Surface treatments — anodising, plasma spraying, vapour deposition: Worked examples

The scenario matrix

Every worked example below is tagged with one cell of this table. Together they touch every cell.

Cell	Process	What makes it tricky	Example
A1	Anodising	Ordinary case: solve for time	Ex 1
A2	Anodising	Solve for a different unknown (current)	Ex 2
A3	Anodising	Degenerate: $η = 1$ (perfect) vs low $η$	Ex 3
A4	Anodising	Limiting/exam twist: what if area doubles? (scaling)	Ex 4
P1	Plasma spray	Ordinary melt-energy for a finite particle mass	Ex 5
P2	Plasma spray	Degenerate: already-hot powder ( $T_{0} \to T_{m}$ )	Ex 6
P3	Plasma spray	Word problem: does the particle melt in flight-time?	Ex 7
V1	Vapour dep.	Ordinary growth-rate, convert to per-hour	Ex 8
V2	Vapour dep.	Zero/degenerate: $Φ = 0$ , and time to reach target thickness	Ex 9
V3	PVD vs CVD	Conceptual exam twist: classify a process, no numbers	Ex 10

Sign note. All physical quantities here (charge, mass, temperature difference up, flux) are non-negative, so there are no "quadrants". The genuine edge cases are zero inputs (Ex 9), efficiency extremes (Ex 3), degenerate temperature gaps (Ex 6), and scaling limits (Ex 4).

Anodising cases

Plasma-spray cases

Figure — Surface treatments — anodising, plasma spraying, vapour deposition

The figure above shows the two-stage heating of a single droplet: a sloped section (temperature climbing, energy $= m c_{p} Δ T$ ) then a flat section (temperature stuck at $T_{m}$ while latent heat $L_{f}$ is poured in to break the solid bonds). Ex 5–7 read straight off this curve.

Worked example Ex 5 — Cell P1: melt energy for a real finite particle

A single spherical $Zr O_{2}$ particle has diameter $30 μ m$ and density $ρ_{p} = 5680 kg/ m^{3}$ . Data: $c_{p} = 580 J k g^{- 1} K^{- 1}$ , $L_{f} = 7 \times 1 0^{5} J/kg$ , $T_{m} = 2700^{\circ} C$ , $T_{0} = 25^{\circ} C$ . Find the total joules to melt it.

Forecast: the particle is tiny, so expect a tiny absolute energy — microjoule scale.

Find the mass. Why? The formula needs $m$ , not diameter. Volume of sphere $V = \frac{4}{3} π r^{3}$ with $r = 15 \times 1 0^{- 6} m$ : $V = \frac{4}{3} π (15 \times 1 0^{- 6})^{3} = 1.4137 \times 1 0^{- 14} m^{3}$ $m = ρ_{p} V = 5680 \times 1.4137 \times 1 0^{- 14} = 8.029 \times 1 0^{- 11} kg$
Apply the two-term formula. Why two terms? Slope (heat up) then flat (melt), from the figure: $Q = m c_{p} (T_{m} - T_{0}) + m L_{f} = m [580 (2675) + 7 \times 1 0^{5}]$
Compute the bracket $= 1.5515 \times 1 0^{6} + 0.7 \times 1 0^{6} = 2.2515 \times 1 0^{6} J/kg$ , then $Q = 8.029 \times 1 0^{- 11} \times 2.2515 \times 1 0^{6} \approx 1.808 \times 1 0^{- 4} J \approx 181 μ J$

Verify: the per-kg value $2.2515 \times 1 0^{6} J/kg$ matches the parent note's $2.25 \times 1 0^{6}$ ✓; absolute energy is microjoules as forecast ✓.

Worked example Ex 7 — Cell P3: word problem — does it melt in flight?

The plasma delivers heat to the Ex-5 particle at a power of $P = 0.30 W$ (0.30 J/s) while it flies for a residence time $τ = 1.0 ms$ . Does it fully melt?

Forecast: it needed $181 μ J$ (Ex 5). In 1 ms at 0.30 W it receives $0.30 \times 1 0^{- 3} = 300 μ J$ . That's more than needed → yes, melts (with margin).

Energy delivered in flight. Why? Compare supply against the demand from Ex 5: $Q_{in} = P τ = 0.30 \times 1.0 \times 1 0^{- 3} = 3.0 \times 1 0^{- 4} J = 300 μ J$
Compare to requirement $Q_{melt} = 181 μ J$ (Ex 5). Since $300 > 181$ , the particle melts.
Margin. Why report it? Practically, extra energy goes to superheat/keep it molten till impact: $ratio = 300/181 \approx 1.66$

Verify: ratio $> 1$ ⇒ melts ✓; consistent with the parent note's conclusion that plasma enthalpy "easily delivers this" ✓. If instead $τ$ were only $0.5 ms$ , $Q_{in} = 150 μ J < 181$ → it would arrive unmelted, forming a defect — the reason Adhesion and Surface Roughness and dwell-time tuning matter.

