12. Steady-State Nonisothermal Reactor Design: Flow Reactors with Heat Exchange

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12. Steady-State Nonisothermal Reactor Design: Flow Reactors with Heat Exchange

Research is to see what everybody else sees, and to think what nobody else has thought.

—Albert Szent-Gyorgyi

Overview. This chapter focuses on chemical reactors with heat exchange. The chapter topics are arranged in the following manner:

Section 12.1 further develops the energy balance for easy application to PFRs and PBRs.
Section 12.2 describes PFRs and PBRs for four types of heat exchanger operations.
1. Constant heat-transfer fluid temperature, T_a
2. Variable fluid temperature T_a with co-current operation
3. Variable fluid temperature T_a with countercurrent operation
4. Adiabatic operation
Section 12.3 gives examples of four types of heat exchanger operations for PFRs and PBRs.
Section 12.4 applies the energy balance to a CSTR.
Section 12.5 shows how a CSTR can operate at different steady-state temperatures and conversions, and how to decide which of these conditions are stable and which are unstable.
Section 12.6 describes one of the most important topics of the entire text, multiple reactions with heat effects with Living Example Problems, which is a unique feature of this textbook.
Section 12.7 discusses radial and axial temperature and concentration gradients.
Section 12.8 Safety: The causes of batch reactor explosions are discussed and hint of things to come when we do a CSI analysis on the Monsanto and T2 Laboratory explosions in Chapter 13.

The Chapter 12 Professional Reference Shelf (R12.4) on the CRE Web site (http://www.umich.edu/~elements/6e/12chap/pdf/sulfuricacid.pdf ) describes a typical nonisothermal industrial reactor and reaction, SO₂ oxidation, and gives many practical details.

12.1 Steady-State Tubular Reactor with Heat Exchange

In this section, we consider a tubular reactor in which heat is either added or removed through the cylindrical walls of the reactor (Figure 12-1). In modeling plug-flow reactors, we shall assume that “there are no radial gradients” in the reactor and that the heat flux through the wall per unit volume of reactor is as shown in Figure 12-1.¹

¹ Radial gradients are discussed in Section 12.7 and in Chapters 17 and 18.

A figure shows a tubular reactor. — **Figure 12-1** Tubular reactor with heat gain or loss.

The figure shows a tubular reactor where heat is added through its central section. The magnitude and temperature of the input stream are labeled as F subscript i 0, T subscript 0. The input Energy is Sigma F subscript i H subscript i. This stream passes through the central section where additional heat, delta Q (over dot) is added. The ambient temperature and the volume before the addition of heat are marked as T subscript a and V respectively. The volume of the central section is marked as delta V. The central section is labeled as T. The heat added through the central section, delta Q equals U delta A (T subscript a minus T) equals Ua (T subscript a minus T) delta V. The final volume is V plus delta V. The magnitude and temperature of the output stream is labeled as F subscript i, T. The output Energy is Sigma F subscript i H subscript i.

12.1.1 Deriving the Energy Balance for a PFR

We will carry out an energy balance on the volume ΔV. There is no work done, that is, ${\dot{W}}_{s} = 0$ , so Equation (11-10) becomes

$\begin{matrix} \begin{matrix} Heat \\ Added \end{matrix} & + & \begin{matrix} \begin{matrix} Energy \end{matrix} \\ In \end{matrix} & - & \begin{matrix} Energy \\ Out \end{matrix} & = 0 \\ Δ \dot{Q} & + & {\begin{matrix} Σ F_{i} H_{i} | \end{matrix}}_{V} & - & {\begin{matrix} Σ F_{i} H_{i} | \end{matrix}}_{V + Δ V} & = 0 & (12-1) \end{matrix}$

The heat flow to the reactor, $Δ \dot{Q}$ , is given in terms of the overall heat-transfer coefficient, U, the heat exchange area, ΔA, and the difference between the ambient temperature, T_a, and the reactor temperature T

$Δ \dot{Q} = U Δ A (T_{a} - T) = U a Δ V (T_{a} - T)$

where a is the heat exchange area per unit volume of reactor. For a tubular reactor

$a = \frac{A}{V} = \frac{π DL}{\frac{π D^{2} L}{4}} = \frac{4}{D}$

where D is the reactor diameter. Substituting for $Δ \dot{Q}$ in Equation (12-1), dividing by ΔV, and then taking the limit as ΔV → 0, we get

$U a (T_{a} - T) - \frac{d Σ (F_{i} H_{i})}{d V} = 0$

Expanding

$\begin{matrix} U a (T_{a} - T) - Σ \frac{{d F}_{i}}{d V} H_{i} - Σ F_{i} \frac{d H_{i}}{d V} = 0 & (12 - 2) \end{matrix}$

From a mole balance on species i, we have

$\begin{matrix} \frac{d F_{i}}{d V} = r_{i} = v_{i} (- r_{A}) & (12-3) \end{matrix}$

Differentiating the enthalpy Equation (11-19) with respect to V

$\begin{matrix} \frac{d H_{i}}{d V} = C_{P_{i}} \frac{d T}{d V} & (12-4) \end{matrix}$

Substituting Equations (12-3) and (12-4) into Equation (12-2), we obtain

$U a (T_{a} - T) - \underset{\overset{⏟}{Δ H_{Rx}}}{Σ v_{i} H_{i}} (- r_{A}) - Σ F_{i} C_{P_{i}} \frac{d T}{d V} = 0$

Rearranging, we arrive at

$\begin{matrix} \frac{d T}{d V} = \frac{\overset{\begin{matrix} Q_{g} \\ \begin{matrix} \begin{matrix} Heat \end{matrix} \\ “ Generated ” \end{matrix} \end{matrix}}{\overset{⏞}{r_{A} Δ H_{Rx}}} - \overset{\begin{matrix} Q_{r} \\ \begin{matrix} \begin{matrix} Heat \end{matrix} \\ “ Removed ” \end{matrix} \end{matrix}}{\overset{⏞}{U a (T - T_{a})}}}{Σ F_{i} C_{P_{i}}} \end{matrix} \begin{matrix} (12-5) \end{matrix}$

which is Equation (T11-1.G) in Table 11-1 on pages 547–549.

This form of the energy balance will also be applied to multiple reactions.

$\begin{matrix} \frac{d T}{d V} = \frac{Q_{g} - Q_{r}}{Σ F_{i} C_{P_{i}}} \end{matrix} \begin{matrix} (T11-1.G) \end{matrix}$

where

$\begin{matrix} Q_{g} = r_{A} Δ H_{Rx} \equiv (- r_{A}) (- Δ H_{Rx}) & (12 - 5 a) \end{matrix}$

$\begin{matrix} Q_{r} = U a (T - T_{a}) & (12 - 5 b) \end{matrix}$

To help remember that Q_r is the “heat” removed from the reacting mixture, we note the driving force is from “T” to “T_a”, that is, Ua (T – T_a).

For exothermic reactions, Q_g will be a positive number. We note that when the heat “generated,” Q_g, is greater than the heat “removed,” Q_r (i.e., Q_g > Q_r), the temperature will increase down the reactor. When Q_r > Q_g, the temperature will decrease down the reactor.

For endothermic reactions ΔH_Rx will be a positive number and therefore the heat generated, Q_g in Equation (12-5a), will be a negative number. The heat removed, Q_r in Equation (12-5b), will also be a negative number because heat is added rather than removed T_a > T. See Chapter 12 Additional Material on the Web site for a sample calculation on endothermic reactions (http://www.umich.edu/~elements/6e/12chap/obj.html#/additional-materials/). This point is also illustrated in Example 12-2 for the case of constant ambient temperature.

We continue development of our algorithm by noting Equation (12-5) is coupled with the mole balances on each species, Equation (12-3). Next, we express r_A as a function of either the concentrations for liquid systems or molar flow rates for gas systems, as described in Chapter 6. We will use the molar flow rate form of the energy balance for membrane reactors and also extend this form to multiple reactions.

We could also write Equation (12-5) in terms of conversion by recalling F_i = F_A0(Θ_i + v_iX) and substituting this expression into the denominator of Equation (12-5).

PFR energy balance

$\begin{matrix} \frac{d T}{d V} = \frac{\overset{Q_{g}}{\overset{⏞}{r_{A} Δ H_{Rx}}} - \overset{Q_{r}}{\overset{⏞}{U a ({T-T}_{a})}}}{F_{A0} (Σ Θ_{i} C_{P_{i}} + Δ C_{p} X)} = \frac{Q_{g} - Q_{r}}{Σ F_{i} C_{i}} \end{matrix} \begin{matrix} (12-6) \end{matrix}$

For a packed-bed reactor dW = ρ_b dV where ρ_b is the bulk density

PFR energy balance

$\begin{matrix} \frac{d T}{d W} = \frac{\overset{Q_{g}^{'}}{\overset{⏞}{r_{A}^{'} Δ H_{Rx}}} - \frac{\overset{Q_{r}^{'}}{\overset{⏞}{U a (T - T_{a})}}}{ρ_{b}}}{Σ F_{i} C_{P_{i}}} \end{matrix} \begin{matrix} (12-7) \end{matrix}$

Equations (12-6) and (12-7) are also given in Table 11-1 as Equations (T11-1.D) and (T11-1.F). As noted earlier, having gone through the derivation to these equations, it will be easier to apply them accurately to CRE problems with heat effects.

12.1.2 Applying the Algorithm to Flow Reactors with Heat Exchange

We continue to use the algorithm described in the previous chapters and simply add a fifth building block, the energy balance.

Gas Phase

If the reaction is in gas phase and pressure drop is included, there are four differential equations that must be solved simultaneously. The differential equation describing the change in temperature with volume (i.e., distance) as we move down the reactor

Energy balance

$\begin{matrix} \frac{d T}{d V} = g ({X,T,T}_{a}) & (A) \end{matrix}$

must be coupled with the mole balance

Mole balance

$\begin{matrix} \frac{d T}{d V} = \frac{- r_{A}}{F_{A0}} = f (X,T, p) & (B) \end{matrix}$

and with the pressure-drop equation

Pressure drop

$\begin{matrix} \frac{d p}{d V} = - h (p,X,T) & (C) \end{matrix}$

and solved simultaneously. If the temperature of the heat-exchange fluid, T_a, varies down the reactor, we must add the energy balance on the heat-exchange fluid. In the next section, we will derive the following equation for co-current heat transfer

Heat exchanger

$\begin{matrix} \frac{{d T}_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{C0} C_{P_{C0}}} & (D) \end{matrix}$

Numerical integration of the coupled differential equations (A) to (D) is required.

along with the equation for countercurrent heat transfer. A variety of software packages (e.g., Polymath) can be used to solve these four coupled differential equations: (A), (B), (C), and (D).

Liquid Phase

For liquid-phase reactions, the rate is not a function of total pressure, so our mole balance is

$\begin{matrix} \frac{d X}{d V} = \frac{- r_{A}}{F_{A0}} = f (X,T) & (E) \end{matrix}$

Consequently, we need to only solve three equations (A), (D), and (E) simultaneously.

12.2 Balance on the Heat-Transfer Fluid

12.2.1 Co-Current Flow

The heat-transfer fluid will be a coolant for exothermic reactions and a heating medium for endothermic reactions. If the flow rate of the heat-transfer fluid is sufficiently high with respect to the heat released (or absorbed) by the reacting mixture, then the heat-transfer fluid temperature will be virtually constant along the length of the reactor. In the material that follows, we develop the basic equations for a coolant to remove heat from exothermic reactions; however, these same equations apply to endothermic reactions where a heating medium is used to supply heat.

We now carry out an energy balance on the coolant in the annulus between R₁ and R₂, and axially between V and V + ΔV, as shown in Figure 12-2. The mass flow rate of the heat-exchange fluid (e.g., coolant) is ${\dot{m}}_{c}$ . We will consider the case when the reactor is cooled and the outer radius of the coolant channel R₂ is insulated. Recall that by convention is the heat added to the system.

A figure shows a cylindrical heat exchanger with inner and outer pipes. — **Figure 12-2** Co-current, double-pipe heat exchanger.

The figure shows a cylindrical heat exchanger where heat is added through the central section. The input stream flows from left to right. The temperature of the input stream through the outer pipe is labeled as T subscript a0. The magnitude and temperature of the input stream through the inner pipe are labeled as F subscript A0 and T subscript 0 respectively. The volume at the two ends of the central section is marked as V on the left and V plus delta V on the right. The inner and outer pipes are heat-transfer fluid reactants and are marked as R1 and R2 respectively. Additional heat, Q (over dot) is supplied through the inner pipe.

For co-current flow, the reactant and the coolant flow in the same direction.

A figure shows a section of the double-pipe heat exchanger. — The figure shows the input and output values of the stream through a section of the double pipe heat exchanger. The mass and enthalpy of the input stream through the outer pipe are labeled as m (over dot) subscript c and H subscript c respectively. The magnitude and temperature of the input stream through the inner pipe are labeled as F subscript A and T respectively. The overall volume difference between the start and end of the double pipe is marked as V and V plus delta 'V' respectively. The mass and enthalpy of the output stream through the outer pipe are labeled as m (overdot) subscript c and H subscript c respectively. The magnitude and temperature of the output stream through the inner pipe are labeled as F subscript A and T respectively.

The energy balance on the coolant in the volume between V and (V + ΔV) is

Coolant energy balance

$\begin{matrix} [\begin{matrix} \begin{matrix} \begin{matrix} Rate of engery \end{matrix} \\ in at V \end{matrix} \end{matrix}] & - & [\begin{matrix} \begin{matrix} \begin{matrix} Rate of energy \end{matrix} \\ out at V + Δ V \end{matrix} \end{matrix}] & + & [\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} Rate of heat added \end{matrix} \\ by conduction through \end{matrix} \end{matrix} \\ the inner wall \end{matrix} \end{matrix}] & = 0 \\ {\dot{m}}_{c} H_{c} |_{V} & - & {\begin{matrix} {\dot{m}}_{c} H_{c} | \end{matrix}}_{V} & + & U a (T - T_{a}) Δ V & = 0 \end{matrix}$

where T_a is the temperature of the heat transfer fluid, that is, coolant, and T is the temperature of the reacting mixture in the inner tube.

Dividing by ΔV and taking limit as ΔV → 0

$\begin{matrix} - {\dot{m}}_{c} \frac{d H_{c}}{d V} +U a (T - T_{a}) = 0 & (12-8) \end{matrix}$

Analogous to Equation (12-4), the change in enthalpy of the coolant can be written as

$\begin{matrix} \frac{{d H}_{c}}{d V} = C_{P_{c}} \frac{{d T}_{a}}{d V} & (12-9) \end{matrix}$

The variation of coolant temperature T_a down the length of reactor is

$\begin{matrix} \frac{{d T}_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{c} C_{P_{c}}} \end{matrix} \begin{matrix} (12-10) \end{matrix}$

The equation is valid whether the heat-transfer fluid is a coolant or a heating medium.

Typical heat-transfer fluid temperature profiles for both exothermic and endothermic reactions when the heat-transfer fluid enters at T_a₀ are shown in Figure 12-3, parts (a) and (b) respectively.

Two graphs depict the fluid temperature profiles for coolant and a heating medium. — **Figure 12-3** Heat-transfer fluid temperature profiles for co-current heat exchanger. **(a)** Coolant. **(b)** Heating medium.

Two graphs compare the volume and temperature. The entry temperature of the heat-transfer fluid is T subscript a 0. In graph (a), the curve starts at the lowest value of the entry temperature, and as the volume increases the temperature increases. This graph indicates an exothermic reaction. In graph (b), the curve starts at the highest value of the entry temperature. Here, as the volume increases the temperature decreases. This graph indicates an an endothermic reaction.

12.2.2 Countercurrent Flow

In countercurrent heat exchange, the reacting mixture and the heat-transfer fluid (e.g., coolant) flow in opposite directions. At the reactor entrance, V = 0, the reactants enter at temperature T₀, and the coolant exits at temperature T_a₂. At the end of the reactor, the reactants and products exit at temperature T, while the coolant enters at T_a₀.

A figure shows a cylindrical heat exchanger with inner and outer pipes where the streams flow in opposite directions. — **Figure 12-4** Countercurrent, double-pipe heat exchanger.

The figure shows the countercurrent flow of the coolants and reactants in a double-pipe heat exchanger. The reactants flow through the inner pipe, from left to right. The magnitude and temperature of the reactants at the input are F subscript A0 and T subscript 0 respectively. The magnitude and temperature of the reactants at the output are F subscript A and T respectively. The value of V at the entry and exit of the reactant is labeled as 0 and V subscript f respectively. The coolants flow through the outer pipe, from right to left. The mass and temperature of the coolants at the input are m (over dot) subscript c and T subscript A 0 respectively. The exit temperature of the coolants is T subscript a2. The central section of the heat exchanger is outlined. The overall volume between the two ends of the central section is marked as V on the left and V plus delta V on the right.

Again, we write an energy balance on the coolant over a differential reactor volume to arrive at

$\begin{matrix} \frac{{d T}_{a}}{d V} = \frac{U a (T_{a} - T)}{{\dot{m}}_{c} C_{P_{c}}} \end{matrix} \begin{matrix} (12-11) \end{matrix}$

At the entrance, V = 0 X = 0 and T_a = T_a₂.

At the exit, V = V_f T_a = T_a₀.

We note that the only difference between Equations (12-10) and (12-11) is a minus sign, that is, (T – T_a) versus (T_a – T).

The solution to a countercurrent flow problem to find the exit conversion and temperature requires a trial-and-error procedure, as shown in Table 12-1.

Trial-and-error procedure required

Two graphs derive the coolant entry temperatures for two different feed entrance temperatures. — **TABLE 12-1 PROCEDURE TO SOLVE FOR THE EXIT CONDITIONS FOR PFRS WITH COUNTERCURRENT HEAT EXCHANGE**

12.3 Examples of the Algorithm for PFR/PBR Design with Heat Effects

We now have all the tools to solve reaction engineering problems involving heat effects in PFRs and PBRs for the cases of both constant and variable coolant temperatures.

Table 12-2 gives the algorithm for the design of PFRs and PBRs with heat exchange: In Case A Conversion is the reaction variable and in Case B Molar-Flow-Rates are the reaction variables. The procedure in Case B must be used to analyze multiple reactions with heat effects.

TABLE 12-2 ALGORITHM FOR PFR/PBR DESIGN FOR GAS-PHASE REACTIONS WITH HEAT EFFECTS

The elementary gas-phase reaction

$A+B \vec{\leftarrow} 2 C$

is to be carried out in a PBR with a co-current heat exchanger.

A. Conversion as the Reaction Variable

1. Mole Balance:

$\begin{matrix} \frac{d X}{d W} = \frac{- r_{A}^{'}}{F_{0}} & (T12-2.1) \end{matrix}$

2. Rate Law:

$\begin{matrix} - r_{A}^{'} = k_{1} (C_{A} C_{B} - \frac{C_{c}^{2}}{K_{C}}) & (T12-2.2) \end{matrix}$

$\begin{matrix} k = k_{1} (T_{1}) \exp [\frac{E}{R} (\frac{1}{T_{1}} - \frac{1}{T})] & (T12-2.3) \end{matrix}$

$\begin{matrix} \begin{matrix} for Δ C_{P} ≅ 0. & K_{C} = K_{C2} (T_{2}) exp [\frac{Δ H_{Rx}^{\circ}}{R} (\frac{1}{T_{2}} - \frac{1}{T})] \end{matrix} & (T12-2.4) \end{matrix}$

Again we note that because δ ≡ 0, K_C ≡ K_e

3. Stoichiometry (gas phase):

$\begin{matrix} C_{A} = \frac{C_{A0 (1 - X)}}{(1+ ε X)} \frac{T_{0}}{T} p & (S4-9) \end{matrix}$

$ε = y_{A0} δ = \frac{1}{3} (2-1-1) = 0$

$\begin{matrix} C_{A} = C_{A0} (1 - X) \frac{T_{0}}{T} p & (T12-2.5) \end{matrix}$

$\begin{matrix} C_{B} = C_{A0} (Θ_{B} - X) \frac{T_{0}}{T} p & (T12-2.6) \end{matrix}$

$\begin{matrix} C_{C} = 2 C_{A0} X \frac{T_{0}}{T} p & (T12-2.7) \end{matrix}$

$\begin{matrix} C_{I} = C_{I0} \frac{T_{0}}{T} p & (T12-2.8) \end{matrix}$

$\begin{matrix} \frac{d p}{d W} = - \frac{α}{2 p} (1+ ε X) \frac{T}{T_{0}} & (5-30) \end{matrix}$

δ = (2 - 1 - 1) = 0, ∴ ε = 0

$\begin{matrix} \frac{d p}{d W} = - \frac{α}{2 p} \frac{T}{T_{0}} & (T12-2.9) \end{matrix}$

An illustration depicts footsteps, and represents the phrase "following the algorithm".

Mole Balance
Rate Law
Stoichiometry
Energy Balance
Parameters
Solution
Analysis

Combine Rate Law and Stoichiometry to find X_e

$\begin{matrix} r_{A}^{'} = k_{1} C_{A0}^{2} [(1 - X) (Θ_{B} - X) - \frac{4 X^{2}}{K_{C}}] {(\frac{T_{0}}{T} p)}^{2} & (T12-2.10) \end{matrix}$

at equilibrium and $- r_{A}^{'} = 0$ and X = X_e

Solving for X_e

$\begin{matrix} X_{e} = \frac{(Θ_{B} + 1) K_{C} - {[{((Θ_{B} + 1) K_{C})}^{2} - 4 K_{C} Θ_{B} (K_{C} - 4)]}^{1 / 2}}{2 (K_{C} - 4)} & (T12-2.11) \end{matrix}$

Because we are solving for the temperature as a function of catalyst weight down the reactor, we can use Equation (T12-2.4) to find the equilibrium constant, K_C, profile and then use Equation (T12-2.11) to find the equilibrium conversion profile, X_e.

4. Energy Balances:

$\begin{matrix} Reactor: \frac{d T}{d W} = \frac{Q_{g}^{'} - Q_{r}^{'}}{Σ F_{i} C_{P_{i}}} & (T12-2.12) \end{matrix}$

$\begin{matrix} Q_{g}^{'} = (- r_{A}^{'}) (- Δ H_{Rx}) = (r_{A}^{'} Δ H_{Rx}) & (T12-2.13) \end{matrix}$

$\begin{matrix} Q_{r}^{'} = (\frac{U a}{ρ_{b}}) (T - T_{a}) & (T12-2.14) \end{matrix}$

$\begin{matrix} Σ F_{i} {C_{P}}_{i} = F_{A0} [C_{P_{A} + θ_{B}} C_{P_{B}} + X Δ C_{p}] & (T12-2.15) \end{matrix}$

$\begin{matrix} Co-Current Coolant: \frac{{d T}_{a}}{d W} = \frac{(\frac{U a}{ρ_{b}}) (T - T_{a})}{{\dot{m}}_{c} C_{P_{cool}}} & (T12-2.16) \end{matrix}$

B. Molar Flow Rates as the Reaction Variable

An illustration depicts footsteps, and represents the phrase "following the algorithm".

1. Mole Balance:

$\begin{matrix} \frac{{d F}_{A}}{d W} = r_{A}^{'} & (T12-2.17) \end{matrix}$

$\begin{matrix} \frac{{d F}_{B}}{d W} = r_{B}^{'} & (T12-2.18) \end{matrix}$

$\begin{matrix} \frac{{d F}_{C}}{d W} = r_{C}^{'} & (T12-2.19) \end{matrix}$

$\begin{matrix} F_{I} = F_{I 0} & (T12-2.20) \end{matrix}$

2. Rate Law (elementary reaction):

$\begin{matrix} - r_{A}^{'} = k_{1} (C_{A} C_{B} - \frac{C_{C}^{2}}{K_{C}}) & (T12-2.2) \end{matrix}$

$\begin{matrix} k = k_{1} (T_{1}) exp [\frac{E}{R} (\frac{1}{T_{1}} - \frac{1}{1})] & (T12-2.3) \end{matrix}$

$\begin{matrix} k_{C} = k_{C2} (T_{2}) exp [\frac{Δ H_{Rx}^{\circ}}{R} (\frac{1}{T_{2}} - \frac{1}{T})] & (T12-2.4) \end{matrix}$

3. Stoichiometry (gas phase):

$\begin{matrix} r_{B}^{'} = r_{A}^{'} & (T12-2.21) \end{matrix}$

$\begin{matrix} r_{C}^{'} = {- 2 r}_{A}^{'} & (T12-2.22) \end{matrix}$

$\begin{matrix} C_{A} = C_{T0} \frac{T_{A}}{F_{T}} \frac{T_{0}}{T} p & (T12-2.23) \end{matrix}$

$\begin{matrix} C_{B} = C_{T0} \frac{T_{B}}{F_{T}} \frac{T_{0}}{T} p & (T12-2.24) \end{matrix}$

$\begin{matrix} C_{C} = C_{T0} \frac{T_{C}}{F_{T}} \frac{T_{0}}{T} p & (T12-2.25) \end{matrix}$

$\begin{matrix} F_{T} = F_{A} + F_{B} + F_{C} + F_{I} & (T12-2.26) \end{matrix}$

$\begin{matrix} \frac{d p}{d w} = \frac{α}{2 p} \frac{F_{T}}{F_{T0}} \frac{T}{T_{0}} & (T12-2.27) \end{matrix}$

4. Energy Balances:

Reactor:

$\begin{matrix} \begin{matrix} \frac{d T}{d W} = \frac{Q_{g}^{'} - Q_{r}^{'}}{Σ F_{i} C_{P_{i}}} = \frac{(r_{A}^{'} Δ H_{Rx}) - (\frac{U a}{ρ_{b}}) (T - T_{a})}{F_{A} C_{P_{A}} + F_{B} C_{P_{B}} + F_{C} C_{P_{C}} + F_{I} C_{P_{I}}} \end{matrix} & (T12-2.28) \end{matrix}$

Heat Exchangers: Same as for A. Conversion as the Reaction Variable

Case A: Conversion as the Independent Variable Example Calculation

5. Parameter Evaluation:

Now we enter all the explicit equations with the appropriate parameter values.

k₁, E, R, C_T0, T_a, T₀, T₁, T₂, K_C2, Θ_B, Θ_I, $Δ H_{Rx}^{°}$ , C_{P_A}, C_{P_B}, C_{C_A}, Ua, ρ_b

Base Case for Parameter Values

$\begin{matrix} \begin{matrix} \begin{matrix} E = 25 kcal/mol \\ Δ H_{Rx} = - 20 kcal/mol \end{matrix} & C_{P_{I}} = 40 cal /mol K \\ k_{1} = 0.004 \frac{{dm}^{6}}{mol \cdot kg \cdot s} @310 K & (\frac{Ua}{ρ_{b}}) = Uarho = 0.5 \frac{cal}{kg \cdot s \cdot K} \\ K_{C2} = 1000 @ 303 K & {\dot{m}}_{c} = 1000 g / s \end{matrix} \end{matrix}$

$\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} α = 0.0002 / kg \end{matrix} \\ F_{A0} = 5 mol/s \end{matrix} \end{matrix} \end{matrix} \\ C_{P_{Cool}} = 0.3 {mol/dm}^{3} \\ C_{P_{A}} = C_{P_{B}} = C_{P_{C}} = 20 cal/mol/K \end{matrix} & \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} C_{P_{Cool}} = 18 cal/g/K \end{matrix} \\ Θ_{B} = 1 \end{matrix} \end{matrix} \\ Θ_{1} = 1 \end{matrix} \end{matrix}$

with initial values T₀ = 330 K, T_a₀ = 320 K, p = 1, and X = 0 at W = 0 and final values: W_f = 5000 kg.

6. Solution:

Equations (T12-2.1)–(T12-2) are entered into the Polymath program along with the corresponding parameter values.

Differential equations

1 d(Ta)/d(W) = Uarho*(T-Ta)/(mc*Cpcool)

2 d(p)/d(W) = -alpha/2*(T/To)/p

3 d(T)/d(W) = (Qg-Qr)/(Fao*sumcp)

4 d(X)/d(W) = -ra/Fao

Explicit equations

1 Hr = -20000

2 thetaB = 1

3 Uarho = 0.5

4 alpha = .0002

5 To = 330

6 Qr = Uarho*(T-Ta)

7 mc = 1000

8 Cpcool = 18

9 Kc = 1000*(exp(Hr/1.987*(1/303-1/T)))

10 Fao = 5

11 thetaI = 1

12 CpI = 40

13 CpA = 20

14 yao = 1/(1+thetaB+thetaI)

15 CpB = 20

16 Cto = 0.3

17 Ea = 25000

18 Xe = ((thetaB+1)*Kc-(((thetaB+1)*Kc)^2-4*(Kc-4)*(Kc*thetaB))^0.5)/(2*(Kc-4))

19 k = .004*exp(Ea/1.987*(1/310-1/T))

20 Cao = yao*Cto

21 Cc = Cao*2*X*p*To/T

22 sumcp = (thetaI*CpI+CpA+thetaB*CpB)

23 Ca = Cao*(1-X)*p*To/T

24 Cb = Cao*(thetaB-X)*p*To/T

25 ra = -k*(Ca*Cb-Cc^2/Kc)

26 Qg = (-ra)*(-Hr)

http://www.umich.edu/~elements/6e/live/chapter12/T12-3/LEP-T12-2.pol

Calculated values of DEQ variables

A graph illustrates the temperature, conversion, and reaction rate profiles inside a reactor. — Three graphs are shown. The first graph shows the temperature profiles of an 'A' plus B reversible 2C exothermic gas-phase reaction. The vertical axis representing the temperature of the gas ranges from 320 to 390 in increments of 7. The horizontal axis representing catalyst (W) in kilogram ranges from 0 to 4500, in increments of 450 kilograms. The initial temperature of the gas inside a reactor is 330.0 for 0 kilograms weight of the catalyst. After that, as the reaction proceeds, the temperature of the gas in the reactor (T) increases sharply from 330.0 to 383.0 at approximately 1350 kilograms of catalyst. From the peak, the temperature of the gas gradually decreases to 327.0 for W equals 4500 kilograms of catalyst. The gradient for temperature T a starts at 0 and gradually increases to 322.0 for W equals 4500 kilograms of catalyst. The second graph shows the conversion of an 'A' plus B reversible 2C exothermic gas-phase reaction. The vertical axis representing the temperature ranges from 0.0 to 1.0 in increments of 0.1. The horizontal axis representing the weight of the catalyst in kilogram ranging from 0 to 4500 in increments of 450. Conversion X increases from 0 at 0 kilograms of catalyst and approaches an equilibrium of 0.32 at approximately 1120 kilograms of catalyst. X increases to 0.5 at W equal 4500 kilograms along with an increase in X e to 0.8 at W equals 4500 kilograms of catalyst. Product P shows a downward curve from 1.0 to approximately 0.2 at W equals 4500 kilograms of catalyst. Note: all mentioned values are approximate. The third graph shows the temperature profiles of an 'A' plus B reversible 2C exothermic gas-phase reaction. The horizontal axis representing the weight in kilograms ranges from 450 to 4500 in intervals of 450. The vertical axis representing the temperature ranges from 0.0E+0 to 8.0E-3. The rate of conversion starts from a 4.0E-4 and increases to 7.8E-3 for a weight of 120 kilograms. Beyond that, the rate decreases gradually and stops after 4050 weight in kilograms. Note: all mentioned values are approximate.

	Variable	Initial value	Final value
1	alpha	0.0002	0.0002
2	Ca	0.1	0.0111092
3	Cao	0.1	0.1
4	Cb	0.1	0.0111092
5	Cc	0	0.0255273
6	CpA	20.	20.
7	CpB	20.	20.
8	Cpcool	18.	18.
9	CpI	40.	40.
10	Cto	0.3	0.3
11	Ea	2.5E+04	2.5E+04
12	Fao	5.	5.
13	Hr	-2.0E+04	-2.0E+04
14	k	0.046809	0.0303238
15	Kc	66.01082	93.4225
16	mc	1000.	1000.
17	p	1.	0.2360408
18	Qg	9.361802	0.0706176
19	Qr	5.	1.615899
20	ra	-0.0004681	-3.531E-06
21	sumcp	80.	80.
22	T	330.	326.2846
23	Ta	320.	323.0528
24	thetaB	1.	1.
25	thetaI	1.	1.
26	To	330.	330.
27	Uarho	0.5	0.5
28	W	0	4500.
29	X	0	0.5346512
30	Xe	0.8024634	0.8285547
31	yao	0.3333333	0.3333333

Analysis:

The temperature profiles are shown in Figure T12-2.1. One notes that due to the exothermic heat of reaction, the temperature of the gas in the reactor increases sharply near the entrance then decreases as the reactants are consumed and the fluid is cooled as it moves down the reactor. Figure T12-2.2 shows the conversion X increases and approaches its equilibrium value X_e at approximately W = 1125 kg of catalyst where the reaction rate approaches zero. At this point the conversion cannot increase further unless X_e increases. The increase in X_e occurs because of the cooling of the reactor’s contents by the heat exchanger thus shifting the equilibrium to the right. By the end of the reactor, the equilibrium conversion has been increased to 82.9% and the conversion in the reactor to 53.4%. One notes from Figure T12.2-2, that owing to the cooling and consumption of reactants the conversion does not increase significantly above a catalyst weight of 2000 kg, even though X is much below X_e (see Problem P12-1_A).

