First slide

Heat engine

Question

N moles of a monoatomic gas is carried round the reversible rectangular cycle ABCDA as shown in the fig. The temperature at A is T₀. The thermodynamic efficiency of the cycle is

Moderate

A

15%
B

50%
C

20%
D

25%

Solution

$\begin{array}{l} Q_{AB} = {NC}_{V} T_{0} \\ Q_{BC} = {NC}_{p} (2 T_{0}) \\ Q_{CD} = - {NC}_{V} (2 T_{0}) \\ Q_{DA} = - {NC}_{p} (T_{0}) \end{array}$
Heat input, $Q_{1} = Q_{AB} + Q_{BC} = {NT}_{0} (C_{V} + 2 C_{p})$
Heat lost, $Q_{2} = {NT}_{0} (2 C_{V} + C_{p})$
$\begin{array}{l} ∴ η = 1 - \frac{Q_{2}}{Q_{1}} = [1 - \frac{(2 C_{v} + C_{p})}{(C_{v} + 2 C_{p})}] \\ = [1 - \frac{(2 + γ)}{(1 + 2 γ)}] = [1 - \frac{(2 + 5 / 3)}{(1 + 2 \times 5 / 3)}] \\ = [1 - \frac{(11 / 3)}{(13 / 3)}] = \frac{2}{13} = \frac{2}{13} \times 100 \approx 15 % \end{array}$

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