A pump was installed in a field to supply water to crops. The duty for this crop is 432 hectares/cumec on field and efficiency of pump is 50%. The sown area of field is 5 hectares. Determine maximum output required (H.P) of pump, if highest water level is 4 meters below highest portion of field. Assume negligible field channel losses.

Area of field to be irrigated = 5 hectares
Duty of water for crop = 432 hectares/cumec
Discharge required for the crop is = (5 / 432) = 1/86.4 cumec
Volume of water lifted per second = 1 / 86.4 cumec
Therefore, weight of water lifted per second = (1 / 86.4) x 9.81 = 0.1135 KN/sec
(unit wt. of water = 9.81 KN/m₃)
Maximum static lift of pump = 4 metres
Work done by the pump in lifting water = 0.1135 x 4 = 0.454 KWatt
The input of the pump = (0.454 / 0.735) = 0.62
(1 metric H.P = 0.735 KWatt)
Output H.P of the pump = (input/?) = (0.62 / 0.5) = 0.31 H.P.