MRMaschinenbaurechner

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 λ per layer or set it manually - the tool returns thermal resistances, heat flow and the temperature profile across the layers, live with every input.

Calculation

Geometry and boundary conditions
Layer build-up (inside/warm to outside/cold)
Layer 1

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. λ values are guide values at room temperature.

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Formulas and fundamentals

Fourier's law

The basis is Fourier's law of steady-state heat conduction. For a plane wall of area A and thickness s with thermal conductivity λ, the heat flow is:

Q = λ·A·ΔT/s

equivalently via the thermal resistance:

R = s/(λ·A) Q = ΔT/R

For a hollow cylinder (pipe) of length L with inner radius ri and outer radius ra:

R = ln(ra/ri)/(2π·λ·L) Q = 2π·λ·L·ΔT/ln(ra/ri)

Multilayer build-up

Several layers act thermally like resistors in series: the individual resistances add up to R_total, and the heat flow is the same in every layer:

Q = ΔT/R_total

The temperature drop across a single layer follows from:

ΔT_i = Q·R_i

the layer with the largest resistance (small λ or large thickness) carries the largest share of the temperature gradient. That is exactly the effect of insulation.

Units and model limits

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 (λ = 0.04 W/(m·K)) with 10 m² area separates two zones with a 20 K temperature difference. The thermal resistance is R = s/(λ·A) = 0.1/(0.04·10) = 0.25 K/W.

This gives the heat flow Q = ΔT/R = 20/0.25 = 80 W, corresponding to a heat flux density of 8 W/m². Halving λ with a better insulation material also halves the heat loss.

For a pipe with ri = 50 mm, ra = 100 mm, length 1 m, λ = 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/(λ·A). For a pipe the area increases outward, so the natural logarithm of the radii appears in the denominator: R = ln(ra/ri)/(2π·λ·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 λ 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). λ 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/(α·A) if required.

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