6. Isothermal Reactor Design: Moles and Molar Flow Rates

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6. Isothermal Reactor Design: Moles and Molar Flow Rates

Don’t let your fears . . .

Get in the way of your dreams

—Anonymous

Overview. In Chapter 5 we used conversion to design a number of isothermal reactors for single reactions. While in many situations writing the mole balances in terms of conversion is an extremely effective strategy, there are many instances where it is more convenient, and in some cases absolutely necessary, to write the mole balance in terms of moles (N_A, N_B) or molar flow rates (F_A, F_B), as shown in Table S1-1 in Chapter 1. In this chapter, we show how to make small changes in our CRE algorithm to analyze these situations. Using our algorithm, we first write a mole balance on each and every species, and second, we need to relate the rates of reaction of each species to one another using the relative rates described in Chapter 3, that is, Equation (3-1).

We will use molar flow rates in our mole balance to analyze

A microreactor with the reaction

2NOCl → 2NO + Cl₂
A membrane reactor used for the dehydrogenation of ethylbenzene

C₆H₅CH₂CH₃ → C₆H₅CH = CH₂ + H₂
A semibatch reactor used for the reaction

CNBr + CH₃NH₂ → CH₃Br + NCNH₂

We will again use mole balances in terms of these variables (N_i, F_i) for multiple reactions in Chapter 8 and for heat effects in Chapters 11–13.

6.1 The Moles and Molar Flow Rate Balance Algorithms

Used for:

Multiple rxns
Membranes
Unsteady state

There are many instances when it is much more convenient to work in terms of the number of moles (N_A, N_B) or molar flow rates (F_A, F_B, etc.) rather than conversion. Membrane reactors and multiple reactions taking place in the gas phase are two such cases where it is absolutely necessary to use molar flow rates rather than conversion. We now modify our CRE algorithm by using concentrations for liquids and molar flow rates for gases as our dependent variables. The main difference between the conversion algorithm and the molar flow rate/concentration algorithm is that, in the conversion algorithm, we needed to write a mole balance on only one species, whereas in the molar flow rate and concentration algorithm, we must write a mole balance on each and every species. This algorithm is shown in Figure 6-1. First, we write the mole balances on all species present, as shown in Step ①. Next, we write the rate law, Step ②, and then we relate the mole balances to one another through the relative rates of reaction, as shown in Step ③. Steps ④ and ⑤ are used to relate the concentrations in the rate law to the molar flow rates. In Step ⑥, all the steps are combined by the ODE solver (e.g., Polymath, MATLAB, Wolfram, and Python).

6.2 Mole Balances on CSTRs, PFRs, PBRs, and Batch Reactors

6.2.1 Liquid Phase

For liquid-phase reactions, the density remains constant and consequently there is no change in either the volume V or the volumetric flow rate, υ = υ₀, during the course of the reaction. Therefore, concentration is the preferred design variable. The mole balances derived in Chapter 1 (Table S1-1) are now applied to each species for the generic reaction

$\begin{matrix} a A + b B \to c C + d D & (2-1) \end{matrix}$

The mole balances are then coupled to one another using the relative rates of reaction

Used to couple the mole balances

$\begin{matrix} \frac{r_{A}}{- a} = \frac{{r_{}}_{B}}{- b} = \frac{r_{C}}{c} = \frac{r_{D}}{d} & (3-1) \end{matrix}$

to arrive at Table 6-1, which gives the balance equations in terms of concentration for the four types of reactors we have been discussing. We see from Table 6-1 that we have only to specify the parameter values for the system (C_A0, υ₀, etc.) and for the rate-law parameters (e.g., k_A, α, β) to solve the coupled ordinary differential equations for either BRs, PFRs, or PBRs, or to solve the coupled algebraic equations for a CSTR.

A design algorithm for isothermal reaction mole balances is shown. — **Figure 6-1** Isothermal reaction design algorithm for mole balances.

The steps involved in the design algorithm for isothermal reaction mole balances are as follows: The first step is to write the mole balance on each species and the example given is partial differentiation of F A over V equals r A, partial differentiation of F B over V equals r B, partial differentiation of F C over V equals r C. The second step is to write rate law in terms of concentration and the example given is negative r A equals k A of C A times C B squared minus C C over K C. The third step is to relate the rate of reaction of each species to one another like negative r A over 1 equals negative r B over 2 equals r C over 1 and the example given is r B equals 2 r A and r C equals negative r A. The fourth step is stoichiometry that is to write the concentrations in terms of molar flow rates for gas-phase reactions and for liquid-phase reactions, concentration is used as the dependent variable. The example given is C A equals C T 0 times F A over F T times T 0 over T times P over P 0, C B equals C T 0 times F B over F T times T 0 over T times P over P 0 with F T equals F A plus F B plus F C plus F 1. The fifth step is to write the gas-phase pressure drop in terms of molar flow rates and the example given is partial differentiation of p over W equals negative alpha over 2 p times T over T 0 times F T over F T 0 with p equals P over P 0. The sixth step is to combine all the previous steps that are from steps 1 to 5. For an isothermal operation, T equals T subscript 0, we can use an ODE solver or a nonlinear equation solver (example-polymath) to combine the previous steps. For example the profiles of molar flow rates, concentration, and pressure.

