Question

An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity kA while the other is fabricated from the material whose thermal conductivity kB is desired. Both rods are attached at one end to a heat source of fixed temperature Tb, are exposed to a fluid of temperature [infinity] T[infinity], and are instrumented with thermocouples to measure the temperature at a fixed distance x1 from the heat source. If the standard material is aluminum, with kA= 200 W/m·K, and measurements reveal values of TA= 75°C and TB= 70°C at x1 for Tb= 100°C and [infinity] T[infinity]= 25°C, what is the thermal conductivity kB of the test material?

Answer 1

Answer: the thermal conductivity of the second material is 125.9 W/m.k

Explanation:

Given that;

The two rods could be approximated as a fins of infinite length.

TA = 75°C,    θA = (TA - T∞) = 75 - 25 = 50°C

TB = 70°C     θB = (TB - T∞) = 70 - 25 = 45°C

Tb = 100°C    θb = (Tb - T∞) = (100 - 25) = 75°C

T∞ = 25°C

KA = 200 W/m · K ,   KB = ?

Now

The temperature distribution for the infinite fins are given by

θ/θb = e^(-mx)

θA/θb= e^-√(hp/A.kA) x 1  --------------1

θB/θb = e^-√(hp/A.kB) x 1---------------2

next we  take the natural logof both sides,  

ln(θA/θb) = -√(hp/A.kA) x 1 ------------3

In(θB/θb) = -√(hp/A.kB) x 1 ------------4

now we divide 3 by 4

[ ln(θA/θb) /in(θB/θb)] = √(KB/KA)

we substitute

 [ In(50/75) /In(45/75)] = √(KB/200)

In(0.6666) / In(0.6) = √KB / √200

-0.4055/-0.5108 = √KB / √200

0.7938 = √KB / 14.14

√KB = 11.22

KB = 125.9 W/m.k

So the thermal conductivity of the second material is 125.9 W/m.k