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+clc;funcprot(0);//EXAMPLE 3.3

+// Initialisation of Variables

+p1=18;..................//Maximum pressure in bar

+t1=410+273;.............//Maximum temperature in Kelvin

+ric=6;.....................//Ratio of isentropic compression

+rie=1.5;.................//Ratio of isothermal expansion

+v1=0.18;..................//Volume of air at the beginning of expansion

+ga=1.4;...................//Degree of freedom of gas

+R=287;.....................//Gas constant in J/kgK

+nc=210;..................//no of working cycles

+//Calculations

+

+t4=t1/(ric^(ga-1));.............//Min temp in K

+t3=t4;

+p4=p1/(ric^ga);..................//Min pressure in bar

+p2=p1/rie;.......................//pressure of gas before isentropic expansion in bar

+p3=p2*((1/6)^ga);.................//Pressure of gas after isentropic expansion in bar

+printf("p1=%f bar \np2=%f bar \np3=%f bar \np4=%f bar \nt1=t2=%f Kelvin \nt3=t4=%f Kelvin \n",p1,p2,p3,p4,t1,t3)

+dels=(p1*10^5*v1*log(rie))/(1000*t1);....................//Change in entropy

+disp(dels,"Change in entropy in kJ/K:")

+qs=t1*dels;.......................//Heat supplied in kJ

+Qr=t4*dels;.......................//Heat rejected in kJ

+eta=(qs-Qr)/qs;............//Efficiency of the cycle

+v3byv1=ric*rie;Vs=(v3byv1-1)*v1;.................//Stroke volume

+pm=((qs-Qr)*10^3)/(Vs*10^5);........//Mean effective pressure of the cycle in bar

+disp(pm,"Mean effective pressure of the cycle in bar:")

+P=(qs-Qr)*(nc/60);.........................//Power of engine

+disp(P,"Mean effective pressure of the cycle in bar:")