The following figures show “representative” profiles that would result from solving the above equations for different parameter values. The reader is encouraged to download the Living Example Problem for Table 12-2 and use Wolfram to vary a number of parameters, as discussed in Problem P12-1_A. Be sure you can explain why these curves look the way they do.

A graph displays the constant Ta in a reversible exothermic reaction inside a reactor with an exchange of heat. — A graph illustrates the temperature profiles in a reversible exothermic reaction. The vertical axis represents the temperature T. The horizontal axis represents the volume V. The constant Ta is a dotted straight line at a constant volume. The curve increases from point T 0 which reaches a peak value and decreases to the lowest value at V. The curve meets the straight line at the end. The graph below illustrates the heat exchange that occurs due to an increase in Xe. The vertical axis represents the conversion X and horizontal axis represents the volume V. It shows two curves where one of the curves increases with the increasing values of X and V. The other curve representing X subscript e, decreases as the volume increases. Both the curve intersect with each other. A graph illustrates the temperature profiles in an an endothermic reaction. The vertical axis represents the temperature T. The horizontal axis represents the volume V. Ta is a constant straight dotted line at a constant volume. The curve that starts from T subscript 0, decreases to the lowest value after which it increases abruptly intersecting the constant line in the end. The graph below illustrates the heat exchange that occurs due to an increase in Xe. The vertical axis represents the conversion X and horizontal axis represents the volume V. It shows two curves where one of the curves increases with the increasing values of X and V. The other curve representing X subscript e, decreases to a certain point as the volume increases. After that, it rises rapidly with the increasing value of V. Both the curve does not meet each other.

Be sure you can explain why these curves look the way they do.

A graph shows the variable temperatures in an exothermic counter current exchange. — A graph illustrates the temperature profiles during an exothermic counter current exchange reaction. The vertical axis represents the temperature. The horizontal axis represents the volume V. Graph for Temperature T is a downward curve at a particular volume and intersects the variable temperature Ta below it at a particular point. It shows two curves where one of the curves representing T increases until it reaches a peak value after which it falls to the lowest value. The other curve representing T subscript 'a' increases to a peak point as the volume increases. After that, it falls rapidly with the increasing value of V. Both the curve intersects with each other at the beginning. The graph below illustrates the countercurrent exchange that occurs due to an increase in Xe. The vertical axis represents the temperature and horizontal axis represents the volume V. It shows two curves where one of the curves representing X, increases with the increasing values of T and V. The other curve representing X subscript e, decreases to a certain point as the volume increases. After that, it rises rapidly with the increasing value of V. Both the curve does not meet each other.

12.3.1 Applying the Algorithm to an Exothermic Reaction

Example 12–1 Butane Isomerization Continued—OOPS!

When plant engineer Maxwell Anthony looked up the vapor pressure at the exit to the adiabatic reactor in Example 11-3, where the temperature is 360 K, he learned the vapor pressure for isobutene was about 1.5 MPa, which is greater than the rupture pressure of the glass vessel the company had hoped to use. Fortunately, when Max looked in the storage shed, he found there was a bank of 10 tubular reactors, each of which was 5 m³. The bank reactors were double-pipe heat exchangers with the reactants flowing in the inner pipe and with Ua = 5,000 kJ/m³·h·K. Max also bought some thermodynamic data from one of the companies he found on the Internet that used Colorimeter experiments to find ΔH_Rx for various reactions. One of the companies had the value of ΔH_Rx for his reaction on sale this week for the low, low price of $25,000.00. For this value of ΔH_Rx the company said it is best to use an initial concentration of A of 1.86 mol/dm³. The entering temperature of the reactants is 305 K and the entering coolant temperature is 315 K. The mass flow rate of the coolant, ${\dot{m}}_{c}$ , is 500 kg/h and the heat capacity of the coolant, C_P_C, is 28 kJ/kg·K. The temperature in any one of the reactors cannot rise above 325 K. Carry out the following analyses with the newly purchased values from the Internet:

A figure shows four tubes connected in parallel. F subscript A D is fed as an input. At the output, conversion 'X' is marked. The phrase, "10 PFR's in parallel" is present below the figure.

Co-current heat exchange: Plot X, X_e, T, T_a, and –r_A, down the length of the reactor.
Countercurrent heat exchange: Plot X, X_e, T, T_a, and –r_A down the length of the reactor.
Constant ambient temperature, T_a: Plot X, X_e, T, and –r_A down the length of the reactor.
Adiabatic operation: Plot X, X_e, T, T_a, and –r_A, down the length of the reactor.
Compare parts (a) through (d) above and write a paragraph describing what you find.

Additional information:

Recall from Example 11-3 that F_T0 = 163 kmol/h and F_A0 = 0.9 F_T0, C_{P_A} = 141 kJ/kmol · K, C_P0 = ΣΘ_iC_P_i = 159 kJ/kmol · K, and data from the company Maxwell got off the Internet are ΔH_Rx = −34500 kJ/kmol with ΔC_P = 0 and C_A0 = 1.86 kmol/m³ and E/R = 7906 K with k = 31.1 h^-1 @ 360 K.

Solution

We shall first solve part (a), the co-current heat exchange case and then make small changes in the Polymath program for parts (b) through (d).

The molar flow rate of A to each of the 10 reactors in parallel

$F_{A0} = (0.9) (163 kmol/h) \times \frac{1}{10} = 14.7 \frac{kmol A}{h}$

The mole balance, rate law, and stoichiometry are the same as in the adiabatic case previously discussed in Example 11-3; that is,

The Algorithm

Mole Balance:

$\begin{matrix} \frac{d X}{d V} = \frac{- r_{A}}{F_{A0}} \end{matrix} \begin{matrix} (E11-3.1) \end{matrix}$

Rate Law and Stoichiometry:

$\begin{matrix} r_{A} = - k C_{A0} [1 - (1 + \frac{1}{K_{C}}) X] \end{matrix} \begin{matrix} (E11-3.7) \end{matrix}$

with

$\begin{matrix} k = 31.1 exp [7906 (\frac{T - 360}{360 T}) h^{- 1}] \end{matrix} \begin{matrix} (E11-3.10) \end{matrix}$

$\begin{matrix} k_{C} = 3.03 exp [\frac{(Δ H_{Rx} / R) (T - 333)}{333 T}] \end{matrix} \begin{matrix} (E11-3.11) \end{matrix}$

Same as Example 11-3

The equilibrium conversion is

$\begin{matrix} X_{e} = \frac{K_{c}}{1 + K_{C}} \end{matrix} \begin{matrix} (T11-3.12) \end{matrix}$

Note: We could substitute for K_C using Equation (E11-3.12) to write the reaction rate as

$\begin{matrix} - r_{A} = k C_{A0} (1 - \frac{X}{X_{e}}) & (E12-1.1) \end{matrix}$

Energy Balance:

The energy balance on the reactor is

$\begin{matrix} \frac{d T}{d V} = \frac{Q_{g} - Q_{r}}{Σ F_{i} C_{P_{i}}} \end{matrix} \begin{matrix} (E12-1.2) \end{matrix}$

$\begin{matrix} Q_{g} = r_{A} Δ H_{Rx} & (E12-1.3) \end{matrix}$

Q_g = r_AΔH_Rx

Q_r = Ua(T - T_a)

$\frac{d T}{d V} = \frac{Q_{g} - Q_{r}}{F_{A0} C_{P0}}$

$\begin{matrix} Q_{r} = U a (T - T_{a}) & (E12-1.4) \end{matrix}$

$\begin{matrix} Σ F_{i} C_{P_{i}} = \underset{\overset{⏟}{C_{P_{0}}}}{F_{A0} Σ Θ_{i} C_{P_{i}}} = F_{A0} C_{P0} & (E12-1.5) \end{matrix}$

Part (a) Co-Current Heat Exchange

We are now going to solve the coupled, ordinary differential and explicit equations (E11-3.1), (E11-3.7), (E11-3.10), (E11-3.11), (E11-3.12), and (E12-1.2), and the appropriate heat-exchange balance using Polymath. After entering these equations we will enter the parameter values. By using co-current heat exchange as our Polymath base case, we only need to change one line in the program for each of the other three cases and solve for the profiles of X, X_e, T, T_a, and –r_A, and not have to reenter the program.

For co-current flow, the balance on the heat transfer fluid is

$\begin{matrix} \frac{d T_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{C} C_{P_{C}}} \end{matrix} \begin{matrix} (E12-1.6) \end{matrix}$

with T₀ = 305 K and T_a = 315 K at V = 0. The Polymath program and solution are shown in Table E12-1.1.

Table E12-1.1 Part (a) CO-CURRENT HEAT EXCHANGE

Differential equations

1 d(Ta)/d(V) = Ua*(T-Ta)/m/Cpc

2 d(X)/d(V) = -ra/Fa0

3 d(T)/d(V) = (Qg-Qr)/(Cpo*Fa0)

Explicit equations

1 Cpc = 28

2 m = 500

3 Ua = 5000

4 deltaH = -34500

5 Qr = Ua*(T-Ta)

6 Ca0 = 1.86

7 Fa0 = 0.9*163*.1

8 Kc = 3.03*exp((deltaH/8.314)*((T-333)/(T*333)))

9 k = 31.1*exp((7906)*(T-360)/(T*360))

10 ra = -k*Ca0*(1-(1+1/Kc)*X)

11 Xe = Kc/(1+Kc)

12 Qg = ra*deltaH

13 Cpo = 159

14 rate = -ra

Calculated values of DEQ variables

	Variable	Initial value	Final value
1	Ca0	1.86	1.86
2	Cpc	28.	28.
3	Cpo	159.	159.
4	deltaH	-3.45E+04	-3.45E+04
5	Fa0	14.67	14.67
6	k	0.5927441	6.80861
7	Kc	9.512006	2.641246
8	m	500.	500.
9	Qg	3.804E+04	4077.238
10	Qr	-5.0E+04	5076.445
11	ra	-1.102504	-0.1181808
12	rate	1.102504	0.1181808
13	T	305.	336.7102
14	Ta	315.	335.6949
15	Ua	5000.	5000.
16	V	0	5.
17	X	0	0.7185996
18	Xe	0.9048707	0.7253687

Figure E12-1.1 shows the profiles for the reactor temperature, T, the coolant temperature, T_a, the conversion, X, the equilibrium conversion, X_e, and the reaction rate, –r_A. Note that at the entrance to the reactor the reactor temperature T is below the coolant temperature, but as reacting mixture moves into and through the reactor the mixture heats up and T becomes greater than T_a. When you load the Poly-math, Python, or Wolfram Living Example Problem (LEP) code on your computer, one of the things you will want to explore is the value of the ambient temperature, T_a, or the entering temperature, T₀, below which the reaction might never take off or “ignite.”

A graph illustrates the temperature, conversion, and reaction rate profiles inside a flow reactor bed with co-current heat exchange. — **Figure E12-1.1** Profiles down the reactor for *co-current heat exchange* **(a)** temperature, **(b)** conversion, **(c)** reaction rate.

Three graphs are shown. The first graph shows the exchange of heat at the entrance to the reactor. The horizontal axis representing the volume in meter squared ranges from 0.00 to 5.00 in intervals of .50 meter cubed. The vertical axis representing the Temperature in Kelvin ranges from 290 to 380 in intervals of 9. Graph for Temperature T starts at 307 at 0 volume and gradually increases to 371 K at an approximate volume of 1.00 meter squared. From there the temperature gradually decreases to 335 K at 5.00 meter squared. Graph for variable temperature T subscript 'a' within the reactor starts at 316 K at 0 meters squared moves parallel to T at 316 K at approximately 0.40 meter squared. T subscript 'a' increases gradually as a slight curve and ends in equilibrium with T at 335 K at 5.00 meter squared. Note: all mentioned values are approximate. The second graph shows the conversion profiles in a flow reactor. The vertical axis representing the Temperature in Kelvin ranges from 0 to 1.0 Kelvin. The horizontal axis representing the volume in meter squared ranges from 0.00 to 5.00 in intervals of 0.50. The vertical axis representing the temperature in Kelvin ranges from 0.0 to 1.0 in intervals of 0.1. The graph for conversion X e starts at 0.9 decreases to 0.4 and meets the increasing curve of X. The curve of X e and X gradually increase and intersect at a temperature of 0.7, volume 4.00 meter squared. Note: all mentioned values are approximate. The third graph shows the reaction rate for a co-current heat exchange inside a nuclear reactor. The horizontal axis represents the volume of reactants in meter cubed ranges from 0 to 5.00 in intervals of 0.50. The vertical axis represents the negative r subscript 'a' in kilo moles per meter cubed h squared ranges from 0 to 30.00, in intervals of 3.0. The reactant concentration starts at 2.0 for a volume of 0.00, gradually increases to 24.0 for a volume of approximately 0.25 meter squared. After that, the concentration decreases as the reaction progresses and reaches 0.00. Note: all mentioned values are approximate.

Analysis: Part (a) Co-Current Exchange: We note that reactor temperature goes through a maximum. Near the reactor entrance, the reactant concentrations are high and therefore the reaction rate is high (cf. Figure E12-1.1(a)) and Q_g > Q_r. Consequently, the temperature and conversion increase with increasing reactor volume while X_e decreases because of the increasing temperature. Eventually, X and X_e come close together (V = 0.95 m³) and the rate becomes very small as the reaction approaches equilibrium. At this point, the reactant conversion X cannot increase unless X_e increases. We also note that when the ambient heat-exchanger temperature, T_a, and the reactor temperature, T, are essentially equal, there is no longer a temperature driving force to cool the reactor. Consequently, the temperature does not change farther down the reactor, nor does the equilibrium conversion, which is only a function of temperature.

Part (b) Countercurrent Heat Exchange

For countercurrent flow, we only need to make two changes in the program. First, multiply the right-hand side of Equation (E12-1.6) by minus one to obtain

$\begin{matrix} \frac{d T_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{C} C_{P_{C}}} & (E12-1.7) \end{matrix}$

Next, we guess T_a at V = 0 and see whether it matches T_a₀ at V = 5 m³. If it doesn’t, we guess again. In this example, we will guess T_a (V = 0) = 340.3 K and see whether T_a = T_a₀ = 315 K at V = 5 m³.

An illustration shows an example of counter current heat exchange. — A cylindrical tube of volume V equals 0 on the left and V equals 5 meters cubed on the right is shown. A current with a temperature 315 K at a conversion rate of X equals 0.76 enters the tube from the left at V equals 5 meters cubed, and exits on the right with a temperature of 340.3 K. A current of temperature 305 K enters the cylindrical tube from the left at volume V equals 0 and exits on the right with a temperature of 318.5 K at a conversion rate of X e equals 0.84.

Images — TABLE E12-1.2 PART **(b)** COUNTERCURRENT HEAT EXCHANGE

Good guess!

At V = 0, we guessed an entering coolant temperature of 340.3 K and found that at V = V_f, the calculation showed that it matched the emerging coolant temperature T_a₀ = 315 K!! (Was that a lucky guess or what?!) The variable profiles are shown in Figure E12-1.2.

A graph illustrates the temperature, conversion, and reaction rate profiles inside a reactor for countercurrent heat exchange. — **Figure E12-1.2** Profiles down the reactor for *countercurrent heat exchange* **(a)** temperature, **(b)** conversion, **(c)** reaction rate.

Three graphs are shown. The first graph shows the temperature at the entrance to the reactor in a heat exchange process. The horizontal axis represents the volume in meter cubed ranges from 0.00 to 5.00 in intervals of .50. The vertical axis representing the temperature in Kelvin ranges from 300 to 400 in intervals of 10. The temperature T starts at 310 at a reactant volume of 0.00 inside the reactor. The temperature gradually increases to 387 for a reactant volume of 0.50. After that, the temperature decreases and ends at 320 for a volume of 5.00 meter squared. Variable temperature T subscript 'a' starts at 340 Kelvin decreases gradually to approximately 315 K for a volume of 5.00 meter squared. T and T subscript 'a' intersect at 341 K at 0.75 meters squared. Note: all mentioned values are approximate. The second graph shows the conversion profiles inside a reactor. The horizontal axis representing the volume in meter cubed ranges from 0.00 to 5.00 in intervals of 0.50. The vertical axis representing the X, X subscript e, ranges from 0.0 to 1.0 in intervals of 0.1. The conversion X starts at 0 K for a reactant volume of 0.00 meter squared. After that, the conversion gradually increases to 0.72 at 5.00 meter squared. X e starts at 0.9 Kelvin for a reactant volume of 0, increases gradually, and meets X at a point 0.35 Kelvin at a reactant volume of 0.50 meter squared. The conversion profiles proceed parallel to one another and do not intersect. Note: all mentioned values are approximate. The third graph depicts the reaction rates inside a reactor. The horizontal axis representing the volume in meter cubed ranges from 0.00 to 5.00 in intervals of 0.50. The vertical axis representing the rate of reaction ranges from 0 to 50 in intervals of 5. The rate of reaction starts at 2 at 0 v, gradually increases as the reaction proceeds and reaches 44 at 0.25 volume meter squared. After that, the reaction rate gradually decreases and reaches 0.00 K for a reactant volume of 5.00 meter squared. Note: all mentioned values are approximate.

We observe that at V = 0.5 m³, X = 0.36 and the rate, –r_A, drops to a low value as X approaches X_e. Because X can never be greater than X_e the reaction will not proceed further unless X_e is increased. In this case, the reactor is cooled, causing the temperature to decrease and resulting in an increase in the equilibrium conversion beyond a reactor volume of V = 0.5 m³.

Analysis: Part (b) Countercurrent Exchange: We note that near the entrance to the reactor, the coolant temperature is above the reactant entrance temperature. However, as we move down the reactor, the reaction generates “heat” and the reactor temperature rises above the coolant temperature. We note that X_e reaches a minimum (corresponding to the reactor temperature maximum) near the entrance to the reactor. At this point (V = 0.5 m³), X cannot increase above X_e. As we move down the reactor, the reactants are cooled and the reactor temperature decreases allowing X and X_e to increase. A higher exit conversion, X, and equilibrium conversion, X_e, are achieved in the countercurrent heat-exchange system than for the co-current system.

Part (c) Constant T_a

For constant T_a, use the Polymath program in part (a), but multiply the right side of Equation (E12-1.6) by zero in the program, that is,

$\begin{matrix} \frac{d T_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{C} C_{P_{C}}} *0 & (E12-1.8) \end{matrix}$

The initial and final values are shown in the Polymath report and the variable profiles are shown in Figure E12-1.3.

A graph illustrates the temperature, conversion, and reaction rate profiles for constant heat-exchange fluid inside a reactor. — **Figure E12-1.3** Profiles down the reactor for *constant heat-exchange fluid temperature T_a*; **(a)** temperature, **(b)** conversion, **(c)** reaction rate.

Three graphs are shown. The first graph shows the temperature of a constant heat-exchanged fluid inside a reactor. The horizontal axis represents the volume in meter cubed ranges from 0.0 to 5.0 in intervals of 0.5. The vertical axis representing the temperature in Kelvin ranges from 300 to 390 in intervals of 9. Temperature T starts at 304 at volume 0, gradually increases to 372 K at a volume of 1.0. After that, the temperature inside the reactor decreases gradually and ends at 318 for a reactant volume of 5.0 meter squared. Variable temperature T subscript a is a straight line starts at a temperature of 317 K for 0.00 volume and ends at 317 K for a volume of 5.00. T a intersects T at 317 K at a volume of approximately 0.25 meter squared. Note: all mentioned values are approximate. The second graph shows the conversion inside a reactor. The horizontal axis represents the volume in meter squared ranges from 0.0 to 5.0 in intervals of 0.5. The vertical axis represents the conversion ranges from 0.0 to 1.0 intervals of 0.1. The curve for conversion X subscript e, starts at 0.9 X at volume 0.00 decreases to 0.44 X and again gradually increases to end at 0.7 X for a volume 5.0. The curve x, starts at 0 gradually increases to 0.5 X for a volume of 1.4. After that, it continues to increase and reaches 0.71 X for a volume of 5.0 meter squared. Note: all mentioned values are approximate. The third graph shows the conversion inside a reactor. The horizontal axis represents the volume in meter squared ranges from 0.0 to 5.0 in intervals of 0.5. The vertical axis represents the rate ranges from negative r subscript A times kilo moles per meter cubed h, ranges from 0.0 to 30.0 intervals of 3. The curve for rate starts at (0, 0), increases to the peak value at (0.8, 23.8), which then decreases to (1.0, 5.0). After that, the curve decreases further and reaches the lowest rate of 0 at 5.0 meters cubed.

We note that for the parameter values chosen, the profiles for T, X, X_e, and –r_A are similar to those for countercurrent flow. For example, as X and X_e become close to each other at V = 1.0 m³, but beyond that point X_e increases because the reactor temperature decreases.

Analysis: Part (c) Constant T_a: When the coolant flow rate is sufficiently large, the coolant temperature, T_a, will be essentially constant. If the reactor volume is sufficiently large, the reactor temperature will eventually reach the coolant temperature, as is the case here. At this exit temperature, which is the lowest achieved in this example (i.e., Part (c)), the equilibrium conversion, X_e, is the largest of the four cases studied in this example.

Part (d) Adiabatic Operation

In Example 11-3, we solved for the temperature as a function of conversion and then used that relationship to calculate k and K_C. An easier way is to solve the general or base case of a heat exchanger for co-current flow and write the corresponding Poly-math program. Next, use Polymath (part (a)), but multiply the parameter Ua by zero, that is,

Ua = 5,000*0

and run the simulation again.

$\frac{d T}{d V} = \frac{Q_{g}}{F_{A0} {C_{}}_{P_{0}}}$ Adiabatic

TABLE E12-1.4 PART (d) ADIABATIC OPERATION

Differential equations

1 d(Ta)/d(V) = Ua*(T-Ta)/(m*Cpc)

2 d(X)/d(V) = -ra/Fa0

3 d(T)/d(V) = (Qg-Qr)/(Cpo*Fa0)

Explicit equations

1 Cpc = 28

2 m = 500

3 Ua = 5000*0

4 deltaH = -34500

5 Qr = Ua*(T-Ta)

6 Ca0 = 1.86

7 Fa0 = 0.9*163*.1

8 Kc = 3.03*exp((deltaH/8.314)*((T-333)/(T*333)))

9 k = 31.1*exp((7906)*(T-360)/(T*360))

10 ra = -k*Ca0*(1-(1+1/Kc)*X)

11 Xe = Kc/(1+Kc)

12 Qg = ra*deltaH

13 Cpo = 159

14 rate = -ra

Calculated values of DEQ variables

	Variable	Initial value	Final value
1	Ca0	1.86	1.86
2	Cpc	28.	28.
3	Cpo	159.	159.
4	deltaH	-3.45E+04	-3.45E+04
5	Fa0	14.67	14.67
6	k	0.5927441	124.2488
7	Kc	9.512006	0.5752071
8	m	500.	500.
9	Qg	3.804E+04	-0.4109228
10	Qr	0	0
11	ra	-1.102504	1.191E-05
12	rate	1.102504	-1.191E-05
13	T	305.	384.2335
14	Ta	315.	315.
15	Ua	0	0
16	V	0	5.
17	X	0	0.3651629
18	Xe	0.9048707	0.3651628

The initial and exit conditions are shown in the Polymath report, while the profiles of T, X, X_e, and –r_A are shown in Figure E12-1.4.

A graph illustrates the temperature, conversion, and reaction rate profiles inside a adiabatic reactor. — **Figure E12-1.4** Profiles down the reactor for *adiabatic reactor*; **(a)** temperature, **(b)** conversion, **(c)** reaction rate.

Three graphs are shown. The first graph shows the adiabatic reactor temperature. The horizontal axis representing the volume in meter cubed ranges from 0.0 to 5.0 in intervals of 0.5. The vertical axis representing the temperature T ranges from 300 to 400 in intervals of 10. The temperature curve starts at 305 for volume 0 increases gradually to 381 for a volume of 1.0. Beyond that, the temperature remains constant at 381 K until equilibrium is reached and ends at 381 for volume 5.0. Note: all mentioned values are approximate. The second graph shows the conversion inside an adiabatic reactor in the presence of increasing temperature. The horizontal axis represents the volume in meter cubed ranges from 0.0 to 5.0 in intervals of 0.5. The vertical axis representing the conversion X ranges from 0.0 to 10 in intervals of 0.1. The curve representing 'x' starts at (0, 0), increases until (1.0, 0.36). The curve representing X subscript e decreases downwards from 0.9 X and meets conversion X at 0.9 meters cubed. Beyond that Xe and X remain constant until the reactor volume 5.0 Note: all mentioned values are approximate. The third graph shows the increase in reaction rate inside an adiabatic reactor. The horizontal axis representing the volume in meter squared ranges from 0 to 5.0 in intervals of 0.5. The vertical axis representing the reaction rate negative r subscript A times kilo mole per meter cubed hour ranges from 0 to 40 in intervals of 4. The reaction rate starts at 2.0 and gradually increases to 36.0 for a volume of 0.75. The reaction rate then gradually decreases to 0.0 for a reactor volume of 1.0. Note: all mentioned values are approximate.

Analysis: Part (d) Adiabatic Operation: Because there is no cooling, the temperature of this exothermic reaction will continue to increase down the reactor until equilibrium is reached, X = X_e = 0.365 at T = 384 K, which is the adiabatic equilibrium temperature. The profiles for X and X_e are shown in Figure E12-1.4(b) where one observes that the adiabatic equilibrium conversion X_e decreases down the reactor because of the increasing temperature until it becomes equal to the reactor conversion (i.e., X ≡ X_e), which occurs at circa 0.9 m³. There is no change in temperature, X or X_e, after this point because the reaction rate is virtually zero and thus the remaining reactor volume serves no purpose.

Finally, Figure E12-1.4(c) shows –r_A increases as we move down the reactor as the temperature increases, reaching a maximum and then decreasing until X and X_e approach each other and the rate becomes virtually zero.

Overall Analysis: This is an extremely important example, as we applied our CRE PFR algorithm with heat exchange to a reversible exothermic reaction. We analyzed four types of heat-exchanger operations. We see that the countercurrent exchanger and the constant T_a cases give the highest conversion and adiabatic operation gives the lowest conversion.

12.3.2 Applying the Algorithm to an Endothermic Reaction

In Example 12-1 we studied the four different types of heat exchanger on an exothermic reaction. In this section we carry out the same study on an endothermic reaction.

Example 12–2 Production of Acetic Anhydride

Jeffreys, in a treatment of the design of an acetic anhydride manufacturing facility, states that one of the key steps is the endothermic vapor-phase cracking of acetone to ketene and methane is²

² G. V. Jeffreys, A Problem in Chemical Engineering Design: The Manufacture of Acetic Anhydride, 2nd ed. London: Institution of Chemical Engineers.

CH₃COCH₃ → CH₂CO + CH₄

The article further states that this reaction is first-order with respect to acetone and that the specific reaction rate is given by

$\begin{matrix} ln k = 34.34 - \frac{34222}{T} & (E12-2.1) \end{matrix}$

where k is in reciprocal seconds and T is in Kelvin. It is desired to feed 7850 kg of acetone per hour to a tubular reactor. The reactor consists of a bank of 1000 one-inch, schedule-40 tubes. We shall consider four cases of heat-exchanger operation. The inlet temperature and pressure are the same for all cases at 1035 K and 162 kPa (1.6 atm) and the entering heating-fluid temperature available is 1250 K. The heat-exchange fluid has a flow rate, ${\dot{m}}_{c}$ , of 0.111 mol/s, with a heat capacity of 34.5 J/mol·K.

A bank of 1000 one-inch, schedule-40 tubes, 1.79 m in length corresponds to 1.0 m³ (0.001 m³/tube = 1.0 dm³/tube) and gives 20% conversion. Ketene is unstable and tends to explode, which is a good reason to keep the conversion low. However, the pipe material and schedule size should be checked to learn whether they are suitable for these temperatures and pressures. In addition, the final design and operating conditions need to be cleared by the safety committee before operation can begin. Also check the Web site, Process Safety Across the Chemical Engineering Curriculum (http://umich.edu/~safeche/index.html).

Case 1 The reactor is operated adiabatically.

Case 2 Constant heat-exchange fluid temperature T_a = 1250 K

Case 3 Co-current heat exchange with T_a₀ = 1250 K

Case 4 Countercurrent heat exchange with T_a₀ = 1250 K

Additional information:

CH₃COCH₃ (A):H°_A(T_R) = –216.67 kJ/mol, C_{P_A} = 163 J/mol · K

CH₂CO (B):H°_B(T_R) = –61.09 kJ/mol, C_{P_B} = 83 J/mol · K

CH₄ (C):H°_C(T_R) = –74.81 kJ/mol, C_{P_C} = 71 J/mol · K

N₂ (I):C_{P_I} = —28.1 J/mol, · K

U = 110 J/S · m² · K

The other parameters are given in Table E12-2.1.

Solution

Let A = CH₃COCH₃, B = CH₂CO, and C = CH₄. Rewriting the reaction symbolically gives us

A → B + C

Gas-phase endothermic reaction examples:

Adiabatic
Heat exchange T_a is constant
Co-current heat exchange with variable T_a
Countercurrent exchange with variable T_a

Algorithm for a PFR with Heat Effects

1. Mole Balance:

$\begin{matrix} \frac{d X}{d V} = \frac{r_{A}}{F_{A0}} & (E12-2.2) \end{matrix}$

2. Rate Law:

$- r_{A} = k C_{A} (E12-2.3)$

Rearranging (E12-2.1)

$\begin{matrix} k = 8.2 \times 10^{14} exp [- \frac{34222}{T}] =3.58 exp [34222 [\frac{1}{1035} - \frac{1}{T}]] & (E12-2.4) \end{matrix}$

3. Stoichiometry (gas-phase reaction with no pressure drop):

$\begin{matrix} \begin{matrix} C_{A} = \frac{C_{A0} (1 - X) T_{0}}{(1 + ϵ X) T} \\ ϵ = y_{A0} δ = 1 (1 + 1 - 1) = 1 \end{matrix} & (E12-2.5) \end{matrix}$

4. Combining yields

$\begin{matrix} - r_{A} = \frac{k C_{A0} (1 - X)}{1 + X} \frac{T_{0}}{T} & (E12-2.5) \end{matrix}$

Before combining Equations (E12-2.2) and (E12-2.6), it is first necessary to use the energy balance to determine T as a function of X.

5. Energy Balance:

Reactor balance

$\begin{matrix} \frac{d T}{d V} = \frac{U a (T_{a} - T) + (r_{A}) [Δ H_{Rx}^{\circ} + Δ C_{P} (T - T_{R})]}{F_{A0} (Σ Θ_{i} C_{P_{i}} + X Δ C_{P})} & (E12-2.7) \end{matrix}$
Heat Exchanger. We will use the heat-exchange fluid balance for co-current flow as our base case. We will then show how we can very easily modify our ODE solver program (e.g., Polymath) to solve for the other cases by simply multiplying the appropriate line in the code by either zero or minus one.

For co-current flow:

$\begin{matrix} \frac{d T_{a}}{d V} = \frac{U a (T - T_{a})}{\dot{m} C_{P_{c}}} & (E12-2.8) \end{matrix}$

6. Calculation of Mole Balance Parameters on a Per Tube Basis:

$\begin{matrix} \begin{matrix} F_{A0} = \frac{7850 kg / h}{58 kg / kmol} \times \frac{1}{1000 Tubes} = 0.135 kmol/h = 0.0376 mol/s \\ C_{A0} = \frac{P_{A0}}{R T} = \frac{162 b kPa}{8.31 \frac{kPa \cdot m^{3}}{kmol \cdot K} (1035 K)} = 0.0188 \frac{kmol}{m^{3}} = 18.8 {mol/m}^{3} \end{matrix} \\ \begin{matrix} \begin{matrix} υ_{0} = \frac{F_{A0}}{C_{A0}} = \frac{0.0376}{0.0188} = 2.0 {dm}^{3} / s, \end{matrix} & V= \frac{1 m^{3}}{1000 tubes} = \frac{0.001 m^{3}}{tube} = \frac{1.0 {dm}^{3}}{tube} \end{matrix} \end{matrix}$

A figure shows four tubes connected in parallel. F subscript A D is fed as an input. At the output, conversion 'X' is marked. The phrase, "bank of 1000 tubes" is present below the figure.

