Question #193161

if a hydroelectric station is 93% efficient. What mass of water must pass through the station each 1.00s to produce 855kW of electricity from a 34.6 meter drop?


1
Expert's answer
2021-05-14T09:41:54-0400

The 855 kW of generated power correspond to

Pi=PoηP_i=\frac{P_o}{\eta}

of input power, or the power of falling water. This power can be found from the work done by water under the influence of gravitation:


Pi=Wt=mght, m=Pitgh=P0tηgh=2573 kg.P_i=\frac{W}{t}=\frac{mgh}{t},\\\space\\ m=\frac{P_it}{gh}=\frac{P_0t}{\eta gh}=2573\text{ kg}.

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