Why does efficiency of heat engine not find by adding up two source value?

Question

Actually I have two questions.

we know the efficiency of heat engine equal {1-(t/T)},where t is sink temperature and T is source temperature.
Now, if two source and one sink reservior are connected by a heat engine(as shown in picture),then if T is taken as sum of two source temperature and that gives wrong answer.

From picture,
t=200 and T=800+400=1200 ,so efficiency is {1-(200/1200)} give wrong answer.
why do we need to find individually?
But also adding up two individual efficiency give wrong answer(100% up efficiency).
If i take heat (from picture Qa+Qb) and work(net work divided by total heat supplied)
instead of temperature then it gives me right answer.

why does the use of heat give me right answer?

Chemomechanics · Answer

Energy (including energy transport caused by a temperature difference, known as heat) is an extensive variable. In other words, if you duplicate a system and its input and output conditions, then all the energy terms will double. Linear superposition of energy fluxes of $dot{Q}_A$ and $dot{Q}_B$, if valid, will produce a flux of $dot{Q}_A+dot{Q}_B$.
In contrast, temperature is an intensive variable. If you duplicate a system at temperature $T$, the new temperature of the dual system is not $2T$ but generally remains $T$. Thus, the effective temperature of an 800 K reservoir and a 400 K reservoir used in conjunction is unlikely to be 1200 K.

Why does efficiency of heat engine not find by adding up two source value?

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