This Wikipedia article states that:

Most steel engines have a thermodynamic limit of 37%.

Is it correct? If so then it is not clear where this number comes from. Is it derived somehow from Carnot's theorem? I tried to google it but found nothing usefull.

If this limit of 37% is wrong then what is correct limit?

  • 1
    $\begingroup$ I'd guess that the limit was indeed computed from the Carnot efficiency $e = 1-T_L/T_H$, where $T_H$ is governed by the maximum temperature at which the engine can operate before material strain becomes a problem, but perhaps someone else can comment on what appropriate $T_L$ and $T_H$ values are for a generic "steel engine." $\endgroup$ Feb 13, 2014 at 0:22
  • $\begingroup$ It's not much of an answer but the Wartsila RTA96-C claims 53% efficiency and their horsepower / fuel consumption per hour numbers seem to back that up. $\endgroup$
    – user88817
    Aug 9, 2015 at 17:08

3 Answers 3


I basically agree with Chase's answer - with some additions. But let's see thoroughly what is behind that sentence in Wikipedia ("Most steel engines have a thermodynamic limit of 37%").

Internal combustion engines are modeled by the Otto cycle rather than by the Carnot cycle. Looking at this link for the Otto cycle (or this great link from MIT), you may observe that in this cycle there are four temperatures involved, and the efficiency is given by

$$ \epsilon_O = 1-\frac{T_4- T_1}{T_3 - T_2}. $$

However, in an ideal Otto cycle we have that $T_4/T_1 = T_3/T_2$, and we get an "effective Carnot form" for the efficiency:

$$ \epsilon_O = 1-\frac{T_1}{T_2}. $$

Now, using the isentropic equations of ideal gases, we obtain the following simple expression

$$ \epsilon_O = 1- \frac{1}{(V_1/V_2)^{\gamma -1}}, $$

where $\gamma$ is the specific heat ratio ($c_p/c_v$)

At the end of the Wikipedia sentence that you quote, there is a reference to a course by the University of Washington (see link here). There it is stated that most current auto engines have compression ratio $V_1/V_2=10$ and the mixture of air, gasoline vapor, CO$_2$, CO and H$_2$O has an effective specific heat ratio of $\gamma=1.27$. Plugging this into our formula we get $\eta_O=0.46=46\% $, which is pretty close to what Chase said. I would call this the theoretical limit for the efficiency of the internal combustion engine!

But how do they get $35-37\%$ instead? Here I have to quote the Uni of Washington link directly:

(...) you get h = 0.46. Multiply this by about 0.75 to account for real cycle effects (such as the time it takes to burn, heat losses to the coolant, and exhaust valves that open before the piston fully reaches bottom position) and you have h = 0.35.

So in conclusion, after a careful look through the different links, I have to agree with Chase (with the above described additions), the efficiency limit dictated by thermodynamics is ~46%, and the 35% limit arrives when additional real-life considerations are taken into account. Although admittedly this "let's multiply by 0.75" approach does not sound very well grounded on its own. But at least it answers your question "where this number comes from".

  • 1
    $\begingroup$ This strange document claims that an IC engine is not subject to the Carnot limit at all. It claims that the IC engine operates on an "open cycle" rather than a "closed cycle" like Carnot, Otto, Diesel, etc. $\endgroup$
    – abalter
    Jul 20, 2014 at 5:47

The combustion temperature of gasoline is about 550K. So the Carnot limit for a gasoline-burning engine outdoors is roughly $$ \epsilon_C = 1-\frac{T_f}{T_i} = 1 - \frac{275 \mathrm{K}}{550 \mathrm{K}} =50\% $$

Here I've taken the output temperature to be near freezing, which is a rudimentary estimate at best (as dmckee points out). This provides us a rough upper limit estimate for the unattainable ideal efficiency.

Since this is much higher than the $37\%$ figure, we can infer that additional considerations have gone into the "thermodynamic limit". In particular, the details of the engine cycle will depend on the engine geometry, stroke pattern etc. Here's a cool website that shows a few P-V diagrams for real engines.

  • 5
    $\begingroup$ I don't think you are routinely getting down anywhere near 275 K for the low temperature limit...the gas comes out of the cylinders still hot. Don't touch the exhaust manifold if the engine is running or has been recently. $\endgroup$ Feb 13, 2014 at 3:07
  • 2
    $\begingroup$ @chase: actually, i followed carefully the thread how they get this 37% figure (see my answer). and even with the Otto cycle efficiency calculation (which actually should include dmckee's comment, as well - i.e., the temperature of the gas after and before the 1:10 compression is given implicitly there) one gets around 46% (so i give +1 to you). what they do then is to multiply by a 0.75 "real life factor" and get 35% that way :). (probably they should have explained in more detail this phenomenological 0.75 factor describing the additional real-life considerations). $\endgroup$ Feb 13, 2014 at 4:36
  • $\begingroup$ @dmckee absolutely agree. But this is just a very rough upper-limit calculation, plus the numbers work out very nicely ;) I'm folding the actual exhaust temperature, back pressure, etc, into the "additional considerations". My point here is that everything else will work to detract from the Carnot limit, which is the unattainable ideal. $\endgroup$
    – chase
    Feb 13, 2014 at 4:53
  • 1
    $\begingroup$ Can burning gasoline heat anythyng to temperature higher than its own combustion temperature? Maybe its a stupid question but I don't know. $\endgroup$
    – Simon
    Feb 14, 2014 at 17:03
  • 2
    $\begingroup$ @Simon yes gasoline routinely heats things up much higher than its combustion temperature. Combustion temperature is threshold for oxidation, but heat released by oxidation can easily heat things to many times that temperature. $\endgroup$
    – mwengler
    Jun 8, 2014 at 15:41

As stated above, the thermodynamic efficiency is based on the temperature of a hot source of energy and a cold sink for the energy after it is used to generate work. Since the temperature of the cold sink is ambient temperature, and can't be changed, the only way to increase the thermodynamic efficiency of an internal combustion engine is to increase the temperature of the hot source of energy. In effect, from a practical standpoint, get rid of the radiator, and build a ceramic engine that can withstand the full heat of combustion with no external cooling. This has been done, and it works as expected from a thermodynamic standpoint ... the thermodynamic efficiency (and hence the gas mileage) immediately goes up dramatically. Unfortunately, ceramics are brittle, difficult to work with, and don't do well when exposed to thermal shock, such as would be expected when the engine is first started and is warming up. So, the mechanical integrity of the ceramic engine is what keeps it from being implemented in practice. IF (and this is a BIG IF) you or someone else can find a way to build a ceramic engine that is also mechanically sound, for a price that is approximately equal to current engine prices, you will undoubtedly become a billionaire in short order.


Not the answer you're looking for? Browse other questions tagged or ask your own question.