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

Question

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

asked Aug 16, 2023 in Irrigation Engineering by Carter (75.2k points)

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

(a) 0.77H.P

(b) 0.85H.P

(c) 0.80H.P

(d) 0.31H.P

This question was addressed to me in my homework.

My doubt stems from Design Capacity for an Irrigation Canal topic in section Sediment Transport and Irrigation Channels Design of Irrigation Engineering

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Greg · Answer 1 · 2023-08-19T06:49:41+0000

Correct answer is (d) 0.31H.P

Easy explanation: 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/η) = (0.62 / 0.5) = 0.31 H.P.

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