Methanol Synthesis Loop with Total Recycle: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2018, Q3

Question A3: Methanol Synthesis Loop with Total Recycle (Part A — equal value)

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. Reaction $CO+2H_2\rightarrow CH_3OH$; single-pass CO conversion 15%; reactor feed ratio $H_2:CO=2:1$; unreacted gas fully recycled (no purge); fresh feed at $35\,{}^\circ\text{C}$, 300 atm; methanol product 6600 kg/hr ($M=32.04$).

Find. (a) the volumetric flow of the fresh feed gas; (b) the recycle ratio (recycle / fresh feed, molar).

Figure A3 — Fresh CO+$H_2$ (1:2) plus recycle enter the reactor at 15% single-pass CO conversion; methanol is condensed out and the unreacted gas is recycled entirely. With no purge, the fresh feed must equal the reaction’s stoichiometric demand.

Approach. Because there is no purge, an overall balance forces 100% net conversion, so the fresh feed is exactly the reaction stoichiometry for the methanol made (part a, via the ideal-gas law); the recycle then follows from the 15% single-pass constraint applied to the reactor (part b).

Methanol production rate. $$\dot n_{CH_3OH}=\frac{6600}{32.04}=206.0\ \text{kmol/hr}.$$
Fresh feed (overall balance, no purge). With the only exit being liquid methanol, an overall carbon/hydrogen balance requires the fresh feed to supply exactly the stoichiometric CO and $H_2$: $$\dot n_{CO,\text{fresh}}=206.0,\quad \dot n_{H_2,\text{fresh}}=2(206.0)=412.0,\quad \dot n_{\text{fresh}}=618.0\ \text{kmol/hr}.$$
Volume of fresh feed (part a, ideal gas at 35 °C, 300 atm). $$\dot V=\frac{\dot n RT}{P}=\frac{(617{,}980\ \text{mol/hr})(8.314)(308.15)}{300\times101{,}325}=\boxed{52.1\ \text{m}^3/\text{hr}}.$$
Reactor throughput from single-pass conversion. Only 15% of the CO fed to the reactor reacts, and all reacted CO becomes methanol, so $$\dot n_{CO,\text{reactor}}=\frac{\dot n_{CH_3OH}}{0.15}=\frac{206.0}{0.15}=1373.3\ \text{kmol/hr},$$ and since the reactor feed keeps $H_2:CO=2:1$, the total reactor feed is $\dot n_{\text{reactor}}=3\,\dot n_{CO,\text{reactor}}=4119.9$ kmol/hr.
Recycle ratio (part b). The recycle is the reactor feed minus the fresh feed: $$\dot n_{\text{recycle}}=4119.9-618.0=3501.9\ \text{kmol/hr},\qquad RR=\frac{\dot n_{\text{recycle}}}{\dot n_{\text{fresh}}}=\frac{3501.9}{618.0}=\boxed{5.67}.$$

Quantity	Result
Methanol produced	206.0 kmol/hr
Fresh feed (CO + $H_2$)	618.0 kmol/hr (206.0 + 412.0)
(a) Fresh-feed volume	52.1 m³/hr (35 °C, 300 atm)
Reactor throughput	4119.9 kmol/hr
(b) Recycle ratio	5.67

Concept & Theory

Topic. Recycle reactor with no purge; overall vs. single-pass conversion.

Key relations. Overall conversion $\to100\%$ when unreacted reactant is fully recycled and only product leaves; single-pass conversion $X=\dot n_{\text{reacted}}/\dot n_{\text{fed to reactor}}$; ideal gas $\dot V=\dot n RT/P$; recycle ratio $RR=\dot n_{\text{recycle}}/\dot n_{\text{fresh}}$.

Why this works. A control volume around the whole loop sees only fresh feed in and methanol out, so at steady state the fresh feed must equal the stoichiometric reactant consumption — independent of how little converts per pass. The low single-pass conversion is then absorbed entirely by a large internal recycle, which the reactor-level balance quantifies.

Common pitfall. Sizing the fresh feed from the 15% single-pass conversion (that governs the recycle, not the make-up), or mixing bases — the fresh-feed volume must be evaluated at the stated 35 °C/300 atm, not at STP.

Source. Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes §4.7 (recycle and purge; overall vs. single-pass conversion).

Question 3 of 6: Methanol Synthesis Loop with Total Recycle

Part A — Process Balances

Question A3: Methanol Synthesis Loop with Total Recycle (Part A — equal value)