14.2 Non-Isothermal Reactor Design and Thermal Runaway

14.2 Non-Isothermal Reactor Design and Thermal Runaway

While textbook problems frequently assume isothermal operation, commercial chemical reactors are seldom isothermal. Industrial reactions involve substantial heats of reaction ($\Delta H_{rxn}$)—ranging from mildly exothermic polymerizations to highly exothermic oxidations, chlorinations, and nitrations exceeding $-200 \text{ kJ/mol}$. Because reaction rates increase exponentially with temperature according to the Arrhenius equation, heat released by reaction accelerates kinetics, releasing more heat and establishing a dangerous positive feedback loop.

Mastering non-isothermal reactor design on the NCEES PE Chemical Exam requires formulating coupled mass and energy balances, calculating the adiabatic temperature rise ($\Delta T_{ad}$), determining thermal stability and multiple steady states using the van Heerden criterion, and mitigating parametric sensitivity and hot spots in cooled tubular reactors.

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1. General Steady-State Energy Balance for Flow Reactors

Applying the First Law of Thermodynamics to an open, steady-state continuous reactor with negligible kinetic and potential energy changes:

$$\dot{Q} - \dot{W}_s + \sum_{in} F_{i,in} H_{i,in} - \sum_{out} F_{i,out} H_{i,out} = 0$$

Neglecting shaft work ($\dot{W}_s \approx 0$) and expressing enthalpies relative to an arbitrary reference temperature $T_R$ where heat of reaction is $\Delta H_{rxn}(T_R)$:

$$\dot{Q} - F_{A0} X \Delta H_{rxn}(T_R) - \sum_{i} F_i C_{p,i} (T - T_0) = 0$$

Where:

• $\dot{Q}$ = rate of heat transfer through cooling coils or jackets ($\text{kW}$ or $\text{Btu/hr}$).
• $F_{A0}$ = molar feed rate of limiting reactant $A$ ($\text{mol/s}$ or $\text{lbmol/hr}$).
• $X$ = fractional conversion of reactant $A$.
• $\Delta H_{rxn}(T_R)$ = enthalpy of reaction at reference temperature $T_R$ ($\text{kJ/mol}$ or $\text{Btu/lbmol}$), defined such that $\Delta H_{rxn} < 0$ for exothermic reactions.
• $C_{p,i}$ = mean molar heat capacity of species $i$ ($\text{J/(mol}\cdot\text{K)}$ or $\text{Btu/(lbmol}\cdot^\circ\text{F)}$).
• $T_0$ = reactor feed inlet temperature.

                    COUPLED NON-ISOTHERMAL DYNAMICS
                    
                       +------------------------+
                       | Mass Balance: X = f(T) |
                       +------------------------+
                                   |  ^
                Arrhenius k(T)     |  |  Heat Generation
                increases rate     v  |  Delta_H_rxn * r_A
                       +------------------------+
                       | Energy Balance: T = f(X)|
                       +------------------------+

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2. Adiabatic Reactor Operation & Adiabatic Temperature Rise

When a reactor is thermally insulated or operates with zero heat transfer ($\dot{Q} = 0$), the operation is adiabatic. All energy released by exothermic reaction must heat the flowing process fluid.

Conversion-Temperature Relationship in Adiabatic Flow

Setting $\dot{Q} = 0$ and assuming constant average heat capacities:

$$T = T_0 + \left[ \frac{(-\Delta H_{rxn}) F_{A0}}{\sum_i F_{i0} C_{p,i}} \right] X = T_0 + \left[ \frac{(-\Delta H_{rxn}) C_{A0}}{\rho C_p} \right] X$$

Where:

• $\rho$ = process fluid density ($\text{kg/m}^3$ or $\text{lb/ft}^3$).
• $C_p$ = specific heat capacity of process mixture ($\text{kJ/(kg}\cdot\text{K)}$ or $\text{Btu/(lb}\cdot^\circ\text{F)}$).

