13.4 Plug Flow Reactors (PFR) and Reactor Combinations

13.4 Plug Flow Reactors (PFR) and Reactor Combinations

Unlike a continuous stirred-tank reactor, where fluid is instantaneously backmixed, an Ideal Plug Flow Reactor (PFR)—also termed a tubular reactor or piston-flow reactor—operates with zero axial mixing. Fluid elements pass through the vessel in an orderly, plug-like manner, each having an identical residence time.

On the NCEES PE Chemical Exam, tubular reactor design requires integrating rate equations along the reactor length, properly accounting for gas-phase expansion or contraction ($\varepsilon$), comparing PFR versus CSTR volume requirements on Levenspiel plots, and selecting optimal configurations (such as CSTR-PFR combinations or recycle loops) for specialized reaction kinetics.

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1. Ideal Plug Flow Reactor (PFR) Design Equations

Consider an ideal tubular reactor with fluid flowing continuously at steady state in the axial direction $z$.

                   Ideal Plug Flow Reactor (PFR)
                   -----------------------------
      Feed In                                                   Product Out
    (F_A0, v_0)   +---------------------------------------+    (F_A, v, X)
    ------------->| ===>   ===>  [ dV ]  ===>   ===>   ==>|---------------->
                  +---------------------------------------+    
                  0                   V                  V_total
                         Differential Slice dV
                         Molar Flow: F_A ---> F_A + dF_A
                         Conversion: X   ---> X + dX

The Ideal PFR Assumptions:

1. Zero Axial Dispersion ($D_z = 0$): There is no backmixing or diffusion of fluid elements in the axial direction of flow. Fluid elements do not overtake or mix with elements ahead or behind them.
2. Complete Radial Homogeneity ($D_r \to \infty$): Fluid velocity, temperature, and species concentrations are perfectly uniform across any cross-sectional slice perpendicular to flow.
3. Uniform Residence Time: Every fluid packet spends the exact same duration traversing the reactor from inlet to outlet.

Derivation of the PFR Mole Balance

Performing a steady-state mole balance on species $A$ across a differential volume element $\Delta V$:

$\left. F_A \right|_V - \left. F_A \right|_{V + \Delta V} + r_A \Delta V = 0$

Dividing by $\Delta V$ and taking the limit as $\Delta V \to 0$:

$-\frac{dF_A}{dV} = -r_A$

Expressing molar flow in terms of conversion ($F_A = F_{A0}(1 - X) \implies dF_A = -F_{A0} dX$):

$F_{A0} \frac{dX}{dV} = -r_A \implies \frac{dX}{dV} = \frac{-r_A}{F_{A0}}$

Separating variables and integrating across the reactor from $V = 0$ ($X = 0$) to total volume $V$ ($X$):

$V = F_{A0} \int_0^X \frac{dX}{-r_A}$

For constant volumetric flow rate ($v = v_0$, constant fluid density), noting $F_{A0} = v_0 C_{A0}$:

$\tau = \frac{V}{v_0} = C_{A0} \int_0^X \frac{dX}{-r_A} = -\int_{C_{A0}}^{C_A} \frac{dC_A}{-r_A}$

[!NOTE] Equivalence of Constant-Density PFR and Batch Reactor:
For constant-density systems, the space time equation for an ideal PFR ($\tau_{PFR} = C_{A0} \int_0^X \frac{dX}{-r_A}$) is mathematically identical to the reaction time equation of an ideal batch reactor ($t_{batch} = C_{A0} \int_0^X \frac{dX}{-r_A}$). An ideal PFR behaves physically like a continuous chain of small batch reactors moving down a pipe.

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2. Analytical Solutions for Constant-Density Tubular Reactors

When density is constant (liquid-phase reactions or gas reactions with no change in moles, temperature, or pressure):

1. Zero-Order Reaction ($-r_A = k$): $\tau = C_{A0} \int_0^X \frac{dX}{k} = \frac{C_{A0} X}{k} \implies X = \frac{k \tau}{C_{A0}}$

2. First-Order Reaction ($-r_A = k C_A = k C_{A0}(1 - X)$): $\tau = C_{A0} \int_0^X \frac{dX}{k C_{A0} (1 - X)} = \frac{1}{k} \ln\left( \frac{1}{1 - X} \right)$ Solving explicitly for conversion $X$ and exit concentration $C_A$: $X = 1 - e^{-k \tau} = 1 - e^{-Da}$ $C_A = C_{A0} e^{-k \tau} = C_{A0} e^{-Da}$

