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authorpriyanka2015-06-24 15:03:17 +0530
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+
+clear;
+clc;
+printf("\tExample 1.2\n");
+x=poly([0],'x');
+k1=372; // Thermal Conductivity of slab,W/m*K
+x1=0.003; // Thickness of slab,m
+x2=0.002;// Thickness of steel,m
+k2=17; // Thermal Conductivity of steel,W/m*K
+T1=400; // Temperature on one side,C
+T2=100;//Temperature on other side,C
+
+Tcu=roots(x+2*x*(k1/x1)*(x2/k2)-(400-100));
+
+//q=k1*(Tcu/x1)=k2*(Tss/x2);
+
+Tss = Tcu*(k1/x1)*(x2/k2); // formula for temperature gradient in steel side
+
+Tcul=T1-Tss;
+Tcur=T2+Tss;
+printf("\t temperature on left copper side is : %.0f C\n",Tcul);
+printf("\t Temperature on right copper side is : %.0f C\n",Tcur);
+q=k2*Tss/(1000*x2); // formula for heat conducted
+printf("\t heat conducted through the wall is : %.0f W\n",q);
+printf("\t our initial approximation was accurate within a few percent.");
+//End \ No newline at end of file