suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. how many kilograms of water must pass through the turbines each second to generate 52.0 mw of electricity? this is a typical value for a small hydroelectric dam.



Answer :

2.64×10^5 kg/s" of water must pass through the turbines each second, assuming a waterfall height of 26 m when the dam is 80% efficient at converting the water's potential energy to electrical energy.

the equation for the efficiency  of potential energy to electrical energy conversion.

[tex]" Efficiency "=(" electrical power ")/(" potential energy/t ")[/tex]

[tex](" potential energy ")/(t)=(" electrical energu ")/(" efficiency ")[/tex]

[tex](mgh)/(t)=(54MW)/(0.8)[/tex]

[tex](mgh)/(t)=67.5MW=67.5*10^6W[/tex]

assuming h = 26 m:

[tex]m/t=67.5*10^6W/9.81m/s^2\cdot26m[/tex]

[tex]m/t=2.64 *10^5kg/s[/tex]

Efficiency is an irrational property of a mechanical system that makes use of thermal energy. Boilers, steam turbines, steam engines, internal combustion engines, furnaces, refrigerators, and other common appliances are included in the list of equipment for which the thermal efficiency is determined.

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