Enter the thickness of the material parallel to heat flow, the thermal conductivity, and the cross-sectional area into the calculator to determine the thermal resistance.
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Thermal Resistance Formula
For steady, one-dimensional heat conduction through a flat material layer, thermal resistance depends on the layer thickness, the cross-sectional area normal to heat flow, and the material's thermal conductivity.
R = \frac{X}{A \cdot k}- R = thermal resistance of the layer
- X = thickness measured in the direction of heat flow
- A = cross-sectional area perpendicular to heat flow
- k = thermal conductivity of the material
A higher thermal resistance means the layer opposes heat flow more strongly. A lower thermal resistance means heat passes through the layer more easily.
Rearranged Forms
Because this calculator can solve for any one missing value, the same relationship can be rearranged as follows:
X = R \cdot A \cdot k
A = \frac{X}{R \cdot k}k = \frac{X}{R \cdot A}How Thermal Resistance Relates to Heat Flow
Once thermal resistance is known, it can be used with the temperature difference across the material to estimate the steady heat-transfer rate.
\dot{Q} = \frac{\Delta T}{R}This means:
- Increasing thickness increases thermal resistance.
- Increasing area decreases thermal resistance.
- Increasing thermal conductivity decreases thermal resistance.
Input Guide
| Input | Meaning | Common Units | Effect on Resistance |
|---|---|---|---|
| Thickness of Material | The distance heat travels through the layer | m, cm, mm, in, ft | More thickness increases resistance |
| Cross-Sectional Area | The area normal to the heat-flow path | m², cm², in², ft² | More area decreases resistance |
| Thermal Conductivity | How readily the material conducts heat | W/(m·K), BTU/(hr·ft·°F) | Higher conductivity decreases resistance |
| Thermal Resistance | Total opposition to heat flow for the entered layer and area | K/W, hr·°F/BTU | Higher values indicate less heat transfer |
How to Use the Calculator
- Enter any three known values.
- Use the material thickness in the same direction the heat is traveling.
- Enter the cross-sectional area that the heat passes through, not the full outside surface area unless they are the same.
- Select the correct unit set for each field.
- Calculate the missing variable and interpret the result as the total resistance for that specific geometry.
Example
A panel is 0.05 m thick, has an area of 2 m², and has a thermal conductivity of 0.04 W/(m·K).
R = \frac{0.05}{2 \cdot 0.04} = 0.625 \text{ K/W}If the temperature difference across the panel is 10 K, the heat-transfer rate is:
\dot{Q} = \frac{10}{0.625} = 16 \text{ W}So under those conditions, about 16 watts of heat flow through the panel.
Multiple Layers
Many real systems contain more than one thermal resistance. When heat passes through layers one after another, the resistances add directly.
R_{total} = R_1 + R_2 + \cdots + R_nIf heat can split into separate parallel paths, the combined resistance is found by adding reciprocals.
\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}When This Calculation Is Useful
Thermal resistance is commonly used when comparing insulation materials, wall sections, pipe insulation, oven panels, machine housings, electronic assemblies, heat shields, and any design where controlling heat flow matters.
Important Assumptions
- The relation above applies to steady conduction through a flat layer.
- The material is treated as uniform through the section.
- The calculation does not automatically include convection, contact resistance, radiation, or internal heat generation.
- Thermal conductivity can vary with temperature, moisture content, density, and material composition.
Common Mistakes
- Using the wrong area for the heat-flow path.
- Entering thickness that is not aligned with heat flow.
- Mixing metric and imperial units without converting them.
- Using a single-layer equation for a multilayer assembly.
- Assuming the conductivity value is constant when operating temperature changes significantly.
