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+clc
+clear
+//INPUT DATA
+l=20;//Stroke in cm
+d=15;//Bore in cm
+N=400;//speed in rpm
+Rc=22;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+n1=1.3;//no of cycles
+n2=1.35;//no of cycles
+g=1.4;//constant
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constnat
+v1=0.003698;//specific volume in m^3/s
+vs=0.00353;//specific volume in m^3/s
+ro=2.68;//cutoff ratio
+
+//CALCULATIONS
+p2=p1*(Rc^(n1));//pressure in kN/m^2
+t2=t1*(Rc^(n1-1));//temperature in K
+t3=t2*(ro);//temperature in K
+p4=p2*((ro/Rc)^n2);//maximum pressure in kN/m^2
+t4=t3*(1/((Rc/ro)^(n2-1)));//maximum temperature in K
+m=p1*v1/(R*t1);//mass flow rate in kg/s
+wn=R*((t3-t2)+((t3-t4)/(n2-1))-((t2-t1)/(n1-1)));//work done in kJ/kg
+pm=wn*m/(vs);//mean effective pressure in kN/m^2
+Qs=cp*(t3-t2);//heat supplied in kJ/kg
+nd=(wn/Qs)*100;//thermal efficiency in percentage
+P=wn*m*N/60;//POWER DEVELOPED
+
+//OUTPUT
+printf('(a)The temperature and pressure at all corner points are \n pressure at point 1 is %3.2f kN/m^2 \n temperature at point 2 is %3.2f K \n temperature at point 3 is %3.2f K \n maximum pressure is %3.2f kN/m^2 \n temperature at point 4 is %3.2f K \n(b)The mean effective pressure is %3.3f kN/m^2 \n (c)Thermal efficiency is %3.2f percentage \n (d)Power developed is %3.2f kJ/s ',p2,t2,t3,p4,t4,pm,nd,P)
+