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authorpriyanka2015-06-24 15:03:17 +0530
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+clc;
+clear;
+//Example 2.5
+printf("Example 2.5\nPage 2.23")
+//Given
+A=1; //let [sq m]
+x1=0.23; //thickness of fir brick layer[m]
+x2=0.115; // [m]
+x3=0.23; //[m]
+T1=1213; //Temperature of furnace [K]
+T2=318; //Temperature of furnace [K]
+dT=T1-T2; //[K]
+k1=6.047; //W/(m.K) (fire brick)
+k2=0.581; //W/(m.K) (insulating brick)
+k3=2.33; //W/(m.K) (building brick)
+Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3)) //Heat lost per unit Area in Watt
+
+R1=(x1/k1) //Thermal resistance
+R1=0.04 //Approximate
+R2=(x2/k2)
+R2=0.2025 //Approximate
+R3=(x3/k3)
+R3=0.1 //Approximate
+Ta=T1-((dT*R1)/(R1+R2+R3))
+Tb=((dT*R3)/(R1+R2+R3))+T2
+Tb=565 //Approximation
+printf("\nAnswer:Heat loss per unit area is %f W=%f J/s\n",Q_by_A,Q_by_A);
+printf("\nAnswer:\n Ta=%f K =Temperature at the interface between fire brick and insulating brick\n Tb=%d K Temperature at the interface between insulating and building brick\n",Ta,Tb)
+