^† For a PBR, use $\frac{{d F}_{A}}{d W} = r_{A}^{'}, \frac{{d F}_{B}}{d W} = r_{B}^{'}, and \frac{{d F}_{C}}{d W} = r_{C}^{'}$ .

Tap water flows to a container placed below it.

TABLE 6-1 MOLE BALANCES FOR LIQUID-PHASE REACTIONS

Batch	$\frac{{d C}_{A}}{d t} = r_{A}$	and	$\frac{{d C}_{B}}{d t} = \frac{b}{a} r_{A}$
CSTR	$V = \frac{υ_{0} (C_{A0} - C_{A})}{{- r}_{A}}$	and	$V = \frac{υ_{0} (C_{B 0} - C_{B})}{{- (b / a) r}_{A}}$
PFR	$υ_{0 \frac{d C_{A}}{d V} =} r_{A}$	and	$υ_{0 \frac{d C_{B}}{d V} = \frac{b}{a}} r_{A}$
PBR	$υ_{0 \frac{{d C}_{A}}{d W} =} r_{A}^{'}$	and	$υ_{0 \frac{{d C}_{B}}{d W} = \frac{b}{a}} r_{A}^{'}$

6.2.2 Gas Phase

The mole balances for gas-phase reactions are given in Table 6-2 in terms of the number of moles (batch) or molar flow rates for the generic rate law for the generic reaction, Equation (2-1). The molar flow rates for each species F_j are obtained from a mole balance on each species (i.e., A, B, C, and D), as given in Table 6-2. For example, for a plug-flow reactor

$\begin{matrix} \frac{d F_{j}}{d V} = r_{j} & (1 -11) \end{matrix}$

Must write a mole balance on each and every species

The generic power-law rate law for species A is

Rate Law

$\begin{matrix} {- r}_{A} = k_{A} C_{A}^{α} C_{B}^{β} & (3 -3) \end{matrix}$

The rate law wrt A is coupled with the equation for relative rates,

\begin{matrix} \frac{r_{A}}{- a} = \frac{r_{B}}{- b} = \frac{r_{C}}{c} = \frac{r_{D}}{d} & (3 - 1) \end{matrix}

Given the rate law, Equation (3-3), for species A, we use Equation (3-1) to substitute for species j, r_j, in Equation (1-11), the PFR mole balance.

To relate concentrations to molar flow rates, recall Equation (4-17), for gas-phase reactions with p = P/P₀

Stoichiometry

$\begin{matrix} C_{j} = C_{T0} \frac{F_{j}}{F_{T}} \frac{T_{0}}{T} p \end{matrix} \begin{matrix} (4 - 17) \end{matrix}$

The pressure-drop equation, Equation (5-28), for isothermal operation (T = T₀) is

$\begin{matrix} \frac{d p}{d W} = \frac{- α}{2 p} \frac{F_{T}}{F_{T0}} \end{matrix} \begin{matrix} (5 - 28) \end{matrix}$

TABLE 6-2 ALGORITHM FOR GAS-PHASE REACTIONS

$a A + b B \to c C + d D$

Mole balances:

BR	PFR	PBR	CSTR
$\frac{d N_{A}}{d t} = r_{A} V$	$\frac{{d F}_{A}}{d V} = r_{A}$	$\frac{{d F}_{A}}{d W} = r_{A}^{'}$	$V = \frac{F_{A0} - F_{A}}{{- r}_{A}}$
$\frac{{d N}_{B}}{d t} = r_{B} V$	$\frac{{d F}_{B}}{d V} = r_{B}$	$\frac{{d F}_{B}}{d W} = r_{B}^{'}$	$V = \frac{F_{B0} - F_{B}}{{- r}_{B}}$
$\frac{{d N}_{C}}{d t} = r_{C} V$	$\frac{{d F}_{C}}{d V} = r_{C}$	$\frac{d F_{C}}{d W} = r_{C}^{'}$	$V = \frac{F_{C0} - F_{C}}{- r_{C}}$
$\frac{{d N}_{D}}{d t} = r_{D} V$	$\frac{{d F}_{D}}{d V} = r_{D}$	$\frac{{d F}_{D}}{d W} = r_{D}^{'}$	$V = \frac{F_{D0} - F_{D}}{{- r}_{D}}$
Also inerts, if any $\frac{{d N}_{1}}{d t} = 0$	$\frac{{d F}_{1}}{d V} = 0$	$\frac{{d F}_{1}}{d W} = 0$	F₁ = F₁₀

We shall continue the algorithm using a PBR as an example.

