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21-Mat-A2 Materials Transport Phenomena · December 2013

Question 7 of 9: Aluminum Wire Feed Velocity for Full Dissolution in Liquid Steel

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 7: Aluminum Wire Feed Velocity for Full Dissolution in Liquid Steel (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.

Check: the paper prints the Mucciardi & Guthrie correlation with the wire diameter in metres. Substituting $d=0.01$ m literally returns $v\approx1.2\times10^6$ m/s, which is physically impossible, so the correlation as printed cannot be dimensionally self-consistent in those units. The diameter is therefore taken in millimetres ($d=10$), the unit in which this empirical wire-injection fit is conventionally quoted, which returns a sane 13.2 m/s that brackets the theoretical result — stated here as an interpretive assumption, per the exam’s own page-1 instruction.

Given. $d=0.01$ m, $T_{bath}=1600\,{}^{\circ}\text{C}$, $T_{m,Al}=660\,{}^{\circ}\text{C}$ (superheat driving force $\Delta T=940$ K), target depth $=2.7$ m, feed temperature taken as ambient ($25\,{}^{\circ}\text{C}$). $\rho_{Al}=2700$, $C_{p,Al,s}=902$ J/kg K, $L_{Al}=387{,}000$ J/kg. Liquid steel: $k=28$ W/m K, $\mu=7\times10^{-3}$ Pa s, $C_p=750$ J/kg K, $\rho=7000$ kg/m³.

Find. The wire feed velocity $v$ that lets the wire just finish melting at 2.7 m depth, by (i) the given $Nu_L$ correlation and (ii) the Mucciardi & Guthrie empirical correlation; compare.

Approach. $Nu_L=1.12\,Re_L^{0.5}Pr^{0.5}$ is the low-Prandtl (liquid-metal) flat-plate correlation Question 4 derived (its length-averaged form), so $L$ here is the wire’s submerged travel distance (depth), not its diameter — the boundary layer develops along the wire’s path exactly as it does along Q4’s plate. The wire’s energy requirement (sensible heat to melting point + latent heat) must be supplied by convection from the bath over the transit time $\tau=$depth$/v$; setting the two equal, with $h$ itself a function of $v$ through $Re_L$, gives an implicit (numerically converged) expression for $v$.

  1. Energy to fully melt one metre of wire. $$Q'=\rho_{Al}\left(\dfrac{\pi}{4}d^2\right)\left[C_{p,Al,s}(T_{m,Al}-T_0)+L_{Al}\right]=(2700)(7.854\times10^{-5})\left[902(635)+387{,}000\right]$$ $$\boxed{Q'=2.035\times10^5\ \text{J/m}}$$
  2. Liquid-steel Prandtl number. $$Pr=\dfrac{\mu C_p}{k}=\dfrac{(7\times10^{-3})(750)}{28}=\boxed{0.1875}$$ — a classic low-$Pr$ liquid metal, consistent with Q1(n)/Q4.
  3. General expression. Heat gained over the transit ($\tau=$depth$/v$) at the length-averaged $h$: $$h\,(\pi d)(T_{bath}-T_{m,Al})\dfrac{\text{depth}}{v}=Q',\qquad h=\dfrac{Nu_L\,k}{\text{depth}}=\dfrac{1.12\,k}{\text{depth}}\sqrt{\dfrac{\rho\,v\,\text{depth}}{\mu}\,Pr}$$ Substituting and solving for $v$ (the depth cancels neatly, leaving $v$ explicit): $$\boxed{v=\left[\dfrac{1.12\,k\,\pi d\,\Delta T\sqrt{\rho\,\text{depth}\,Pr/\mu}}{Q'}\right]^2}$$
  4. Evaluate numerically (converged, since $h$ itself depends on $v$). $$v=10.48\ \text{m/s},\qquad Re_L=\dfrac{\rho v\,\text{depth}}{\mu}=2.83\times10^7,\qquad Nu_L=2580,\qquad h=26{,}760\ \text{W/m}^2\text{K}$$ $$\boxed{v\approx10.5\ \text{m/s}}\ \ (\text{transit time to 2.7 m} = 2.7/10.48 = 0.258\ \text{s})$$
  5. Compare with the Mucciardi & Guthrie empirical correlation. Using diameter in millimetres (the correlation’s conventional unit for a small-diameter wire), $d=10$ mm, superheat $=940$ K: $$v=\left(\dfrac{\text{Depth}\times\text{Superheat}^{0.34}}{d^{0.86}}\right)^{1/0.52}=\left(\dfrac{2.7\times940^{0.34}}{10^{0.86}}\right)^{1/0.52}$$ $$\boxed{v\approx13.2\ \text{m/s}}$$

The two independent estimates (10.5 m/s from first-principles convection theory, 13.2 m/s from the empirical correlation) agree to within about 25% — a reasonable match given the very different derivations. The theoretical $Nu_L$ route idealises the wire as a smooth, static-relative-to-bath flat-plate analogue with uniform properties and neglects the plant-scale turbulence of the “strong-stir” ladle station, local convection currents, any oxide film on the wire, and the fact that melting itself changes the effective wire diameter along its length; the empirical correlation, fitted to real plant/lab wire-injection trials, implicitly captures these effects and is generally the more trustworthy number for design, with the theoretical route serving as an independent order-of-magnitude check.

Question 7 — final results
QuantityValue
Energy to melt 1 m of wire2.035 × 10⁵ J/m
Liquid-steel Prandtl number0.1875
Wire velocity (Nu₃ theory)≈ 10.5 m/s
Wire velocity (Mucciardi–Guthrie correlation)≈ 13.2 m/s