Carbon diffusion through FCC and BCC iron: 22-Mec-B8 Engineering Materials, May 2018, Q6

Question 6: Carbon diffusion through FCC and BCC iron (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. Carbon diffusing interstitially through iron in its two allotropic forms, under one and the same driving gradient:

Given data
Quantity	Symbol	Value
Diffusion coefficient in FCC iron (austenite)	D_FCC	1 × 10⁻⁷ cm²/s
Diffusion coefficient in BCC iron (ferrite)	D_BCC	4 × 10⁻⁵ cm²/s
Concentration gradient (same in both cases)	dc/dx	5 × 10²⁰ atoms/cm³ per cm

Find. The flux of carbon atoms — the number crossing unit area per unit time — through each lattice, and hence how much faster carbon moves through one than the other.

Carbon flux through the two iron lattices, drawn on a logarithmic axis because the two answers are 400 times apart and would otherwise be unreadable on one scale. Underneath, the structural reason: BCC iron is the less densely packed of the two, and it is the openness of the diffusion path, not the size of the resting site, that sets the mobility.

Approach. Apply Fick’s first law to each lattice in turn. Since the question fixes the same concentration gradient in both cases, the two fluxes are in the ratio of the two diffusion coefficients, and the arithmetic is a single multiplication for each.

State Fick’s first law and fix the sign convention. The steady-state flux of a diffusing species is proportional to its concentration gradient, $$J = -D\,\frac{dc}{dx}$$ where $J$ is in atoms per square centimetre per second, $D$ in cm²/s and $dc/dx$ in atoms per cubic centimetre per centimetre. The minus sign records that atoms move down the gradient, from rich to lean; the question asks how many atoms diffuse, so the magnitudes are what is reported.
Check the units of the given gradient. The question quotes the gradient as “5 × 10²⁰ atoms per cubic centimetre”, which is a concentration, not a gradient. A gradient must carry one further inverse length, so the value is read as $$\frac{dc}{dx} = 5\times 10^{20}\ \frac{\text{atoms/cm}^3}{\text{cm}} = 5\times 10^{20}\ \text{atoms/cm}^{4}$$ Making that reading explicit is worth doing: it is the only ambiguity in the problem, and the units of the answer depend on it.
Compute the flux through FCC iron. Substituting the austenite value, $$J_{FCC} = D_{FCC}\frac{dc}{dx} = \left(1\times 10^{-7}\ \tfrac{\text{cm}^2}{\text{s}}\right)\left(5\times 10^{20}\ \tfrac{\text{atoms}}{\text{cm}^4}\right)$$ $$\boxed{\ J_{FCC} = 5\times 10^{13}\ \text{atoms}\,\text{cm}^{-2}\text{s}^{-1}\ }$$
Compute the flux through BCC iron. By the same substitution with the ferrite value, $$J_{BCC} = \left(4\times 10^{-5}\right)\left(5\times 10^{20}\right)$$ $$\boxed{\ J_{BCC} = 2\times 10^{16}\ \text{atoms}\,\text{cm}^{-2}\text{s}^{-1}\ }$$
Form the ratio and read the physical meaning. Dividing one by the other, and noting that the gradient cancels because the question made it common to both cases, $$\frac{J_{BCC}}{J_{FCC}} = \frac{D_{BCC}}{D_{FCC}} = \frac{4\times 10^{-5}}{1\times 10^{-7}} = 400$$ Carbon therefore crosses BCC iron four hundred times as fast as it crosses FCC iron at the same driving gradient. Put in tangible terms, one square millimetre of interface passes about 1.8 × 10¹⁵ carbon atoms per hour through austenite and 7.2 × 10¹⁷ through ferrite.
Explain the result from the two crystal structures. The face-centred cubic lattice has an atomic packing factor of 0.74 and the body-centred cubic lattice only 0.68, so BCC iron is the more open of the two. Its largest interstitial hole is nevertheless the smaller: the BCC tetrahedral site takes a sphere of radius 0.291R against 0.414R for the FCC octahedral site. Those two facts are not in conflict, and holding both is the whole insight of the question. Diffusion is a sequence of hops between sites, and the constriction the carbon atom must squeeze through on the way is far more open in BCC iron, so the activation energy per hop is much lower (about 80 kJ/mol in ferrite against 148 kJ/mol in austenite) and the mobility correspondingly higher.
Note the counterpart fact about solubility, which is the practical consequence. Because the resting site in BCC iron is so tight, ferrite dissolves at most 0.022 wt % carbon while austenite dissolves up to 2.11 wt %, a factor of nearly a hundred the other way. Carbon therefore moves quickly but is held sparingly in ferrite, and moves slowly but is held generously in austenite. This pairing is why carburising is done in the austenite field, where there is somewhere for the carbon to go, and why the martensite transformation — which must outrun carbon diffusion entirely — needs quench rates high enough to beat the far more mobile carbon of the low-temperature structure.

