Given:

Two stage single acting reciprocating comp
$V_1=0.2 m^3/s, T_1=16^o, P_1=0.1M,P_a=1 ×10^5 Pa$

$P_3=0.7mPa=7×10^5 Pa,n=1.25 , N=600 rpm $

To Find:$ P_2= ?,V_{LP}=?,V_{HP}=?,Power,Q_(\frac{i}{c})$

Solution:

Assume optimum intermediate pressure

$P_2=\sqrt(P_1.P_3 )=\sqrt(0.1 ×0.7) $

=$0.264 MPa=2.64×10^5 Pa$

$V_{LP}=V_1=0.2 m^3/s$

$V_{LP}=V_1=\frac{0.2 ×60}{600}=0.02 m^3$

PV=mRT

$\frac{PV}{RT}=m →constant $

$\frac{P_1 V_1}{RT_1}=\frac{P_2 V_5}{RT_5}$

$\frac{1 ×10^5 ×0.02}{287 ×289}=\frac{2.64 ×10^5 ×V_5}{287 ×289}$

$V_5=7.57 ×10^{-3} m^3=0.07575m^3/s$

$V_{HP}=V_5=7.57 ×10^{-3} m^3$

**Power** =$\frac{n}{n-1}[ {P_1 V_1 [(\frac{P_2}{P_1})^\frac{n-1}{n}- 1]+ P_2 V_5 [(\frac{P_3}{P_2})^\frac{n-1}{n}- 1]}]$

=$\frac{1.25}{0.25} [{10^5×0.2[(\frac{2.64}{1})^\frac{0.25}{1.25}- 1]+ 2.64 ×10^5×0.07575[(\frac{7}{2.64})^\frac{0.25}{1.25}- 1]}]$

=**42.857** KW

**Heat rejected by air in inter cooler**

$Q_(\frac{i}{c})=mC_p (T_2-T_1) $

$m=\frac{P_1 V_1}{RT_1}=\frac{10^5×0.2}{287 ×289}=0.2411 kg/s$

$\frac{T_2}{T_1} =(\frac{P_2}{P_1})^\frac{n-1}{n}$

$T_2=T_1 \frac{P_2}{P_1}^\frac{n-1}{n}=289\frac{2.64}{1}^\frac{0.25}{1.25}$

$T_2=350.93 k$

$ Q_{\frac{i}{c}}=0.2411 ×1005 ×(350.93-289)$

$Q_{\frac{i}{c}}=15.006 KW $