Water falls down a $500.0 \,m$ shaft to reach a turbine which generates electricity. ................ $m^3$ water must fall per second in order to generate $1.00 \times 10^9 \,W$ of power ? (Assume $50 \%$ efficiency of conversion and $\left.g=10 \,ms ^{-2}\right)$
KVPY 2016, Medium
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(b)
Power output, $P=50 \%$ of potential energy of water
Received by generator per second
$=\frac{50}{100} \times(\frac{m g h}{t})$
$=05 \times\left(\frac{V}{t}\right) \rho \times g \times h \quad \dot(i)$
Here, $P=1 \times 10^9 \,W$,
$g=10 \,ms ^{-2}, \rho=1000 \,kgm ^{-3}$
and $h=500 \,m$.
Substituting these values in Eq. $(i)$, we get
Volume flow rate of water $=\frac{V}{t}$
$=\frac{1 \times 10^9}{0.5 \times 1000 \times 10 \times 500}=400 \,m ^3 s ^{-1}$
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