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At a hydroelectric power plant, the water pressure head is at a height of 300 m and thewater flow available is 100 m^3 s^(-1). If the turbine generator efficiency is 60%, estimatethe electric power available from the plant (g = 9.8 ms^(-2)). |
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Answer» SOLUTION :Hydroelectric power = pressure X RATE of FLOW of water ` = (h rho g) (av) = 300 xx 9.8 xx 10^3 xx (100)` Electric power ` = 60/100 xx 300 xx 9.8 xx 10^3 xx 100 = 176 MW` |
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