Calculating the Efficiency of an Ericsson Cycle
An Ericsson cycle with air as the working fluid is operating between a low temperature of 70°C and a high temperature of 450°C, a low pressure of 2 bar, and a pressure compression ratio of 8 (so that the high pressure is 16 bar). Assume that at these conditions air can be considered to be an ideal gas with C_{ P }^{*} constant. Compute the properties at each point in each of these cycles and the cycle efficiencies.
We first compute the energy flows in each step of the process. In this analysis the process is at steady state so that all the time derivatives are equal to zero and the mass flow rate \dot{M} (actually we will use the molar flow rate \dot{N} ) is constant throughout the process.
1 → 2 Through the regenerative heat exchanger
Energy balance:
2 → 3 Through the isothermal turbine
Energy balance:
Entropy balance:
0=\dot{N} \underline{S}_{2}-\dot{N} \underline{S}_{3}+\frac{\dot{Q}_{ T }}{T}or
\frac{\dot{Q}_{ T }}{\dot{N}}=T_{2}\left(\underline{S}_{3}-\underline{S}_{2}\right)=T_{2}\left[C_{ P }^{*} \ln \frac{T_{3}}{T_{2}}-R \ln \frac{P_{3}}{P_{2}}\right]=-R T_{2} \ln \frac{P_{3}}{P_{2}} \equiv-R T_{2} \ln K_{ T }where K_{ T }=P_{3} / P_{2} is the compression ratio of the turbine. Consequently,
\frac{\dot{W}_{23}}{\dot{N}}=\underline{H}_{3}-\underline{H}_{2}-\frac{\dot{Q}_{ T }}{\dot{N}}=C_{ P }^{*}\left(T_{3}-T_{2}\right)+R T_{2} \ln K_{ T }=R T_{2} \ln K_{ T } 3 → 4 Through the regenerative heat exchanger
Energy balance:
However,
\dot{Q}_{12}=-\dot{Q}_{34}which implies \text { (since } C_{ P }^{*} \text { is constant) that } T_{2}-T_{1}=T_{3}-T_{4}. Also, since both the compressor and turbine are isothermal \text { (so that } T_{2}=T_{3} \text { and } T_{4}=T_{1} \text { ) }, the temperature decrease of stream 1 exactly equals the temperature increase of stream 3 for this case of an ideal gas of constant heat capacity.
4 → 1 Isothermal compressor
As for the isothermal turbine, we find
The efficiency of this cycle is
\eta=\frac{-\dot{W}_{\text {out }}}{\dot{Q}_{ T }}=\frac{-\left(\dot{W}_{23}+\dot{W}_{41}\right)}{\dot{Q}_{ T }}=\frac{\left[R T_{2} \ln \frac{P_{3}}{P_{2}}+R T_{1} \ln \frac{P_{1}}{P_{4}}\right]}{R T_{2} \ln \frac{P_{3}}{P_{2}}}Since each of the streams in the heat exchanger operates at constant pressure, it follows that P_{3}=P_{4} \text { and } P_{2}=P_{1}. Therefore,
</span>\eta=\frac{\left[R T_{2} \ln \frac{P_{3}}{P_{2}}+R T_{1} \ln \frac{P_{1}}{P_{3}}\right]}{R T_{2} \ln \frac{P_{3}}{P_{2}}}=\frac{T_{2}-T_{1}}{T_{2}}<span class="fontstyle0">
As T_{2} \text { is the high temperature of the cycle and } T_{1} is the low temperature, this efficiency is exactly equal to the Carnot efficiency. (Indeed, it can be shown that for any cycle, if the two heat transfers to the surroundings occur at different but constant temperatures, as is the case here with an isothermal compressor and an isothermal turbine, the efficiency of the cycle will be that of a Carnot cycle.)
Therefore, the efficiency of the Ericsson cycle considered here is
\eta=\frac{(450+273.15)-(70+273.15)}{(450+273.15)}=0.525Note that the efficiency is dependent on the temperatures of the various parts of the cycle, but not on the pressures or the compression ratio. However, the pressure ratio does affect the heat and work flows in each part of the cycle.