Question

A thermodynamic cycle with an ideal gas as working fluid is shown below.


If the specific heats of the working fluid are constant and the value of specific heat ratio $\gamma$ is $1.4$, the thermal efficiency $\left(\text{in}\%\right)$ of the cycle is

• A. $21$
• B. $42$
• C. $2.1$
• D. $0.21$

Answer 1

The correct option is A $21$


For process $2-3$:
$Q_{2-3}=(U_3-U_2)+W_{2-3}...\left(1\right)$

Since, the volume is constant so work done, $W_{2-3}=0$. From $\left(1\right)$,

$Q_{2-3}=(U_3-U_2)+0$

$\Rightarrow Q_{2-3}=m{c}_v(T_3-T_2)$

From ideal gas equation, $PV=mRT$

$\Rightarrow Q_{2-3}=m{c}_v(\frac{p_3{V}_3}{mR}-\frac{p_2{V}_2}{mR})$

$\Rightarrow Q_{2-3}=\frac{c_v}{R}(p_3{V}_3-p_2{V}_2)$

Substituting, $c_p-c_v=R$ and the values of pressure and volume from $\text{P-V}$ diagram,

$Q_{2-3}=\frac{c_v}{c_p-c_v}(400\times 1-100\times 1)$

$\Rightarrow Q_{2-3}=\frac{1}{\gamma -1}(400-100)$

$\Rightarrow Q_{2-3}=\frac{300}{1.4-1}=750\text{kJ}$

Similarly for process $1-2:$

$Q_{1-2}=(U_2-U_1)+W_{1-2}$

$\Rightarrow Q_{1-2}=m{c}_v(T_2-T_1)+p(V_2-V_1)$

$\Rightarrow Q_{1-2}=m{c}_v[\frac{p_2{V}_2}{mR}-\frac{p_1{V}_1}{mR}]+p(V_2-V_1)$

$\Rightarrow Q_{1-2}=\frac{c_v}{R}[p(V_2-V_1)+p(V_2-V_1\left)\right]$

$\Rightarrow Q_{1-2}=(V_2-V_1)\{\frac{c_v}{R}+1\}$

$\Rightarrow Q_{1-2}=p(1-V_1)\{\frac{c_v}{c_p-c_v}+1\}(\because V_2=1m^3)$

$\Rightarrow Q_{1-2}=100(1-V_1)\{\frac{1}{\gamma -1}+1\}...\left(2\right)$

Now, calculating $V_1$ by using,

$p_3{V}_3^{\gamma }=p_1{V}_1^{\gamma }$

$\Rightarrow V_1={\left[\frac{p_3{V}_3^{\gamma }}{p_1}\right]}^{\frac{1}{\gamma }}$

Substituting the values from figure,

$\Rightarrow V_1={[\frac{400}{100}\times 1]}^{\frac{1}{1.4}}=2.692{\text{m}}^3$

Substituting the value of $V_1$ in $\left(1\right)$, we have

$Q_{1-2}=100(1-2.692)\times$ $\{\frac{1}{1.4-1}+1\}$

$\therefore Q_{1-2}=-592.2\text{kJ}$

Minus sign represent heat is rejected in this process.

There is no heat transfer in process $3-1$ as it is reversible adiabatic process.

So, efficiency of the cycle,

$\eta =1-\frac{Q_{reject}}{Q_{added}}$

$\Rightarrow \eta =1-\frac{Q_{1-2}}{Q_{2-3}}$

Substituting the values in the equation,

$\eta =1-\frac{592.2}{750}$

$\therefore \eta =0.2104=21.04\%$

Hence, option $\left(A\right)$ is the correct answer.