☀️ Radiation Heat Transfer
Calculate the net radiation heat transfer between an object and its surroundings using the Stefan-Boltzmann Law.
☀️ Radiation Heat Transfer
Calculate the net radiation heat transfer between an object and its surroundings using the Stefan-Boltzmann Law.
✅ Calculation Result
Radiation Process
Calculator Description
Radiation heat transfer is heat carried as electromagnetic waves (infrared) without needing a transmitting medium. It dominates solar heating, furnaces, radiators and thermal-shielding design, and unlike convection or conduction it depends sensitively on the fourth power of temperature.
What this calculator finds
It computes the net radiation heat transfer q between a body and its surroundings using the Stefan–Boltzmann law. Given area, emissivity and the two absolute temperatures, it returns the heat transferred per unit time.
Why it matters
- Evaluating heat loss and insulation design for high-temperature equipment (boilers, furnaces)
- Estimating solar gain and building-envelope radiation loads
- Assessing passive radiative cooling of electronics and batteries
Formula
Stefan-Boltzmann Law
This is the net radiation from a grey body exposed to large surrounding walls. It is the difference between the radiation the body emits and the radiation it absorbs from the surroundings, proportional to the difference of the fourth powers of the two absolute temperatures.
- q — Net radiation heat transfer [W or Btu/hr]
- ε — Emissivity (0–1; 1 for a black body, ≈0 for a mirror)
- σ — Stefan-Boltzmann constant (5.67×10⁻⁸ W/(m²·K⁴) or 0.1714×10⁻⁸ Btu/(hr·ft²·°R⁴))
- A — Surface area [m² or ft²]
- T₁ — Absolute temperature of the body [K or °R]
- T₂ — Absolute temperature of surroundings [K or °R]
How the formula works
- Heat transfer grows sharply with temperature difference, especially at high temperature (fourth-power dependence).
- Emissivity near 1 (dark surfaces) radiates strongly; polished metals radiate little.
- It is proportional to area A, so reducing exposed area reduces radiative loss.
Assumptions & scope
This relation assumes the body is completely enclosed by the surroundings, both surfaces are isothermal, and the view factor is 1 (a large enclosure). For exchange between two finite surfaces, a view factor and effective emissivity must be included.
Worked example
For a steel plate with ε = 0.9 and A = 1 m² at T₁ = 500 K in surroundings T₂ = 300 K: with σ = 5.67×10⁻⁸ and (500⁴ − 300⁴) ≈ 5.44×10¹⁰, q = 0.9 × 5.67×10⁻⁸ × 1 × 5.44×10¹⁰ ≈ 2776 W.
Useful Tips
Practical tips
- A polished-metal (low-emissivity) surface or insulating blanket sharply cuts radiative loss.
- Always enter absolute temperature (K or °R); using Celsius directly gives wrong results.
- Radiation and convection occur together, so total loss is their sum.
Limitations & cautions
- Not directly applicable to radiation between two finite surfaces away from the large-enclosure assumption.
- Emissivity varies with wavelength, temperature and surface condition — verify literature values for precision.
- This ideal grey-body model ignores absorption by the intervening gas/vapor.