7. Calculation of Energy Balance Parameters:

Thermodynamics:

$Δ_{Rx}^{°} (T_{R}) : At 298 K$ , using the standard heats of formation

$\begin{matrix} Δ H_{Rx}^{\circ} (T_{R}) & = & H_{B}^{\circ} (T_{R}) + H_{C}^{\circ} (T_{R}) - H_{A}^{\circ} (T_{R}) \\ = & (- 61.09) + (- 74.81) - (- 216.67) kJ / mol \\ = & 80.77 k J / mol \end{matrix}$
ΔC_p : Using the mean heat capacities

ΔC_p = C_{P_B} + C_{P_C} – C_{P_A} = (83 + 71 – 163) J/mol · K

ΔC_p = – 9 J/mol · K

$Sum C_{P_{i}} Θ_{i} : Θ_{i} C_{P_{i}} = C_{P_{A}} + Θ_{I} C_{P_{I}} = 163 + 28.1 Θ_{I}, {(\frac{J}{mol \cdot K})}_{}$

Use Wolfram or Python to vary Θ_I in the LEP.

Heat Exchange:

Energy balance. The heat-transfer area per unit volume of pipe is

$\begin{matrix} a = \frac{π D L}{(π D^{2} / 4) L} = \frac{4}{D} = \frac{4}{0.0266 m} = 150 m^{- 1} \\ U = 110 {J/m}^{2} \cdot s \cdot K \end{matrix}$

Combining the overall heat transfer coefficient with the area yields

Ua = 16500 J/m³ · s · K

TABLE E12-2.1 SUMMARY OF PARAMETER VALUES

	Parameter Values
$Δ H_{Rx}^{°} (T_{R}) = 80.77 kJ/mol$	$C_{P_{I}} = 28.1 \frac{J}{mol \cdot K}$	T₀ = 1035 K
F_A0 = 0.0376 mol/s	ΔC_P = =9 J/mol·K	T_R = 298 K
C_PA = 163 J/mol A/K	C_A0 = 18.8 mol/m³	${\dot{m}}_{C} = 0.111 mol/s$
C_{P_Cool} ≡ C_Pc = 34.5 J/mol/K	Ua = 16500 J/m³·s·K	V_f = 0.001 m³

We will solve for all four cases of heat-exchanger operation for this endothermic reaction example in the same way we did for the exothermic reaction in Example 12-1. That is, we will write the Polymath equations for the case of co-current heat exchange and use that as the base case. As noted in the LEP for this example one should use Wolfram or Python to explore the simulation. We will then manipulate the different terms in the heat-transfer fluid balance (Equations 12-10 and 12-11) to solve for the other cases. In the four cases that follow we will take Θ_I = 0, but on the Wolfram and Python LEPs we allow the reader to vary Θ_I. We will start with the adiabatic case where we multiply the heat transfer coefficient in the base case by zero.

Case 1 Adiabatic

We are going to start with the adiabatic case first to show the dramatic effects of how the reaction dies out as the temperature drops. In fact, we are now going to extend the length of each tube to make the total reactor volume 5 dm³ in order to observe this effect of a reaction dying out as well as showing the necessity of adding a heat exchanger. For the adiabatic case, we simply multiply the value of Ua in our Polymath program by zero. No other changes are necessary. For the adiabatic case, the answer will be the same whether we use a bank of 1000 reactors, each a 1-dm³ reactor, or one of 1 m³. To illustrate how an endothermic reaction can virtually die out completely, let’s extend the single-pipe volume from 1 to 5 dm³.

Ua = 16500*0

The Polymath program is shown in Table E12-2.2. Figure E12-2.1 shows the graphical output.

Adiabatic endothermic reaction in a PFR

TABLE E12-2.2 POLYMATH PROGRAM AND OUTPUT FOR ADIABATIC OPERATION

Differential equations

1 d(X)/d(V) = -ra/Fao

2 d(T)/d(V) = (Qg-Qr)/(Fao*(Cpa+X*delCp))

3 d(Ta)/d(V) = Ua*(T-Ta)/(mc*Cpc)

Explicit equations

1 To = 1035

2 Ua = 16500*0

3 Fao = .0376

4 Cpa = 163

5 delCp = -9

6 Cao = 18.8

7 ra = -Cao*3.58*exp(34222*(1/To-1/T))*(1-X)*(To/T)/(1+X)

8 deltaH = 80770+delCp*(T-298)

9 Qg = ra*deltaH

10 Qr = Ua*(T-Ta)

11 mc = .111

12 Cpc = 34.5

13 rate = -ra

Calculated values of DEQ variables

	Variable	Initial value	Final value
1	Cao	1.88	1.88
2	Cpa	163.	163.
3	Cpc	34.5	34.4
4	delCp	-9.	-9.
5	deltaH	7.414E+04	7.414E+04
6	Fao	0.0376	0.0376
7	mc	0.111	0.111
8	Qg	-4.99E+06	-2.79E+04
9	Qr	0	0
10	ra	-67.304	-0.3704982
11	rate	67.304	0.3704982
12	T	1035.	904.8156
13	Ta	1250.	1250.
14	To	1035.	1035.
15	Ua	0	0
16	V	0	0.005
17	X	0	0.2817744

A graph depicts the adiabatic conversion and temperature and reaction rate profiles inside a reactor. — **Figure E12-2.1** *Adiabatic* conversion and temperature **(a)**, and reaction rate **(b)** profiles.

Two graphs are shown. The first graph shows the adiabatic conversion and temperature of an an endothermic reaction inside a reactor. The horizontal axis representing the volume (in dm cubed) ranges from 0.5 to 4 in intervals of 0.5. The vertical axis on the left representing the temperature in Kelvin ranges from 900 to 1150 in intervals of 50. The vertical axis on the right representing the conversion X ranges from 0.1 to 0.5 in intervals of 0.1. The conversion curve X starts at (0.2, 900), which increases with the increasing volume and reaches (4, 0.28). The temperature curve starts at (0.2, 1045), decreases rapidly as the volume increases and reaches (4, 0). Note: all mentioned values are approximate. The second graph shows the reaction rate inside a reactor. The horizontal axis representing the volume in dm cubed ranges from 0.00 to 5.00 in intervals of 1.00. The vertical axis representing the rate, negative r subscript A times mole per meters cubed times second, ranges from 0.0 to 68.0 in intervals of 6.8. The reaction rate starts at 59.8 and gradually decreases with a drop in temperature. The rate dies out after 3.5 dm cubed. Note: the mentioned values are approximate.

Analysis: Case 1 Adiabatic Operation: As temperature drops, so does k and hence the rate, –r_A, drops to an insignificant value. Note that for this adiabatic endothermic reaction, the reaction virtually dies out after 3.5 dm³, owing to the large drop in temperature, and we observe very little conversion is achieved beyond this point. One way to increase the conversion would be to add a diluent such as nitrogen, which could supply the sensible heat for this endothermic reaction. The Wolfram LEP allows the reader to vary the nitrogen in the feed. However, if too much diluent is added, the concentration, and hence the rate, will be quite low. On the other hand, if too little diluent is added, the temperature will drop and virtually extinguish the reaction. How much diluent to add is left as an exercise. Figures E12-2.1(a) and (b) give the reactor volume as 5 dm³ in order to show the reaction “dying out.” However, because the reaction is nearly complete near the entrance to the reactor, that is, –r_A ≅ 0, we are going to study and compare the heat-exchange systems in a 1-dm³ reactor (0.001 m³) in each of the next three cases.

An illustration depicting the death of a reaction in a humorous manner. The reaction rate is shown to be buried and a cross labeled RIP is placed next to the buried reaction rate.

Case 2 Constant heat-exchange fluid temperature, T_a

We make the following changes in our program on line 3 of the base case (a)

$\begin{matrix} \begin{matrix} \frac{{d T}_{a}}{d V} = \frac{U_{a} (T - T_{a})}{\dot{m} C_{P_{c}}} * 0 \\ U a = 16500 {J/m}^{3} / s/K \end{matrix} \\ {and V}_{f} = 0.001 m^{3} \end{matrix}$

The Polymath Program is shown in Table E12-2.3 and the profiles for T, X, and –r_A are shown in Figure E12-2.2, parts (a), (b), and (c) respectively.

TABLE E12-2.3 POLYMATH PROGRAM AND OUTPUT FOR CONSTANT T_a

Differential equations

1 d(X)/d(V) = -ra/Fao

2 d(T)/d(V) = (Qg-Qr)/(Fao*(Cpa+X*delCp))

3 d(Ta)/d(V) = Ua*(T-Ta)/(mc*Cpc)*0

Explicit equations

1 To = 1035

2 Ua = 16500

3 Fao = .0376

4 Cpa = 163

5 delCp = -9

6 Cao = 18.8

7 ra = -Cao*3.58*exp(34222*(1/To-1/T))*(1-X)*(To/T)/(1+X)

8 deltaH = 80770+delCp*(T-298)

9 Qg = ra*deltaH

10 Qr = Ua*(T-Ta)

11 mc = .111

12 Cpc = 34.5

13 rate = -ra

	Variable	Initial value	Final value
1	Cao	1.88	1.88
2	Caa	163.	163.
3	Cpc	34.5	34.5
4	delCp	-9.	-9.
5	deltaH	7.414E+04	7.343E+04
6	Fao	0.0376	0.0376
7	mc	0.111	0.111
8	Qg	-4.99E+06	-1.211E+06
9	Qr	-3.548E+06	-2.242E+06
10	ra	-67.304	-16.48924
11	rate	67.304	16.48924
12	T	1035.	1114.093
13	Ta	1250.	1250.
14	To	1035.	1035.
15	Ua	1.65E+04	1.65E+04
16	V	0	0.001
17	X	0	0.9508067

Endothermic reaction profiles

A graph illustrates the temperature, conversion, and reaction rate profiles for constant heat exchanger fluid temperature. — **Figure E12-2.2** *Profiles for constant heat exchanger fluid temperature*, *T_a*; **(a)** temperature, **(b)** conversion, **(c)** reaction rate.

Three graphs are shown. The first graph shows the temperature inside a reactor at constant heat exchange. The horizontal axis representing the volume in dm cubed ranges from 0.0 to 1.0 in intervals of 0.1. The vertical axis representing the temperature ranges from 1030.0 to 1120.0 in increments of 9. The temperature curve starts at 1037 K for a reactor volume of 0.0, drops to 1030 K at 0.8 volume, gradually increases to 1114 K at 1.0 volume. Note: the mentioned values are approximate. The second graph shows the conversion inside a 1 dm cubed reactor. The horizontal axis representing the volume in dm cubed ranges from 0.0 to 1.0 in intervals of 0.1. The vertical axis representing the conversion X, ranges from 0.0 to 1.0 in intervals of 0.1. The curve starts at 0.0 gradually increases to 0.94 for a reactor volume of 1.0 dm cubed. Note: all mentioned values are approximate. The third graph shows the conversion inside a 1 dm cubed reactor. The horizontal axis representing the volume in dm cubed ranges from 0.0 to 1.0 in intervals of 0.1. The vertical axis representing the rate negative r subscript A times mole per meter cubed times second ranges from 10.0 to 70.0 in intervals of 6. The curve starts at 68.0 at 0.0, gradually decreases to 40.0 for a reactor volume of 0.3 dm cubed, and ends at 18.0 at 1.0. Note: all mentioned values are approximate.

Analysis: Case 2 Constant T_a: Just after the reactor entrance, the reaction temperature drops as the sensible heat from the reacting fluid supplies the energy for the endothermic reaction. This temperature decrease in the reactor also causes the rate of reaction to decrease. As we move farther down the reactor, the reaction rate drops further as the reactants are consumed. Beyond V = 0.08 dm³, the heat supplied by the constant T_a heat exchanger becomes greater than that “consumed” by the endothermic reaction and the reactor temperature rises. In the range between V = 0.2 dm³ and V = 0.6 dm³, the rate decreases slowly owing to the depletion of reactants, which is counteracted, to some extent, by the increase in temperature and hence the rate constant k. Consequently, we are eventually able to achieve an exit conversion of 95%.

Case 3 Co-Current Heat Exchange

The energy balance on a co-current exchanger is

$\frac{{d T}_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{C} C_{P_{C}}}$

with T_a₀ = 1250 K at V = 0.

The Polymath Program is shown in Table E12-2.4 and the variable profiles for T, T_a, X, and –r_A are shown in Figure E12-2.3, parts (a), (b), and (c) respectively. Because the reaction is endothermic, T_a needs to start off at a high temperature at V = 0.

TABLE E12-2.4 POLYMATH PROGRAM AND OUTPUT FOR CO-CURRENT EXCHANGE

Differential equations

1 d(X)/d(V) = -ra/Fao

2 d(T)/d(V) = (Qg-Qr)/(Fao*(Cpa+X*delCp))

3 d(Ta)/d(V) = Ua*(T-Ta)/(mc*Cpc)

Explicit equations

1 To = 1035

2 Ua = 16500

3 Fao = .0376

4 Cpa = 163

5 delCp = -9

6 Cao = 18.8

7 ra = -Cao*3.58*exp(34222*(1/To-1/T))*(1-X)*(To/T)/(1+X)

8 deltaH = 80770+delCp*(T-298)

9 Qg = ra*deltaH

10 Qr = Ua*(T-Ta)

11 mc = .111

12 Cpc = 34.5

13 rate = -ra

Calculated values of DEQ variables

	Variable	Initial value	Final value
1	Cao	1.88	1.88
2	Cpa	163.	163.
3	Cpc	34.5	34.5
4	delCp	-9.	-9.
5	deltaH	7.414E+04	7.459E+04
6	Fao	0.0376	0.0376
7	mc	0.111	0.111
8	Qg	-4.99E+06	-3.654E+05
9	Qr	-3.548E+06	-1.881E+05
10	ra	-67.304	-4.899078
11	rate	67.304	4.899078
12	T	1035.	984.8171
13	Ta	1250.	996.215
14	To	1035.	1035.
15	Ua	1.65E+04	1.65E+04
16	V	0	0.001
17	X	0	0.456201

Analysis: Case 3 Co-Current Exchange: In co-current heat exchange, we see that the heat-exchanger fluid temperature, T_a, decreases rapidly initially and then continues to decrease along the length of the reactor as it supplies the energy to the heat drawn by the endothermic reaction. Eventually T_a decreases to the point where it approaches T and the rate of heat exchange is small; as a result, the temperature of the reactor, T, continues to decrease, as does the rate, resulting in a small conversion. Because the reactor temperature for co-current exchange is lower than that for Case 2 constant T_a, the reaction rate will be lower. As a result, significantly less conversion will be achieved than in the case of constant heat-exchange temperature T_a.

Endothermic reaction profiles

A graph shows the temperature, conversion, and reaction rate profiles inside a reactor for an an endothermic reaction with co-current exchange of heat. — **Figure E12-2.3** *Profiles down the reactor for an endothermic reaction with co-current heat exchange*; (a) temperature, (b) conversion, and (c) reaction rate.

Three graphs are shown. The first graph shows the temperature profiles during an an endothermic reaction inside a reactor of volume 1 dm cubed. The horizontal axis representing the volume in dm cubed ranges from 0.0 to 1.0 in intervals of 0.1. The vertical axis representing the temperature in Kelvin ranges from 980 to 1250 in intervals of 27. The temperature T starts at 1034 K for a volume 0.0 and gradually decreases to end at 981 K at 1 dm cubed. Variable temperature T subscript a starts at 1250 Kelvin decreases gradually and ends at 1000 Kelvin for a reactor volume of 1.0 dm cubed. Note: the mentioned values are approximate. The second graph shows the conversion inside a reactor with a co-current exchange of heat. The horizontal axis representing the volume in dm cubed ranges from 0.0 to 1.0 in intervals of 0.1. The vertical axis representing the conversion X ranges from 0.00 to 0.46 in intervals of 0.04. Conversion X starts from 0.00 and increases gradually as a slight curve to end at 0.44 for a reactor volume of 1.0 dm cubed. Note: the mentioned values are approximate. The third graph shows the reaction rate inside a reactor. The horizontal axis representing the volume in dm cubed ranges from 0.0 to 1.0 in intervals of 0.1. The vertical axis representing the rate negative r subscript A times mole per meter cubed times second ranges from 0.00 to 70.00 in intervals of 7. Rate starts on the y-axis from 66 and gradually decreases to end at 7.00 for a reactor volume of 1 dm cubed. Note: the mentioned values are approximate.

Case 4 Countercurrent Heat Exchange

A figure depicts the countercurrent heat exchange in a reactor. — The figure shows a cylindrical tube of volume V equals 0 on the left and V equals 0.001 meters cubed on the right. The countercurrent heat exchange inside the cylinder is shown to flow in opposite directions. Temperature 1250 Kelvin enters the top of the cylinder from the right and exits the top of the cylinder from the left at 995 Kelvin. Temperature 1035 Kelvin enters the center of the cylinder from the left and exits the center of the cylinder on the left at 1034.7 Kelvin.

For countercurrent exchange, we first multiply the rhs of the co-current heat-exchanger energy balance by –1, leaving the rest of the Polymath program in Table 12-2.5 the same.

$\frac{{d T}_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{C} C_{P_{C}}}$

Next, guess T_a (V = 0) = 995.15 K to obtain T_a0 = 1250 K at V = 0.001m³. (Don’t you believe for a moment 995.15 K was my first guess.) Once this match is obtained as shown in Table E12-2.5, we can report the profiles shown in Figure E12-2.4.

Good guess!

TABLE E12-2.5 POLYMATH PROGRAM AND OUTPUT FOR COUNTERCURRENT EXCHANGE

Differential equations

1. d(X)/d(V) = -ra/Fao

2. d(T)/d(V) = (Qg-Qr)/(Fao*(Cpa+X*delCp))

3. d(Ta)/d(V) = -Ua*(T-Ta)/(mc*Cpc)

Explicit equations are the same as Case 3 Co-Current Heat Exchange.

Endothermic reaction profiles

Analysis: Case 4 Countercurrent Exchange: At the front of the reactor where the reactants enter, that is, V = 0, the reaction takes place very rapidly, drawing energy from the sensible heat of the gas and causing the gas temperature to drop because the heat exchanger cannot supply energy at an equal or greater rate to that being drawn by the endothermic reaction. Additional “heat” is lost at the entrance in the case of countercurrent exchange because the temperature of the exchange fluid, T_a, is below the entering reactor temperature, T. One notes there is a minimum in the reaction rate, –r_A, profile that is rather flat. In this flat region, the rate is “virtually” constant between V = 0.2 dm³ and V = 0.8 dm³, because the increase in k caused by the increase in T is balanced by the decrease in rate brought about by the consumption of reactants. Just past the middle of the reactor, the rate begins to increase slowly as the reactants become depleted and the heat exchanger now supplies energy at a rate greater than the reaction draws energy and, as a result, the temperature eventually increases. This lower temperature coupled with the consumption of reactants causes the rate of reaction to be low in the plateau, resulting in a lower conversion than either the co-current or constant T_a heat exchange cases.

Analysis Summary of 4 Cases. The exit temperatures and conversion are shown in the Summary Table E12-2.6.

TABLE E12-2.6 SUMMARY TABLE

Heat Exclusion	X	T(K)	T_a(K)
Adiabatic	0.28	905	---
Constant T_a	0.95	1114	1250
Co-Current	0.456	985	996
Countercurrent	0.35	1034	995

For this endothermic reaction, the highest conversion is obtained for constant T_a and the lowest conversion is for an adiabatic reaction. These two extremes correspond to the greatest and least amount of heat transfer to the reacting fluid.

AspenTech: Example 12-2 has also been formulated in AspenTech and can be downloaded on your computer directly from the CRE Web site.

A screenshot shows the menu of AspenTech taken from the CRE Website. — Screenshot of the software application AspenTech is shown. It shows the menu bar that includes the following tabs. Mixers or splitters, separations, heat exchangers, columns, reactors, pressure changers, manipulations, solids, user models, and retired. The tab 'reactor' is selected. Icons such as R Stoic, RYield, REquil, RGibbs, RCSTR, RPlug, and RBatch are displayed under this tab. A Dropdown is provided for each of the options. The material streams drop-down is found at the left corner of the screen.

12.4 CSTR with Heat Effects

In this section we apply the general energy balance (Equation (11-22)) to a CSTR at steady state. We then present example problems showing how the mole and energy balances are combined to design reactors operating adiabatically and non-adiabatically.

In Chapter 11 the steady-state energy balance was derived as

$\begin{matrix} \dot{Q} - {\dot{W}}_{s} - F_{A0} Σ Θ_{i} C_{p_{i}} (T - T_{i 0}) - [Δ H_{Rx}^{\circ} (T_{R}) + Δ C_{P} (T - T_{R})] F_{A0} X = 0 \end{matrix} \begin{matrix} (11-28) \end{matrix}$

Recall that Ẇ_s is the shaft work, that is, the work done by a stirrer or mixer in the CSTR on the reacting fluid inside the CSTR. Consequently, because the convention that done by the system on the surroundings is positive, the CSTR stirrer work will be a negative number, for example, ${\dot{W}}_{s} = - 1, 000 J/S$ . (See Problem P12-6_B, a California Professional Engineers’ Exam Problem.)

Note: In many calculations the CSTR mole balance derived in Chapter 2

(F_A0X = –r_AV)

These are the forms of the steady-state balance we will use.

will be used to replace the term following the brackets in Equation (11-28), that is, (F_A0X) will be replaced by (–r_AV) to arrive at Equation (12-12).

Rearranging yields the steady-state energy balance

$\begin{matrix} \dot{Q} - {\dot{W}}_{s} - F_{A0} Σ Θ_{i} C_{p_{i}} (T - T_{i 0}) + (r_{A} V) (Δ H_{Rx}) = 0 \end{matrix} \begin{matrix} (12-12) \end{matrix}$

Although the CSTR is well mixed and the temperature is uniform throughout the reaction vessel, these conditions do not mean that the reaction is carried out isothermally. Isothermal operation occurs when the feed temperature is identical to the temperature of the fluid inside the CSTR.

The $\dot{Q}$ Term in the CSTR

12.4.1 Heat Added to the Reactor, $\dot{Q}$

Figure 12-6 shows the schematics of a CSTR with a heat exchanger. The heat-transfer fluid enters the exchanger at a mass flow rate (e.g., kg/s) at a temperature T_a₁ and leaves at a temperature T_a₂. The rate of heat transfer from the exchanger to the reactor fluid at temperature T is³

³ Information on the overall heat transfer coefficient may be found in J. R. Welty, G. L. Rorrer, and D. G. Foster, Fundamentals of Momentum Heat and Mass Transfer, 6th ed. New Jersey: Wiley, 2015, p. 370.

$\begin{matrix} \dot{Q} = \frac{UA (T_{a 1} - T_{a 2})}{l n [(T - T_{a 1}) / (T - T_{a 2})]} & (12-13) \end{matrix}$

The figure shows the operation of a CSTR tank reactor with a heat exchanger. — **Figure 12-6** CSTR tank reactor with heat exchanger.

A schematic representation of a CSTR reactor with a heat exchanger is shown. Here, T subscript 0 and F subscript A 0 is marked in the inlet pipe. Transfer fluid, T subscript a1 enters the exchanger at a flow rate of m (overdot) subscript c and leaves at T subscript a subscript 2. The rate of change of temperature inside the reactor is T. The cross-sectional diagram of a stirred tank reactor is displayed. The vertical columns on the interior of the tank are marked as a fine-type baffle and the outer frame is marked conventional jacket. Horizontal rods on the outer of the tank is marked as half-pipe jacket.

The following derivations, based on a coolant (exothermic reaction), apply also to heating mediums (endothermic reaction). As a first approximation, we assume a quasi-steady state operation for the coolant flow and neglect the accumulation term (i.e., dT_a/dt = 0). An energy balance on the heat-exchanger fluid entering and leaving the exchanger is

For exothermic reactions (T>T_a₂>T_a₁)

For endothermic reactions (T₁_a>T₂_a>T)

Energy balance on heat-exchanger fluid

$\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} [\begin{matrix} \begin{matrix} \begin{matrix} Rate of \\ energy \end{matrix} \\ in \end{matrix} \\ by flow \end{matrix}] & - \end{matrix} & [\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} Rate of \\ energy \end{matrix} \end{matrix} \\ out \end{matrix} \end{matrix} \\ by flow \end{matrix} \end{matrix}] \end{matrix} & - \end{matrix} & [\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} Rate of \end{matrix} \\ heat transfer \end{matrix} \end{matrix} \\ f r o m exchanger \end{matrix} \end{matrix} \\ to reactor \end{matrix} \end{matrix}] \end{matrix} & = 0 \end{matrix} & (12-14) \end{matrix}$

$\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} {\dot{m}}_{c} C_{P_{c}} (T_{a 1} - T_{R}) & - \end{matrix} & {\dot{m}}_{c} \end{matrix} C_{P_{c}} (T_{a 2} - T_{R}) & - \frac{UA (T_{a 1} - T_{a 2})}{1 n [(T - T_{a 1}) / (T - T_{a 2})]} = 0 \end{matrix} & (12-15) \end{matrix}$

where C_{P_c} is the heat capacity of the heat exchanger fluid and T_R is the reference temperature. Simplifying gives us

$\begin{matrix} \begin{matrix} \dot{Q} = {\dot{m}}_{c} C_{P_{c}} (T_{a 1} - T_{a 2}) = \frac{UA (T_{a 1} - T_{a 2})}{1 n [(T - T_{a 1}) / (T - T_{a 2})]} = 0 \end{matrix} & (12-16) \end{matrix}$

Solving Equation (12-16) for the exit temperature of the heat-exchanger fluid yields

$\begin{matrix} T_{a 2} = T - (T - T_{a 1}) exp (\frac{- UA}{{\dot{m}}_{c} C_{P_{c}}}) \end{matrix} \begin{matrix} (12-17) \end{matrix}$

From Equation (12-16)

$\begin{matrix} \dot{Q} = {\dot{m}}_{c} C_{P_{c}} (T_{a 1} - T_{a 2}) & (12-18) \end{matrix}$

Substituting for T_a₂ in Equation (12-18), we obtain

$\begin{matrix} \dot{Q} = {\dot{m}}_{c} C_{P_{c}} {(T_{a 1} - T) [1 - exp (\frac{- UA}{{\dot{m}}_{c} C_{P_{c}}})]} \end{matrix} \begin{matrix} (12-19) \end{matrix}$

Heat transfer to a CSTR

For large values of the heat-exchanger fluid flow rate, ${\dot{m}}_{c}$ , the exponent will be small and can be expanded in a Taylor series (e^–x = 1 – x + . . .) where second-order terms are neglected in order to give

$\dot{Q} = {\dot{m}}_{c} C_{P_{c}} (T_{a 1} - T) [1 - (\frac{UA}{{\dot{m}}_{c} C_{P}})]$

Then

Valid only for large heat-transfer fluid flow rates!!

$\begin{matrix} \dot{Q} = UA (T_{a} - T) \end{matrix} \begin{matrix} (12-20) \end{matrix}$

where T_a1 ≌ T_a₂ = T_a.

With the exception of processes involving highly viscous materials such as in Problem P12-6_B, a California Professional Engineers’ Exam Problem, the work done by the stirrer can usually be neglected. Setting ${\dot{W}}_{s}$ equal to zero in Equation (11-27), neglecting ΔC_P, in ΔH_Rx substituting for $\dot{Q}$ , and rearranging, we have the following relationship between conversion and temperature in a CSTR:

$\begin{matrix} \frac{UA}{F_{A0}} (T_{a} - T) - Σ Θ_{i} C_{p_{i}} (T - T_{0}) - Δ H_{Rx}^{\circ} X = 0 & (12-21) \end{matrix}$

Solving for X

$\begin{matrix} X = \frac{\frac{UA}{F_{A0}} (T - T_{a}) + Σ Θ_{i} C_{P_{i}} (T - T_{0})}{[- Δ H_{Rx}^{\circ} (T_{R})]} \end{matrix} \begin{matrix} (12-22) \end{matrix}$

Equation (12-22) is coupled with the mole balance equation

$\begin{matrix} V = \frac{F_{A0} X}{- r_{A} (X,T)} \end{matrix} \begin{matrix} (12-23) \end{matrix}$

to design CSTRs.

To help us more easily see the effect of the operating parameters of T, T₀, and T_a, we collect terms and define three parameters: C_P₀, κ, and T_c.

We now will further rearrange Equation (12-21) after letting

$\begin{matrix} C_{P_{0}} = {Σ Θ}_{i} C_{P_{i}} \end{matrix}$

then

$C_{P_{0}} (\frac{UA}{F_{A0} C_{P_{0}}}) T_{a} + C_{P_{0}} T_{0} - C_{P_{0}} (\frac{UA}{F_{A0} C_{p_{0}}} + 1) T - Δ H_{Rx}^{\circ} X = 0$

Let κ and T_C be non-adiabatic parameters defined by

$\begin{matrix} \begin{matrix} κ = \frac{U A}{F_{A0} C_{P_{0}}} \end{matrix} & and & \begin{matrix} T_{c} = \frac{κ T_{a} + T_{0}}{1 + κ} \end{matrix} \end{matrix}$

Non-adiabatic CSTR heat exchange parameters: κ and T_c

Then

$\begin{matrix} - X Δ H_{Rx}^{\circ} = C_{P_{0}} (1 + k) (T - T_{c}) & (12-24) \end{matrix}$

The parameters κ and T_c are used to simplify the equations for non-adiabatic operation. Solving Equation (12-24) for conversion X

$\begin{matrix} X = \frac{C_{P_{0}} (1 + k) (T - T_{c})}{- Δ H_{Rx}^{\circ}} \end{matrix} \begin{matrix} (12-25) \end{matrix}$

Solving Equation (12-24) for the reactor temperature

$\begin{matrix} T = T_{c} + \frac{(- Δ H_{Rx}^{\circ}) (X)}{C_{P_{0}} (1 + k)} \end{matrix} \begin{matrix} (12-26) \end{matrix}$

Table 12-3 shows three ways to specify the design of a CSTR. This procedure for nonisothermal CSTR design will be illustrated by considering a first-order irreversible liquid-phase reaction. To solve CSTR problems of this type, we have three variables X, T, and V and we specify one and then solve for the other two values. The algorithm for working through either cases A (X specified), B (T specified), or C (V specified) is shown in Table 12-3. Its application is illustrated in Example 12-3.

TABLE 12-3 WAYS TO SPECIFY THE SIZING OF A CSTR

The energy balance for Continuous stirred-tank reactor with heat exchange in various forms is depicted. — The depiction has three columns A, B, and C in which A specifies X to find V and T, B specifies T to find X and V, and C specifies V to find X and T. In column A, to find V and T, first the value of T is determined with the given equation T equals T subscript c plus negative delta H R x degrees times X over C P 0 of 1 plus K. In this equation the value of k can be determined by using the equation k equals A e to the power of negative E over R T, then the next step is to calculate negative r A of (X, T). The value of V can be determined by the below-given equation V equals F A0 times X over negative r A. In column B, for finding the value of X and V, First, the value of X E B is calculated with the given equation X E B equals C P 0 times 1 plus k times T minus T c over negative delta H degrees R x. All the other steps are followed as in column 1. In column 3, for finding the values of X and T, at first, a graph is drawn between X E B vs T by using the equation (12-25), the graph resembles an increasing straight-line from T0, then the equation (12-23) is solved whereas the equation for X M B is given as X M B equals tau A exponential of negative E over R T over 1 plus tau A exponential of negative E over R T, then a graph is plotted for X E B and X M B as a function of T, in which the graph resembles a curve for X M B and a straight line for X E B.

Forms of the energy balance for a CSTR with heat exchange

Production, uses, and economics

Example 12–3 Production of Propylene Glycol in an Adiabatic CSTR

Propylene glycol is produced by the hydrolysis of propylene oxide:

The formation of propylene glycol is represented. — Ethylene oxide (C3H6O) is a 3-carbon compound with an Oxygen group connecting the first 2 carbons. This ethylene oxide and water react in the presence of sulfuric acid (H2SO4). The H plus and OH minus ions of the water molecule combine with the first and the second carbons of the oxide to form propylene glycol (C3H8O2).

