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Heat is transferred to a heat engine (runs on a thermodynamic cycle) from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, (a) identify the thermal reservoirs, (b) determine the net power output, and (c) the thermal efficiency for this heat engine.

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Answer:

37.5%

Explanation:

We define first our values, that are,

[tex]Q_H=80MW\\Q_c=50MW[/tex]

We can now calculate the net power output, that is

[tex]W=Q_H-Q_c = (80-50) = 30MW[/tex]

Thermal efficiency is given by

[tex]\eta = \frac{W}{Q_H}\\\eta = \frac{30}{80}\\\eta = 0.375\\\eta = 37.5%[/tex]

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