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+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 14")
+//Considering two very long slender rods of the same diameter but of different materials.
+//The base of each rod is maintained at 100°C while the surfaces of the rod are exposed to 20°C
+//By traversing length of each rod with a thermocouple it was observed that tempratures of rod were equal at the position xA=0.15m and xB=0.075 from base.
+xA=0.15;
+xB=0.075;
+//Thermal conductivity of rod A is known to be kA=72 W/(m*K)
+kA=72;
+//In case of a very long slender rod we use the tip boundary condition thetaL=0 as L--->infinity
+//Therfore we can write for the locations where the tempratures are equal thetab*e^(-mA*xA)=thetab*e^(-mB*xB) or xA/xB=mB/mA,Again mB/mA=(kA/kB)^0.5
+//So kB=kA*(xB/xA)^2
+//The thermal conductivity of Rod B iskB
+disp("The thermal conductivity of Rod B kB in W/(m*K) is ")
+kB=kA*(xB/xA)^2
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