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A turbine generates 1.64 MW. Assuming th...

A turbine generates 1.64 MW. Assuming the efficiency to be 80% and head of water used 1663 m, calculate the rate of flow of water.

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`P=1.64xx10^(6)W`
Gravitational head = 1663 m
`eta=80%` available power = `0.8xx1.64xx10^(6)W`
i.e. `P'=1.312xx10^(6)W=(("mgh"))/(t)`
i.e. `(m)/(t)=(1.312xx10^(6))/(gh)=(1.312xx10^(6))/(9.8xx1663)`
`(m)/(t)=80.5kgs^(-1)or(V)/(t)=((80.5)/(10^(3)))m^(3)s^(-1)`
i.e. `(V)/(t)=80.5xx10^(-3)m^(3)s^(-1)=80.5" litres "s^(-1)`.
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