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authorpriyanka2015-06-24 15:03:17 +0530
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+clc;funcprot(0);//EXAMPLE 20.28
+// Initialisation of Variables
+p1=1;.............//Intake pressure in bar
+p4=36;........//Final pressure in bar
+n=1.25;.........//Compression index
+R=0.287;.......//Gas constant in kJ/kgK
+t1=300;..........//Intake temperature in K
+ns=3;...........//No of stages
+v=15;..........//Volume of air delivered in m^3
+//Calculations
+p2=p1*((p4/p1)^(1/ns));
+p3=p2*((p4/p1)^(1/ns));
+printf("\n\nIntermediate pressures\n\n p2=%f\n p3=%f\n\n",p2,p3)
+t2=t1*((p4/p1)^(((n-1)/n)*(1/ns)));....//Delivery temperature in K
+m=p1*10^5*v/(R*1000*t1);...........//Mass of air handled per min in kg
+Wt=((n/(n-1))*m*R*(1/60)*(t2-t1)*ns);........//Total work done in three stages
+disp(Wt,"Indicated power required in kW:")