13.3 Batch Reactors and Continuous Stirred-Tank Reactors (CSTR)

13.3 Batch Reactors and Continuous Stirred-Tank Reactors (CSTR)

Chemical reactor design is founded on the general mole balance for species $A$ across any control volume:

$$\text{In} - \text{Out} + \text{Generation} = \text{Accumulation}$$ $$F_{A0} - F_A + \int_V r_A dV = \frac{dn_A}{dt}$$

Where:

• $F_{A0}$ = molar flow rate of species $A$ entering the reactor ($\text{mol/s}$ or $\text{kmol/h}$).
• $F_A$ = molar flow rate of species $A$ leaving the reactor ($\text{mol/s}$ or $\text{kmol/h}$).
• $r_A$ = rate of generation of species $A$ per unit volume ($\text{mol}/(\text{L}\cdot\text{s})$).
• $n_A$ = total moles of species $A$ within the control volume.
• $V$ = reactor volume ($\text{L}$ or $\text{m}^3$).

The two fundamental backmixed reactor archetypes tested on the NCEES PE Chemical Exam are the Ideal Batch Reactor (unsteady-state, closed system) and the Continuous Stirred-Tank Reactor (CSTR) (steady-state, open, continuous-flow system).

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1. Ideal Batch Reactor Design and Kinetics

An ideal batch reactor operates as a closed system. Reactants are charged into a vessel at time $t = 0$, thoroughly mixed, and allowed to react over a specified residence time. No fluid enters or leaves during the reaction ($F_{A0} = 0, F_A = 0$).

                  Ideal Batch Reactor (Unsteady State)
                  -------------------------------------
                       Motor & Impeller Stirrer
                                  |
                                [===]
                                  |
                           +-------------+
                           |  Uniform    |
                           |  Liquid     |   dn_A / dt = r_A * V
                           |  Mixture    |   No feed in, no product out
                           |             |   during reaction period
                           +-------------+

General Mole Balance and Conversion

Setting inlet and outlet streams to zero in the general balance:

$$r_A V = \frac{dn_A}{dt} \implies -r_A V = -\frac{dn_A}{dt}$$

Defining fractional conversion $X$ based on initial moles charged ($n_{A0}$):

$$n_A = n_{A0} (1 - X) \implies dn_A = -n_{A0} dX$$

Substituting $dn_A$ into the balance yields the general batch reactor design equation:

$$t = n_{A0} \int_0^X \frac{dX}{(-r_A) V}$$

Constant-Volume Liquid Systems ($V = V_0$)

For liquid-phase reactions and constant-volume gas reactions, the fluid volume remains constant throughout the cycle ($V = V_0$). Factoring $V$ out and noting $C_{A0} = n_{A0} / V_0$:

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

Analytical Solutions for Constant-Volume Batch Kinetics

1. Zero-Order Reaction ($-r_A = k$):

   $$t = C_{A0} \int_0^X \frac{dX}{k} = \frac{C_{A0} X}{k}$$

2. First-Order Reaction ($-r_A = k C_A = k C_{A0}(1 - X)$):

   $$t = C_{A0} \int_0^X \frac{dX}{k C_{A0} (1 - X)} = \frac{1}{k} \int_0^X \frac{dX}{1 - X} = \frac{1}{k} \ln\left( \frac{1}{1 - X} \right) = \frac{1}{k} \ln\left( \frac{C_{A0}}{C_A} \right)$$

3. Second-Order Reaction ($2A \to \text{Products}$, $-r_A = k C_A^2 = k C_{A0}^2(1 - X)^2$):

   $$t = C_{A0} \int_0^X \frac{dX}{k C_{A0}^2 (1 - X)^2} = \frac{1}{k C_{A0}} \int_0^X \frac{dX}{(1 - X)^2} = \frac{1}{k C_{A0}} \left( \frac{X}{1 - X} \right) = \frac{1}{k} \left( \frac{1}{C_A} - \frac{1}{C_{A0}} \right)$$

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2. Batch Cycle Time, Downtime, and Industrial Productivity Sizing

In industrial practice (e.g., pharmaceutical synthesis or polymer compounding), a batch reactor does not produce continuously. Sizing a batch facility to satisfy an annual production quota requires analyzing the full batch cycle time ($t_{cycle}$):

$$t_{cycle} = t_{reaction} + t_{dead}$$ $$t_{dead} = t_{charge} + t_{heat/cool} + t_{discharge} + t_{clean}$$

Where $t_{dead}$ represents non-productive downtime spent charging reagents, heating or cooling the vessel, transferring products, and cleaning or sterilizing between batches.

