21-Mat-A2 Materials Transport Phenomena · December 2015
Question 2 of 8: Continuous-Caster Tundish Grade Change (PFR + CSTR in Series)
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).
Find. The elapsed time, measured from the moment the new (40 ppm) heat begins entering the tundish, for the tundish outlet carbon content to fall to within 10% of the 40 ppm target (i.e. $C(t)\le 44$ ppm).
Fig. 2 — combined-model tundish: plug-flow zone in series with a well-mixed zone.
Approach. Model the tundish, per the given volume split, as a plug-flow segment feeding a well-mixed (CSTR) segment in series; the PFR delays the step change by a pure time-lag $\tau_p$, after which the CSTR relaxes exponentially toward the new feed composition with time constant $\tau_m$.
Mean residence times. Total residence time $\tau=V/Q=10/1=10$ min. Split by volume fraction:
$$\tau_p=f_p\tau=0.20(10)=2\ \text{min},\qquad \tau_m=f_m\tau=0.80(10)=\boxed{8\ \text{min}}$$
Behaviour for $t<\tau_p$. The plug-flow segment is still discharging steel that entered before the switch, so the CSTR's feed — and hence the tundish outlet — remains at the old composition: $C(t)=700$ ppm for $0\le t<\tau_p=2$ min.
CSTR step response for $t\ge\tau_p$. At $t=\tau_p$ the plug-flow outlet composition switches instantly (ideal step, by the PFR's own definition) to $C_{new}=40$ ppm, becoming the CSTR's feed. The CSTR, which has been at steady state $C_0=700$ ppm throughout $0\le t<\tau_p$, then obeys the first-order mixing equation $V_m\,dC/dt=Q(C_{new}-C)$, giving
$$C(t)=C_{new}+(C_0-C_{new})\exp\!\left(-\dfrac{t-\tau_p}{\tau_m}\right),\qquad t\ge\tau_p$$
Solve for the 10%-of-spec time. Target $C(t)=1.10\,C_{new}=1.10(40)=44$ ppm:
$$44=40+(700-40)\exp\!\left(-\dfrac{t-2}{8}\right)\ \Rightarrow\ \exp\!\left(-\dfrac{t-2}{8}\right)=\dfrac{4}{660}=6.06\times10^{-3}$$
$$-\dfrac{t-2}{8}=\ln(6.06\times10^{-3})=-5.106\ \Rightarrow\ t-2=40.8\ \text{min}$$
$$\boxed{t\approx 42.8\ \text{min}\ (\text{about }0.71\ \text{h})\ \text{after the new heat starts entering the tundish}}$$
Quantity
Value
Plug-flow lag, $\tau_p$
2 min
Well-mixed time constant, $\tau_m$
8 min
Time to reach 44 ppm (10% of the 40 ppm spec)
≈ 42.8 min
Check
“Within 10% of final spec (of 40 ppm)” is read here as within 10% of the target value, i.e. $C(t)\le44$ ppm — not 10% of the total 660 ppm composition swing. The alternative (90%-of-the-way-through-the-transient) reading would instead require $C(t)-40=0.10(660)=66$, i.e. $C=106$ ppm, reached after only $t\approx2+8\ln(660/66)=2+18.4\approx20.4$ min — roughly half the answer above. State the interpretation used, since both are defensible from the wording.