3. Second-Order Reaction ($2A \to \text{Products}$, $-r_A = k C_A^2 = k C_{A0}^2(1 - X)^2$): $\tau = C_{A0} \int_0^X \frac{dX}{k C_{A0}^2 (1 - X)^2} = \frac{1}{k C_{A0}} \left( \frac{X}{1 - X} \right)$ $X = \frac{k C_{A0} \tau}{1 + k C_{A0} \tau} = \frac{Da}{1 + Da}$

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3. Variable-Density Gas-Phase Systems and Expansion Factor ($\varepsilon$)

In gas-phase reactions where the total number of moles changes (e.g., $A \to 2B$), the volumetric flow rate changes continuously as the reaction proceeds down the tube. Sizing the reactor assuming constant volumetric flow introduces massive design errors.

Equation of State and Volumetric Flow Rate

For ideal gases, volumetric flow rate $v$ at any point in the reactor is related to molar flow by $v = F_T \left(\frac{R T}{P}\right)$:

$v = v_0 (1 + \varepsilon X) \left( \frac{P_0}{P} \right) \left( \frac{T}{T_0} \right)$

Where:

• $v_0$ = entering volumetric flow rate.
• $P_0, T_0$ = entering pressure and temperature.
• $P, T$ = local pressure and temperature.
• $\varepsilon$ = fractional change in total volume between zero conversion and complete conversion ($X = 1.0$).

Calculation of the Expansion Factor ($\varepsilon$)

For a reaction $a A + b B \longrightarrow c C + d D$, the stoichiometric change in moles per mole of $A$ reacted is:

$\delta = \frac{d + c - b - a}{a}$

If the feed contains mole fraction $y_{A0}$ of reactant $A$ (with the balance being inerts or non-limiting reactants):

$\varepsilon = y_{A0} \cdot \delta = y_{A0} \left( \frac{d + c - b - a}{a} \right) = \frac{V_{X=1} - V_{X=0}}{V_{X=0}}$

Reactant Concentration Under Variable Volume

Under isothermal ($T = T_0$) and isobaric ($P = P_0$) conditions:

$C_A = \frac{F_A}{v} = \frac{F_{A0} (1 - X)}{v_0 (1 + \varepsilon X)} = C_{A0} \left( \frac{1 - X}{1 + \varepsilon X} \right)$

Design Equation for First-Order Gas Reaction with Expansion

Substituting $C_A$ into the PFR design integral:

$V = F_{A0} \int_0^X \frac{dX}{k C_A} = \frac{F_{A0}}{k C_{A0}} \int_0^X \left( \frac{1 + \varepsilon X}{1 - X} \right) dX = \frac{v_0}{k} \int_0^X \left( \frac{1 + \varepsilon X}{1 - X} \right) dX$

Using the algebraic identity $\frac{1 + \varepsilon X}{1 - X} = \frac{(1 + \varepsilon) - \varepsilon(1 - X)}{1 - X} = \frac{1 + \varepsilon}{1 - X} - \varepsilon$:

$V = \frac{v_0}{k} \left[ (1 + \varepsilon) \ln\left( \frac{1}{1 - X} \right) - \varepsilon X \right]$

   Physical Impact of Expansion Factor (epsilon):
   ---------------------------------------------------------------------------
   epsilon > 0 (Expansion, e.g. A -> 2B):
     Volume increases with X -> Gas velocity increases -> Residence time drops
     Requires LARGER reactor volume than constant-density prediction.

   epsilon < 0 (Contraction, e.g. 2A -> B):
     Volume decreases with X -> Gas velocity slows -> Residence time rises
     Requires SMALLER reactor volume than constant-density prediction.

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4. Performance Comparison: CSTR vs. PFR and Levenspiel Plot Analysis

A cornerstone of reaction engineering is comparing the volume required for a CSTR versus a PFR to achieve the exact same conversion under identical feed and temperature conditions.

           Levenspiel Plot: (1 / -r_A) versus Conversion (X)

   1 / (-r_A)
     ^
     |                     / (1 / -r_A) Curve (for normal order n > 0)
     |                    / 
     |                   *-------------------+
     |                  /|                   |  CSTR Volume = F_A0 * [X / (-r_A)_exit]
     |                 / |                   |  (Area of full RECTANGLE)
     |                /  |   PFR Volume =    | 
     |               /   |   F_A0 * INT(dX/-r_A)
     |              /    |   (Area UNDER     |
     |             /     |    the curve)     |
     +------------*------+-------------------+-----> Conversion (X)
     0           X_exit                      X_exit

Graphical Analysis on a Levenspiel Plot:

• PFR Volume: Equals $F_{A0}$ multiplied by the area under the curve of $1/(-r_A)$ from $0$ to $X$.
• CSTR Volume: Equals $F_{A0}$ multiplied by the area of the rectangle of height $1/(-r_A)_{exit}$ and width $X$.

