Subtopic: Sources of power on the farm – human, animal, mechanical, electrical, wind, solar and biomass
Question Statement:
A solar photovoltaic system comprising solar photovoltaic array, inverter and a motor-pump unit is installed for supplying drinking water in a village. There are 24 modules in the array and each module contains 36 number of cells of size 104×104 mm with a conversion efficiency of 12.8%. The global solar radiation incident normally on the cells is 945 W m⁻². The power consumed in lifting the water is found to be 435 W. If the pump-motor unit efficiency is 45%, the efficiency of the inverter is
Options:
(A) 56.21%
(B) 69.42%
(C) 80.25%
(D) 85.52%
Official Answer Key:
D
Step-by-Step Solution & Explanation
Total cell area = 24 modules × 36 cells × (0.104×0.104 m²) = 864 × 0.010816 = 9.345 m². Array (DC) power output = Area × Irradiance × cell efficiency = 9.345 × 945 × 0.128 = 1130.4 W. Electrical power needed at motor input = hydraulic (lifting) power / pump-motor efficiency = 435/0.45 = 966.7 W. Inverter efficiency = (power delivered to motor)/(array DC power) = 966.7/1130.4 = 85.52%.
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