Skip to content
ENExplore Calculators

Conduction Heat Transfer Calculator

Calculate conductive heat flow through a material using Fourier’s law.

About Conduction Heat Transfer Calculator

Conductive heat flow is inversely proportional to thickness, so doubling insulation thickness halves the conductive loss. Thermal resistance is thickness divided by conductivity. Because the relationship is inverse rather than linear in reduction, the first 50 mm of insulation saves far more than the second 50 mm.

Formula

Conduction Heat Transfer

Q̇ = k A ΔT ÷ L (Fourier’s law)

k
Thermal conductivity of the material
L
Thickness in the direction of heat flow

Worked example

With the default values below, Conduction Heat Transfer Calculator returns Conductive Heat Flow of 80.000. Change any input above and every figure here is recalculated in your browser.

Inputs used

Thermal conductivityminimum 0 W/m K
0.04 W/m K
Cross-sectional areaminimum 0 m2
10 m2
Material thicknessminimum 0 m
0.1 m
Temperature differenceminimum 0 K
20 K

Results

Conductive Heat Flow
80.000 W
Thermal Resistance
2.5000 m2 K/W
Equivalent U-Value
0.4000 W/m2 K
Heat Flux
8.000 W/m2
Thickness Needed to Halve the Loss
0.2000 m

Step by step

  1. 1Values used
  2. 2Conduction Heat Transfer
  3. 3Conductive Heat Flow= 80.000 W
  4. 4Thermal Resistance= 2.5000 m2 K/W
  5. 5Equivalent U-Value= 0.4000 W/m2 K
  6. 6Heat Flux= 8.000 W/m2
  7. 7Thickness Needed to Halve the Loss= 0.2000 m

Assumptions and limits

  • The result is derived only from the 4 inputs listed above. Anything not among them is not part of the calculation.
  • In particular, this calculator does not take inflation, tax, fees or charges or compounding frequency into account.
  • Figures are rounded for display to 4 decimal places, so totals built from several rounded parts can differ slightly from the unrounded arithmetic.

How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.

Frequently Asked Questions

How is Conduction Heat Transfer calculated?

Q̇ = k A ΔT ÷ L (Fourier’s law) Conductive heat flow is inversely proportional to thickness, so doubling insulation thickness halves the conductive loss. Thermal resistance is thickness divided by conductivity.

Why does Conduction Heat Transfer matter?

Because the relationship is inverse rather than linear in reduction, the first 50 mm of insulation saves far more than the second 50 mm.

What values do I need to enter?

This calculator takes 4 inputs: Thermal conductivity, Cross-sectional area, Material thickness, Temperature difference. The pre-filled defaults are a realistic starting point — replace them with figures from your own environment for a result you can act on.

Related Calculators

Conduction Heat Transfer Calculator | Calcrivo