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+clear;
+clc;
+disp(' Example 14.3');
+
+// aim : To determine 
+// (a) the intermediate pressures
+// (b) the effective swept volume of the LP cylinder
+// (c) the temperature and the volume of air delivered per stroke at 15 bar
+// (d) the work done per kilogram of air
+
+// given values
+d = 450*10^-3;//  bore , [m]
+L = 300*10^-3;//  stroke, [m]
+cl = .05;//  clearence
+P1 = 1; // intake pressure, [bar]
+T1 = 273+18;// intake temperature, [K]
+P4 = 15;// final delivery pressure, [bar]
+n = 1.3;//  compression and expansion index
+R = .29;// gas constant, [kJ/kg K]
+
+// solution
+// (a)
+k=(P4/P1)^(1/3); 
+// hence
+P2 = k*P1;//  intermediare pressure, [bar]
+P3 = k*P2;//  intermediate pressure, [bar]
+
+mprintf('\n (a) The intermediate pressure is P2 =  %f  bar\n',P2);
+mprintf('\n     The intermediate pressure is  P3=  %f  bar\n',P3);
+
+// (b)
+SV = %pi*d^2/4*L;//  swept volume of LP cylinder, [m^3]
+// hence
+V7 = cl*SV;// volume, [m^3]
+V1 = SV+V7;// volume, [m^3]
+// also
+P7 = P2;
+P8 = P1;
+V8 = V7*(P7/P8)^(1/n);//  volume, [m^3]
+ESV = V1-V8;// effective swept volume of LP cylinder, [m^3]
+
+mprintf('\n (b) The effective swept volume of the LP cylinder is  =  %f litres\n',ESV*10^3);
+
+// (c)
+T9 = T1;
+P9 = P3;
+T4 = T9*(P4/P9)^((n-1)/n);// delivery temperature, [K]
+// now using P4*(V4-V5)/T4=P1*(V1-V8)/T1
+V4_minus_V5 = P1*T4*(V1-V8)/(P4*T1);// delivery volume, [m^3]
+ 
+mprintf('\n (c) The delivery temperature is =  %f  C\n',T4-273);
+mprintf('\n      The delivery volume is =  %f  litres\n',V4_minus_V5*10^3);
+
+//  (d)
+
+W = 3*n*R*T1*((P2/P1)^((n-1)/n)-1)/(n-1);//  work done/kg ,[kJ]
+mprintf('\n (d) The work done per kilogram of air is =  %f  kJ\n',W);
+ 
+// End