Rates:

Rate Law

$- r_{A}^{'} = k_{A} C_{A}^{α} C_{B}^{β}$

Relative Rates

$\frac{r_{A}^{'}}{- a} = \frac{r_{B}^{'}}{- b} = \frac{r_{C}^{'}}{c} = \frac{r_{D}^{'}}{d}$

then

$\begin{matrix} \begin{matrix} r_{B}^{'} = \frac{b}{a} r_{A}^{'} & r_{C}^{'} = - \frac{c}{a} r_{A}^{'} \end{matrix} & r_{D}^{'} = - \frac{d}{d} r_{A}^{'} \end{matrix}$
Stoichiometry:

Concentrations

$\begin{matrix} \begin{matrix} C_{A} = C_{T0} \frac{F_{A} T_{0}}{F_{T} T} p & C_{B} = C_{T0} \frac{F_{B} T_{0}}{F_{T} T} \end{matrix} p \\ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} C_{C} = C_{T0} \frac{F_{C} T_{0}}{F_{T} T} p \end{matrix} & C_{D} = C_{T0} \frac{F_{D} T_{0}}{F_{T} T} \end{matrix} p \end{matrix} \\ \frac{d p}{d W} = \frac{- α}{2 p} \frac{F_{T}}{F_{T0}} \frac{T}{T_{0}}, p = \frac{P}{P_{0}} \end{matrix} \end{matrix}$

Gas phase

Total molar flow rate: F_T = F_A+F_B+F_C+F_D+F₁
Combine:
- Appropriate reactor mole balance on each species
- Rate law
- Concentration for each species
- Pressure-drop equation
Evaluate:
1. Specify and enter parameter values: k_A, C_T0, α,β,T₀,a,b,c,d
2. Specify and enter entering molar flow rates: F_A0 F_B0, F_C0, F_D0, and final volume, V_final
Use an ODE solver.

Many times we will let the ODE solver replace Step 4, Combine.

The total molar flow rate is given by the sum of the flow rates of the individual species

$F_{T} = Σ_{j = 1}^{n} F_{j}$

When species A, B, C, D, and inert I are the only species present, then

F_T = F_A + F_B + F_C + F_D + F_I

We now combine all the preceding information, as shown in Table 6-2.¹

¹ View the YouTube video made by the chemical reaction engineering students at the University of Alabama, titled Find Your Rhyme. Videos can be accessed directly from the home page of the CRE Web site, www.umich.edu/~elements/6e/index.html.

6.3 Application of the PFR Molar Flow Rate Algorithm to a Microreactor

A photo of a micro reactor is shown in Figure P5-21_B on page 225.

Advantages of microreactors

Microreactors have emerged as a useful technology in CRE. They are characterized by their high surface-area-to-volume ratios in their microstructured regions that contain tubes or channels. A typical channel width might be 100 μm with a length of 20000 μm (2 cm). The resulting high surface-area-to-volume ratio (ca. 10000 m²/m³) reduces or even eliminates heat and mass transfer resistances often found in larger reactors. Consequently, surface-catalyzed reactions can be greatly facilitated, hot spots in highly exothermic reactions can be minimized, and in many cases highly exothermic reactions can be carried out virtually isothermally. These features provide the opportunity for microreactors to be used to study the intrinsic kinetics of reactions. Another advantage of microreactors is their use in the production of toxic or explosive intermediates where a leak or microexplosion for a single unit will do minimal damage because of the small quantities of material involved. Other advantages include shorter residence times and narrower residence time distributions.

Figure 6-2 shows (a) a microreactor with a heat exchanger and (b) a microplant with reactor, valves, and mixers. Heat, $\dot{Q}$ , can be added or taken away by the fluid flowing perpendicular to the reaction channels, as shown in Figure 6-2(a). Production in microreactor systems can be increased simply by adding more units in parallel. For example, the catalyzed reaction

$\begin{matrix} \begin{matrix} R - {CH}_{2} {OH + ½ O}_{2} & \underset{Ag}{\begin{matrix} \to \end{matrix}} \end{matrix} & R - CHO + \end{matrix} H_{2} O$

required only 32 microreaction systems in parallel to produce 2000 tons/yr of acetate!

Microreactors are also used for the production of specialty chemicals, combinatorial chemical screening, lab-on-a-chip, and chemical sensors. In modeling microreactors, we will assume they are either in plug flow for which the mole balance is

$\begin{matrix} \frac{{d F}_{A}}{d V} = r_{A} & (1 -12) \end{matrix}$

A figure of a microreactor and a microplant. — **Figure 6-2** Microreactor (a) and microplant (b). (Photo courtesy of Ehrfeld, Hessel, and Löwe, *Microreactors: New Technology for Modern Chemistry*, Weinheim, Germany: Wiley-VCH.)

