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+clc
+h = 17.4 // Convective heat transfer coefficient in W/m^2K
+K = 52.2 // Thermal conductivity in W/mK
+t = 120 // Heat reservoir wall temperature in degree celcius
+t0 = 35 // Ambient temperature in degree celcius
+L = 0.4 // Lenght of rod in m
+b  = .050 // width of rod in mm
+H = .050 // Heigth of rod in mm
+
+printf("\n Example 18.4\n")
+l= L/2
+A = b*H
+m = sqrt(4*h*b/(K*b*H))
+t1 = (t-t0)/cosh(m*l) + t0 // Midway temperature of rod
+Q1 = 2*5.12*K*A*(t-t0)*tanh(m*l) // Heat loss rate 
+printf("\n Midway temperature of rod is %f degree Celcius",t1)
+printf("\n Heat loss rate is %fW",Q1)
+//The answers vary due to round off error
+