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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+p1=1;//pressure in atm
+t3=1700;//temperature in K
+Rc=7;//compression ratio
+R=0.287;//gas constant
+cv=0.7234;//calorific value
+
+//CALCULATIONS
+t2=t1*(Rc^(1.4-1));//temperature in K
+p2=p1*(Rc^(1.4));//pressure in atm
+p3=p2*t3/(t2);//pressure in atm
+t4=t3/((Rc)^(1.4-1));//temperature in K
+p4=p3/(Rc^(1.4));//pressure in atm
+no=1-((1/Rc)^(1.4-1))*100;//Thermal efficiency in percentage
+Rwo=1-((t1/t3)*((Rc)^(1.4-1)));//work ratio
+v1=R*t1/p1;//specific volume in m^3/kg
+wn=(cv*(t3-t2))-((cv*(t4-t1)));//net work
+pm=(wn/(v1*(1-(1/Rc))));//mean effective pressure in Bar
+
+//OUTPUT
+printf('(a)pressure at state point 2 is %3.2f atm \n temperature at point 2 is %3.2f K \n pressure at state point 3 is %3.2f atm \n temperature at point 4 is %3.2f K \n pressure at state point 3 is %3.2f atm \n (b)Thermal efficiency is %3.2f percentage \n (c)work ratio is %3.5f \n (d)mean effective pressure is %3.2f Bar',p2,t2,p3,t4,p4,no,Rwo,pm)
+