Editing Thermal conduction (section)

==Conductance==
Writing
<math display="block"> U = \frac{k}{\Delta x}, </math>
where {{mvar|U}} is the conductance, in W/(m<sup>2</sup> K).

Fourier's law can also be stated as:
<math display="block"> \frac{\Delta Q}{\Delta t} = U A\, (-\Delta T).</math>

The reciprocal of conductance is resistance, <math>\big. R</math> is given by:
<math display="block"> R = \frac{1}{U} = \frac{\Delta x}{k} = \frac{A\, (-\Delta T)}{\frac{\Delta Q}{\Delta t}}.</math>

Resistance is additive when several conducting layers lie between the hot and cool regions, because {{math|''A''}} and {{math|''Q''}} are the same for all layers. In a multilayer partition, the total conductance is related to the conductance of its layers by:
<math display="block"> R = R_1+ R_2 + R_3 + \cdots</math> or equivalently <math display="block">\frac{1}{U} = \frac{1}{U_1} + \frac{1}{U_2} + \frac{1}{U_3} + \cdots</math>

So, when dealing with a multilayer partition, the following formula is usually used:
<math display="block"> \frac{\Delta Q}{\Delta t} = \frac{A\,(-\Delta T)}{\frac{\Delta x_1}{k_1} + \frac{\Delta x_2}{k_2} + \frac{\Delta x_3}{k_3}+ \cdots}.</math>

For heat conduction from one fluid to another through a barrier, it is sometimes important to consider the conductance of the [[thin film]] of fluid that remains stationary next to the barrier. This thin film of fluid is difficult to quantify because its characteristics depend upon complex conditions of [[turbulence]] and [[viscosity]]—but when dealing with thin high-conductance barriers it can sometimes be quite significant.

===Intensive-property representation===
The previous conductance equations, written in terms of [[Intensive and extensive properties|extensive properties]], can be reformulated in terms of [[Intensive and extensive properties|intensive properties]]. <!-- material was posted as a revision suggestion for the article --> Ideally, the formulae for conductance should produce a quantity with dimensions independent of distance, like [[Ohm's law]] for electrical resistance, <math>R = V/I\,\!</math>, and conductance, <math> G = I/V \,\!</math>.

From the electrical formula: <math>R = \rho x / A </math>, where ''ρ'' is resistivity, ''x'' is length, and ''A'' is cross-sectional area, we have <math>G = k A / x \,\!</math>, where ''G'' is conductance, ''k'' is conductivity, ''x'' is length, and ''A'' is cross-sectional area.

For heat,
<math display="block"> U = \frac{k A} {\Delta x}, </math>
where {{mvar|U}} is the conductance.

Fourier's law can also be stated as:
<math display="block"> \dot{Q} = U \, \Delta T, </math>
analogous to Ohm's law, <math> I = V/R </math> or <math> I = V G .</math>

The reciprocal of conductance is resistance, ''R'', given by:
<math display="block"> R = \frac{\Delta T}{\dot{Q}}, </math>
analogous to Ohm's law, <math> R = V/I .</math>

The rules for combining resistances and conductances (in series and parallel) are the same for both heat flow and electric current.

===Cylindrical shells===
Conduction through cylindrical shells (e.g. pipes) can be calculated from the internal radius, <math>r_1</math>, the external radius, <math>r_2</math>, the length, <math>\ell</math>, and the temperature difference between the inner and outer wall, <math>T_2 - T_1</math>.

The surface area of the cylinder is <math>A_r = 2 \pi r \ell</math>

When Fourier's equation is applied:
<math display="block">\dot{Q} = -k A_r \frac{dT}{dr} = -2 k \pi r \ell \frac{dT}{dr}</math>
and rearranged:
<math display="block">\dot{Q} \int_{r_1}^{r_2} \frac{1}{r} \, dr = -2 k \pi \ell \int_{T_1}^{T_2} dT</math>
then the rate of heat transfer is:
<math display="block">\dot{Q} = 2 k \pi \ell \frac{T_1 - T_2}{\ln (r_2 /r_1)}</math>
the thermal resistance is:
<math display="block">R_c = \frac{\Delta T}{\dot{Q}}= \frac{\ln (r_2 /r_1)}{2 \pi k \ell}</math>
and <math display="inline">\dot{Q} = 2 \pi k \ell r_m \frac{T_1-T_2}{r_2-r_1}</math>, where <math display="inline">r_m = \frac{r_2-r_1}{\ln (r_2 /r_1)}</math>. It is important to note that this is the log-mean radius.

===Spherical===
The conduction through a spherical shell with internal radius, <math>r_1</math>, and external radius, <math>r_2</math>, can be calculated in a similar manner as for a cylindrical shell.

The [[surface area]] of the sphere is: <math>A = 4\pi r^2.</math>

Solving in a similar manner as for a cylindrical shell (see above) produces:
<math display="block">\dot{Q} = 4 k \pi \frac{T_1 - T_2}{1/{r_1}-1/{r_2}} = 4 k \pi \frac{(T_1 - T_2) r_1 r_2}{r_2-r_1}</math>