12.2 Ideal Reactor Design

Ideal reactor design forms the core of chemical reaction engineering. To analyze and design chemical reactors, engineers use idealized models that represent the limiting behaviors of mixing and flow. On the FE Chemical exam, you will be expected to apply the design equations for three fundamental reactor types: the Batch Reactor, the Continuous Stirred Tank Reactor (CSTR), and the Plug Flow Reactor (PFR). Understanding the underlying assumptions and mathematical formulations of these reactors is essential for solving design, sizing, and conversion problems.

The General Mole Balance Equation

All reactor design equations are derived from a general mole balance on a control volume. For any species $i$ (such as reactant A): $\text{Rate In} - \text{Rate Out} + \text{Rate of Generation} = \text{Rate of Accumulation}$ $F_{i0} - F_i + \int r_i dV = \frac{dN_i}{dt}$ where $F_{i0}$ is the molar flow rate of species $i$ entering the control volume ($\text{mol/s}$), $F_i$ is the molar flow rate leaving ($\text{mol/s}$), $r_i$ is the rate of generation of species $i$ per unit volume ($\text{mol}/(\text{L}\cdot\text{s})$), $V$ is the reactor volume ($\text{L}$), and $N_i$ is the number of moles of species $i$ inside the reactor.

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1. The Batch Reactor

A batch reactor is a closed system with no inflow or outflow of chemical species ($F_{\text{A}0} = F_{\text{A}} = 0$). It is assumed to be perfectly mixed, meaning concentration and temperature are spatially uniform. The reaction is unsteady-state, with concentrations changing over time.

• Mole Balance: $r_{\text{A}} V = \frac{dN_{\text{A}}}{dt}$
• In terms of Conversion ($X_{\text{A}}$): Substituting $N_{\text{A}} = N_{\text{A}0}(1 - X_{\text{A}})$ and differentiating gives $dN_{\text{A}} = -N_{\text{A}0} dX_{\text{A}}$. Substituting this back into the mole balance yields the design equation: $t = N_{\text{A}0} \int_0^{X_{\text{A}}} \frac{dX_{\text{A}}}{-r_{\text{A}} V}$
• Constant-Volume Batch Reactor: For liquid-phase reactions (where density and volume are constant), $V = V_0$, which simplifies the equation to: $t = C_{\text{A}0} \int_0^{X_{\text{A}}} \frac{dX_{\text{A}}}{-r_{\text{A}}} = \int_{C_{\text{A}}}^{C_{\text{A}0}} \frac{dC_{\text{A}}}{-r_{\text{A}}}$ For a first-order reaction ($-r_{\text{A}} = k C_{\text{A}}$), integrating gives the classic relationship: $t = \frac{1}{k} \ln\left(\frac{C_{\text{A}0}}{C_{\text{A}}}\right) = \frac{1}{k} \ln\left(\frac{1}{1-X_{\text{A}}}\right)$

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2. The Continuous Stirred Tank Reactor (CSTR)

A CSTR is an open, steady-state system ($\frac{dN_i}{dt} = 0$). It is assumed to be perfectly mixed, meaning the fluid composition and temperature are uniform throughout the vessel. A key consequence of this perfect mixing is that the concentration of any species inside the reactor is identical to its concentration in the exit stream. Consequently, the reaction rate $-r_{\text{A}}$ is evaluated entirely at the exit concentration $C_{\text{A}}$ (or exit conversion $X_{\text{A}}$).

• Design Equation: Under steady-state, the accumulation term is zero, and the volume is constant. The mole balance simplifies to: $F_{\text{A}0} - F_{\text{A}} + r_{\text{A}} V = 0$ Rearranging and expressing in terms of conversion ($F_{\text{A}} = F_{\text{A}0}(1 - X_{\text{A}})$) gives the CSTR sizing equation: $V = \frac{F_{\text{A}0} X_{\text{A}}}{-r_{\text{A}}}$ If the volumetric flow rate $v_0$ is constant ($v = v_0$), we can write: $V = \frac{v_0(C_{\text{A}0} - C_{\text{A}})}{-r_{\text{A}}}$

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3. The Plug Flow Reactor (PFR)

A PFR (also called a tubular reactor) is an open, steady-state system. Fluid passes through the reactor as a series of coherent plugs. It is assumed that there is no mixing in the axial direction (along the length of the reactor), but complete mixing in the radial direction (across the cross-section). As a result, the concentration of reactants decreases continuously along the length of the reactor, and the reaction rate varies with position.

• Design Equation: A mole balance is performed on a differential volume slice $dV$: $\frac{dF_{\text{A}}}{dV} = r_{\text{A}}$ In terms of conversion, since $F_{\text{A}0}$ is constant, we have $V = F_{\text{A}0} \int_0^{X_{\text{A}}} \frac{dX_{\text{A}}}{-r_{\text{A}}}$. For constant volumetric flow rate ($v = v_0$), this can be written as: $V = v_0 \int_{C_{\text{A}}}^{C_{\text{A}0}} \frac{dC_{\text{A}}}{-r_{\text{A}}}$

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Space Time ($\tau$) and Space Velocity ($\text{SV}$)

Space time ($\tau$) is defined as the time required to process one reactor volume of feed measured at entering conditions: $\tau = \frac{V}{v_0}$ Space velocity ($\text{SV}$) is the reciprocal of space time, representing the number of reactor volumes of feed that can be processed per unit time: $\text{SV} = \frac{1}{\tau} = \frac{v_0}{V}$ These variables normalize reactor sizing with respect to throughput. GHSV (Gas Hourly Space Velocity) and LHSV (Liquid Hourly Space Velocity) are common variants used in industry.