Carbon flux through the two iron lattices
Quantity	Symbol	Value
Flux through FCC iron (austenite)	J_FCC	5 × 10¹³ atoms cm⁻² s⁻¹
Flux through BCC iron (ferrite)	J_BCC	2 × 10¹⁶ atoms cm⁻² s⁻¹
Same fluxes in SI units	—	5 × 10¹⁷ and 2 × 10²⁰ atoms m⁻² s⁻¹
Ratio of the two fluxes	J_BCC/J_FCC	400
Atomic packing factor, FCC / BCC	APF	0.74 / 0.68
Maximum carbon solubility, austenite / ferrite	—	2.11 wt% / 0.022 wt%

Check: the question states the concentration gradient in units of atoms per cubic centimetre, which is a concentration rather than a gradient; the solution assumes the intended quantity is 5 × 10²⁰ atoms cm⁻³ per cm, since Fick’s law cannot otherwise be applied and the resulting flux units are then the conventional ones. The calculation is also a steady-state one: Fick’s first law presumes the gradient does not change with time, whereas a real carburising treatment has a gradient that flattens as the case grows, which is Fick’s second law and a different calculation. Finally, the two diffusion coefficients as quoted must belong to different temperatures, because the two phases are stable in different temperature ranges; the comparison the question asks for is therefore a structural one, not one made at a single temperature.

Topic. Steady-state interstitial diffusion, Fick’s first law, and the effect of crystal structure on diffusivity.

Key relations. Fick’s first law J = −D(dc/dx); the Arrhenius form of the diffusion coefficient D = D₀exp(−Q/RT); atomic packing factors of 0.74 (FCC) and 0.68 (BCC); interstitial hole radii of 0.414R (FCC octahedral) and 0.291R (BCC tetrahedral); Fick’s second law ∂c/∂t = D∂²c/∂x² for the non-steady case, whose error-function solution governs carburising depth.

Why this works. Carbon is small enough to sit between the iron atoms rather than replacing them, so it moves by hopping from one interstitial site to the next without needing a vacancy to be present. The rate of hopping is set by the energy barrier the atom must cross at the narrowest point of its path, and that barrier — not the size of the site it rests in — is what the diffusion coefficient measures. Because BCC iron is the less densely packed structure, its diffusion paths are wider, its activation energy is roughly half that of austenite, and its diffusivity is orders of magnitude greater. Once the two diffusivities are known, Fick’s first law converts them into a flux directly, and since the question holds the gradient common to both cases the flux ratio simply reproduces the diffusivity ratio.

Common pitfall. Assuming that the more open lattice must also dissolve more carbon, and so predicting that ferrite is the higher-solubility phase; the opposite is true, and the distinction between mobility and solubility is the point of the question. The other standard error is unit confusion: mixing cm and m in the same substitution moves the answer by four orders of magnitude, because flux carries an inverse area.

Source. Askeland & Wright, Ch. 5 (atom and ion movements in materials — Fick’s laws, interstitial diffusion); Callister & Rethwisch, Ch. 5 (diffusion) and Ch. 3 (crystal structures, APF and interstitial sites); Shackelford, Ch. 5.

Question 6 of 8: Carbon diffusion through FCC and BCC iron

Question 6: Carbon diffusion through FCC and BCC iron (20 marks)