The Adiabatic Temperature Rise ($\Delta T_{ad}$)

The adiabatic temperature rise ($\Delta T_{ad}$) represents the total theoretical temperature change if the limiting reactant were $100\%$ converted ($X = 1.0$) under complete adiabatic conditions:

$$\Delta T_{ad} \equiv \frac{(-\Delta H_{rxn}) C_{A0}}{\rho C_p} = \frac{(-\Delta H_{rxn}) y_{A0}}{\sum \theta_i C_{p,i}}$$

Thus, at any intermediate conversion $X$:

$$T = T_0 + \Delta T_{ad} \cdot X$$

Reversible Exothermic Reactions: The Thermodynamic Ceiling

For reversible exothermic reactions (such as ammonia synthesis or sulfur dioxide oxidation):

• According to Le Chatelier's principle, increasing temperature shifts equilibrium toward reactants, reducing the maximum thermodynamic equilibrium conversion ($X_{eq}$).
• However, kinetic rate constants increase with temperature.
• In an adiabatic PFR, as conversion proceeds, temperature rises along the operating line $T = T_0 + \Delta T_{ad} X$. The reaction proceeds rapidly at first, but slows as it approaches the falling equilibrium conversion curve $X_{eq}(T)$, eventually reaching a complete thermodynamic standstill.
• To achieve high conversions industrially, reactors employ multistage adiabatic beds with interstage cooling (heat exchangers or cold-shot gas quenching).

   Reversible Exothermic Reaction: Equilibrium Ceiling
   Conversion (X)
     ^ 
  1.0|   \ Equilibrium Curve X_eq(T)
     |    \ 
     |     \     / Operating Line: T = T_0 + Delta_T_ad * X
     |      \   / 
     |       \ / <--- Equilibrium Pinch (Reaction Rate -> 0)
     |        x
     |       / \
     +----------------------------------------> Temperature (T)
            T_0

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3. CSTR Heat Balances: Heat Generation $G(T)$ vs. Heat Removal $R(T)$

In a non-adiabatic continuous stirred-tank reactor equipped with an internal cooling coil or external heat transfer jacket, the steady-state thermal balance is conceptualized by plotting heat generated ($G(T)$) versus heat removed ($R(T)$) as functions of temperature.

Heat Generation Curve: $G(T)$

The total rate of heat released by reaction is:

$$G(T) = (-\Delta H_{rxn}) \cdot F_{A0} \cdot X(T)$$

For an irreversible first-order liquid-phase reaction where $X = \frac{\tau k(T)}{1 + \tau k(T)}$ and $k(T) = A e^{-E / RT}$:

$$G(T) = (-\Delta H_{rxn}) F_{A0} \left[ \frac{\tau A e^{-E / RT}}{1 + \tau A e^{-E / RT}} \right]$$

Behavior of $G(T)$:

• As $T \to 0$: $k(T) \to 0 \implies X \to 0 \implies G(T) \to 0$.
• In intermediate temperature ranges: $k(T)$ increases exponentially, causing $G(T)$ to curve sharply upward.
• As $T \to \infty$: $k(T) \to \infty \implies X \to 1.0$. The reaction becomes feed-limited, and $G(T)$ asymptotes to the maximum ceiling $(-\Delta H_{rxn}) F_{A0}$.
• Therefore, $G(T)$ is an S-shaped (sigmoidal) curve.

Heat Removal Line: $R(T)$

Heat is removed from the CSTR via two mechanisms: sensible heating of the flowing fluid from $T_0$ to $T$, and heat transfer across jacket/coil area $A_c$ to coolant at temperature $T_a$:

$$R(T) = \dot{m} C_p (T - T_0) + U A_c (T - T_a)$$

Factoring temperature $T$:

$$R(T) = (\dot{m} C_p + U A_c) T - (\dot{m} C_p T_0 + U A_c T_a)$$

Defining the combined heat removal parameter $C_R = \dot{m} C_p + U A_c$ and the weighted coolant/feed temperature $T_c'$:

$$T_c' = \frac{\dot{m} C_p T_0 + U A_c T_a}{\dot{m} C_p + U A_c}$$

Yields the linear relation:

$$R(T) = (\dot{m} C_p + U A_c) (T - T_c')$$

Behavior of $R(T)$:

• $R(T)$ is a strictly straight line with a constant positive slope equal to $(\dot{m} C_p + U A_c)$ and an x-intercept at $T_c'$.

