23-Chem-A4 Chemical Reactor Engineering · May 2015
Question 3 of 5: Triangular RTD — Dispersion and Tanks-in-Series Models
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2015 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; the designated Fogler textbook (any edition), unit-conversion / mathematical tables (CRC Handbook) and a non-communicating programmable calculator are permitted. Five questions are printed and any four constitute a complete paper (each worth 20 marks); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the batch / MFR / PFR design equations are quoted and applied. Property look-ups not printed on the paper (molar masses, the gas constant) are stated explicitly in each Given block as permitted open-book references.
Reference texts: H. S. Fogler, Elements of Chemical Reaction Engineering (4th/5th ed., Prentice Hall) — the designated open-book text: batch/CSTR/PFR design equations, the stoichiometric table with expansion factor $\varepsilon$ for gas-phase reactions with a change in moles, and reversible-reaction kinetics; O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — non-ideal flow (dead-zone / bypass models), the dispersion and tanks-in-series RTD models, and rate-equation determination from a differential (mixed) catalytic reactor; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 3: Triangular RTD — Dispersion and Tanks-in-Series Models (20 marks: a 10, b 10)
Given. First-order reaction; a PFR of the same mean residence time would reach $X=0.999$. Symmetrical triangular RTD, base $8\le t\le12$ min, so $a=4$ min and $\bar t=\tau=10$ min. Variance $\sigma^2=a^2/24$. Feed $C_{A0}=1000$ (arbitrary units).
Find. The reactor outlet concentration (and conversion) for the first-order reaction using (a) the axial-dispersion model and (b) the tanks-in-series model.
Figure 3 — The measured RTD is a symmetric triangle spanning $t=8$ to $12$ min, centred at the mean $\bar t=10$ min; its base $a=4$ min gives the variance $\sigma^2=a^2/24$ used to fit both non-ideal-flow models.
Approach. Convert the PFR performance to the rate group $k\tau$, characterise the spread of the RTD by its dimensionless variance, then apply the two standard first-order non-ideal models (small-dispersion and $N$ equal tanks) to find the outlet.
Rate group from the ideal-PFR performance. For a first-order PFR, $1-X=e^{-k\tau}$, so $k\tau=\ln\!\dfrac{1}{1-0.999}=\ln(1000)=6.908.$ This fixes the kinetics at the common mean residence time $\tau=\bar t=10$ min.
Spread of the RTD. The triangle’s variance is $\sigma^2=a^2/24=4^2/24=0.667$ min², so the dimensionless variance is $\sigma_\theta^2=\sigma^2/\bar t^2=0.667/100=0.006667.$ The distribution is narrow, so both non-ideal models predict only a slight shortfall from plug flow.
(a) Axial-dispersion model. For small dispersion, $\sigma_\theta^2=2(D/uL)$, so $D/uL=0.003333.$ Levenspiel’s first-order small-dispersion result gives$$\frac{C_A}{C_{A0}}=\exp\!\Big[-k\tau+(k\tau)^2\frac{D}{uL}\Big]=\exp[-6.908+6.908^2(0.003333)]=1.17\times10^{-3}.$$Hence $C_A=1000(1.172\times10^{-3})=\boxed{1.17}$ and $X=0.99883.$
(b) Tanks-in-series model. Match the variance with $N=1/\sigma_\theta^2=1/0.006667=150$ equal CSTRs. For $N$ tanks in series with a first-order reaction,$$\frac{C_A}{C_{A0}}=\Big(1+\frac{k\tau}{N}\Big)^{-N}=\Big(1+\frac{6.908}{150}\Big)^{-150}=1.167\times10^{-3},$$so $C_A=\boxed{1.17}$ and $X=0.99883$ — essentially identical to the dispersion model.
Compare with plug flow. Ideal plug flow would give $C_A=1000e^{-6.908}=1.0$ ($X=0.999$). The modest RTD spread raises the outlet to about 1.17, i.e. it costs only $\sim0.02$ percentage points of conversion — a narrow triangular RTD is very nearly plug-flow.
