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Can I make a heat pump beat the Carnot efficiency? Why is the Carnot process the most efficient one?

If I have a heatpump that is sphere shaped, and cascaded in layers like a onion can I beat Carnot efficiency? Heat would be transferred and concentrated from the outermost layer that is in contact with ambient air, and brought towards the center to heat a fluid.

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If we assume that the second law is true (which all patent examiners do) then it is not possible to beat the Carnot efficiency. – Fabian Mar 11 '13 at 19:36
i was going to make it out of peltier cells, 100 layers, each layer with a small delta-t. Ambient heat and high grade electrical energy end up in the centre of the device. – gary Mar 11 '13 at 20:00
I was going to ripple the current through each peltier cell in a method that turns the peltier cell into a near maxwells demon. – gary Mar 12 '13 at 18:09
I have to make a presentation to some venture capitalists about the device. – gary Mar 16 '13 at 14:41
it's a fake. Not even Maxwell's daemon can beat the second law. – Fabian Mar 16 '13 at 14:46


The laws of thermodynamics have nothing to do with specific engineering designs - they state properties of the time-evolution of certain observables of physical systems, independent of design. Nor are they empirical laws - they are derived rigorously from first principles that no one doubts.

You can no more beat the Carnot efficiency by clever construction than you can build a time machine if only you had enough gears. Just as no amount of pistons and lasers could ever change the rest mass of the electron, no amount of Peltier cells can undo physics.

Thousands of people have thought along similar lines - "maybe if I just tweak things, my design will work." When they actually do build their physically impossible device and claim it works, the fundamental flaw is always one of two varieties: (1) improper calorimetry misses sources or losses of energy, or (2) the temperature reservoirs are not really at the temperatures the inventors suspect. Without more detailed design specs, I can't say where your reasoning fails, but doing that analysis is beyond the scope of physics anyway.

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Due to the first and second law of thermodynamics: No.

The Carnot cylce is more efficient for a high temperature gradient of cold and hot reservoir. The efficiency of the Carnot cyle $\eta = 1 - \frac{T_{cold}}{T_{hot}}$ is the upper limit of efficiency.

E.g.: $T_{cold} = 600\,K$ and $T_{cold} = 300\,K$ yields a high efficency of 0.5. New 2008 Otto Motor techniques (Split cyle engine, german wiki) yield $\eta\approx 0.4$. The most efficient process would be to increase the hot Temperature und insolate versus the cold reservoir. Cooling $T_{cold}$ to near absolute zero Kelvin is harder and therefore not the way to engineer.

Remark: laws of thermodynamics

First law of thermodynamics: The increase in internal energy of a closed system is equal to the difference of the heat supplied to the system and the work done by it. Using it's mathematic form $dU = \delta Q + \delta W$ allows derivation of carnot efficiency $\eta$.

Second law links the loss of $Q$ to the increased entropy $S$ of the environment.

These laws both underline the carnot efficiency as a limit.

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"E.g.: $T_{cold} = 30\,°C$ and $T_{cold} = 300\,°C$ yields a high efficency of 0.9." No, it doesn't. You must express these temperatures on an absolute scale. $T_{cold} = 30\,\text{K}$ and $T_{cold} = 300\,\text{K}$ gets a 90% efficiency. As written your situation gets a bit less than 50%. – dmckee Mar 11 '13 at 21:57
Good point. The fraction demands Kelvin. – Stefan Bischof Mar 11 '13 at 22:02
A slight correction: it's not due to the "first principle" of thermodynamics but to the second law of thermodynamics. – Nathaniel Mar 12 '13 at 2:03
@Nathaniel Edited principle -> law. I was confused with the german "Erster Hauptsatz". Thanks. Untill now I wasn't aware of link Carnot <-> second law. Feel free to link the second law to the carnot efficiency. I was not capable of. – Stefan Bischof Mar 12 '13 at 7:23
@Stefan the Carnot limit follows very simply from the second law. Consider removing an amount of heat $Q$ from a reservoir at $T_H$. This decreases its entropy by $Q/T_H$. However, simultaneously you add a different amount of heat $Q'$ to the cold reservoir, which increases its entropy by $Q'/T_C$. You can extract $W=Q-Q'$ as work. The second law is obeyed if the total entropy increase $Q'/T_C-Q/T_H$ is positive, i.e. $(Q-W)/T_C-Q/T_H>0$, which you can rearrange to get $W<(1-T_C/T_H)Q$. – Nathaniel Mar 12 '13 at 10:42

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