For Normal Reaction Orders ($n > 0$):

For any reaction where rate increases with reactant concentration (all orders $n > 0$), the reaction rate is highest at the inlet and decreases continuously as conversion rises.

• In a PFR, the fluid experiences high rates throughout the beginning and middle of the reactor.
• In a CSTR, the entire reactor operates at the lowest rate, $(-r_A)_{exit}$.
• Therefore, for all positive reaction orders:

$V_{CSTR} > V_{PFR} \quad (\text{for } n > 0)$

The ratio $V_{CSTR} / V_{PFR}$ expands rapidly as conversion increases:

At $99%$ conversion for a second-order reaction, a single CSTR must be 100 times larger than an equivalent PFR!

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5. Autocatalytic Reactions and Optimal Reactor Sequencing

While $V_{CSTR} > V_{PFR}$ holds for normal kinetics, an important exception occurs for autocatalytic reactions, where a reaction product acts as a catalyst:

$A + P \longrightarrow P + P$ $-r_A = k C_A C_P = k C_{A0}^2 (1 - X) (\theta_P + X)$

Where $\theta_P = C_{P0} / C_{A0}$ is the ratio of product to reactant in the entering feed.

   Autocatalytic Levenspiel Curve & Optimal Sequencing:

   1 / (-r_A)
     ^
     |  \ (High 1/-r_A at inlet due to low CP0)
     |   \ 
     |    \       CSTR Area        PFR Area
     |     \     [Rectangle]     [Under Curve]
     |      \+----------------+  / 
     |       |                | / 
     |       |  CSTR operates |/ 
     |       +----------------* (MINIMUM: Maximum Rate)
     |                        |\ 
     |                        | \ 
     +------------------------+--+------------------> X
     0                       X_opt               X_final

The Autocatalytic Rate Profile:

• At $X = 0$, $C_P$ is near zero, so the rate is extremely slow ($1/(-r_A)$ is very large).
• As product $P$ forms, the rate accelerates, reaching a maximum where $\frac{d(-r_A)}{dX} = 0$ (at $X_{opt} = \frac{1 - \theta_P}{2} \approx 0.50$). At this point, $1/(-r_A)$ reaches a minimum.
• Beyond $X_{opt}$, reactant $A$ is depleted, causing the rate to decline and $1/(-r_A)$ to rise again.

Optimal Reactor Configuration:

1. First Stage (CSTR): From $X = 0$ to $X_{opt}$, use a CSTR operating at the point of maximum reaction rate. On the Levenspiel plot, the CSTR rectangle operating at the minimum of $1/(-r_A)$ is far smaller than the huge area under the curve of a PFR.
2. Second Stage (PFR): From $X_{opt}$ to final high conversion $X_{final}$, use a PFR. In this region, rate decreases with conversion, so the area under the curve is much smaller than a second CSTR rectangle.

[!TIP] Rule for Optimal Sequencing:
Whenever the Levenspiel curve ($1/(-r_A)$ vs. $X$) has a negative slope ($d(-r_A)/dX > 0$, rate increases with conversion), a CSTR is superior. Whenever the curve has a positive slope ($d(-r_A)/dX < 0$, rate decreases with conversion), a PFR is superior.

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6. Recycle Plug Flow Reactors

A Recycle PFR recycles a portion of the product stream leaving the tubular reactor back to the inlet.

                       Recycle Plug Flow Reactor
                       -------------------------
                      Recycle Stream (R * v_exit, X_exit)
                     +-----------------------------------+
                     |                                   |
       v_0, C_A0     v                                   |     v_exit, C_A,exit
       ------------>(+)---> [ PFR: Volume V ] ---------->+----> (1 - X_exit)
                    Mix     Inlet to PFR: X_in'          Split

The Recycle Ratio ($R$)

The recycle ratio ($R$) is defined as the volumetric flow rate returned to the reactor divided by the net volumetric flow rate leaving the system:

$R \equiv \frac{v_{recycle}}{v_{exit}}$

Conversion Entering the Reactor ($X_{in}'$)

Molecular mixing at the inlet tee combines fresh feed ($X = 0$) with recycle fluid ($X = X_{exit}$):

