Heat conduction through a wall
Solve steady-state conduction through a single layer or a two-layer wall. Thermal resistance, heat flux, the U-value and the interface temperature all come out of the same series network.
| Layer | k (W/m·K) | Thickness | R (K/W) | ΔT across |
|---|---|---|---|---|
| Layer 1 | 0.720 | 200 mm | 0.0231 | 2.500 K |
| Layer 2 | 0.040 | 100 mm | 0.2083 | 22.500 K |
| Total | — | 300 mm | 0.2315 | 25.000 K |
Resistances add in series. The layer with the highest R takes the largest share of the temperature difference — which is why insulation works.
Q = k·A·ΔT/L · steady state, one dimension
R = L/(kA) · series resistances add
q″ = Q/A = ΔT/R″ · R″ = L/k per unit area
T_i = T_hot − Q·R₁ · linear drop across each layer
A clear path from inputs to results.
Steady conduction through a wall is Ohm’s law for heat: a temperature difference drives a flow through a resistance that depends on thickness, area and conductivity. This engine builds that series network for one or two layers and reports the heat flow, the flux, the U-value and the temperature at the interface.
How to use
- Enter the conductivity of the first layer. Brick is about 0.72 W/m·K; mineral wool about 0.04.
- Set its thickness in millimetres.
- Add a second layer, or set its thickness to zero for a single-layer wall.
- Enter the wall area.
- Set the hot-side and cold-side temperatures.
- Read the heat flow and check the interface temperature for condensation risk.
The formula
Reference: Incropera & DeWitt, "Fundamentals of Heat and Mass Transfer", ch. 3; ISO 6946 for building U-values.
Worked example
200 mm brick plus 100 mm mineral wool over 12 m² (default inputs)
- Layer 1: k = 0.72 W/m·K, 200 mm.
- Layer 2: k = 0.04 W/m·K, 100 mm.
- Area 12 m², 20 °C inside, −5 °C outside.
- 1
R₁ = 0.20 / (0.72 × 12) = 0.0231 K/W.
- 2
R₂ = 0.10 / (0.04 × 12) = 0.2083 K/W.
- 3
Total R = 0.2315 K/W, so Q = 25 / 0.2315 = 108 W.
- 4
Flux = 108 / 12 = 9.0 W/m²; U-value = 1 / (0.2315 × 12) = 0.36 W/m²K.
- 5
Interface temperature: 20 − 108 × 0.0231 = 17.5 °C.
108 W total · 9.0 W/m² flux · U-value 0.36 W/m²K · interface at 17.5 °C.
Interpreting the result
The insulation layer holds 90% of the thermal resistance despite being half the thickness of the brick. That is the whole point of insulation: resistance scales inversely with conductivity, and the conductivity ratio here is 18:1. The interface temperature of 17.5 °C is close to the inside face, which means the dew point will be reached inside the insulation rather than on the structural wall — the correct place for it to happen.
Assumptions & limitations
- One-dimensional steady-state conduction. Corners and thermal bridges need a 2D or 3D model.
- Surface convection films are excluded. Real walls add roughly 0.12–0.17 m²K/W of internal and external film resistance.
- Contact resistance between layers is ignored, which overstates the real heat flow slightly.
- Conductivity is constant. Mineral wool varies with temperature and moisture; wet insulation conducts far better than dry.
- No radiation across air gaps and no moisture transport are modelled.
Frequently asked questions
Why is R divided by area in this model?
Because thermal resistance as used here is absolute (K/W), so it includes the wall area. Building R-values are quoted per unit area (m²K/W), which is why the U-value row multiplies back by area.
Does thickness matter more than conductivity?
Both enter identically — resistance is L/k. Doubling thickness halves the heat flow; halving conductivity does the same. Insulation wins in practice because it improves k by a factor of 20, while thickness can only double or triple.
What about a cavity?
A ventilated cavity is better modelled as a surface resistance than as a conduction layer. A sealed air cavity conducts and convects, so it needs an effective conductivity of about 0.15–0.2 W/m·K, not 0.026.
What U-value should a wall have?
New-build standards vary: UK Part L requires around 0.18 W/m²K for walls, passive house around 0.15, and older buildings are often 1.0 or worse. Lower is better.
Why is the interface temperature useful?
Because if it drops below the dew point, water condenses inside the wall. Placing insulation on the outside keeps the structural layer warm and dry; placing it on the inside can push the dew point into the structure.
Does this apply to pipes as well as walls?
Not directly. Cylindrical conduction uses a logarithmic resistance, R = ln(r₂/r₁)/(2πkL), because the area changes with radius.
Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.