Library / Heat transfer and thermal systems / HTS-06
Heat transfer and thermal systems
Symmetric slab with uniform heat generation
Temperature in a slab of half-thickness L, with both faces held at Ts and uniform volumetric heating q‴, measured from the centreline.
- Ref
- HTS-06
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- T [K]
In a slab of half-thickness L with both faces at Ts and uniform heating q‴, the temperature is a parabola, T(x) = Ts + q‴(L² − x²)/(2k), with its maximum at the centre. The heat generated in each half leaves through its face, so the face flux is q‴L. Use the half-thickness, not the full thickness.
Temperature in a slab of half-thickness L, with both faces held at Ts and uniform volumetric heating q‴, measured from the centreline. Steady one-dimensional conduction with constant conductivity.
Calculator
- Centreline temperature, Ts + q‴L²/(2k)340 K
- Centreline rise above the faces40 K
- Heat flux leaving each face, q‴L20,000 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.
- At x = 0.01 m the slab is at 330 K, and the centre reaches 340 K, a rise of 40 K above the faces.
- Each face removes 20,000 W/m²; doubling the thickness would quadruple the centre rise but only double the face flux.
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 | 30,300 | 30,000 | +10% | +0.909 % |
| L Half-thickness | 0.01 to 0.2 m | 300 | 4,290 | 3,990 | +10% | +2.55 % |
| k Thermal conductivity | 0.1 to 500 W/(m K) | 300.3 | 1,800 | 1,500 | +10% | −0.826 % |
| Ts Face temperature | 250 to 1,000 K | 280 | 1,030 | 750 | +10% | +9.09 % |
| x Distance from the centre | 0 to 0.01 m | 330 | 340 | 10 | +10% | −0.636 % |
Sweep of Half-thickness
| Half-thickness (m) | Temperature at x (K) |
|---|---|
| 0.01 | 300 |
| 0.03375 | 403.9 |
| 0.0575 | 620.6 |
| 0.08125 | 950.2 |
| 0.105 | 1,392 |
| 0.1288 | 1,948 |
| 0.1525 | 2,616 |
| 0.1763 | 3,396 |
| 0.2 | 4,290 |
Sweep one variable
Variables and how each one contributes
Temperature at x
What the equation gives you. Everything else feeds it.
Face temperature
The temperature of both faces, in K. Raises the temperature everywhere by the same amount. Raising it by 10 % (to 330) moves the result by +9.09 %, from 330 to 360 K.
Volumetric heat generation
The uniform volumetric heat generation, in W/m³. Raises the temperature rise above the faces in direct proportion. Raising it by 10 % (to 1,100,000) moves the result by +0.909 %, from 330 to 333 K.
Half-thickness
The half-thickness of the slab, in m. A thicker slab raises the temperature rise with the square of L. Raising it by 10 % (to 0.022) moves the result by +2.55 %, from 330 to 338.4 K.
Distance from the centre
The distance from the centre plane, in m. Moving towards the face lowers the temperature, to Ts at x = L. Raising it by 10 % (to 0.011) moves the result by −0.636 %, from 330 to 327.9 K.
Thermal conductivity
The thermal conductivity, in W/(m K). A higher conductivity lowers the temperature rise in inverse proportion. Raising it by 10 % (to 5.5) moves the result by −0.826 %, from 330 to 327.3 K.
Worked example with the values above
- Geometry term: L² − x² = 0.02² − 0.01² = 3×10−4 m².
- Rise above the face: q‴(L² − x²)/(2k) = 1,000,000 × 3×10−4 / (2 × 5) = 30 K.
- T = 300 + 30 = 330 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