The air is expanded at constant pressure to 0.09 m3 a polytrophic process with n=1.5 is then carried out, followed by a constant temperature process which completes the cycle. All the process are reversible. Sketch the cycle on pressure-volume diagram and find the heat received and heat rejected in the cycle. Take R=0.287 KJ/KgK, Cr = 0.713 KJ/KgK.

Given: $P_1$=7 bar,$T_1$=206℃,$V_1$=0.03$m^3$,$V_2$=0.09$m^3$,n=1.5

To find mass of the air we will use the relation,

$P_1 V_1=mRT_1$ m=$\frac{(P_1 V_1)}{(RT_1 )}$=$\frac{(7×10^5×0.03)}{(287×479)}$=0.1527Kg

From

$P_2 V_2=mRT_2 T_2$=$\frac{(P_2 V_2)}{mR}$=$\frac{(7×10^5×0.09)}{(0.1527×287)}$=1437.53K

Also

$P_2 V_2^1.4=P_3 V_3^1.4$

And

$\frac{T_2}{T_3}$ =$\frac{p_2}{p_3 }^\frac{1.4-1}{1.4}$

But $T_3$=$T_1$ as 1 and 3 are on an isothermal line.

$\frac{1437.53}{479}$=$\frac{7}{p_3}^\frac{0.4}{1.4}$

$p_3$=0.581 bar

Now,

$p_3 V_3$=$mRT_3$

$0.581×10^5×V_3$=0.1527×287×479 …