EngineeringFormulaThermodynamics
Heat Conduction Calculator
Calculate steady one-dimensional heat conduction rate, heat flux, conductivity, area, temperature difference, or thickness using Fourier's law.
Heat conduction through a flat layer is commonly estimated with Fourier's law. This calculator covers two closely related plane-wall forms: total heat-transfer rate, Q̇ = k A ΔT / L, and heat flux, q'' = k ΔT / L. Use Heat Rate mode when total wall area matters and you want the overall watts crossing the wall. Use Heat Flux mode when you want watts per unit area through the material itself.
k · Material property describing how readily heat conducts through the wall.
A · Cross-sectional area normal to the heat-flow direction.
ΔT · Magnitude of the temperature drop across the conduction path.
L · Length of the conduction path through the layer.
Plane-wall, steady one-dimensional conduction only. If you need resistance-form results directly, use the Thermal Resistance Calculator instead.
Solution
Enter the required values to calculate heat transfer rate.
Q̇ = k A ΔT / L
Formula Sheet
- Q̇Heat Transfer Rate
- q''Heat Flux
- kThermal Conductivity
- AArea
- ΔTTemperature Difference
- LWall Thickness
Variables & Units
| Symbol | Variable | Description | Common Units |
|---|---|---|---|
| Q̇ | Heat Transfer Rate | Total steady heat-flow rate through the full wall area. | W, kW, Btu/h |
| q'' | Heat Flux | Heat-transfer rate per unit area through the wall. | W/m², kW/m², W/cm², Btu/(h·ft²) |
| k | Thermal Conductivity | Material property describing how readily heat conducts through the wall. | W/(m·K), Btu/(h·ft·°F) |
| A | Area | Cross-sectional area normal to the heat-flow direction. | cm², m², in², ft² |
| ΔT | Temperature Difference | Magnitude of the temperature drop across the conduction path. | K, °C, °F |
| L | Wall Thickness | Length of the conduction path through the layer. | mm, cm, m, in |
How to Use This Calculator
- 01Choose the calculation mode first. Use Heat Rate for total watts through the wall, or Heat Flux for watts per unit area.
- 02Select which variable to solve for. The calculator then shows only the inputs needed for that form of Fourier's law.
- 03Enter positive magnitudes for conductivity, area, temperature difference, thickness, heat-transfer rate, and heat flux. This page does not track sign direction with the negative sign from differential Fourier's law.
- 04Use a temperature difference, not an absolute temperature. A change of 20 °C equals a change of 20 K for conduction calculations. Use the material table as a starting point, then replace it with manufacturer data for final work.
How the Formula Works
For a steady one-dimensional plane wall with uniform area and constant thermal conductivity, Fourier's law reduces to Q̇ = k A ΔT / L. Heat transfer increases when conductivity, area, or temperature difference increases, and it decreases when the conduction path gets thicker.
Heat flux is the heat-transfer rate per unit area: q'' = Q̇ / A = k ΔT / L. That makes Heat Flux mode useful when you want a wall-performance quantity independent of panel size, while Heat Rate mode is the better choice when total wall area is part of the design problem.
Worked Example 01
Heat transfer rate through insulation
Known
- Thermal Conductivity (k): 0.04 W/(m·K)
- Area (A): 10 m²
- Temperature Difference (ΔT): 25 K
- Wall Thickness (L): 0.1 m
Formula
Q̇ = k A ΔT / L
Substitution
Q̇ = 0.04 × 10 × 25 / 0.1
Result
Q̇ = 100 W
A 100 mm insulation layer over 10 m² with a 25 K temperature drop conducts 100 W in this steady plane-wall approximation.
Worked Example 02
Required thickness for a target heat leak
Known
- Heat Transfer Rate (Q̇): 100 W
- Thermal Conductivity (k): 0.04 W/(m·K)
- Area (A): 10 m²
- Temperature Difference (ΔT): 25 K
Formula
L = k A ΔT / Q̇
Substitution
L = 0.04 × 10 × 25 / 100
Result
L = 0.1 m
To limit the steady conduction rate to 100 W under those conditions, the required thickness is 0.1 m.