Over 900 million pounds of propylene glycol were produced in 2010 and the selling price was approximately $0.80 per pound. Propylene glycol makes up about 25% of the major derivatives of propylene oxide. The reaction takes place readily at room temperature when catalyzed by sulfuric acid.

You are the engineer in charge of an adiabatic CSTR producing propylene glycol by this method. Unfortunately, the reactor is beginning to leak, and you must replace it. (You told your boss several times that sulfuric acid was corrosive and that mild steel was a poor material for construction. He wouldn’t listen.) There is a nice-looking, bright, shiny overflow CSTR of 300-gal capacity standing idle in Professor Köttlov’s storage shed at his mountain vacation home. It is glass-lined, and you would like to use it.

We are going to work this problem in lb_m, s, ft³, and lb-moles rather than g, mol, and m³ in order to give the reader more practice in working in both the English and metric systems. Why?? Many plants still use the English system of units.

You are feeding 2500 lb_m/h (43.04 lb-mol/h) of propylene oxide (P.O.) to the reactor. The feed stream consists of (1) an equivolumetric mixture of propylene oxide (46.62 ft³/h) and methanol (46.62 ft³/h), and (2) water containing 0.1 wt % H₂SO₄. The volumetric flow rate of water is 233.1 ft³/h, which is 2.5 times the methanol–P.O. volumetric flow rate. The corresponding molar feed rates of methanol and water are 71.87 and 802.8 lb-mol/h, respectively. The water–propylene oxide–methanol mixture undergoes a slight decrease in volume upon mixing (approximately 3%), but you neglect this decrease in your calculations. The temperature of both feed streams is 58°F prior to mixing, but there is an immediate 17°F temperature rise upon mixing of the two feed streams caused by the heat of mixing. The entering temperature of all feed streams is thus taken to be 7°F (Figure E12-3.1).

The figure shows the manufacturing of propylene glycol in a CSTR. — **Figure E12-3.1** Propylene glycol manufacture in a CSTR.

A CSTR is shown in which the following are marked: The input, here is propylene oxide (F subscript A0) and methanol (F subscript M0), belonging to the first feed stream. Water (F subscript B0), belongs to the second feed streams at 58 degrees Fahrenheit. The temperature, T subscript 0 increases to 75 degrees Fahrenheit when it enters the tank reactor, on mixing. The temperature of the coolant is approximately 85 degrees Fahrenheit. The volume of the fluid inside the reactor is 300 gallons. The output values of T and X are to be found.

The Furusawa Engineering Team in Japan state that under conditions similar to those at which you are operating, the reaction is apparent first-order in propylene oxide concentration and apparent zero-order in excess of water with the specific reaction rate⁴

⁴ T. Furusawa, H. Nishimura, and T. Miyauchi, J. Chem. Eng. Jpn., 2, 95.

k = Ae^–E/RT = 16.96 × 10¹² (e ^–32400/RT) h^–1

The units of E are Btu/lb-mol and T is in °R.

There is an important constraint on your operation. Propylene oxide is a rather low-boiling-point substance. With the mixture you are using, the company’s safety team feels that you cannot exceed an operating temperature of 125°F, or you will lose too much propylene oxide by vaporization through the vent system.

Can you use the idle CSTR as a replacement for the leaking one if it will be operated adiabatically?
If so, what will be the expected conversion of propylene oxide to glycol?

Solution

(All data used in this problem were obtained from the CRC Handbook of Chemistry and Physics unless otherwise noted.) Let the reaction be represented by

A + B → C

where

A is propylene oxide (C_{P_A} = 35 Btu/lb-mol · °F)⁵

⁵ C_{P_A} and C_{P_C} are estimated from the observation that the great majority of low-molecular-weight oxygen-containing organic liquids have a mass heat capacity of 0.6 cal/g · °C ± 15%.

B is water (C_{P_B} = 18 Btu/lb-mol· °F)

C is propylene glycol (C_{P_C} = 46 Btu/lb-mol · °F)

M is methanol (C_{P_M} = 19.5 Btu/lb-mol· °F)

In this problem, neither the exit conversion nor the temperature of the adiabatic reactor is given. By application of the mole and energy balances, we can solve two equations with two unknowns (X and T), as shown on the right-hand pathway in Table 12-3. Solving these coupled equations, we determine the exit conversion and temperature for the glass-lined reactor to see whether it can be used to replace the present reactor.

Mole Balance and Design Equation:

F_A0 – F_A + r_AV = 0

The design equation in terms of X is

$\begin{matrix} V = \frac{F_{A0} X}{- r_{A}} & (E12-3.1) \end{matrix}$
Rate Law:

$\begin{matrix} - r_{A} = {k C}_{A} & (E12-3.2) \end{matrix}$

k = 16.96 10¹² exp[–32400/R/T] h^–1
Stoichiometry (liquid phase, (υ = υ₀):

$\begin{matrix} C_{A} = C_{A0} (1 - X) & (E12-3.3) \end{matrix}$
Combining yields

$\begin{matrix} V = \frac{F_{A0} X}{k C_{A0} (1 - X)} = \frac{υ_{0} X}{k (1 - X)} & (E12-3.4) \end{matrix}$

Solving for X as a function of T and recalling that τ = V/υ₀ gives

$\begin{matrix} X_{MB} = \frac{τ k}{1 + τ k} = \frac{{τ A_{e}}^{- E / RT}}{{1 + τ A_{e}}^{- E / RT}} \end{matrix} \begin{matrix} (E12-3.5) \end{matrix}$

This equation relates temperature and conversion through the mole balance.

Two equations, two unknowns
The energy balance for this adiabatic reaction in which there is negligible energy input provided by the stirrer is

$\begin{matrix} X_{EB} = \frac{Σ Θ_{i} C_{P_{i}} (T - T_{i 0})}{- [Δ H_{Rx}^{\circ} (T_{R}) + Δ C_{p} (T - T_{R})]} \end{matrix} \begin{matrix} (E12-3.6) \end{matrix}$

This equation relates X and T through the energy balance. We see that two equations, Equations (E12-3.5) and (E12-3.6), and two unknowns, X and T, must be solved to find the conversion where X_EB = X_MB = X.
Calculations:

Rather than putting all those numbers in the mole and heat balance equations yourself, you can outsource this task, for a small fee, to Sven Köttlov Consulting Company, located on the third floor of the downtown Market Center Building (called the “MCB”) in Riça, Jofostan. The results of our outsourcing are given below.
1. Evaluate the mole balance terms (C_A0, Θ_i, τ): The total liquid volumetric flow rate entering the reactor is
  
  $\begin{matrix} \begin{matrix} υ_{0} & = υ_{A0} + υ_{MD} + υ_{B0} \\ = 46.62 + 46.62 + 233.1 = 326.3 {ft}^{3} / h \\ V & = 300 gal = {40.1 ft}^{3} \end{matrix} & (E12-3.7) \end{matrix}$
  
  $\begin{matrix} τ = \frac{V}{υ_{0}} = \frac{40.1 {ft}^{3}}{326.3 {ft}^{3} / h} = 0.123 h & (E12-3.8) \end{matrix}$
  
  $\begin{matrix} \begin{matrix} C_{A0} = \frac{F_{A0}}{υ_{0}} = \frac{43.0 1b-mol/h}{326.3 {ft}^{3} /h} \\ = 0.132 {1b-mol/ft}^{3} \end{matrix} & (E12-3.9) \end{matrix}$
  
  I know these are tedious calculations, but someone’s gotta know how to do them.
  
  For methanol: $Θ_{M} = \frac{F_{m0}}{F_{A0}} = \frac{71.87 1b-mol/h}{43.0 1b-mol/h} = 1.67$
  
  For water: $Θ_{M} = \frac{F_{B 0}}{F_{A0}} = \frac{802.8 1b-mol/h}{43.0 1b-mol/h} = 18.65$
  
  The conversion calculated from the mole balance, X_MB, is found from Equation (E12-3.5)
  
  $X_{MB} = \frac{(16.96 \times 10^{12} h^{- 1}) (0.1229 h) exp (- 32400 / 1.987 T)}{1 + (16.96 \times 10^{12} h^{- 1}) (0.1229 h) exp (- 32400 / 1.987 T)}$
  
  Plot X_MB as a function of temperature.
  
  $\begin{matrix} X_{MB} = \frac{(2.084 \times 10^{12}) exp (- 16306 / T)}{1 + (2.084 \times 10^{12}) exp (- 16306 / T)}, {T is in}^{\circ} R \end{matrix} \begin{matrix} (E12-3.10) \end{matrix}$
2. Evaluating the energy balance terms
  
  (1) Heat of reaction at temperature T
  
  $\begin{matrix} \begin{matrix} Δ H_{Rx} (T) = Δ H_{Rx}^{\circ} (T_{R}) + Δ C_{P} (T - T_{R}) \\ Δ C_{P} = C_{P_{C}} - C_{P_{C}} - C_{P_{A}} = 46 - 18 - 35 = - 7 {Btu/lb-mol/}^{\circ} F \end{matrix} & (11-26) \end{matrix}$
  
  $\begin{matrix} {Δ H}_{Rx} = - 36400 - 7 (T - T_{R}) & (E12-3.11) \end{matrix}$
  
  (2) Heat capacity term
  
  $\begin{matrix} \begin{matrix} Σ Θ_{i} C_{P_{i}} & = C_{P_{A}} + Θ_{B} C_{P_{B}} + Θ_{M} C_{P_{M}} \\ = 35 + (18.65) (18) + (1.67) (19.5) \\ = 403.3 Btu/lb-mol \cdot^{\circ} F \end{matrix} & (E12-3.12) \end{matrix}$
  
  $\begin{matrix} \begin{matrix} T_{0} & = T_{00} + Δ T_{mix} = 58^{\circ} F + 17^{\circ} F = 75^{\circ} F \\ = 535^{\circ} R \\ T_{R} & = 68^{\circ} F = 528^{\circ} R \end{matrix} & (E12-3.13) \end{matrix}$
  
  The conversion calculated from the energy balance, X_EB, for an adiabatic reaction is given by Equation (11-29)
  
  $\begin{matrix} X_{EB} = \frac{Σ Θ_{i} C_{P_{i}} (T - T_{i 0)}}{Δ H_{Rx}^{\circ} (T_{R}) + Δ C_{P} (T - T_{R})} & (11-29) \end{matrix}$
  
  Substituting all the known quantities into the energy balance gives us
  
  $X_{EB} = \frac{(403.3 Btu/lb-mol \cdot^{\circ} F) {(T - 535)}^{\circ} F}{- [- 36400 - 7 (T - 528)] Btu/lb-mol}$
  
  $\begin{matrix} X_{EB} = \frac{403.3 {(T - 535)}^{}}{36400 + 7 (T - 528)]} \end{matrix} \begin{matrix} (E12-3.14) \end{matrix}$
  
  Adiabatic CSTR
Solving: There are a number of different ways to solve these two simultaneous algebraic equations (E12-3.10) and (E12-3.14). The easiest way is to use the Polymath nonlinear equation solver. However, to give insight into the functional relationship between X and T for the mole and energy balances, we shall obtain a graphical solution. Here, X is plotted as a function of T for both the mole (X_MB) and energy (X_EB) balances, and the intersection of the two curves gives the solution where both the mole and energy balance solutions are satisfied, that is, X_EB = X_MB. In addition, by plotting these two curves we can learn whether there is more than one intersection (i.e., multiple steady states) for which both the energy balance and mole balance are satisfied. If numerical root-finding techniques were used to solve for X and T, it would be quite possible that you would find only one root when there is actually more than one. If Polymath were used, you could learn whether multiple roots exist by changing your initial guesses in the nonlinear equation solver. We shall discuss multiple steady states further in Section 12-5. We now choose T and then calculate X_MB and X_EB (Table E12-3.1). The calculations for X_MB and X_EB are plotted in Figure E12-3.2. The virtually straight line corresponds to the energy balance, Equation (E12-3.14), and the curved line corresponds to the mole balance, Equation (E12-3.10). We observe from this plot that the only intersection

TABLE E12-3.1 CALCULATIONS OF X_EB AND X_MB AS A FUNCTION OF T

T (°R)	X_MB (Eq. (E12-3.10))	X_EB (Eq. (E12-3.14))
535	0.108	0.000
550	0.217	0.166
565	0.379	0.330
575	0.500	0.440
585	0.620	0.550
595	0.723	0.656
605	0.800	0.764
615	0.860	0.872
625	0.900	0.980

The reactor cannot be used because it will exceed the specified maximum temperature of 585°R.

A graph shows the conversions curves, X subscript EB and X subscript MB. — **Figure E12-3.2** The conversions X_EB and X_MB as a function of temperature.

In the graph, the vertical axis representing conversion, X ranges from 0.1 to 1.0 in increments of 0.1. The horizontal axis representing T (Rankine scale) ranges from 535 to 635 in increments of 10. Two curves are shown for X subscript EB and X subscript MB. An increasing line representing X subscript EB is drawn from (535, 0) to (613, 0.83). An increasing curve is drawn from (535, 0.1) and (613, 0.83). The curve starts at (535, 0.1) and increases until it reaches (645, 0.9). The curves intersect at (613, 0.83). Dashed lines are drawn from this point to both the axes. Note: all values are approximate.

An icon for an LEP slider is shown.

point is at 83% conversion and 613=R. At this point, both the energy balance and mole balance are satisfied. Because the temperature must remain below 125°F (585°R), we cannot use the 300-gal reactor as it is now.

Analysis: After using Equations (E12-3.10) and (E12-3.14) to make a plot of conversion as a function of temperature, we see that there is only one intersection of X_EB(T) and X_MB(T), and consequently only one steady state. The exit conversion is 83% and the exit temperature (i.e., the reactor temperature) is 613°R (153°F), which is above the acceptable limit of 585°R (125°F) and we thus cannot use the CSTR operating at these conditions. Be sure to go to the LEP 12-3 and use Wolfram to see the X_MB and X_EB versus T lines change along with their intersection change as you change the parameters in the slider box.

Ouch! Looks like our plant will not be able to be completed and our multimillion-dollar profit has flown the coop. But wait, don’t give up, let’s ask reaction engineer Maxwell Anthony to fly to our company’s plant in the country of Jofostan to look for a cooling coil heat exchanger that we could place in the reactor. See what Max found in Example 12-4.

Example 12–4 CSTR with a Cooling Coil

Fantastic! Max has located a cooling coil in an equipment storage shed in the small, mountainous village of Ölofasis, Jofostan, for use in the hydrolysis of propylene oxide discussed in Example 12-3. The cooling coil has 40 ft² of cooling surface and the cooling-water flow rate inside the coil is sufficiently large that a constant coolant temperature of 85°F can be maintained. A typical overall heat-transfer coefficient for such a coil is 100 Btu/h·ft²·°F. Will the reactor satisfy the previous constraint of 125°F maximum temperature if the cooling coil is used?

Solution

If we assume that the cooling coil takes up negligible reactor volume, the conversion calculated as a function of temperature from the mole balance is the same as that in Example 12-3, Equation (E12-3.10).

Combining the mole balance, stoichiometry, and rate law, we have, from Example 12-3

$\begin{matrix} X_{EB} = \frac{τ k}{1 + τ k} = \frac{(2.084 \times 10^{12}) exp (- 16306 /T)}{1 + (2.084 \times 10^{12}) exp (- 16306 /T)} \end{matrix} \begin{matrix} (E12-3.10) \end{matrix}$

T is in °R.
Energy balance: Neglecting the work by the stirrer, we combine Equations (11-27) and (12-20) to write

$\begin{matrix} \frac{UA(T_{a} - T)}{F_{A0}} - X [Δ H_{Rx}^{\circ} (T_{R}) + Δ C_{P} (T - T_{R}] = Σ Θ_{i} C_{P_{i}} (T - T_{0}) & (E12-4.1) \end{matrix}$

Solving the energy balance for X_EB yields

$\begin{matrix} X_{EB} = \frac{Σ Θ_{i} C_{P_{i}} (T - T_{0}) + [UA (T - T_{a}) / F_{A0}]}{- [Δ H_{Rx}^{\circ} (T_{R}) + Δ C_{P} (T - T_{R})]} \end{matrix} \begin{matrix} (E12-4.2) \end{matrix}$

The cooling-coil term in Equation (E12-4.2) is

$\begin{matrix} \frac{UA}{F_{A0}} = (100 \frac{Btu}{h \cdot {ft}^{2} \cdot^{\circ} F}) \frac{(40 {ft}^{2})}{(43.04 lb-mol/h)} = \frac{92.9 Btu}{1b-mol \cdot^{\circ} F} & (E12-4.3) \end{matrix}$

Recall that the cooling temperature is

T_a = 85°F = 545°R

The numerical values of all other terms of Equation (E12-4.2) are identical to those given in Equation (E12-3.13), but with the addition of the heat exchange term, X_EB becomes

$\begin{matrix} X_{EB} = \frac{403.3 (T - 535) + 92.9 (T - 545)}{36400 + 7 (T - 528)} \end{matrix} \begin{matrix} (E12-4.4) \end{matrix}$

We now have two equations, Equations (E12-3.10) and (E12-4.4), and two unknowns, X and T, which we can solve with Polymath. Recall Examples 4-5 and 8-6 to review how to solve nonlinear, simultaneous equations of this type with Polymath. (See Problem P12-1_A(f) on pages 661–662 to plot X versus T on Figure E12-3.2.) We could generate Figure E12-3.2 by “fooling” Polymath to plot X and T, as explained in the tutorial on the Web site (http://www.umich.edu/~elements/6e/software/Polymath_fooling_tutorial.pdf ).

A table is shown for non-linear and explicit equations along with variable values. — TABLE E12-4.1 POLYMATH: CSTR WITH HEAT EXCHANGE

The Polymath program and solution to these two Equations (E12-3.10) for X_MB, and (E12-4.4) for X_EB, are given in Table E12-4.1. The exiting temperature and conversion are 103.7°F (563.7°R) and 36.4%, respectively, that is,

$\begin{matrix} T = 564 ° R and X = 0.36 \end{matrix}$

Be sure to go to the LEP 12-3 and use Wolfram or Python to see how X_MB and X_EB vary as you change the parameters in the slider box.

Analysis: We are grateful to the people of the village of Ölofasis in Jofostan for their help in finding this shiny new heat exchanger. By adding heat exchange to the CSTR, the X_MB(T) curve is unchanged but the slope of the X_EB(T) line in Figure E12-3.2 increases and intersects the X_MB curve at X = 0.36 and T = 564°R. This conversion is low! We could try to reduce the cooling by increasing T_a or T₀ to raise the reactor temperature closer to 585°R, but not above this temperature. The higher the temperature in this irreversible reaction, the greater the conversion.

We will see in the next section that there may be multiple exit values of conversion and temperature (multiple steady states, MSS) that satisfy the parameter values and entrance conditions.

12.5 Multiple Steady States (MSS)

In this section, we consider the steady-state operation of a CSTR in which a first-order reaction is taking place. An excellent experimental investigation that demonstrates the multiplicity of steady states was carried out by Vejtasa and Schmitz.⁶ They studied the reaction between sodium thiosulfate and hydrogen peroxide

⁶ S. A. Vejtasa and R. A. Schmitz, AIChE J., 16 (3), 415.

2Na₂S₂O₃ + 4H₂O₂ → Na₂S₃O₆ + Na₂SO₄ + 4H₂O

in a CSTR operated adiabatically. The multiple steady-state temperatures were examined by varying the flow rate over a range of space times, τ.

To illustrate the concept of MSS, reconsider the X_MB(T) curve, Equation (E12-3.10), shown in Figure E12-3.2, which has been redrawn and shown as dashed lines in Figure E12-3.2A. Now consider what would happen if the volu-metric flow rate ν₀ is increased (τ decreased) just a little. The energy balance line, X_EB(T), remains unchanged, but the mole balance line, X_MB, moves to the right, as shown by the curved, solid line in Figure E12-3.2A. This shift of X_MB(T) to the right results in the X_EB(T) and X_MB(T) intersecting three times, X_MB = X_EB, indicating three possible steady-state conditions at which the reactor could operate.

When more than one intersection occurs, there is more than one set of conditions that satisfy both the energy balance and mole balance (i.e., X_EB = X_MB); consequently, there will be multiple steady states at which the reactor may operate. These three steady states are easily determined from a graphical solution, but only one could show up in the Polymath equation solver solution. Thus, when using the Polymath nonlinear equation solver, we need to either choose different initial guesses to find whether there are other solutions that exist for which X_MB = X_EB. Or We could also fool Polymath to obtain the

The graphs of X subscript MB at different time intervals is shown. — **Figure E12-3.2A** Plots of X_EB(T) and X_MB(T) for different spaces times τ.

In the graph, the vertical axis representing X ranges from 0.1 to 0.8 in increments of 0.1 and the horizontal axis representing T (degrees R) ranges from 535 to 655 in increments of 10. When the temperature increases, the line representing X subscript E B increases linearly with the increasing values of X. An increasing curve representing X subscript MB is drawn along the line, from (0, 0.1), and intersects the line at (555, 0.1), (605, 0.45), and (645, 0.65). A dashed curve, parallel to X subscript MB is drawn from X equals 0.18. Note: all values are approximate.

plot in Figure E12-3.2A by using the ODE solver and setting $\frac{d T}{d t} = 0.1$ and then plot X_MB and X_EB versus T, as in Example 12-3. Tutorials on “How to Fool Polymath” and generate G(T) and R(T) plots can be found at http://www.umich.edu/~elements/6e/software/Polymath_fooling_tutorial.pdf.

We begin by recalling Equation (12-24), which applies when one neglects shaft work and ΔC_P (i.e., ΔC_P = 0 and therefore $Δ H_{Rx} = Δ H_{Rx}^{°}$ )

$\begin{matrix} - X Δ H_{Rx}^{\circ} = C_{P0} (1 + κ) (T - T_{c}) & (12-24) \end{matrix}$

where

$\begin{matrix} C_{P0} = Σ Θ_{i} C_{P_{i}} \end{matrix} \begin{matrix} (12 - 26 a) \end{matrix}$

$\begin{matrix} \begin{matrix} κ = \frac{UA}{C_{P0} F_{A0}} \end{matrix} \end{matrix} \begin{matrix} (12 - 26 b) \end{matrix}$

“Fooling” Polymath

and

$\begin{matrix} T_{c} = \frac{T_{0} F_{A0} C_{P0} + {UAT}_{a}}{UA + C_{P0} F_{A0}} = \frac{κ T_{a} + T_{0}}{1 + κ} \end{matrix} \begin{matrix} (12-27) \end{matrix}$

Using the CSTR mole balance $X = \frac{- r_{A} V}{F_{A0}}$ , Equation (12-24) may be rewritten as

$\begin{matrix} (- r_{A} V/ F_{A0}) (- Δ H_{Rx}^{\circ}) = C_{P0} (1 + κ) (T - T_{c}) \end{matrix} \begin{matrix} (12-28) \end{matrix}$

The left-hand side is referred to as the heat-generated term

G(T) = Heat-generated term

$\begin{matrix} G (T) = (- Δ H_{Rx}^{\circ}) (- r_{A} V/ F_{A0}) \end{matrix} \begin{matrix} (12-29) \end{matrix}$

The right-hand side of Equation (12-28) is referred to as the heat-removed term (by flow and heat exchange) R(T)

R(T) = Heat-removed term

$\begin{matrix} R (T) = C_{P0} (1 - κ) (T - T_{c}) \end{matrix} \begin{matrix} (12 - 30) \end{matrix}$

To study the multiplicity of steady states, we shall plot both R(T) and G(T) as a function of temperature on the same graph and analyze the circumstances under which we will obtain multiple intersections of R(T) and G(T).

12.5.1 Heat-Removed Term, R(T )

Vary Entering Temperature. From Equation (12-30), we see that R(T) increases linearly with temperature, with slope and intercept T_c. As the entering temperature T₀ is increased, the line retains the same slope but shifts to the right as the intercept T_c increases, as shown in Figure 12-7.

Heat-removed curve R(T)

Vary Non-adiabatic Parameter κ. If one increases κ by either decreasing the molar flow rate, F_A0, or increasing the heat-exchange area, A, the slope increases and for the case of T_a < T₀ the ordinate intercept moves to the left, as shown in Figure 12-8.

A graph presents the data for the heat-removed line. — **Figure 12-8** Variation of heat-removed line with κ (κ = UA/C_P0F_A0).

The graph is plotted for R(T) against T. The horizontal axis represents T and the horizontal axis represents R of T. A vertical line representing k equals infinity is drawn at T subscript a. Three increasing lines are drawn between T subscript a and T subscript 0. An increasing line representing k equals 0 is drawn at T subscript 0. The increase in heat is noted from right to left.

$\begin{matrix} κ = \frac{UA}{C_{P0} F_{A0}} \\ T_{c} = \frac{T_{0} + κ T_{a}}{1 + κ} \end{matrix}$

On the other hand, if T_a < T₀, the intercept will move to the right as κ increases.

12.5.2 Heat-Generated Term, G(T )

The heat-generated term, Equation (12-29), can be written in terms of conversion. (Recall that X = –r_AV/F_A0.)

$\begin{matrix} G (T) = (- Δ H_{Rx}^{\circ}) X & (12-31) \end{matrix}$

To obtain a plot of heat generated, G(T), as a function of temperature, we must solve for X as a function of T using the CSTR mole balance, the rate law, and stoichiometry. For example, for a first-order liquid-phase reaction, the CSTR mole balance becomes

$V = \frac{F_{A0} X}{k C_{A}} = \frac{υ_{0} C_{A0} X}{k C_{A0} (1 - X)}$

Solving for X yields

$\begin{matrix} X = \frac{τ k}{1 + τ k} & (5-8) \end{matrix}$

First-order reaction

Substituting for X in Equation (12-31), we obtain

$\begin{matrix} G (T) = \frac{- Δ H_{Rx}^{\circ} τ k}{1 + τ k} & (12-32) \end{matrix}$

Finally, substituting for k in terms of the Arrhenius equation, we obtain

$\begin{matrix} G (T) = \frac{- Δ H_{Rx}^{\circ} {τ A e}^{- E/RT}}{1 + {τ A e}^{- E/RT}} \end{matrix} \begin{matrix} (12-33) \end{matrix}$

Note that equations analogous to Equation (12-33) for G(T) can be derived for other reaction orders and for reversible reactions simply by solving the CSTR mole balance for X. For example, for the second-order liquid-phase reaction

Second-order reaction

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

the corresponding heat-generated term is

$\begin{matrix} G (T) = \frac{- Δ H_{Rx}^{\circ} [(2 τ C_{A0} {A e}^{- E/RT} + 1)] - \sqrt{4 τ C_{A0} {A e}^{- E/RT} + 1}}{2 τ C_{A0} t {A e}^{- E/RT}} \end{matrix} \begin{matrix} (12-34) \end{matrix}$

Let’s now return to our first-order reaction, Equation (12-33), and examine the behavior of the G(T) curve. At very low temperatures, the second term in the denominator of Equation (12-33) for the first-order reaction can be neglected, so that G(T) varies as

Low T

$G (T) = - Δ H_{Rx}^{\circ} {τ A e}^{- E/RT}$

(Recall that ${Δ H}_{Rx}^{°}$ means that the standard heat of reaction is evaluated at T_R.)

At very high temperatures, the second term on the right side in the denominator is large and dominates in Equation (12-33) so that the τk in the numerator and denominator cancel, and G(T) is reduced to

High T

$G (T) = - Δ H_{Rx}^{\circ}$

G(T) is shown as a function of T for two different activation energies, E, in Figure 12-9. If the flow rate is decreased or the reactor volume increased so as to increase τ, the heat-generated term, G(T), changes, as shown in Figure 12-10.

A graph is shown in which the variation curve with activation energy is plotted against temperature. — **Figure 12-9** Variation of G(T) curve with activation energy.

A graph is shown in which the vertical axis represents G(T) and horizontal axis represents temperature. Two increasing curves representing low and high energies are shown. The curve representing high energy starts from the origin and the curve representing low energy starts from a point in the horizontal axis. The curve traces an almost similar path and the gap between the curves reduce as it increases.

A graph is shown in which the variation curve with space time is plotted against temperature. — **Figure 12-10** Variation of G(T) curve with space time.

The vertical axis represents G(T) and horizontal axis represents temperature. Four increasing curves are shown. Three of the curves start from the origin and the fourth starts from a point in the horizontal axis and run almost parallel to each other. Tau is increasing from the left to right.

Heat-generated curves, G(T)

Can you combine Figures 12-10 and 12-8 to explain why a Bunsen burner flame goes out when you turn up the gas flow to a very high rate? As an exercise, plot R(T) verses G(T) for different values of T₀ and κ.

12.5.3 Ignition–Extinction Curve

The points of intersection of R(T) and G(T) give us the temperature at which the reactor can operate at steady state. This intersection is the point at which both the energy balance, R(T), and the mole balance, G(T), are satisfied. Suppose that we begin to operate our reactor at some relatively low temperature, T₀₁. If we construct our G(T) and R(T) curves, illustrated by curve y = G(T) and line a = R(T), respectively, in Figure 12-11, we see that there will be only one point of intersection, point 1. From this point of intersection, one can find the steady-state temperature in the reactor, T_s₁, by following a vertical line down to the T-axis and reading off the temperature, T_s₁, as shown in Figure 12-11.

Ignition: The point of no return

If one were now to increase the entering temperature to T₀₂, the G(T) curve, y, would remain unchanged, but the R(T) curve would move to the right, as shown by line b in Figure 12-11, and will now intersect the G(T) at point 2 and also be tangent at point 3. Consequently, we see from Figure 12-11 that there are two steady-state temperatures, T_s₂ and T_s₃, that can be realized in the CSTR for an entering temperature T₀₂. If the entering temperature is further increased to T₀₃, the R(T) curve, line c (Figure 12-12), intersects the G(T) curve three times and there are three steady-state temperatures, T_s₄, T_s₅, and T_s₆. As we continue to increase T₀, we finally reach line e, in which there are only two steady-state temperatures, a point of tangency at T_s₁₀ and an intersection at T_s₁₁. If the entering temperature is slightly increased beyond T₀₅, say to T₀₆, then the point of tangency disappears and there is a jump in reactor temperature, from T_s₁₀ up to T_s₁₂. This jump is called the ignition point for the reaction. By further increasing T₀, we reach line f, corresponding to T₀₆, in which we have only one reactor temperature that will satisfy both the mole and energy balances, T_s₁₂. For the six entering temperatures, we can form Table 12-4, relating the entering temperature to the possible reactor operating temperatures.

By plotting steady-state reactor temperature T_s as a function of entering temperature T₀, we obtain the well-known ignition–extinction curve shown in

Both the mole and energy balances are satisfied at the points of intersection or tangency.

A graph is shown in which the R(T) and G(T) curves are plotted against temperature. — **Figure 12-11** Finding multiple steady states with T₀ varied.

A graph is shown to find the steady values with variations in temperature. Two increasing lines representing a and b are drawn, from points T subscript 01 and T subscript 02. A curve representing y equals G of T is drawn across temperature, intersecting the line "a" at one point and at two points of line b. The intersections are marked as 1, 2, and 3. Vertical lines are drawn from points 1, 2, and 3 to the horizontal axis and these are marked T subscript S1, T subscript S2, and T subscript 3. Here, line "a" is marked as R(T) and y equals G(T).

A graph is shown to find the multiple steady values with variations in temperature. — **Figure 12-12** Finding multiple steady states with T₀ varied.

A graph is shown in which the R(T) and G(T) curves are plotted against temperature. Six increasing lines representing a, b, c, d, e, and f are drawn across temperature T, and they start from T subscript 01, T subscript 02, T subscript 03, T subscript 04, T subscript 05, and T subscript 06. An increasing s-curve, represented by y is drawn such that it intersects the lines at 12 points. Point 1 is marked on line "a;" 2 and 3 are marked on b; 4, 5, and 6 are marked on line c; 7, 8, and 9 are marked on line d; 10 and 11 are marked on line e; and 12 is marked on line f.

TABLE 12-4 MULTIPLE STEADY-STATE TEMPERATURES

Entering Temperature	Reactor Temperatures
T₀₁			T_s1
T₀₂		T_s₂		T_s₃
T₀₃	T_s₄		T_s₅		T_s₆
T₀₄	T_s₇		T_s₈		T_s₉
T₀₅		T_s₁₀		T_s₁₁
T₀₆			T_s₁₂

Figure 12-13. From this figure, we see that as the entering temperature T₀ is increased, the steady-state temperature T_s increases along the bottom line until T₀₅ is reached. Any fraction-of-a-degree increase in temperature beyond T₀₅ the point of tangency disappears and the steady-state reactor temperature T_s will jump up from T_s10 to T_s11, as shown in Figure 12-13. The temperature at which this jump occurs is called the ignition temperature. That is, we must exceed a certain feed temperature, T₀₅, to operate at the upper steady state where the temperature and conversion are higher. The ignition point temperature is sometimes also called the onset temperature for a runaway reaction or the point of no return, especially when the upper steady state is at a very high temperature.