Sizing Equations for Production Quotas

Let:

• $\dot{P}$ = required production rate of product ($\text{mol/day}$ or $\text{kg/yr}$).
• $t_{op}$ = operating availability of the plant per year (e.g., $300\text{ days/yr} \times 24\text{ h/day} = 7,\!200\text{ h/yr}$).

The number of batches that can be executed per year is:

$$N_{batches} = \frac{t_{op}}{t_{cycle}}$$

The moles of reactant $A$ that must be processed per batch is:

$$n_{A0} = \frac{\dot{P} \cdot t_{cycle}}{X \cdot t_{op}}$$

The required reactor fluid volume is:

$$V = \frac{n_{A0}}{C_{A0}} = \frac{\dot{P} \cdot t_{cycle}}{C_{A0} \cdot X \cdot t_{op}}$$

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3. Continuous Stirred-Tank Reactor (CSTR) Fundamentals

The Continuous Stirred-Tank Reactor (CSTR), also called a backmix reactor, is an agitated vessel with continuous feed and effluent streams.

                    Continuous Stirred-Tank Reactor (CSTR)
                    -------------------------------------
                       Feed In (F_A0, v_0, C_A0)
                                  |
                                  v
                           +-------------+
                           |  PERFECT    |  
                           |  MIXING     |-----> Effluent Out (F_A, v, C_A)
                           |  (-r_A)_exit|       C_A,exit = C_A,vessel
                           +-------------+       (-r_A)_exit = (-r_A)_vessel

The CSTR Ideal Modeling Assumptions:

1. Perfect Spatial Homogeneity: The agitation is sufficiently intense that the fluid inside the tank is perfectly mixed. Temperature, concentration, and reaction rate are uniform throughout the vessel volume.
2. Effluent Identity: Because mixing is instantaneous, the fluid stream exiting the vessel has the exact same composition, temperature, and reaction rate as the bulk fluid inside the vessel: $$C_{A,exit} = C_{A,vessel}, \qquad (-r_A)_{exit} = (-r_A)_{vessel}$$
3. Steady-State Operation: There is no accumulation of mass or energy ($dn_A / dt = 0$).

Derivation of the CSTR Design Equation

Applying the steady-state mole balance ($F_{A0} - F_A + r_A V = 0$):

$$F_{A0} - F_A = (-r_A) V$$

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

$$V = \frac{F_{A0} X}{(-r_A)_{exit}}$$

In terms of volumetric flow rate $v_0$ ($\text{L/s}$ or $\text{m}^3\text{/h}$) for constant fluid density ($v = v_0$):

$$F_{A0} = v_0 C_{A0}, \qquad F_A = v_0 C_A$$ $$V = \frac{v_0 (C_{A0} - C_A)}{(-r_A)_{exit}}$$

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4. Space Time ($\tau$), Space Velocity ($SV$), and Damköhler Number ($Da$)

In continuous reactor analysis, operating performance is normalized against fluid throughput.

Space Time ($\tau$)

Space time ($\tau$) represents the time required to process one reactor volume of feed measured at entering conditions:

$$\tau \equiv \frac{V}{v_0} = \frac{C_{A0} X}{(-r_A)_{exit}} = \frac{C_{A0} - C_A}{(-r_A)_{exit}}$$

Where:

• $V$ = active reactor volume ($\text{m}^3$ or $\text{L}$).
• $v_0$ = volumetric feed flow rate ($\text{m}^3\text{/s}$ or $\text{L/min}$).
• $\tau$ = space time (units of time: $\text{s}$, $\text{min}$, or $\text{h}$).