$X_{in}' = \frac{R}{R + 1} X_{exit}$

Design Equation for a Recycle PFR

Integrating the tubular reactor balance from $X_{in}'$ to $X_{exit}$:

$V = F_{A0} (R + 1) \int_{\frac{R}{R+1} X_{exit}}^{X_{exit}} \frac{dX}{-r_A}$

Limiting Behaviors of the Recycle Reactor:

• No Recycle ($R = 0$): $X_{in}' = 0$, and $(R + 1) = 1$. The equation collapses to an ideal PFR: $V = F_{A0} \int_0^{X_{exit}} \frac{dX}{-r_A}$
• Infinite Recycle ($R \to \infty$): The fluid recirculates so rapidly that concentration gradients vanish. The equation converges identically to an ideal CSTR: $\lim_{R \to \infty} V = \frac{F_{A0} X_{exit}}{(-r_A)_{exit}}$

Industrial Uses of Recycle Reactors:

1. Autocatalytic seeding: Returns product $P$ to the feed to eliminate the slow induction period.
2. Exothermic heat management: Recycled fluid dilutes the inlet reactant concentration and absorbs heat, moderating temperature spikes.
3. Catalytic boundary layer control: Maintains high superficial fluid velocities through fixed catalyst beds to minimize external film mass transfer resistance without requiring large external fresh feeds.

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7. Summary Comparison Table: PFR, CSTR, and Recycle Reactor Performance

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8. Step-by-Step Worked Numerical Example: Gas-Phase Cracking Reactor Design with Expansion

Problem Statement

An industrial gaseous hydrocarbon $A$ is thermally cracked according to the stoichiometry:

$A_{(g)} \longrightarrow 2 B_{(g)} + C_{(g)}$

The reaction is elementary first order with rate constant $k = 0.250\text{ s}^{-1}$ at $700\text{ K}$ and $4.00\text{ atm}$. The feed consists of pure $A$ ($y_{A0} = 1.00$) entering at $F_{A0} = 40.0\text{ mol/s}$. The target conversion is $80.0%$ ($X = 0.800$). The reactor operates isothermally and isobarically.

Calculate:

1. The expansion factor $\varepsilon$.
2. The entering volumetric flow rate $v_0$ in $\text{m}^3\text{/s}$ and entering concentration $C_{A0}$ in $\text{mol/m}^3$.
3. The required PFR volume ($V_{PFR}$) accounting for gas expansion.
4. The PFR volume that would be calculated if the engineer erroneously assumed constant density ($\varepsilon = 0$), and the resulting percentage under-sizing error.
5. The required volume for a single CSTR ($V_{CSTR}$) under identical operating conditions, and the volume ratio $V_{CSTR} / V_{PFR}$.

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Step 1: Calculate the Expansion Factor ($\varepsilon$)

Stoichiometry: $1 A \to 2 B + 1 C$.

$\delta = \frac{\sum \nu_{products} - \sum \nu_{reactants}}{\nu_A} = \frac{(2 + 1) - 1}{1} = \frac{3 - 1}{1} = +2.00$

Because the feed is pure $A$ ($y_{A0} = 1.00$):

$\varepsilon = y_{A0} \cdot \delta = 1.00 \times (+2.00) = \mathbf{+2.00}$

Every mole of $A$ converted produces three moles of product gas, resulting in a $200%$ increase in total gas volume at complete conversion.

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Step 2: Calculate Entering Concentration and Volumetric Flow Rate

Using the ideal gas law with $R = 8.2057\times 10^{-5}\text{ m}^3\cdot\text{atm}/(\text{mol}\cdot\text{K})$:

$C_{A0} = \frac{P_0}{R T_0} = \frac{4.00\text{ atm}}{(8.2057\times 10^{-5}\text{ m}^3\cdot\text{atm}/(\text{mol}\cdot\text{K})) \times 700\text{ K}} = \frac{4.00}{0.057440} = \mathbf{69.638\text{ mol/m}^3} \quad (0.06964\text{ mol/L})$ $v_0 = \frac{F_{A0}}{C_{A0}} = \frac{40.0\text{ mol/s}}{69.638\text{ mol/m}^3} = \mathbf{0.57440\text{ m}^3\text{/s}} \quad (574.4\text{ L/s})$

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Step 3: Calculate PFR Volume with Gas Expansion

Applying the analytical first-order variable-volume PFR equation:

$V_{PFR} = \frac{v_0}{k} \left[ (1 + \varepsilon) \ln\left( \frac{1}{1 - X} \right) - \varepsilon X \right]$

