# How do I convert $W/(mK)$, $W/m^2$ and $W/(m^2K)$ to the same “dimensionality” and unit?

How do I convert $$W/(mK)$$, $$W/m^2$$ and $$W/(m^2K)$$ to the same "dimensionality" and unit?

$$W/(mK)$$ is thermal conductivity. $$W/m^2$$ is heat flow density related to one unit. $$W/(m^2K)$$ is the heat transfer value.

By performing arithmetic on them?

Particularly,

I have a sum of integrals where each term is multiplied by a constant in one of the given units. And I need to be able to compute the sum so that the units "agree".

So as an example consider some heat system governed by:

$$\int a \space f \space ds, \int b \space g \space ds, \int c\space h \space ds$$

And particularly I want to make these satisfy equilibrium so that e.g.

$$-\int a \space f \space ds -\int b \space g \space ds +\int c\space h \space ds=0$$

where $$a,b,c$$ have the given different units respectively and $$f,g,h$$ are some functions. The integrals can be computed, but how to make the units agree?

• In what context? – probably_someone Sep 23 '18 at 13:57
• You are asking for help in a step you are trying to do, but we cannot help unless we know the full context of what you are trying to do. – CR Drost Sep 23 '18 at 13:58
• @probably_someone Heat transfer? I have a sum of integrals where each term is multiplied by a constant in one of the given units. And I need to be able to compute the sum so that the units "agree". – mavavilj Sep 23 '18 at 13:58
• @mavavilj Which specific quantities are associated with each of those dimensions? It may be that the relation between them is nontrivial. – probably_someone Sep 23 '18 at 13:59
• @probably_someone Does it really matter what $a,b,c$ are, if I have $a W/(mK)$, $b W/m^2$ and $c W/(m^2K)$? – mavavilj Sep 23 '18 at 14:01

For an interface consisting of a barrier with thermal conductivity $$k$$, area $$A$$, and thickness $$\ell$$, the total heat $$\dot{Q}$$ flowing through the barrier by conduction (in W) is determined by $$\dot{Q}=kA\frac{\Delta T}{\ell}$$.

For the same interface, the heat transfer value $$\alpha$$ (more commonly known as the heat transfer coefficient) is calculated using $$\alpha=\frac{\dot{Q}}{\Delta T}=\frac{kA}{\ell}$$.

For the same interface, the heat flow density $$\dot{q}$$ (in W/m^2) is determined by $$\dot{q}=\frac{\dot{Q}}{A}=\frac{k\Delta T}{\ell}=\frac{\alpha\Delta T}{A}$$.

For the same physical system, these are just three different ways to describe the way that heat flows across an interface. At best, you need more information about the specific physical system you're studying in order to convert between them (since, for example, barrier with different lengths and areas will have different conversion factors between heat flow density, heat transfer coefficient, and thermal conductivity). If you intend to convert all of these to the same unit and add them together, it's likely you're doing something that doesn't have any physical meaning, as you're basically adding the same thing three times. Any of these quantities alone can be used to calculate the total heat flow across an interface.

But since the question does not contain enough information about the physical system to ascertain exactly what it is you're trying to do with this sum of integrals, it's hard to say much more than that.

• "Any of these quantities alone can be used to calculate the total heat flow across an interface.". Does this imply that they can be converted to each other? – mavavilj Sep 23 '18 at 14:31
• @mavavilj If you have a particular physical system that you're studying, and you know the material and geometric details of that physical system, then yes. However, in such a situation, I have no idea why you would want to add the results of all three of them together. – probably_someone Sep 23 '18 at 14:33
• I also wonder, if it's necessary to consider the units to "have to be compatible each other". Couldn't one think that "well they're all watts in some space". So then one could add only the watts together, regardless of whether, they're $/$ $mK$, $m^2$ or $m^2K$? – mavavilj Sep 23 '18 at 14:51
• @mavavilj The answer to that is emphatically no, if you want your expression to have physical meaning. For example, it would make no sense to add together torque and energy, even though they have the same units (force times distance), because the two quantities have different physical meaning. – probably_someone Sep 23 '18 at 15:04
• I found this reference, mathicse.epfl.ch/files/content/sites/mathicse/files/… ,which on p. 24/28, seems to solve something involving $k (du/dn)+ \alpha u = \alpha u_s$ where $\alpha$ is $W/m^2 K$, while $k$ is $W/mK$. So how do they consider those to be summable? – mavavilj Sep 23 '18 at 15:06