Calculating Effective Coefficient of Thermal Conductivity

AI Thread Summary
The discussion centers on calculating the effective coefficient of thermal conductivity, with a focus on the incorrect formula initially presented, which does not account for plate thickness. Participants clarify that effective thermal conductivity should be approached similarly to electrical resistances in parallel, emphasizing the need to consider thickness and area correctly. The conversation highlights the importance of using the right formulas for different scenarios, specifically for stacked layers of materials. Participants also discuss the significance of units in calculations to avoid confusion. Ultimately, the thread provides insights into the correct methodology for determining effective thermal conductivity in various contexts.
songoku
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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
 
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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?
 
Arjan82 said:
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
 
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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
$$\frac{1}{R_{total}}=\frac{1}{R_{brass}}+\frac{1}{R_{copper}}=1+8=9\Rightarrow R_{total}=\frac{1}{9}$$
 
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Delta2 said:
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
 
songoku said:
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.
 
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Ah sorry total confusion lol, I corrected my post i think now it is correct.
 
Arjan82 said:
That formula is definitely wrong
Impossible to say since none of the variables are defined in post #1.
Arjan82 said:
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.
Delta2 said:
He also said something about resistances in parallel
Clearly they are in series here, so resistances add conventionally and conductances add harmonically.
songoku said:
I am also confused about the unit of the final answer
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.
 
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haruspex said:
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
 
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songoku said:
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}##
 
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haruspex said:
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
 
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