Runaway reaction?

If a reactor were operating at T_s12 and we began to cool the entering temperature down from T₀₆, the steady-state reactor temperature, T_s3, would eventually be reached, corresponding to an entering temperature, T₀₂. Any slight decrease below T₀₂ the point of tangency on line b in Figure 12-12 would disappear and the temperature would drop the steady-state reactor temperature to immediately the lower steady-state value T_s₂. Consequently, T₀₂ is called the extinction temperature.

The middle points 5 and 8 in Figures 12-12 and 12-13 represent unstable steady-state temperatures. Consider the heat-removed line d in Figure 12-12, along with the heat-generated curve y, which are replotted in Figure 12-14. If we were operating at the middle steady-state temperature T_s₈, for example, and a

A graph is shown in which the temperature ignition-extinction curve is plotted. — **Figure 12-13** Temperature ignition–extinction curve.

The reactor temperature is plotted against entering temperature. The vertical axis depicts T subscript S1, T subscript S, and T subscript S12 and the horizontal axis depicts entries between T subscript 01 and T subscript 06. An increasing line representing lower steady rates is drawn from T subscript 01 to T subscript 05 containing points 1, 2, 4, 7, and 10. An increasing line representing upper steady states is drawn parallel to the line representing lower steady states, from T subscript 02 to T subscript 06, containing points 3, 6, 9, 11, and 12. Point 5 is marked between points 4 and 6 at T subscript 03 and point 8 is marked between 7 and 9 at T subscript 04. Points 3, 5, 8, and 10 are connected using a dashed line. A line is drawn from point 3 to 2 and another line is drawn from 10 to 11, and the direction is marked in the anti-clockwise direction. 2 is marked as the extinction temperature and 10 is marked as the ignition temperature.

A graph is shown in which multiple steady temperatures are marked. — **Figure 12-14** Stability of multiple steady-state temperatures.

A graph is shown in which its vertical axis represents R(T) and G(T) and the horizontal axis represents Temperature. An increasing line d initiating from the origin is drawn and an increasing, s-curve is drawn such that they intersect at three points, marked 7, 8, and 9. Four increasing vertical lines, marked 4, 3, 2, and 1 are present across temperature T. Three dashed vertical lines are drawn at T subscript S7, T subscript S8 and T subscript S9. It is noted that G is greater than R between T subscript 04 and T subscript S7; R is greater than G between T subscript S7 and T subscript S8; G is greater than T subscript S8 and T subscript S9; and R is greater than G beyond T subscript S9.

Point of No Return

The values of G(T) and R(T) are plotted against temperature. — A graph is shown in which an increasing s-curve intersects the straight line. The point of intersection is marked as point of no return, onset for runaway.

Beyond the onset point/temperature, if no action is taken the reaction will proceed to its maximum temperature.

pulse increase in reactor temperature occurred, we would find ourselves at the temperature shown by vertical line , between points 8 and 9. We see that along this vertical line , the heat-generated curve, y ≡ G(T), is greater than the heat-removed line d ≡ R(T), that is, (G > R). Consequently, the temperature in the reactor would continue to increase until point 9 is reached at the upper steady state. On the other hand, if we had a pulse decrease in temperature from point 8, we would find ourselves on a vertical line between points 7 and 8. Here, we see that the heat-removed curve d is greater than the heat-generated curve y (R > G), so the temperature will continue to decrease until the lower steady state is reached. That is, a small change in temperature either above or below the middle steady-state temperature, T_s8, will cause the reactor temperature to move away from this middle steady state. Steady states that behave in this manner are said to be unstable.

In contrast to these unstable operating points, there are stable operating points. Consider what would happen if a reactor operating at T_s9 were subjected to a pulse increase in reactor temperature indicated by line in Figure 12-14. We see that the heat-removed line d is greater than the heat-generated curve y (R > G), so that the reactor temperature will decrease and return to T_s9. On the other hand, if there is a sudden drop in temperature below T_s9, as indicated by line , we see the heat-generated curve y is greater than the heat-removed line d (G > R), and the reactor temperature will increase and return to the upper steady state at T_s9. Consequently, T_s9 is a stable steady state.

Next, let’s look at what happens when the lower steady-state temperature at T_s7 is subjected to pulse increase to the temperature shown as line in Figure 12-14. Here, we again see that the heat removed, R, is greater than the heat generated, G, so that the reactor temperature will drop and return to T_s7. If there is a sudden decrease in temperature below T_s7 to the temperature indicated by line , we see that the heat generated is greater than the heat removed (G > R), and that the reactor temperature will increase until it returns to T_s7. Consequently, T_s7 is a stable steady state. A similar analysis could be carried out for temperatures T_s1, T_s2, T_s4, T_s6, T_s11, and T_s12, and one would find that reactor temperatures would always return to locally stable steady-state values when subjected to both positive and negative fluctuations.

While these points are locally stable, they are not necessarily globally stable. That is, a large perturbation in temperature or concentration may be sufficient to cause the reactor to fall from the upper steady state (corresponding to high conversion and temperature, such as point 9 in Figure 12-14), to the lower steady state (corresponding to low temperature and conversion, point 7).

12.6 Nonisothermal Multiple Chemical Reactions

Most reacting systems involve more than one reaction and do not operate isothermally. This section is one of the most important, if not the most important, sections of the book. It ties together all the previous chapters to analyze multiple reactions that do not take place isothermally.

12.6.1 Energy Balance for Multiple Reactions in Plug-Flow Reactors

In this section we give the energy balance for multiple reactions. We begin by recalling the energy balance for a single reaction taking place in a PFR, which is given by Equation (12-5)

$\begin{matrix} \frac{d T}{d V} = \frac{(- r_{A}) [- Δ H_{Rx} (T)] - U a (T - T_{a})}{Σ_{j = 1}^{m} F_{j} C_{p_{j}}} & (12-5) \end{matrix}$

When we have multiple reactions occurring, we have to account for, and sum up, all the heats of reaction in the reactor for each and every reaction. For q multiple reactions taking place in a PFR where there are m species, it is easily shown that Equation (12-5) can be generalized to (cf. Problem P12-1(j))

Energy balance for multiple reactions

$\begin{matrix} \frac{d T}{d V} = \frac{Σ_{i=1}^{q} (- r_{ij}) [- {Δ H}_{Rxij} (T)] - U a (T - T_{a})}{Σ_{j=1}^{m} F_{j} C_{p_{j}}} \end{matrix} \begin{matrix} (12-35) \end{matrix}$

i = Reaction number

j = Species

Note that we now have two subscripts on the heat of reaction, the reaction number “i” and species “j”. The heat of reaction for reaction i must be referenced to the same species in the rate, r_ij, by which ΔH_Rxij is multiplied, which is

$\begin{matrix} [- r_{ij}] [- Δ H_{Rxij}] = [\frac{Moles of j reaction in reaction i}{Volume \cdot time}] \times [\frac{Joulesed “released” in reaction i}{Moles of j reacted in reaction i}] \\ = [\frac{Joules “released” in reaction i}{Volume \cdot time}] & (12-36) \end{matrix}$

where again we note the subscript j refers to the species, the subscript i refers to the particular reaction, q is the number of independent reactions, and m is the number of species. We are going to let

$\begin{matrix} Q_{R} = Σ_{i=l}^{q} (- r_{ij}) [- Δ H_{Rxij} (T)] \end{matrix}$

and

$\begin{matrix} Q_{r} = U a (T - T_{a}) \end{matrix}$

Then Equation (12-35) becomes

$\begin{matrix} \frac{d T}{d V} = \frac{Q_{g} - Q_{r}}{Σ_{j=l}^{m} F_{j} C_{P_{j}}} \end{matrix} \begin{matrix} (12-37) \end{matrix}$

Equation (12-37) represents a nice compact form of the energy balance for multiple reactions.

12.6.1A Series Reactions in a PFR

Consider the following series reaction sequence carried out in a PFR:

$\begin{matrix} \begin{matrix} Re action 1: & A \overset{k_{1}}{\to} B \end{matrix} \\ \begin{matrix} Re action 1: & B \overset{k_{2}}{\to} C \end{matrix} \end{matrix}$

The PFR energy balance becomes

One of the major goals of this text is that the reader will be able to solve multiple reactions with heat effects, and this section shows how!

$\begin{matrix} Q_{g} = (- Δ H_{Rx1A}) (- r_{1 A}) + (- r_{2 B}) (- Δ H_{Rx2 B}) \\ Q_{r} = U a (T - T_{a}) \end{matrix}$

Substituting for Q_g and –Q_r in Equation (12-37) we obtain

$\begin{matrix} \frac{d T}{d V} = \frac{Ua (T - T_{a}) + (- r_{1 A}) (- Δ H_{Rx1A}) + (- r_{2B}) (- Δ H_{Rx2B})}{F_{A} C_{P_{A}} + F_{B} C_{P_{B}} + F_{C} C_{P_{c}}} & (12-38) \end{matrix}$

where ΔH_Rx1A = [J/mol of A reacted in reaction 1] and

ΔH_Rx2B = [J/mol of B reacted in reaction 2].

12.6.1B Parallel Reactions in a PFR

We will now give three examples of multiple reactions with heat effects: Example 12-5 discusses parallel reactions, Example 12-6 discusses series reactions, and Example 12-7 discusses complex reactions.

Example 12-5 Parallel Reactions in a PFR with Heat Effects

The following gas-phase reactions occur in a PFR:

$\begin{matrix} \begin{matrix} Reaction: 1 & A \overset{k_{1}}{\to} B & - r_{1 A} = k_{1 A} C_{A} \end{matrix} & (E12-5.1) \end{matrix}$

$\begin{matrix} \begin{matrix} Reaction: 2 & 2 A \overset{k_{2}}{\to} C & - r_{2 A} = k_{2 A} C_{A}^{2} \end{matrix} & (E12-5.2) \end{matrix}$

A figure representing LEP sliders is shown. Sliders are represented as squares on horizontal lines.

Pure A is fed at a rate of 100 mol/s, at a temperature of 150=C and at a concentration of 0.1 mol/dm³. Neglect pressure drop and determine the temperature and molar flow rate profiles down the reactor.

Additional information:

Δ_H_Rx1A = – 20000 J/(mol = of A reacted in reaction 1)

Δ_H_Rx2A = – 60000 J/(mol = of A reacted in reaction 2)

$\begin{matrix} C_{P_{A}} = 90J/mol \cdot^{\circ} C & k_{1 A} = 10 exp [\frac{E_{1}}{R} (\frac{1}{300} - \frac{1}{T})] s^{- 1} \\ C_{P_{B}} = 90 J/mol \cdot^{\circ} C & E_{1} /R = 4000 K \\ C_{P_{C}} = 180 J/mol \cdot^{\circ} C & k_{2 A} = 0.09 exp [\frac{E_{2}}{R} (\frac{1}{300} - \frac{1}{T})] \frac{{dm}^{3}}{mol \cdot S} \\ U a = 4000 {J/m}^{3} \cdot s \cdot^{\circ} C & E_{0} / R=9000K \\ T_{a} = 100^{\circ} C (Constant) \end{matrix}$

Solution

0. Number Each Reaction: This step was given in the problem statement.

$\begin{matrix} (1) & A \overset{k_{1 A}}{\to} B \\ (2) & 2 A \overset{k_{2 A}}{\to} C \end{matrix}$

Mole Balances:

$\begin{matrix} \frac{d F_{A}}{d V} = r_{A} & (E12-5.3) \end{matrix}$

$\begin{matrix} \frac{d F_{B}}{d V} = r_{B} & (E12-5.4) \end{matrix}$

$\begin{matrix} \frac{d F_{C}}{d V} = r_{C} & (E12-5.5) \end{matrix}$
Rates:

Rate laws

$\begin{matrix} r_{1 A} = - k_{1 A} C_{A} & (E12-5.1) \end{matrix}$

$\begin{matrix} r_{2A} = - k_{1A} C_{A}^{2} & (E12-5.2) \end{matrix}$

Relative rates

$\begin{matrix} \begin{matrix} Reaction 1: & \frac{r_{1 A}}{- 1} = \frac{r_{1 B}}{1}; \end{matrix} & r_{1 B} \end{matrix} = - r_{1 A} = k_{1 A} C_{A}$

$\begin{matrix} \begin{matrix} Reaction 2: & \frac{r_{2A}}{- 2} = \frac{r_{2B}}{1}; \end{matrix} & r_{2 C} \end{matrix} = - \frac{1}{2} r_{2 A} = \frac{k_{2 A}}{2} C_{A}^{2}$

Net rates

$\begin{matrix} r_{A} = r_{1 A} + r_{2 A} = k_{1 A} C_{A} - k_{2 A} C_{A}^{2} & (E12-5.6) \end{matrix}$

$\begin{matrix} r_{B} = r_{1 B} = k_{1 A} C_{A} & (E12-5.7) \end{matrix}$

$\begin{matrix} r_{C} = r_{2 C} = \frac{1}{2} k_{2 A} C_{A}^{2} & (E12-5.8) \end{matrix}$
Stoichiometry (gas-phase but ΔP = 0, i.e., p = 1):

$\begin{matrix} C_{A} = C_{T0} (\frac{F_{A}}{F_{T}}) (\frac{T_{0}}{T}) & (E12-5.9) \end{matrix}$

$\begin{matrix} C_{B} = C_{T0} (\frac{F_{B}}{F_{T}}) (\frac{T_{0}}{T}) & (E12-5.10) \end{matrix}$

$\begin{matrix} C_{C} = C_{T0} (\frac{F_{C}}{F_{T}}) (\frac{T_{0}}{T}) & (E12-5.11) \end{matrix}$

$\begin{matrix} F_{T} = F_{A} + F_{B} + F_{C} & (E12-5.12) \end{matrix}$

$\begin{matrix} k_{1 A} = 10 exp [4000 (\frac{1}{300} - \frac{1}{T}) s^{- 1}] & (E12-5.13) \end{matrix}$

(T in K)

$\begin{matrix} k_{2A} = 0.09 exp [9000 (\frac{1}{300} - \frac{1}{T})] \frac{{dm}^{3}}{mol \cdot s} & (E12-5.14) \end{matrix}$

Selectivity:

$\begin{matrix} {\tilde{S}}_{B/C} = \frac{F_{B}}{F_{C}} & (E12-5.15) \end{matrix}$

Major goal of CRE: Analyze Multiple Reactions with Heat Effects
Energy Balance:

The PFR energy balance (cf. Equation (12-35))

$\begin{matrix} \frac{d T}{d V} = \frac{U a (T_{a} - T) + (- r_{1 A}) (- Δ H_{Rx1A}) + (- r_{2 A}) (- Δ H_{Rx2A})}{F_{A} C_{P_{A}} + F_{B} C_{P_{B}} + F_{C} C_{P_{C}}} & (E12-5.16) \end{matrix}$

becomes

$\begin{matrix} \frac{d T}{d V} = \frac{4000 (373 - T) + (- r_{1 A}) (2000) + (- r_{2 A}) (60000)}{{90 F}_{A} + 90 F_{B} + 180 F_{C}} & (E12-5.17) \end{matrix}$
Evaluation:

The Polymath program and its graphical outputs are shown in Table E12-5.1 and Figures E12-5.1 and E12-5.2.

TABLE E12-5.1 POLYMATH PROGRAM

Differential equations

1 d(Fa)/d(V) = r1a+r2a

2 d(Fb)/d(V) = -r1a

3 d(Fc)/d(V) = -r2a/2

4 d(T)/d(V) = (Qg-Qr)/(90*Fa+90*Fb+180*Fc)

Explicit equations

1 Qr = 4000*(T-373)

2 To = 423

3 k1a = 10*exp(4000*(1/300-1/T))

4 k2a = 0.09*exp(9000*(1/300-1/T))

5 Cto = 0.1

6 deltaH1 = -20000

7 deltaH2 = -60000

8 Ft = Fa+Fb+Fc

9 Ca = Cto*(Fa/Ft)*(To/T)

10 r2a = -k2a*Ca^2

11 Cb = Cto*(Fb/Ft)*(To/T)

12 Cc = Cto*(Fc/Ft)*(To/T

13 r1a = -k1a*Ca

14 Qg = (-r1a)*(-deltaH1)+(-r2a)*(-deltaH2)

Calculated values of DEQ variables

	Variable	Initial value	Final value
1	Ca	0.1	2.069E-09
2	Cb	0	0.0415941
3	Cc	0	0.016986
4	Cto	0.1	0.1
5	Fa	100.	2.738E-06
6	Fb	0	55.04326
7	Fc	0	22.47837
8	Ft	100.	77.52163
9	k1a	482.8247	2.426E+04
10	k2a	553.0557	3.716E+06
11	Qg	1.297E+06	1.003708
12	Qr	2.0E+05	1.396E+06
13	r1a	-48.28247	-5.019E-05
14	r2a	-5.530557	-1.591E-11
15	T	423.	722.0882
16	To	423.	423.
17	V	0	1.

Why does the temperature go through a maximum value?

A trend graph of Temperature in kelvin against Volume is shown. — **Figure E12-5.1** Temperature profile.

A trend graph shows the volume corresponding to the rise in temperature. The horizontal axis represents the volume V in decimeter cubed and ranges from 0.0 to 1.0 in increments of 0.2. The vertical axis represents the temperature T in Kelvin (K) and ranges from 350 to 850 in increments of 100. The trend curve begins from a temperature of 430 K when the corresponding Volume is 0. The temperature then goes up to 500 K when the corresponding volume is 0.4 decimeter cubed. The temperature is at a maximum of 800 K when the volume is 0.5 decimeter cubed. From there the temperature comes down to 750 K for a volume of 1.0 decimeter cubed. It is inferred from the graph that the volume increases as the temperature increases or vice-versa till a maximum point. Note that the values are approximate.

A trend graph of molar flow rates against volume is shown. — **Figure E12-5.2** Profile of molar flow rates F_A, F_B and F_C

A trend graph shows the change in volume in correspondence to the molar flow rates. The horizontal axis represents the volume V in decimeter cubed and ranges from 0.2 to 1.0 in increments of 0.2. The vertical axis represents the molar flow rate F Subscript i in mol per second and ranges from 0 to 100 in increments of 20. There are three trend curves each representing different molar flow rates such as F subscript A, F subscript B, and F subscript C. The curve representing F Subscript A begins at the value 0.5 on the x- axis when the corresponding molar flow rate is 0 mol per second. The Volume then reduces to 0.2 for which the molar flow value is 80 mol per second. The Volume finally reaches the value 0.0 for which the molar flow rate is 100 mol per second. The curve representing F Subscript C is a dotted curve and it begins at the value 0.0 on the x-axis when the corresponding molar flow is 0 mol per second. The Volume then increases and reaches a mid value of 0.4 when the corresponding molar flow rate is 60 mol per second. The curve remains constant from this point. The Volume then increases to 1.0 but the molar value remains the same. The curve representing F Subscript C begins at the value 0.1 when the corresponding molar flow is 0 mol per second. The Volume then increases and reaches the mid value of 0.4 when the corresponding molar flow rate is 20 mol per second. The curve remains constant from this point. The Volume then reaches up to 1.0 but the molar flow rate remains the same. The curves representing F Subscript A and F Subscript B intersect at the point (0.4, 40) and the curves representing F Subscript A and F Subscript C intersect at the point (0.4, 20). Note that the values are approximate.

Analysis: From Figure E12-5.1 we see that the temperature increases slowly up to a reactor volume of 0.4 dm³ then suddenly jumps up (ignites) to a temperature of 850 K. Correspondingly, from Figure E12-5.2 the reactant molar flow rate decreases sharply at this point. The reactant is virtually consumed by the time it reaches a reactor volume V = 0.45 dm³; beyond this point,Q_r > Q_g, and the reactor temperature begins to drop. In addition, the selectivity ${\tilde{S}}_{B/ C} = F_{B} / F_{C} = 55/22.5 = 2.44$ remains constant after this point. If a high selectivity is required, then the reactor should be shortened to V = 0.3 dm³, at which point the selectivity is ${\tilde{S}}_{B/ C} = 21.9 /2.5 = 8.9.$ .

12.6.2 Energy Balance for Multiple Reactions in a CSTR

Recall that for the steady-state mole balance in a CSTR with a single reaction [–F_A0X = r_AV], and that $Δ H_{Rx} (T) = Δ H_{Rx}^{\circ} + Δ C_{P} (T - T_{R})$ , so that for T₀ = T_i₀ Equation (11-27) may be rewritten as

$\begin{matrix} \dot{Q} - {\dot{W}}_{s} - F_{A0} Σ Θ_{j} C_{P_{j}} (T - T_{0}) + [Δ H_{Rx} (T)] [r_{A} V] = 0 & (11-27A) \end{matrix}$

Again, we must account for the “heat generated” by all the reactions in the reactor. For q multiple reactions and m species, the CSTR energy balance becomes

$\begin{matrix} \dot{Q} - {\dot{W}}_{s} - F_{A0} Σ_{j=1}^{m} Θ_{j} C_{P_{j}} (T - T_{0}) + V Σ_{i=l}^{q} r_{ij} Δ H_{Rxij} (T) = 0 \end{matrix} \begin{matrix} (12-39) \end{matrix}$

Energy balance for multiple reactions in a CSTR

Substituting Equation (12-20) for $\dot{Q}$ , neglecting the work term, and assuming constant heat capacities and large coolant flow rates ${\dot{m}}_{c}$ , Equation (12-39) becomes

$\begin{matrix} U A (T_{a} - T) - F_{A0} Σ_{j=1}^{m} C_{P_{j}} Θ_{j} (T - T_{0}) + V Σ_{i=l}^{q} r_{ij} Δ H_{Rxij} (T) = 0 \end{matrix} \begin{matrix} (12-40) \end{matrix}$

For the two parallel reactions described in Example 12-5, the CSTR energy balance is

$\begin{matrix} U A (T_{a} - T) - F_{A0} Σ_{j=1}^{m} Θ_{j} C_{P_{j}} (T - T_{0}) + {Vr}_{1 A} Δ H_{Rx1 A} (T) + {Vr}_{2 A} Δ H_{Rx2 A} (T) = 0 & (12-41) \end{matrix}$

Major goal of CRE

As stated previously, one of the major goals of this text is to have the reader solve problems involving multiple reactions with heat effects (cf. Problems P12-23_B, P12-24_B, P12-25_C, and P12-26_C). That’s exactly what we are doing in the next two examples!

12.6.3 Series Reactions in a CSTR

Example 12-6 Multiple Reactions in a CSTR

The elementary liquid-phase reactions

$A \overset{k_{1}}{\to} B \overset{k_{2}}{\to} C$

take place in a 10-dm³ CSTR. What are the effluent concentrations for a volumetric feed rate of 1000 dm³/min at a concentration of A of 0.3 mol/dm³? The inlet temperature is 283 K.

Additional information:

$\begin{matrix} \begin{matrix} \begin{matrix} C_{P_{A}} = C_{P_{B}} = C_{P_{C}} = 200 J/mol \cdot K \\ k_{1} = 3.3 {min}^{- 1} at 300 {K, with E}_{1} = 9900 cal/mol \end{matrix} \\ k_{2} = 4.58 {min}^{- 1} at 500 {K, with E}_{2} = 27000 cal/mol \\ \begin{matrix} \begin{matrix} Δ H_{Rx1 A} = 55000 J/mol A \end{matrix} & UA = 40000 J/min \cdot {K with T}_{a} = 57^{\circ} \end{matrix} C \end{matrix} \\ Δ H_{Rx2B} = 71500J/mol B \end{matrix}$

Solution

The Algorithm:

0. Number Each Reaction:

$\begin{matrix} \begin{matrix} Re action (1) & A \overset{k_{1}}{\to} B \end{matrix} \\ \begin{matrix} Re action (2) & B \overset{k_{1}}{\to} C \end{matrix} \end{matrix}$

1. Mole Balance on Every Species:

Species A: Combined mole balance and rate law for A

$\begin{matrix} V = \frac{F_{A0} - F_{A}}{- r_{A}} = \frac{υ_{0} [C_{A0} - C_{A}]}{- r_{1 A}} = \frac{υ_{0} [C_{A0} - C_{A}]}{k_{1} C_{A}} & (E12-6.1) \end{matrix}$

Solving for C_A gives us

$\begin{matrix} C_{A} = \frac{C_{A0}}{1 + {τ k}_{1}} & (E12-6.2) \end{matrix}$

Species B: Combined mole balance and rate law for B

$\begin{matrix} V = \frac{0 - C_{B} υ_{0}}{- r_{B}} = \frac{C_{B} υ_{0}}{r_{B}} & (E12-6.3) \end{matrix}$

An illustration depicts footsteps, and represents the phrase "following the algorithm".

2. Rates:

The reactions follow elementary rate laws

$\begin{matrix} (a) Laws & (b) Relative Rates & (\begin{matrix} C \end{matrix}) Net Rates \\ r_{1 A = - k_{1 A} C_{A} \equiv - k_{1} C_{A}} & r_{1 B} = - r_{1 A} & r_{A} = r_{1 A} \\ r_{2 B = - k_{2 B} C_{B} \equiv - k_{2} C_{B}} & r_{2 C} = - r_{2 B} & r_{B} = r_{1 B} + r_{2 B} \end{matrix}$

3. Combine:

Substituting for r_1B and r_2B in Equation (E12-6.3) gives

$\begin{matrix} V = \frac{C_{B} υ_{0}}{k_{1} C_{A} - k_{2} C_{B}} & (E12-6.4) \end{matrix}$

Solving for C_B yields

$\begin{matrix} C_{B} = \frac{τ k_{1} C_{A}}{1 + τ k_{2}} = \frac{τ k_{1} C_{A0}}{(1 + τ k_{1}) (1 + τ k_{2})} & (E12-6.5) \end{matrix}$

$\begin{matrix} - r_{1 A} = k_{1} C_{A} = \frac{k_{1} C_{A0}}{1 + τ k_{1}} \end{matrix} \begin{matrix} (E12-6.6) \end{matrix}$

$\begin{matrix} - r_{2 B} = k_{2} C_{B} = \frac{k_{2} τ k_{1} C_{A0}}{(1 + τ k_{1}) + ((1 + τ k_{2}))} \end{matrix} \begin{matrix} (E12-6.7) \end{matrix}$

4. Energy Balances:

Applying Equation (12-41) to this system gives

$\begin{matrix} [r_{1A} Δ H_{Rx1A} + r_{2B} Δ H_{Rx2B}] V - UA (T - T_{a}) - F_{A0} C_{P_{A}} (T - T_{0}) = 0 & (E12-6.8) \end{matrix}$

Substituting for F_A0 = υ₀C_A0, r_1A, and r_2B and rearranging, we have

$\begin{matrix} \overset{G(T)}{\overset{⏞}{[- \frac{Δ H_{Rx1A} τ k_{1}}{1 + τ k_{1}} - \frac{τ k_{1} τ k_{2} Δ H_{Rx2B}}{(1 + τ k_{1}) (1 + τ k_{2})}]}} = \frac{R (T)}{c_{P_{A}} (1 + κ) [T - T_{e}]} & (E12-6.9) \end{matrix}$

$\begin{matrix} G (T) = [- \frac{Δ H_{Rx1A} τ k_{1}}{1 + τ k_{1}} - \frac{τ k_{1} τ k_{2} Δ H_{Rx2B}}{(1 + τ k_{1}) (1 + τ k_{2})}] \end{matrix} \begin{matrix} (E12-6.10) \end{matrix}$

$\begin{matrix} R (T) = C_{P_{A}} (1 + κ) [T - T_{c}] \end{matrix} \begin{matrix} (E12-6.11) \end{matrix}$

5. Parameter Evaluation

$\begin{matrix} κ = \frac{UA}{F_{A0} C_{P_{A}}} = \frac{40000 J/min \cdot K}{(0.3 {mol/dm}^{3}) (1000 {dm}^{3} /min) 200 J/mol \cdot K} = 0.667 & (E12-6.12 \end{matrix}$

$\begin{matrix} T_{e} = \frac{T_{0} + κ T_{a}}{1 + κ} = \frac{283 + (0.666) (330)}{1 + 0.667} = 301.8 K & (E12-6.13) \end{matrix}$

We are going to generate G(T) and R(T) by "fooling" Polymath to first generate T as a function of a dummy variable, t. We then use our plotting options to convert T(t) to G(T) and R(T). The Polymath program to plot R(T) and G(T) vs. T is shown in Table E12-6.1, and the resulting graph is shown in Figure E12-6.1 (http://www.umich.edu/~elements/5e/tutorials/Polymath_fooling_tutorial_Example-12-6.pdf).

Fooling Polymath

Incrementing temperature in this manner is an easy way to generate R(T) and G(T) plots.

When F = 0, then G(T) = R(T) and the steady states can be found.

Wow!

Five (5) multiple steady states!

A trend graph of heat-removed and heat-generated against the temperature in Kelvin is given. — **Figure E12-6.1** Heat-removed and heat-generated curves.

The graph represents the data of heat-removed and heat-generated corresponding to the temperature. The horizontal axis represents the temperature in Kelvin and ranges from 250,000 to 750,000 in increments of 100,000. The vertical axis represents the values of heat-removed - R(T) and heat-generated - G(T) and ranges from 0.0 to 15.0 in increments of 5.0. The heat-removed R(T) is represented in a dotted straight line whereas the heat-generated G(T) is represented as a curve. Both the dotted line and curve are intertwined and there steady state points at which the dotted line and the curve touches each other. The points 1, 3, and 5 are stable steady states whereas the points 2 and 4 are unstable steady states. The temperature and the respective R(T) and G(T) values at these steady states are as follows: Steady state 1: Temperature - 310,000 K and R(T)/G(T) - 0.284/0.016, Steady state 2: Temperature - 363,000 K and R(T)/G(T) - 0.189/0.111, Steady state 3: Temperature - 449,000 K and R(T)/G(T) - 0.033/0.267, Steady state 4: Temperature - 558,000 K and R(T)/G(T) - 0.0041/0.167, Steady state 5: Temperature - 677,000 K and R(T)/G(T) - 0.00087/0.0053. Note that the values are approximate.

TABLE E12-6.2 EFFLUENT CONCENTRATIONS AND TEMPERATURES

SS	T(K)	C_A (mol/dm³)	C_B (mol/dm³)	C_C (mol/dm³)
1	310	0.284	0.016	0
2	363	0.189	0.111	0
3	449	0.033	0.267	0.00056
4	558	0.0041	0.167	0.129
5	677	0.00087	0.0053	0.294

Analysis: Wow! We see that five steady states (SS) exist!! The exit concentrations and temperatures listed in Table E12-6.2 were determined from the tabular output of the Polymath program. Steady states 1, 3, and 5 are stable steady states, while 2 and 4 are unstable. The selectivity at steady state 3 is ${\tilde{S}}_{B/C} = \frac{0.0053}{0.294} = 477$ , while at steady state 5 the selectivity is ${\tilde{S}}_{B/C} = \frac{0.0053}{0.294} = 0.018$ and is far too small. Consequently, to operate at a large selectivity we either have to operate at steady state 3 or find a different set of operating conditions. What do you think of the value of tau, that is, τ = 0.01 min? Is it a realistic number? (See Problem P12-1_A (h).)

12.6.4 Complex Reactions in a PFR

Example 12-7 Complex Reactions with Heat Effects in a PFR

We will use the reactions that we discussed in Chapter 8 that were coded with dummy names, A, B, C, and D for reasons of national security.

The following complex gas-phase reactions follow elementary rate laws

$\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} (1) & A + 2 B \to C \end{matrix} & - r_{1 A} = k_{1 A} C_{A} C_{B}^{2} \end{matrix} & Δ H_{Rx1 B} = - 15000 cal/mol B \end{matrix} \\ \begin{matrix} \begin{matrix} \begin{matrix} (2) & 2 A + 3C \to D \end{matrix} & - r_{2C} = k_{2C} C_{A}^{2} C_{C}^{2} \end{matrix} & Δ H_{Rx2 A} = - 10000 cal/mol A \end{matrix} \end{matrix}$

and take place in a PFR. The feed is stoichiometric for reaction (1) in A and B with F_A0 = 5 mol/min. The reactor volume is 10 dm³ and the total entering concentration is C_T0 = 0.2 mol/dm³. The entering pressure is 100 atm and the entering temperature is 300 K. The coolant flow rate is 50 mol/min and the entering coolant fluid has a heat capacity of C_{P_C0} = 10 cal/mol · K and enters at a temperature of 325 K.

