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 while the other is fabricated from the material whose thermal conductivity is desired. Both rods are attached at one end to a heat source of fixed temperature , are exposed to a fluid of temperature T[infinity], and are instrumented with thermocouples to measure the temperature at a fixed distance from the heat source. If the standard material is aluminum, with 200 W/m·K, and measurements reveal values of TA= 75°C and Tb= 55°C at xi for Tb= 100°C and T[infinity]= 25°C, what is the thermal conductivity of the test material?

Explanation:

Given:

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

TA = 75 0C θA = (TA - T∞) = 75 - 25 = 50 0C

TB = 55 0C     θB = (TB - T∞) = 55 - 25 = 30 0C

Tb = 100 0C   θb = (Tb - T∞) = (100 - 25) = 75 0C

KA = 200 W/m · K

T∞ = 25 0C

Solution:

The temperature distribution for the infinite fins are given by

θ/θb=e⁻mx

θA/θb= e-√(hp/A.kA) x1    (1)

 θB/θb = e-√(hp/A.kB) x1(2)

Taking natural log on both sides we get,

Ln(θA/θb) = -√(hp/A.kA) x1 (3)

Ln(θB/θb) = -√(hp/A.kB) x1 (4)

Dicving (3) and (4) we get

[ Ln(θA/θb) /Ln(θB/θb)] = √(KB/KA)

 [ Ln(50/75) /Ln(30/75)] = √(KB/200)

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answer:

a

explanation: