 506 kW.
Problem 8.12
If the pump having the performance curve shown is operated at a speed of 1500 rpm,
what will be the maximum possible head developed?
Solution:
CH =
∆Hg
D2n
2
Since CH will be the same for the maximum head condition, then
∆H ∝ n
2
or H1500 = H1000 ·
1500
1000!2
= 102 ft · 2.25 = 229.5 ft.