Question

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 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 = 180 W/m ⋅ K, and measurements reveal values of TA = 77°C and TB = 62°C at x1 for Tb = 100°C and T[infinity] = 25°C, what is the thermal conductivity kB of the test material?

Answer 1

Step-by-step explanation:

Given the following ;

• TA = 77°C and TB = 62°C at x1, KA = 180w/mk

• Tb = 100°C and T[infinity] = 25°C

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

• TA = 77 0C, θA = (TA - T∞) = 77 - 25 = 52 0C

• TB = 62 0C , θB = (TB - T∞) = 62 - 25 = 37 0C

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

• KA = 180 W/m · K

• T∞ = 25 0C

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,

• In(θA/θb) = -√(hp/A.kA) x1 ...................(3)

• In(θB/θb) = -√(hp/A.kB) x1 .....................(4)

• Dividing (3) by (4) ;

• [ In(θA/θb) / In(θB/θb)] = √(KB/KA)

• [ In(52/75) / In(37/75)] = √(KB/180)

• 0.51834 = √(KB/180)

• KB = 48.36W/m. K, Hence The thermal conductivity of the second material is KB = 48.36 W/m.K