The first figure depicts a microreactor. Microreactor is cube like structure with different layers. Each layer has rectangular slots built in it. The rectangular slots in the adjacent layers are in different direction. It allows the flow in two different direction (which includes the flow of heat Q). The second figure presents a microplant. A number of circular slots are present on the surface. Small square boxes are present above that with knobes above the boxes. Some wires are attached to the square boxes through the knobes.

or in laminar flow, in which case we will use the segregation model discussed in Chapter 17. For the plug-flow case, the algorithm is described in Figure 6-1.

Example 6–1 Gas-Phase Reaction in a Microreactor—Molar Flow Rates

The NFPA label for Chlorine (Cl) is given. — The blue diamond indicates a health hazard, the red diamond indicates a fire hazard, the white diamond indicates a specific hazard, and the red diamond indicates the reactivity. For chlorine, the health hazard is 4 which is of deadly, the fire hazard is 0, the reactivity is 0 that is stable, and the specific hazard is OX that indicates an oxidizer.

The NFPA label for Nitric oxide (NO) is given. — The blue diamond indicates a health hazard, the red diamond indicates a fire hazard, the white diamond indicates a specific hazard, and the red diamond indicates the reactivity. For Nitric oxide, the health hazard is 3 which is of extreme danger, the fire hazard is 0, the reactivity is 3 that indicates a violent chemical change, and the specific hazard is OX that indicates an oxidizer.

NOCl

Nitrous oxide (NO) gas is used by a number of dentists on their patients (the author being one) to eliminate pain during drilling and tooth extraction. Nitrous oxide can be produced by the gas-phase second-order reaction.

$2 NOCl \to 2 NO + {Cl}_{2}$

is to be carried out at 425°C and 1641 kPa (16.2 atm). Pure NOCl is to be fed, and the reaction follows an elementary rate law.² It is desired to produce 20 tons of NO per year in a microreactor system using a bank of 10 microreactors in parallel. Each microreactor has 100 channels with each channel 0.2 mm square and 250 mm in length, giving a volume of a single channel to be 10^–5dm³.

² J. B. Butt, Reaction Kinetics and Reactor Design, 2nd ed. New York: Marcel Dekker, 2001, p. 153.

A diagrammatic representation of a microreactor with 12 channels on its one phase, is shown. An arrow passes through the reactor.

Plot and analyze the molar flow rates of NOCl, NO, and Cl₂ as a function of volume down the length of the reactor.
What is the reactor volume necessary to achieve 85% conversion of NOCl?

Additional Information

To produce 20 tons per year of NO at 85% conversion would require a feed rate of 0.0226 mol/s of NOCl, or 2.26 × 10^–5 mol/s per channel. The reaction-rate constant is

$k = 0.29 \frac{{dm}^{3}}{mol \cdot s} at 500 K with E = 24 \frac{kcal}{mol}$

Solution

For one channel

The process of conversion of Nitrosyl chloride is shown. — The flow of particles through the microreactor is shown. An inward arrow is marked towards the left of microreactor indicating the variable FA0 and an outward arrow is marked towards the right of microreactor indicating the variable FB. Here, F A O equals 22.6 mu mole over 's' and F B equals 19.2 mu mole over S, x equals 0.85, and volume is unknown.

Find V.

Although this particular problem could be solved using conversion, we shall illustrate how it can also be solved using molar flow rates as the variable in the mole balance. Why do we do this? We do this to give practice using molar flow rates as the variables in order to help prepare the reader for the more complex problems, such as multiple reactions, where conversion cannot be used as the design variable. However, at the end we can use molar flow rate F_A to calculate the conversion X.

We first write the reaction in symbolic form and then divide by the stoichio-metric coefficient of the limiting reactant, NOCl.

$\begin{matrix} Chemistry: & 2 NOCl {\to 2NO + Cl}_{2} \\ Symbolic Form: & 2 A \to 2 B + C \\ Limiting Reactant: & A \to B + \frac{1}{2} C \end{matrix}$

Mole balances on species A, B, and C:

$\begin{matrix} \frac{{d F}_{A}}{d V} = r_{A} & (E6-1.1) \end{matrix}$

$\begin{matrix} \frac{{d F}_{B}}{d V} = r_{B} & (E6-1.2) \end{matrix}$

$\begin{matrix} \frac{{d F}_{C}}{d V} = r_{C} & (E6-1.3) \end{matrix}$
Rates:
1. Rate Law
  
  $\begin{matrix} - r_{A} = k C_{A}^{2}, with k = 0.29 \frac{{dm}^{3}}{mol \cdot s} at 500 K & (E6-1.4) \end{matrix}$
2. Relative Rates
$\begin{array}{l} \frac{r_{A}}{- 1} = \frac{r_{B}}{1} = \frac{r_{C}}{\frac{1}{2}} \\ \begin{matrix} r_{B} = {- r}_{A} \\ {\begin{matrix} r_{C} = \frac{1}{2} r \end{matrix}}_{A} \end{matrix} \end{array}$
Stoichiometry: Gas phase with T = T₀ and P = P₀, then ${υ = υ}_{0} \frac{F_{T}}{F_{T0}}$ Concentration Gas Phase