Parameters

$k_{1 A} = 40 {(\frac{{dm}^{3}}{mol})}^{2} {/min at 300 K with E}_{1} = 8000 cal/mol$

$k_{2 C} = 2 {(\frac{{dm}^{3}}{mol})}^{4} {/min at 300 K with E}_{2} = 12000 cal/mol$

$\begin{matrix} C_{P_{A}} = 10 cal/mol/K & U a = 80 \frac{cal}{m^{3} \cdot min \cdot K} \\ C_{P_{B}} = 12 cal/mol/K & T_{a0} = 325 K \\ C_{P_{C}} = 14 cal/mol/K & {\dot{m}}_{c} = 50 mol/min \\ C_{P_{C}} = 16 cal/mol/K & C_{P_{C0}} = 10 cal/mol/K \end{matrix}$

Plot F_A, F_B, F_C, F_D, p, T, and T_a as a function of V for

(a) Co-current heat exchange

(b) Countercurrent heat exchange

(c) Constant T_a

(d) Adiabatic operation

Solution

Gas Phase PFR No Pressure Drop (p = 1)

1. Mole Balances:

$\begin{matrix} \begin{matrix} (1) & \frac{d F_{A}}{d V} = r_{A} & (F_{A0} = 5 mol/min) \end{matrix} & (E12-7.1) \end{matrix}$

$\begin{matrix} \begin{matrix} (2) & \frac{d F_{B}}{d V} = r_{B} & (F_{B0} = 10 mol/min) \end{matrix} & (E12-7.2) \end{matrix}$

$\begin{matrix} \begin{matrix} (3) & \frac{d F_{C}}{d V} = r_{C} & V_{f} = {10dm}^{3} \end{matrix} & (E12-7.3) \end{matrix}$

$\begin{matrix} \begin{matrix} (4) & \frac{d F_{D}}{d V} = r_{D} \end{matrix} & (E12-7.4) \end{matrix}$

An illustration depicts footsteps, and represents the phrase "following the algorithm".

2. Rates:

2a. Rate Laws

$\begin{matrix} \begin{matrix} (5) & r_{1 A} = - k_{1 A} C_{A} C_{B}^{2} \end{matrix} & (E12-7.5) \end{matrix}$

$\begin{matrix} \begin{matrix} (6) & r_{2C} = - k_{2C} C_{A}^{2} C_{C}^{2} \end{matrix} & (E12-7.6) \end{matrix}$

2b. Relative Rates

$\begin{matrix} \begin{matrix} (7) & r_{1B} \end{matrix} = {2r}_{1 A} & (E12-7.7) \end{matrix}$

$\begin{matrix} \begin{matrix} (8) & r_{1 C} \end{matrix} = r_{1A} & (E12-7.8) \end{matrix}$

$\begin{matrix} \begin{matrix} (9) & r_{2A} \end{matrix} = \frac{2}{3} r_{2C} = \frac{2}{3} k_{2 C} C_{A}^{2} C_{C}^{3} & (E12-7.9) \end{matrix}$

$\begin{matrix} \begin{matrix} (10) & r_{2D} \end{matrix} =- \frac{1}{3} r_{2 C} = \frac{1}{3} k_{2C} C_{A}^{2} C_{C}^{3} & (E12-7.10) \end{matrix}$

2c. Net Rates of reaction for species A, B, C, and D are

$\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} (11) & r_{A} \end{matrix} = r_{1 A} + r_{2A} = - k_{1 A} C_{A} C_{B}^{2} - \frac{2}{3} k_{2C} C_{A}^{2} C_{C}^{3} & (E12-7.11) \end{matrix} \\ \begin{matrix} \begin{matrix} (12) & r_{B} \end{matrix} = r_{1 B} = - 2 k_{1 A} C_{A} C_{B}^{2} & (E12-7.12) \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} (13) & r_{C} = r_{1C} + r_{2C} = k_{1A} C_{A} C_{A}^{2} C_{B}^{2} - k_{2C} C_{A}^{2} C_{C}^{3} \end{matrix} & (E12-7.13) \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} (14) & r_{D} \end{matrix} = r_{2D} = \frac{1}{3} k_{2C} C_{A}^{2} C_{C}^{2} & (E12-7.14) \end{matrix} \end{matrix}$

3. Selectivity: ${\tilde{S}}_{C/D} = F_{C} {/F}_{D}$

At V = 0, F_D = 0 causing S_C/D to go to infinity. Therefore, we set S_C/D = 0 between V = 0 and a very small number, say, V = 0.0001 dm³ to prevent the ODE solver from crashing.

$\begin{matrix} \begin{matrix} (15) & S_{C/D} = if (V>0.0001) then (\frac{F_{C}}{F_{D}}) else (0) \end{matrix} & (E12-7.15) \end{matrix}$

Major goal of CRE

4. Stoichiometry:

$\begin{matrix} \begin{matrix} (16) & C_{A} = C_{T0} (\frac{F_{A}}{F_{T}}) P (\frac{T_{0}}{T}) \end{matrix} & (E12-7.16) \end{matrix}$

$\begin{matrix} \begin{matrix} (17) & C_{B} = C_{T0} (\frac{F_{B}}{F_{T}}) P (\frac{T_{0}}{T}) \end{matrix} & (E12-7.17) \end{matrix}$

$\begin{matrix} \begin{matrix} (18) & C_{C} = C_{T0} (\frac{F_{C}}{F_{T}}) P (\frac{T_{0}}{T}) \end{matrix} & (E12-7.18) \end{matrix}$

$\begin{matrix} \begin{matrix} (19) & C_{D} = C_{T0} (\frac{F_{D}}{F_{T}}) P (\frac{T_{0}}{T}) \end{matrix} & (E12-7.19) \end{matrix}$

Neglect pressure drop

$\begin{matrix} \begin{matrix} (20) & P \end{matrix} = 1 & (E12-7.20) \end{matrix}$

$\begin{matrix} \begin{matrix} (21) & F_{T} \end{matrix} = F_{A} + F_{B} + F_{C} + F_{D} & (E12-7.21) \end{matrix}$

5. Parameters:

$\begin{matrix} \begin{matrix} (22) & k_{1 A} \end{matrix} = 40 exp [\frac{E_{1}}{R} (\frac{1}{300} - \frac{1}{T})] {({dm}^{3} / mol)}^{2} / min & (E12-7.22) \end{matrix}$

$\begin{matrix} \begin{matrix} (23) & k_{2C} \end{matrix} = 2 exp [\frac{E_{2}}{R} (\frac{1}{300} - \frac{1}{T})] {({dm}^{3} / mol)}^{4} / min & (E12-7.23) \end{matrix}$

(24) C_A0 = 0.2 mol/dm³

(25) R= 1.987 cal/mol/K

(26) E₁ = 8,000 cal/mol

(27) E₂ = 12,000 cal/mol

Other parameters are given in the problem statement, that is, Equations (28) to (34).

(28) Cp_A, (29) Cp_B, (30) Cp_c, (31) ${\dot{m}}_{c}$ , (32) $Δ H_{Rx1B}^{\circ}$ ,

(33) $Δ H_{Rx2A}^{\circ}$ , (34) Cp_C0

6. Energy Balance:

Recalling Equation (12-37)

$\begin{matrix} \begin{matrix} (\begin{matrix} 35 \end{matrix}) & \frac{d T}{d V} = \frac{Q_{g} - Q_{r}}{Σ F_{j} C_{P_{j}}} \end{matrix} & (E12-7.35) \end{matrix}$

The denominator of Equation (E12-7.35) is

$\begin{matrix} \begin{matrix} (36) & Σ F_{j} \end{matrix} C_{p_{j}} = F_{A} C_{p_{A}} + F_{B} C_{p_{B}} + F_{C} C_{p_{C}} + F_{D} C_{p_{D}} & (E12-7.36) \end{matrix}$

The “heat removed” term is

$\begin{matrix} \begin{matrix} (37) & Q_{r} = U a (T - T_{a}) \end{matrix} & (E12-7.37) \end{matrix}$

The “heat generated” is

$\begin{matrix} \begin{matrix} (38) & Q_{r} = Σ r_{i j} {Δ H}_{Rxij} = r_{1 B} {Δ H}_{Rx1B} + r_{2B} {Δ H}_{Rx2B} \end{matrix} & (E12-7.38) \end{matrix}$

(a) Co-current heat exchange

The heat exchange balance for co-current exchange is

$\begin{matrix} \begin{matrix} (39) & \frac{{d T}_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{C} C_{P_{C0}}} \end{matrix} & (E12-7.39) \end{matrix}$

Part (a) Co-current flow: Plot and analyze the molar flow rates, and the reactor and coolant temperatures as a function of reactor volume. The Polymath code and output for co-current flow are shown in Table E12-7.1.

Co-current heat exchange

Solution

Two graphs of Temperature in K against Volume in decimeter cubed is given. — **Figure E12-7.1** Profiles for co-current heat exchange; (a) temperature (b) molar flow rates. *Note*: The molar flow rate FD is very small and is essentially the same as the bottom axis.

Two trend graphs on representing the example of co-current heat exchange due to change in temperature and another due to change in molar flow rate is shown. The horizontal axis of the first graph represents the Volume in dm cubed and ranges from 1 to 10 in increments of 1. The vertical axis of the graph represents temperature in Kelvin and ranges from 300 to 840 in increments of 60. There are two trend curves in representing T and another representing T Subscript a. The trend curve of reactor temperature at begins at 300K when V is at 0.3 dm cubed and it goes up to a maximum of 886K when V is at 6 dm cubed. T then comes down to 540K when V is at 10 dm cubed. The reactor temperature T Subscript a begins at 310K when V is at 0 and it join T at 540K when V is at 10 dm cubed. The horizontal axis of the second graph represents the Volume in dm cubed and ranges from 1 to 10 in increments of 1. The vertical axis of the graph represents the molar flow rate F Subscript i in mol per min and ranges from 0 to 10 in increments of 1. There are three trend curves each representing F Subscript A, F Subscript B, and F Subscript C. The curve representing F Subscript A begins with a molar flow rate of 5 mol per min when V is at 0 and it then goes down to 0.5 mol per min when V is at 10 dm cubed. The curve representing F Subscript B begins with the highest molar flow rate of 10 mol per min when V is at 0 and it then goes down to 1 mol per min when V is at 10 dm cubed. The curve representing F Subscript C begins with a molar flow rate of 0 when V is at 0.5 dm cubed and it then goes up to 5 mol per min when V is at 10 dm cubed. Note that the values are approximate.

The temperature and molar flow rate profiles are shown in Figure E12-7.1.

Analysis: Part (a): We also note that the reactor temperature, T, increases when Q_g > Q_r and reaches a maximum, T = 886 K, at approximately V = 6 dm³. After that, Q_r > Q_g the reactor temperature decreases and approaches T_a at the end of the reactor. For co-current heat exchange, the selectivity ${\tilde{S}}_{C/D} = \frac{4.59}{0.0035} = 1258$ is really quite good.

Part (b) Countercurrent heat exchange: Solution: We will use the same program as part (a), but will change the sign of the heat-exchange balance and guess T_a at V = 0 to be 507 K and see whether the value gives us a T_a0 of 325 K at V = V_f.

$\frac{{d T}_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{C} C_{P_{Cool}}}$

We find our guess of 507 K matches T_a0 = 325 K. Are we lucky or what?!

The Polymath Program is shown in Table E12-7.2.

Countercurrent heat exchange

Two graphs of Temperature in K against Volume in dm cubed is given. — **Figure E12-7.2** Profiles for countercurrent heat exchange; (a) temperature (b) molar flow rates.

Two trend graphs on representing the example of co-current heat exchange due to change in temperature and another due to change in molar flow rate is shown. The horizontal axis of the first graph represents the Volume in dm cubed and ranges from 1 to 10 in increments of 1. The vertical axis of the graph represents temperature in Kelvin and ranges from 300 to 2000 in increments of 90. There are two trend curves in representing T and another representing T Subscript a. The trend curve of reactor temperature T begins at 300K when V is at 0.2 dm cubed and it goes up to a maximum of 1100K when V is at 2 dm cubed. T then comes down to 310K when V is at 10 dm cubed. The reactor temperature T Subscript a begins at 410K when V is at 0 and it join T at 310K when V is at 10 dm cubed. The horizontal axis of the second graph represents the Volume in dm cubed and ranges from 1 to 10 in increments of 1. The vertical axis of the graph represents the molar flow rate F Subscript i in mol per min and ranges from 0 to 10 in increments of 1. There are four trend curves each representing F Subscript A, F Subscript B, F Subscript C, and F Subscript D. The curve representing F Subscript A begins with a molar flow rate of 5 mol per min when V is at 0 and it then goes down to 0.3 mol per min when V is at 10 dm cubed. The curve representing F Subscript B begins with the highest molar flow rate of 10 mol per min when V is at 0 and it then goes down to 1 mol per min when V is at 10 dm cubed. The curve representing F Subscript C begins with a molar flow rate of 0 when V is at 0.5 dm cubed and it then goes up to 5 mol per min when V is at 10 dm cubed. The curve representing F Subscript D remains flat and lies on the x-axis. Note that the values are approximate.

Analysis: Part (b): For countercurrent exchange, the coolant temperature reaches a maximum at V = 1.3 dm³ while the reactor temperature reaches a maximum at V = 2.1 dm³. The reactor with a countercurrent exchanger reaches a maximum reactor temperature of 1100 K, which is greater than that for the co-current exchanger, (i.e., 930 K). Consequently, if there is a concern about additional side reactions occurring at this maximum temperature of 1100 K, one should use a co-current exchanger or if possible use a large coolant flow rate to maintain constant T_a in the exchanger. In Figure 12-7.2(a) we see that the reactor temperature approaches the coolant entrance temperature at the end of the reactor. The selectivity for the countercurrent systems, ${\tilde{S}}_{C/ D} = 1181$ , is slightly lower than that for the co-current exchange.

Part (c) Constant T_a: Solution: To solve the case of constant heating-fluid temperature, we simply multiply the right-hand side of the heat-exchanger balance by zero, that is,

$\frac{{d T}_{a}}{d V} = \frac{U a (T - T_{a})}{{\dot{m}}_{C} C_{P}} * 0$

and use Equations (E12-7.1)–(E12-7.39). The Polymath Progam is shown in Table E12-7.3 and the temperature and molar flow rate profiles are shown in Figures E12-7.3(a) and E12-7.3(b), respectively.

Constant T_a

Two graphs plotting temperature and flow rate values for volume is shown. — **Figure E12-7.3** Profiles for constant T_a; (a) temperature (b) molar flow rates.

Two trend graphs representing the example of co-current heat exchange due to change in temperature and another due to change in molar flow rate is shown. The horizontal axis of the first graph represents the Volume in decimeter cubed and ranges from 1 to 10 in increments of 1. The vertical axis of the graph represents temperature in Kelvin and ranges from 300 to 900 in increments of 60. There is a trend line and a trend curve, one representing T and another representing T Subscript a respectively. The trend line of reactor temperature T begins at 310 K when V is at 0 and this line remains constant. The reactor temperature T Subscript a begins at 300 K when V is at 0 and it goes up to 840 K when V is at 6 decimeter cubed. The horizontal axis of the second graph represents the Volume in decimeter cubed and ranges from 1 to 10 in increments of 1. The vertical axis of the graph represents the molar flow rate F Subscript i in mol per minute and ranges from 0 to 10 in increments of 1. There are three trend curves each representing F Subscript A, F Subscript B, F Subscript C, and F Subscript D represents a line. The curve representing F Subscript A begins with a molar flow rate of 5 mol per minute when V is at 0 and it then goes down to 0.5 mol per min when V is at 10 decimeter cubed. The curve representing F Subscript B begins with the highest molar flow rate of 10 mol per minute when V is at 0 and it then goes down to 1 mol per minute when V is at 10 decimeter cubed. The curve representing F Subscript C begins with a molar flow rate of 0 when V is at 0 and it then goes up to 5 mol per minute when V is at 10 decimeter cubed. The curve representing F Subscript D remains flat and lies on the x-axis. The curve for F subscript c intersects F subscript A and F subscript B at (5,2.5) and (5.5, 3.5) respectively. Note that the values are approximate.

Analysis: Part (c): For constant T_a, the maximum reactor temperature, 837 K, is less than either co-current or countercurrent exchange while the selectivity, ${\tilde{S}}_{C/D} = 1861$ is greater than either co-current or countercurrent exchange. Consequently, one should investigate how to achieve sufficiently high mass flow of the coolant in order to maintain constant T_a.

Part (d) Adiabatic: To solve for the adiabatic case, we simply multiply the overall heat transfer coefficient by zero.

Ua = 80*0

The Polymath Program is shown in Table E12-7.4 while the temperature and molar flow rate profiles are shown in Figures E12-7.4(a) and E12-7.4(b), respectively.

Adiabatic operation

Analysis: Part (d): For the adiabatic case, the maximum temperature, which is the exit temperature, is higher than the other three exchange systems, and the selectivity is the lowest with ${\tilde{S}}_{C/D} = 700$ . At this high temperature, the occurrence of unwanted side reactions certainly is a concern.

Overall Analysis Parts (a) to (d): Suppose the maximum temperature in each of these cases is outside the safety limit of 750 K for this system. Problem P12-1_A (i) asks how you can keep the maximum temperature below 750 K.

12.7 Radial and Axial Temperature Variations in a Tubular Reactor

In the previous sections, we have assumed that there were no radial variations in velocity, concentration, temperature, or reaction rate in the tubular and packed-bed reactors so that the axial profiles could be determined using an ordinary differential equation (ODE) solver. However, we consider both radial and axial variations in our mole and energy balance, we obtain partial differential equations (PDEs), which cannot be solved with Polymath and Wolfram. Consequently, it was decided to move this discussion on radial variation to Chapter 18 where we use the PDE solver COMSOL to explore two-dimensional problems of this type.

12.8 And Now... A Word from Our Sponsor—Safety 12 (AWFOS–S12 Safety Statistics)^†

^† Also see http://umich.edu/~safeche.

12.8.1 The Process Safety Across the Chemical Engineering Curriculum Web site

The Process Safety Across the Chemical Engineering Curriculum Web site (http://umich.edu/~safeche/index.html) provides a safety module for every core course in chemical engineering. A module consists of viewing a Chemical Safety Board (CSB) video, carrying out a safety analysis of the incident and doing a calculation related to a specific course. The Web site includes tutorials on a number of safety algorithms such as the BowTie Diagram (Section 6.7), the NFPA Diamond (Section 2.7), and the process Safety Pyramid (Section 9.5). This Web site was developed by the author of this text and the tutorials are also given at the end of the various chapters in a section titled “A Word From Our Sponsor—Safety.” The CRE safety modules include the explosions at Monsanto, Synthron, and T2 Laboratories. The CRE and safety Web sites are open to all as they have no passwords or other to be accessed.

12.8.2 Safety Statistics

Scaling up exothermic chemical reactions can be very tricky. Tables 12-5 and 12-6 give reactions that have resulted in accidents and their causes, respectively.⁷ The reader should review the case histories of these reactions to learn how to avoid similar accidents.

⁷ Courtesy of J. Singh, Chemical Engineering, 92 (1997) and B. Venugopal, Chemical Engineering, 54 (2002).

TABLE 12-5 INCIDENCE OF BATCH PROCESS ACCIDENTS

Process Type	Number of Incidents in United Kingdom, 1962–1987
Polymerization	64
Nitration	15
Sulfurization	13
Hydrolysis	10
Salt formation	8
Halogenation	8
Alkylation (Friedel-Crafts)	5
Amination	4
Diazolization	4
Oxidation	2
Esterification	1
Total:	134

Source: Courtesy of J. Singh, Chemical Engineering, 92 (1997).

TABLE 12-6 CAUSES OF BATCH REACTOR ACCIDENTS IN TABLE 12-5

Cause	Contribution, %
Lack of knowledge of reaction chemistry	20
Problems with material quality	9
Temperature-control problems	19
Agitation problems	10
Mis-charging of reactants or catalyst	21
Poor maintenance	15
Operator error	5

Source: B. Venugopal, Chemical Engineering, 54 (2002).

Runaway reactions are the most dangerous in reactor operation, and a thorough understanding of how and when they could occur is part of the chemical reaction engineer’s responsibility. The reaction in Example 12-7 could be thought of as running away. Recall that as we moved down the length of the reactor, none of the cooling arrangements could keep the reactor from reaching an extremely high temperature (e.g., 800 K). In Chapter 13, we study case histories of two runaway reactions. One is the nitroaniline explosion discussed in Example E13-2 and the other is Example E13-6, concerning the explosion at T2 Laboratories. See Example 13-7 and videos on T2 Laboratories’ explosion and what caused it to happen (https://www.csb.gov/t2-laboratories-inc-reactive-chemical-explosion/ and https://www.youtube.com/watch?v=C561PCq5E1g).

There are many resources available for additional information on reactor safety and the management of chemical reactivity hazards. Guidelines for managing chemical reactivity hazards and other fire, explosion, and toxic release hazards are developed and published by the Center for Chemical Process Safety (CCPS) of the American Institute of Chemical Engineers. CCPS books and other resources are available at www.aiche.org/ccps. For example, the book Essential Practices for Managing Chemical Reactivity Hazards, written by a team of industry experts, is also provided free of charge by CCPS on the site at https://app.knovel.com/web/toc.v/cid:kpEPMCRH02/viewerType:toc/. A concise and easy-to-use software program that can be used to determine the reactivity of substances or mixtures of substances, the Chemical Reactivity Worksheet, is provided by the National Oceanic and Atmospheric Administration (NOAA) for free on its Web site, www.noaa.gov.

12.8.3 Additional Resources CCPS and SAChE

The Safety and Chemical Engineering Education (SAChE) program was formed in 1992 as a cooperative effort between the AIChE, CCPS, and engineering schools to provide teaching materials and programs that bring elements of process safety into the education of undergraduate and graduate students studying chemical and biochemical products and processes. The SAChE Web site (www.sache.org) has a great discussion of reactor safety with examples as well as information on reactive materials. These materials are also suitable for training purposes in an industrial setting.

The following instruction modules are available on the SAChE Web site (www.sache.org).

Chemical Reactivity Hazards: This Web-based instructional module contains about 100 Web pages with extensive links, graphics, videos, and supplemental slides. It can be used either for classroom presentation or as a self-paced tutorial. The module is designed to supplement a junior or senior chemical engineering course by showing how uncontrolled chemical reactions in industry can lead to serious harm, and by introducing key concepts for avoiding unintended reactions and controlling intended reactions.
Runaway Reactions: Experimental Characterization and Vent Sizing: This instruction module describes the ARSST and its operation, and illustrates how this instrument can easily be used to experimentally determine the transient characteristics of runaway reactions, and how the resulting data can be analyzed and used to size the relief vent for such systems. The theory along with an example behind the ARSST is given on the CRE Web site under Additional Material for Chapter 13 and the LEPs for Chapter 13, that is, PRS Example R13-1.
Rupture of a Nitroaniline Reactor: This case study demonstrates the concept of runaway reactions and how they are characterized and controlled to prevent major losses.
Seveso Accidental Release Case History: This presentation describes a widely discussed case history that illustrates how minor engineering errors can cause significant problems; problems that should not be repeated. The accident was in Seveso, Italy, in 1976. It was a small release of a dioxin that caused many serious injuries.

Note: Access is restricted to SAChE members and universities.

Membership in SAChE is required to view these materials. Virtually all U.S. universities and many non-U.S. universities are members of SAChE—contact your university SAChE representative, listed on the SAChE Web site, or your instructor or department chair to learn your university’s username and password. Companies can also become members—see the SAChE Web site for details.

Certificate Program

SAChE also offers several certificate programs that are available to all chemical engineering students. Students can study the material, take an online test, and receive a certificate of completion. The following two certificate programs are of value for reaction engineering:

Runaway Reactions: This certificate focuses on managing chemical reaction hazards, particularly runaway reactions.
Chemical Reactivity Hazards: This is a Web-based certificate that provides an overview of the basic understanding of chemical reactivity hazards.

Many students are taking the certificate test online and put the fact that they successfully obtained the certificate on their résumés.

More information on safety is given in the Summary Notes and Professional Reference Shelf on the Web. Particularly study the use of the ARSST to detect potential problems. These will be discussed in Chapter 13 Professional Reference Shelf R13.1 on the CRE Web site.

Closure. Virtually all reactions that are carried out in industry involve heat effects. This chapter provides the basis to design reactors that operate at steady state and involve heat effects. To model these reactors, we simply add another step to our algorithm; this step is the energy balance. Here, it is important to understand how the energy balance was applied to each reactor type so that you will be able to describe what would happen if you changed some of the operating conditions (e.g., the initial or entering temperature, T₀) in order to determine whether those changes result in unsafe conditions such as a runaway reaction. The Living Example Problems (especially T12-2) and the ICG module will help you achieve a high level of understanding of reactors and reactions with heat effects. Another major goal after studying this chapter is to be able to design reactors that have multiple reactions taking place under nonisothermal conditions. Work through Problem 12-26_C to make sure you have achieved this goal. An industrial example that provides a number of practical details is included as an appendix to this chapter. We close with a brief discussion on safety and list some resources where you can obtain more information.

Summary

1. For single reactions, the energy balance on a PFR/PBR in terms of molar flow rate is

$\begin{matrix} \frac{d T}{d V} = \frac{(r_{A}) [Δ H_{Rx} (T)] - U a (T - T_{a})}{Σ F_{i} C_{P_{i}}} = \frac{Q_{g} - Q_{r}}{Σ F_{i} C_{P_{i}}} \end{matrix} \begin{matrix} (S12-1) \end{matrix}$

In terms of conversion

$\begin{matrix} \frac{d T}{d V} = \frac{(r_{A}) [Δ H_{Rx} (T)] - U a (T - T_{a})}{F_{A0} (Σ Θ_{j} C_{P_{i}} + X Δ C_{P})} = \frac{Q_{g} - Q_{r}}{F_{A0} (Σ Θ_{j} C_{P_{j}} + X Δ C_{p})} \end{matrix} \begin{matrix} (S12-2) \end{matrix}$

2. The temperature dependence of the specific reaction rate is given in the form

$\begin{matrix} k (T) = k (T_{1}) exp [\frac{E}{R} (\frac{1}{T_{1}} - \frac{1}{T})] = k {(T}_{1}) exp [\frac{E}{R} (\frac{T - T_{1}}{{TT}_{1}})] \end{matrix} \begin{matrix} (S12-3) \end{matrix}$

3. The temperature dependence of the equilibrium constant is given by van’t Hoff’s equation for ΔC_p = 0

$\begin{matrix} K_{P} (T) = K_{P} (T_{2}) exp [\frac{Δ H_{Rx}^{\circ}}{R} (\frac{1}{T_{2}} - \frac{1}{T})] \end{matrix} \begin{matrix} (S12-4) \end{matrix}$

4. Neglecting changes in potential energy, kinetic energy, and viscous dissipation, and for the case of no work done on or by the system, large coolant flow rates (m_c), and all species entering at the same temperature, the steady-state CSTR energy balance for single reactions is

$\begin{matrix} \frac{UA}{F_{A0}} (T_{a} - T) - X [Δ H_{Rx}^{\circ} (T_{R}) + Δ C_{P} (T - T_{R})] = Σ Θ_{j} C_{P_{j}} (T - T_{i 0}) \end{matrix} \begin{matrix} (S12-5) \end{matrix}$

5. Multiple steady states

$\begin{matrix} G (T) = (- Δ H_{Rx}^{\circ}) (\frac{- r_{A} V}{F_{A0}}) = (- Δ H_{Rx}^{\circ}) (X) \end{matrix} \begin{matrix} (S12-6) \end{matrix}$

$\begin{matrix} R (T) = C_{P0} (1 + κ) (T - T_{c}) \end{matrix} \begin{matrix} (S12-7) \end{matrix}$

A trend graph plots heat-removed and heat-generated against the temperature (in kelvin) is shown. — The graph represents the data of heat-removed and heat-generated corresponding to the temperature. The horizontal axis represents the temperature and the vertical axis represents the heat-removed - R(T) and heat-generated - G(T). The heat-removed R(T) is represented using a straight line whereas the heat-generated G(T) is represented as a curve. Both the line and curve are intertwined and there are steady state points at which the dotted line and the curve intersects each other. These points are referred to as the point of no return Onset for runaway. Also, G of T equals negative delta H subscript RX degrees.

where $κ = \frac{UA}{C_{p_{0}} F_{A0}} {and T}_{c} = \frac{κ T_{a} + T_{0}}{1 + κ}$

6. When q multiple reactions are taking place and there are m species

$\begin{matrix} \frac{d T}{d V} = \frac{Σ_{i = 1}^{q} (r_{ij}) [Δ H_{Rx i j} (T)] - U a (T - T_{a})}{Σ_{j = 1}^{m} F_{j} C_{P j}} = \frac{Q_{g} - Q_{r}}{Σ_{j = 1}^{m} F_{j} C_{P j}} \end{matrix} \begin{matrix} (S12-8) \end{matrix}$

CRE Web Site Materials (http://www.umich.edu/~elements/6e/12chap/obj.html#/)

An illustration shows the CRE website materials including the useful links for living example problems, extra help, additional materials, professional reference shelf, computer simulation problem statements, and COMSOL and Evaluation.

Interactive Computer Games that Talk to Each Other

A screenshot of an animated computer screen titled Heat Effects 1 is shown. — An example of interactive computer games is shown. The left most corner of the screen shows a chemical reaction between A and B showing the reaction parameters such as irreversible and reversible reactions. Next to this there is a trend graph of conversion versus temperature next to which there is an image of a thermometer and a scale that shows the molar flow rate value as 100. Below this the parameters of the heat of reaction such as exothermic and endothermic is shown. Also, given are the heat exchange parameters.

Heat Effects I (http://www.umich.edu/~elements/6e/icm/heatfx1.html)

A screenshot of an animated computer screen titled Heat Effects 2 is shown. — An example of interactive computer games is shown. The screen shows a trend graph of conversion versus temperature. The values marked on the graph are given on the left corner of the graph. The horizontal axis of the graph represents V times 10 to the power negative 1 and ranges from 0.0 to 1.0 in increments of 0.1. The vertical axis of the graph represents x and ranges from 0.0 to 1.0. Below the graph, the graph type is mentioned, which is endothermic and irreversible. Different values for various variables are given below that.

Heat Effects II (http://www.umich.edu/~elements/6e/icm/heatfx2.html)

The derivation of the user-friendly energy balances are color coded and derived in the tutorials in Heat Effects I and Heat Effects II. Here you see animation of the various terms as they move around the screen and talk to each other to arrive at the final form of the balance equation.

A screenshot shows the tutorial displayed on the CRE website. — A screenshot of a window showing the example of Production of Acetic Anhydride is shown. There is a table given with details about the production. At the bottom of the window there are a few material streams given that are used in the production. Options for reactors displayed include RStoic, RYield, REquil, RGibbs, RCSTR, RPlug, and RBatch. Tabs are present to select options for Mixers or splitters, separators, heat exchanges, columns, reactors, pressure changers, manipulators, solids, user models, and retired models.

A step-by-step AspenTech tutorial is given on the CRE Web site.

See Example 12-2 (http://www.umich.edu/~elements/6e/software/aspen-example12-2.html). Formulated in AspenTech: Download AspenTech directly from the CRE Web site.

Questions, Simulations, and Problems

The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.

The representations for the difficulty levels A, B, C, and D are presented. The difficulty level A is represented by a circle, B is represented by a square, C is represented by a rhombus, and D is represented by two rhombuses.

In each of the questions and problems, rather than just drawing a box around your answer, write a sentence or two describing how you solved the problem, the assumptions you made, the reasonableness of your answer, what you learned, and any other facts that you want to include. See Preface Section G.2 for additional generic parts (x), (y), and (z) to the home problems.

Before solving the problems, state or sketch qualitatively the expected results or trends.

Questions

Q12-1_A QBR (Question Before Reading). Before reading this chapter, identify the differences in applying the energy balance for co-current heat exchange and counter current heat exchange.

Q12-2_A i>clicker. Go to the Web site (http://www.umich.edu/~elements/6e/12chap/iclicker_ch12_q1.html) and view five i>clicker questions. Choose one that could be used as is, or a variation thereof, to be included on the next exam. You also could consider the opposite case: explaining why the question should not be on the next exam. In either case, explain your reasoning.