   Multiple Steady States: Heat Generation G(T) vs Removal R(T)
   Heat Rate
     ^
     |                        R(T) [Straight Line, Slope = m_dot*Cp + UA]
     |                       /       
     |                      /   /---\ G(T) [S-Shaped Sigmoidal]
     |                     /   /     \
     |                    /   /       ----- Ceiling: (-Delta_H)*F_A0
     |                   /   /  
     |               (3)*   /         (3) = Ignited State (Stable)
     |                 /   /          (2) = Saddle Point (Unstable!)
     |                / * (2)         (1) = Extinguished State (Stable)
     |               / / 
     |           (1)* /  
     |             / / 
     +------------*---------------------------> Temperature (T)
                 T_c'

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4. Multiple Steady States and the van Heerden Stability Criterion

At steady state, heat generation must equal heat removal: $G(T) = R(T)$. Because $G(T)$ is sigmoidal and $R(T)$ is linear, they can intersect at one, two, or three distinct operating temperatures.

The Three Steady States

When three intersections exist ($T_1 < T_2 < T_3$):

1. Point 1 ($T_1$, Extinguished State): Low temperature, low conversion ($X < 10\%$). The reaction proceeds sluggishly, generating minimal heat.
2. Point 2 ($T_2$, Intermediate State): Moderate temperature, intermediate conversion. This state is physically unstable.
3. Point 3 ($T_3$, Ignited State): High temperature, high conversion ($X > 90\%$). The reaction is vigorous and nearly complete.

The van Heerden Stability Criterion

To determine whether an operating steady state is stable against transient perturbations, consider an infinitesimal temperature disturbance $dT$:

• If temperature increases to $T + dT$, the reactor is stable only if heat removal exceeds heat generation ($R > G$), cooling the reactor back to $T$.
• If temperature decreases to $T - dT$, the reactor is stable only if heat generation exceeds heat removal ($G > R$), warming the reactor back to $T$.

Mathematically, stability requires that the slope of the removal line be steeper than the slope of the generation curve at the intersection point:

$$\mathbf{\frac{dR}{dT} > \frac{dG}{dT}} \quad (\text{Stable Steady State})$$ $$\frac{dR}{dT} < \frac{dG}{dT} \quad (\text{Unstable Steady State})$$

Ignition, Extinction, and Hysteresis

Shifting feed temperature $T_0$ or coolant temperature $T_a$ translates the removal line $R(T)$ horizontally:

• Ignition: Starting at low temperature ($T_1$), gradually raising $T_0$ shifts $R(T)$ to the right until the line is tangent to the lower knee of $G(T)$. Any further heating causes a sudden, discontinuous jump from the low branch up to the ignited state $T_3$ (ignition temperature, $T_{ign}$).
• Extinction: Starting from the ignited state ($T_3$), cooling the reactor shifts $R(T)$ to the left until it becomes tangent to the upper knee of $G(T)$. Further cooling causes an abrupt crash back to the extinguished state $T_1$ (extinction temperature, $T_{ext}$).
• Because $T_{ext} < T_{ign}$, the system exhibits thermal hysteresis.

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5. Plug Flow Reactors with Heat Exchange & Thermal Runaway

In a tubular plug flow reactor with active wall cooling (e.g., shell-and-tube packed reactor with molten salt or boiling water in the jacket):

$$\frac{dT}{dV} = \frac{(-\Delta H_{rxn}) (-r_A) - U a (T - T_a)}{\sum F_i C_{p,i}}$$

Where $a$ is the heat transfer area per unit reactor volume ($a = 4 / D_{tube}$ for circular tubes).