Space Velocity ($SV$)

Space velocity ($SV$) is the reciprocal of space time, measuring the number of reactor volumes of feed that can be processed per unit time:

$$SV \equiv \frac{1}{\tau} = \frac{v_0}{V}$$

Common industrial metrics include:

• LHSV (Liquid Hourly Space Velocity): Volumetric liquid feed rate at $60^\circ\text{F}$ divided by catalyst/reactor volume ($\text{h}^{-1}$).
• GHSV (Gas Hourly Space Velocity): Volumetric gas feed rate at standard conditions (STP: $0^\circ\text{C}, 1\text{ atm}$) divided by reactor volume ($\text{h}^{-1}$).

The Damköhler Number ($Da$)

The Damköhler number ($Da$) is a dimensionless parameter expressing the ratio of characteristic chemical reaction rate to convective fluid transport rate:

$$Da \equiv \frac{\text{Reaction Rate}}{\text{Convective Flow Rate}} = \frac{(-r_{A0}) V}{F_{A0}} = \frac{(-r_{A0}) \tau}{C_{A0}}$$

• For a first-order reaction ($-r_A = k C_A$): $$Da = k \tau$$
• For a second-order reaction ($-r_A = k C_A^2$): $$Da = k C_{A0} \tau$$

Significance of $Da$:

• $Da \ll 0.1$: Slow reaction / short residence time; conversion is very low ($X \approx Da$).
• $Da \gg 10$: Fast reaction / long residence time; conversion approaches complete equilibrium or stoichiometric completion ($X \to 1.0$).

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5. Analytical Solutions for CSTR Performance

1. First-Order Reaction in a CSTR

Substituting $-r_A = k C_A = k C_{A0}(1 - X)$ into the space time definition:

$$\tau = \frac{C_{A0} X}{k C_{A0} (1 - X)} = \frac{X}{k (1 - X)}$$

Multiplying by $k$ yields $Da = k \tau = \frac{X}{1 - X}$. Rearranging explicitly for conversion $X$:

$$X = \frac{k \tau}{1 + k \tau} = \frac{Da}{1 + Da}$$

In terms of effluent concentration $C_A$:

$$C_A = C_{A0} (1 - X) = \frac{C_{A0}}{1 + k \tau} = \frac{C_{A0}}{1 + Da}$$

2. Second-Order Reaction in a CSTR ($2A \to \text{Products}$, $-r_A = k C_A^2$)

Substituting into the CSTR design equation:

$$\tau = \frac{C_{A0} - C_A}{k C_A^2} \implies k \tau C_A^2 + C_A - C_{A0} = 0$$

Applying the quadratic formula and retaining the physically meaningful positive root:

$$C_A = \frac{-1 + \sqrt{1 + 4 k \tau C_{A0}}}{2 k \tau} = \frac{-1 + \sqrt{1 + 4 Da}}{2 Da} C_{A0}$$

Conversion $X$ is:

$$X = 1 - \frac{C_A}{C_{A0}} = \frac{(1 + 2 Da) - \sqrt{1 + 4 Da}}{2 Da}$$

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6. Continuous Stirred-Tank Reactors in Series (CSTR Cascades)

A major disadvantage of a single CSTR is the backmixing penalty: the entire reactor operates at the lowest possible reactant concentration ($C_{A,exit}$), resulting in the lowest possible reaction rate and requiring a large volume. Connecting multiple smaller CSTRs in series mitigates this penalty.

                  CSTR Cascade in Series (N Tanks)
                  --------------------------------
      v_0, C_A0 ---> [ Tank 1 ] ---> [ Tank 2 ] ---> ... ---> [ Tank N ] ---> v_0, C_AN
                      V_1, C_A1       V_2, C_A2               V_N, C_AN

First-Order Reaction in $N$ Equal-Volume CSTRs

Consider a train of $N$ identical CSTRs in series, each having volume $V_i = V_{total} / N$ and individual space time $\tau_i = V_i / v_0 = \tau_{total} / N$.

