Understanding the heat conduction calculator — concept and principle
This calculator applies Fourier's law to find how much heat flows through a flat layer of material, such as an insulation board, a wall or a window pane, when one side is warmer than the other. It also returns the layer's thermal resistance and its U-value, the numbers used to compare insulation and estimate heat loss.
How it is calculated
For steady one-dimensional conduction through a flat slab: Q = k · A · ΔT / d.
- k: thermal conductivity of the material in W/m·K (how easily it conducts heat).
- A: area in m² through which heat passes.
- ΔT: temperature difference between the two faces in K (a difference of 1 K equals a difference of 1 °C).
- d: thickness, entered in mm and converted to metres (d = t/1000).
The layer's thermal resistance is R = d/k (m²K/W), and its conductance per unit area is U = k/d = 1/R (W/m²K), so Q = U · A · ΔT.
The principle
Fourier's law states that heat flux is proportional to the temperature gradient: q = −k dT/dx. In a flat, uniform slab with fixed face temperatures and no internal heat sources, the temperature falls linearly through the thickness, so the gradient is simply ΔT/d. Doubling the thickness halves the gradient and therefore halves the heat flow; doubling k or the area doubles it. As with Ohm's law, layers in series simply add their resistances.
Worked example
- Inputs: k = 0.04 W/m·K (typical of mineral wool or polystyrene), A = 10 m², ΔT = 20 K, t = 100 mm.
- d = 100/1000 = 0.1 m.
- Q = 0.04 × 10 × 20 / 0.1 = 80 W.
- R = 0.1 / 0.04 = 2.5 m²K/W.
- U = 0.04 / 0.1 = 0.4 W/m²K (check: 0.4 × 10 × 20 = 80 W).
| Thickness (mm) | Heat flow (W) |
|---|---|
| 25 | 320 |
| 50 | 160 |
| 75 | 106.6666667 |
| 100 | 80 |
| 125 | 64 |
| 150 | 53.33333333 |
| 200 | 40 |
How to read the result
Q is the continuous heat flow in watts, so 80 W running for 24 hours is 1.92 kWh. A larger R or a smaller U means better insulation. Typical conductivities: still air about 0.026, insulation foams and wools roughly 0.02–0.04, timber around 0.1–0.2, brick about 0.6–1, concrete about 1–2, steel around 50 and copper about 400 W/m·K.
Common mistakes
- Entering the thickness in metres when the field expects millimetres.
- Using a material's k from a different temperature or moisture condition.
- Forgetting surface (air film) resistances and other layers when calculating a whole wall.
- Using it for changing (transient) conditions.
Frequently asked questions
What is the difference between thermal conductivity and U-value?
Conductivity k is a material property independent of thickness. The U-value describes a specific layer or assembly of a given thickness, including all its layers.
How do I combine several layers?
Add the R values (d/k) of each layer plus the surface resistances, then take U = 1/R_total.
Sources: Incropera, DeWitt, Bergman and Lavine, Fundamentals of Heat and Mass Transfer, Wiley · ISO 6946 — Building components and building elements: Thermal resistance and thermal transmittance