Library / Heat transfer and thermal systems / HTS-35
Heat transfer and thermal systems
Effective radiant surroundings temperature
Equivalent black-body temperature seen by a small diffuse receiver surrounded by isothermal black surfaces, weighted by view factors.
- Ref
- HTS-35
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- Tr [K]
A small receiver surrounded by isothermal black surfaces feels an effective radiant temperature Tr = (ΣFᵢTᵢ⁴)^¼, a view-factor-weighted fourth-power mean, so that the irradiation is σTr⁴. The view factors must add to one; here the third surface takes the remainder F₃ = 1 − F₁ − F₂. Reflective grey surroundings need radiosities, not blackbody emission.
Equivalent black-body temperature seen by a small diffuse receiver surrounded by isothermal black surfaces, weighted by view factors. Surface 3 takes the remainder of the view, F₃ = 1 − F₁ − F₂.
Calculator
- Remaining view factor, F₃ = 1 − F₁ − F₂0
- Irradiation from the surroundings, σTr⁴707.4 W/m²
- View-weighted arithmetic mean, ΣFT325 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.
- The receiver sees an effective radiant temperature of 334.2 K, from a view of 25 % at 400 K and 75 % at 300 K.
- The simple weighted average of the temperatures would give 325 K, which is lower, because hotter surfaces count more in the fourth-power mean.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (K) | Local step | Change in result |
|---|---|---|---|---|---|---|
| T2 Temperature of surface 2 | 100 to 2,000 K | 283.7 | 1,861 | 1,578 | +10% | +5.23 % |
| T1 Temperature of surface 1 | 100 to 2,000 K | 279.5 | 1,415 | 1,135 | +10% | +5.48 % |
| F1 View factor to surface 1 | 0 to 0.25 | 300 | 334.2 | 34.2 | +10% | — |
| F2 View factor to surface 2 | 0.5 to 0.75 | 334.2 | 334.2 | 0 | +10% | — |
| T3 Temperature of surface 3 (view F₃ = 1 − F₁ − F₂) | 100 to 2,000 K | 334.2 | 334.2 | 0 | +10% | 0 % |
Sweep of Temperature of surface 1
| Temperature of surface 1 (K) | Effective radiant temperature (K) |
|---|---|
| 100 | 279.5 |
| 337.5 | 310.7 |
| 575 | 427.5 |
| 812.5 | 582.4 |
| 1,050 | 746.1 |
| 1,288 | 912.4 |
| 1,525 | 1,080 |
| 1,763 | 1,247 |
| 2,000 | 1,415 |
Sweep one variable
Variables and how each one contributes
Effective radiant temperature
What the equation gives you. Everything else feeds it.
View factor to surface 1
The view factor to surface 1. A larger share of the view to a hotter surface raises Tr. Raising it by 10 % (to 0.275) gives no valid result at the other default values.
Temperature of surface 1
The temperature of surface 1, in K. Raises Tr in proportion to its view factor weight. Raising it by 10 % (to 440) moves the result by +5.48 %, from 334.2 to 352.5 K.
View factor to surface 2
The view factor to surface 2. A larger share of the view to a hotter surface raises Tr. Raising it by 10 % (to 0.825) gives no valid result at the other default values.
Temperature of surface 2
The temperature of surface 2, in K. Raises Tr in proportion to its view factor weight. Raising it by 10 % (to 330) moves the result by +5.23 %, from 334.2 to 351.7 K.
Temperature of surface 3 (view F₃ = 1 − F₁ − F₂)
The temperature of surface 3, which sees the remaining view. Matters in proportion to the remaining view factor F₃. Raising it by 10 % (to 330) moves the result by +0 %, from 334.2 to 334.2 K.
Worked example with the values above
- Remaining view: F₃ = 1 − 0.25 − 0.75 = 0.
- Weighted fourth powers: 6.4×109 + 6.075×109 + 0 = 1.2475×1010 K⁴.
- Tr = (that)^¼ = 334.2 K.
Assumptions and limits
- Opaque, diffuse and grey surfaces, isothermal with uniform properties.
- A transparent, non-participating gap, and absolute temperatures in kelvin.
Common mistakes
- Using °C in the fourth-power terms.
- Using the plate formula when the surfaces are not large compared with the gap.
- Forgetting that an emissivity of 1 means a perfect black surface and 0 a perfect reflector.
Statistics and uncertainty
Emissivity values are uncertain by 10 % to 30 % for real surfaces, and temperature errors are multiplied by four in the fourth-power law, so a 2 % error in absolute temperature gives roughly 8 % in flux.
Where it is used
- Furnaces, cryostats, ovens and spacecraft thermal design.
- Choosing shield layers and estimating radiant exposure.
Maths you need
- Stefan–Boltzmann law: E = σT⁴, with σ = 5.670×10⁻⁸ W/(m² K⁴)
- Emissivity: ε, the ratio of emission to that of a blackbody
- View factor: the fraction of radiation leaving one surface that reaches another