Compute heat transfer rates, thermal resistance, and visualize temperature gradients for three fundamental modes of heat transfer. Includes reference tables, real-world examples, and detailed theory.
Heat transfer is the thermal energy in transit due to a temperature difference. The three fundamental modes — conduction, convection, and radiation — govern energy exchange in everything from electronic cooling to planetary climate. This calculator implements the core governing equations with interactive visualization, enabling engineers, students, and researchers to explore thermal systems quantitatively.
The Fourier rate equation (conduction): Q = −k · A · (dT/dx)
Newton's law of cooling (convection): Q = h · A · (Ts − T∞)
Stefan–Boltzmann law (radiation): Q = ε · σ · A · (T₁4 − T₂4)
where σ = 5.670374419 × 10−8 W/(m²·K⁴)
Conduction follows Fourier's law: the heat flux is proportional to the temperature gradient. For a plane wall of thickness L and cross-sectional area A, with thermal conductivity k assumed constant, the steady-state heat transfer rate is:
The thermal resistance is R = L / (k · A), so that Q = ΔT / R. This electrical–thermal analogy is powerful for analyzing composite walls and series/parallel configurations.
Convection occurs at a solid–fluid interface. The local heat flux is governed by Newton's law: q″ = h · (Ts − T∞). The convective heat transfer coefficient h depends on fluid properties, flow regime, and geometry. It can range from ~5 W/(m²·K) for natural convection in air to over 10,000 W/(m²·K) for boiling water.
Radiation is the emission of electromagnetic waves from a surface due to its temperature. The net radiative exchange between two blackbodies is proportional to the difference of the fourth powers of absolute temperatures. For real surfaces, emissivity ε (0–1) accounts for the surface's efficiency as a thermal radiator.
| Material | k (W/(m·K)) | Typical Use | Emissivity ε (approx.) |
|---|---|---|---|
| Copper | 401 | Heat exchangers, electrical conductors | 0.03 – 0.05 (polished) |
| Aluminum | 237 | Fins, automotive radiators | 0.04 – 0.08 (polished) |
| Steel (carbon) | 50 | Structural components, pipes | 0.70 – 0.80 (oxidized) |
| Glass | 0.8 | Windows, solar collectors | 0.90 – 0.95 |
| Brick | 0.7 | Building walls | 0.85 – 0.90 |
| Wood (oak) | 0.15 | Furniture, flooring | 0.85 – 0.90 |
| Fiberglass insulation | 0.04 | Thermal insulation in buildings | — |
| Air (still) | 0.026 | Natural convection, insulation | — |
| Water (liquid) | 0.60 | Heat transfer fluids | — |
| Scenario | h (W/(m²·K)) | Example |
|---|---|---|
| Natural convection – air | 2 – 25 | Room heating, electronics cooling |
| Forced convection – air | 25 – 250 | Fan cooling, HVAC ducts |
| Forced convection – water | 100 – 1000 | Water-cooled heat exchangers |
| Boiling water | 1000 – 10,000 | Boilers, kettle |
| Condensing steam | 5000 – 15,000 | Condensers, power plants |
An architect is designing a multi‑layer wall for a passive house. The wall consists of: 20 mm plaster (k=0.5), 150 mm fiberglass insulation (k=0.04), and 100 mm brick (k=0.7). Using the thermal resistance calculator, the total R‑value is computed: Rtotal = (0.02/0.5) + (0.15/0.04) + (0.10/0.7) = 0.04 + 3.75 + 0.143 ≈ 3.93 K/W. For a temperature difference of 25°C and wall area of 30 m², the heat loss is: Q = 25 / 3.93 ≈ 6.36 W per m² of wall area, or about 191 W for the entire wall. This analysis demonstrates the dominant role of insulation thickness in reducing energy consumption.