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A material (A) of width 10cm is generating heat at a rate of 5 kw/m3 when curren

ID: 1818434 • Letter: A

Question

A material (A) of width 10cm is generating heat at a rate of 5 kw/m3 when current is passed through it. An unknown material (B) and another material (C) are attached serially to it. You have set up an experiment to determine the thermal conductivity of material B. Material B is 15 cm thick. Material C is also 15 cm thick and has a thermal conductivity of 10 W/mK. The thermal conductivity of material A is 5 W/mK. The outer surface of Material C has been measured to be 30degreeC. The outer surfaces arc cooled with a fluid at 25degreeC and a heat transfer coefficient of 25degreeC and a heat transfer coefficient of 25W/m2K. Determine surface temperature of Material A that is exposed to the fluid. Determine the other surface temperature of Material A Determine the thermal conductivity of Material B. Determine the maximum temperature in the setup

Explanation / Answer

assuming 1-dimensional and steady state

since heat is generating in material A ,so two heat fluxes will come out of its two sides.
let q1 from right face and q2 from left face. so the same flux q1 will come out from outer face of material C.let denote the surface temperatures be T1 ,T2 , T3 and T4 taking from left side.
at T4 surface T4=30°C so q2= 25*(30-25) = 125

also q2 = 10*(T3-30)/0.15 which gives T3

for material A whatever heat is generated must be dissipated out through both of its surfaces so

5000*0.1*area = (q1+q2)*area , it gives q2

at surface 1 q2 = 25*(T1- 25) , it gives T1

for material A

5*d2T/dx2 + 5000 = 0 , gives T= -2500x2 -c1x +c2

boundary conditions : at x=0 , T =T1   

at x = 0 , -5*dT/dx = q2 , gives c1 and c2

so temperature at x= 0.1 is calculated =T2

and for material B k*(T2-T3)/0.15 = q1 , gives k

using temperature profile temperature ditribution in material A is found then maximum temperature can also be found.

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