Problem — Stirling engine

Problem

An ideal Stirling engine operates between $T_c = 500\;\text{K}$ and $T_f = 300\;\text{K}$ with two isotherms and two isochores. With $n = 0{,}1\;\text{mol}$ of monatomic ideal gas, $V_1 = 0{,}5\;\text{L}$, $V_2 = 1{,}5\;\text{L}$, calculate the ideal $\eta$ (with regenerator), the heat absorbed $Q_c$, and the work per cycle.

Solution

Ideal efficiency with regenerator. The two isochores exchange heat, but in an ideal Stirling engine the regenerator stores the heat rejected during the cooling isochore and returns it during the heating isochore: those two exchanges cancel internally. Only the two isotherms remain, exactly as in Carnot, so the efficiency reaches the Carnot limit:

$\eta = 1 - \frac{T_f}{T_c} = 1 - \frac{300}{500} = \ev{0{,}40 \ (40\%)}$

Heat absorbed $Q_c$. This is exchanged during the hot isothermal expansion at $T_c$. For an isothermal process of an ideal gas, $Q = L = nRT\ln(V_2/V_1)$:

$Q_c = nRT_c \ln\!\frac{V_2}{V_1} = 0{,}1 \times 8{,}31 \times 500 \times \ln\!\frac{1{,}5}{0{,}5}$

With $\ln 3 \approx 1{,}099$:

$Q_c = 415{,}5 \times 1{,}099 \approx \ev{457\;\text{J}}$

Work per cycle. From the efficiency, $L = \eta\, Q_c$:

$L = 0{,}40 \times 457\;\text{J} \approx \ev{183\;\text{J}}$

The ideal Stirling engine with regenerator thus matches Carnot ($40\%$). This is the reason for the recurring interest in the Stirling engine: an externally-heated, silent cycle, in principle as efficient as the perfect theoretical cycle. The practical difficulties of the real regenerator, however, limit its actual output.

Collegamenti

Argomenti: Macchine termiche Concetti: Rendimento · Trasformazioni termodinamiche · Legge dei gas perfetti Competenze: Risoluzione di un ciclo termodinamico Metodi: Diagramma P-V e ln P - ln V Oggetti: Gas ideale · Macchina termica