1.3 Recycle, Bypass, and Purge Operations

1.3 Recycle, Bypass, and Purge Operations

In industrial chemical manufacture, single-pass reactor conversions are frequently constrained by thermodynamic equilibrium, catalyst selectivity, or thermal runaway limits. To achieve economic viability, chemical plants route unreacted materials back to the reactor via recycle loops, bypass streams around equipment for fine dynamic control, and bleed purge streams to prevent inert accumulation. On the NCEES PE Chemical Exam, problems involving looped flowsheets are among the most heavily weighted calculation questions. Mastering boundary definitions and the relationships connecting overall conversion, single-pass conversion, and purge rates is essential.

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Engineering Roles of Flowsheet Loops

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Single-Pass Conversion vs. Overall Conversion

A critical distinction tested on the PE exam is the mathematical difference between conversion across the reactor vessel alone versus conversion across the entire manufacturing plant envelope:

1. Overall System Conversion ($X_{\text{ov}}$)

Evaluated across the Overall Plant Control Envelope (considering only fresh feed entering and net product/waste exiting):

$X_{\text{ov}} = \frac{\dot{n}_{A, \text{fresh feed}} - \dot{n}_{A, \text{net loss}}}{\dot{n}_{A, \text{fresh feed}}}$

If unreacted reactant is completely separated from the product stream and recycled with zero purge losses, the overall conversion is 100%, even if the reactor achieves only a $10%$ conversion per pass!

2. Single-Pass Reactor Conversion ($X_{\text{sp}}$)

Evaluated across the Reactor Control Envelope (considering the mixed feed entering the reactor and the gross effluent exiting):

$X_{\text{sp}} = \frac{\dot{n}_{A, \text{reactor feed}} - \dot{n}_{A, \text{reactor effluent}}}{\dot{n}_{A, \text{reactor feed}}}$

Mathematical Linking Equation

Let $F_0$ be the fresh feed of reactant $A$, $R_A$ be the recycle molar rate of $A$, and $F_1 = F_0 + R_A$ be the total reactant entering the reactor. Assuming all product is separated and only $A$ is recycled without purge:

$\text{Reactant } A \text{ reacted in reactor} = X_{\text{sp}} \cdot F_1 = X_{\text{sp}} (F_0 + R_A)$

At steady state, the reactant consumed in the reactor must identically equal the net reactant consumed across the entire plant ($X_{\text{ov}} \cdot F_0$):

$X_{\text{sp}} (F_0 + R_A) = X_{\text{ov}} \cdot F_0$

Solving for the Recycle Ratio of reactant $A$ ($R_A / F_0$):

$\frac{R_A}{F_0} = \frac{X_{\text{ov}} - X_{\text{sp}}}{X_{\text{sp}}}$

If $X_{\text{ov}} = 1.00$ ($100%$ overall conversion):

$\frac{R_A}{F_0} = \frac{1 - X_{\text{sp}}}{X_{\text{sp}}} = \frac{1}{X_{\text{sp}}} - 1$

Exam Rule: If single-pass conversion is $X_{\text{sp}} = 0.20$ (20%), the recycle ratio must be $(1 - 0.20) / 0.20 = 4.0$. Four moles of reactant must be recycled for every one mole of fresh feed introduced.

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The Cardinal Purge Rule & Inert Accumulation

When a fresh feed contains even a minute concentration of an inert species (such as argon in air, or methane in synthesis gas), closing the recycle loop without a bleed stream violates steady-state physics.

$\frac{dM_{\text{inert}}}{dt} = \dot{m}_{\text{inert, in}} - 0 > 0$

Inert continuously accumulates within the loop, increasing system pressure, blanketing catalyst active sites, and eventually forcing an emergency shutdown.

To establish steady state ($dM_{\text{inert}}/dt = 0$), a continuous purge stream must be vented from the recycle loop. Because the purge stream is split from the recycle stream, its intensive composition is identical to the recycle stream:

$x_{\text{inert, purge}} = x_{\text{inert, recycle}}$

[!IMPORTANT] The Cardinal Purge Rule: At steady state, the total mass/molar rate of inert entering in the fresh feed must identically equal the total mass/molar rate of inert exiting in the purge stream!

$\dot{n}_{\text{inert, fresh feed}} = \dot{n}_{\text{inert, purge}}$ $F_{\text{fresh}} \cdot z_{\text{inert, fresh}} = F_{\text{purge}} \cdot x_{\text{inert, purge}}$

Solving for the required continuous purge flow rate: $F_{\text{purge}} = \frac{F_{\text{fresh}} \cdot z_{\text{inert, fresh}}}{x_{\text{inert, purge}}}$

The Economic Trade-Off of Purge Streams

The purge stream cannot discriminate between inerts and unreacted feedstocks; it bleeds gas at the prevailing recycle loop composition. Consequently, valuable reactants are co-purged at a molar rate equal to $F_{\text{purge}} (1 - x_{\text{inert, purge}})$.

