23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2017
Question 1 of 6: HDA Process — Purge/Recycle Balance for Benzene from Toluene
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
16-Chem-A1, May 2017 — three-hour, open-book examination. The paper is in two parts: Part A (Questions 1–3, process mass & energy balances) and Part B (Questions 4–6, chemical thermodynamics). Candidates answer two questions from each part; all six are worked below as a complete study resource.
Reference texts (this subject). Felder & Rousseau, Elementary Principles of Chemical Processes, 4th ed. (mass & energy balances, combustion, recycle/purge); Smith, Van Ness & Abbott, Introduction to Chemical Engineering Thermodynamics, 8th ed. (fugacity, activity coefficients, generalized correlations, equations of state); property data from the paper's appended Smith–Van Ness tables (App. B critical constants; App. E Lee/Kesler charts).
Question 1: HDA Process — Purge/Recycle Balance for Benzene from Toluene (Part A)
Find. (a) the purge fraction $\alpha$ as a function of the loop methane fraction $y$; (b) the reactor-effluent molar composition when $y = 0.4$.
Figure 1.1 — HDA flowsheet. The H₂-rich vapour (less the purge) and the recovered toluene both return to the reactor inlet as recycle streams; only the fresh feed, purge and net products cross the system boundary.
Approach. Fix the reaction turnovers from the benzene rate and selectivity, close the overall CH₄ and H₂ balances with the makeup impurity, and eliminate the makeup rate to express $\alpha$ in $y$.
Selectivity and toluene reacted. The consistent secondary reaction is benzene dimerisation $2\,\text{C}_6\text{H}_6 \rightleftharpoons \text{C}_{12}\text{H}_{10} + \text{H}_2$ (the equation printed for it in the exam is a transcription slip). Selectivity splits the reacted toluene between benzene and diphenyl:
$$S = 1 - \frac{0.0036}{(1-0.75)^{1.544}} = 1 - \frac{0.0036}{0.11756} = 0.9694, \qquad R_{\text{tol}} = \frac{265}{S} = 273.4\ \text{kmol/hr}$$
By-product and reactant turnovers. Each reacted toluene makes one CH₄; diphenyl consumes benzene two-for-one and releases H₂:
$$n_{\text{diphenyl}} = \tfrac12 R_{\text{tol}}(1-S) = 4.19, \qquad n_{\text{CH}_4}^{\text{gen}} = R_{\text{tol}} = 273.4, \qquad n_{\text{H}_2}^{\text{cons}} = R_{\text{tol}} - n_{\text{diphenyl}} = 269.2\ \text{kmol/hr}$$
Reactor feed and effluent hydrogen. Conversion sets the toluene fed; the 5:1 ratio sets the inlet H₂; the effluent H₂ (all of which reports to the vapour) is inlet less net consumption:
$$n_{\text{tol}}^{\text{fed}} = \frac{R_{\text{tol}}}{X} = 364.5,\quad n_{\text{H}_2}^{\text{in}} = 5\,n_{\text{tol}}^{\text{fed}} = 1822.5,\quad n_{\text{H}_2}^{\text{eff}} = 1822.5 - 269.2 = 1553.3\ \text{kmol/hr}$$
(a) The $\alpha$–$y$ relation. The vapour $V$ is H₂ + CH₄ with methane fraction $y$, so $n_{\text{H}_2}^{\text{eff}} = (1-y)V$. Writing the overall CH₄ balance (methane leaves only in the purge $\alpha V$; makeup carries CH₄ at $1/19$ of its H₂) and eliminating the makeup rate gives
$$\boxed{\;\alpha = \frac{n_{\text{CH}_4}^{\text{gen}} + n_{\text{H}_2}^{\text{cons}}/19}{n_{\text{H}_2}^{\text{eff}}\left[\dfrac{y}{1-y} - \dfrac{1}{19}\right]} = \frac{287.5}{1553.3\left[\dfrac{y}{1-y} - \dfrac{1}{19}\right]}\;}$$
The $1/19$ terms carry the 5 mol% CH₄ impurity ($0.05/0.95 = 1/19$); the purge fraction falls as $y$ rises, because tolerating more methane in the loop shrinks the purge needed to reject it.
(b) Composition at $y = 0.4$. Evaluate the relation and the vapour rate, then assemble the effluent (liquid components are the net benzene, unreacted toluene and diphenyl):
$$\alpha = \frac{287.5}{1553.3\,(0.6667 - 0.0526)} = 0.301, \qquad V = \frac{1553.3}{0.6} = 2588.8\ \text{kmol/hr}$$
