Efficiency of Two Carnot Engines in Series

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SUMMARY

The discussion centers on the efficiency of two Carnot engines operating in series, specifically how their efficiencies combine. The overall efficiency is determined solely by the temperatures of the heat reservoirs, represented by the formula ε = 1 - Tc/Th. The participant initially misinterpreted the series arrangement as yielding improved efficiency, mistakenly concluding that an infinite number of engines could achieve 100% efficiency. However, it was clarified that the efficiency remains constant regardless of the number of engines, as it is dependent only on the highest and lowest temperatures involved.

PREREQUISITES
  • Understanding of Carnot engines and their efficiency formula
  • Knowledge of thermodynamic temperature scales (Th and Tc)
  • Familiarity with the concept of heat transfer in thermodynamic systems
  • Basic algebra for manipulating equations
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  • Explore the implications of operating multiple Carnot engines in series
  • Investigate real-world applications of Carnot engines and their limitations
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Students of thermodynamics, engineers working on heat engine design, and anyone interested in the principles of energy efficiency in thermal systems.

Archduke
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Homework Statement



There are two Carnot engines in series, so that the heat exhaust from the first engine drives the second. Find the overall efficiency of the arrangement which contains only the efficiences of the individual engines.

Homework Equations



\epsilon = \frac{W}{Q_{input}}

The Attempt at a Solution



OK, I thought this was really easy, but it just doesn't seem right.

\epsilon = \frac{W_{1}+W_{2}}{Q_{1}} = \frac{(Q_{1} - Q_{2}) + (Q_{2} - Q_{3})}{Q_{1}} = 1 - \frac{Q_{3}}{Q_{1}}

But, surely that's saying that if you put two carnots in series that you get improved efficiency (As Q_{3} < Q_{2}), so if you put an infinite amount of carnot engines in series, you'll be able to have a 100% efficient heat engine!

So, I've either cured the world's energy problems :rolleyes:, or I've messed up somewhere! Any hints as to where I've gone wrong? Thanks!
 

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Archduke said:
\epsilon = \frac{W_{1}+W_{2}}{Q_{1}} = \frac{(Q_{1} - Q_{2}) + (Q_{2} - Q_{3})}{Q_{1}} = 1 - \frac{Q_{3}}{Q_{1}}

But, surely that's saying that if you put two carnots in series that you get improved efficiency (As Q_{3} < Q_{2}), so if you put an infinite amount of carnot engines in series, you'll be able to have a 100% efficient heat engine!
You have not made any error. The efficiency of a single Carnot heat engine operating between temperatures Th and Tc is 1-Qc/Qh = 1 - Tc/Th.

You have just shown that you get the same efficiency if you put two Carnot engines in series between these two temperatures, (or any number of Carnot engines between these temperatures). The overall efficiency depends only upon these temperatures (i.e the highest and lowest temperatures involved).

The Carnot efficiency approaches 100% as Th gets arbitrarily large or Tc gets arbitrarily close to 0. But I don't see why this raises a problem. It is the same whether there are one or a zillion Carnot engines operating over that temperature range.

AM
 
Oh, yes, of course! The output heat from Engine One running in series isn't the same as the output heat from the engine running alone. I see what I've done; thanks for clearing it up for me.
 

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