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21-Mat-A2 Materials Transport Phenomena · December 2013

Question 2 of 9: FeO Reduction by Carbon via a Gas-Halo Intermediate

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2013 — Met-A2, Metallurgical Rate Phenomena. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator, one double-sided aid sheet permitted; candidates were told to state any interpretive assumptions. Candidates answer Question 1 plus any four of Questions 2–9 — all nine are solved below for completeness. All questions are of equal value (20 marks each, five questions = 100%).

Reference texts: Geankoplis, C. J., Transport Processes and Separation Process Principles — mass/heat/momentum transfer fundamentals (Fick's/Newton's/Fourier's laws, boundary-layer correlations); Szekely, J. & Themelis, N. J., Rate Phenomena in Process Metallurgy — gas-halo diffusion, ladle/tundish fluid flow, wire injection; Turkdogan, E. T., Fundamentals of Steelmaking — BOF/EAF heat and mass balances; Callister, W. D., Materials Science and Engineering — TTT/CCT diagrams and phase transformations; Incropera, F. P. & DeWitt, D. P., Fundamentals of Heat and Mass Transfer — liquid-metal (low-Pr) convection correlations.

Question 2: FeO Reduction by Carbon via a Gas-Halo Intermediate (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. Reaction pair: at the droplet/char surface, $C_{(droplet)} + CO_2 \rightarrow 2CO$; at the slag-halo interface, $CO + FeO \rightarrow Fe + CO_2$ — net $C + FeO \rightarrow Fe + CO$, with CO₂ shuttling across the halo as the diffusing intermediate. $T=1600\,{}^{\circ}\text{C}=1873\,\text{K}$, gas pressure in halo $P=1.2$ atm, halo thickness $\delta=120\,\mu\text{m}$, $D_{CO}=D_{CO_2}=1.0\times10^{-3}\ \text{m}^2/\text{s}$, bulk (slag-side) $y_{CO_2}=6\%$, droplet-side $C_{CO_2}\approx0$ (fast interfacial reaction). Fe smelting rate $=4$ t/hr, slag $=40$ t at 6 wt% FeO, $\rho_{slag}=3300\ \text{kg/m}^3$, average particle diameter $d_p=1$ cm.

Find. Total char + droplet surface area $A$ sustaining the observed smelting rate, and the resulting dispersed-phase volumetric loading in the slag.

Fe-2%C droplet 6% FeO slag gas halo, δ = 120 μm droplet: C + CO₂ → 2CO slag interface: CO + FeO → Fe + CO₂
CO₂/CO diffuse across the thin (120 μm) gas halo separating the char/droplet surface from the slag; CO₂ is the species consumed at the droplet and regenerated at the slag interface.

Approach. Since $\delta\ll d_p$, treat the halo as a flat stagnant film: Fick’s law gives the CO₂ flux from the slag-side (6%) to the droplet-side ($\approx0$) interface; this flux fixes the molar rate of FeO reduction per unit area (1:1 stoichiometry through the coupled reactions), which combined with the observed Fe production rate gives the required total interfacial area, then the particle count and dispersed volume.

  1. Total gas concentration and CO₂ driving force in the halo. $C_{tot}=\dfrac{P}{RT}=\dfrac{1.2\times101325}{8.314\times1873}=7.808\ \text{mol/m}^3$. Bulk (slag-side) $C_{CO_2}=0.06\times7.808=0.4685\ \text{mol/m}^3$; droplet-side $\approx0$.
  2. Fick’s-law flux across the thin film. $N_{CO_2}=\dfrac{D\,\Delta C}{\delta}=\dfrac{(1.0\times10^{-3})(0.4685)}{120\times10^{-6}}$ $$\boxed{N_{CO_2}=3.904\ \text{mol/(m}^2\text{s)}}$$ This flux equals the molar rate of both carbon consumption at the droplet and FeO reduction at the slag interface per unit area (each mole of CO₂ delivered drives one net mole of FeO→Fe).
  3. Observed molar Fe production rate. $\dot n_{Fe}=\dfrac{4000\ \text{kg/hr}\times1000}{56\ \text{g/mol}\times3600\ \text{s/hr}}=19.84\ \text{mol/s}$.
  4. Required total interfacial area. Substituting into $\dot n_{Fe}=N_{CO_2}\,A$: $$A=\dfrac{19.84}{3.904}\Rightarrow\boxed{A=5.08\ \text{m}^2}$$ of combined char + droplet surface must be present in the slag at any instant.
  5. Particle count and dispersed volume. Per 1 cm-diameter sphere: $a_p=\pi d_p^2=3.142\times10^{-4}\ \text{m}^2$, $v_p=\tfrac{\pi}{6}d_p^3=5.236\times10^{-7}\ \text{m}^3$. $N=A/a_p=5.08/3.142\times10^{-4}=1.62\times10^4$ particles; $V_p=N\,v_p=1.62\times10^4\times5.236\times10^{-7}=8.47\times10^{-3}\ \text{m}^3$.
  6. Volumetric loading. $V_{slag}=\dfrac{40{,}000}{3300}=12.12\ \text{m}^3$. $$\boxed{\%\,\text{loading}=\dfrac{V_p}{V_{slag}}\times100=\dfrac{8.47\times10^{-3}}{12.12}\times100=0.070\%}$$ — a very dilute dispersion, consistent with a foaming slag carrying suspended reacting particles rather than a packed bed.
Question 2 — final results
QuantityValue
CO₂ flux across halo3.90 mol/(m²·s)
Required char + droplet area5.08 m²
Particle count (1 cm dia.)≈ 1.62 × 10⁴
Dispersed volumetric loading0.070 vol%