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diff --git a/1808/CH2/EX2.20/Chapter2_Example20.sce b/1808/CH2/EX2.20/Chapter2_Example20.sce
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+clc
+clear
+//INPUT DATA
+mf=0.4;//flue flow rate in g/s
+ma=5.6;//air flow rate in g/s
+hbc=1.8;//gasoline with H/C ratio
+//a CH1.87+b O2+c N2=13 CO2 +2.8 CO +0.933 H2+ d H2O+ 83.267 N2 ;//EXHAUST GAS COMPOSITION
+c=83.267;//Composition of N2
+b=22.069;//Composition of O2
+a=15.8;//Composition of CH1.87
+d=13.84;//Composition of H2O
+
+//CALCULATIONS
+xa=mf/ma;//air fuel ratio
+//CH1.87+1.4675(O2+3.773N2)=CO2+0.935 H2O+ 5.536 N2 ;//CHEMICAL COMPOSITION
+xs=(12*1+1*hbc)/(202+86);//air fuel ratio
+x=(xa/xs);//Equivlent ratio from the fuel and air flow rate
+
+//15.8 CH1.87+22.069 O2+83.267 N2=13 CO2 +2.8 CO +0.937 H2+ 13.84 H2O+ 83.267 N2 ;//EXHAUST GAS COMPOSITION
+//CH1.87+1.397 O2+5.27 N2=0.823 CO2 +0.177 CO +0.059 H2+ 0.876 H2O+ 5.27 N2 ;//CALCULATED EXHAUST GAS COMPOSITION
+xa1=(d)/(1.397*32+5.27*28);//air fuel ratio
+x1=xa1/xs;//Equivalent ratio of calculated exhaust gas composition
+
+//OUTPUT
+printf('(i)The Equivlent ratio from the fuel and air flow rate is %3.3f \n (ii)TheEquivalent ratio of calculated exhaust gas composition is %3.3f',x,x1)
+
+
+
+
+
+