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

Question

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?

Answer 1

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.

Answer 2

I had the same issue recently but figured it out. The confusion comes from the units (m in the divider) and the intuition that to get the total conductivity you should multiply the coefficient by thickness, but in reality you should divide by thickness (because thicker wall = less heat transfer).

The units of conductivity is actually not $W/(mK)$, but $Wm/(m^2K)$. If you multiply it by area (m^2), you get watt-metre -- if it's 1 Watt and 1 metre of thickness, it's 1. If it's 1 Watt and 2 metres, it's twice more of conductivity.

1. To get $W/(m^2K)$ you should simply divide $W/(mK)$ by thickness (m).

2. For $W/$m^2 you need to know actual temperature delta. Multiply the conductivity ($W/(mK)$) by delta T and divide by material thickness, and you'll get your $W/(m^2)$.