Question

An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to __________.

Answer 1

The above questions can be solved as well:

Step 1: percentage of the efficiency,$\eta =(W/Q)\times 100$

Step 2: Work $=$Area of $\mathrm{ABCD}$

$=\left(2P_{\mathit{0}}\right)/\left(V_{\mathit{0}}\right)$

Step 3: Heat $=Q_{AB}+Q_{BC}$(input)

Step 4:

${\mathrm{Q}}_{\mathrm{AB}}=$isochoric process

$$\begin{gathered} Q_{AB}=n{C}_V\Delta T\mathit{}\therefore PV=nRT \\ \mathit{}\mathit{}\mathit{}\Rightarrow n{C}_v(T_B–T_A)T=PV/nR \\ \mathit{}\mathit{}\mathit{}\mathit{}{Q}_{AB}=1\times (3/2)k(3P_{\mathit{0}}{V}_{\mathit{0}}–P_{\mathit{0}}{V}_{\mathit{0}})\Rightarrow 3P_{\mathit{0}}{V}_{\mathit{0}} \end{gathered}$$

$Q_{BC}=$isobaric process

$$\begin{gathered} \mathit{}{Q}_{BC}\mathit{}=n{C}_p\Delta T \\ \mathit{}\mathit{}\mathit{}\mathit{}\Rightarrow 1\times (5/2)k(6P_{\mathit{0}}{V}_{\mathit{0}}–3P_{\mathit{0}}{V}_{\mathit{0}})\Rightarrow 7.5P_{\mathit{0}}{V}_{\mathit{0}} \end{gathered}$$

Step 5: Finally, on calculating efficiency

$$\begin{gathered} \mathit{}\eta \mathit{}\mathit{=}\mathit{}\mathit{(}W\mathit{/}Q\mathit{)}\mathit{\times }\mathit{100} \\ \mathit{}\eta \mathit{}=2P_{\mathit{0}}{V}_{\mathit{0}}/[3P_{\mathit{0}}{V}_{\mathit{0}}+7.5P_{\mathit{0}}{V}_{\mathit{0}}]\times 100 \\ \eta \mathit{}\mathit{}\approx 19\% \end{gathered}$$

Therefore, the percentage efficiency of the engine close to $19\%$