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Graphical Comparison: Levenspiel Plots

A Levenspiel plot graphs $\frac{F_{\text{A}0}}{-r_{\text{A}}}$ (or $\frac{1}{-r_{\text{A}}}$) on the y-axis against conversion $X_{\text{A}}$ on the x-axis.

• CSTR Volume: The CSTR design equation is $V = X_{\text{A}} \cdot \left(\frac{F_{\text{A}0}}{-r_{\text{A}}}\right)$. Visually, the CSTR volume is equal to the area of a rectangle of width $X_{\text{A}}$ and height $\frac{F_{\text{A}0}}{-r_{\text{A}}}$ evaluated at the exit conversion.
• PFR Volume: The PFR design equation is $V = \int_0^{X_{\text{A}}} \frac{F_{\text{A}0}}{-r_{\text{A}}} dX_{\text{A}}$. Visually, the PFR volume is the area under the curve of $\frac{F_{\text{A}0}}{-r_{\text{A}}}$ from $X_{\text{A}} = 0$ to the exit conversion.

For normal, positive-order kinetics (where reaction rate decreases as reactant concentration decreases), the term $\frac{F_{\text{A}0}}{-r_{\text{A}}}$ increases with conversion. In this case, the CSTR rectangle always encompasses a larger area than the PFR integral. Therefore, to achieve the same conversion, a CSTR will always require a larger volume than a PFR: $V_{\text{CSTR}} > V_{\text{PFR}} \quad \text{(for } n > 0\text{)}$

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Worked Example: CSTR vs. PFR Sizing

Problem: A liquid-phase reaction $\text{A} \rightarrow \text{B}$ follows first-order kinetics with a rate constant $k = 0.05\text{ min}^{-1}$. The feed stream enters at a volumetric flow rate $v_0 = 10\text{ L/min}$ with an initial concentration $C_{\text{A}0} = 2.0\text{ mol/L}$. Calculate the reactor volume required to achieve $80%$ conversion ($X_{\text{A}} = 0.80$) in:

1. A CSTR
2. A PFR

Solution: First, compute the inlet molar flow rate of A: $F_{\text{A}0} = v_0 C_{\text{A}0} = 10\text{ L/min} \cdot 2.0\text{ mol/L} = 20\text{ mol/min}$

At $80%$ conversion:

• Exit conversion $X_{\text{A}} = 0.80$
• Exit concentration $C_{\text{A}} = C_{\text{A}0}(1 - X_{\text{A}}) = 2.0(1 - 0.80) = 0.4\text{ mol/L}$

1. CSTR Sizing

The reaction rate at the exit of the CSTR is evaluated at the exit concentration: $-r_{\text{A}} = k C_{\text{A}} = 0.05\text{ min}^{-1} \cdot 0.4\text{ mol/L} = 0.02\text{ mol/(L}\cdot\text{min)}$ Now, apply the CSTR design equation: $V_{\text{CSTR}} = \frac{F_{\text{A}0} X_{\text{A}}}{-r_{\text{A}}} = \frac{20\text{ mol/min} \cdot 0.80}{0.02\text{ mol/(L}\cdot\text{min)}} = 800\text{ L}$ Alternatively, using space time: $\tau_{\text{CSTR}} = \frac{C_{\text{A}0} - C_{\text{A}}}{-r_{\text{A}}} = \frac{2.0 - 0.4}{0.02} = 80\text{ min}$ $V_{\text{CSTR}} = \tau_{\text{CSTR}} \cdot v_0 = 80\text{ min} \cdot 10\text{ L/min} = 800\text{ L}$

2. PFR Sizing

For a PFR, substitute the first-order rate law into the design equation: $V_{\text{PFR}} = F_{\text{A}0} \int_0^{X_{\text{A}}} \frac{dX_{\text{A}}}{k C_{\text{A}0}(1 - X_{\text{A}})} = \frac{F_{\text{A}0}}{k C_{\text{A}0}} \int_0^{X_{\text{A}}} \frac{dX_{\text{A}}}{1 - X_{\text{A}}}$ Since $F_{\text{A}0} / C_{\text{A}0} = v_0$: $V_{\text{PFR}} = \frac{v_0}{k} \left[ -\ln(1 - X_{\text{A}}) \right]_0^{0.80} = \frac{v_0}{k} \ln\left(\frac{1}{1 - X_{\text{A}}}\right)$ Substitute the values: $V_{\text{PFR}} = \frac{10\text{ L/min}}{0.05\text{ min}^{-1}} \ln\left(\frac{1}{1 - 0.80}\right) = 200 \cdot \ln(5) \approx 200 \cdot 1.6094 = 322\text{ L}$

Comparison:

$V_{\text{CSTR}} = 800\text{ L} > V_{\text{PFR}} = 322\text{ L}$ As expected, the CSTR requires more than double the volume of the PFR to achieve the same level of conversion. This is because the CSTR operates entirely at the low exit concentration ($0.4\text{ mol/L}$), whereas the PFR operates at higher concentrations (ranging from $2.0$ down to $0.4\text{ mol/L}$) throughout most of its volume, leading to a higher average rate of reaction.