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Heat transfer and thermal systems
Resistance of a hollow sphere
Thermal resistance of a complete spherical shell with inner radius ri and outer radius ro, for steady radial conduction with constant conductivity k and no internal generation.
- Ref
- HTS-01
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- R [K/W]
Radial conduction through a complete spherical shell has a resistance R = (1/ri − 1/ro)/(4πk). It is the sphere equivalent of the plane-wall L/(kA), with the area growing as r². The resistance approaches a finite limit, 1/(4πk ri), as the outer radius grows, because the heat flow spreads over ever larger spheres.
Thermal resistance of a complete spherical shell with inner radius ri and outer radius ro, for steady radial conduction with constant conductivity k and no internal generation.
Calculator
- Heat rate for a 100 K temperature difference62.83 W
- Limiting resistance for an infinitely thick shell, 1/(4πk ri)3.183 K/W
- Conduction shape factor, 4π/(1/ri − 1/ro)1.257 m
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.
- A shell from 0.05 m to 0.1 m resists heat flow with 1.592 K/W, so a 100 K difference drives 62.83 W through it.
- Even an infinitely thick shell would only reach 3.183 K/W, so insulation beyond a few inner radii adds little to a small sphere.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (K/W) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k Thermal conductivity | 0.01 to 500 W/(m K) | 0.001592 | 79.58 | 79.58 | +10% | −9.09 % |
| ri Inner radius | 0.01 to 0.099 m | 0.01608 | 14.32 | 14.31 | +10% | −18.2 % |
| ro Outer radius | 0.1 to 1 m | 1.592 | 3.024 | 1.432 | +10% | +9.09 % |
Sweep of Outer radius
| Outer radius (m) | Thermal resistance (K/W) |
|---|---|
| 0.1 | 1.592 |
| 0.2125 | 2.434 |
| 0.325 | 2.693 |
| 0.4375 | 2.819 |
| 0.55 | 2.894 |
| 0.6625 | 2.943 |
| 0.775 | 2.978 |
| 0.8875 | 3.004 |
| 1 | 3.024 |
Sweep one variable
Variables and how each one contributes
Thermal resistance
What the equation gives you. Everything else feeds it.
Inner radius
The inner radius of the shell, in m. A larger inner radius shrinks the gap and the inner area grows, so the resistance falls. Raising it by 10 % (to 0.055) moves the result by −18.2 %, from 1.592 to 1.302 K/W.
Outer radius
The outer radius of the shell, in m. A thicker shell raises the resistance, with falling returns because it approaches a limit. Raising it by 10 % (to 0.11) moves the result by +9.09 %, from 1.592 to 1.736 K/W.
Thermal conductivity
The thermal conductivity of the shell, in W/(m K). A higher conductivity lowers the resistance in inverse proportion. Raising it by 10 % (to 0.55) moves the result by −9.09 %, from 1.592 to 1.447 K/W.
Worked example with the values above
- Radius term: 1/ri − 1/ro = 1/0.05 − 1/0.1 = 10 1/m.
- Denominator: 4πk = 4π × 0.5 = 6.2832 W/(m K).
- R = 10 / 6.2832 = 1.5915 K/W.
Assumptions and limits
- Steady state, constant properties unless the page states a temperature law, and no internal heat generation.
- One-dimensional conduction in the stated geometry, with no edge or end losses.
Common mistakes
- Using diameters where radii are needed.
- Using °C in a formula that needs an absolute temperature, or leaving a temperature difference out of the heat rate.
- Applying a plane-wall formula to a curved wall when the thickness is not small compared with the radius.
Statistics and uncertainty
Thermal conductivities are rarely known better than 5 % to 10 %, and contact resistances can vary by a factor of two between nominally identical joints. Treat results as estimates unless the inputs were measured.
Where it is used
- Insulation sizing for vessels, pipes and walls.
- Estimating interface and layer temperatures in composite constructions.
Maths you need
- Fourier’s law: q″ = −k dT/dx
- Thermal resistance: R = ΔT/Q, in K/W
- Series and parallel paths: resistances add in series; conductances add in parallel