Problem — Ranking four Carnot engines

Problem

Four heat engines operate between a hot source $T_c$ and a cold one $T_f$; all are ideal Carnot engines. Rank the efficiencies $\eta = 1 - T_f/T_c$ from smallest to largest.

• (A) $T_c = 600\;\text{K}$, $T_f = 300\;\text{K}$
• (B) $T_c = 1000\;\text{K}$, $T_f = 400\;\text{K}$
• (C) $T_c = 2000\;\text{K}$, $T_f = 300\;\text{K}$
• (D) $T_c = 400\;\text{K}$, $T_f = 300\;\text{K}$

Solution

Formula. For every Carnot engine $\eta = 1 - T_f/T_c$. Let’s compute case by case:

$\eta_A = 1 - \frac{300}{600} = 0{,}50 \qquad \eta_B = 1 - \frac{400}{1000} = 0{,}60$

$\eta_C = 1 - \frac{300}{2000} = 0{,}85 \qquad \eta_D = 1 - \frac{300}{400} = 0{,}25$

Ranking (from smallest to largest):

$0{,}25 < 0{,}50 < 0{,}60 < 0{,}85 \quad\Rightarrow\quad \ev{\eta_D < \eta_A < \eta_B < \eta_C}$

Efficiency grows the smaller the ratio $T_f/T_c$ is: C wins because its hot source is much higher than the cold one; D loses because the two temperatures are almost equal.

Links

Topics: Macchine termiche Concepts: Ciclo di Carnot · Rendimento Skills: Ragionamento di ranking Objects: Macchina termica