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authorpriyanka2015-06-24 15:03:17 +0530
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+clc;
+clear;
+printf("\t\t\tChapter2_example1\n\n\n");
+// determination of the heat flow through a composite wall
+T3=-10; // temperature of inside wall in degree Fahrenheit
+T0=70; // temperature of outside wall in degree Fahrenheit
+dT=T0-T3; // overall temperature difference
+// values of thermal conductivity in BTU/(hr.ft.degree Rankine) from appendix table B3
+k1=0.38; // brick masonry
+k2=0.02; // glass fibre
+k3=0.063; // plywood
+dx1=4/12; // thickness of brick layer in ft
+dx2=3.5/12; // thickness of glass fibre layer in ft
+dx3=0.5/12; // thickness of plywood layer in ft
+A=1; // cross sectional area taken as 1 ft^2
+R1=dx1/(k1*A); // resistance of brick layer in (hr.degree Rankine)/BTU
+R2=dx2/(k2*A); // resistance of glass fibre layer in (hr.degree Rankine)/BTU
+R3=dx3/(k3*A); // resistance of plywood layer in (hr.degree Rankine)/BTU
+printf("\nResistance of brick layer is %.3f (hr.degree Rankine)/BTU",R1);
+printf("\nResistance of glass fibre layer is %.1f (hr.degree Rankine)/BTU",R2);
+printf("\nResistance of plywood layer is %.3f (hr.degree Rankine)/BTU",R3);
+qx=(T0-T3)/(R1+R2+R3);
+printf("\nHeat transfer through the composite wall is %.2f BTU/hr",qx);