Evaluate each term for $\varepsilon = 2.00$ and $X = 0.800$:

$1 + \varepsilon = 1 + 2.00 = 3.00$ $\ln\left( \frac{1}{1 - 0.800} \right) = \ln(5.00) = 1.60944$ $(1 + \varepsilon) \ln\left( \frac{1}{1 - X} \right) = 3.00 \times 1.60944 = 4.82831$ $\varepsilon X = 2.00 \times 0.800 = 1.60000$ $\text{Integral Bracket} = 4.82831 - 1.60000 = 3.22831$

Now calculate volume:

$V_{PFR} = \frac{0.57440\text{ m}^3\text{/s}}{0.250\text{ s}^{-1}} \times 3.22831 = 2.2976\text{ m}^3 \times 3.22831 = \mathbf{7.417\text{ m}^3} \quad (7,\!417\text{ L})$

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Step 4: Compare with Constant-Density PFR Sizing

If expansion were ignored ($\varepsilon = 0$):

$V_{\varepsilon=0} = \frac{v_0}{k} \ln\left( \frac{1}{1 - X} \right) = \frac{0.57440}{0.250} \times 1.60944 = 2.2976 \times 1.60944 = \mathbf{3.698\text{ m}^3}$

Under-sizing error:

$\text{Error} = \frac{V_{PFR} - V_{\varepsilon=0}}{V_{PFR}} \times 100\% = \frac{7.417 - 3.698}{7.417} \times 100\% = \mathbf{50.1\%}$

Neglecting gas expansion cuts the required reactor volume in half! The actual reactor would achieve far less than $80%$ conversion because the expanding gas accelerates through the tube, drastically cutting residence time.

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Step 5: Calculate Single CSTR Volume and Volume Ratio

For a CSTR with variable volume, the exit concentration is:

$C_{A,exit} = C_{A0} \left( \frac{1 - X}{1 + \varepsilon X} \right) = 69.638 \times \left( \frac{1 - 0.800}{1 + 2.00(0.800)} \right) = 69.638 \times \left( \frac{0.200}{2.600} \right) = 5.3568\text{ mol/m}^3$ $(-r_A)_{exit} = k C_{A,exit} = 0.250\text{ s}^{-1} \times 5.3568\text{ mol/m}^3 = 1.3392\text{ mol}/(\text{m}^3\cdot\text{s})$

CSTR volume:

$V_{CSTR} = \frac{F_{A0} X}{(-r_A)_{exit}} = \frac{40.0\text{ mol/s} \times 0.800}{1.3392\text{ mol}/(\text{m}^3\cdot\text{s})} = \frac{32.0}{1.3392} = \mathbf{23.895\text{ m}^3} \quad (23,\!895\text{ L})$

Volume comparison ratio:

$\frac{V_{CSTR}}{V_{PFR}} = \frac{23.895\text{ m}^3}{7.417\text{ m}^3} = \mathbf{3.22}$

The CSTR must be $3.22$ times larger than the PFR to deliver the same conversion.

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9. Critical PE Exam Traps & Pitfalls

Trap 1: Omitting Inerts when Computing $\varepsilon$
Remember $\varepsilon = y_{A0} \delta$. If feed contains $40%$ reactant $A$ and $60%$ nitrogen diluent, and $\delta = +2$, then $\varepsilon = 0.40 \times 2 = +0.80$, NOT $+2.0$. Inerts do not react, so they dampen the overall fractional expansion of the gas stream.

Trap 2: Using the Constant-Density Integral for Gas Reactions
On tubular gas reactor problems, immediately check if moles change: $\sum \nu_{products} \neq \sum \nu_{reactants}$. If moles change, you cannot use $\tau = (1/k)\ln[1/(1-X)]$. You must use the integrated form containing $(1 + \varepsilon) \ln[1/(1-X)] - \varepsilon X$.

Trap 3: Claiming PFR is Always Smaller Than CSTR
While true for orders $n > 0$, this is false for autocatalytic kinetics ($A + P \to 2P$) or reactions with product inhibition. At low conversions, $1/(-r_A)$ decreases with conversion, so a CSTR requires less volume than a PFR. Be alert for autocatalytic kinetics on the PE exam!

Trap 4: Inverting the Recycle Inlet Conversion Formula
In recycle reactors, the mixed feed conversion entering the reactor is $X_{in}' = \frac{R}{R+1} X_{exit}$. A common algebra error is writing $X_{in}' = \frac{1}{R+1} X_{exit}$. When $R = 0$ (no recycle), $X_{in}'$ must equal $0$.