the cycle efficiency and mean effective pressure.

$p_1=0.1 MPa,T_1=35+273=308K,r=\frac{v_1}{v_2} =8:1,\text{heat supplied}=2100KJ/Kg$

Adiabatic compression process 1-2:

$\frac{T_2}{T_1} =\frac{v_1}{v_2}^(γ-1)=r^(γ-1)=8^(1.4-1)$

$T_2$=2.297×308=707.476K

$p_1 v_1^γ=p_2 v_2^γ$

$\frac{p_2}{p_1} =\frac{v_1}{v_2}^γ=8^1.4=18.379$

$p_2$=18.379×1=18.379 bar

Constant volume process 2-3:

Heat added during the process,

$c_v (T_3-T_2 )=2100$

$0.72(T_3-707.476)=2100 $

$T_3=3624.143K$

Also,

$\frac{p_2}{T_2} =\frac{p_3}{T_3}$

$p_3=\frac{p_2 T_3}{T_2}$

=$\frac{18.379×3624.143}{707.476}$

$p_3$=94.14 bar

Thermal efficiency,

$η_th=1-\frac{1}{r^(γ-1)}$

=$1-\frac{1}{8^(1.4-1)}$

$η_th$=.5674=56.74%

∴pressure ratio,$r_p$=$\frac{p_3}{p_2}$ =$\frac{94.14}{18.379}$=5.122

The mean effective pressure is given by,

$p_m=\frac{p_1 r[(r^(γ-1)-1)(r_p-1) ]}{(γ-1)(r-1)}$

=$\frac{(1×8[{8^(1.4-1)-1}{5.122-1} ]}{(1.4-1)(8-1)}$

$p_m$=15.28bar

1.243457723×10^-3 this is my answer

= (1×88(1.4−1)−1(5.122−1(8−1)/(1.4-1)(8-1)