Inward flux of carbon during carburizing, from Fick's first law: 22-Mec-B8 Engineering Materials, Undated paper, Q2

Question 2: Inward flux of carbon during carburizing, from Fick's first law (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A steel held in the austenite (FCC) field with a carburizing atmosphere at its surface, the case profile idealised as linear:

Given data
Quantity	Symbol	Value
Carbon content at the surface	x₁	3.8 atomic %
Carbon content 2.5 mm below the surface	x₂	1.0 atomic %
Separation of the two planes	Δx	2.5 mm = 2.5 × 10⁻³ m
Diffusivity of carbon in FCC iron	D	2.9 × 10⁻¹¹ m²/s
Lattice parameter of FCC (γ) iron, assumed	a	0.36 nm = 3.6 × 10⁻¹⁰ m
Atoms per FCC unit cell	—	4

Find. The steady inward flux of carbon atoms across the case, in atoms per square metre per second.

The idealised carbon profile through the case, and the lattice the carbon has to move through. The linear gradient the question prescribes is the whole of Fick’s first law once the atomic percentages have been converted into atoms per cubic metre using the FCC iron atom density; the interstitial site shown in red is the octahedral hole through which each carbon atom hops.

Approach. Fick's first law needs concentrations expressed as atoms per unit volume, not as percentages, so the atomic percentages must first be converted using the atom density of the FCC iron lattice; the gradient is then the difference in those concentrations divided by the 2.5 mm separation, and the flux follows directly.

Establish the atom density of the iron lattice. A face-centred cubic cell contains four atoms — eight corners shared eight ways plus six faces shared two ways — so $$N_{Fe} = \frac{4}{a^{3}} = \frac{4}{(3.6\times10^{-10})^{3}} = \frac{4}{4.6656\times10^{-29}}$$ $$N_{Fe} = 8.573\times10^{28}\ \text{Fe atoms/m}^{3}$$ Carbon occupies the octahedral interstices between those iron atoms and does not displace them, so this figure is the reference against which the atomic percentages are converted.
Convert the two atomic percentages into carbon concentrations. Atomic per cent means carbon atoms as a fraction of all atoms present, so for an atom fraction $x$ the carbon concentration is $C = N_{Fe}\,x/(1-x)$. At the surface, $$C_1 = 8.573\times10^{28}\times\frac{0.038}{0.962} = 3.387\times10^{27}\ \text{C atoms/m}^{3}$$ and at 2.5 mm depth, $$C_2 = 8.573\times10^{28}\times\frac{0.010}{0.990} = 8.660\times10^{26}\ \text{C atoms/m}^{3}$$ The surface carries roughly four times as much carbon as the plane 2.5 mm beneath it, and it is that imbalance which drives the diffusion.
Form the concentration gradient. With the profile assumed linear, $$\frac{dC}{dx} = \frac{C_2 - C_1}{x_2 - x_1} = \frac{8.660\times10^{26} - 3.387\times10^{27}}{2.5\times10^{-3}}$$ $$\frac{dC}{dx} = -1.008\times10^{30}\ \text{atoms}/(\text{m}^{3}\!\cdot\!\text{m})$$ The gradient is negative because carbon becomes scarcer with depth; keeping that sign is what makes the next step give an inward, positive flux.
Apply Fick's first law. $$J = -D\frac{dC}{dx} = -(2.9\times10^{-11})(-1.008\times10^{30})$$ $$\boxed{\ J = 2.92\times10^{19}\ \text{carbon atoms}/(\text{m}^{2}\!\cdot\!\text{s})\ }$$ The minus sign in Fick's law encodes the physical fact that matter moves down the concentration gradient; with a negative gradient the flux comes out positive, that is, directed inward from the surface, which is what the question asks for.
Express the same flux as a mass rate, for a feel for the number. Dividing by Avogadro's number and multiplying by the atomic mass of carbon, $$J_m = \frac{(2.92\times10^{19})(12.011)}{6.022\times10^{23}} = 5.83\times10^{-4}\ \text{g}/(\text{m}^{2}\!\cdot\!\text{s})$$ that is 0.583 milligrams of carbon crossing each square metre every second, or about 50 grams per square metre per day. For a part with 0.1 m² of surface held for eight hours, that is some 1.7 g of carbon absorbed — a plausible figure for a commercial carburizing cycle, which is the sanity check worth making before writing the answer down.