• Specifying a lower allowable inert level in the reactor loop requires a higher purge rate, which increases raw material loss and lowers overall conversion.
• Specifying a higher allowable inert level reduces feedstock loss, but requires a larger reactor volume, larger heat exchangers, and higher recycle gas compressor power.

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Bypass Operations & The Lever-Arm Rule

In bypass configurations, a portion of the feed stream ($B$) bypasses a processing unit (such as an evaporator, chiller, or chemical treater) and mixes with the unit effluent ($E_{\text{out}}$) to produce final product ($P$):

$F_{\text{fresh}} = E_{\text{in}} + B$ $P = E_{\text{out}} + B$

For binary concentration blending across the downstream mixing junction:

$B \cdot x_B + E_{\text{out}} \cdot x_E = P \cdot x_P$

Substituting $P = B + E_{\text{out}}$ yields the classic lever-arm rule:

$\frac{B}{E_{\text{out}}} = \frac{x_E - x_P}{x_P - x_B}$

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Comprehensive Worked Numerical Example: Methanol Synthesis Loop

Problem Statement

A continuous methanol synthesis plant is fed $1,000\text{ kmol/h}$ of fresh synthesis gas (Stream 0) containing:

• Carbon Monoxide ($\text{CO}$): $32.0\text{ mol%}$
• Hydrogen ($\text{H}_2$): $64.0\text{ mol%}$ (exact 2:1 stoichiometric ratio to $\text{CO}$)
• Methane ($\text{CH}_4$, inert): $4.0\text{ mol%}$

The overall gas-phase reaction is:

$\text{CO} + 2\text{H}_2 \rightleftharpoons \text{CH}_3\text{OH}$

Process Specifications:

1. Fresh feed combines with a recycle stream (Stream 5) to form the mixed reactor feed (Stream 1).
2. The catalytic reactor operates with a single-pass conversion of carbon monoxide of $X_{\text{sp}} = 0.200$ ($20.0%$).
3. Reactor effluent (Stream 2) enters a condenser-separator where $100%$ of the synthesized methanol is condensed and recovered as pure liquid product (Stream 3).
4. The non-condensable gas leaving the separator (Stream 4) contains unreacted $\text{CO}$, $\text{H}_2$, and inert $\text{CH}_4$. This gas is divided into a recycle stream (Stream 5) and a continuous purge stream (Stream 6).
5. The purge stream flow rate is adjusted to maintain the concentration of methane in the recycle gas at exactly $20.0\text{ mol% }\text{CH}_4$.

Calculate:

1. The molar flow rate of the continuous purge stream (Stream 6, kmol/h).
2. The molar production rate of liquid methanol (Stream 3, kmol/h) and the overall plant conversion of $\text{CO}$.
3. The molar flow rate of the recycle gas (Stream 5, kmol/h) and the recycle ratio ($R / F_0$).
4. The molar flow rate and composition of the gas entering the reactor (Stream 1).

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Step 1: Overall Process Boundary Envelope

Draw a control volume around the entire plant. Crossing the boundary are:

• Input: Fresh Feed ($F_0 = 1,000\text{ kmol/h}$).
• Outputs: Pure liquid methanol ($N_3$), and Purge gas ($P_6$).

Apply the Cardinal Purge Rule to inert methane: $\text{Methane In with Fresh Feed} = F_0 \times z_{\text{CH}_4, 0} = 1,000 \times 0.040 = 40.0\text{ kmol/h}$

At steady state, all entering methane must leave in the purge stream: $\dot{n}_{\text{CH}_4, 6} = 40.0\text{ kmol/h}$

Given that the purge stream contains $20.0\text{ mol% }\text{CH}4$ ($y{\text{CH}_4, 6} = 0.200$):

$P_6 = \frac{\dot{n}_{\text{CH}_4, 6}}{y_{\text{CH}_4, 6}} = \frac{40.0\text{ kmol/h}}{0.200} = \mathbf{200.0\text{ kmol/h}}$

The required continuous purge rate is 200.0 kmol/h.

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Step 2: Purge Stream Composition & Methanol Production

Because $\text{CO}$ and $\text{H}_2$ enter in an exact 1:2 stoichiometric ratio ($320\text{ kmol/h CO}$ and $640\text{ kmol/h H}_2$) and react in an exact 1:2 ratio, they maintain a 1:2 ratio throughout the entire loop.

