Compute thermal resistance (R = L / (k·A)) for conductive heat transfer. Evaluate temperature difference or heat flow, visualize thermal gradient, and access material database. Essential for electronics cooling, building insulation, and mechanical design.
Thermal resistance (Rth) quantifies a material's opposition to conductive heat flow. Based on Fourier's law of heat conduction, the steady-state thermal resistance for a plane wall is given by:
\[ R = \frac{L}{k \cdot A} \quad \left[ \frac{K}{W} \right] \]
where L = thickness, k = thermal conductivity, A = cross-sectional area.
The heat equation:\[ Q = \frac{\Delta T}{R} \quad \text{or} \quad \Delta T = Q \cdot R \]
This fundamental relationship is crucial in thermal management of electronics (heat sinks, interface materials), building insulation (R-value), aerospace thermal control, and industrial heat exchangers. Lower thermal resistance implies better heat transfer.
Fourier's law states that heat flux q = -k ∇T. For one-dimensional conduction through a flat wall, q = k (ΔT)/L. Since Q = q·A, we obtain Q = k·A·(ΔT)/L → ΔT = Q·(L/(k·A)). The proportionality factor L/(k·A) defines the thermal resistance R. Multiple layers in series sum resistances: Rtotal = Σ Ri. This calculator handles single-layer conduction; for complex stacks, use the series resistance principle.
Typical values: Copper (k ~ 401 W/m·K), Aluminum (~205), Air (~0.026), Epoxy (~0.2). High k materials are excellent conductors; low k materials are thermal insulators.
| Material | Thermal Conductivity k [W/(m·K)] | Typical Applications |
|---|---|---|
| Copper (pure) | 401 | Heat pipes, CPU coolers, high-end heatsinks |
| Aluminum (6061) | 167-205 | Extruded heatsinks, LED cooling |
| Stainless Steel | 15 | Structural components, vacuum chambers |
| Fiberglass insulation | 0.04 | Building walls, HVAC ducts |
| Air (20°C, 1 atm) | 0.026 | Natural convection gaps, thermal breaks |
| Thermal grease (typical) | 3-8 | Interface between CPU and heatsink |
A power MOSFET dissipates 30 W. It is mounted on an aluminum heat spreader (thickness 3 mm, area 0.005 m², k=205 W/m·K). Thermal resistance of the spreader: R = 0.003/(205×0.005) = 2.93 K/W. Temperature rise across the spreader: ΔT = 30 × 2.93 ≈ 88 K. This indicates that without additional heatsink fins or fan, the temperature rise is critical. Engineers then add a heatsink with lower R to keep junction temperature below limit. The calculator allows rapid iteration: increasing area to 0.02 m² reduces R to 0.73 K/W and ΔT to 21.9 K – a viable improvement.
In construction, thermal resistance per unit area (R-value = L/k) is expressed in (m²·K)/W or imperial (hr·ft²·°F)/Btu. Our tool computes absolute thermal resistance R (K/W) which, multiplied by area, gives R-value per area. For building codes, high R-values reduce energy consumption. Understanding conduction resistance is the first step towards comprehensive thermal analysis including convection and radiation.