Combine thermal resistances in series

Find the unknown interface temperature

Two layers share the 60 K difference between the 80 °C and 20 °C outer surfaces. The interface temperature is unknown. Treating both layers as if each spanned the full 60 K would describe incompatible boundaries. Instead use the fact that the same rate crosses both layers at steady state.

The control changes the thickness and conductivity of layer 2. Layer 1 remains 20 mm thick with k₁ = 0.2 W/(m K). Both layers share area A = 0.5 m², with perfect contact and no lateral leakage. Each selected setting describes a new steady condition.

Express each layer as a resistance

Rearrange the slab relation P = kAΔT/L to ΔT = PR, where R = L/(kA). R is thermal resistance in K/W: the temperature difference required per watt through the layer. A larger R means a smaller rate for the same face-temperature difference.

Do not confuse R with heat capacity C in J/K. Nor is this whole-assembly resistance the same quantity as an area-normalized insulation value in m² K/W. Here area is already included. The OpenStax conduction reference provides the slab relation; this lesson uses it to construct an original two-layer energy account.

For layer 1, R₁ = 0.020/(0.2 × 0.5) = 0.2 K/W. At the defaults, R₂ = 0.030/(0.1 × 0.5) = 0.6 K/W.

Add drops, not heat rates

The total temperature drop equals the sum across layers: Th − Tc = PR₁ + PR₂. Therefore P = (Th − Tc)/(R₁ + R₂). Series resistances add because the same energy-transfer rate passes through successive layers.

At the defaults, Rtotal = 0.8 K/W, so P = 60/0.8 = 75 W. Layer 1’s drop is PR₁ = 15 K and layer 2’s is PR₂ = 45 K. The interface temperature is 80 − 15 = 65 °C, also 20 + 45 = 65 °C. Both calculations must agree.

The larger resistance receives the larger share of the total temperature drop. The drop ratio is R₂/R₁ = 3. This does not mean layer 2 carries three times the heat rate or stores three times the energy; neither claim follows from a steady resistance account.

Test a tempting shortcut

Hold conductivity at 0.1 W/(m K). At 20 mm, layer 2 has R₂ = 0.4 K/W. Total R is 0.6 K/W and P is 100 W, with interface at 60 °C. Double that layer’s thickness to 40 mm: R₂ doubles to 0.8 K/W, but total R rises only to 1.0 K/W. The new rate is 60 W, with interface at 68 °C.

The whole-assembly rate falls to 60%, not 50%. Layer 1’s resistance did not double. The interface gets hotter because a smaller heat rate needs a smaller drop across the unchanged first layer. Capture both cases and explain each change from the resistance ledger.

Increasing layer 2 conductivity at fixed thickness reduces its resistance and increases the total rate. Equal conductivities produce equal temperature slopes because the same area and rate apply. If layer 2 is also 20 mm thick with k₂ = 0.2 W/(m K), both resistances are 0.2 K/W: the rate is 150 W, both drops are 30 K and the interface is 50 °C.

Keep the boundaries and path arrangement explicit

Adding a layer in series increases total resistance under these assumptions. A gap or a metal connection that creates another route for heat is a different geometry: parallel paths share boundary temperatures and their rates add. This widget does not calculate such a thermal bridge.

Real outer surface temperatures can differ from nearby air temperatures because convection and radiation also act at surfaces. Those effects require additional boundary relations. Imperfect contact can produce a temperature jump. Temperature-dependent conductivity can curve a profile. The current piecewise-linear graph does not estimate any of these effects.

Changing the order of these two constant-property layers would leave total resistance and rate unchanged, but it would change the interface temperature because a different resistance lies next to the hot surface. This conclusion depends on the ideal assumptions; it is not a rule for every composite material or boundary condition.

Transfer to three layers

Three series layers have resistances 0.1, 0.3 and 0.2 K/W. The outer surfaces are held at 70 °C and 10 °C. Calculate the common rate and temperatures after the first and second layers. Then add another 0.4 K/W layer and predict the new rate.

Check the resistance-network transfer

Total resistance is 0.6 K/W, so P = 60/0.6 = 100 W. Drops are 10, 30 and 20 K. The interfaces are 60 °C and 30 °C, ending at 10 °C. Adding 0.4 K/W in series gives total 1.0 K/W and P = 60 W. Do not retain the old interface temperatures: all drops must be recalculated with the new common rate.