The temperature-entropy (T-S) diagram of a re | vedclass

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Question

(A) The area under the $T-S$ curve represents the heat exchanged in a process.For the given cycle, the heat absorbed $(Q_{in})$ is the area under the

Vedclass · Accepted Answer

$0.33$

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(A) The area under the $T-S$ curve represents the heat exchanged in a process.For the given cycle, the heat absorbed $(Q_{in})$ is the area under the top slanted line from $S_0$ to $2S_0$.The area is a rectangle plus a triangle: $Q_{in} = (T_0 	imes S_0) + \frac{1}{2} 	imes (2T_0 - T_0) 	imes (2S_0 - S_0) = T_0 S_0 + \frac{1}{2} T_0 S_0 = 1.5 T_0 S_0$.The heat rejected $(Q_{out})$ is the area under the bottom horizontal line from $2S_0$ to $S_0$ at temperature $T_0$: $Q_{out} = T_0 	imes (2S_0 - S_0) = T_0 S_0$.The efficiency $\eta$ is given by $\eta = 1 - \frac{Q_{out}}{Q_{in}}$.Substituting the values: $\eta = 1 - \frac{T_0 S_0}{1.5 T_0 S_0} = 1 - \frac{1}{1.5} = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.33$.

List-$I$	List-$II$
$A$. $T_1 = 500 \text{ K}, T_2 = 300 \text{ K}$	$i$. $0.2$
$B$. $T_1 = 500 \text{ K}, T_2 = 350 \text{ K}$	$ii$. $0.3$
$C$. $T_1 = 800 \text{ K}, T_2 = 400 \text{ K}$	$iii$. $0.4$
$D$. $T_1 = 450 \text{ K}, T_2 = 360 \text{ K}$	$iv$. $0.5$

The temperature-entropy $(T-S)$ diagram of a reversible engine cycle is given in the figure. Its efficiency is

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