Results for the carburizing flux
Quantity	Symbol	Value
Iron atom density, FCC	N_Fe = 4/a³	8.573 × 10²⁸ m⁻³
Carbon concentration at the surface	C₁	3.387 × 10²⁷ m⁻³
Carbon concentration at 2.5 mm	C₂	8.660 × 10²⁶ m⁻³
Concentration gradient	dC/dx	−1.008 × 10³⁰ m⁻⁴
Inward flux of carbon	J = −D dC/dx	2.92 × 10¹⁹ atoms/(m²·s)
Same flux as a mass rate	J_m	5.83 × 10⁻⁴ g/(m²·s) = 0.583 mg/(m²·s)

Check: the question gives concentrations in atomic per cent but no lattice parameter, so a value must be assumed to convert them into atoms per cubic metre. The standard figure for FCC γ-iron at carburizing temperature, a = 0.36 nm, is used here and is stated in the answer as an assumption. Two conventions for “atomic per cent” are in circulation for an interstitial solute: the rigorous one used above, C = N_Fex/(1−x), and the dilute approximation C = N_Fex. The approximation gives J = 2.79 × 10¹⁹ atoms/(m²·s), which is 4.8 per cent lower — the same answer to one significant figure, and well inside the uncertainty carried by the assumed lattice parameter. The gradient is also idealised as constant, which the question instructs; a real carburized case follows an error-function profile whose gradient is steepest at the surface, so the true surface flux is higher than this average.

Topic. Steady-state interstitial diffusion, Fick's first law, and the conversion of composition units into a volumetric concentration.

Key relations. Fick's first law J = −D dC/dx; atom density of a cubic lattice N = (atoms per cell)/a³, giving 4/a³ for FCC and 2/a³ for BCC; conversion of an atom fraction for an interstitial solute, C = N_hostx/(1−x); the Arrhenius temperature dependence D = D₀exp(−Q/RT); and, for the non-steady case the question deliberately sets aside, Fick's second law with its error-function solution and the case-depth scaling x ∝ √(Dt).

Why this works. Carbon is a small atom — about 0.077 nm radius against iron's 0.124 nm — so it does not substitute for iron atoms but sits in the octahedral holes between them. Moving from one hole to the next needs no vacancy, only enough thermal energy to squeeze past the surrounding iron atoms, which is why interstitial diffusion is orders of magnitude faster than substitutional diffusion at the same temperature and why carburizing is industrially practical at all. Carburizing is done in the austenite field for two compounding reasons: FCC iron dissolves far more carbon than BCC ferrite (up to 2.1 wt % against 0.02 wt %), so a steep surface concentration can be established, and the temperature required to reach austenite also raises D by orders of magnitude through the Arrhenius term. Fick's first law then says only that atoms drift down the concentration gradient at a rate set by how easily they hop, which is precisely what D measures.

Common pitfall. Substituting the percentages straight into Fick's law and reporting a flux in “per cent per metre”, which is dimensionally meaningless — the law requires concentration as amount per unit volume. The second pitfall is losing the minus sign and reporting an outward flux, and the third is using the BCC atom count 2/a³, or ferrite's smaller lattice parameter, when the question has explicitly said the steel is face-centred cubic.

Source. Askeland & Wright, The Science and Engineering of Materials, 7th ed., Ch. 5 (atom and ion movements: Fick's first and second laws, interstitial diffusion, carburizing); Callister & Rethwisch, Ch. 5.3–5.5; Shackelford, Ch. 5.

Question 2 of 8: Inward flux of carbon during carburizing, from Fick's first law

Question 2: Inward flux of carbon during carburizing, from Fick's first law (20 marks)