Hot-Spot Formation along the Tube Length

• At the reactor inlet ($z = 0$), reactant concentration $C_A$ is at its maximum, causing high reaction rate $(-r_A)$ and intense heat release.
• If heat generation initially exceeds heat removal capacity: $(-\Delta H_{rxn})(-r_A) > U a (T - T_a)$, the fluid temperature rises sharply ($dT/dV > 0$).
• As $A$ is consumed along the tube, $(-r_A)$ decreases while the driving force $(T - T_a)$ increases.
• Eventually, heat removal catches up to heat generation: $(-\Delta H_{rxn})(-r_A) = U a (T - T_a)$, where $dT/dV = 0$. This peak temperature is the reactor hot spot ($T_{max}$).
• Beyond the hot spot, heat removal dominates, and fluid temperature drops toward $T_a$.

   Temperature Profile in a Cooled Tubular Reactor: Hot Spot
   Temp (T)
     ^
     |               Hot Spot (T_max, dT/dV = 0)
     |                   /\ 
     |                  /  \ 
     |                 /    \ 
     |   T_0          /      \------------------ Coolant Temp (T_a)
     |    *----------/        
     |   
     +----------------------------------------> Tube Length (z)

Parametric Sensitivity

A reactor exhibits parametric sensitivity when a minor perturbation in an operational parameter (such as a $2^\circ\text{C}$ rise in feed temperature $T_0$, a $1\%$ increase in feed concentration $C_{A0}$, or a small drop in coolant flow) causes a disproportionate, catastrophic spike in the hot-spot temperature (often $\Delta T_{hotspot} > 50-100^\circ\text{C}$). When temperature spirals out of control, the condition is termed thermal runaway.

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6. Summary Comparison Table: Reactor Thermal Regimes

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7. Comprehensive Step-by-Step Worked Numerical Example: CSTR Thermal Stability

Problem Statement

An exothermic liquid-phase isomerization reaction $A \to B$ takes place in an ideal $1.50 \text{ m}^3$ CSTR equipped with an internal cooling coil. The reaction is first-order in $A$ with rate constant:

$$k(T) = 2.00 \times 10^9 \exp\left( -\frac{6,500}{T} \right) \text{ min}^{-1} \quad (T \text{ in Kelvin})$$

Process Data:

• Feed flow rate: $v_0 = 0.050 \text{ m}^3/\text{min}$ ($50.0 \text{ L/min}$)
• Inlet reactant concentration: $C_{A0} = 2.00 \text{ kmol/m}^3$ ($2,000 \text{ mol/m}^3$)
• Molar feed rate: $F_{A0} = v_0 C_{A0} = 0.050 \times 2.00 = 0.100 \text{ kmol/min} = 1.667 \text{ mol/s}$
• Heat of reaction: $\Delta H_{rxn} = -90,000 \text{ kJ/kmol}$ ($-90.0 \text{ kJ/mol}$)
• Fluid density: $\rho = 1,000 \text{ kg/m}^3$
• Fluid heat capacity: $C_p = 4.00 \text{ kJ/(kg}\cdot\text{K)}$
• Volumetric heat capacity: $\rho C_p = 4,000 \text{ kJ/(m}^3\cdot\text{K)}$
• Total fluid heat capacity rate: $\dot{m} C_p = v_0 \rho C_p = 0.050 \times 4,000 = 200.0 \text{ kJ/(min}\cdot\text{K)}$
• Feed temperature: $T_0 = 300.0 \text{ K}$
• Cooling coil overall heat transfer capacity: $U A_c = 300.0 \text{ kJ/(min}\cdot\text{K)}$
• Coolant temperature: $T_a = 285.0 \text{ K}$

Calculate:

1. The adiabatic temperature rise $\Delta T_{ad}$.
2. The heat removal slope and zero-removal intercept temperature $T_c'$.
3. The heat removal rate $R(T)$ and heat generation rate $G(T)$ at candidate operating temperatures $T = 310.0 \text{ K}$, $T = 330.0 \text{ K}$, and $T = 360.0 \text{ K}$.
4. Identify whether a steady state exists near $T = 330.0 \text{ K}$, and verify its thermal stability using the van Heerden criterion.

