Question

Common Data: 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 (%) of the cycle is

• A. $42.6$
• B. $40.9$
• C. $21$
• D. $59.7$

Answer 1

The correct option is C $21$


Process $2-3$
$Q_{2-3}=(U_3-U_2)+W_{2-3}$
$Q_{2-3}=(U_3-U_2)+0$
$=m{c}_v(T_3-T_2)$
$=m{c}_v(\frac{p_3{V}_3}{mR}-\frac{p_2{V}_2}{mR})$
$=\frac{c_v}{R}(p_3{V}_3-p_2{V}_2)$
$=\frac{c_v}{c_p-c_v}(400\times 1-100\times 1)$
$Q_{2-3}=\frac{1}{\gamma -1}(400-100)$
$=\frac{300}{1.4-1}=\frac{300}{0.4}=750kJ$
$Q_{1.2}=(U_2-U_1)+W_{1-2}$
$=m{c}_v[\frac{p_2{V}_2}{mR}-\frac{p_1{V}_1}{mR}]+p(V_2-V_1)$
$=\frac{c_v}{R}\left[p\right(V_2-V_1\left)\right]+p(V_2-V_1)$
$=p(V_2-V_1)\{\frac{c_v}{R}+1\}$
$=p(1-V_1)\{\frac{c_v}{c_p-c_v}+1\}[\because V_2=1m^3]$
$=100(1-V_1)\{\frac{1}{\gamma -1}+1\}$
Calculating $V_1$
$p_3{V}_3^{\gamma }=p_1{V}_1^{\gamma }$
$\therefore V_1={\left[\frac{p_3.V_3^{\gamma }}{p_1}\right]}^{\frac{1}{\gamma }}$
$={[\frac{400}{100}\times 1]}^{\frac{1}{1.4}}$
$Q_{1-2}=100[1-2.692]\times \{\frac{1}{0.4}+1\}$
$=-592.2kJ$
(minus sign represent heat is rejected). There is no transfer in process $3-1$ as it is reversible adiabatic proces.
Effeiency of cycle.
$\eta =1-\frac{Q_{reject}}{Q_{added}}=1-\frac{592.2}{750}$
$=0.2014=21.04$%