# Inteface thermal conductivity

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 March 10, 2002, 06:28 Inteface thermal conductivity #1 ado Guest   Posts: n/a Dear friends, My question is related to the inteface thermal conductivity between two adjacent materials. One is chip and the other is glass covering the chip completly. Within the chip there is a constant heat generation and in the glass no heat generation. The conduction model is supposed 2D (cylindrical coordinates)..my question is how can we determine the 2 effective thermal conductivities at the intefaces in radial and axial directions ???? As an example in Pakankar it is given as Kint=2*K1*K2/(K1+K2) where Kint,K1 and K2 are the thermal conductivities at the interface, the material 1 (chip for example) and material2 (glass for example)..but in presence of heat generation or heat source the interface thermal conducitivities ( radial and axial directions) should be different from this relation isn't it??? Waiting for you answers and helps Regards, Amir

 March 10, 2002, 22:53 Re: Inteface thermal conductivity #2 mukhopadhyay Guest   Posts: n/a A good point. Yes, you are right - you can't use that type of formulation. Just make a cv energy balance.

 March 11, 2002, 03:16 Re: Inteface thermal conductivity #3 Neale Guest   Posts: n/a Don't bother with that and use two independent control volumes at the interface, perform a heat flux balance through the interface and use the additional requirement that the control volume temperatures be equal. That way there is no need to blend the conductivities. Neale

 March 11, 2002, 04:15 Re: Inteface thermal conductivity #4 Henry Eatsink Guest   Posts: n/a Use FLOTHERM and it's ability to define explicitly an interfacial resistance in terms of K m^2/W.

 March 11, 2002, 20:46 Re: Inteface thermal conductivity #5 Sebastien Perron Guest   Posts: n/a I assume you are using an FVM method. This is another way of computing the interface conductivity. The formula you wrote was only valid for a constant mesh spacing. This procedure is good for any coordinate system, only change the expressions for the numerical flux according to the flux in your coordinate system if different: 1) First you assume that the flux on either side of the interface is equal (conservativity of the flux): K1 (T1-T)/D1 = - K2 (T2-T)/D2 (1) 2) express T (the temperature at the volume interface) in function of the others: T = [ (K2/D2)*T2 + (K1/D1)*T1 ] / [ K2/D2 + K1/D1 ] 3) replace the expression for T in either side of equation (1) and make it equal to the flux across the interface: K12 (T2-T1)/(d1+d2) where K12 is the conductivity your looking for, after some manipualtion you get the final expression for it: K12 = (d1+d2)K1K2/(d1K2+d2K1) Good luck.

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