Library / Heat transfer and thermal systems / HTS-10
Heat transfer and thermal systems
Maximum temperature of an internally heated annulus
Peak temperature at the insulated inner surface of a heated annulus whose outer surface is held at To, for uniform generation q‴ and uniform conductivity.
- Ref
- HTS-10
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- Tmax [K]
For a heated annulus with an insulated inner surface at ri and the outer surface held at To, all the heat leaves through the outer surface, so the maximum temperature is at the inner radius: Tmax = To + (q‴/4k)[ro² − ri² − 2ri² ln(ro/ri)]. As ri → 0 it reduces to the solid cylinder result q‴ro²/(4k).
Peak temperature at the insulated inner surface of a heated annulus whose outer surface is held at To, for uniform generation q‴ and uniform conductivity. Steady radial conduction.
Calculator
- Temperature rise above the outer surface32.27 K
- Heat generated per metre of length, q‴π(ro² − ri²)3,770 W/m
- Heat flux at the outer surface, q‴(ro² − ri²)/(2ro)15,000 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 surface runs 32.27 K above the outer surface, at 332.3 K.
- The annulus sheds 3,770 W per metre through its outer surface alone, since the inner surface is insulated.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (K) | Local step | Change in result |
|---|---|---|---|---|---|---|
| q‴ Volumetric heat generation | 1,000 to 1e9 W/m³ | 300 | 32,570 | 32,270 | +10% | +0.971 % |
| ro Outer radius | 0.04 to 0.3 m | 332.3 | 4,672 | 4,339 | +10% | +3.91 % |
| k Thermal conductivity | 0.1 to 500 W/(m K) | 300.3 | 1,914 | 1,613 | +10% | −0.883 % |
| To Outer-surface temperature | 250 to 1,000 K | 282.3 | 1,032 | 750 | +10% | +9.03 % |
| ri Inner radius (insulated) | 0.005 to 0.039 m | 300.1 | 373.6 | 73.45 | +10% | −1.63 % |
Sweep of Inner radius (insulated)
| Inner radius (insulated) (m) | Maximum temperature (K) |
|---|---|
| 0.005 | 373.6 |
| 0.00925 | 363.2 |
| 0.0135 | 351.1 |
| 0.01775 | 338.6 |
| 0.022 | 326.9 |
| 0.02625 | 316.5 |
| 0.0305 | 308.3 |
| 0.03475 | 302.6 |
| 0.039 | 300.1 |
Sweep one variable
Variables and how each one contributes
Maximum temperature
What the equation gives you. Everything else feeds it.
Outer-surface temperature
The outer-surface temperature, in K. Raises the maximum temperature by the same amount. Raising it by 10 % (to 330) moves the result by +9.03 %, from 332.3 to 362.3 K.
Volumetric heat generation
The uniform volumetric heat generation, in W/m³. Raises the temperature rise above the outer surface in direct proportion. Raising it by 10 % (to 1,100,000) moves the result by +0.971 %, from 332.3 to 335.5 K.
Outer radius
The outer radius, in m. A larger outer radius raises the rise roughly with its square. Raising it by 10 % (to 0.044) moves the result by +3.91 %, from 332.3 to 345.3 K.
Inner radius (insulated)
The inner (insulated) radius, in m. A larger inner radius makes the wall thinner, so the rise falls. Raising it by 10 % (to 0.022) moves the result by −1.63 %, from 332.3 to 326.9 K.
Thermal conductivity
The thermal conductivity, in W/(m K). A higher conductivity lowers the rise in inverse proportion. Raising it by 10 % (to 5.5) moves the result by −0.883 %, from 332.3 to 329.3 K.
Worked example with the values above
- Bracket: ro² − ri² − 2ri² ln(ro/ri) = 0.0016 − 4×10−4 − 5.5452×10−4 = 6.4548×10−4 m².
- Rise: q‴/(4k) × bracket = 50,000 × 6.4548×10−4 = 32.274 K.
- Tmax = 300 + 32.274 = 332.27 K.
Assumptions and limits
- Uniform volumetric generation and constant conductivity in a steady state.
- The stated boundary condition holds exactly, for example a fixed surface temperature or an insulated inner surface.
Common mistakes
- Mixing the factor 2, 4 or 6 between the slab, cylinder and sphere.
- Using the half-thickness or the full thickness in the wrong place.
- Forgetting that generation is per unit volume, not per unit area.
Statistics and uncertainty
Volumetric heating is often known only to 10 % to 20 %, and the peak temperature scales directly with it. Use a margin on predicted centreline or hot-spot temperatures.
Where it is used
- Hot-spot checks in busbars, fuel rods, heated plates and spherical particles.
- Finding the surface temperature a coolant must hold for a given peak.
Maths you need
- Volumetric generation: q‴ in W/m³
- Steady conduction equation: k∇²T + q‴ = 0
- Symmetry: zero gradient at the centre