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+// Example 13_11
+clc;funcprot(0);
+// Given data
+T_1=800;// R
+T_4=530;// R
+T_3=T_4;// R
+p_4=14.7;// psia
+p_3=p_4;// psia
+p_2s=p_3;// psia
+m=1.00*10^-3;// lbm of air
+R=53.34;// ft.lbf/lbm.R
+k=1.4;// The specific heat ratio
+
+// Calculation
+// (a)
+V_4=(m*R*T_4)/(p_4*144);// ft^3
+V_1=V_4;// ft^3
+p_1=(m*R*T_1)/(V_1*144);// psia
+// (b)
+T_2s=T_1*(p_2s/p_1)^((k-1)/k);// R
+CR=T_2s/T_3;// The isentropic compression ratio
+// (c)
+n_T_L=(1-((k*T_3*(CR-1))/(T_1-T_4)))*100;// The Lenoir cold ASC thermal efficiency in %
+printf("\n(a)The combustion pressure,p_1=%2.1f psia \n(b)The isentropic compression ratio,CR=%1.2f \n(c)The Lenoir cold ASC thermal efficiency,n_T=%1.2f percentage",p_1,CR,n_T_L);
+// The answer vary due to round off error