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A power plant running at 39 % efficiency generates 330 MW of electric power. Part A At what rate (in MW) is heat energy exhausted to the river that cools the plant

Sagot :

516.154 megawatts of heat are exhausted to the river that cools the plant.

By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river ([tex]Q_{out}[/tex]), in megawatts:

[tex]Q_{out} = Q_{in} - W[/tex]

[tex]Q_{out} = \left(\frac{1}{\eta}-1 \right)\cdot W[/tex](1)

Where:

  • [tex]\eta[/tex] - Efficiency.
  • [tex]W[/tex] - Electric power, in megawatts.

If we know that [tex]\eta = 0.39[/tex] and [tex]W = 330\,MW[/tex], then the energy rate exhausted to the river is:

[tex]Q_{out} = \left(\frac{1}{0.39}-1 \right)\cdot (330\,MW)[/tex]

[tex]Q_{out} = 516.154\,MW[/tex]

516.154 megawatts of heat are exhausted to the river that cools the plant.

We kindly to check this question on first law of thermodynamics: https://brainly.com/question/3808473