Vapour-deposition cases

Worked example Ex 8 — Cell V1: growth rate → per hour

A PVD source delivers $Φ = 2 \times 1 0^{19} atoms m^{- 2} s^{- 1}$ of Ti, $M = 0.0479 kg/mol$ , $ρ = 4500 kg/ m^{3}$ , $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ . Find $r$ in µm/hr.

Forecast: flux is double the parent's $1 \times 1 0^{19}$ (which gave 0.64 µm/hr). Rate scales linearly with $Φ$ → expect ~1.3 µm/hr.

Apply the flux formula. Why divide by $N_{A} ρ$ ? $Φ M / N_{A}$ = kg per m² per s; dividing by density (kg/m³) leaves m/s = thickness rate: $r = \frac{Φ M}{N _{A} ρ} = \frac{2 \times 1 0 ^{19} \times 0.0479}{6.022 \times 1 0 ^{23} \times 4500}$
Compute: numerator $= 9.58 \times 1 0^{17}$ ; denominator $= 2.71 \times 1 0^{27}$ : $r = 3.535 \times 1 0^{- 10} m/s$
Convert to per hour. Why? Human-scale reporting: $\times 3600$ : $r = 3.535 \times 1 0^{- 10} \times 3600 = 1.273 \times 1 0^{- 6} m/hr \approx 1.27 μ m/hr$

Verify: exactly double the parent's 0.64 µm/hr (since $Φ$ doubled and $r \propto Φ$ ) ✓; still sub-few-µm/hr, why PVD is a thin-film method ✓.

Worked example Ex 9 — Cell V2: zero input + time to target

(a) The shutter closes: $Φ = 0$ . What is $r$ ? (b) With the Ex-8 flux running, how long to reach a $3 μ m$ TiN-class film?

Forecast: (a) no atoms arriving → zero growth (film paused). (b) at ~1.27 µm/hr, 3 µm takes a bit over 2 hours.

(a) Set $Φ = 0$ . Why? Growth is entirely driven by arriving flux: $r = \frac{0 \cdot M}{N _{A} ρ} = 0 m/s$ The film thickness holds constant — the degenerate "nothing happens" case.
(b) Time = thickness ÷ rate. Why? $r$ is constant, so $t = d / r$ with $d = 3 \times 1 0^{- 6} m$ , $r = 3.535 \times 1 0^{- 10} m/s$ : $t = \frac{3 \times 1 0 ^{- 6}}{3.535 \times 1 0 ^{- 10}} = 8487 s \approx 2.36 hr$

Verify: (a) $r = 0$ trivially ✓. (b) cross-check via rate: $2.36 hr \times 1.273 μ m/hr = 3.00 μ m$ ✓. This slowness is exactly why thick heat-shields use plasma spray, not PVD (see Thermal Barrier Coatings).

Worked example Ex 10 — Cell V3: conceptual exam twist (classify, no numbers)

A process bubbles $TiC l_{4}$ vapour plus $N_{2}$ and $H_{2}$ over a blade heated to 900 °C; on the surface they form TiN and release HCl gas. PVD or CVD? Balance the reaction.

Forecast: a reaction happens on the hot surface (new bonds, byproduct gas) → smells like CVD.

Test the one-word criterion. Why? PVD = physical evaporation only; CVD = a chemical reaction. Here reactants become new compounds (TiN + HCl) → CVD.
Balance. Why balance? Atom conservation is the check an examiner wants: $TiC l_{4} + \frac{1}{2} N_{2} + 2 H_{2} \to TiN + 4 HCl$
Count atoms. Ti: 1=1. N: $2 \times \frac{1}{2} = 1 = 1$ . Cl: 4=4. H: $2 \times 2 = 4 = 4$ . ✓

Verify: every element balances ✓; the presence of a genuine reaction (not mere condensation) confirms CVD, matching the parent's TiN example.

Recall Which cell was hardest for you?

The genuinely tricky ones are the limits: $η \to 0$ (bath never finishes, Ex 3), $T_{0} \to T_{m}$ (only latent heat, Ex 6), $Φ \to 0$ (growth pauses, Ex 9). If a formula behaves sanely at its extremes, you can trust it in the middle.

Connections

Faraday's Laws of Electrolysis — the charge bookkeeping behind Ex 1–4
Latent Heat and Phase Change — the two-term melt energy of Ex 5–7
Thermal Barrier Coatings — why plasma spray (thick) beats PVD (thin) here
Adhesion and Surface Roughness — unmelted particles (Ex 7) ruin bonding
Corrosion and Passivation — the anodised oxide these examples grow
Aluminium Alloys in Aerospace — the substrate being anodised