$\begin{matrix} C_{A j} = C_{T0} \frac{F_{j}}{F_{T}} p \frac{T_{0}}{T} & (4-17) \end{matrix}$

$\begin{matrix} {p = (1 - α_{p} V)}^{1 / 2} & (5-35) \end{matrix}$

Side Note: Calculations in the Chapter 6 Expanded Material on the Web site (http://www.umich.edu/~elements/6e/06chap/alpha-P-calculation-Example-6-1.pdf ) using Equation (5-36) show that the pressure-drop parameter, α_p, for each channel to

be $α_{p} = \frac{{{4 f}_{F} G}^{2}}{A_{C ρ 0} P_{0} D} = 3.47$ cm^–3. For a channel volume of V = 0.01 cm³ the pressure ratio p is

p = (1 – (3.47)(0.01))^1/2 ≅ 1

Therefore, we will neglect pressure drop, that is, P = P₀, p = 1.

Applying Equation (4-17) to species A, B, and C, for isothermal operation (T = T₀), and for no pressure drop, P = P₀, (p = 1), the concentrations are

$\begin{matrix} C_{A} = C_{T0} \frac{F_{A}}{F_{T}}, C_{B} & = & C_{T0} \frac{F_{B}}{F_{T}}, C_{c} = C_{T 0} \frac{F_{C}}{F_{T}} & (E6-1.5) \\ with F_{T} & = & F_{A} + F_{B} + F_{C} \end{matrix}$
Combine: The rate law in terms of molar flow rates is

${{- r}_{A} = {k C}_{T0}^{2} (\frac{F_{A}}{F_{T}})}^{2}$

combining all

$\begin{matrix} \frac{d F_{A}}{d V} = - k C_{T0}^{2} {(\frac{F_{A}}{F_{T}})}^{2} & (E6-1.6) \end{matrix}$

$\begin{matrix} \frac{d F_{B}}{d V} = k C_{T0}^{2} {(\frac{F_{A}}{F_{T}})}^{2} & (E6-1.7) \end{matrix}$

$\begin{matrix} \frac{d F_{C}}{d V} = \frac{k}{2} C_{T0}^{2} {(\frac{F_{A}}{F_{T}})}^{2} & (E6-1.8) \end{matrix}$

We can also calculate the conversion X,

$\begin{matrix} X = (1 - \frac{F_{A}}{F_{A0}}) \end{matrix}$
Evaluate:

$C_{T0} = \frac{P_{0}}{R T_{0}} = \frac{(1641 kPa)}{(8.314 \frac{kPa \cdot {dm}^{3}}{mol \cdot K}) 698 K} = 0.286 \frac{mol}{{dm}^{3}} = \frac{0.286 mmol}{{cm}^{3}}$

When using Polymath or another ODE solver, one does not have to actually combine the mole balances, rate laws, and stoichiometry, as was done in the Combine step previously in Chapter 5. The ODE solver will do that for you. Thanks, ODE solver!

The Polymath program and output are shown in Table E6-1.1 and Figure E6-1.1. Note that explicit Equation #6 in the Polymath program calculates the reaction-rate constant k at the specified temperature of 425°C (i.e., 698 K) for you.

A graphical representation of the molar flow rate in microreactor. — **Figure E6-1.1** Profiles of microreactor molar flow rates.

The graph shows the flow rate in micromole per second. The x-axis represents the volume (in dm cube) ranging from 0 to 1.0 exponential minus 5 and the y-axis ranges from 0 to 25. The output curve for F subscript a originates from the point (0, 23) and decreases along x-axis up to the point (1.0 exponential minus 5, 5). The output curve of F subscript b starts from the origin and increases along the x-axis up to the point (1.0 exponential minus 5, 20). The output curve of F subscript c starts from the origin and increases along the x-axis up to the point (1.0 exponential minus 5, 10).

Note that we see not much product is formed in the last part of the reactor, that is, beyond V = 4 × 10^–6 dm³.

The LEP sliders icon is shown. — TABLE E6-1.1 POLYMATH PROGRAM

Analysis: This gas-phase reaction in a PFR example could just as easily have been solved using conversion as a basis. However, membrane reactors and multiple reactions cannot be solved using conversion. You will note we just wrote out the equations in Steps 1–5 of our reaction algorithm (Table 6-2) and then typed them directly into our ODE solver, Polymath, to obtain the molar flow rate profiles shown in Figure E6-1.1. Notice the profiles change rapidly near the reactor entrance and then there is very little change after 4 × 10^–6 dm³ down the reactor. Other interesting variables you will want to plot when you download this program from the Living Example Problem (i.e., LEP) file are the total molar flow rate, F_T; the concentrations of the reacting species, C_A, C_B, and C_C (for C_B and C_C you will need to type in two additional equations); and the rates –r_A, r_B, and r_C. This task can be easily outsourced to the ChE students of Jofostan University in Riça for a small stipend.

