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+clear;
+clc;
+printf("\t\t\tExample Number 8.10\n\n\n");
+// heat transfer reduction with parallel plate shield
+// Example 8.10 (page no.-413) 
+// solution
+
+E1 = 0.3;// emissivity of first plane
+E2 = 0.8;// emissivity of second plane
+E3 = 0.04;// emissivity of shield
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// the heat transfer without the shield is given by 
+// q_by_A = sigma*(T1^4-T2^4)/((1/E1)+(1/E2)-1) = 0.279*sigma*(T1^4-T2^4)
+// where T1 is temperature of first plane and T2 is temperature of second plane
+// the radiation network for the problem with the shield in place is shown in figure (8-32) (page no.-410). 
+// the resistances are 
+R1 = (1-E1)/E1;
+R2 = (1-E2)/E2;
+R3 = (1-E3)/E3;
+// the total resistance with the shield is 
+R = R1+R2+R3;
+// and the heat transfer is 
+// q_by_A = sigma*(T1^4-T2^4)/R = 0.01902*sigma*(T1^4-T2^4)
+printf("so the heat tranfer is reduced by %f percent",((0.279-0.01902)/0.279)*100);
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