Library / Heat transfer and thermal systems / HTS-09
Heat transfer and thermal systems
Hottest location in a slab with unequal face temperatures
Position of the temperature maximum in a heated slab with different face temperatures Tl and Tr, measured from the left face.
- Ref
- HTS-09
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- xmax [m]
With unequal face temperatures Tl and Tr and uniform heating, the temperature is a parabola whose peak is at xmax = L/2 + k(Tr − Tl)/(q‴L), measured from the left face. The hotter face pulls the peak towards itself. If this stationary point falls outside the slab, the maximum is at the hotter face, so the result is clipped to the range 0 to L.
Position of the temperature maximum in a heated slab with different face temperatures Tl and Tr, measured from the left face. If the stationary point falls outside the slab, the hotter face is the maximum.
Calculator
- Unclipped stationary position0.07 m
- Offset of the peak from the centre0.02 m
- Heat generated per unit area, q‴L10,000 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 temperature peaks 0.07 m from the left face, 0.02 m to the right of the centre, towards the hotter face.
- A result of exactly 0 or 0.1 m would mean the maximum sits on a face, which happens when the face temperature difference outweighs the internal heating.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| L Slab thickness | 0.001 to 1 m | 0.001 | 0.502 | 0.501 | +10% | +4.55 % |
| Tl Left-face temperature | 250 to 1,000 K | 0 | 0.1 | 0.1 | +10% | −42.9 % |
| Tr Right-face temperature | 250 to 1,000 K | 0 | 0.1 | 0.1 | +10% | +42.9 % |
| q‴ Volumetric heat generation | 1,000 to 1e9 W/m³ | 0.05 | 0.1 | 0.05 | +10% | −2.6 % |
| k Thermal conductivity | 0.1 to 500 W/(m K) | 0.0502 | 0.1 | 0.0498 | +10% | +2.86 % |
Sweep of Right-face temperature
| Right-face temperature (K) | Position of maximum temperature (m) |
|---|---|
| 250 | 0 |
| 343.8 | 0.09375 |
| 437.5 | 0.1 |
| 531.3 | 0.1 |
| 625 | 0.1 |
| 718.8 | 0.1 |
| 812.5 | 0.1 |
| 906.3 | 0.1 |
| 1,000 | 0.1 |
Sweep one variable
Variables and how each one contributes
Position of maximum temperature
What the equation gives you. Everything else feeds it.
Slab thickness
The full slab thickness, in m. A thicker slab reduces the shift of the peak from the centre, so the position moves nearer L/2. Raising it by 10 % (to 0.11) moves the result by +4.55 %, from 0.07 to 0.07318 m.
Thermal conductivity
The thermal conductivity, in W/(m K). A higher conductivity makes the face temperatures matter more, shifting the peak further from the centre. Raising it by 10 % (to 11) moves the result by +2.86 %, from 0.07 to 0.072 m.
Left-face temperature
The left-face temperature, in K. A hotter left face moves the peak towards the left. Raising it by 10 % (to 330) moves the result by −42.9 %, from 0.07 to 0.04 m.
Right-face temperature
The right-face temperature, in K. A hotter right face moves the peak towards the right. Raising it by 10 % (to 352) moves the result by +42.9 %, from 0.07 to 0.1 m.
Volumetric heat generation
The uniform volumetric heat generation, in W/m³. More heating moves the peak back towards the centre. Raising it by 10 % (to 110,000) moves the result by −2.6 %, from 0.07 to 0.06818 m.
Worked example with the values above
- Shift from the centre: k(Tr − Tl)/(q‴L) = 10 × 20 / (100,000 × 0.1) = 0.02 m.
- Unclipped position: L/2 + shift = 0.05 + 0.02 = 0.07 m.
- Clip to 0 … 0.1: xmax = 0.07 m.
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