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+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.13\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.13  (page no. 479) 
+// Solution
+
+//Now,in problem 9.12,
+//An air-standard Diesel engine
+rc=16; //Compression Ratio Rc=v2/v3
+v4byv3=2; //Cutoff ratio=v4/v3
+k=1.4; //with the cycle starting at 14 psia and 100 F  //It is apparent incerease in compression ratio yields an increased cycle efficiency
+T2=100; //Unit:F //temperature 
+T5=1018; //Unit:F //Found in 9.12 //The temperature of the exhaust of the cycle //Unit:R
+ndiesel=0.614 //Efficiency of the diesel engine //Found in 9.12
+//Now,in problem 9.13,
+cp=0.24; //Unit:Btu/(lbm*R) //Specific heat constant for constant pressure process
+cv=0.172; //Unit:Btu/(lbm*R) //Specific heat constant for constant volume process
+
+Qr=cv*(T5-T2); //Heat rejected //Unit:Btu/lbm
+//ndeisel=1-(Qr/qin); //Efficiency=ndeisel //qin=heat added
+qin=Qr/(1-ndiesel); //Unit:Btu/lbm
+J=778; //J=Conversion factor
+networkout=J*(qin-Qr); //(ft*lbf)/lbm //Net work out per pound of gas
+printf("Net work out per pound of gas is %f (ft*lbf)/lbm\n",networkout);
+//The mean effective pressure is net work divided by (v2-v3):
+mep=networkout/((16-1)*144); //1ft^2=144 in^2 //Unit:psia  //The mean effective pressure
+printf("The mean effective pressure is %f psia",mep);