Q12-3_A Review Figure 12-13. Use this figure to write a few sentences (or at least draw an analogy) explaining why, when you strike the head of a safety match slowly on its pumice with little pressure, it may heat up a little, but does not ignite, yet when you put pressure on it and strike it rapidly, it does ignite. Thanks to Oscar Piedrahita, Medellín, Colombia.

A sketch of an opened safety match box is shown.

Q12-4_A Read over the problems at the end of this chapter. Make up an original problem that uses the concepts presented in this chapter. To obtain a solution:

(a) Make up your data and reaction.

(b) Use a real reaction and real data. See Problem P5-1_A for guidelines.

(c) Prepare a list of safety considerations for designing and operating chemical reactors. (See www.sache.org.) The August 1985 issue of Chemical Engineering Progress may be useful for part (c).

Q12-5_A Go to the LearnChemE screencast link for Chapter 12 (http://www.umich.edu/~elements/6e/12chap/learn-cheme-videos.html). View one or more of the screencast 5- to 6-minute videos and write a two-sentence evaluation. What would you suggest changing in the video to increase the comment from Recommended to Highly Recommended?

Q12-6_A What did the section And Now… A Word From Our Sponsor point out or emphasize that was not done in the other AWFOS sections? What were the top two takeaway points in Chapter 12’s AWFOS?

Computer Simulations and Experiments

We will use the Living Example on the CRE Web site extensively to carry out simulations. Why carry out simulations to vary the parameters in the Living Example Problems? We do it in order to

Get a more intuitive feel reactor system.
Gain insight about the most sensitive parameters (e.g., E, K_C) and how they affect outlet conditions.
Learn how reactors are affected by different operating conditions.
Simulate dangerous situations such as potential runaway reactions.
Compare the model and parameters with experimental data.
Optimize the reaction system.

P12-1_B (a) LEP Table 12-2: Exothermic Reaction with Heat Exchange

Download the Polymath, MATLAB, Python, or Wolfram codes for the algorithm and data given in Table T12-2 for the exothermic gas phase reversible reaction

A + B ⇄ C

given on the Web in the Living Example Problems (LEPs).

Vary the following parameters in the ranges shown in parts (i) through (xi). Write a paragraph describing the trends you find for each parameter variation and why they look the way they do. Use the base case for parameters not varied. The feedback from students on this problem is that one should use Wolfram on the Web in the LEP T12-2 as much as possible to carry out the parameter variations. For each part, write two or more sentences describing the trends.

Wolfram and Python

(i) This is a Stop and Smell the Roses Simulation. View the base case X, X_e, and T profiles and explain why the conversion (X and X_e) and temperature profiles look the way they do.

(ii) The slider parameters Ua/rho, Θ_I, F_A0, and ΔH_Rx all can be moved to make the X and X_e profiles to come together and to separate. What is the reason these sliders affect the profiles similarly?

(iii) Starting with the base case, determine which parameters when changed only a small amount most dramatically affects the conversion and temperature profiles.

(iv) What parameters separate X and X_e the most?

(v) Finally, write at least three conclusions about what you found in your experiments (i) through (iv).

Polymath

(vi) Vary T₀: 310 K ≤ T₀ ≤ 350 K and write a conclusion.

(vii) Vary T_a: 300 K ≤ T_a ≤ 340 K and write a conclusion.

Repeat (i) for countercurrent coolant flow.

Hint: In analyzing the trends it might be helpful to plot X, X_e, and p as a function of W on the same graph and T and T_a as a function of W on the same graph.

(viii) Repeat this problem for the case of constant T_a and adiabatic operation, and describe the most dramatic affect you find.

(b) Example 12-1: Butane Isomerization

Wolfram and Python Co-Current

This is another Stop and Smell the Roses Simulation.

(i) Describe the differences of the X, X_e, and T profiles between the four cases of heat exchange, co-current, counter current, constant T_a, and adiabatic.

(ii) Explain why the temperature (T and T_a) and conversion (X and X_e) look the way they do, and explain what happens to the X, X_e, T, and T_a profiles as you move slider y_A0.

(iii) What parameter value brings T and T_a profiles close together?

(iv) What parameter when varied separates X and X_e the most?

(v) What parameter keeps the X and X_e profiles the closest together?

(vi) Write at least three conclusions about what you found in your experiments (i) through (v).

Polymath Co-Current

(vii) What is the entering value of the temperature T_a₀ of the heat exchanger fluid below which the reaction will never “ignite”?

(viii) Vary some of the other parameters and see whether you can find unsafe operating conditions.

(ix) Plot Q_r and T_a as a function of V necessary to maintain isothermal operation.

Wolfram and Python Countercurrent

(x) Explain why T, T_a, X, and X_e profiles look the way they do for the base case.

(xi) Vary Ua and T_a₀ and describe what you find.

(xii) Vary y_A0 and then one of the other parameters and describe what you find.

(xiii) Describe how Q_r and Q_g and their intersection change when you vary the molar flow rates of coolant, inert and y_A0.

(xiv) Write at least three conclusions about what you found in your experiments (x) through (xiii).

Polymath Countercurrent

(xv) Describe and explain what happens to the X and X_e profiles as you vary ΔH_Rx.

(xvi) Describe what happens to the temperature profile T and T_a as you vary F_A0.

(xvii) Compare the variations in the profiles for X, X_e, T, and T_a when you change the parameter values for all four cases: adiabatic, countercurrent exchange, co-current exchange, and constant T_a. What parameters when changed only a small amount dramatically change the profiles?

(xviii) The heat exchanger is designed for a maximum temperature of 370 K. Which parameter will you vary so that at least 75% conversion is still achieved while maintaining temperature under a safe limit?

Wolfram and Python Constant T_a

(xix) Vary y_A0 and one other parameter of your choice, and describe how the X, X_e, and T profiles change.

(xx) Vary Ua and T_a and describe what you find.

(xxi) Vary ${Δ H}_{Rx}^{°}$ and then one of the other parameters (e.g., y_A0) and describe what you find.

(xxii) Write at least three conclusions about what you found in your experiments (xix)–(xxi).

Wolfram and Python Adiabatic Operation

(xxiii) Why do the shape of the profiles change in the way they do as you change y_A0?

(xxiv) Write a set of three conclusions, one of which compares adiabatic operation with the other three modes of heat transfer (e.g., co-current).

(c) Example 12-2: Production of Acetic Anhydride-Endothermic Reaction Wolfram and Python

Co-Current

The badge of the member of the hall of fame.

(i) Explain why the temperatures (T and T_a) and conversions (X and X_e) look the way they do for the base case slider values.

(ii) What parameter value when increased or decreased causes the reaction to die out the most quickly near the reactor entrance?

(iii) Which parameter, when varied most, drastically changes the profiles?

Adiabatic

(iv) How does the conversion change as heat capacity of A is increased?

Constant T_a

(v) You find that conversion has decreased after 6 months of operation. You checked the flow rate and material properties and found that these values have not changed. Which parameter would have changed?

Countercurrent

(vi) Explain why the temperatures (T and T_a) and conversions (X and X_e) look the way they do for the base case slider values.

(vii) Discuss the most profound differences between the four heat exchange modes (e.g., co-current, adiabatic) for this endothermic reaction.

(viii) Write two conclusions of what you learned from your experiments (i) through (vii).

Polymath

(ix) Let Q_g = r_AΔH_Rx and Q_r = Ua (T – T_a), and then plot Q_g and Q_r on the same figure as a function of V.

(x) Repeat (vi) for V = 5 m³.

(xi) Plot Q_g, Q_r, and –r_A versus V for all four cases on the same figure and describe what you find.

(xii) For each of the four heat exchanger cases, investigate the addition of an inert I with a heat capacity of 50 J/mol · K, keeping F_A0 constant and letting the other inlet conditions adjust accordingly (e.g., ε).

(xiii) Vary the inert molar flow rate (i.e., Θ_I, 0.0 < Θ_I < 3.0 mol/s). Plot X and analyze versus Θ_I.

(xiv) Finally, vary the heat-exchange fluid temperature T_a₀ (1000 K < T_a₀ < 1350 K). Write a paragraph describing what you find, noting interesting profiles or results.

(d) Example 12-2: AspenTech Formulation. Go to the Aspen LEP and repeat Example 12-2 using AspenTech.

(e) Example 12-3: Production of Propylene Glycol in a CSTR

Wolfram and Python

(i) Vary activation energy (E) and find the values of E for which there are at least two solutions.

(ii) Vary the flow rate of F_B0 to find the temperature at which the conversion is 0.8.

(iii) What is the operating range of inlet temperatures such that at least one steady state solution exists while maintaining the reactor temperature below 640°R? Describe how your answers would change if the molar flow rate of methanol were increased by a factor of 4.

(iv) Vary V and find the points of (1) tangency and (2) intersection of X_EB and X_MB.

(v) Write a set of conclusions based on your experiments (i) through (iv).

(f) Example 12-4: CSTR with a Cooling Coil

Wolfram and Python

(i) Explore how variations in the activation energy (E) and in the heat of reaction, at reference temperature ( $Δ H_{Rx}^{\circ}$ ), affect the conversions X_EB and X_MB.

(ii) What variables most affect the intersection of and the points of tangency of X_EB and X_MB?

(iii) Find a value of E and of $Δ H_{Rx}^{\circ}$ that increase conversion to more than 80% while maintaining the constraint of 125°F maximum temperature.

(iv) Write two conclusions about what you found in your experiments (i) through (iii).

Polymath

(v) The space time was calculated to be 0.01 minutes. Is this realistic? Decrease ν₀ and both reaction rate constants by a factor of 100 and describe what you find.

(vi) Use Figure E12-3.2 and Equation (E12-4.4) to plot X versus T to find the new exit conversion and temperature.

(vii) Other data on the Jofostan chemical engineering Web site show $Δ H_{Rx}^{\circ} = - 38700$ Btu/lb-mol and ΔC_P = 29 Btu/lb-mol/°F. How would these values change your results?

(viii) Make a plot of conversion as a function of heat exchanger area. [0 < A < 200 ft²] and write a conclusion.

(g) Example 12-5: Parallel Reactions in a PFR with Heat Effects

Wolfram and Python

(i) Vary the sliders for the rate constants k_1A0 and k_2A0 and describe what you find.

(ii) What should be the initial concentration of A so that the company can produce equal flow rates of B and C?

(iii) Describe how the profiles for T, F_A, F_B, and F_C change as you move the sliders. Comment specifically on the drastic changes that occur when you change the overall heat transfer coefficient.

(iv) What set of conditions give you the greatest selectivity, ${\tilde{S}}_{B/C}$ at the exit and what is the corresponding conversion.

(v) Write at least three conclusions about what you found in your experiments (i) through (iv).

Polymath

(vi) Why is there a maximum in temperature? How would your results change if there is a -pressure drop with α = 1.05 dm^–3?

(vii) What if the reaction is reversible with K_C= 10 at 450 K?

(viii) How would the selectivity change if Ua is increased? Decreased?

(h) Example 12-6: Multiple Reactions in a CSTR (Use the LEP)

Wolfram and Python

(i) What is the minimum value of T_a that causes the reactor temperature to jump to the single, upper steady-state value?

(ii) What is the operating range for the entering temperature T₀ for which there are only three steady states?

(iii) Vary UA between 10,000 and 80,000, and describe how the number of steady-state solutions changes.

(iv) Write a set of conclusions from carrying out experiments (i) through (iii).

(i) Example 12-7: Complex Reactions with Heat Effects in a PFR—Safety

Wolfram and Python

(i) Co-current: Explore the effect of the coolant flow rate ${\dot{m}}_{c}$ on the reactant temperature and coolant temperature profiles, and describe what you find.

(ii) Co-current: Explore the effect of Ua on the selectivity and describe what you find.

(iii) Constant T_a: How does the overall heat transfer coefficient affect the product flow rate distribution? Should it be kept minimum or maximum? Explain.

(iv) Constant T_a: Which are the variables that have virtually no effect on the profiles? Explain the reason.

(v) Adiabatic: You want to save capital cost by using a smaller volume reactor, that is, 5 dm³ instead of 10 dm³. Which parameters will you vary to achieve exit reactant and product flow rates same as base case?

(vi) Describe how the selectivity, ${\tilde{S}}_{C/D}$ changes as you change the parameters and operating conditions. What two variable slider(s) affect ${\tilde{S}}_{C/D}$ the most?

(vii) Write a set of conclusions of what you learned in experiments (i) through (vi).

Polymath

(viii) Plot Q_g and Q_r as a function of V. How can you keep the maximum temperature below 700 K? Would adding inerts help, and if so what should the flow rate be if C_{P_I} = 10 cal/mol/K?

(ix) Look at the figures. What happened to species D? What conditions would you suggest to make more species D?

(x) Make a table of the temperature (e.g., maximum T, T_a) and molar flow rates at two or three volumes, comparing the different heat-exchanger operations.

(xi) Why do you think the molar flow rate of C does not go through a maximum? Vary some of the parameters to learn whether there are conditions where it goes through a maximum. Start by increasing F_A0 by a factor of 5.

(xii) Include pressure in this problem. Vary the pressure-drop parameter (0 < αρ_b < 0.0999 dm^–3) and describe what you find and write a conclusion.

(j) Review the steps and procedure by which we derived Equation (12-5) and then, by analogy, derive Equation (12-35).

(k) CRE Web site SO₂ Example PRS-R12.4-1. Download the SO₂ oxidation LEP R12-1. How would your results change if (1) the catalyst particle diameter were cut in half? (2) The pressure were doubled? At what particle size does pressure drop become important for the same catalyst weight, assuming the porosity doesn’t change? (3) You vary the initial temperature and the coolant temperature? Write a paragraph with at least two conclusions describing what you find.

The badge of the member of the hall of fame.

(l) SAChE. Go to the SAChE Web site, www.sache.org. Your instructor or department chair should have the username and password to enter the SAChE Web site in order to obtain the modules with the problems. On the left-hand menu, select “SAChE Products.” Select the “All” tab and go to the module titled, “Safety, Health and the Environment” (S, H & E). The problems are for KINETICS (i.e., CRE). There are some example problems marked “K” and explanations in each of the above S, H & E selections. Solutions to the problems are in a different section of the site. Specifically look at: Loss of Cooling Water (K-1), Runaway Reactions (HT-1), Design of Relief Values (D-2), Temperature Control and Runaway (K-4) and (K-5), and Runaway and the Critical Temperature Region (K-7). Go through the K problems and write a paragraph on what you have learned.

P12-2_B (a) Example 12-2 Production of Acetic Anhydride Endothermic Reaction

Case 1 Adiabatic

(i) Vary inlet flow rate of inert and observe the conversion and temperature profiles. Describe what you find.

(ii) Find inlet temperature, T₀, for which the reaction rate at V = 0.001 m³ is 0.1 times the rate at inlet. Note the conversion at the reactor exit for this temperature.

(iii) After varying two other parameters, write a set of conclusions from your experiments in (i) and (ii).

Case 2 Constant T_a

(iv) To reduce operating cost, the inlet temperature of heating fluid can be reduced. Can you find a minimum temperature of heating fluid at which exit conversion is virtually 100% for constant T_a.

(v) Find maximum value of the inlet concentration for which the conversion at exit is virtually 100% constant T_a.

(vi) Write a set of conclusions from your experiments in (iv) and (v).

P12-3_B OEQ (Old Exam Question). Computer Experiments on a PFR for Reaction in Table 12-2.2. In order to develop a greater understanding of the temperature effects in PBRs, download the Living Example Problem Table 12-2, LEP T12-2, from the Web site to learn how changing the different parameters changes the conversion and temperature profiles.

This is a Sherlock Holmes Problem. Starting with the base case, one, and only one, parameter was varied between its maximum and minimum values to give the following figures. You need to decide what operating variable (e.g., T₀) property value (e.g., C_{P_A}), or engineering variable (e.g., $\frac{U_{a}}{ρ_{b}}$ ) was varied. Choose from the following:

$k_{1}, {E, R, C}_{T0}, T_{a}, T_{0}, T_{1}, T_{2}, K_{C2}, Θ_{B}, Θ_{1}, Δ H_{Rx}^{\circ}, C_{P_{A}}, C_{P_{B}}, C_{P_{C}}, U a, ρ_{a}$

A set of trend graphs shows a design of the steady-state isothermal rector for flow reactors and heat exchange. — 10 Graphs are shown, each displaying minimum or maximum values and two curves are plotted for X and X subscript e. The horizontal axis of each graph represents the weight W in kilogram and the vertical axis of each graph represents X and X Subscript e. It is observed that X subscript e shows a higher value that X for almost all the cases. 5 graphs show intersecting points of X and X subscript e. In 4 graphs, X subscript e shows a nearly constant trend, while x is closer to the horizontal axis (low value).

Interactive Computer Games

P12-4_A Download the Interactive Computer Games (ICG) from the CRE Web site. Play the game, and then record your performance number for the module, which indicates your mastery of the material. Note: For simulation (b), only do the first three reactors, as reactors 4 and above do not work because of the technician tinkering with them.

(a) ICG Heat Effects Basketball 1 Performance # ________________.

(b) ICG Heat Effects Simulation 2 Performance # ________________.

Before attempting to play the ICG/Interactive Computer Games simulations, be sure to first go through the reviews to see the equations talking to each other.

Problems

P12-5_C OEQ (Old Exam Question). Safety Problem. The following is an excerpt from The Morning News, Wilmington, Delaware (August 3, 1977): “Investigators sift through the debris from blast in quest for the cause [that destroyed the new nitrous oxide plant]. A company spokesman said it appears more likely that the [fatal] blast was caused by another gas—ammonium nitrate—used to produce nitrous oxide.” An 83% (wt) ammonium nitrate and 17% water solution is fed at 200°F to the CSTR operated at a temperature of about 510°F. Molten ammonium nitrate decomposes directly to produce gaseous nitrous oxide and steam. It is believed that pressure fluctuations were observed in the system and, as a result, the molten ammonium nitrate feed to the reactor may have been shut off approximately 4 min prior to the explosion.

Assume that at the time the feed to the CSTR stopped, there was 500 lb_m of ammonium nitrate in the reactor. The conversion in the reactor is believed to be virtually complete at about 99.99%.

Additional information (approximate but close to the real case):

$\begin{matrix} Δ H_{Rx}^{\circ} & = & - 336 {Btu/lb}_{m} {ammonium nitrate at 500}^{\circ} F (constant) \\ C_{P} & = & 0.38 {Btu/lb}_{m} ammonium nitrate \cdot^{\circ} F \\ C_{P} & = & 0.47 {Btu/lb}_{m} of steam \cdot^{\circ} F \\ - r_{A} V & = & k C_{A} V = k \frac{M}{V} V = k M ({lb}_{m} / h) \end{matrix}$

where M is the mass of ammonium nitrate in the CSTR (lb_m) and k is given by the relationship below.

T (°F)	510	560
k (h^-1)	0.307	2.912

The enthalpies of water and steam are

$\begin{matrix} H_{w} (200^{\circ} F) = 168 {Btu/lb}_{m} \\ H_{g} (500^{\circ} F) = 1202 {Btu/lb}_{m} \end{matrix}$

(a) Can you explain the cause of the blast? Hint: See Problem P13-3_B.

(b) If the feed rate to the reactor just before shutoff was 310 lb_m of solution per hour, what was the exact temperature in the reactor just prior to shutdown? Hint: Plot Q_r and Q_g as a function of temperature on the same plot.

(c) How would you start up or shut down and control such a reaction? Hint: See Problem P13-2_B.

(d) Explore this problem and describe what you find. For example, add a heat exchanger UA (T – T_a), choose values of UA and T_a, and then plot R(T) versus G(T)?

(e) Discuss what you believe to be the point of the problem. The idea for this problem originated from an article by Ben Horowitz.

P12-6_B OEQ (Old Exam Question)—Taken from California Professional Engineer’s Exam. The endothermic liquid-phase elementary reaction

A + B → 2C

An outline of the map of California is shown.

proceeds, substantially, to completion in a single steam-jacketed, continuous-stirred reactor (Table P12-6_B). From the following data, calculate the steady-state reactor temperature:

Reactor volume: 125 gal

Steam jacket area: 10 ft²

Jacket steam: 150 psig (365.9°F saturation temperature)

Overall heat-transfer coefficient of jacket, U: 150 Btu/h · ft² · °F

Agitator shaft horsepower: 25 hp

Heat of reaction, $Δ H_{Rx}^{\circ} = + 20000$ Btu/lb-mol of A (independent of temperature)

TABLE P12-6_B FEED CONDITIONS AND PROPERTIES

	Component
	A	B	C
Feed (lb-mol/hr)	10.0	10.0	0
Feed temperature (°F)	80	80	—
Specific heat (Btu/lb-mol·°F)^*	51.0	44.0	47.5
Molecular weight	128	94	111
Density (lb_m/ft³)	63.0	67.2	65.0

^* Independent of temperature. (Ans: T = 199°F)

(Courtesy of the California Board of Registration for Professional & land surveyors.)

P12-7_B OEQ (Old Exam Question). Use the data in Problem P11-4_A for the following reaction. The elementary, irreversible, organic liquid-phase reaction

A + B → C

is carried out in a flow reactor. An equal molar feed in A and B enters at 27°C, and the volumetric flow rate is 2 dm³/s and C_A0 = 0.1 kmol/m³.

Additional information:

$\begin{matrix} H_{A}^{\circ} (273 K) = - 20 kcal/mol, & H_{B}^{\circ} (273 K) = - 15 kcal/mol, & H_{C}^{\circ} (273 K) = - 41 kcal/mol \\ C_{P_{A}} = C_{P_{B}} = 15 cal/mol \cdot K & C_{P_{C}} = 30 cal/mol \cdot K \\ k = 0.01 \frac{{dm}^{3}}{mol \cdot s} at 300 K & E = 10000 cal/mol \\ U a = 20 {cal/m}^{3} / s/ K & {\dot{m}}_{C} = 50_{g /s} \\ T_{a 0} = 450 K & C_{P_{Cool}} = 1 cal/g/K \end{matrix}$

(a) Calculate the conversion when the reaction is carried out adiabatically in one 500-dm³ CSTR and then compare the results with the two adiabatic 250-dm³ CSTRs in series.

The reversible reaction (part (d) of Problem P11-4_A) is now carried out in a PFR with a heat exchanger. Plot and then analyze X, X_e, T, T_a, Q_r, Q_g, and the rate, –r_A, for the following cases:

(b) Constant heat-exchanger temperature T_a

(c) Co-current heat exchanger T_a (Ans: At V = 10 m³ then X = 0.36 and T = 442 K)

(d) Countercurrent heat exchanger T_a (Ans: At V = 10 m³ then X = 0.364 and T = 450 K)

(e) Adiabatic operation

(f) Make a table comparing all your results (e.g., X, X_e, T, T_a). Write a paragraph describing what you find.

(g) Plot Q_r and T_a as a function of V necessary to maintain isothermal operation.

P12-8_A The gas-phase reversible reaction as discussed in Problem P11-7_B

A ⇄ B

is now carried out under high pressure in a packed-bed reactor with pressure drop. The feed consists of both inerts I and species A with the ratio of inerts to the species A being 2 to 1. The entering molar flow rate of A is 5 mol/min at a temperature of 300 K and a concentration of 2 mol/dm³. Work this problem in terms of volume. Hint: $V = W/ ρ_{B}, r_{A} = ρ_{B} r_{A}^{'}$

Additional information:

$\begin{matrix} F_{A0} = 5.0 mol/min & T_{0} = 300 K & Δ H_{Rx} = - 20000 cal/mol & {\begin{matrix} α ρ \end{matrix}}_{b} = 0.02 {dm}^{- 3} \\ C_{A0} = 2 {mol/dm}^{3} & T_{1} = 300 K & K_{C} = 1000 at 300 K & Coolant \\ C_{1} = 2 C_{A0} & k_{1} = 0.1 {min}^{- 2} at 300 K & C_{{P_{B}}_{}} = 160 cal/mol/K & {\dot{m}}_{C} = 50 mol/min \\ C_{P_{I}} = 18 cal/mol/K & U a = 150 {cal/dm}^{3} / min/K & p_{B} = 1.2 {kg/dm}^{3} & C_{P_{Cool}} = 20 cal/mol/K \\ C_{P_{A}} = 160 cal/mol/ K & T_{a 0} = 300 K \\ E = 10000 cal/mol & V = 40 {dm}^{3} \end{matrix}$

Plot and then analyze X, X_e, T, T_a, and the rate (–r_A) profiles in a PFR for the following cases. In each case, explain why the curves look the way they do.

(a) Co-current heat exchange

(b) Countercurrent heat exchange (Ans: When V = 20 dm³ then X = 0.86 and X_e = 0.94)

(c) Constant heat-exchanger temperature T_a

(d) Compare and contrast each of the above results and the results for adiabatic operation (e.g., make a plot or a table of X and X_e obtained in each case).

(e) Vary some of the parameters, for example, (0 < Θ_I < 10) and describe what you find.

(f) Plot Q_r and T_a as a function of V necessary to maintain isothermal operation.

P12-9_A Algorithm for reaction in a PBR with heat effects and pressure drop

The elementary gas-phase reaction

A + B ⇄ 2C

in Problem P11-8_B is now continued and carried out in packed-bed reactor. The entering molar flow rates are F_A0 = 5 mol/s, F_B0 = 2F_A0, and F_I = 2F_A0 with C_A0 = 0.2 mol/dm³. The entering temperature is 325 K and a coolant fluid is available at 300 K.

Additional information:

$\begin{matrix} C_{P_{A}} = C_{P_{B}} = C_{P_{C}} = 20 cal/mol/ K & k = 0.0002 \frac{{dm}^{6}}{kg \cdot mol \cdot s} @300K \\ C_{P_{A}} = 18 cal/mol/K & α = 0.00015 {kg}^{- 1} & U a = 320 \frac{Cal}{s \cdot m^{3} \cdot K} \\ E = 25 \frac{kcal}{mol} & {\dot{m}}_{C} = 18 mol/s & ρ_{b} = 1400 \frac{kg}{m^{3}} \\ Δ H_{Rx} = - 20 \frac{kcal}{mol} @ 298 K & C_{P_{Cool}} = 18 cal/mol (coolant) \\ K_{C} = 1000 @ 305 K \end{matrix}$

Plot X, X_e, T, T_a, and –r_A down the length of the PFR for the following cases:

(a) Co-current heat exchange

(b) Countercurrent heat exchange

(c) Constant heat-exchanger temperature T_a

(d) Compare and contrast your results for (a), (b), and (c) along with those for adiabatic operation and write a paragraph describing what you find.

P12-10_B Use the data and reaction in Problems P11-4_A and P12-7_B for the following reaction:

A+B → C+D

(a) Plot and then analyze the conversion, Q_r, Q_g, and temperature profiles up to a PFR reactor volume of 10 dm³ for the case when the reaction is reversible with K_C = 10 m³/kmol at 450 K. Plot and then analyze the equilibrium conversion profile.

(b) Repeat (a) when a heat exchanger is added, Ua = 20 cal/m³/s/K, and the coolant temperature is constant at T_a = 450 K.

(c) Repeat (b) for both a co-current and a countercurrent heat exchanger. The coolant flow rate is 50 g/s, C_P_c = 1 cal/g · K, and the inlet coolant temperature is T_a₀ = 450 K. Vary the coolant rate $(10 < {\dot{m}}_{c} < 1000_{g / s})$ .

(d) Plot Q_r and T_a as a function of V necessary to maintain isothermal operation.

(e) Compare your answers to (a) through (d) and describe what you find. What generalizations can you make?

(f) Repeat (c) and (d) when the reaction is irreversible but endothermic with ${Δ H}_{Rx}^{°} = 6000 cal/mol$ . Choose T_a₀ = 450 K.

P12-11_B OEQ (Old Exam Question). Use the reaction data in Problems P11-4_A and P12-7_B for the case when heat is removed by a heat exchanger jacketing the reactor. The flow rate of coolant through the jacket is sufficiently high that the ambient exchanger temperature is constant at T_a = 50°C.

A+B → C

(a) (1) Plot and then analyze the temperature conversion, Q_r, and Q_g profiles for a PBR with where

$\frac{U a}{ρ_{b}} = 0.08 \frac{J}{s \cdot kg-cat \cdot K}$

where

ρ_b = bulk density of the catalyst (kg/m³)

a = heat-exchange area per unit volume of reactor (m²/m³)

U = overall heat transfer coefficient (J/s · m² · K)

(2) How would the profiles change if Ua/=_b were increased by a factor of 3000?

(3) If there is a pressure drop with α = 0.019 kg^–1?

(b) Repeat part (a) for co-current and countercurrent flow and adiabatic operation with ${\dot{m}}_{c} = 0.2 kg/s$ , C_{P_c} = 5000 J/kg K and an entering coolant temperature of 50°C.

(c) Find X and T for a “fluidized” CSTR with 80 kg of catalyst.

$\begin{matrix} U A = 500 \frac{J}{s \cdot K}, & ρ_{b} \end{matrix} = 1 {kg/m}^{3}$

(d) Repeat parts (a) and (b) for W = 80.0 kg, assuming a reversible reaction with a reverse specific reaction rate of

$\begin{matrix} k_{r} = 0.2 exp [\frac{E_{r}}{R} (\frac{1}{450} - \frac{1}{T})] (\frac{{dm}^{6}}{kg-cat \cdot mol \cdot s}); & E_{r} = 51.4 kJ/mol \end{matrix}$

A diagram of an adiabatic reactor is shown. The reactor has one opening on top and another opening at the bottom. The middle portion of the reactor consists of fluidized catalyst pellets.

Vary the entering temperature, T₀, and describe what you find.

(e) Use or modify the data in this problem to suggest another question or calculation. Explain why your question requires either critical thinking or creative thinking. See Preface Section G and http://www.umich.edu/~scps.

P12-12_C Derive the energy balance for a packed-bed membrane reactor. Apply the balance to the reaction in Problem P11-5_A

A ⇄ B + C

for the case when it is reversible with K_C = 1.0 mol/dm³ at 300 K. Species C diffuses out of the membrane with k_C = 1.5 s^–1.

(a) Plot and then analyze the concentration profiles for different values of K_C when the reaction is carried out adiabatically.

(b) Repeat part (a) when the heat transfer coefficient is Ua = 30 J/s·kg-cat·K with T_a = 50°C.

P12-13_B OEQ (Old Exam Question). Circle the correct answer.

(a) The elementary reversible isomerization of A to B was carried out in a packed-bed reactor. The following profiles were obtained:

Profiles obtained for isomerization of A to B in a packed-bed reactor is shown. — Two trend graph are shown. The horizontal axis of the first graph represents the weight W in Kg and the vertical axis of the first graph represents X. A trend curve is drawn on the graph from the origin and traces the path of the point 0.4 on the y-axis. Two dotted lines are drawn passing through the curve from the x-axis one for 4 kilograms and another for 10 kilograms. The horizontal axis of the second graph represents the weight W in Kilogram and the vertical axis of the second graph represents T and ranges from 500 to 600 in increments of 100. A trend curve is drawn on the graph from the point 570 on the y-axis and comes down to the point 500. Two dotted lines are drawn passing through the curve from the x-axis one for 4 kilograms and another for 10 kilograms.

If the total entering volumetric flow rate remains constant, the addition of inerts to the feed stream will most likely

A) Increase conversion.

B) Decrease conversion.

C) Have no effect.

D) Insufficient information to tell

(b)

Which of the following statements are true?

A) The above reaction could be adiabatic.

B) The above reaction could be exothermic with constant cooling temperature.

C) The above reaction could be endothermic with constant heating temperature.

D) The above reaction could be second order.

(c) The conversion is shown below as a function of catalyst weight down a PBR.

Which of the following statements are false?

A) The reaction could be first-order endothermic and carried out adiabatically.

B) The reaction could be first-order endothermic and reactor is heated along the length with T_a being constant.

C) The reaction could be second-order exothermic and cooled along the length of the reactor with T_a being constant.

D) The reaction could be second-order exothermic and carried out adiabatically.

E) The reaction could be irreversible.

To view more conceptual problems similar to (a)–(c) above go to http://www.umich.edu/~elements/6e/12chap/iclicker_ch12_q1.html.

P12-14_A OEQ (Old Exam Question). The irreversible reaction

A+B → C+D

is carried out adiabatically in a CSTR. The “heat generated” G(T) and the “heat removed” R(T) curves are shown in Figure P12-14_A.

A trend graph of heat-removed and heat-generated against the temperature in Celsius is shown. — **Figure P12-14_A** Heat removed R(T) and heat “generated” G(T) curves.

