Efficiency of Two Carnot Engines in Series

AI Thread Summary
The discussion revolves around the efficiency of two Carnot engines operating in series, where the heat exhaust from the first engine powers the second. The overall efficiency is derived from the individual efficiencies of the engines, leading to a formula that suggests improved efficiency with multiple engines. However, it is clarified that the overall efficiency is solely dependent on the highest and lowest temperatures of the system, not on the number of engines. As the temperatures approach certain limits, the Carnot efficiency can approach 100%, but this does not imply that adding more engines will yield infinite efficiency. The misunderstanding is resolved by recognizing that the output heat from the first engine is not the same as if it were operating alone.
Archduke
Messages
59
Reaction score
1

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!
 

Attachments

  • Carnot in Series.PNG
    Carnot in Series.PNG
    3.6 KB · Views: 2,355
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top