The figure below is based on material presented in an MIT OpenCoursware program on thermodynamics on the web. They were comparing a Carnot two-phase power cycle to a Rankine cycle with superheating (only the Carnot cycle shown here).

I have issues with the isentropic compression process $a-b$ in diagram. In typical depictions of this cycle this process is shown in the two-phase region where the vapor phase is what is being compressed from the condenser temperature into saturated liquid at the boiler temperature.

In this diagram, however, the isentropic compression starts with saturated liquid at the condenser temperature and ends with liquid at the boiler temperature. This doesn’t seem possible. I would think compressing water isentropically should have an insignificant effect on its temperature. For example, isentropically compressing saturated liquid water from 10 KPa to liquid at 10 MPa increases the temperature only about 5 C (based on the properties of compressed liquid water).

Is process $a-b$ possible, and, if so, how? enter image description here


1 Answer 1


Process a-b corresponds to an isentropic (constant $s$) increase of $T$.

For an incompressible liquid, $T$ and $s$ are directly related ($\text{d}s = c_p \text{d}T/T$), so there is no way of changing one without the other and process a-b is impossible.

For a real liquid, you could theoretically add heat (which tends to increase $s$) while simultaneously compressing (which tends to decrease $s$), but it would be impractical - the pressure would get unmanageably high really quickly.

  • $\begingroup$ You have confirmed what I believed to be the case. This is why I felt process a-b should have been in the two-phase region. The increase in temperature increases s but the compression of the vapor decreases s by the same amount. Looks like MIT got this one wrong. Thanks. $\endgroup$
    – Bob D
    Oct 11, 2018 at 13:59

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