The graph represents the data of heat-removed and heat-generated corresponding to the temperature. The horizontal axis represents the temperature in degree Celsius and ranges from 200 to 450 in increments of 25. The vertical axis represents the values of heat-removed - R(T) and heat-generated - G(T) and ranges from 0 to 12000 in increments of 2000. The heat-removed R(T) is represented in a dotted straight line whereas the heat-generated G(T) is represented as a curve. Both the dotted lines and curve are intertwined and there are steady state points at which the dotted line and the curve intersect each other. The temperature and the respective R(T) and G(T) values at different steady states are as follows: Steady state 1: Temperature - 240 degree Celsius and G(T),R(T) - 1000 cal/mol, Steady state 2: Temperature - 330 degree Celsius and G(T),R(T) - 6400 cal/mol, Steady state 3: Temperature - 410 degree Celsius and G(T),R(T) - 1190 cal/mol. Note that the values are approximate.

(a) What is the ΔH_Rx of the reaction?

(b) What are the inlet ignition and extinction temperatures?

(c) What are all the temperatures in the reactor corresponding to the inlet ignition and extinction temperatures?

(d) What are the conversions at the ignition and extinction temperatures?

P12-15_B The first-order, irreversible, exothermic liquid-phase reaction

A → B

is to be carried out in a jacketed CSTR. Species A and an inert I are fed to the reactor in equimolar amounts. The molar feed rate of A is 80 mol/min.

Additional information:

$\begin{matrix} Heat capacity of the inert: 30 cal/mol \cdot^{\circ} C & τ = 100 min \\ Heat capacity of A and B: 20 cal/mol \cdot^{\circ} C & Δ H_{Rx}^{\circ} = - 7500 cal/mol \\ UA: 8000 cal/min \cdot^{\circ} C & k = 6.6 \times 10^{- 3} {min}^{- 1} at 350 K \\ Ambient temperature, T_{a} : 300 K & E = 40000 cal/mol \cdot K \end{matrix}$

(a) What is the reactor temperature for a feed temperature of 450 K?

(b) Plot and then analyze the reactor temperature as a function of the feed temperature.

(c) To what inlet temperature must the fluid be preheated for the reactor to operate at a high conversion? What are the corresponding temperature and conversion of the fluid in the CSTR at this inlet temperature?

(d) Suppose that the fluid inlet temperature is now heated 5°C above the reactor temperature in part (c) and then cooled 20°C, where it remains. What will be the conversion?

(e) What is the inlet extinction temperature for this reaction system? (Ans: T₀ = 87°C)

P12-16_B The elementary reversible liquid-phase reaction

A reversible reaction of A and B is shown. Where A is the reactant and B is the product or vice-versa.

takes place in a CSTR with a heat exchanger. Pure A enters the reactor.

(a) Derive an expression (or set of expressions) to calculate G(T) as a function of the heat of reaction, equilibrium constant, temperature, and so on. Show a sample calculation for G(T) at T = 400 K.

(b) What are the steady-state temperatures? (Ans: 310, 377, 418 K)

(c) Which steady states are locally stable?

(d) What is the conversion corresponding to the upper steady state?

(e) Vary the ambient temperature T_a and make a plot of the reactor temperature as a function of T_a, identifying the ignition and extinction temperatures.

(f) If the heat exchanger in the reactor suddenly fails (i.e., UA = 0), what would be the conversion and the reactor temperature when the new upper steady state is reached? (Ans: 431 K)

(g) What heat exchanger product, UA, will give the maximum conversion?

(h) Write a question that requires critical thinking and then explain why your question requires critical thinking. Hint: See Preface Section G.

(i) What is the adiabatic blowout flow rate, υ₀?

(j) Suppose that you want to operate at the lower steady state. What parameter values would you suggest to prevent runaway, for example, the upper SS?

Additional information:

$\begin{matrix} U A = 3600 cal/min \cdot K & E/R = 20000 K \\ C_{P_{A}} = C_{P_{B}} = 40 cal/mol \cdot K & V = 10 {dm}^{3} \\ Δ H_{Rx}^{\circ} = - 80000 cal/mol A & υ_{0} = 1 {dm}^{3} / \min \\ K_{C} = 100 at 400 k & F_{A0} = 10 mol/min \\ k = 1 {min}^{- 1} at 400 K \\ Ambient temperature, T_{a} = 37^{\circ} & Feed temperature, T_{0} = 37^{\circ} C \end{matrix}$

P12-17_B OEQ (Old Exam Question). The reversible liquid-phase reaction

A ⇄ B

is carried out in a 12-dm³ CSTR with heat exchange. Both the entering temperature, T₀, and the heat exchange fluid, T_a, are at 330 K. An equal molar mixture of inerts and A enter the reactor.

(a) Choose a temperature, T, and carry out a calculation to find G(T) to show that your calculation agrees with the corresponding G(T) value on the curve shown in Figure P12-17_B at the temperature you choose.

(b) Find the exit conversion and temperature from the CSTR. X = _____ T = _____ Answers

(c) What entering temperature T₀ would give you the maximum conversion? T₀ = _____ X = _____

(d) What would the exit conversion and temperature be if the heat-exchange system failed (i.e., U = 0)?

(e) Can you find the inlet ignition and extinction temperatures? If yes, what are they? If not, go on to the next problem.

(f) Use Preface Section G to ask another question.

Additional information:

The G(T) curve for this reaction is shown in Figure P12-17_B.

$\begin{matrix} C_{P_{A}} = C_{P_{B}} = 100 {cal/mol/K, C}_{P_{I}} = 150 cal/mol/K & k = 0.001 h^{- 1} at 300 K with E=30000 cal/mol \\ F_{A0} = {10 mol/h, C}_{A0} = {1 mol/dm}^{3}, υ_{0} = 10 {dm}^{3} /h & K_{C} = 5, 000, 000 at 300 K \\ Δ H_{Rx} = - 42000 cal/mol & UA = 5000 cal/h/K \end{matrix}$

A graph plots heat-removed against the temperature (in kelvin). — **Figure P12-17_B** Heat removed curve, G(T), for a reversible reaction.

The graph represents the data of heat removed corresponding to the temperature. The horizontal axis represents the temperature in degree Celsius and ranges from 290 to 440 in increments of 15. The vertical axis represents the values of heat removed - G(T) and ranges from 0 to 40500 in increments of 4500. The curve begins at a G(T) value of 0 when the temperature is 290 K. It then goes up and reaches the maximum of 36050 cal/mol when the temperature is 365K. From there, the value drops down to 0 again when the temperature is 440 K. Note that the values are approximate.

P12-18_C The elementary gas-phase reaction

2A ⇄ C

is carried out in a packed-bed reactor. Pure A enters the reactor at a 450-K flow rate of 10 mol/s, and a concentration of 0.25 mol/dm³. The PBR contains 90 kg of catalyst and is surrounded by a heat exchanger for which cooling fluid is available at 500 K. Compare the conversion achieved for the four types of heat exchanger operation: adiabatic, constant T_a, co-current flow, and countercurrent flow.

Additional information:

$\begin{matrix} α = 0.019 /kg-cat & C_{P_{C}} = 20 J/mol/K \\ U a / ρ_{b} = 0.8 J/kg-cat \cdot s \cdot K & F_{A 0} = 10 mol/h \\ Δ H_{Rx} = - 20000 J/mol & C_{A0} = {1 mol/dm}^{3} \\ C_{P_{A}} = 10 J/mol \cdot K & υ_{0} = 10 {dm}^{3} / h \end{matrix}$

P12-19_C A reaction is to be carried out in the packed-bed reactor shown in Figure P12-19_C.

A diagram of a packed-bed reactor with a heat exchange is shown. The instrument is in the shape of a cylinder and there is tube placed inside and it carries the catalyst. Directional arrows are drawn to represent the heat transfer. — **Figure P12-19_C** PFR with heat exchange.

The reactants enter the annular space between an outer insulated tube and an inner tube containing the catalyst. No reaction takes place in the annular region. Heat transfer between the gas in this packed-bed reactor and the gas flowing countercurrently in the annular space occurs along the length of the reactor. The overall heat-transfer coefficient is 5 W/m² · K. Plot the conversion and temperature as a function of reactor length for the data given in Problem P12-7_B.

P12-20_B OEQ (Old Exam Question). The reaction

$A + B \vec{\leftarrow} 2 C$

is carried out in a packed-bed reactor. Match the following temperature and conversion profiles for the four different heat-exchange cases: adiabatic, constant T_a, co-current exchange, and countercurrent exchange.

Six trend graphs represent different heat-exchange cases for a reaction that is carried out in a packed-bed reactor. — For the different heat-exchange cases that occurs in a reaction carried out in a packed-bed reactor six different trend graphs are shown. The first two graphs have their horizontal axis representing W and ranges from 0 to 4350 and the vertical axis ranges from 0.00 to 1.00. The two graphs have three trend curves each representing P, X Subscript e, and X. The third graph has its horizontal axis representing W and ranges from 0 to 4350 and the vertical axis ranges from 0.00 to 0.70. There are two trend curves on the graph each representing X and X Subscript e. The third graph has its horizontal axis representing W and ranges from 0 to 4350 and the vertical axis ranges from 0.00 to 0.70. There are two trend curves on the graph each representing X and X Subscript e. The fourth graph has its horizontal axis representing W and ranges from 0 to 4350 and the vertical axis ranges from 0.00 to 1.00. There are three trend curves on the graph each representing P, X and X Subscript e. The fifth graph has the horizontal axis representing W and ranges from 0 to 4350 and the vertical axis ranges from 320 to 370. There are two trend curves drawn on the graph. The trend curve representing T begins at the value 330 on the y-axis when W is 0 and goes up to the highest value of 375 on the y-axis when W is 870. It then drops down to the value of 340 when W is 4350. The trend curve representing T Subscript a begins with a value of 320 on the y-axis when W is 0 and then goes up and joins T at the value of 340 when W is 4350. The sixth graph has the horizontal axis representing W and ranges from 0 to 4350 and the vertical axis ranges from 320 to 380. There is a trend curve drawn on the graph. The trend curve representing T begins at the value 330 on the y-axis when W is 0 and goes up to the highest value of 374 on the y-axis when W is 870. From there the y-axis value of the curve remains constant while the final W value is 4350. The seventh graph has the horizontal axis representing W and ranges from 0 to 4350 and the vertical axis ranges from 315 to 380. There are two trend curves drawn on the graph. The trend curve representing T begins at the value 330 on the y-axis when W is 0 and goes up to the highest value of 373 on the y-axis when W is 870. It then drops down to the value of 328 when W is 4350. The trend curve representing T Subscript a begins at the value 341 on the y-axis when W is 0 and then goes down and joins T at the value of 328 when W is 4350. The eighth graph has the horizontal axis representing W and ranges from 0 to 4350 and the vertical axis ranges from 315 to 360. There is a trend curve drawn on the graph. The trend curve representing T begins at the value 330 on the y-axis when W is 0 and goes up to the highest value of 359 on the y-axis when W is 870. From there the curve drops down to the value of 324 while the final W value is 4350. Note that the values are approximate.

(a) Figure 1 matches Figure ___

(b) Figure 2 matches Figure ___

(c) Figure 3 matches Figure ___

(d) Figure 4 matches Figure ___

P12-21_B OEQ (Old Exam Question). Also Hall of Fame Problem. The irreversible liquid-phase reactions

$\begin{matrix} Reaction(1) & A + B \to 2 C & r_{1 C} = k_{1 C} C_{A} C_{B} \\ Reaction(2) & 2 B + C \to D & r_{2 D} = k_{2 D} C_{B} C_{C} \end{matrix}$

are carried out in a PFR with heat exchange. The temperature profiles shown in Figure P12-21_B were obtained for the reactor and the coolant stream:

Two graphs represents the temperature profiles for the reactor and coolant steam in a packed-bed reactor. — **Figure P12-21_B** Reactant temperature T and coolant temperature *T_a* profiles.

Two graphs, one plots Reactant temperature (T) against the volume (V) and another plots the coolant temperature (T Subscript a) against the volume (V) are shown. The horizontal axis of the first graph represents V in meter cubed and the vertical axis of the first graph ranges from 400 to 550 in increments of 50. There is a curve drawn on the graph from the point 430 on the y-axis from where it goes up to 500 and reaches a midpoint. From there it drops down to the value of 400. The points where the values of y-axis meets the x-axis is marked using dotted lines. The lowest point of the curve is said to be 400 whereas the highest point is said to be 500. The horizontal axis of the second graph represents V in meter cubed and the vertical axis of the second graph ranges from 300 to 375 in increments of 25. There is a curve drawn on the graph from the point 300 on the y-axis from where it goes up to 325 and reaches the midpoint. From there it further goes up to the value of 350. The points where the values of y-axis meets the x-axis is marked using straight lines. The lowest point of the curve is observed to be 400 whereas the highest point is observed to be 500. And the mid value 325 is marked using a dotted line. Note that the values are approximate.

The concentrations of A, B, C, and D were measured at the point down the reactor where the liquid temperature, T, reached a maximum, and they were found to be C_A = 0.1, C_B = 0.2, C_C = 0.5, and C_D = 1.5, all in mol/dm³. The product of the overall heat transfer coefficient and the heat-exchanger area per unit volume, Ua, is 10 cal/s · dm³ · K. The entering molar flow rate of A is 10 mol/s.

Additional information:

$\begin{matrix} C_{P_{A}} = C_{P_{B}} = C_{P_{C}} = 30 cal/mol/K & C_{P_{D}} = 90 cal/mol/K, C_{P_{I}} = 100 cal/mol/ K \\ Δ H_{Rx1 A}^{\circ} = - 50000 cal/molA & k_{1 C} = 0.043 \frac{{dm}^{3}}{mol \cdot S} at 400 K \\ Δ H_{Rx2 B}^{\circ} = + 5000 cal/molB & k_{2D} = 0.4 \frac{{dm}^{3}}{mol \cdot S} at 500 K, with \frac{E}{R} = 5000 K \end{matrix}$

(a) What is the activation energy for Reaction (1)?

P12-22_B The following elementary reactions are to be carried out in a PFR with a heat exchange with constant T_a:

$\begin{matrix} 2A + B \to C & Δ H_{Rx1 B} = - 10 \frac{kJ}{mol B} \\ A \to D & Δ H_{Rx2A} = + 10 \frac{kJ}{mol A} \\ B + 2 C \to E & Δ H_{Rx3C} = - 20 \frac{kJ}{mol C} \end{matrix}$

The reactants all enter at 400 K. Only A and B enter the reactor. The entering concentration of A and B are 3 molar and 1 molar at a volumetric flow rate of 10 dm³/s.

Additional information:

$\begin{matrix} U a = {100J/dm}^{3} / S/K & C_{P_{A}} = 10 J/mol/K \\ k_{1 A} (400 K) = 1.0 {(\frac{{dm}^{3}}{mol})}^{2} / S & C_{P_{B}} = 20 J/mol/K \\ k_{2 A} (400 K) = {1.333}_{S^{- 1}} & C_{P_{C}} = 40 J/mol/K \\ k_{3 B} (400 K) = 2 {(\frac{{dm}^{3}}{mol})}^{2} / S & C_{P_{D}} = 20 J/mol/K \\ C_{P_{E}} = 100 J/mol/K \end{matrix}$

What constant coolant temperature, T_a, is necessary such that at the reactor entrance, that is, V = 0, $\frac{d T}{d V} = 0$ ?

P12-23_B The complex gas-phase reactions are elementary

$\begin{matrix} (1) & 2 A \vec{\leftarrow} 2 B & - r_{1A} = k_{1 A} [C_{A}^{2} - \frac{C_{B}}{K_{C_{A}}}] & Δ H_{Rx1A} = - 20 kJ/mol A \\ (2) & 2 B + A \to C & r_{2 C} = k_{2 C} [C_{B}^{2} C_{A}] & Δ H_{Rx2B} = + 30 kJ/mol B \end{matrix}$

and carried out in a PFR with a heat exchanger. Pure A enters at a rate of 5 mol/min, a concentration of 0.2 mol/dm³, and temperature 300 K. The entering temperature of an available coolant is 320 K.

Additional information:

k_1A1 = 50 dm³/mol · min@305 K with E₁ = 8000 J/mol

k_2C2 = 4000 dm⁹/mol³ · min@310 K with E₂ = 4000 J/mol

K_CA = 10 dm³/mol@315 K

Note: This is the equilibrium constant with respect to A in reaction 1 when using van’t Hoff’s equation, that is, (S12-4),

$\begin{matrix} U a = 200 \frac{J}{{dm}^{3} \cdot min K} & C_{P_{Cool}} = 10 /mol/K & C_{P_{B}} = 80 J/mol/K & C_{P_{C}} = 100 J/mol/K \\ {\dot{m}}_{C} = 50 g /min & C_{P_{A}} = 20 J/mol/K & R = 8.31 J/mol/K \end{matrix}$

The reactor volume is 10 dm³.

(a) Plot (F_A, F_B, F_C) on one graph and (T and T_a) on another along the length of the reactor for adiabatic operation, heat exchange with constant T_a, and co-current and countercurrent heat exchange with variable T_a. Only turn in a copy of your code and output for co-current exchange.

Adiabatic operation

(b) What is the maximum temperature and at what reactor volume is it reached?

(c) At what reactor volume is the flow rate of B a maximum, and what is F_Bmax at this value?

Constant Ta

(d) What is the maximum temperature and at what reactor volume is it reached?

(e) At which reactor volume is the flow rate of B a maximum, and what is F_Bmax at this volume?

Co-current exchange

(f) At what reactor volume does T_a become greater than T? Why does it become greater?

Countercurrent exchange

(g) At what reactor volume does T_a become greater than T?

Hint: Guess T_a at entrance around 350 K.

P12-24_B The elementary liquid-phase reactions

(1) A+ 2B → 2C

(2) A+C→ 2D

are carried out adiabatically in a 10 dm³ PFR. After streams A and B mix, species A enters the reactor at a concentration of C_A0 = 2 mol/dm³ and species B at a concentration of 4 mol/dm³. The entering volumetric flow rate is 10 dm³/s.

Assuming you could vary the entering temperature between 300 K and 600 K, what entering temperture would you recommend to maximize the concentration of species C exiting the reactor? (±25°K). Assume all species have the same density.

Additional information:

$\begin{matrix} C_{P_{A}} = C_{P_{B}} = 20 {cal/mol/K,C}_{P_{C}} = 60 {cal/mol/K,C}_{P_{D}} = 80 cal/mol/K \\ Δ H_{Rx1A} = 20000 cal/mol A, Δ H_{Rx2A} = - 10000 cal/mol A \\ k_{1A} = 0.001 \frac{{dm}^{6}}{{mol}^{2} \cdot S} a t 300 K with E = 5000 cal/mol \\ k_{2 A} = 0.001 \frac{{dm}^{3}}{mol \cdot S} a t 300 K with E = 7500 cal/mol \end{matrix}$

P12-25_C OTHEQ (Old Take Home Exam Question). Multiple reactions with heat effects. Xylene has three major isomers, m-xylene (A), o-xylene (B), and p-xylene (C). When m-xylene (A) is passed over a Cryotite catalyst, the following elementary reactions are observed. The reaction to form p-xylene is irreversible:

A chemical reaction shows the three isomers of xylene. — Xylene is a 6-membered aromatic compound with two alkyl groups. The ortho isomer shows the presence of alkyl group in positions 1 and 2 of the aromatic ring, which interchanges to take the meta form (alkyl groups in 1 and 3). The rate constants of the forward and backward reactions are k1 and k2. The conversion of ortho to para form (alkyl positions 1 and 4) is irreversible with rate constant k3.

The feed to the reactor is pure m-xylene (A). For a total feed rate of 2 mol/min and the reaction conditions below, plot the temperature and the molar flow rates of each species as a function of catalyst weight up to a weight of 100 kg.

(a) Plot the concentrations of each of xylenes down the length (i.e., V) of a PBR.

(b) Find the lowest concentration of o-xylene achieved in the reactor.

(c) Find the maximum concentration of o-xylene in the reactor.

(d) Repeat part (a) for a pure feed of o-xylene (B). What is the maximum concentration of meta-xylene and where does it occur in the reactor?

(e) Vary some of the system parameters and describe what you learn.

(f) What do you believe to be the point of this problem?

Additional information⁸

⁸ Obtained from inviscid pericosity measurements.

All heat capacities are virtually the same at 100 J/mol · K.

$\begin{matrix} C_{T0} = {2 mol/dm}^{3} & K_{C} = 10 exp [4.8 (430 / T - 1.5)] \\ Δ H_{Rx10}^{\circ} = - 1800 J/mol \circ - xylene & T_{0} = 330 K T_{a} = 500 K \\ Δ H_{Rx30}^{\circ} = - 1100 J/mol \circ - xylene & k_{2} = k_{1} / K_{C} \\ k_{1} = 0.5 exp [2 (1 - 320 /T)] {dm}^{3} /kg-cat \cdot min, (T is in K) & U a / ρ_{b} = 16 J/kg-cat \cdot \min \cdot^{\circ} C \\ k_{3} = 0.005 exp {[4.6 (1- (460 /T))]} {dm}^{3} /kg-cat \cdot min & W = 100 kg \end{matrix}$

P12-26_C OTHEQ (Old Take Home Exam Question). Comprehensive problem on multiple reactions with heat effects. Styrene can be produced from ethylbenzene by the following reaction:

$\begin{matrix} ethylbenzene \leftrightarrow styrene + H_{2} & (1) \end{matrix}$

However, several irreversible side reactions also occur:

$\begin{matrix} \begin{matrix} ethylbenzene \to benzene + ethylene & (2) \end{matrix} \\ \begin{matrix} \begin{matrix} ethylbenzene + H_{2} \to toluene + methane \end{matrix} & (3) \end{matrix} \end{matrix}$

(J. Snyder and B. Subramaniam, Chem. Eng. Sci., 49, 5585 (1994)). Ethylbenzene is fed at a rate of 0.00344 kmol/s to a 10.0-m³ PFR (PBR), along with inert steam at a total pressure of 2.4 atm. The steam/ethylbenzene molar ratio is initially, that is, parts (a) to (c), 14.5:1 but can be varied.

Given the following data, find the exiting molar flow rates of styrene, benzene, and toluene along with ${\tilde{S}}_{St/BT}$ for the following inlet temperatures when the reactor is operated adiabatically:

The badge of the member of the hall of fame.

(a) T₀ = 800 K

(b) T₀ = 930 K

(c) T₀ = 1100 K

(d) Find the ideal inlet temperature for the production of styrene for a steam/ethylbenzene ratio of 58:1. Hint: Plot the molar flow rate of styrene versus T₀. Explain why your curve looks the way it does.

(e) Find the ideal steam/ethylbenzene ratio for the production of styrene at 900 K. Hint: See part (d).

(f) It is proposed to add a countercurrent heat exchanger with Ua = 100 kJ/m³/min/K, where T_a is virtually constant at 1000 K. For an entering stream to ethylbenzene ratio of 20, what would you suggest as an entering temperature? Plot the molar flow rates and ${\tilde{S}}_{St/BT}$ .

(g) What do you believe to be the major points of this problem?

(h) Ask another question or suggest another calculation that can be made for this problem.

Note: Whenever I teach Chemical Reaction Engineering, I always assign this problem.

Additional information

Heat capacities
Methane	68 J/mol · K
Ethylene	90 J/mol·K
Benzene	201 J/mol · K
Toluene	249 J/mol · K
Styrene	273 J/mol · K
Ethylbenzene	299 J/mol·K
Hydrogen	30 J/mol · K
Steam	40 J/mol · K

$\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} ρ = 2137 {kg/m}^{3} of pellet \\ ϕ = 0.4 \end{matrix} \\ Δ H_{Rx1EB}^{\circ} = 118000 kJ/kmol ethylbenzene \end{matrix} \\ Δ H_{Rx2EB}^{\circ} = 105200 kJ/kmol ethylbenzene \end{matrix} \\ Δ H_{Rx3EB}^{\circ} = 53900 kJ/kmol ethylbenzene \end{matrix}$

$\begin{matrix} K_{P1} = exp {b_{1} + \frac{b_{2}}{T} + b_{3} \ln (T) + [(b_{4} T + b_{5}) T + b_{6}] T} atm with \\ b_{1} = - 17.34 & b_{4} = 2.314 \times 10^{- 10} K^{- 3} \\ b_{2} = 1.302 \times 10^{4} K & b_{5} = 1.302 \times 10^{- 6} K^{- 2} \\ b_{3} = 5.051 & b_{6} = - 4.931 \times 10^{- 3} K^{- 1} \end{matrix}$

The kinetic rate laws for the formation of styrene (St), benzene (B), and toluene (T), respectively, are as follows (EB = ethylbenzene).

$\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} r_{1 St} = ρ (1 - ϕ) \exp (- 0.08539 - \frac{10925 K}{T}) (P_{E B} - \frac{P_{S t} P_{H_{2}}}{K_{p 1}}) \end{matrix} & ({kmol/m}^{3} \cdot s) \end{matrix} \\ \begin{matrix} \begin{matrix} r_{2 B} = ρ (1 - ϕ) \exp (13.2392 - \frac{25000 K}{T}) (P_{E B}) \end{matrix} & ({kmol/m}^{3} \cdot s) \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} r_{3 T} = ρ (1 - ϕ) \exp (0.2961 - \frac{11000 K}{T}) (P_{E B} P_{H_{2}}) \end{matrix} & ({kmol/m}^{3} \cdot s) \end{matrix} \end{matrix}$

The temperature T is in Kelvin and P_i is in atm.

P12-27_B OTHEQ (Old Take Home Exam Question). The liquid-phase, dimer-quadmer series addition reaction

4 A → 2A₂ → A₄

can be written as

$\begin{matrix} 2A \to A_{2} & - r_{1 A} = k_{1 A} C_{A}^{2} & Δ H_{Rx1A} = - 32.5 \frac{kcal}{mol A} \\ {2A}_{2} \to A_{4} & - r_{2 A_{2}} = k_{2 A_{2}} C_{A_{2}}^{2} & Δ H_{Rx2 A_{2}} = - 27.5 \frac{kcal}{{mol A}_{2}} \end{matrix}$

and is carried out in a 10-dm³ PFR. The mass flow rate through the heat exchanger surrounding the reactor is sufficiently large so that the ambient temperature of the exchanger is constant at T_a = 315 K. The reactants enter at a temperature T₀, of 300 K. Pure A is fed to the rector at a volumetric flow rate of 50 dm³/s and a concentration of 2 mol/dm³.

(a) Plot, compare, and analyze the profiles F_A, F_A2, and F_A4 down the length of the reactor up to 10 dm³.

(b) The desired product is A₂ and it has been suggested that the current reactor may be too large. What reactor volume would you recommend to maximize F_A2?

(c) What operating variables (e.g., T₀, T_a) would you change and how would you change them to make the reactor volume as small as possible and to still maximize F_A2? Note any opposing factors in maximum production of A₂. The ambient temperature and the inlet temperature must be kept between 0°C and 177°C.

Additional information:

$\begin{matrix} \begin{matrix} \begin{matrix} k_{1 A} = 0.6 \frac{{dm}^{3}}{mol \cdot S} {at 300 K with E}_{1} = 4000 \frac{cal}{mol} \\ k_{2 A_{2}} = 0.35 \frac{{dm}^{3}}{mol \cdot S} {at 320 K with E}_{2} = 5000 \frac{cal}{mol} \end{matrix} \\ C_{P_{A}} = 25 \frac{cal}{molA \cdot K}, C_{P_{A_{2}}} = 50 \frac{cal}{{molA}_{2} \cdot K}, C_{P_{A_{4}}} = 100 \frac{cal}{mol A_{4} \cdot K} \end{matrix} \\ U a = 1000 \frac{cal}{{dm}^{3} \cdot S \cdot K} \end{matrix}$

Turn in your recommendation of reactor volume to maximize and the molar flow rate at this maximum.

Supplementary Reading

1. An excellent development of the energy balance is presented in

R. ARIS, Elementary Chemical Reactor Analysis. Upper Saddle River, NJ: Prentice Hall, 1969, Chaps. 3 and 6. JOSEPH FOGLER, A Reaction Engineer’s Handbook of Thermochemical Data. Riça, Jofostan: Jofostan Press, (2025).

2. Safety

CENTER FOR CHEMICAL PROCESS SAFETY (CCPS), Guidelines for Chemical Reactivity Evaluation and Application to Process Design. New York: American Institute of Chemical Engineers (AIChE), 1995.

DANIEL A. CROWL and JOSEPH F. LOUVAR, Chemical Process Safety: Fundamentals with Applications, 3rd ed. Upper Saddle River, NJ: Prentice Hall, 2011.

G. A. MELHEM and H. G. FISHER, International Symposium on Runaway Reactions and Pressure Relief Design. New York: Center for Chemical Process Safety (CCPS) of the American Institute of Chemical Engineers (AIChE) and The Institution of Chemical Engineers, 1995.

See the Center for Chemical Process Safety (CCPS) Web site, www.aiche.org/ccps.

3. A number of example problems dealing with nonisothermal reactors may or may not be found in

THORNTON W. BURGESS, The Adventures of Jerry Muskrat. New York: Dover Publications, Inc., 1914.

G. F. FROMENT and K. B. BISCHOFF, Chemical Reactor Analysis and Design, 3rd ed. New York: Wiley, 2010.

S. M. WALAS, Chemical Reaction Engineering Handbook of Solved Problems. Amsterdam: Gordon and Breach, 1995. See the following solved problems: Problem 4.10.1, page 444; Problem 4.10.08, page 450; Problem 4.10.09, page 451; Problem 4.10.13, page 454; Problem 4.11.02, page 456; Problem 4.11.09, page 462; Problem 4.11.03, page 459; Problem 4.10.11, page 463.

4. A review of the multiplicity of the steady state and reactor stability is discussed by

D. D. PERLMUTTER, Stability of Chemical Reactors. Upper Saddle River, NJ: Prentice Hall, 1972.

5. The heats of formation, (H_i T), Gibbs free energies, (G_i T_R), and the heat capacities of various compounds can be found in

DON W. GREEN and ROBERT H. PERRY, Perry’s Chemical Engineers’ Handbook, 8th ed. (Chemical Engineers Handbook). New York: McGraw-Hill, 2008.

DAVID R. LIDE, CRC Handbook of Chemistry and Physics, 90th ed. Boca Raton, FL: CRC Press, 2009.

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Table of Contents for 12. Steady-State Nonisothermal Reactor Design: Flow Reactors with Heat Exchange

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12. Steady-State Nonisothermal Reactor Design: Flow Reactors with Heat Exchange

12.1 Steady-State Tubular Reactor with Heat Exchange

12.1.1 Deriving the Energy Balance for a PFR

12.1.2 Applying the Algorithm to Flow Reactors with Heat Exchange

12.2 Balance on the Heat-Transfer Fluid

12.2.1 Co-Current Flow

12.2.2 Countercurrent Flow

12.3 Examples of the Algorithm for PFR/PBR Design with Heat Effects

12.3.1 Applying the Algorithm to an Exothermic Reaction

12.3.2 Applying the Algorithm to an Endothermic Reaction

12.4 CSTR with Heat Effects

12.4.1 Heat Added to the Reactor, Q˙

12.5 Multiple Steady States (MSS)

12.5.1 Heat-Removed Term, R(T )

12.5.2 Heat-Generated Term, G(T )

12.5.3 Ignition–Extinction Curve

12.6 Nonisothermal Multiple Chemical Reactions

12.6.1 Energy Balance for Multiple Reactions in Plug-Flow Reactors

12.6.1A Series Reactions in a PFR

12.6.1B Parallel Reactions in a PFR

12.6.2 Energy Balance for Multiple Reactions in a CSTR

12.6.3 Series Reactions in a CSTR

12.6.4 Complex Reactions in a PFR

12.7 Radial and Axial Temperature Variations in a Tubular Reactor

12.8 And Now... A Word from Our Sponsor—Safety 12 (AWFOS–S12 Safety Statistics)†

12.8.1 The Process Safety Across the Chemical Engineering Curriculum Web site

12.8.2 Safety Statistics

12.8.3 Additional Resources CCPS and SAChE

Summary

CRE Web Site Materials (http://www.umich.edu/~elements/6e/12chap/obj.html#/)

Questions, Simulations, and Problems

Questions

Computer Simulations and Experiments

Interactive Computer Games

Problems

Supplementary Reading

Table of Contents for
12. Steady-State Nonisothermal Reactor Design: Flow Reactors with Heat Exchange

12.4.1 Heat Added to the Reactor, $\dot{Q}$

12.8 And Now... A Word from Our Sponsor—Safety 12 (AWFOS–S12 Safety Statistics)^†