Library / Heat transfer and thermal systems / HTS-33
Heat transfer and thermal systems
Radiative flux through a single floating shield
Net radiant flux between two large parallel plates with one floating opaque shield between them.
- Ref
- HTS-33
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- q″ [W/m²]
One floating shield between two large parallel plates, with the same emissivity εs on both faces, adds two surface resistances and a gap. The flux becomes q″ = σ(T₁⁴ − T₂⁴)/(1/ε₁ + 2/εs + 1/ε₂ − 2). For black plates and a black shield it halves the exchange, and low-emissivity shields cut it much more. Several shields reduce it roughly in proportion to their number plus one.
Net radiant flux between two large parallel plates with one floating opaque shield between them. Both shield faces have the same emissivity εs and temperature, with no conduction through supports.
Calculator
- Flux without the shield992.3 W/m²
- Flux reduction factor with the shield0.1
- Shield temperature (from the plate-1 side)360.3 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.
- With the shield the flux is 99.23 W/m², against 992.3 W/m² without it, a reduction to 10 %.
- The shield settles at about 360.3 K, between the plate temperatures.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W/m²) | Local step | Change in result |
|---|---|---|---|---|---|---|
| T1 Temperature of plate 1 | 100 to 2,000 K | −45.36 | 90,680 | 90,730 | +10% | +67.9 % |
| T2 Temperature of plate 2 | 100 to 2,000 K | −90,580 | 144.6 | 90,730 | +10% | −21.5 % |
| εs Emissivity of the shield (both faces) | 0.02 to 1 | 9.923 | 496.2 | 486.2 | +10% | +10 % |
| ε1 Emissivity of plate 1 | 0.05 to 1 | 34.22 | 99.23 | 65.01 | +10% | — |
| ε2 Emissivity of plate 2 | 0.05 to 1 | 34.22 | 99.23 | 65.01 | +10% | — |
Sweep of Emissivity of the shield (both faces)
| Emissivity of the shield (both faces) | Net radiant flux (W/m²) |
|---|---|
| 0.02 | 9.923 |
| 0.1425 | 70.7 |
| 0.265 | 131.5 |
| 0.3875 | 192.3 |
| 0.51 | 253 |
| 0.6325 | 313.8 |
| 0.755 | 374.6 |
| 0.8775 | 435.4 |
| 1 | 496.2 |
Sweep one variable
Variables and how each one contributes
Net radiant flux
What the equation gives you. Everything else feeds it.
Temperature of plate 1
The temperature of plate 1, in K. Raises the net flux as the fourth power. Raising it by 10 % (to 440) moves the result by +67.9 %, from 99.23 to 166.6 W/m².
Temperature of plate 2
The temperature of plate 2, in K. Raises the opposing term and lowers the net flux. Raising it by 10 % (to 330) moves the result by −21.5 %, from 99.23 to 77.92 W/m².
Emissivity of plate 1
The emissivity of plate 1. A higher emissivity raises the flux. Raising it by 10 % (to 1.1) gives no valid result at the other default values.
Emissivity of plate 2
The emissivity of plate 2. A higher emissivity raises the flux. Raising it by 10 % (to 1.1) gives no valid result at the other default values.
Emissivity of the shield (both faces)
The emissivity of both faces of the shield. A lower shield emissivity cuts the flux strongly. Raising it by 10 % (to 0.22) moves the result by +10 %, from 99.23 to 109.2 W/m².
Worked example with the values above
- Fourth-power difference: σ(T₁⁴ − T₂⁴) = 992.32 W/m².
- Resistance term: 1/ε₁ + 2/εs + 1/ε₂ − 2 = 10.
- q″ = 992.32 / 10 = 99.232 W/m².
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 the formula for several shields: each extra shield adds its own terms.
- Using a shield with different emissivities on its two faces.
- Forgetting that a floating shield conducts no heat through supports.
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