We may realize that since heat transfer due to conduction and convection are linear with respect to temperature differences, a series of related temperatures and heat transfers can be solved as a unified linear system. Notably, the method to follow is only applicable with there is no heat generation throughout the media.

Thermal Resistances

Helpful for this is observe that our conduction and convection equations are analogous to Ohm’s law, where voltage drop is analogous to temperature differences, heat transfer is analogous to current, and thus, various heat transfer properties may be considered as thermal resistances. Notably, this analysis method may not be applied as well to radiation, because it heat transfer due to radiation is non-linearly related to temperature differences, so that the effective thermal resistance of radiation is dependent on temperature.

$$ i=\frac VR~~\approx~~\dot Q_{cond}=\frac{T_1-T_0}{\frac t{kA}}\approx\dot Q_{conv}=\frac{T_s-T_\infin}{\frac 1{hA_s}} $$

$$ R_{cond}=\frac t{kA}~~~~~~R_{conv}=\frac 1{hA_s}~~~~~~R_{rad}=\frac{1}{\varepsilon\sigma A(T_s^2+T_{surr}^2)(T_s+T_{surr})} $$

Finally, recall Kirchhoff’s Current Law, which states that the sum of currents out of a node must equal zero. Similarly, is the first law of thermodynamics, the conservation of energy indicates that for a steady state system, a rate balance of heat transfer must exist. The sum of heat transfers out of a temperature node must equal zero.

$$ \Sigma\dot Q_i=0 $$

For a steady state conduction and convection system, we may draw an analogous thermal circuit and apply Ohm’s law and Kirchhoff’s current law to solve for heat transfers between various temperature nodes, and the temperature of various nodes. When successive regions have differing thermal resistances, these resistances are in series. When parallel parts of a surface have different thermal resistances, these resistances are in parallel. These nodes may indicate temperatures on alternate sides of a conducting solid, on either side of a contact between two solids (which has a characteristic contact resistance), surface temperatures, and convection fluid temperatures.

Generalized Solution Technique

We may apply a generalized linear algebra technique to solving these problems with the linear system $\bold{AT=B}$.

1. Given a temperature boundary condition for a node $i$, we simply set the temperature at that node.
   1. $A_{i,i}=1$
   2. $B_i=T_{bi}$.
2. Given a heat transfer boundary condition at node $i$, we set set the sum of heat transfers into the node to be equal to our boundary heat flux.
   1. $A_i=[~\cdots, -\frac 1{R_{i,i-1}},~ \Sigma\frac 1{R_{i,k}},~ -\frac 1{R_{i,i+1}},~\cdots~]$
   2. $B_{i}=\dot Q_{bi}$
3. Given a normal node with no boundary heat flux, we set the sum of heat transfers into the node to be equal to zero.
   1. $A_i=[~\cdots, -\frac 1{R_{i,i-1}},~ \Sigma\frac 1{R_{i,k}},~ -\frac 1{R_{i,i+1}},~\cdots~]$
   2. $B_{i}=0$

Thus, the system of temperatures is obtained by the solution to the linear system.

$$ \bold T=\bold A^{-1}\bold B $$