Library / Heat Transfer / HTR-03
Heat Transfer
Net thermal radiation
Net heat radiated by a grey surface to large surroundings.
- Ref
- HTR-03
- Rev
- v1
- Field
- Heat Transfer
- Result
- Q̇net [W]
Diffuse grey surface facing large isothermal black surroundings, with view factor one. Temperatures are absolute. σ = 5.6704 × 10⁻⁸ W/(m² K⁴).
Hot surfaces glow in the infrared; the heat they lose by radiation rises with the fourth power of their temperature.
Calculator
- Emitted power εσAT42,468 W
- Radiation coefficient hr8.435 W/(m²·K)
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Doubling the absolute surface temperature multiplies the emitted power by 16.
- Low-emissivity coatings can cut radiative loss by an order of magnitude.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Ts Surface temperature | 250 to 1,500 K | −404.3 | 487,200 | 487,600 | +10% | +67.9 % |
| A Surface area | 0.01 to 100 m² | 8.435 | 84,350 | 84,340 | +10% | +10 % |
| Tsur Surroundings temperature | 100 to 500 K | −3,557 | 2,458 | 6,015 | +10% | −21.5 % |
| ε Emissivity | 0.05 to 1 | 99.23 | 1,985 | 1,885 | +undefined% | — |
Sweep of Surface temperature
| Surface temperature (K) | Net radiated heat (W) |
|---|---|
| 250 | −404.3 |
| 406.3 | 1,845 |
| 562.5 | 8,870 |
| 718.8 | 24,950 |
| 875 | 55,730 |
| 1,031 | 108,200 |
| 1,188 | 190,900 |
| 1,344 | 313,500 |
| 1,500 | 487,200 |
Sweep one variable
Variables and how each one contributes
Net radiated heat
What the equation gives you. Everything else feeds it.
Emissivity
Surface emissivity (polished metal 0.05, paint 0.9). Directly proportional.
Surface area
Radiating area. Directly proportional.
Surface temperature
Absolute surface temperature. Fourth power: the strongest driver.
Surroundings temperature
Absolute temperature of the surroundings. Lowers the net loss.
Worked example with the values above
- Ts⁴ − Tsur⁴ = 2.56×1010 − 8.1×109 = 1.75×1010 K⁴.
- εσA = 0.85 × 5.670×10⁻⁸ × 2 = 9.64×10−8.
- Qnet = 1,687 W.
Assumptions and limits
- Grey diffuse surface.
- Large black surroundings with view factor one.
Common mistakes
- Using °C.
- Using the emissivity of a polished surface for an oxidised one.
Statistics and uncertainty
Emissivity values are uncertain by ±10–20%, and the T⁴ dependence amplifies temperature errors fourfold.
Where it is used
- Infrared signature and cooling estimates.
- Furnace, engine and electronics enclosure heat loss.
Maths you need
- Absolute temperature: kelvin
- Emissivity: 0 to 1