D.60 Analysis of the ideal gas Carnot cycle

Refer to figure D.4 for the physical device to be analyzed. The refrigerant circulating through the device is an ideal gas with constant specific heats, like a thin gas of helium atoms. Chapter 11.14 will examine ideal gases in detail, but for now some reminders from introductory classical physics classes about ideal gasses must do. The internal energy of the gas is $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mIC_VT$ where $mI$ is its mass and $C_v$ is a constant for a gas like helium whose atoms only have translational kinetic energy. Also, the ideal gas law says that $PV$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mIRT$, where $P$ is the pressure, $V$ the volume, and the constant $R$ is the gas constant, equal to the universal gas constant divided by the molar mass.

The differential version of the first law, energy conservation, (11.11), says that

\begin{displaymath}
{\rm d}E= \delta Q - P{\,\rm d}V
\end{displaymath}

or getting rid of internal energy and pressure using the given expressions,

\begin{displaymath}
mIC_v {\,\rm d}T = \delta Q - mIRT \frac{{\rm d}V}{V}.
\end{displaymath}

Figure D.4: Schematic of the Carnot refrigeration cycle.
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Now for the transitions through the heat exchangers, from 1 to 2 or from 3 to 4 in figure D.4, the temperature is approximated to be constant. The first law above can then be integrated to give the heat added to the substance as:

\begin{displaymath}
Q_{\rm {L}}= mIRT_{\rm {L}} \left(\ln V_2 - \ln V_1\right)...
...
Q_{\rm {H}}=-mIRT_{\rm {H}} \left(\ln V_4 - \ln V_3\right).
\end{displaymath}

Remember that unlike $Q_{\rm {L}}$, $Q_{\rm {H}}$ is taken positive if it comes out of the substance.

On the other hand, for the transitions through the adiabatic turbine and compressor, the heat $\delta{Q}$ added is zero. Then the first law can be divided through by $T$ and integrated to give

\begin{displaymath}
mI C_v \left(\ln T_{\rm {H}} - \ln T_{\rm {L}}\right)
= -mIR \left(\ln V_3 - \ln V_2\right)
\end{displaymath}


\begin{displaymath}
mI C_v \left(\ln T_{\rm {L}} - \ln T_{\rm {H}}\right)
= -mIR \left(\ln V_1 - \ln V_4\right)
\end{displaymath}

Adding these two expressions shows that

\begin{displaymath}
\ln V_3 - \ln V_2 + \ln V_1 - \ln V_4 = 0
\quad \Longrightarrow\quad
\ln V_3 - \ln V_4 = \ln V_2 - \ln V_1
\end{displaymath}

and plugging that into the expressions for the exchanged heats shows that $Q_{\rm {H}}$$\raisebox{.5pt}{$/$}$$T_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Q_{\rm {L}}$$\raisebox{.5pt}{$/$}$$T_{\rm {L}}$.