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Step 1: Adiabatic Temperature Rise

$$\Delta T_{ad} = \frac{(-\Delta H_{rxn}) C_{A0}}{\rho C_p} = \frac{90,000 \text{ kJ/kmol} \times 2.00 \text{ kmol/m}^3}{4,000 \text{ kJ/(m}^3\cdot\text{K)}} = \frac{180,000}{4,000} = \mathbf{45.0 \text{ K}}$$

If cooling completely fails and the reactor reaches full conversion ($X = 1.0$), the maximum adiabatic temperature is $T_{max, ad} = T_0 + \Delta T_{ad} = 300.0 + 45.0 = \mathbf{345.0 \text{ K}}$.

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Step 2: Heat Removal Function $R(T)$

The total slope of the removal line is:

$$\text{Slope} = \dot{m} C_p + U A_c = 200.0 + 300.0 = \mathbf{500.0 \text{ kJ/(min}\cdot\text{K)}}$$

The effective weighted coolant/feed temperature $T_c'$ is:

$$T_c' = \frac{\dot{m} C_p T_0 + U A_c T_a}{\dot{m} C_p + U A_c} = \frac{(200.0 \times 300.0) + (300.0 \times 285.0)}{500.0} = \frac{60,000 + 85,500}{500.0} = \frac{145,500}{500.0} = \mathbf{291.0 \text{ K}}$$

Thus, the heat removal equation is:

$$R(T) = 500.0 \cdot (T - 291.0) \text{ kJ/min}$$

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Step 3: Space Time and Heat Generation Function $G(T)$

The reactor space time is:

$$\tau = \frac{V}{v_0} = \frac{1.50 \text{ m}^3}{0.050 \text{ m}^3/\text{min}} = \mathbf{30.0 \text{ min}}$$

The maximum possible heat generation rate at complete conversion ($X = 1.0$) is:

$$G_{max} = (-\Delta H_{rxn}) F_{A0} = 90,000 \text{ kJ/kmol} \times 0.100 \text{ kmol/min} = \mathbf{9,000.0 \text{ kJ/min}}$$

At any temperature $T$:

$$k(T) = 2.00 \times 10^9 \exp\left( -\frac{6,500}{T} \right)$$ $$X(T) = \frac{\tau k(T)}{1 + \tau k(T)} = \frac{30.0 k(T)}{1 + 30.0 k(T)}$$ $$G(T) = 9,000.0 \cdot X(T) \text{ kJ/min}$$

Now, evaluate $k(T)$, $X(T)$, $G(T)$, and $R(T)$ across the candidate temperatures:

At $T = 310.0 \text{ K}$:

• $\frac{6,500}{310.0} = 20.9677 \implies \exp(-20.9677) = 7.831 \times 10^{-10}$
• $k(310) = 2.00 \times 10^9 \times 7.831 \times 10^{-10} = 1.566 \text{ min}^{-1}$
• $\tau k = 30.0 \times 1.566 = 46.98$
• $X = \frac{46.98}{1 + 46.98} = \frac{46.98}{47.98} = 0.9792$
• $G(310) = 9,000.0 \times 0.9792 = 8,812.8 \text{ kJ/min}$
• $R(310) = 500.0 \times (310.0 - 291.0) = 500.0 \times 19.0 = 9,500.0 \text{ kJ/min}$
• Here $R > G$ (net cooling of $-687.2 \text{ kJ/min}$).

At $T = 330.0 \text{ K}$:

• $\frac{6,500}{330.0} = 19.6970 \implies \exp(-19.6970) = 2.7834 \times 10^{-9}$
• $k(330) = 2.00 \times 10^9 \times 2.7834 \times 10^{-9} = 5.567 \text{ min}^{-1}$
• $\tau k = 30.0 \times 5.567 = 167.0$
• $X = \frac{167.0}{1 + 167.0} = 0.9940$
• $G(330) = 9,000.0 \times 0.9940 = 8,946.0 \text{ kJ/min}$
• $R(330) = 500.0 \times (330.0 - 291.0) = 500.0 \times 39.0 = 19,500.0 \text{ kJ/min}$
• At high temperature, heat removal vastly exceeds heat generation ($R \gg G$).

