Heat Conduction Calculator for a Plane Wall

Technical article

Understand Heat Conduction Calculator for a Plane Wall

This calculator finds the missing quantity among heat flow P, thermal conductivity k, area A, temperature difference ΔT and wall thickness L from the other four. Using a thermos flask as an example, it explains why a vacuum is the most effective way to nearly eliminate heat conduction.

What does this quantity describe?

Steady heat flow through a plane, homogeneous wall is P = k·A·ΔT/L. It grows with area and temperature difference, falls as wall thickness increases, and is smaller the lower the material's thermal conductivity k.

A thermos flask keeps drinks hot for hours because it nearly eliminates heat conduction: a vacuum between its double walls contains no matter particles to carry heat via collisions. Since conduction requires a carrier medium, it drops to essentially zero in a vacuum — what remains is the small heat loss through the narrow contact points of the walls near the neck, plus thermal radiation, further reduced by a mirrored inner surface.

Formula and variables

P = k · A · ΔT / L

P = k · A · ΔT / L
L = k · A · ΔT / P
ΔT = P · L / (k·A)

Symbol / input	Meaning
Heat flow P	The steady-state heat flow lost through the wall (not a heat quantity in joules but a power in watts in steady state). Computed as a result from k·A·ΔT/L.
Thermal conductivity k	The thermal conductivity of the wall material – how well or poorly it conducts heat. Source: the datasheet; typical values mineral wool ≈ 0.035–0.045, concrete ≈ 1.0–2.1, steel ≈ 50–60 W/(m·K). Small values mean good insulation.
Area A	The wall area through which the heat flows, normal to the flow direction. Source: calculate it from a building drawing, or measure it directly.
Temperature difference ΔT	The magnitude of the temperature difference between the warm and cold side of the wall (inside minus outside, as a positive number). Source: two temperature measurements, or a design specification.
Wall thickness L	The wall thickness in the direction of heat flow. Source: a building plan, or a measurement.

Choose the inputs correctly

Enter three of the four quantities to find the fourth. Use a k value matching the actual wall material; common insulators such as fiberglass, wool or goose down reach low values because they trap air pockets, benefiting from air's low thermal conductivity.

How to use the calculator

Select the target quantity and enter the three known values with units. For multi-layer walls, thermal resistances (L/k) of the individual layers add; calculate the combined resistance first before finding heat flow in that case.

Worked example

Given k = 0.04 W/(m·K), A = 10 m², ΔT = 20 K and L = 0.8 m, substitution gives P = 0.04 · 10 · 20 / 0.8 = 10 W.

Understand the result and units

A heat flow of 10 W through this wall is low for these dimensions and this temperature difference — a sign of a well-insulated construction. With thinner walls or a material of higher thermal conductivity, the same temperature difference would cause a considerably larger heat loss.

Heat flow in watts (W), thermal conductivity in W/(m·K), area in m², temperature difference in kelvin (equivalent to a °C difference), and wall thickness in metres.

Typical applications

Heat conduction calculations are used for building insulation, designing heat sinks and heat-exchanger walls, insulating pipework, and estimating heat losses from technical vessels.

Assumptions, limits and common mistakes

The formula applies to steady, one-dimensional conduction through a homogeneous, plane wall without surface heat-transfer resistances. Real components with thermal bridges, multiple material layers, or unsteady temperature profiles need more detailed treatment, e.g. via a U-value including surface resistances.

Common mistake: Do not equate heat conduction with a component's total heat loss — convection and radiation at surfaces add further loss for real components and aren't captured by the pure conduction coefficient k. Also, for multi-layer walls, don't use a single k-value for the whole wall; correctly add the layer resistances instead.

Frequently asked questions

Why does a thermos flask keep drinks hot for so long?

Because the vacuum between its double walls nearly eliminates conduction — without a carrier medium, no heat can transfer via particle collisions. A mirrored inner surface further reduces the remaining radiative heat loss.

Why do materials like fiberglass or down insulate so well?

Because they trap many small air pockets, benefiting from still air's low thermal conductivity, rather than being especially poor-conducting solid materials themselves.

How do I calculate heat flow through a multi-layer wall?

First add the thermal resistances L/k of each individual layer into a combined resistance, then calculate heat flow using that combined value.