The temperature and pressure at the end of suction stroke are 57°C and 1 bar respectively. Determine the maximum pressure in the cycle. The adiabatic index of compression is 1.36. The variation of specific heat at constant volume heat addition with respect to temperature is expressed as $C_v=0.7117+2.1×10^{-4}T.$ If the $C_v$ remains constant at 0.7117kJ/kgK, together with adiabatic index, what would be the changes in the maximum pressure?

$$(p_2 v_2)^n= (p_1 v_1)^n \hspace{1cm} ∴ p_2=1(10)^{1.36} =22.91 bar$$

Now $\frac{p1v1}{T1} = \frac{p2v2}{T2} \hspace{1cm} ∴T2=(57+273)×22.91× \frac{1}{10}=756.03 K$

Average temperature during combustion of charge = (T3+T2)/2

Mean specific heat of product during combustion $Cv_{mean} = 0.7117+2.1x10^{-4} [(T3+T2)/2]$

Assume air in cylinder is 1 kg

Heat added per kg of air= 48000/15 = 3200 kJ

Mass of charge per kg of air= 1 + 1/15 = 16/15 kg

Q = Cv x mass of charge x (T3-T2)

$3200 = \Big[0.7117+2.1x10^{-4} \Big(\frac{T3+756.03}{2}\Big)\Big] ×\frac{16}{15}×[T3-756.03]$

Solving we get T3 = 3375.05 K

Therefore max. pressure in cycle p3 = p2(T3/T2) = 102.27 bar

For constant specific heat 3200 = 0.7117 x (16/15) x (T3 - 756.03)

T3= 4971.3 K

Therefore max. pressure in cycle p3 = 150.64 bar

Thus the max. pressure has increased by 48.38 bar for constant specific heat.