Mining Methods and Machinery · Mining Machinery · 2 marks · Numerical answer
A pump raises 320 m³/hour through 180 m head at overall efficiency 0.70. Mine-water unit weight is 10.20 kN/m³. It works 14 h/day for 270 days and 20 h/day for 95 days. Find annual electrical consumption in GWh, to two decimals.
Input power isγQH/η with Q=320/3600 m³/s, giving 10.20×(320/3600)×180/0.70=233.143 kW. Annual hours are 270×14+95×20=5680. Energy=233.143×5680 kWh=1.32425 GWh, or 1.32.
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