I What Is the Connection Between Exergy Heat and Losses?

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Exergy heat, represented by the formula (1-Ta/Tc)*Q, indicates the maximum work obtainable from a heat transfer process between two reservoirs at different temperatures. This same expression can also reflect losses, as higher Tc values lead to greater heat loss to the atmosphere, thus reducing efficiency. The Carnot efficiency, defined by (1-Ta/Tc), quantifies the maximum fraction of absorbed heat that can be converted into work, while also suggesting the minimum expected losses. Therefore, while the formula highlights potential work, it simultaneously illustrates the inherent losses in the system. Understanding these concepts requires considering the entropy changes within the entire system.
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Hello !

I am having a bit of trouble understanding something about exergy.

On one hand, I read that (1-Ta/Tc)*Q (exergy heat) is the maximum work given a heat transfer and a reservoir's temperature.

But from the other hand, I read that this exact same (1-Ta/Tc)*Q represents losses. ( Which i can also understand, if Tc is high, then the losses to the atmosphere will be higher due to the higher Heat transfer).

Does the two affirmations are true ? How ?

Thank you !
 
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Ta represents the atmospheric temperature (cold reservoir) and Tc represents the higher reservoir temperature. The expression (1-Ta/Tc) is what is called the Carnot efficiency. It's value is between 0 and 1 and it represents the maximum fraction of the heat energy absorbed at the the higher temperature Tc that can be used to do work so the maximum work is as you have written it. It does in general give you an idea of the minimum loss you can expect given by Ta/Ta but your expression gives the maximum work not losses. If Tc is very high, you can expect to do more work since the engine can only lose heat to the cold reservoir(assumed to have constant Ta).

If you are interested, a full explanation (e.g. on wikipedia) requires considering the changes in entropy of the entire system.
 
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