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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.

Correct Answer: 0.31H.P
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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(1) Duty of utmost good faith, i.e. uberrimae fidei.
(2) Duty of carry on business to greatest common advantage.
(3) Duty to render true accounts and full information.
(4) Duty to indemnify for fraud.
(5) Duty to indemnify for willful neglect.
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