Library / Heat transfer and thermal systems / HTS-49
Heat transfer and thermal systems
Quasi-steady planar solidification front
Thickness of a solid layer growing from a cold wall at Tw into liquid at its melting point Tm, from the initial thickness s0.
- Ref
- HTS-49
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- s [m]
For a planar solid layer growing from a cold wall at Tw into liquid at its freezing point Tm, the quasi-steady conduction balance gives s(t) ≈ √[s₀² + 2ks(Tm − Tw)t/(ρLf)], where s₀ is the initial thickness. It is accurate for a small Stefan number, cp(Tm − Tw)/Lf, and ignores heat transfer from the liquid and contact resistance. The thickness grows as √t.
Thickness of a solid layer growing from a cold wall at Tw into liquid at its melting point Tm, from the initial thickness s0. Valid for small Stefan number, with constant solid properties and no liquid heat transfer or contact resistance.
Calculator
- Thickness gained, s − s₀0.00291 m
- Thickness from a zero start, √(2ks(Tm − Tw)t/(ρLf))0.008165 m
- Latent energy released per unit area, ρLf(s − s₀)873,000 J/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.
- After 1,000 s the solid layer is 12.91 mm thick, having grown 2.91 mm.
- Because the thickness varies as √t, the layer needs four times as long to double its growth, so freezing slows as the layer insulates the wall.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| s0 Initial solid thickness | 0 to 0.5 m | 0.008165 | 0.5001 | 0.4919 | +10% | +6.11 % |
| t Time | 0 to 1,000,000 s | 0.01 | 0.2584 | 0.2484 | +10% | +1.98 % |
| Lf Latent heat of fusion | 10,000 to 500,000 J/kg | 0.01183 | 0.04583 | 0.03399 | +10% | −1.84 % |
| Tw Cold-wall temperature | 200 to 273 K | 0.01 | 0.03276 | 0.02276 | +10% | — |
| ks Solid conductivity | 0.1 to 10 W/(m K) | 0.01017 | 0.02082 | 0.01065 | +10% | +1.98 % |
| Tm Melting temperature | 273 to 300 K | 0.01291 | 0.02295 | 0.01004 | +10% | +78.4 % |
| ρ Solid density | 500 to 3,000 kg/m³ | 0.01106 | 0.01528 | 0.00422 | +10% | −1.84 % |
Sweep of Time
| Time (s) | Solid thickness (m) |
|---|---|
| 0 | 0.01 |
| 125,000 | 0.09183 |
| 250,000 | 0.1295 |
| 375,000 | 0.1584 |
| 500,000 | 0.1828 |
| 625,000 | 0.2044 |
| 750,000 | 0.2238 |
| 875,000 | 0.2417 |
| 1,000,000 | 0.2584 |
Sweep one variable
Variables and how each one contributes
Solid thickness
What the equation gives you. Everything else feeds it.
Initial solid thickness
The initial solid thickness, in m. A thicker start raises the final thickness. Raising it by 10 % (to 0.011) moves the result by +6.11 %, from 0.01291 to 0.0137 m.
Solid conductivity
The solid conductivity, in W/(m K). A higher conductivity grows the layer faster. Raising it by 10 % (to 2.2) moves the result by +1.98 %, from 0.01291 to 0.01317 m.
Melting temperature
The melting temperature, in K. A larger temperature difference to the wall grows the layer faster. Raising it by 10 % (to 300.3) moves the result by +78.4 %, from 0.01291 to 0.02304 m.
Cold-wall temperature
The cold-wall temperature, in K. A colder wall grows the layer faster. Raising it by 10 % (to 294.8) gives no valid result at the other default values.
Time
The time, in s. The thickness grows with the square root of time. Raising it by 10 % (to 1,100) moves the result by +1.98 %, from 0.01291 to 0.01317 m.
Solid density
The solid density, in kg/m³. A denser solid has more latent heat per volume, so growth is slower. Raising it by 10 % (to 1,100) moves the result by −1.84 %, from 0.01291 to 0.01267 m.
Latent heat of fusion
The latent heat of fusion, in J/kg. A higher latent heat slows growth. Raising it by 10 % (to 330,000) moves the result by −1.84 %, from 0.01291 to 0.01267 m.
Worked example with the values above
- Growth term: 2k(Tm − Tw)t/(ρLf) = 2 × 2 × 5 × 1,000 / (1,000 × 300,000) = 6.6667×10−5 m².
- Add s₀²: 1×10−4 + 6.6667×10−5 = 1.6667×10−4 m².
- s = √(that) = 0.01291 m.
Assumptions and limits
- Constant properties in each phase, a single melting temperature and no losses unless stated.
- Steady or quasi-steady balances, with absolute temperatures where logarithms or powers are used.
Common mistakes
- Using it when the Stefan number is not small.
- Forgetting that heat from superheated liquid also has to be removed.
- Using the liquid’s properties in place of the solid’s.
Statistics and uncertainty
Latent heats and specific heats are tabulated to a few per cent, but real storage materials melt over a range and lose heat, so real systems typically deliver 5 % to 20 % less than the ideal values.
Where it is used
- Sizing thermal storage, ice banks and phase-change units.
- Checking exchanger data, mixing streams and cooling times.
Maths you need
- Latent heat: energy per kilogram to change phase
- Heat-capacity rate: C = ṁcp
- Second law: entropy generation cannot be negative