Thermal Resistance Network
Theory: heat flow as a circuit
Overview
I want to estimate a house wall's insulation performance. Can I use just the insulation thickness? There are studs too.
Ignoring studs will make you optimistic. A thermal resistance network treats heat flow like a circuit: temperature difference as voltage, heat flow as current, and gypsum board, insulation, plywood and surface heat transfer as resistances in series. Where studs and insulation sit side by side, the layer is in parallel. How you connect the parallel part changes the answer, so the trick is to calculate two ways and look at the range.
Basic relations
$t$ is layer thickness, $k$ conductivity, $h$ surface heat transfer coefficient, $U$ the overall transmittance (heat per m² per K) and $A$ area. Resistances add in series; their reciprocals add in parallel.
Two methods for the parallel part
$f$ is the stud area fraction. The parallel-path method ($U_{pp}$) assumes the stud and insulation paths run separately from inside to outside. The isothermal-planes method ($U_{ip}$) assumes uniform temperature across each layer interface, freely allowing lateral flow. The true value lies between.
The analogy between Fourier and Ohm
Fourier formulated the law of heat conduction in 1822, and Ohm published his law of electrical conduction in 1827 — reportedly modelling his thinking on Fourier's theory of heat. Because heat and current obey equations of the same form, mid-20th-century engineers even built analogue computers that solved thermal problems with electrical circuits, wiring resistors and capacitors to reproduce the temperature response of walls and buildings. Today thermal resistance networks are indispensable for quick electronics-cooling estimates.
Worked example: a framed wall
Layer resistances
| Layer | Condition | Resistance (m²·K/W) |
|---|---|---|
| Inside surface | h = 7.7 W/(m²·K) | 0.130 |
| Gypsum board | 12.5 mm, k = 0.25 | 0.050 |
| Insulation | 100 mm, k = 0.04 | 2.500 |
| Plywood | 12 mm, k = 0.13 | 0.092 |
| Outside surface | h = 25 W/(m²·K) | 0.040 |
Without studs the total is 2.81 m²·K/W and U = 0.356 W/(m²·K); insulation provides about 90% of the resistance.
With studs
| Studs | Area fraction | Parallel path | Isothermal planes |
|---|---|---|---|
| Wood (k = 0.13) | 10% | 0.413 | 0.425 |
| Wood (k = 0.13) | 15% | 0.441 | 0.458 |
| Steel (k = 50, thin web) | 0.25% | 0.363 | 1.089 |
| Steel (k = 50, thin web) | 1% | 0.384 | 2.010 |
For wood studs the two methods nearly agree, but for steel they differ threefold.
Wood conducts only about three times better than insulation, so lateral concentration of heat is weak and either assumption is fine — averaging to 0.42–0.45 is enough. Steel conducts over a thousand times better, so heat is drawn sideways into the thin steel web. Where the truth lies between the two assumptions can't be settled by a 1D network; such thermal bridges need 2D or 3D conduction analysis. And even wood studs raise U by about 20% over the stud-free wall — not something to overlook.
Converting to heat loss
For 100 m² of wall and a 20 K indoor–outdoor difference, the stud-free wall (0.356) loses about 710 W and the 15% wood-stud wall (mean 0.450) about 900 W. Differences in U translate directly into heating load.
Use and limits
Where networks work well
- Early design: quick comparisons of materials and thicknesses.
- Electronics cooling paths: junction → case → TIM → heat sink → air in series, to find the dominant resistance.
- Checking detailed analysis: confirm FEA results are the right order of magnitude.
Limits
- Where lateral flow is strong, as at thermal bridges, the two methods diverge widely.
- Temperature-dependent properties or coefficients require iteration.
- Transients require adding heat capacities (capacitors) to the network.
“More insulation, still condensation”
Insulation was thickened, yet stripes of black mould appear on part of the wall — often a thermal bridge where steel columns or fixings carry heat to the inside. The inside surface over the bridge runs colder, drops below the dew point and condenses. A network may show an acceptable overall U-value, but local surface temperature needs 2D analysis. Treat heat-loss calculation and condensation checks as separate problems with different accuracy needs.
Common mistakes
Mistakes and fixes
| Mistake | Effect | Fix |
|---|---|---|
| Ignoring studs | Heat loss about 20% low | Include as a parallel layer |
| Using only one method | Error size unknown | Bound with both |
| Handling thermal bridges by network | Wrong loss and condensation | 2D/3D analysis |
| Forgetting surface resistances | U overestimated | Include inside and outside surface resistances |
| Ignoring contact resistance | Electronics temperatures low | Add interface resistances |
I'd like to try thermal resistance calculations myself.
Try the building heat loss calculator for building heat loss and the 1D heat conduction tool and 2D heat conduction tool for 1D and 2D conduction. Related topics are covered in composite walls, multi-layer walls, thermal contact resistance and critical insulation radius.
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