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+clc;
+clear;
+//Example 4.20
+sigma=5.67*10^-8 //[W/sq m.K^4]
+T1=750 //[K]
+T2=500 //[K]
+e1=0.75;
+e2=0.5;
+//Heat transfer without shield :
+
+Q_by_a=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1) //[W/sq m]
+
+//Heat transfer with shield:
+R1=(1-e1)/e1 //Resistance 1
+
+F13=1;
+R2=1/F13 //Resistance 2
+
+e3=0.05
+R3=(1-e3)/e3 //Resistance 3
+
+R4=(1-e3)/e3 //Resistance 4
+
+F32=1;
+R5=1/F32 //Resistance 5
+
+R6=(1-e2)/e2 //Resistance 6
+
+Total_R=R1+R2+R3+R4+R5+R6 //Total resistance
+
+Q_by_as=sigma*(T1^4-T2^4)/Total_R //[W/sq m]
+
+Red=(Q_by_a-Q_by_as)*100/Q_by_a //Reduciton in heat tranfer due to shield
+
+printf("\n Reduction in heat transfer rate as a result of radiaiotn shield is %f percent",Red);