6.4 Membrane Reactors

Membrane reactors can be used to increase conversion when the reaction is thermodynamically limited, as well as to increase the selectivity when multiple reactions are occurring. Thermodynamically limited reactions are reactions where the equilibrium lies far to the left (i.e., reactant side) and as a result there is little conversion. If the reaction is exothermic, increasing the temperature will only drive the reaction further to the left, and decreasing the temperature will move the equilibrium to the right, but could result in a reaction rate so slow that there is very little conversion. If the reaction is endothermic, increasing the temperature will move the reaction to the right to favor a higher conversion; however, for many reactions these higher temperatures cause the catalyst to become deactivated.

By having one of the products pass through the membrane, we drive the reaction toward completion.

The term membrane reactor describes a number of different types of reactor configurations that contain a membrane. The membrane can either provide a barrier to certain components while being permeable to others, prevent certain components such as particulates from contacting the catalyst, or contain reactive sites and be a catalyst in itself. Like reactive distillation, the membrane reactor is another technique for driving reversible reactions to the right toward completion in order to achieve very high conversions. These high conversions can be achieved by having one of the reaction products diffuse out through a semipermeable membrane surrounding the reacting mixture. As a result, the reverse reaction will not be able to take place, and the reaction will continue to proceed to the right toward completion.

Two of the main types of catalytic membrane reactors are shown in Figure 6-3. The reactor in Figure 6-3(b) is called an inert membrane reactor with catalyst pellets on the feed side (IMRCF). Here, the membrane is inert and serves as a barrier to the reactants and some of the products. The reactor in Figure 6-3(c) is a catalytic membrane reactor (CMR). The catalyst is deposited directly on the membrane, and only specific reaction products are able to exit the permeate side. As an example, consider the reversible reaction where

$\begin{matrix} C_{6} H_{12} & ⇄ & 3 H_{2} + C_{6} H_{6} \\ A & ⇄ & 3 B + C \end{matrix}$

H₂ diffuses through the membrane, while C₆H₆ does not.

the hydrogen molecule is small enough to diffuse through the small pores of the membrane, while C₆H₁₂ and C₆H₆ cannot. Consequently, the reaction continues to proceed to the right even for a small value of the equilibrium constant.

A schematic representation of membrane reactors is shown. — The diagram indicates three membranes horizontally arranged in a continuous order. Two reactants such as hydrogen (H2) and Cyclohexane (C6H12) is passed through the membrane. The hydrogen molecule disperses through the membrane whereas the cyclohexane does not diffuse through the membrane. Anyway, C6H12 proceeds towards the right that is in between the second and third membranes.

An illustration of ceramic reactors and a cross-sectional view of IMRCF and CMR is shown. — **Figure 6-3** Membrane reactors. (a) Photo of ceramic reactors, (b) cross section of IMRCF, (c) cross section of CMR, (d) schematic of IMRCF for mole balance.

Four figures are shown. Figure 'a' presents the outer view of a catalytic membrane resembling a cylindrical structure. Figure 'b' is the Cross-section of the IMRCF reactor showing the feed side of the reactor surrounded by the permeate whereas H2 acts as the permeate and C6H12 acts as the feed. The feed and the permeate are separated by an inert membrane. The feed is filled with effluent (C6H6) and closely packed catalyst particles. Figure 'c' is the cross-section of the CMR reactor showing the feed side of the reactor surrounded by the permeate whereas H2 acts as the permeate and C6H12 acts as the feed. The feed and the permeate are separated by catalyst membrane which is directly deposited. The feed is filled with effluent (C6H6). The schematic diagram of IMRCF is shown in figure 'd.' It shows that the reactor is filled with affluent and closely packed catalyst particles. The inputs and outputs of the reactor are marked as F A, F B, and F C. The outer layer of the feed is marked as a membrane. The flow of hydrogen species B from the top and downsides of the reactor is marked as R B. The differential volume (del V) is marked within the feed that is the space between the catalyst particles. The total volume is denoted as volume plus differential volume.

Hydrogen, species B, flows out through the sides of the reactor as it flows down the reactor while the other reactants and products cannot leave until they exit the end of the reactor.

$\begin{matrix} V = \frac{W}{ρ_{b}} \end{matrix}$

In analyzing membrane reactors, we only need to make a small change to the algorithm shown in Figure 6-1. We shall choose the reactor volume rather than catalyst weight as our independent variable for this example. The catalyst weight, W, and reactor volume, V, are easily related through the bulk catalyst density, ρ_b (i.e., W = ρ_bV). We first consider the chemical species that stay within the reactor and do not diffuse out through the side, namely A and C, as shown in Figure 6-3(d). The steady state mole balances on A and C are derived in Example 6-2 to give the usual differential form found in Table S1-1 in Chapter 1.

