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Q. In a hydroelectric power station, the water is flowing at $2\, ms ^{-1}$ in the river, which is $100\, m$ wide and $5\, m$ deep. The maximum power output from the river is

Work, Energy and Power

Solution:

Given, area of river, $A=100\, m \times 5\, m$
$=500\, m ^{2}$
Density of water, $\rho=10^{3} kg / m ^{3}$ and $v=2\, ms ^{-1}$
$\therefore$ Mass of water flowing per second, $m=A \rho v$
$=500 \times 10^{3} \times 2=10^{6} kg / s$
Power of power station, $P=$ Kinetic energy of water flowing per second
$=\frac{1}{2} m v^{2}=\frac{1}{2} \times 10^{6} \times 2^{2}$
$=2 \times 10^{6} W =2\, MW$