Carnot engine – a realistic version

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.6.

The main problem with an engine that follows a Carnot cycle is that the two isothermal stages in the cycle proceed very slowly, since we are attempting to transfer heat between two systems that are almost at the same temperature. One way of making the cycle go a bit faster is to make the temperature of the working substance significantly different from that of the reservoir where it absorbs, and later expels, heat. That is, if the system absorbs heat {Q_{h}} from a hot reservoir at temperature {T_{h}}, then the temperature of the working substance (typically a gas) when it absorbs heat is {T_{hw}<T_{h}}. Similarly, at the other isothermal stage where heat {Q_{c}} is expelled to the cold reservoir at temperature {T_{c}}, the temperature of the gas is {T_{cw}>T_{c}}.

To make things simple, we’ll assume that the rate of heat transfer is the same at both the hot and cold reservoirs, and is proportional to the temperature difference between the gas and the reservoir. That is

\displaystyle \frac{Q_{h}}{\Delta t} \displaystyle = \displaystyle K\left(T_{h}-T_{hw}\right)\ \ \ \ \ (1)
\displaystyle \frac{Q_{c}}{\Delta t} \displaystyle = \displaystyle K\left(T_{cw}-T_{c}\right) \ \ \ \ \ (2)

where {K} is a constant and {\Delta t} is taken to be the same for both cases (that is, the durations of both isothermal stages in the cycle are the same). From this, we get the relation

\displaystyle \frac{Q_{h}}{T_{h}-T_{hw}}=\frac{Q_{c}}{T_{cw}-T_{c}} \ \ \ \ \ (3)

 

If the only entropy that is created in the cycle is along the two isothermal stages (no entropy is generated along the adiabatic stages) then, since the state of the engine is the same at the end of the cycle as it was at the start, the gas must have expelled exactly the same amount of entropy when expelling heat to the cold reservoir as it absorbed when absorbing heat from the hot reservoir. That is

\displaystyle \frac{Q_{h}}{T_{hw}}=\frac{Q_{c}}{T_{cw}} \ \ \ \ \ (4)

 

so

\displaystyle Q_{c}=Q_{h}\frac{T_{cw}}{T_{hw}} \ \ \ \ \ (5)

Combining this with 3 gives

\displaystyle \frac{1}{T_{h}-T_{hw}} \displaystyle = \displaystyle \frac{T_{cw}}{T_{hw}\left(T_{cw}-T_{c}\right)}\ \ \ \ \ (6)
\displaystyle T_{cw} \displaystyle = \displaystyle \frac{T_{c}T_{hw}}{2T_{hw}-T_{h}} \ \ \ \ \ (7)

If the time required for the two adiabatic steps is much less than that for the two isothermal steps, we can work out the power output of the engine. The work is produced over a time interval of {2\Delta t} and is

\displaystyle \mathcal{P}=\frac{W}{2\Delta t} \displaystyle = \displaystyle \frac{1}{2\Delta t}\left(Q_{h}-Q_{c}\right)\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle \frac{K}{2}\left(T_{h}+T_{c}-T_{hw}-T_{cw}\right)\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle \frac{K}{2}\left(T_{h}+T_{c}-T_{hw}-\frac{T_{c}T_{hw}}{2T_{hw}-T_{h}}\right) \ \ \ \ \ (10)

We can maximize the power output for given values of {T_{h}} and {T_{c}} by varying {T_{hw}}. Taking the derivative we get

\displaystyle \frac{d\mathcal{P}}{dT_{hw}}=\frac{K}{2}\left[-1-\frac{T_{c}}{2T_{hw}-T_{h}}+\frac{2T_{c}T_{hw}}{\left(2T_{hw}-T_{h}\right)^{2}}\right]=0 \ \ \ \ \ (11)

This can be solved for {T_{hw}} by multiplying through by {\left(2T_{hw}-T_{h}\right)^{2}} and expanding the terms in the numerator. This results in

\displaystyle \frac{K\left(-4T_{hw}^{2}+4T_{h}T_{hw}+T_{c}T_{h}-T_{h}^{2}\right)}{2\left(2T_{hw}-T_{h}\right)^{2}}=0 \ \ \ \ \ (12)

Solving the quadratic equation and taking the positive root gives

\displaystyle T_{hw}=\frac{1}{2}\left(T_{h}+\sqrt{T_{h}T_{c}}\right) \ \ \ \ \ (13)

Substituting this into 7 gives

\displaystyle T_{cw}=\frac{1}{2}\left(T_{c}+\sqrt{T_{h}T_{c}}\right) \ \ \ \ \ (14)

To find the efficiency we have, using 4

\displaystyle e \displaystyle = \displaystyle 1-\frac{Q_{c}}{Q_{h}}\ \ \ \ \ (15)
\displaystyle \displaystyle = \displaystyle 1-\frac{T_{cw}}{T_{hw}}\ \ \ \ \ (16)
\displaystyle \displaystyle = \displaystyle 1-\frac{T_{c}+\sqrt{T_{h}T_{c}}}{T_{h}+\sqrt{T_{h}T_{c}}}\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle 1-\frac{\left(T_{c}+\sqrt{T_{h}T_{c}}\right)\left(T_{h}-\sqrt{T_{h}T_{c}}\right)}{T_{h}^{2}-T_{h}T_{c}}\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle 1-\frac{T_{h}T_{c}+\left(T_{h}-T_{c}\right)\sqrt{T_{h}T_{c}}-T_{h}T_{c}}{T_{h}\left(T_{h}-T_{c}\right)}\ \ \ \ \ (19)
\displaystyle \displaystyle = \displaystyle 1-\sqrt{\frac{T_{c}}{T_{h}}} \ \ \ \ \ (20)

For a coal-fired steam turbine with {T_{h}=600^{\circ}\mbox{ C}=873\mbox{ K}} and {T_{c}=25^{\circ}\mbox{ C}=298\mbox{ K}}, this gives an efficiency of

\displaystyle e=0.416 \ \ \ \ \ (21)

This is very close to the actual efficiency of about 40% for a real coal-burning power plant. The ‘ideal’ Carnot efficiency for these temperatures is

\displaystyle e=1-\frac{T_{c}}{T_{h}}=0.659 \ \ \ \ \ (22)

4 thoughts on “Carnot engine – a realistic version

  1. Nikos

    Below eq. (7) it is stated: “The work is produced over a time interval of 2Δt”. Why is that? In an increment of time Δt the engine produces W = Qh-Qc. Why are we saying that in this time increment the work is W/2 instead of W? What am I missing?

    Reply
    1. gwrowe Post author

      {\Delta t} is defined after eqn 2 as the time for each isothermal stage, and we’re assuming that the two adiabatic stages are much shorter than the isothermals, so the total time for one cycle is {2\Delta t}.

      Reply
      1. Nikos

        It actually makes sense also from the point of view of spending half of the work to transfer the heat from the hot reservoir to the working substance and half of it for expelling the heat to the cold reservoir during that Δt.

        Reply

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