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Plug Flow Reactor Volume (First-Order Reaction)

Sizes a plug flow reactor needed to achieve a target conversion for a first-order irreversible reaction at constant density.

Chemical EngineeringReaction EngineeringReactor Design

Plug Flow Reactor (PFR) CalculatorFirst‑Order Reaction

V = (FA0 / k) · ln(1 / (1 – X))
Select what to solve for — enter the other three values, then click Check
Solve for:
mol/s
s⁻¹
Conversion (X)
Low (<0.5) Moderate (0.5–0.8) High (0.8–0.95) Very High (>0.95)
V = (FA0 / k) · ln(1 / (1 – X)) · Typical conversion: 0–1 (0–100%)

Interpretation

PFR volume for first‑order: V = (F_A0 / k) · ln(1/(1−X)). Example: F_A0=10 mol/min, k=0.5 min⁻¹, X=0.8 → V ≈ 32.2 L.

V = (F_A0 / k) * ln(1 / (1 - X))
Plug Flow Reactor Volume (First-Order Reaction)

Variables

SymbolQuantityUnit
VReactor volumem3
F_A0Molar feed rate of reactant Amol/s
kFirst-order rate constant1/s
XTarget fractional conversion

What it means

For an isothermal plug flow reactor (PFR) with a first‑order reaction A → products, the design equation gives the reactor volume V required to achieve a conversion X. The equation is V = (F_A0 / k) ln(1/(1 − X)), where F_A0 is the inlet molar flow of A, k is the rate constant, and X is the conversion. This expression is derived from the mole balance and the rate law. It shows that as conversion approaches 1, the required volume goes to infinity because the reaction rate becomes very slow at low concentrations. In practice, PFRs are used for high‑throughput, continuous production. The equation is also applicable to other reaction orders with different integral forms. It is essential for sizing tubular reactors, which are common in the petrochemical and polymer industries. Understanding the design equation for PFRs is a core competency in chemical reaction engineering.

Worked example

PFR Volume – Two Examples

Real‑World
Scenario: k = 0.1 s⁻¹, F_A0 = 10 mol/s, X = 0.8. Find PFR volume V.
ParameterValue
F_A010 mol/s
k0.1 s⁻¹
X0.8
1V = (F_A0/k)×ln(1/(1-X)) = (10/0.1)×ln(1/0.2) = 100×1.609 ≈ 160.9 m³
Result V ≈ 161 m³ ✓ Large
Scenario: F_A0 = 5, k = 0.05, X = 0.9. Compute V.
ParameterValue
F_A05
k0.05
X0.9
1V = (5/0.05)×ln(1/0.1) = 100×2.303 = 230.3 m³
Result V ≈ 230 m³ ✓ Higher conversion
Key insight: PFR volume depends on feed, k, and conversion; higher X needs larger V.

Common mistakes

  • First‑order reaction: The rate law must be −r_A = k C_A. If the order is different, the integral changes.
  • Constant density: This equation assumes constant volumetric flow (ρ constant). For gas‑phase with pressure drop, adjust.
  • Molar flow F_A0: Inlet molar flow of reactant A, in mol/s or kmol/h – consistent with k and V.
  • Rate constant k: Units depend on reaction order; for first order, s⁻¹ (if using C in mol/m³).
  • Conversion X: Must be between 0 and 1; at X=1, volume is infinite.

Applications

The design equation for a plug flow reactor (PFR) with first‑order reaction is V = (F_A0 / k) · ln(1/(1−X)). This relationship allows engineers to calculate the required reactor volume for a given conversion, or the conversion achievable with a given volume. PFRs are characterised by no axial mixing and are often used for fast reactions and high‑temperature processes. This equation assumes constant density and isothermal operation. Engineers apply it to design tubular reactors, catalytic cracking units, and polymerisation systems. By understanding the PFR equation, professionals can optimise reactor dimensions, select appropriate operating conditions, and integrate reactors with other process units to achieve overall process objectives.

  • Design of tubular reactors for gas‑phase and liquid‑phase reactions
  • Sizing of catalytic reactors (e.g., ammonia synthesis, cracking)
  • Optimisation of reaction conversion with respect to volume and temperature
  • Scale‑up of PFRs from laboratory data
  • Comparison with CSTR performance for given kinetics

Frequently Asked Questions

Q01What is the design equation for a PFR for a first‑order reaction?
A01

For a first‑order reaction with constant density, the required volume is V = (F_A0 / k) · ln(1/(1 – X)). This is derived from the PFR mole balance: dX/dV = –r_A/F_A0 = k·C_A0·(1–X)/F_A0, integrated.

Q02What are the assumptions of this formula?
A02

  • First‑order reaction: –r_A = k·C_A.
  • Constant density (no volumetric change).
  • Isothermal operation (k constant).
  • Ideal plug flow (no axial mixing).

Q03What are the common mistakes when using this formula?
A03

  • Applying it to reactions of different order.
  • Using it when the density changes (gas‑phase with mole change).
  • Ignoring the effect of temperature on k.
  • Using outlet concentration instead of inlet in the integration.

Q04How does the required volume scale with conversion?
A04

As X approaches 1, ln(1/(1–X)) grows rapidly, so the volume increases asymptotically. High conversion requires disproportionately large reactors.

Q05What is the effect of the rate constant k on the volume?
A05

Volume is inversely proportional to k. A higher k (faster reaction) reduces the required volume.

Q06How do you convert this to a volumetric basis?
A06

Using F_A0 = C_A0·v₀, the equation becomes V = (v₀ / k) · ln(1/(1–X)). This is the familiar form for a first‑order PFR.

Q07What is the difference between the PFR and CSTR volume for the same conversion?
A07

For a first‑order reaction, the PFR volume is always less than the CSTR volume for the same conversion (because the PFR operates at higher concentrations). The ratio V_PFR/V_CSTR = ln(1/(1–X)) / X.

Q08What are the practical applications of this formula?
A08

Preliminary sizing of tubular reactors for first‑order reactions, often used in environmental engineering (e.g., UV disinfection) and polymerisation.