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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+clear
+//INPUT DATA
+v1=0.5;//volume in m^3
+p1=1;//pressure in bar
+t1=303;//temperature in K
+p2=12;//pressure in bar
+Qs=250;//heat is added in kJ
+wc=200;//working cycles in cycles/min
+v2=0.085;//volume in m^3
+m=1;//mass of air
+cv=0.7243;//calorific value
+
+//CALCULATIONS
+Rc=(p2/p1)^(1/1.4);//compression ratio
+t2=t1*((Rc)^(1.4-1));//temperature in K
+pc=((v2/(v1-v2))*100);//percentage clearance
+t3=(Qs/(m*cv))+t2;//temperature in K
+t4=((1/Rc)^(1.4-1))*t3;//temperature in K
+Qr=m*cv*(t4-t1);//heat rejected in kJ/kg
+no=((Qs-Qr)/Qs)*100;//thermal efficiency in percentage
+pm=((Qs-Qr)/(v1-v2));//mean effective pressure
+p=((Qs-Qr)*wc)/60;//power developed in kJ/s
+
+//OUTPUT
+printf('(a)percentage clearance is %3.2f percentage \n (b)the thermal efficiency is %3.2f percentage \n (c)mean effective pressure is %3.2f \n (d)power developed is %3.2f kJ/s',pc,no,pm,p)
+