Heat Conduction Calculator (Fourier)
Calculate the steady-state heat flow through multilayer plane walls and pipes using Fourier's law. Enter geometry and layer build-up, choose the thermal conductivity lambda per layer or set it manually – the tool returns thermal resistances, heat flow and the temperature profile across the layers, live with every input.
Heat Conduction Calculator (Fourier)
Model: steady-state, one-dimensional heat conduction after Fourier with thermal resistances in series. Convective heat transfer and thermal radiation are not included; the temperature difference applies across the entire layer build-up. lambda values are guide values at room temperature.
Results
Calculating …
Formulas and fundamentals
The basis is Fourier's law of steady-state heat conduction. For a plane wall of area A and thickness s with thermal conductivity lambda, the heat flow is Q = lambda·A·dT/s, equivalently via the thermal resistance R = s/(lambda·A) as Q = dT/R. For a hollow cylinder (pipe) of length L with inner radius ri and outer radius ra, R = ln(ra/ri)/(2π·lambda·L) and therefore Q = 2π·lambda·L·dT/ln(ra/ri).
Several layers act thermally like resistors in series: the individual resistances add up to R_ges, and the heat flow Q = dT/R_ges is the same in every layer. The temperature drop across a single layer follows from dT_i = Q·R_i – the layer with the largest resistance (small lambda or large thickness) carries the largest share of the temperature gradient. That is exactly the effect of insulation.
Internally the tool works in SI units (metres, W/(m·K), kelvin). Convective heat transfer at the surfaces and thermal radiation are not included; for a pure conduction calculation with a given temperature difference across the build-up this is the governing part.
Worked example
A 100 mm thick layer of mineral wool (lambda = 0.04 W/(m·K)) with 10 m² area separates two zones with a 20 K temperature difference. The thermal resistance is R = s/(lambda·A) = 0.1/(0.04·10) = 0.25 K/W.
This gives the heat flow Q = dT/R = 20/0.25 = 80 W, corresponding to a heat flux density of 8 W/m². Halving lambda with a better insulation material also halves the heat loss.
For a pipe with ri = 50 mm, ra = 100 mm, length 1 m, lambda = 0.04 and a 50 K difference, R = ln(2)/(2π·0.04·1) = 2.76 K/W and Q = 18.1 W. Unlike the plane wall, the area grows with the radius, which is why the logarithmic term appears.
Frequently asked questions
What is the difference between a plane wall and a pipe?
For a plane wall the cross-sectional area is constant and the thermal resistance is R = s/(lambda·A). For a pipe the area increases outward, so the natural logarithm of the radii appears in the denominator: R = ln(ra/ri)/(2π·lambda·L). For thin pipes with small ra/ri the result approaches the plane wall.
Why do thermal resistances add up?
Because in a steady-state, one-dimensional build-up the heat flow through every layer is the same – analogous to electric current through resistors in series. The temperature difference is distributed proportionally across the layers, and the total resistance is the sum of the individual resistances.
Which lambda values are provided?
Common guide values at room temperature: insulation such as mineral wool (0.04), EPS (0.035) or PUR (0.025), building materials such as wood (0.13) or concrete (2.1), and metals such as steel (50), stainless steel (15) or copper (400). lambda is temperature dependent; for precise calculations the value can be entered manually.
Are convection and radiation included?
No. The tool treats pure heat conduction with a given temperature difference across the layer build-up. Convective heat transfer at the surfaces and radiative exchange must be added as additional resistances 1/(alpha·A) if required.
Related tools
Insulated Pipe Heat Loss
Calculate the heat loss of insulated pipes online: pipe wall and insulation as a series of cylindrical thermal resistances, heat flow per length and total, surface temperature with touch-protection indicator and radial temperature profile.
Convection Coefficient
Calculate the convective heat transfer coefficient α for forced pipe flow online: Reynolds, Prandtl and Nusselt number, Dittus-Boelter and Gnielinski, laminar Nu = 3.66, heat flow Q = α·A·ΔT and report PDF.
Material properties
Look up strength and material properties of common engineering materials.