In the purge stream (total $200.0\text{ kmol/h}$):

• Methane: $40.0\text{ kmol/h}$ ($20.0\text{ mol%}$)
• Remaining syngas ($\text{CO} + \text{H}_2$): $200.0 - 40.0 = 160.0\text{ kmol/h}$
• $\text{CO}$ in purge: $\dot{n}{\text{CO}, 6} = \frac{160.0}{3} = \mathbf{53.33\text{ kmol/h}}$ ($y{\text{CO}, 6} = 0.2667$)
• $\text{H}2$ in purge: $\dot{n}{\text{H}2, 6} = \frac{2 \times 160.0}{3} = \mathbf{106.67\text{ kmol/h}}$ ($y{\text{H}_2, 6} = 0.5333$)

Now perform an overall Carbon balance to find liquid methanol production ($N_3$): $\text{Carbon In with Fresh Feed} = \dot{n}_{\text{CO}, 0} + \dot{n}_{\text{CH}_4, 0} = 320.0 + 40.0 = 360.0\text{ kmol C/h}$ $\text{Carbon Out in Purge} = \dot{n}_{\text{CO}, 6} + \dot{n}_{\text{CH}_4, 6} = 53.33 + 40.0 = 93.33\text{ kmol C/h}$ $\text{Methanol Produced } (N_3) = 360.0 - 93.33 = \mathbf{266.67\text{ kmol/h}}$

Overall conversion of carbon monoxide across the plant:

$X_{\text{ov}} = \frac{\dot{n}_{\text{CO}, 0} - \dot{n}_{\text{CO}, 6}}{\dot{n}_{\text{CO}, 0}} = \frac{320.0 - 53.33}{320.0} = \frac{266.67}{320.0} = \mathbf{0.8333\text{ (83.33\%)}}$

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Step 3: Reactor Feed & Recycle Stream Calculations

From the single-pass conversion specification ($X_{\text{sp}} = 0.200$): $\text{Methanol produced in reactor} = \text{CO reacted} = X_{\text{sp}} \times \dot{n}_{\text{CO}, 1}$ $266.67\text{ kmol/h} = 0.200 \times \dot{n}_{\text{CO}, 1} \implies \dot{n}_{\text{CO}, 1} = \frac{266.67}{0.200} = \mathbf{1,333.33\text{ kmol/h}}$

Perform a $\text{CO}$ balance around the fresh feed / recycle mixing junction: $\dot{n}_{\text{CO}, 1} = \dot{n}_{\text{CO}, 0} + \dot{n}_{\text{CO}, 5}$ $1,333.33 = 320.0 + \dot{n}_{\text{CO}, 5} \implies \dot{n}_{\text{CO}, 5} = 1,333.33 - 320.0 = \mathbf{1,013.33\text{ kmol/h}}$

Because the recycle stream has the exact same intensive composition as the purge stream ($y_{\text{CO}, 5} = 0.2667 = 4/15$):

$R_5 = \frac{\dot{n}_{\text{CO}, 5}}{y_{\text{CO}, 5}} = \frac{1,013.333}{4/15} = \mathbf{3,800.0\text{ kmol/h}}$

Recycle ratio relative to fresh feed: $\frac{R_5}{F_0} = \frac{3,800.0\text{ kmol/h}}{1,000.0\text{ kmol/h}} = \mathbf{3.80}$

Total non-condensable gas leaving condenser (Stream 4): $G_4 = R_5 + P_6 = 3,800.0 + 200.0 = 4,000.0\text{ kmol/h}$

• Recycle fraction of loop gas: $3,800 / 4,000 = 95.0%$
• Purge fraction of loop gas: $200 / 4,000 = 5.0%$

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Step 4: Flowsheet Stream Summary Table

Verification Check:

• Reactor feed: $F_0 + R_5 = 1,000.0 + 3,800.0 = 4,800.0\text{ kmol/h}$ (Matches Stream 1 exactly).
• Reactor effluent: In reaction $\text{CO} + 2\text{H}_2 \rightarrow \text{CH}_3\text{OH}$, $\Delta n = -2$ moles per mole of methanol formed. Moles contracted $= 2 \times 266.67 = 533.33\text{ kmol/h}$. Effluent $= 4,800.0 - 533.33 = 4,266.67\text{ kmol/h}$ (Matches Stream 2 exactly).
• Methane into reactor: $40.0\text{ (fresh)} + 760.0\text{ (recycle)} = 800.0\text{ kmol/h}$. Concentration entering reactor: $800 / 4,800 = 16.67\text{ mol%}$.