$\begin{matrix} \frac{d F_{A}}{d V} = r_{A} \end{matrix} \begin{matrix} (1 - 11) \end{matrix}$

The mole balance on C is carried out in an identical manner to A, and the resulting equation is

$\begin{matrix} \frac{{d F}_{C}}{d V} = r_{C} \end{matrix} \begin{matrix} (6-1) \end{matrix}$

However, the mole balance on B (H₂) must be modified because hydrogen leaves through both the sides of the reactor and at the end of the reactor.

First, we shall perform mole balances on the volume element ΔV shown in Figure 6-3(d). The mole balance on hydrogen (B) is over a differential volume ΔV shown in Figure 6-3(d) and it yields the following.

Balance on B in the catalytic bed:

$\begin{matrix} [\begin{matrix} In \\ by flow \end{matrix}] & - & [\begin{matrix} Out \\ by flow \end{matrix}] & - & [\begin{matrix} Out \\ by diffusion \end{matrix}] & + & [Generation] & = & [Accumulation] \\ \overset{︷}{\begin{matrix} F_{{B|}_{V}} \end{matrix}} & - & \overset{︷}{\begin{matrix} F_{{B|}_{V + Δ V}} \end{matrix}} & - & \overset{︷}{\begin{matrix} R_{B} Δ V \end{matrix}} & + & \overset{︷}{\begin{matrix} r_{B} Δ V \end{matrix}} & = & 0 (6 - 2) \end{matrix}$

Now there are two “OUT” terms for species B.

where R_B is the molar rate of B leaving through the sides of the reactor per unit volume of reactor (mol/m³·s). Dividing by ΔV and taking the limit as ΔV → 0 gives

$\begin{matrix} \frac{{d F}_{B}}{d V} = r_{B} - R_{B} \end{matrix} \begin{matrix} (6-3) \end{matrix}$

An illustration depicts the transport of B through the membrane. — A diagram shows a rectangular membrane. Two arrows are present on either sides of the membrane which represents B. A curved arrow line is drawn across the membrane from one side to another. The starting point of the arrow line on one side of the membrane represents C subscript B and the arrow on the other side represents C subscript BS.

The rate of transport of B out through the membrane R_B is the product of the molar flux of B normal to the membrane, W_B (mol/m²/s), and the surface area per unit volume of reactor, a (m²/m³). The molar flux of B, W_B in (mol/m²/s) out³ through the sides of the reactor is the product of the mass transfer coefficient, $k_{C}^{'}$ (m/s), and the concentration driving force across the membrane.

$\begin{matrix} W_{B} = k_{C}^{'} (C_{B} - C_{BS}) & (6-4) \end{matrix}$

³ See Chapter 14, section 14.4, for further elaboration on the mass transfer coefficient and its correlations in the literature.

Here, $k_{C}^{'}$ is the overall mass transfer coefficient in (m/s) and C_BS is the concentration of B in the sweep gas channel (mol/m³). The overall mass transfer coefficient accounts for all resistances to transport: the tube-side resistance of the membrane, the membrane itself, and on the shell- (sweep gas) side resistance. Further elaboration of the mass transfer coefficient and its correlations can be found in the literature and in Chapter 14. In general, this coefficient can be a function of the membrane and fluid properties, the fluid velocity, and the tube diameters.

To obtain the rate of removal of B per unit volume of reactor, R_B (mol/m³/s), we need to multiply the flux through the membrane, W_B (mol/m²·s), by the membrane surface area per volume of reactor, a (m²/m³); that is

$\begin{matrix} R_{B} = W_{B} a = k_{C}^{'} a (C_{B} - C_{BS}) & (6-5) \end{matrix}$

The membrane surface area per unit volume of reactor is

$a = \frac{Area}{Volume} = \frac{π D L}{\frac{{π D}^{2}}{4} L} = \frac{4}{D}$

Letting $k_{C} = k_{C}^{'}$ a and assuming that the flow rate of the inert sweep gas is sufficiently high to keep the concentration in the sweep gas at essentially zero (i.e., C_BS ≈ 0), we obtain

Rate of B out through the sides.

$\begin{matrix} R_{B} = k_{C} C_{B} \end{matrix} \begin{matrix} (6-6) \end{matrix}$

where the units of k_C are s^–1.

More detailed modeling of the transport and reaction steps in membrane reactors is beyond the scope of this text but can be found in Membrane Reactor Technology.⁴ The salient features, however, can be illustrated by the following example. When analyzing membrane reactors, we must use molar flow rates because expressing the molar flow rate of B in terms of conversion will not account for the amount of B that has left the reactor through the sides.

⁴ R. Govind and N. Itoh, eds., Membrane Reactor Technology, AIChE Symposium Series 85, no. 268; T. Sun and S. Khang, Ind. Eng. Chem. Res., 27, 1136.

