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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+p1=100;//Pressure in kPa
+t1=30;//temperature in Degree C
+t2=800;//temperature in Degree C
+g=1.4;//constant
+Rc=5;//compression ratio
+Qs=900;//heat supplied in kJ/kg
+nr=0.75;//regenerator efficiency in percentage
+
+
+//CALCULATIONS
+Qs1=R*(t2+273)*log(Rc)+(1-nr)*cv*(t2-t1);//heat supplied in kJ/kg
+m=Qs/Qs1;//mass flow rate in kg/min
+wn=(m/60)*R*log(Rc)*(t2-t1);//net work done in kW
+ns=(wn/(Qs/60))*100;//Thermal efficiency in percentage
+vs=((m)*R*(t1+273)*(1-(1/Rc)))/(p1*60);//swept volume in m^3/s
+pm=wn/vs;//mean effective pressure in kN/m^2
+P=wn*1;//Power developed by the engine in kW
+
+//OUTPUT
+printf('(i)The net work done is %3.2f kW \n (ii)Thermal efficiency is %3.2f percentage \n (iii)mean effective pressure is %3.2f kN/m^2 \n (iv)Power developed by the engine is %3.2f kW ',wn,ns,pm,P)
+