An 800 MW electric power plant

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An 800 MW electric power plant operates at 30% efficiency, meaning that 30% of the energy input is converted to electricity, while the remaining 70% is lost as waste heat. To calculate the waste heat discharged per second, one must convert the electrical output into heat units, considering the efficiency. The discussion includes a light-hearted exchange about the meaning of efficiency and references to a nebula. The key focus remains on understanding the conversion of power units to determine waste heat. Ultimately, the calculation of waste heat is essential for assessing the plant's environmental impact.
Thankful_LV
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An 800 MW electric power plant has an efficiency of 30%. It loses its waste heat in large cooling towers. Approximately how much waste heat is discharged per second?"
 
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Thankful_LV said:

The Attempt at a Solution


Always nice to see this.



But if it's 30% efficient, what does that mean?
 
Welcome to PF!

Thankful_LV said:
An 800 MW electric power plant has an efficiency of 30%. It loses its waste heat in large cooling towers. Approximately how much waste heat is discharged per second?"

Hi Thankful_LV ! Welcome to PF! :smile:

Hint: MW are electric units.

How do you convert them into heat units? :smile:
 
Thankful_LV, don't be a zhirgo galvas, my dear Livonian friend!
 
Irid said:
… zhirgo galvas …

:smile: Isn't that the name of a beautiful nebula? :smile:
 
tiny-tim said:
:smile: Isn't that the name of a beautiful nebula? :smile:

Indeed it is! That's exactly what I had in mind.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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