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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?

Respuesta :

nuhulawal20

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