For Tank 1:

$$C_{A1} = \frac{C_{A0}}{1 + k \tau_i}$$

For Tank 2:

$$C_{A2} = \frac{C_{A1}}{1 + k \tau_i} = \frac{C_{A0}}{(1 + k \tau_i)^2}$$

Extending by induction to the $N$-th tank:

$$C_{AN} = \frac{C_{A0}}{(1 + k \tau_i)^N} = \frac{C_{A0}}{\left( 1 + k \frac{\tau_{total}}{N} \right)^N}$$

The overall conversion across the entire cascade is:

$$X_N = 1 - \frac{C_{AN}}{C_{A0}} = 1 - \frac{1}{(1 + k \tau_i)^N}$$

Convergence to Plug Flow as $N \to \infty$

From calculus, the definition of the natural exponential is $\lim_{N \to \infty} \left( 1 + \frac{k \tau}{N} \right)^N = e^{k \tau}$. Therefore:

$$\lim_{N \to \infty} C_{AN} = C_{A0} e^{-k \tau_{total}}$$

As the number of tanks in series approaches infinity, the performance of a CSTR cascade converges identically to that of an ideal Plug Flow Reactor (PFR)! In practice, a cascade of just 3 to 5 CSTRs achieves over $85\%$ to $90\%$ of the volume savings of an ideal PFR while retaining individual vessel agitation, temperature control, and slurry-handling capability.

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7. Summary Comparison Table: Batch vs. CSTR vs. CSTR Cascades

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8. Step-by-Step Worked Numerical Example: Reactor Sizing Comparison for Industrial Synthesis

Problem Statement

A specialty ester is synthesized via an irreversible liquid-phase first-order reaction:

$$A \longrightarrow \text{Products}$$

The feed enters at concentration $C_{A0} = 2.00\text{ mol/L}$ with volumetric flow rate $v_0 = 10.0\text{ L/min}$ ($0.0100\text{ m}^3\text{/min}$). At the reaction temperature, the rate constant is $k = 0.0500\text{ min}^{-1}$. The target conversion is $85.0\%$ ($X = 0.850$).

Calculate:

1. The required volume ($V$) and space time ($\tau$) for a single CSTR.
2. The reaction time ($t_{rxn}$) for a batch reactor, and the total cycle time ($t_{cycle}$) assuming charging, heating, discharging, and cleaning downtime is $t_{dead} = 45.0\text{ min}$.
3. The batch reactor volume ($V_{batch}$) required to match the continuous plant capacity ($F_{A0} = 20.0\text{ mol/min} = 28,\!800\text{ mol/day}$) operating $24.0\text{ h/day}$.
4. The required individual volume ($V_i$) and total volume ($V_{total}$) for two identical CSTRs in series, and the percentage volume reduction achieved relative to the single CSTR.

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Step 1: Single CSTR Sizing

For $85.0\%$ conversion:

$$C_{A,exit} = C_{A0}(1 - X) = 2.00\text{ mol/L} \times (1 - 0.850) = 0.300\text{ mol/L}$$ $$(-r_A)_{exit} = k C_{A,exit} = 0.0500\text{ min}^{-1} \times 0.300\text{ mol/L} = 0.0150\text{ mol}/(\text{L}\cdot\text{min})$$

Using the CSTR design equation:

$$\tau = \frac{C_{A0} X}{(-r_A)_{exit}} = \frac{2.00\text{ mol/L} \times 0.850}{0.0150\text{ mol}/(\text{L}\cdot\text{min})} = \frac{1.70}{0.0150} = \mathbf{113.33\text{ min}} \quad (1.889\text{ h})$$ $$V_{CSTR} = v_0 \cdot \tau = 10.0\text{ L/min} \times 113.33\text{ min} = \mathbf{1,\!133.3\text{ L}} \quad (1.133\text{ m}^3)$$

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Step 2: Batch Reaction Time and Cycle Time

For a constant-volume batch reactor with first-order kinetics:

$$t_{rxn} = \frac{1}{k} \ln\left( \frac{1}{1 - X} \right) = \frac{1}{0.0500\text{ min}^{-1}} \ln\left( \frac{1}{1 - 0.850} \right) = 20.0 \times \ln(6.6667) = 20.0 \times 1.89712 = \mathbf{37.94\text{ min}}$$

Total batch cycle time:

$$t_{cycle} = t_{rxn} + t_{dead} = 37.94\text{ min} + 45.00\text{ min} = \mathbf{82.94\text{ min}} \quad (1.382\text{ h})$$