Let us find the precise intersection where $G(T) = R(T)$: At $T = 308.6 \text{ K}$:

• $R(308.6) = 500.0 \times (308.6 - 291.0) = 500.0 \times 17.6 = 8,800.0 \text{ kJ/min}$
• At $T = 308.6 \text{ K}$, $6500/308.6 = 21.06286 \implies k = 1.424 \text{ min}^{-1} \implies \tau k = 42.71 \implies X = 0.9771 \implies G = 9,000 \times 0.9771 = 8,794 \text{ kJ/min} \approx R$.

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Step 4: van Heerden Stability Check

At the operating steady state ($T \approx 308.6 \text{ K}$):

• The slope of the heat removal line is constant: $\frac{dR}{dT} = \mathbf{500.0 \text{ kJ/(min}\cdot\text{K)}}$.
• To evaluate $\frac{dG}{dT}$: $$\frac{dG}{dT} = G_{max} \frac{dX}{dT} = G_{max} \frac{\tau \frac{dk}{dT}}{(1 + \tau k)^2} = G_{max} \frac{\tau k \left( \frac{E}{R T^2} \right)}{(1 + \tau k)^2}$$ At $T = 308.6 \text{ K}$, $\frac{E}{R T^2} = \frac{6,500}{(308.6)^2} = \frac{6,500}{95,234} = 0.06825 \text{ K}^{-1}$. With $\tau k = 42.71$: $$\frac{dG}{dT} = 9,000.0 \times \frac{42.71 \times 0.06825}{(1 + 42.71)^2} = 9,000.0 \times \frac{2.915}{(43.71)^2} = 9,000.0 \times \frac{2.915}{1,910.6} = 9,000.0 \times 0.001526 = \mathbf{13.73 \text{ kJ/(min}\cdot\text{K)}}$$

Comparing slopes:

$$\frac{dR}{dT} = 500.0 \gg \frac{dG}{dT} = 13.73 \text{ kJ/(min}\cdot\text{K)}$$

Because $\frac{dR}{dT} > \frac{dG}{dT}$, the steady state is exceptionally stable. Any minor temperature surge creates strong net cooling, driving the reactor back to equilibrium.

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8. Critical PE Exam Traps & Pitfalls

Trap 1: Forgetting Solvent and Inert Heat Capacities in $\Delta T_{ad}$
In dilute liquid-phase reactions, the heat capacity in the denominator of $\Delta T_{ad} = (-\Delta H_{rxn}) C_{A0} / (\rho C_p)$ is that of the entire mixture (essentially the solvent). A classic exam blunder is calculating $\rho C_p$ using only the tiny molar mass and heat capacity of solute $A$, yielding absurdly inflated temperature rises in the thousands of degrees.

Trap 2: Assuming the Middle Steady State is Usable
When an exothermic CSTR exhibits three steady states ($T_1 < T_2 < T_3$), candidate engineers often select $T_2$ because it provides a moderate conversion without extreme temperatures. However, $T_2$ is fundamentally unstable ($dG/dT > dR/dT$). A controller cannot hold a reactor at an open-loop unstable saddle point without dynamic closed-loop feedback.

Trap 3: Inverting the van Heerden Criterion
Always remember: Cooling must respond faster than heating. Therefore, $\frac{dR}{dT} > \frac{dG}{dT}$ is required for stability. If the generation curve is steeper than the cooling line, a positive perturbation $\Delta T$ generates more incremental heat than the jacket can remove, causing runaway.

Trap 4: Neglecting Coolant Temperature Changes along Jackets
In tubular reactors, assuming the coolant temperature $T_a$ is constant is only valid for boiling water jackets (where phase change fixes $T$). For single-phase liquid coolants (like cooling water or heat transfer oils), the coolant warms up along the reactor, which degrades heat transfer driving force $(T - T_a)$ and can turn a stable reactor into a runaway hot spot.