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

$V_3=0.3613 m^3$

i) The heat received in the cycle:

Applying firs law to the constant pressure process 1-2,

Q=∆U+W

W=$∫_1^2pdV$

=p($V_2-V_1 $)

=7×10^5 (0.09-0.03)

=42000J or 42 KJ

Q=$mC_v$ ($T_2-T_1 $)+42

=0.1527×0.713(1437.53-479)+42

=146.36KJ

i.e., Heat received= 146.36kJ

Applying first law to reversible polytropic process 2-3

Q=∆U+W

But W=$\frac{p_2 V_2-p_1 V_1}{(n-1)}$

=mR$\frac{(T_2-T_3 )}{(n-1)}$

=$\frac{(0.1527×287×(1437.53-479))}{(1.4-1)}$

=105.01KJ

$Q=mC_v (T_3-T_2 )$

=0.1527×0.713×(479-1437.53)+105.01

=0.6499KJ

Therefore total heat received in the cycle=146.36+0.6499=147KJ ii) The heat rejected in the cycle:

applying first law to reversible isothermal process 3-1 Q=∆U+W

W=$p_3 V_3$ log_e $\frac{V_1}{V_3}$

=0.581×0.3613 log_e $\frac{0.03}{0.3613}$

=-0.52 KJ
Q=$mC_v (T_1-T_3 )+W$

=0-0.52

=-0.52KJ

Hence heat rejected in the cycle = 0.52KJ