Bernoulli's principle states that an increase in the kinetic energy of a fluid occurs simultaneously with a decrease in pressure for isentropic processes. What then limits us from converting all of the gas's pressure into kinetic energy? Could not this kinetic energy then be used to perform useful work (e.g. move turbine)? This appears contrary to Carnot efficiency but I'm failing to see a very simple limit on what prevents us from being able to continually transform the pressure into work. Cannot a nozzle or similar device convert all this pressure into directed kinetic energy?

The one simple limit I can think of is shock transitions where the mach number goes from greater than 1 to below 1 (i.e. not an isentropic process). But since we're only going to higher and higher mach numbers by decreasing pressure and increasing velocity, we should be able to avoid this limit. Therefore, is it not possible to isentropically decrease pressure arbitrarily low at least until quantum mechanical effects become important?

Perhaps a better way to define my question is through a thermodynamic cycle:

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One can define the enthalpies at each position of a thermodynamic cycle (i.e. $h_1, h_2, h_3, h_4$) and the overall thermal efficiency as follows:

$\eta_{th} = \frac {W_{out} - W_{in}}{Q_{in}} = \frac {(h_2 - h_3) - (h_1 - h_4)}{h_2 - h_1} = \frac {\eta_N h_2 - (h_1 - h_4)}{h_2 - h_1} $

where $\eta_N = \frac{h_2 - h_3}{h_2}$ is the enthalpy extraction ratio. We know that $\eta_{th}$ cannot exceed $1 - T_{cold}/T_{hot}$ by Carnot efficiency, but can we make any generalized statements about the relationship between $\eta_N$ and $\eta_{th}$? For example, must $\eta_N$ always be less than $\eta_{th}$? Must $\eta_N$ always be greater than $\eta_{th}$? Or can no general statement be made?

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    $\begingroup$ We don't need to invoke the intermediate step of converting to kinetic energy first to run a turbine. We typically convert the compressional energy of the high-pressure gas directly to work in a turbine. The decrease In enthalpy of the gas at nearly constant entropy is equal to the shaft work delivered by the turbine. $\endgroup$ Commented Dec 12, 2018 at 22:36
  • $\begingroup$ @ChesterMiller Is not the compressional energy equivalent to the total pressure? I agree that isentropic decrease in enthalpy is equal to the work delivered to the turbine. But is there any limit on this decrease in enthalpy? For example, can a gas of enthalpy, $H$, lose all of its enthalpy in a turbine and have this all converted into work? $\endgroup$
    – Mathews24
    Commented Dec 12, 2018 at 23:12
  • $\begingroup$ Well, of course the gas is also cooling. It would be undesirable to condense too much vapor to liquid in the turbine. So this would certainly be one limitation. Another limitation would be: once the pressure becomes pretty low, you can’t get much more work out of it. $\endgroup$ Commented Dec 12, 2018 at 23:20
  • $\begingroup$ @ChesterMiller I've tried to clarify my question; I am essentially seeking an intuitive understanding and mathematical statement of what limits enthalpy extraction, $\eta_N$, compared to a fixed thermal efficiency, $\eta_{th}$ (i.e. fixed $T_{cold}$ and $T_{hot}$). $\endgroup$
    – Mathews24
    Commented Dec 13, 2018 at 0:03
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    $\begingroup$ Sorry. I don't know what else I can add. $\endgroup$ Commented Dec 13, 2018 at 4:06

1 Answer 1


Could not this kinetic energy then be used to perform useful work (e.g. move turbine)?

Yes, and that's exactly how turbines work.

But when we operate as a cycle, it turns out there is a limit to how much net work can be produced. Think about it: at the exit of the turbine the gas is cold and at low pressure. To complete the cycle we must compress the gas and also heat it. By the time we return the gas to its initial state we will have consumed some work. The Carnot cycle tells us what's the maximum net work we can extract from this process.


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