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21-Mat-A2 Materials Transport Phenomena · December 2015

Question 5 of 8: Hot-Strip-Mill Transfer-Bar Radiative Cooling and Annual Capacity

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

Notes on this paper

National Exams — December 2015 — 10-Met-A2 Metallurgical Rate Phenomena. Three-hour, open-book exam; any non-communicating calculator permitted. Candidates answer Question 1 (compulsory) plus any four of Questions 2–8; the five answered count equally (20 marks each). All eight are solved below for completeness. Candidates were told to state any interpretive assumptions where doubt exists.

Reference texts: Geiger, G. H. & Poirier, D. R., Transport Phenomena in Materials Processing (TMS) — the primary reference for the momentum-, heat- and mass-transfer analyses in Questions 3, 6, 7 and 8; Szekely, J., Evans, J. W. & Sohn, H. Y., Rate Phenomena in Process Metallurgy — reactor-network and interfacial mass-transfer modelling (Questions 2, 3); Welty, J. R. et al., Fundamentals of Momentum, Heat and Mass Transfer — dimensional analysis and pipe-friction data (Questions 4, 7); Gaskell, D. R., Introduction to the Thermodynamics of Materials — Fe–C phase relations (Question 1h).

Question 5: Hot-Strip-Mill Transfer-Bar Radiative Cooling and Annual Capacity (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.

QuantitySymbolValue
Steel density$\rho$7450 kg/m$^3$
Specific heat$C_p$0.45 kJ/kg·K = 450 J/kg·K
Emissivity$\varepsilon$0.8
Exit (initial) temperature$T_0$1590 K
Minimum entry (final) temperature$T_f$1422 K
Minimum inter-slab lag$t_{avail}$5 s
Coiling speed$V_{coil}$20 m/s
Final strip thickness$\delta_{strip}$2.3 mm
Slab (strip) width$w$1.21 m
Current transfer-bar thickness$\delta_0$32 mm

Find. The critical transfer-bar thickness set by the 5 s / 1422 K constraint, and the mill's new theoretical annual tonnage.

Approach. Write an unsteady, two-sided (top+bottom), radiation-only lumped-capacitance energy balance for the slab, integrate it to get cooling time as a function of thickness, evaluate it at the current 32 mm gauge and at the stated 5 s window, then compute the mill's fixed downstream (coiler-limited) mass throughput.

  1. Radiative cooling energy balance. For a slab of thickness $\delta$ radiating from both faces (area $A$, volume $A\delta$), with the surroundings taken as a cold background (no back-radiation, per the stated assumptions): $$\rho A\delta C_p\,\dfrac{dT}{dt}=-2A\varepsilon\sigma T^4\ \Rightarrow\ \dfrac{dT}{T^4}=-\dfrac{2\varepsilon\sigma}{\rho\delta C_p}\,dt$$ Integrating from $(T_0,0)$ to $(T,t)$: $$\boxed{t(\delta)=\dfrac{\rho\delta C_p}{6\varepsilon\sigma}\left(\dfrac{1}{T^3}-\dfrac{1}{T_0^3}\right)}$$
  2. Current 32 mm transfer bar — implied cooling time to 1422 K. $$t(0.032)=\dfrac{(7450)(0.032)(450)}{6(0.8)(5.67\times10^{-8})}\left(\dfrac{1}{1422^3}-\dfrac{1}{1590^3}\right)=\boxed{39.0\ \text{s}}$$ This is the time a 32 mm bar takes to radiate its way down from 1590 K to exactly 1422 K — nearly 8× longer than the stated 5 s minimum lag, so the current gauge carries a large thermal margin.
  3. Critical thickness for a 5 s window. Solving Step 1 for $\delta$ at $t=5$ s: $$\delta_{crit}=\dfrac{6\varepsilon\sigma\,t_{avail}}{\rho C_p\left(1/T_f^3-1/T_0^3\right)}=\boxed{4.10\ \text{mm}}$$ Because thinner sections have more radiating area per unit thermal mass, they cool faster: any bar thinner than $\delta_{crit}\approx4.1$ mm would drop below 1422 K within a 5 s transit, while any bar at or above it (including the current 32 mm gauge, by a wide margin) stays compliant. The 5 s/1422 K thermal constraint therefore sets a floor on gauge, not a ceiling — the superintendent's proposed route to more tonnage (longer slabs at the existing 32 mm gauge) is not thermally constrained at all.
  4. Mill's fixed downstream throughput. The finishing train's own speed/gauge limits (20 m/s at 2.3 mm) set the mass flow rate irrespective of transfer-bar thickness (mass is conserved from transfer bar through to coiled strip): $$\dot m=\rho\,w\,\delta_{strip}\,V_{coil}=(7450)(1.21)(0.0023)(20)=\boxed{414.7\ \text{kg/s}}$$
  5. Annual capacity (24 h/day, 365 d/yr, no shutdowns). $$\text{Annual tonnage}=\dot m\times(3600\times24\times365)=414.7\times3.156\times10^7\ \text{s}=\boxed{1.31\times10^7\ \text{tonnes/yr}\ (\approx13.1\ \text{Mt/yr})}$$
QuantityValue
Cooling time, current 32 mm bar≈ 39.0 s (8× the 5 s minimum)
Critical (minimum-viable) thickness for a 5 s transit≈ 4.10 mm
Mass flow rate (coiler-limited)414.7 kg/s
Theoretical annual capacity≈ 1.31×10&sup7; tonnes/yr
Check
The 5 s figure is read here as the time budget available for radiative cooling between rougher exit and F1 entry (the value the question pairs directly with the temperature constraint). Under the governing physics, that reading necessarily produces a minimum viable gauge (thinner sections cool faster, not slower), not a maximum — there is no thickness at which pure two-sided radiative loss over a fixed time budget would push a bar below spec by being too thick. The annual capacity is set entirely by the fixed downstream coiling speed/gauge (independent of transfer-bar thickness, since final strip dimensions and speed are unchanged); it is not increased by the thickness finding itself, but the finding does confirm the thermal constraint does not block the superintendent's length-increase plan at the mill's current gauge.