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Q. At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is $100 \,m ^{3} \,s ^{-1}$. If the turbine generator efficiency is $60 \%,$ the electric power available from the plant is $\left(g=9.8\, m s ^{-2}\right)$

Alternating Current

Solution:

Hydroelectric power $=\frac{P \cdot E}{t}=\frac{mgh}{t}=\rho\left(\frac{V}{t}\right) gh$
Electric power = Efficiency $\times$ hydroelectric power
$=0.6 \times\left(10^{3} \times 100 \times 9.8 \times 300\right)$
$=176 \,MW$