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diff --git a/2090/CH2/EX2.8/Chapter2_example8.sce b/2090/CH2/EX2.8/Chapter2_example8.sce
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+clc
+clear
+//Input data
+rc=15//Compression ratio
+p1=1//Pressure at which compression begins in bar
+T1=27+273//Temperature in K
+pm=60//Maximum pressure in bar
+h=2//Heat transfered to air at constant volume is twice that at constant pressure
+g=1.4;//Isentropic index
+Cv=0.718;//specific heat at constant volume for air in kJ/kgK
+Cp=1.005;//specific heat at constant pressure for air in kJ/kgK
+R=0.287;//Real gas constant in kJ/kgK
+
+//Calculations
+T2=(T1*rc^(g-1))//Temperature in K
+p2=(p1*rc^g)//Pressure in bar
+T3=(T2*(pm/p2))//Temperature in K
+T4=(Cv*(T3-T2))/(2*Cp)+T3//Temperature in K
+b=(T4/T3)//Cut-off ratio
+T5=(T4*(b/rc)^(g-1))//Temperature in K
+p5=(p1*(T5/T1))//Pressure in bar
+Q1=(Cv*(T3-T2))+(Cp*(T4-T3))//Heat supplied per unit mass in kJ/kg
+Q2=Cv*(T5-T1)//Heat rejected per unit mass in kJ/kg
+W=(Q1-Q2)//Workdone in kJ/kg
+n=(W/Q1)*100//Air standard efficiency
+Vs=((1*R*1000*T1)/(p1*10^5))*(1-1/rc)//Swept volume in m^3/kg
+pmean=((W*1000)/Vs)/10^5//Mean-effective pressure in bar
+
+//Output
+printf('(a) The pressures and temperatures at the cardinal points of the cycle are \n T2 = %3.0f K  p2 = %3.1f bar \n T3 = %3.0f K  p3 = %3.0f bar \n T4 = %3.0f K  p4 = %3.0f bar \n T5 = %3.0f K  p5 = %3.2f bar \n (b) The cycle efficiency is %3.0f percent \n (c) The mean effective pressure of the cycle is %3.2f bar',T2,p2,T3,pm,T4,pm,T5,p5,n,pmean)