Library / Heat transfer and thermal systems / HTS-21
Heat transfer and thermal systems
Plane-wall transient temperature by the first eigenmode
Dimensionless temperature in a wall of half-thickness L cooled or heated by convection on both faces, using only the first term.
- Ref
- HTS-21
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- θ/θi
For a plane wall of half-thickness L with convection on both faces, the first-term solution is θ/θi ≈ C₁ cos(λx/L) exp(−λ²Fo). The eigenvalue λ is the first root of λ tan λ = Bi on (0, π/2), with Bi = hL/k, and C₁ = 4 sin λ/(2λ + sin 2λ). The calculator finds λ from the Biot number. The first term is generally accurate for Fo > 0.2.
Dimensionless temperature in a wall of half-thickness L cooled or heated by convection on both faces, using only the first term. The root λ is found from λ tan λ = Bi on (0, π/2) with Bi = hL/k; the first term is generally used for Fo > 0.2.
Calculator
- First eigenvalue, λ0.7854
- Coefficient, C₁ = 4 sin λ/(2λ + sin 2λ)1.1
- Centre-plane ratio, C₁ exp(−λ²Fo)0.5937
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 Fo = 1 the point at x = 0 m has lost 40.63 % of its initial excess temperature, leaving a ratio of 0.5937.
- The surface lags the ambient less than the centre does, by the factor cos λ = 0.7071; as Bi grows towards infinity, λ tends to π/2 and the surface follows the ambient at once.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| Fo Fourier number αt/L² | 0 to 5 | 0.05035 | 1.1 | 1.05 | +10% | −5.98 % |
| Bi Biot number hL/k (L = half-thickness) | 0.01 to 100 | 0.1133 | 0.9917 | 0.8784 | +10% | −3.95 % |
| x Distance from the centre plane | 0 to 0.01 m | 0.4198 | 0.5937 | 0.1739 | +10% | 0 % |
| L Half-thickness | 0.01 to 0.1 m | 0.5937 | 0.5937 | 0 | +10% | 0 % |
Sweep of Fourier number αt/L²
| Fourier number αt/L² | Temperature ratio |
|---|---|
| 0 | 1.1 |
| 0.625 | 0.7482 |
| 1.25 | 0.5089 |
| 1.875 | 0.3461 |
| 2.5 | 0.2354 |
| 3.125 | 0.1601 |
| 3.75 | 0.1089 |
| 4.375 | 0.07404 |
| 5 | 0.05035 |
Sweep one variable
Variables and how each one contributes
Temperature ratio
What the equation gives you. Everything else feeds it.
Biot number hL/k (L = half-thickness)
The Biot number hL/k, with L the half-thickness. A larger Biot number gives a larger λ, so the temperature falls faster. Raising it by 10 % (to 0.8639) moves the result by −3.95 %, from 0.5937 to 0.5703.
Distance from the centre plane
The distance from the centre plane, in m. Moving towards the surface lowers the temperature ratio, since the face cools first. Raising it by 10 % (to 0) moves the result by +0 %, from 0.5937 to 0.5937.
Half-thickness
The half-thickness of the wall, in m. A larger half-thickness moves a fixed x closer to the centre in the ratio x/L, so the ratio rises. Raising it by 10 % (to 0.011) moves the result by +0 %, from 0.5937 to 0.5937.
Fourier number αt/L²
The Fourier number αt/L². More time lowers the ratio exponentially. Raising it by 10 % (to 1.1) moves the result by −5.98 %, from 0.5937 to 0.5582.
Worked example with the values above
- Root of λ tan λ = 0.7854: λ = 0.785398; C₁ = 1.1002.
- Decay: exp(−λ²Fo) = exp(−0.61685 × 1) = 0.53964; position factor cos(λx/L) = 1.
- θ/θi = 1.1002 × 1 × 0.53964 = 0.59372.
Assumptions and limits
- Constant properties, a uniform initial temperature and convection to a fixed ambient on both sides for the wall solutions.
- For lumped models, a body with small internal temperature differences and constant conductances.
Common mistakes
- Using the full thickness instead of the half-thickness in Bi and Fo.
- Using the first-term form at Fo below about 0.2.
- Giving x outside the range 0 to L.
Statistics and uncertainty
First-term wall solutions are accurate to within about 2 % for Fo above 0.2, and lumped formulas need a small Biot number, below about 0.1, to stay within a few per cent.
Where it is used
- Cooling and heating of slabs, plates and electronics enclosures.
- Estimating warm-up and settling times.
Maths you need
- Biot number: Bi = hL/k
- Fourier number: Fo = αt/L²
- Time constant: τ = C/G, capacitance over conductance