How can exergy heat be both maximum work and losses?

[PHstud]

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 !

 

[Mgcini Keith Phuthi]

T_a represents the atmospheric temperature (cold reservoir) and T_c represents the higher reservoir temperature. The expression (1-T_a/T_c) 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 T_c 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 T_a/T_a 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.