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 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?

Answer 1

Answer:

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)