Worked Example 03
Heat flux through a thin layer
Known
- Thermal Conductivity (k): 0.5 W/(m·K)
- Temperature Difference (ΔT): 40 K
- Wall Thickness (L): 0.02 m
Formula
q'' = k ΔT / L
Substitution
q'' = 0.5 × 40 / 0.02
Result
q'' = 1000 W/m²
This wall conducts 1000 W of heat per square meter under the given steady one-dimensional conditions.
Worked Example 04
Back-solve thermal conductivity from measured heat flux
Known
- Heat Flux (q''): 1000 W/m²
- Temperature Difference (ΔT): 40 K
- Wall Thickness (L): 0.02 m
Formula
k = q'' L / ΔT
Substitution
k = 1000 × 0.02 / 40
Result
k = 0.5 W/(m·K)
Measured heat-flux data can be rearranged to estimate conductivity when the thickness and temperature drop are known.
Applications
- 01Estimating watts conducted through insulation, panels, plates, and building layers
- 02Checking heat flux through walls, liners, and flat thermal barriers
- 03Back-solving required thickness, conductivity, or allowable temperature drop in preliminary thermal design
- 04Screening whether insulation thickness changes have a meaningful effect before doing a full U-value or heat-loss calculation
Representative Thermal Conductivities
| Material | Thermal Conductivity | Use note |
|---|---|---|
| Copper | 390 W/(m·K) | High-conductivity metal; verify alloy and temperature. |
| Aluminum | 200 W/(m·K) | Common heat-spreader material. |
| Steel | 50 W/(m·K) | Varies significantly by grade. |
| Water | 0.6 W/(m·K) | Conduction only; moving water also convects. |
| Brick | 0.4-0.5 W/(m·K) | Representative masonry range. |
| Wood | 0.2 W/(m·K) | Strongly depends on species and moisture. |
| Cork | 0.04 W/(m·K) | Insulation-like material. |
| Air | 0.026 W/(m·K) | Still air only; gaps can include convection. |
Assumptions
- 01The wall is modeled as steady, one-dimensional conduction through a uniform flat layer.
- 02Thermal conductivity is treated as constant over the temperature range.
- 03The page works with positive magnitudes and does not represent directional sign with the negative gradient form.
Where This Model Stops
- 01Does not model cylindrical or spherical conduction, fins, contact resistance, internal heat generation, or multidimensional heat spreading.
- 02Does not include convection or radiation boundary resistances unless you convert the problem into an equivalent conduction-only screening estimate.
- 03Use the Thermal Resistance Calculator if you want resistance-form results directly rather than Fourier-law heat rate or heat flux.
- 04Does not model pipes or cylinders directly; for pipe insulation, use a cylindrical conduction method rather than treating the pipe wall as a flat slab.
References
- [1]16.2 Introduction to Conduction
MIT Unified Engineering Notes
States the plane-wall Fourier-law forms for total heat-transfer rate and heat flux, and defines thermal conductivity and heat flux units.
- [2]16.3 Steady-State One-Dimensional Conduction
MIT Unified Engineering Notes
Develops the steady one-dimensional conduction model and applies it to a plane slab.
- [3]NIST Guide to the SI, Appendix B.9
National Institute of Standards and Technology
Provides official conversion factors for heat flow rate, heat-flux-related energy-per-area-time units, and thermal conductivity.
Frequently Asked Questions
What is the difference between heat transfer rate and heat flux?
Heat transfer rate Q̇ is the total watts crossing the whole wall, while heat flux q'' is watts per unit area. They are related by q'' = Q̇/A.
How is this different from the Thermal Resistance Calculator?
This page is centered on Fourier's law and explicitly exposes heat flux as well as total heat rate. The Thermal Resistance Calculator expresses the same plane-wall physics in resistance form R = L/(kA) and is better when you want resistance directly.
Can I use Celsius for the temperature difference?
Yes. For temperature differences, 1 °C change equals 1 K change, so either is fine. Fahrenheit differences are larger units and are converted appropriately by the calculator.
Can I use this for pipe insulation?
Only as a rough screening estimate for thin, nearly flat sections. Cylindrical conduction uses a logarithmic radius term, so a pipe-wall or pipe-insulation design should use a cylindrical conduction method when accuracy matters.