# Effective coefficient of thermal conductivity for the brass-copper interface layer

songoku
Homework Statement:
A brass plate with a thickness of 1.00 cm is placed over a copper plate with a thickness of 2.00 cm. The coefficient of thermal conductivity of brass is 100 W/m C and of copper is 400 W/m C. What is the effective coefficient of thermal conductivity for the brass-copper layer?
Relevant Equations:
Q/t = k . A . ΔT / d
I know k is thermal conductivity but my teacher never told me about effective coefficient of thermal conductivity. I tried googling and found:
$$k_{effective}=\frac{\Sigma{k.A}}{\Sigma A}$$

But I don't know the area to used that information. Is there another approach to do this question?

Thanks

Delta2

Arjan82
That formula is definitely wrong since it doesn't involve the thickness of the plates, which it should. And, indeed, the effective thermal conductivity is not dependent on the area.

There is a good analogy with electric resistances in parallel. The conductivity times the thickness is the reciprocal of the resistance and now you need to find the effective resistance of the unit. Do you know how to do that?

songoku
That formula is definitely wrong since it doesn't involve the thickness of the plates, which it should. And, indeed, the effective thermal conductivity is not dependent on the area.

There is a good analogy with electric resistances in parallel. The conductivity times the thickness is the reciprocal of the resistance and now you need to find the effective resistance of the unit. Do you know how to do that?
Conductivity of brass x thickness of brass = 100 x 1 x 10-2 = 1 W/oC

Conductivity of copper x thickness of copper = 400 x 2 x 10-2 = 8 W/oC

Since the conductivity times thickness is the reciprocal of resistance, it means that
$$\frac{1}{R_{brass}}=1 ~\text{and}~ \frac{1}{R_{copper}}=\frac{1}{8}$$

So the total R will be:
$$\frac{1}{R_{total}}=1+8$$

$$R_{total}=9$$

Is my working even correct? And I am also confused about the unit of the final answer

Thanks

Last edited:
Homework Helper
Gold Member
I am not sure I can agree with what @Arjan82 says, however if we take what he says as correct, then in your work it should be
$$\frac{1}{R_{total}}=\frac{1}{R_{brass}}+\frac{1}{R_{copper}}=1+8=9\Rightarrow R_{total}=\frac{1}{9}$$

Last edited:
songoku
songoku
I am not sure I can agree with what @Arjan82 says, however if we take what he says as correct, then in your work it should be
$$\frac{1}{R_{total}}=\frac{1}{R_{brass}}+\frac{1}{R_{copper}}=1+\frac{1}{8}=\frac{9}{8}\Rightarrow R_{total}=\frac{8}{9}$$
But he said "The conductivity times the thickness is the reciprocal of the resistance" ?

Maybe you have another way to approach this question? Thanks

Homework Helper
Gold Member
But he said "The conductivity times the thickness is the reciprocal of the resistance" ?

Maybe you have another way to approach this question? Thanks
He also said something about resistances in parallel so i take it to be that we add them as resistances in parallel not in series as you did.

songoku
Homework Helper
Gold Member
Ah sorry total confusion lol, I corrected my post i think now it is correct.

Homework Helper
Gold Member
2022 Award
That formula is definitely wrong
Impossible to say since none of the variables are defined in post #1.
The conductivity times the thickness is the reciprocal of the resistance
No, you would not be wanting to multiply conductivity by thickness. Increasing the thickness does not increase the heat flow. You would generally divide by thickness and multiply by cross sectional area.
Heat flow through a sample is proportional to temperature difference, thermal conductivity and cross sectional area, and inversely proportional to thickness: ##\dot Q=-\frac{\Delta\Theta k A}L##.
In the electrical context, conductivity is the reciprocal of resistivity, and resistance is resistivity x length / area. That's all consistent.
He also said something about resistances in parallel
Clearly they are in series here, so resistances add conventionally and conductances add harmonically.
So quote units throughout the calculation.
Often that can alert you to a misquoted formula, but in this case the correct conductivity x area / length and the incorrect conductivity x length produce the same dimension.

Last edited:
songoku
songoku
Impossible to say since none of the variables are defined in post #1.
k is the thermal conductivity of each material and A is cross-sectional area of the material

Is the information given by the question enough to solve it? Thanks

Homework Helper
Gold Member
2022 Award
k is the thermal conductivity of each material and A is cross-sectional area of the material
Then the formula might be right, but not for the context of your question.
(It is right for thin layers of insulation, all the same thickness, laid adjacent to each other, not on top of each other. A rather peculiar scenario. Can you provide a link?)
For stacked layers of the same cross section but different thicknesses it would be ##\frac{\Sigma L_i}{k_{eff}}=\Sigma\frac{ L_i}{k_i}##

songoku
songoku
Then the formula might be right, but not for the context of your question.
(It is right for thin layers of insulation, all the same thickness, laid adjacent to each other, not on top of each other. A rather peculiar scenario. Can you provide a link?)
For stacked layers of the same cross section but different thicknesses it would be ##\frac{\Sigma L_i}{k_{eff}}=\Sigma\frac{ L_i}{k_i}##
I am sorry for late reply

https://physics.stackexchange.com/q...ive-thermal-conductivity-multi-layer-cylinder

That's where I got the formula, but as you said it is actually for another case and I just thought I could use that formula

Thank you very much for the help and explanation Arjan82, Delta2, haruspex

Delta2