Which one's better?

Classical Mechanics Level 3

Heat engines are cyclic devices which produce work by exchanging heat with two thermal reservoirs. Thermal reservoirs are bodies having high heat capacity, so even when they gain or lose heat their temperature remains constant.

There are two heat engines A and B .

A works between two thermal reservoirs having temperatures T H = 600 K { T }_{ H }=600K , T L = 300 K { T }_{ L }=300K .

B works between two thermal reservoirs having temperatures T H = 1000 K { T }_{ H }=1000K , T L = 400 K { T }_{ L }=400K .

Subscripts H H and L L indicate high temperature reservoir and low temperature reservoir respectively.

It was found out that both heat engines, when working, have the same thermal efficiency of 30 30 %. Thermal efficiency of a heat engine is the ratio of work output to the heat input.

You are an engineer and you have been sent by your company to review the performances of the two heat engines and replace only that heat engine which is working poorly out of the two. What do you report to your superiors?

Guess what, I Resign! Replace A Don't replace anything . Both are equally good Replace B

Rahul Badenkal by Rahul Badenkal , 26, India

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1 solution

Rahul Badenkal
Jul 22, 2015

The highest efficiency a heat engine can have while operating between two reservoirs is η m a x = 1 − T L T H { \eta }_{ max }=1-\frac { { T }_{ L } }{ { T }_{ H } } . And such a heat engine having this maximum efficiency is known as Reversible heat engine. The simplest example of a reversible heat engine is Carnot engine.

1.) Therefore the maximum efficiency heat engine A can have is:

η m a x = 1 − 300 600 { \eta }_{ max }=1-\frac { 300 }{ 600 }

η m a x = 0.5 o r 50 { \eta }_{ max }=0.5\quad or\quad 50 %

But because of irreversibilities it is only able to give 30 30 %. So it is only able to achieve 30 50 × 100 = 60 \frac { 30 }{ 50 } \times 100=60 % of it's potential.

2.) The maximum efficiency heat engine B can have is:

η m a x = 1 − 400 1000 { \eta }_{ max }=1-\frac { 400 }{ 1000 }

η m a x = 0.6 o r 60 { \eta }_{ max }=0.6\quad or\quad 60 %

But because of irreversibilities it is only able to give 30 30 %. So it is only able to achieve 30 60 × 100 = 50 \frac { 30 }{ 60 } \times 100=50 % of it's potential.

Since A's actual efficiency is more closer to it's ideal efficiency than B, we can say that A is performing better than B or we should replace B.

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