Library / Heat transfer and thermal systems / HTS-32
Heat transfer and thermal systems
Radiative heat exchange between concentric spheres
Net radiation from the outside of an inner sphere (surface 1) to the inside of an outer sphere (surface 2), with A = 4πr² and diffuse grey opaque surfaces across a non-participating gap.
- Ref
- HTS-32
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- Q [W]
For a small sphere inside a larger concentric one, the net radiant exchange is Q = A₁σ(T₁⁴ − T₂⁴)/[1/ε₁ + (A₁/A₂)(1/ε₂ − 1)], with A = 4πr². The area ratio (r₁/r₂)² weights the outer surface’s resistance, so a large outer sphere behaves almost like a black enclosure regardless of its own emissivity.
Net radiation from the outside of an inner sphere (surface 1) to the inside of an outer sphere (surface 2), with A = 4πr² and diffuse grey opaque surfaces across a non-participating gap.
Calculator
- Inner-sphere area, 4πr₁²0.1257 m²
- Area ratio, (r₁/r₂)²0.25
- Net flux at the inner sphere, Q/A₁1,371 W/m²
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 inner sphere loses 172.3 W by radiation, a flux of 1,371 W/m² on its own surface.
- With an outer sphere twice the radius, the outer emissivity only enters with a weight of 0.25, so improving the outer coating has little effect.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W) | Local step | Change in result |
|---|---|---|---|---|---|---|
| T1 Inner-sphere temperature | 100 to 2,000 K | −25.34 | 50,650 | 50,670 | +10% | +53.3 % |
| T2 Outer-sphere temperature | 100 to 2,000 K | −50,470 | 197.6 | 50,670 | +10% | −6.91 % |
| r1 Radius of the inner sphere | 0.01 to 0.19 m | 1.936 | 482.1 | 480.2 | +10% | +18.2 % |
| ε1 Emissivity of the inner sphere | 0.05 to 1 | 19.14 | 310.1 | 291 | +10% | +8.79 % |
| ε2 Emissivity of the outer sphere | 0.05 to 1 | 57.43 | 193.8 | 136.4 | +10% | +2.06 % |
| r2 Radius of the outer sphere | 0.2 to 1 m | 172.3 | 192.9 | 20.57 | +10% | +1.97 % |
Sweep of Emissivity of the inner sphere
| Emissivity of the inner sphere | Net radiant heat rate (W) |
|---|---|
| 0.05 | 19.14 |
| 0.1688 | 62.77 |
| 0.2875 | 104 |
| 0.4062 | 143 |
| 0.525 | 179.9 |
| 0.6438 | 214.9 |
| 0.7625 | 248.2 |
| 0.8812 | 279.9 |
| 1 | 310.1 |
Sweep one variable
Variables and how each one contributes
Net radiant heat rate
What the equation gives you. Everything else feeds it.
Radius of the inner sphere
The radius of the inner sphere, in m. A larger inner sphere radiates over a larger area, raising the heat rate. Raising it by 10 % (to 0.11) moves the result by +18.2 %, from 172.3 to 203.7 W.
Radius of the outer sphere
The radius of the outer sphere, in m. A larger outer sphere weights the outer emissivity less, slightly raising the heat rate. Raising it by 10 % (to 0.22) moves the result by +1.97 %, from 172.3 to 175.7 W.
Inner-sphere temperature
The temperature of the inner sphere, in K. Raises the heat rate as the fourth power. Raising it by 10 % (to 550) moves the result by +53.3 %, from 172.3 to 264.1 W.
Outer-sphere temperature
The temperature of the outer sphere, in K. Raises the opposing term and lowers the heat rate. Raising it by 10 % (to 330) moves the result by −6.91 %, from 172.3 to 160.4 W.
Emissivity of the inner sphere
The emissivity of the inner sphere. A higher emissivity raises the heat rate. Raising it by 10 % (to 0.55) moves the result by +8.79 %, from 172.3 to 187.4 W.
Emissivity of the outer sphere
The emissivity of the outer sphere. A higher emissivity raises the heat rate, less so when the outer sphere is large. Raising it by 10 % (to 0.55) moves the result by +2.06 %, from 172.3 to 175.8 W.
Worked example with the values above
- Inner area: A₁ = 4π × 0.1² = 0.12566 m²; area ratio 0.25.
- Resistance term: 1/ε₁ + (A₁/A₂)(1/ε₂ − 1) = 2.25.
- Q = 387.63 / 2.25 = 172.28 W.
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