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Step 3: Batch Reactor Volume for Equivalent Throughput

Daily production throughput required:

$$\dot{N}_A = v_0 C_{A0} \times 1,\!440\text{ min/day} = 10.0\text{ L/min} \times 2.00\text{ mol/L} \times 1,\!440 = 28,\!800\text{ mol/day}$$

Number of batches possible per 24-hour operating day:

$$N_{batches} = \frac{1,\!440\text{ min/day}}{82.94\text{ min/batch}} = 17.362\text{ batches/day}$$

Moles of $A$ charged per batch:

$$n_{A0} = \frac{28,\!800\text{ mol/day}}{17.362\text{ batches/day}} = 1,\!658.8\text{ mol/batch}$$

Required batch vessel working volume:

$$V_{batch} = \frac{n_{A0}}{C_{A0}} = \frac{1,\!658.8\text{ mol}}{2.00\text{ mol/L}} = \mathbf{829.4\text{ L}} \quad (0.829\text{ m}^3)$$

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Step 4: Two Identical CSTRs in Series Sizing

For two equal-volume CSTRs in series:

$$\frac{C_{A2}}{C_{A0}} = 1 - X = 1 - 0.850 = 0.150$$ $$\frac{C_{A2}}{C_{A0}} = \frac{1}{(1 + k \tau_i)^2} = 0.150 \implies (1 + k \tau_i)^2 = \frac{1}{0.150} = 6.6667$$ $$1 + k \tau_i = \sqrt{6.6667} = 2.5820 \implies k \tau_i = 1.5820$$ $$\tau_i = \frac{1.5820}{0.0500\text{ min}^{-1}} = \mathbf{31.64\text{ min}}$$

Individual tank volume:

$$V_i = v_0 \cdot \tau_i = 10.0\text{ L/min} \times 31.64\text{ min} = \mathbf{316.4\text{ L}}$$

Total cascade volume:

$$V_{total} = 2 \times V_i = 2 \times 316.4\text{ L} = \mathbf{632.8\text{ L}}$$

Percentage volume reduction compared to the single CSTR:

$$\text{Reduction} = \frac{V_{CSTR} - V_{total}}{V_{CSTR}} \times 100\% = \frac{1,\!133.3 - 632.8}{1,\!133.3} \times 100\% = \frac{500.5}{1,\!133.3} \times 100\% = \mathbf{44.2\%}$$

Splitting the single CSTR into just two equal-volume stages eliminates over $44\%$ of the required total reactor volume!

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9. Critical PE Exam Traps & Pitfalls

Trap 1: Evaluating CSTR Rate at Inlet or Average Conditions
Because an ideal CSTR is completely backmixed, fluid elements mix instantaneously. The reaction rate is strictly evaluated at the exit concentration ($C_{A,exit}$), NOT at $C_{A0}$ and NOT at the arithmetic average $(C_{A0} + C_A)/2$. Evaluating at average conditions will drastically underestimate the required CSTR size.

Trap 2: Omitting Downtime in Batch Sizing
When sizing an industrial batch plant, never calculate vessel volume from $t_{rxn}$ alone. Clean-in-place (CIP), steam-in-place (SIP), charging, and discharging operations frequently take longer than the chemical reaction itself. Always use $t_{cycle} = t_{rxn} + t_{dead}$.

Trap 3: Confusing Cascade Space Time with Individual Tank Space Time
In a CSTR cascade equation $C_N = C_{A0} / (1 + k \tau_i)^N$, the variable $\tau_i$ is the space time of one individual tank ($V_i / v_0$), NOT the total cascade space time. The total space time is $\tau_{total} = N \tau_i$. Substituting $\tau_{total}$ into the denominator produces errors greater than $300\%$.

Trap 4: Confusing Space Time ($\tau$) with True Mean Residence Time ($\bar{t}$)
Space time is defined strictly as $\tau \equiv V / v_0$ based on entering volumetric flow rate. For constant-density liquid systems, space time equals the mean residence time ($\tau = \bar{t}$). However, for gas-phase reactions where temperature, pressure, or total moles change, volumetric flow varies ($v \neq v_0$), meaning space time $\tau$ diverges from true mean residence time.