Thermal conduction formula
Fourier law of conduction gives the rate heat flows through a solid. It rises with the material conductivity, the area, and the temperature difference, and falls with thickness, which is the basis of all insulation design.
Variables
| Q | Heat flow rate | W |
| k | Thermal conductivity | W/m·K |
| A | Cross-sectional area | m² |
| ΔT | Temperature difference | K or °C |
| L | Thickness | m |
Rearranged
Q = A × ΔT / R
Worked example
An insulating wall (k = 0.04) is 10 m², 0.1 m thick, with 20 degrees across it.
Apply Fourier law: 0.04 × 10 × 20 / 0.1 = 80.
Heat flow is proportional to conductivity, area, and temperature difference, and inversely proportional to thickness. Doubling the insulation thickness halves the loss. The ratio of thickness to conductivity is the thermal resistance, or R-value, used to rate insulation.
Need conductivity values?
See the Thermal Conductivity Chart for metals and materials.
Fourier law of conduction
Fourier law states that heat flows from hot to cold at a rate set by the temperature gradient and the material conductivity. A good conductor like copper moves heat readily; an insulator like foam resists it. The same law, applied across a wall, gives the steady heat loss and underlies every insulation and heat-sink calculation.
Thermal resistance and R-value
Rewriting the law, heat flow equals area times temperature difference divided by thermal resistance, where resistance is thickness over conductivity. This R-value adds up for layered walls just like electrical resistors in series, which is why insulation is rated by R-value: a higher number means less heat gets through.
FAQ
What is the thermal conduction formula?
Fourier law: heat flow equals conductivity times area times temperature difference divided by thickness, Q = k A delta T / L.
What is an R-value?
The thermal resistance of a layer, equal to its thickness divided by its conductivity. Higher R-values mean better insulation and less heat flow.
How does thickness affect heat loss?
Heat flow is inversely proportional to thickness, so doubling the insulation thickness halves the conductive heat loss.