Can you also explain the two in plain words ?


The two are confusingly similar. Heat transfer coefficient is given by: $$ h = \frac{q}{\Delta T} $$ where $q$ is heat flux. This corresponds to the ratio of heat flux to the temperature difference between two points. Thermal conductivity is often given by: $$ k = -\left|\frac{\mathbf{q}}{\nabla T}\right| $$ i.e. the ratio between the heat flux vector, and the temperature gradient vector (I've assumed the material is isotropic here). The difference is that $h$ is a property of an object or system, whereas $k$ is a property of material, the two can be easily related in 1D: $$ h=kl $$ where $l$ is the length of the object.

  • $\begingroup$ how do you find the lenght $l$ in $h=kl$? and can you suggest a book so that I can study this relation? $\endgroup$
    – math
    Apr 20 '15 at 8:16
  • $\begingroup$ In one dimensional thermodynamics, it is simply the length of your object (normally envisaged as a rod or suchlike). You might like to consider something like this for further reading. $\endgroup$ Apr 20 '15 at 10:48
  • $\begingroup$ Why is it $h=kl$ and not $k=hl$ and $k/l=h$? Because $k/l=h$ is what wikipedia gives. Also, the units would seem to suggest that it's $k/l=h$, as @mark jay had pointed out, and which is corroborated by wikipedia, which agrees that $h=W⋅K^{−1}⋅m^{−2}$ and $k=W⋅K^{−1}⋅m^{−1}$ $\endgroup$
    – phlaxyr
    Sep 17 '20 at 23:48

the heat transfer coefficient (h) is equal to the thermal conductivity (k) divided by the thickness of the object. $$ h=\frac{k}{l} $$ the units for h are $[W/m^2 K]$

the units for k are $[W/mK]$


Thermal conductivity is applied in conduction of heat through media(Fourier's Law)

Heat transfer coefficient is applied in convection of heat through a fluid (Newton's Law).

  • $\begingroup$ It might be better to highlight (i.e., quote) the relevant section of the Wikipedia links into the post $\endgroup$
    – Kyle Kanos
    Nov 11 '19 at 13:17
  • $\begingroup$ Heat transfer coefficient applies for convection or conduction. $\endgroup$
    – JMac
    Nov 11 '19 at 14:00

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