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Q & A: Ideal Engines

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Most recent answer: 01/14/2013
Q:
We cannot make a heat engine with 100% efficiency, but can we make one with 99% efficiency?
- Yaju Arya (age 15)
Dayanand Model School, Jalandhar, Punjab, India
A:
Heat engines are governed by the laws of thermodynamics. If you work through the math of it, you get the the maximum efficiency depends on the temperatures of the hot and cold reservoir. We show that the maximum possible efficiency = 1- Tc/Th where Tc and Th are the reservoir temperatures in kelvin, so in principle we can get any efficiency we like (as long as its less than 100%) just by picking Th and Tc to be very different.

The problem is that in order to get this maximum efficiency things generally must happen very slowly. [The reason for this is that at some points in the cycle heat flows from a hotter region to a colder one due to the temperature difference. However, maximum efficiency is only reached as that temperature difference becomes small, which makes the heat flow slow./ mw] 

Adam

Here's another problem. On earth, the waste heat will be dumped somewhere with a temperature near TC= 300K. To get efficiency of 99%, even in slow operation, would then require TH > 30,000K. There just aren't any materials that can withstand that sort of temperature.  Mike W.

(published on 10/22/2007)

Follow-Up #1: Carnot efficiency

Q:
I thought for reasons I can't easily recall that the maximum efficiency of any heat engine could only be 40% called a Carnot engine.
- Devon (age 26)
Lansing
A:
That 1-Tc/Th that Adam wrote about is the Carnot efficiency. The reasons engine efficiencies don't exceed 40% in most typical applications are the ones we mentioned. Say the heat is dumped to a typical ambient environment  at Tc=300K. To get Carnot efficiency above 40%, you need for Th to be above 500K. That's perfectly possible, but you can't get a whole lot hotter than that without running out of suitable materials. If you then remember that the real efficiency is always less than the Carnot efficiency, both due to friction and the heat flow problem we mentioned above, it's rare to find examples much better than 40%.

I'm surprised, given the extraordinary sophistication of many of your questions, that you aren't familiar with the reason for the Carnot limit. As heat Q flows out of the hot reservoir, its entropy S goes down by Q/Th. (That's by definition of T.)  According to the second law of thermodynamics, no process decreases net S. So there must be a heat flow to the cold reservoir of at least (Q/Th)Tc. That's energy not available to do work. So of the Q energy drawn from the hot source, what's left to do work is at most Q(1-Tc/Th).

Mike W.

(published on 01/14/2013)

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