Conduction heat transfer with varying cross section

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To calculate the conduction heat transfer (Q) between two stadium-shaped parts with different thicknesses and temperatures, the heat transfer equations for each part must be established. The first part has a temperature of 1020°C and thickness d1, while the second part has a temperature of 20°C and thickness d2. The equations Q1 = (-K*A1*(1020-T))/d1 and Q2 = (-K*A2*(T-20))/d2 represent the heat transfer for each part, where T is the temperature at the junction. Since Q1 and Q2 are equal, T can be eliminated between the two equations to find a single expression for Q. This approach allows for the calculation of heat transfer across the two materials effectively.
imselva
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Conduction_Case_12.jpg

How to modify this Formula? and arrive at the solution.
Q = -K(Th-Tc)A/thickness
 
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I don't understand the geometry you are trying to specify.
 
The geomentry consists of two parts in complete contact with each other.
The first part is a Stadium shape with 5 mm thickness and a height of 39.4mm.
The second part is also Stadium shape with 0.9mm thickness and a height of 64.8mm.

The Bottom of the first part is at 1020°C and the top for the second part is 20°C.
I want to calculate the conduction heat Transfer between the two different temperatures. How to find Q?
 
View attachment Unbenannt.PNGThe temperature T1 at the top Surface is 1020°C and T2 at the Bottom Surface is 20°C and K is the thermal conductivity of the material.

What is the conduction heat Transfer Q. ?
 
Let T be the temperature at the junction between the two parts. Q is going to be the same through each of the two parts. In terms of T, what are the two heat conduction equations for each of the two parts?
 
Q1 = (-K*A1*(1020-T))/d1
Q2 = (-K*A2*(T-20))/d2
 
imselva said:
Q1 = (-K*A1*(1020-T))/d1
Q2 = (-K*A2*(T-20))/d2
Q1 and Q2 are both Q. Eliminate T between the two equations.
 
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