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

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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.

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mavavilj

Particular interests in computational modelling of useful things and related algorithms. Also, musical DSP. B. Sc. in Information Technology (math, cs, stats minor) and doing a M. Sc. in Computational Sciences. Have thought about making online notes some day due to finding others' blogs on e.g. machine learning etc. sometimes very useful: https://someonlinemathnotes.blogspot.com/

Updated on March 18, 2022

Comments

  • mavavilj
    mavavilj over 1 year

    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?

    • probably_someone
      probably_someone about 5 years
      In what context?
    • CR Drost
      CR Drost about 5 years
      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.
    • mavavilj
      mavavilj about 5 years
      @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".
    • probably_someone
      probably_someone about 5 years
      @mavavilj Which specific quantities are associated with each of those dimensions? It may be that the relation between them is nontrivial.
    • mavavilj
      mavavilj about 5 years
      @probably_someone Does it really matter what $a,b,c$ are, if I have $a W/(mK)$, $b W/m^2$ and $c W/(m^2K)$?
    • probably_someone
      probably_someone about 5 years
      @mavavilj I'm not asking for the values of the quantities, but rather their names. For example, is the quantity with units $W/m^2$ a radiative intensity? The reason I ask is because there are many different quantities which share the same units and are not interconvertible. For example, torque and energy both have units of $N\cdot m$.
    • Admin
      Admin about 5 years
      @mavavilj: yes, it matters what they are. You can trivially just invent values which have the appropriate dimensions (so multiplying by $1\,\mathrm{mK/m^2}$ will make the first quantity compatible with the second) to make it 'work' but that's probably just meaningless, because you need to know what the scale factor is (probably it is not $1$).
    • Deschele Schilder
      Deschele Schilder about 3 years
      You get the last unit by dividing the first through m. You get the second unit by dividing the first through K/m.
  • mavavilj
    mavavilj about 5 years
    "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?
  • probably_someone
    probably_someone about 5 years
    @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.
  • mavavilj
    mavavilj about 5 years
    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 $m^2K$?
  • probably_someone
    probably_someone about 5 years
    @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.
  • mavavilj
    mavavilj about 5 years
    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?
  • probably_someone
    probably_someone about 5 years
    Because one is multiplying $u$ and the other is multiplying the derivative of $u$ with respect to distance. $dn$ has units of distance, so the units agree.
  • mavavilj
    mavavilj about 5 years
    So you say that $k(du/dn)$ is of $W/m^2K$? But why are the units for $du/dn$ $1/m$?
  • probably_someone
    probably_someone about 5 years
    @mavavilj Yes, that is the case, because the units of $dn$ are $m$. It's just like the difference between position $x$ and velocity $v=dx/dt$. $dx$ has units of $m$ and $dt$ has units of $s$, so $v$ has units of $m/s$.