Repeat Problem 12.29 when the intercooler brings the air to T_3= 320 K. The corrected formula for the optimal pressure is P _2=\left[ P _1 P _4\left( T _3 / T _1\right)^{ n /( n -1)}\right]^{1 / 2} see Problem 9.241, where n is the exponent in the assumed polytropic process.
The polytropic process has n = k (isentropic) so n/(n – 1) = 1.4/0.4 = 3.5
P _2=400 \sqrt{(320 / 290)^{3.5}}=475.2 \,kPa
C.V. C1: s _2= s _1 \Rightarrow P _{ r 2}= P _{ r 1}\left( P _2 / P _1\right)=0.9899 \times(475.2 / 100)
\begin{gathered}=4.704 \Rightarrow T _2=452 K , h _2=453.75 \\- w _{ C 1}= h _2- h _1=453.75-290.43= 1 6 3 . 3 ~ k J / k g\end{gathered}
C.V. Cooler: q _{\text {OUT }}= h _2- h _3=453.75-320.576= 1 3 3 . 2 \,k J / k g
C.V. C2: s _4= s _3 \Rightarrow P _{ r 4}= P _{ r 3}\left( P _4 / P _3\right)=1.3972 \times(1600 / 475.2)=4.704
\begin{gathered}\Rightarrow T _4= T _2=452 \,K , \quad h _4=453.75 \\- w _{ C 2}= h _4- h _3=453.75-320.576= 1 3 3 . 2 \,k J / k g\end{gathered}