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

Question 5: Hot-Strip-Mill Transfer-Bar Radiative Cooling — Maximum Slab Thickness 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.

Find. The gauge (thickness) of transfer bar the 5 s / 1422 K thermal constraint permits, and in which direction it is bounded; the mill's new theoretical annual tonnage; and a brief justification for neglecting through-thickness gradients, natural convection, roller conduction and back-radiation.

Approach. Write an unsteady, two-sided (top+bottom), radiation-only lumped-capacitance energy balance for a slab element of thickness $\delta$, integrate it to get cooling time as a function of thickness, then invert it at the 5 s window to get the critical gauge — and check which side of that gauge is actually feasible, since the question's word "maximum" presumes a ceiling; separately compute the mill's fixed downstream (coiler-limited) mass throughput for the annual-capacity figure.

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 (back-radiation neglected, justified in Step 5): $$\rho A\delta C_p\,\dfrac{dT}{dt}=-2A\varepsilon\sigma T^4\ \Rightarrow\ \dfrac{dT}{T^4}=-\dfrac{2\varepsilon\sigma}{\rho\delta C_p}\,dt$$ Integrating from $(T_0,0)$ to $(T,t)$: $$\boxed{t(\delta)=\dfrac{\rho\delta C_p}{6\varepsilon\sigma}\left(\dfrac{1}{T^3}-\dfrac{1}{T_0^3}\right)}$$ Every term in this balance is per unit plan area of slab, so the only geometric variable it contains is the thickness $\delta$ — and $t$ is linear and increasing in $\delta$, which is what decides the direction of the bound in Step 4.
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: $$\delta_{crit}=\dfrac{6\varepsilon\sigma\,t_{avail}}{\rho C_p\left(1/T_f^3-1/T_0^3\right)}=\boxed{4.10\ \text{mm}}$$ This is the one gauge the given data actually pins down: it is the thickness whose cooling time is exactly the 5 s window.
4. Which side of $\delta_{crit}$ is feasible — the constraint is a floor, not a ceiling. Because $t(\delta)$ is linear and increasing in $\delta$ (Step 1), a thinner section carries more radiating area per unit thermal mass and therefore cools faster. Any bar thinner than $\delta_{crit}\approx4.10$ mm drops below 1422 K inside the 5 s window; every bar at or above it — including the current 32 mm gauge, by a factor of nearly 8 — stays compliant, and stays more compliant the thicker it gets. There is no way for a slab to be too thick to hold 1422 K over a fixed radiative-cooling window, so the constraint admits no finite maximum: the question's word "maximum" is the wrong sense for the physics it supplies. $$\boxed{\delta_{crit}=4.10\ \text{mm is the critical gauge; the 5 s/1422 K constraint sets it as a }\textbf{minimum}\text{ viable thickness, not a maximum — any }\delta\ge4.10\ \text{mm is admissible, and the current 32 mm transfer bar clears it with an }8\times\text{ margin}}$$ Reporting the 4.10 mm figure together with the direction of the inequality is the honest answer: the superintendent's proposal to lengthen the slabs is not limited by this radiative-cooling constraint at all, and any real cap on transfer-bar gauge or length would come from non-thermal factors outside this problem's scope — rougher-stand draft schedule, reheat-furnace hearth length, coiler mandrel/coil-weight rating, or crane and handling limits — none of which are given here.
5. Justifying the four neglected mechanisms.
   • Temperature gradients across the slab (lumped-capacitance assumption). A rough Biot-number check, $Bi=h_{rad}(\delta/2)/k_{steel}$ with a linearized radiative coefficient $h_{rad}\approx4\varepsilon\sigma T^3\approx4(0.8)(5.67\times10^{-8})(1500)^3\approx 610$ W/m$^2$K and $k_{steel}\approx30$ W/m·K, gives $Bi\approx610(0.016)/30\approx0.33$ — small enough (well under 1) that a single representative temperature through the 32 mm thickness is a reasonable exam-level approximation, though not a negligible correction in a rigorous design calculation.
   • Natural convection from the slab surfaces. Free-convection coefficients in air are typically $h_{conv}\sim10$–$30$ W/m$^2$K, versus the effective radiative coefficient of order 600 W/m$^2$K estimated above at $\sim$1500 K (radiation scales as $T^4$, so it dominates overwhelmingly at these temperatures) — convection is at most a few percent correction and is safely dropped.
   • Conduction into the support rollers. Contact between the moving bar and each roller is both brief (rolling contact, not stationary) and confined to a small fraction of the bar's total surface area at any instant, compared with the large, continuously-radiating top and bottom faces — the conductive heat sink is small in both area and duration relative to the radiative loss.
   • Back-radiation from the plant. The net radiative exchange is $\propto(T_{slab}^4-T_{surr}^4)$; with $T_{surr}\sim300$ K versus $T_{slab}\sim1500$ K, $T_{surr}^4/T_{slab}^4\sim(300/1500)^4=0.0016$, under 0.2% of the slab's own emissive term — negligible.
6. 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 or length (mass is conserved from transfer bar through to coiled strip): $$\dot m=\rho\,w\,\delta_{strip}\,V_{coil}=(7450)(1.21)(0.0023)(20)=\boxed{414.7\ \text{kg/s}}$$
7. Annual capacity (24 h/day, 365 d/yr, no shutdowns). $$\text{Annual tonnage}=\dot m\times(3600\times24\times365)=414.7\times3.156\times10^7\ \text{s}=\boxed{1.31\times10^7\ \text{tonnes/yr}\ (\approx13.1\ \text{Mt/yr})}$$