23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2013
Question 3 of 7: Reactor with Recycle and Purge (Inert Build-up)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2013 — 04-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Where property data are needed they are stated explicitly in each Given block. Units follow the paper (mixed SI and US customary).
Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material & energy balances; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — VLE, residual properties, reaction equilibrium; supporting data from Perry's Chemical Engineers' Handbook (9th ed.) and the NIST Chemistry WebBook.
Question 3: Reactor with Recycle and Purge (Inert Build-up) (equal value)
Given. Reaction A → B. Fresh feed $F$: 1.5 mol% inert. Combined reactor feed: 4.5 mol% inert. Recycle ratio $R/F=3.0$. Separator removes all B into product $P$; purge $S$ and recycle $R$ are a split of the same inert-bearing overhead, so they share one composition. Inert is chemically inert and leaves only in the purge.
Find. (a) $S/F$; (b) whether inert mole fraction is higher in $R$ or in the reactor effluent; (c) the rationale for purging the recycle loop rather than the effluent.
Figure 3 — Fresh feed mixes with recycle, reacts (A → B), the separator drops all B as product, and the inert-bearing overhead is split into recycle $R$ and purge $S$.
Approach. Because the purge and recycle share one composition, an inert balance around the mixing point fixes that composition, and an overall inert balance (in via feed = out via purge) then gives $S/F$. Parts (b) and (c) are qualitative consequences of the same inert bookkeeping.
Inert balance around the mixing point. Let $y_I$ be the inert mole fraction in the recycle (= purge). Take a basis $F=1$, so $R=3$. Inert into the combined reactor feed:
$$0.045\,(F+R) = 0.015\,F + y_I\,R \;\Rightarrow\; 0.045(4) = 0.015 + 3\,y_I.$$
Solving, $3\,y_I = 0.18-0.015 = 0.165$, so
$$y_I = 0.055 = \boxed{5.5\%\ \text{inert in the recycle/purge}}.$$
Overall inert balance $\Rightarrow$ purge rate. Inert enters only in $F$ and, since it neither reacts nor leaves in the (B + A) product, exits only in the purge:
$$0.015\,F = y_I\,S = 0.055\,S \;\Rightarrow\; \frac{S}{F} = \frac{0.015}{0.055} = \boxed{0.273}.$$
(b) Where is inert more concentrated? The reactor effluent contains A, B and inert; the separator then removes all of B. Taking B out shrinks the total moles while leaving the inert untouched, so the inert mole fraction rises. Hence inert is greater in the recycle stream than in the reactor effluent — the separation concentrates it.
(c) Why purge the recycle loop, not the effluent? The reactor effluent still holds all the valuable product B and carries inert at a lower concentration. Purging there would (i) throw away product B and (ii) demand a much larger purge to bleed the same amount of inert. Purging the post-separator overhead — where B is gone and inert is most concentrated — removes the required inert with the smallest loss of reactant A and zero loss of product. That is exactly why the purge is placed on the recycle loop.
