Design equations for batch, CSTR and PFR

Mole balances and design equations for ideal batch, CSTR and PFR reactors, space time, gas-phase volume change and the Levenspiel plot, with sizing examples.

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Why it matters

The three ideal reactors — batch, continuous stirred tank (CSTR, mixed flow) and plug flow (PFR) — are the building blocks of every reactor design. Their design equations turn a rate law into a reaction time or reactor volume, and every later topic (sizing comparisons, series arrangements, recycle, non-isothermal operation, non-ideal flow) is built on them.

Key ideas

Start from a mole balance. For species A in any volume: in − out + generation = accumulation, i.e. FA0 − FA + ∫rA·dV = dNA/dt, where rA is the rate of formation of A (negative for a reactant), so −rA is the rate of disappearance.

Conversion. XA = moles of A reacted per mole of A fed. Batch: NA = NA0·(1 − XA). Flow: FA = FA0·(1 − XA). Working in XA lets the same rate data be used for every reactor type.

Ideal batch reactor. Closed, perfectly mixed, composition uniform in space but changing with time. Used for small volumes, multi-product plants and kinetic studies. Real production time also includes charging, heating, cooling, discharging and cleaning.

Ideal CSTR (mixed flow). Perfect mixing: the exit stream has exactly the composition and temperature of the tank contents. The whole reactor therefore operates at the low exit concentration, so the rate is evaluated at exit conditions. It is an algebraic balance — no integration.

Ideal PFR. Fluid moves as a series of plugs with no axial mixing and complete radial uniformity. Composition changes continuously along the length, so the balance is written over a differential slice dV and integrated. Every fluid element spends exactly the same time τ in the reactor.

Space time and space velocity. τ = V/v0 = CA0·V/FA0 is the time to process one reactor volume of feed measured at inlet conditions; space velocity s = 1/τ. For constant-density systems τ equals the mean residence time; for gas reactions with volume change it does not.

Variable density. For gas-phase reactions with a change in moles, v = v0·(1 + εA·XA) (isothermal, isobaric), where εA = (V at XA = 1 − V at XA = 0)/(V at XA = 0). For A → 2R with pure A, εA = 1; with 50 % inerts, εA = 0.5. Then CA = CA0·(1 − XA)/(1 + εA·XA).

Graphical view (Levenspiel plot). Plot FA0/(−rA) against XA. The PFR volume is the area under the curve from 0 to XA; the CSTR volume is the rectangle of width XA and height FA0/(−rA) at the exit. For a rate that falls with conversion (normal kinetics, n > 0) the rectangle is larger, so a CSTR needs more volume than a PFR for the same duty.

Formulas

t = NA0·∫(0→XA) dXA / [(−rA)·V] (batch); constant volume: t = CA0·∫(0→XA) dXA/(−rA) = −∫(CA0→CA) dCA/(−rA)

  • t: reaction time (s); NA0: initial moles of A (mol); V: volume (m³).

V = FA0·XA / (−rA)exit, τ = V/v0 = CA0·XA/(−rA)exit (CSTR)

  • FA0: molar feed rate of A (mol/s); v0: volumetric feed rate (m³/s); CA0 in mol/m³; −rA in mol/m³·s.

V = FA0·∫(0→XA) dXA/(−rA), τ = CA0·∫(0→XA) dXA/(−rA) (PFR)

First order, constant density: kτ = ln[1/(1 − XA)] (PFR and batch with t in place of τ); kτ = XA/(1 − XA) (CSTR)

Second order (−rA = k·CA²), constant density: k·CA0·τ = XA/(1 − XA) (PFR, batch); k·CA0·τ = XA/(1 − XA)² (CSTR)

First order with volume change (PFR): kτ = (1 + εA)·ln[1/(1 − XA)] − εA·XA; (CSTR): kτ = XA·(1 + εA·XA)/(1 − XA)

  • εA: fractional volume change at complete conversion (dimensionless).

Worked examples

Example 1 (standard). A liquid-phase first-order reaction A → R has k = 0.2 min⁻¹. Feed: v0 = 0.5 m³/min. Find the CSTR and PFR volumes for XA = 0.75.

  1. CSTR: kτ = XA/(1 − XA) = 0.75/0.25 = 3 → τ = 3/0.2 = 15 min.
  2. V = v0·τ = 0.5 × 15 = 7.5 m³.
  3. PFR: kτ = ln[1/(1 − 0.75)] = ln 4 = 1.386 → τ = 6.93 min.
  4. V = 0.5 × 6.93 = 3.47 m³.

CSTR: 7.5 m³; PFR: 3.47 m³ — the mixed reactor is more than twice as large.

Example 2 (GATE level, gas phase with expansion). Pure gaseous A reacts by A → 2R, first order, k = 0.05 s⁻¹, in an isothermal, isobaric PFR. Feed v0 = 0.01 m³/s. Find V for XA = 0.8.

  1. εA = (2 − 1)/1 = 1.
  2. kτ = (1 + εA)·ln[1/(1 − XA)] − εA·XA = 2 × ln 5 − 0.8 = 2 × 1.6094 − 0.8 = 2.4189.
  3. τ = 2.4189/0.05 = 48.38 s.
  4. V = v0·τ = 0.01 × 48.38 = 0.484 m³.

V ≈ 0.484 m³. Ignoring expansion would give kτ = ln 5 and V = 0.322 m³ — about one-third too small, because expanding gas speeds through the reactor.

