Conduction Heat Transfer
What this calculator does
This conduction heat transfer calculator estimates one-dimensional steady heat flow through a material layer using Fourier's law. It is a practical tool for quick wall, slab, panel, and insulation estimates when you know conductivity, area, thickness, and temperature difference.
Conduction is one of the core ways heat moves through solids. If one side of a material is hotter than the other, energy flows from hot to cold. The rate depends on how conductive the material is, how much area participates, how thick the path is, and how large the temperature difference is.
Inputs explained
- Thermal conductivity k: Enter the material conductivity in W/(m K) or the unit expected by the page.
- Area A: Enter the heat-transfer area normal to the conduction path.
- Thickness L: Enter the distance heat must travel through the material.
- Temperatures: Enter the hot-side and cold-side temperatures so the page can compute delta T.
How it works / method
The page applies the simplest steady one-dimensional conduction relationship. It computes the temperature difference between the two sides and then scales that difference by conductivity and area, divided by thickness. The result is a heat-transfer rate and a heat-flux value so you can compare both total transfer and transfer per unit area.
Formula used
Q = (k x A x delta T) / L; heat flux = Q / A
This is Fourier's law in a very simplified steady-state form. It assumes a uniform layer, constant material conductivity, one-dimensional flow, and no thermal contact resistance or internal heat generation.
Practical note: Real heat transfer problems can be more complicated than a single-layer conduction estimate. Material properties, unit consistency, contact quality, moisture, and multidimensional effects can all change the real answer.
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Heat Flux: --
Step-by-step example
Suppose a panel has k = 0.04 W/(m K), area = 10 m^2, thickness = 0.1 m, a hot-side temperature of 30 C, and a cold-side temperature of 10 C.
- Enter 0.04 for thermal conductivity.
- Enter 10 for area and 0.1 for thickness.
- Enter 30 for the hot side and 10 for the cold side so delta T is 20 C.
- The calculator estimates the steady conduction heat-transfer rate through the panel.
- If thickness doubles, the heat-transfer rate is roughly cut in half under the same assumptions.
Use cases
- Quick insulation sanity checks for walls, roofs, coolers, and enclosures.
- Comparing candidate materials with different conductivity values.
- Teaching the direct proportionalities in Fourier's law.
- Estimating heat-transfer rates before moving into layered or transient models.
Assumptions and limitations
- The page assumes steady one-dimensional conduction and a single effective conductivity value.
- It does not model radiation, convection, thermal bridges, moisture migration, or contact resistance.
- Thermal conductivity can vary with temperature and material orientation, which this compact tool does not resolve.
- Units must be consistent. Incorrect area, thickness, or conductivity units will produce misleading outputs.
Use this calculator for quick estimates and screening. For multilayer assemblies, transient heating, or real construction details, a more complete thermal model is appropriate.
Frequently Asked Questions
Use Fourier's law in its simplest one-dimensional form: Q = kAΔT/L. Q is heat rate in watts, k is thermal conductivity in W/(m·K), A is cross-sectional area in m2, ΔT is the temperature difference across the slab in kelvin or Celsius, and L is thickness in metres. So a brick wall (k ≈ 0.7) of 4 m2, 0.2 m thick, with 20°C inside and 0°C outside, conducts Q = 0.7 × 4 × 20 / 0.2 = 280 W. Simple, steady-state, and surprisingly powerful for first estimates.
The rate form is Q = kAΔT/L watts. If you want heat flux instead — that is, watts per square metre — drop the area and write q" = kΔT/L. Both forms come from Fourier's law, which says heat flows down a temperature gradient at a rate proportional to thermal conductivity. Always check your units before plugging in: a thickness in centimetres or a conductivity in cal/(cm·s·°C) will give an answer that looks reasonable but is wrong by orders of magnitude.
Treat the wall as a steady-state conductor. Find the conductivity k of each layer (brick, insulation, drywall), the thickness L of each, and the temperature difference between inside and outside. For a single-layer wall, Q = kAΔT/L gives watts of loss. For a composite wall, sum the thermal resistances: Rtotal = ΣLi/ki, then Q = AΔT/Rtotal. Multiply by hours and you get energy in watt-hours. This is the bones of every home insulation calculation, and a good calculator just hides the algebra.
Thermal conductivity has units of watts per metre per kelvin — W/(m·K), sometimes written W/m/K or W·m-1·K-1. It tells you how many watts cross a one-metre cube of material when the two opposite faces differ by one kelvin. Copper is around 400, aluminium 235, water 0.6, dry air 0.026, polyurethane foam 0.025. Mixing units — calories, BTUs, inches — is the most common mistake junior researchers make. Stick to SI for engineering, and convert only at the end if needed.
Heat flux is the rate per unit area, so q" = kΔT/L in W/m2. For one-dimensional steady-state conduction this is Fourier's law without the area term. Example: a 5 cm slab of insulation with k = 0.04 W/(m·K) and 25°C across it gives q" = 0.04 × 25 / 0.05 = 20 W/m2. Multiply by total area to get total heat. Flux is the cleaner quantity to compare materials, because it normalises out the area and reveals the intrinsic resistance.
Steady-state means the temperature at every point inside the material has stopped changing with time. Heat is still flowing through, but the temperature profile is fixed. Mathematically, dT/dt = 0, so Fourier's law reduces to Q = kAΔT/L. Most textbook problems and most building-energy calculations assume this, because it makes the maths tractable. Real walls are never quite there — morning sun and night cooling shift the profile constantly — but over a day-night cycle, steady-state gives a reliable average.
Thermal resistance is R = L/(kA), measured in K/W. Once you have R for each layer of a composite wall, add them in series exactly as you would electrical resistors: Rtotal = R1 + R2 + R3. Then heat flow is Q = ΔT/Rtotal. For a 0.1 m brick layer (k = 0.7) over 4 m2, R = 0.1/(0.7×4) = 0.036 K/W. Add the insulation layer's resistance, and the wall behaves as one combined unit. This is the basis of R-value ratings used in construction.
Higher k means heat moves through faster for the same temperature difference and geometry. Copper (k ≈ 400) shifts heat about 10,000 times faster than dry air (k ≈ 0.026). That is why heat sinks are aluminium, frying pans are copper-bottomed, and insulation is foam or fibreglass. In Q = kAΔT/L, k sits as a multiplier, so doubling it doubles the heat rate. When you are choosing materials — for a wall, a circuit board, a thermos — you are mostly choosing the right k.
Sources & references
- NASA Glenn - Heat transfer fundamentals
- NASA Glenn - Specific heat capacity
- NASA Glenn - Earth atmosphere equation
- NASA Glenn - Air pressure fundamentals
- NIST Chemistry WebBook - Water phase change data
- NIST WTT-Pro - Boiling point from vapor pressure
- NIST - Fused quartz thermal expansion reference
- NIST - Example thermal conductivity tables