21-Mat-A2 Materials Transport Phenomena · December 2013
Question 3 of 9: Hot-Strip-Mill Transfer-Bar Rundown and Annual Capacity
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 3: Hot-Strip-Mill Transfer-Bar Rundown and Annual Capacity (20 marks)
Check: the source gives no holding-table length, so the transit/cooling window on the entry table is taken directly as the stated “minimum lag of 5 s between each slab” (a flat 5 s radiative-cooling budget), per the exam’s own instruction to state any interpretive assumption. Under this reading the rundown constraint bounds thickness from below (a thinner bar has less thermal mass per radiating area and so cools faster) — the computed value is reported as the critical/threshold thickness the constraint permits. Note also that the paper’s own data block already fixes the slab at 0.61 m thick and 1.21 m wide, so the phrase “maximum thickness of slab” cannot be asking for a slab dimension that the question itself supplies; it is read here as the transfer-bar gauge that the rundown constraint bounds. The 1.21 m width is used as given for the tonnage calculation.
Given. $\rho=7450\ \text{kg/m}^3$, $C_p=450\ \text{J/kg K}$, $\varepsilon=0.8$, $\sigma=5.67\times10^{-8}\ \text{W/m}^2\text{K}^4$, $T_0=1590\ \text{K}$, floor $T_1=1422\ \text{K}$, cooling window $\Delta t=5$ s (both faces radiate, back-radiation ignored). Finishing exit: gauge $2.3$ mm at $V=20\ \text{m/s}$, strip width $w=1.21$ m.
Find. (i) The critical transfer-bar thickness $t^*$ for which 5 s of two-sided radiative cooling exactly reaches 1422 K from 1590 K; (ii) the theoretical annual tonnage at the fixed 2.3 mm/20 m/s exit condition.
Approach. Lumped-capacitance transient radiation balance ($Bi$ not required since both faces are isothermal by symmetry) integrates in closed form to a $1/T^3$ relation in thickness and time; the annual capacity follows directly from mass conservation at the fixed finishing-exit condition (width × gauge × speed × density), independent of upstream thickness.
Radiative cooling ODE. Per unit plan area, both faces radiating with no back-flux: $\rho C_p t\dfrac{dT}{dt}=-2\varepsilon\sigma T^4$. Separating and integrating from $T_0$ to $T_1$ over $\Delta t$: $$\dfrac{1}{T_1^3}-\dfrac{1}{T_0^3}=\dfrac{6\varepsilon\sigma\,\Delta t}{\rho C_p\,t}$$
Solve for the critical thickness. $$t^*=\dfrac{6\varepsilon\sigma\,\Delta t}{\rho C_p\left(1/T_1^3-1/T_0^3\right)}=\dfrac{6(0.8)(5.67\times10^{-8})(5)}{(7450)(450)\left(1/1422^3-1/1590^3\right)}$$ $$\boxed{t^*=4.10\ \text{mm}}$$ Any transfer bar at or above this thickness survives the 5 s window above the 1422 K floor; the current 32 mm bar has a large margin — cooling it forward the same way for 5 s gives $T(5\text{s})=1564\ \text{K}$, 142 K clear of the floor, confirming the constraint does not itself cap how far the operation could go.
Annual capacity — mass conservation at the fixed exit condition. Since the coiled gauge (2.3 mm) and coiling speed (20 m/s) are the downstream product specification, the steady mass flow is fixed there regardless of upstream transfer-bar thickness: $$\dot m=\rho\,w\,t_{final}\,V_{final}=(7450)(1.21)(0.0023)(20)$$ $$\boxed{\dot m=414.7\ \text{kg/s}=1493\ \text{t/hr}}$$
Theoretical annual tonnage (24 h/day, no shutdowns). $$\dot m\times(8760\ \text{hr}\times3600\ \text{s/hr})=414.7\times3.154\times10^7$$ $$\boxed{\approx13.1\times10^6\ \text{tonnes/yr}\ (13.1\ \text{Mt/yr})}$$ — an aspirational ceiling; the real mill’s achieved tonnage is set by actual uptime, well below this zero-downtime theoretical figure.
