Thermal Resistance and Rods in Series and in Parallel

Kelvin per watt

Fourier's law for a uniform lagged rod, \(H = kA\,\Delta T/L\), can be turned round. Write it as \(\Delta T = H \times L/(kA)\). The factor \(L/(kA)\) depends only on the rod: its length, its area and its material. It tells how many kelvin are needed to drive one watt through the rod. This factor is the thermal resistance of the rod.

The copper rod of the last lesson, 1 m long and 1 cm² in section with \(k = 400\), has \(R_\text{th} = 1/(400 \times 10^{-4}) = 25\) K/W. So a difference of 100 K drives \(100/25 = 4\) W through it, as before. A long rod, a thin rod or a poor conductor has a large resistance. The dimensions of \(R_\text{th}\) are \([\text{M}^{-1}\,\text{L}^{-2}\,\text{T}^{3}\,\text{K}]\).

Rods joined end to end

Now join the copper rod end to end with the steel rod of the same size. Put the free copper end at 100 °C and the free steel end at 0 °C, and lag both. The rods are in series. In the steady state no heat can pile up at the joint. So the heat current leaving the copper is the heat current entering the steel: the same \(H\) flows through both.

For copper and steel, \(R = 25 + 200 = 225\) K/W, so \(H = 100/225 = 0.444\) W. The drop across the copper is \(HR_1 = 0.444 \times 25 = 11.1\) K. So the junction temperature is \(100 - 11.1 = 88.9\) °C. The steel, eight times the resistance, takes eight times the drop, 88.9 K. For rods of equal size, \(T_j = 100\,k_1/(k_1 + k_2)\).

Try the slider in the figure above. At \(k = 50\), steel, the junction sits high at 88.9 °C and nearly all the drop is across the steel. At \(k = 400\) the rods are alike and the junction is at 50 °C. At \(k = 800\) the second rod is the better conductor, and the junction falls to 33.3 °C. In every case the arrows for the two rods are equal.

Two rods of equal length \(L\) and equal area in series behave like one rod of length \(2L\) with an equivalent conductivity \(k\). Setting \(2L/(kA) = L/(k_1A) + L/(k_2A)\) gives \(k = 2k_1k_2/(k_1 + k_2)\). For \(k_1 = 400\) and \(k_2 = 100\) W/(m K) this is 160 W/(m K), much nearer the poorer conductor.

Rods side by side

Now place the copper rod and the steel rod side by side, as in Fig 3.1. Each runs from the body at 100 °C to the body at 0 °C. These rods are in parallel. Each has the full 100 K across it. So each carries its own current, as if the other were not there: copper 4 W and steel 0.5 W. Together they carry 4.5 W.

For the pair, \(R = 100/4.5 = 22.2\) K/W. The formula agrees: \(1/R = 1/25 + 1/200\) gives \(R = 22.2\) K/W. The pair's resistance is smaller than either rod's, because a second path always helps. For two rods of the same length, the equivalent conductivity is \(k = (k_1A_1 + k_2A_2)/(A_1 + A_2)\). For equal areas it is the plain average: \((400 + 50)/2 = 225\) W/(m K).

Where does that average come from? Both rods have the same \(\Delta T\) and the same length \(L\), so \(H = k_1A_1\,\Delta T/L + k_2A_2\,\Delta T/L\). One rod of area \(A_1 + A_2\) carrying the same \(H\) needs \(k(A_1 + A_2) = k_1A_1 + k_2A_2\). So each conductivity counts in proportion to the area it has.

Networks of rods

Real problems join rods into networks. Two rules settle them. First, in the steady state every junction passes on all the heat it receives, so the currents into a junction add up to zero. Second, every rod carries \(\Delta T/R\) from its hot end to its cold end. Suppose three identical rods meet at a point, with their far ends at 90 °C, 60 °C and 0 °C. The currents into the junction must sum to zero, so its temperature is the average, 50 °C. Then 40 units and 10 units flow in, and 50 units flow out.

Symmetry often does the work. In Fig 3.2 the paths ABD and ACD are alike, so B and C are at the same temperature. With no temperature difference, the cross rod BC carries no heat, and it could be removed. The network is then two paths of two rods each, in parallel. With each rod of 25 K/W, each path is 50 K/W and carries 2 W, 4 W in all.