Example 3 (batch). A second-order liquid reaction 2A → products has −rA = k·CA² with k = 0.5 m³/kmol·min and CA0 = 2 kmol/m³. Time for 90 % conversion: t = XA/[k·CA0·(1 − XA)] = 0.9/(0.5 × 2 × 0.1) = 9 min (reaction time only; add down-time for the cycle).

Common mistakes

  • Evaluating the CSTR rate at inlet conditions — it must be at exit conditions.
  • Using FA0 (mol/s) where v0 (m³/s) is needed: check that V comes out in m³.
  • Treating τ as the mean residence time for gas reactions with volume change.
  • Forgetting the εA terms for gas-phase reactions with mole change, or counting inerts wrongly in εA.
  • Using t = τ for a batch reactor but forgetting loading/cleaning time in production-rate problems.

For GATE CH

Expect NAT questions on CSTR or PFR volume (or τ) for first- or second-order kinetics, batch time to reach a conversion, εA for a feed with inerts, and Levenspiel-plot area/rectangle reasoning from tabulated −rA vs XA data. Memorise the first- and second-order design forms for all three reactors and practise reading FA0/(−rA) graphs.

Quick check

  1. In a CSTR, at which concentration is the rate evaluated?
  2. Write τ for a PFR, first order, constant density.
  3. Find εA for A → 3R with a feed of 50 % A and 50 % inerts.
  4. On a Levenspiel plot, what represents the PFR volume?

Answers: 1. the exit (tank) concentration; 2. τ = (1/k)·ln[1/(1 − XA)]; 3. εA = (0.5·3 + 0.5 − 1)/1 = 1.0; 4. the area under FA0/(−rA) versus XA from 0 to the exit conversion.

Interview questions

All Chemical Reaction Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.Explain the design equation for a Continuous Stirred Tank Reactor (CSTR).Concept

    A CSTR is assumed perfectly mixed, so the exit stream has the same composition as the tank contents and the rate is uniform throughout. The steady-state mole balance on A is FA0 − FA + rA·V = 0, which with conversion becomes V = FA0·XA/(−rA), with −rA evaluated at exit conditions. In terms of space time, τ = V/v0 = CA0·XA/(−rA). Because it is algebraic, no integration is needed, but the whole reactor operates at the low exit concentration.

  2. 2.Describe the design equation for a Plug Flow Reactor (PFR) and its assumptions.Concept

    The design equation for a PFR is derived from the assumption that there is no mixing in the axial direction, and the reaction occurs as the reactants flow through the reactor. The equation is: dF/dV = r, where F is the molar flow rate, V is the reactor volume, and r is the reaction rate. This equation is used to calculate the reactor volume required for a specific conversion, assuming ideal plug flow conditions.

  3. 3.Why is a CSTR often used for liquid-phase reactions?Application

    A CSTR is often used for liquid-phase reactions because it provides excellent mixing, ensuring uniform temperature and concentration throughout the reactor. This is particularly important for reactions that are sensitive to temperature or concentration changes. Additionally, CSTRs are easier to control and scale up for industrial applications compared to other reactor types.

  4. 4.What happens if the flow rate in a PFR is increased?Application

    If the flow rate in a PFR is increased, the residence time of the reactants in the reactor decreases. This can lead to lower conversion rates because the reactants have less time to react. However, the effect on conversion also depends on the reaction kinetics and the specific reaction being carried out.

  5. 5.How does the reaction order affect the design of a batch reactor?Application

    The reaction order affects the rate at which reactants are consumed and products are formed in a batch reactor. For zero-order reactions, the rate is constant, while for first-order reactions, the rate depends linearly on the concentration of one reactant. Higher-order reactions have more complex dependencies. Understanding the reaction order is crucial for determining the time required to achieve a desired conversion.

  6. 6.Calculate the volume of a liquid-phase CSTR needed to achieve 80% conversion for a first-order reaction with k = 0.5 s⁻¹ and a volumetric feed rate of 2 L/s.Numerical

    For a first-order reaction in a CSTR at constant density, kτ = XA/(1 − XA). With XA = 0.8, kτ = 0.8/0.2 = 4, so τ = 4/0.5 = 8 s. The volume is V = v0·τ = 2 L/s × 8 s = 16 L (0.016 m³). Note that you need the volumetric flow (or FA0 together with CA0), not just a molar flow, to get a volume.

  7. 7.For a PFR, how does the conversion change along the length of the reactor?Concept

    In a PFR, the conversion increases along the length of the reactor as the reactants are continuously consumed. The rate of conversion depends on the reaction kinetics and the concentration of reactants. Unlike a CSTR, the concentration gradient in a PFR is not uniform, leading to varying reaction rates along the reactor length.

  8. 8.What are the advantages of using a batch reactor for polymerization reactions?Application

    Batch reactors are advantageous for polymerization reactions because they allow for precise control over reaction time, temperature, and mixing. This is important for achieving the desired molecular weight and polymer properties. Additionally, batch reactors are flexible and can be used for small-scale production or research and development purposes.

  9. 9.A batch reactor is used for a second-order reaction with an initial concentration of 1 mol/L. If the rate constant is 0.1 L/mol·s, how long will it take to reach 50% conversion?Numerical

    For a second-order reaction, the integrated rate law is: 1/[A] - 1/[A]₀ = kt. At 50% conversion, [A] = 0.5 mol/L. Substituting the values: 1/0.5 - 1/1 = 0.1 * t, solving for t gives t = 10 s.

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