According to the DOE, 10 trillion Btu/yr could be saved by using membrane reactors.

Example 6–2 Membrane Reactor

According to the Department of Energy (DOE), an energy saving of 10 trillion Btu per year could result from the use of catalytic membrane reactors as replacements for conventional reactors for dehydrogenation reactions such as the dehydrogenation of ethylbenzene to styrene

A reaction representing the dehydrogenation of ethyl benzene to styrene is shown. — Ethylbenzene is a benzene with CH2CH3 attached to one carbon atom. After dehydrogenation, a double bond is formed for the ethane group (becomes ethene) attached and hydrogen molecule (H2) is released in the process.

and of butane to butene:

C_{4} H_{10} ⇄ C_{4} H_{8} + H_{2}

Ethyl Benzene

Styrene

The dehydrogenation of propane is another reaction that has proven successful with a membrane reactor.⁵

$C_{3} H_{8} ⇄ C_{3} H_{6} + H_{2}$

⁵ J. Membrane Sci., 77, 221 (1993).

All the preceding elementary dehydrogenation reactions described above can be represented symbolically as

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

and will take place on the catalyst side of an IMRCF. The equilibrium constant for this reaction is quite small at 227°C (e.g., K_C = 0.05 mol/dm³). The membrane is permeable to B (e.g., H₂) but not to A and C. Pure gaseous A enters the reactor at 8.2 atm and 227°C (C_T0 = 0.2 mol/dm³) at a molar flow rate of 10 mol/min.

The rate of diffusion of B out of the reactor per unit volume of reactor, R_B, is proportional to the concentration of B (i.e., R_B = k_CC_B).

Perform differential mole balances on A, B, and C to arrive at a set of coupled differential equations to solve.
Plot and analyze the molar flow rates of each species as a function of reactor volume.
Calculate the conversion of A at V = 500 dm³.

Additional information: Even though this reaction is a gas–solid catalytic reaction, we will use the bulk catalyst density in order to write our balances in terms of reactor volume rather than catalyst weight (recall $- r_{A} = - r_{A}^{'} ρ_{b}$ and W = ρ_bV). For the bulk catalyst density of ρ_b = 1.5 g/cm³ and a 2-cm inside-diameter tube containing the catalyst pellets, the specific reaction rate, k, and the transport coefficient, k_C, are k = 0.7 min^–1 and k_C = 0.2 min^–1, respectively. As a first approximation, we will neglect pressure drop, that is, p = 1.

Solution

(a) As discussed earlier, we choose reactor volume rather than catalyst weight as our independent variable for this example. The catalyst weight, W, and reactor volume, V, are easily related through the bulk catalyst density, ρ_b, (i.e., W = ρ_bV). First, we shall perform mole balances on the volume element ΔV shown in Figure 6-3(d).

Mole balances:

Balance on A in the catalytic bed:

$\begin{matrix} [\underset{by flow}{In}] & - & [\underset{by flow}{Out}] & + & [Generation] & = & [Accumulation] \\ \begin{matrix} \overset{︷}{F_{A} |_{V}} \end{matrix} & - & \overset{︷}{\begin{matrix} F_{A} |_{V + Δ V} \end{matrix}} & + & \overset{︷}{\begin{matrix} r_{A} Δ V \end{matrix}} & = & 0 \end{matrix}$

Mole balance on each and every species

Dividing by ΔV and taking the limit as ΔV → 0 gives

$\begin{matrix} \frac{{d F}_{A}}{d V} = r_{A} \end{matrix} \begin{matrix} (E6-2.1) \end{matrix}$

Balance on B in the catalytic bed:

The balance on B is given by Equation (6-3)

$\begin{matrix} \frac{{d F}_{B}}{d V} = r_{B} - R_{B} \end{matrix} \begin{matrix} (E6-2.2) \end{matrix}$

where R_B is the molar flow of B out through the membrane per unit volume of the reactor.

The mole balance on C is carried out in an identical manner to A, and the resulting equation is

$\begin{matrix} \frac{{d F}_{C}}{d V} = r_{C} \end{matrix} \begin{matrix} (E6-2.3) \end{matrix}$
Rates:

Rate Law

$\begin{matrix} {- r}_{A} = k (C_{A} - \frac{C_{B} C_{C}}{K_{C}}) \end{matrix} \begin{matrix} (E6-2.4) \end{matrix}$

Relative Rates

$\begin{matrix} \frac{r_{A}}{- 1} = \frac{r_{B}}{1} = \frac{r_{C}}{1} & (E6-2.5) \end{matrix}$

$\begin{matrix} \begin{matrix} r_{B} & = & - r_{A} \\ r_{C} & = & - r_{A} \end{matrix} \end{matrix} \begin{matrix} (E6-2.6) \\ (E6-2.7) \end{matrix}$
Transport out of the reactor: We apply Equation (6-5) for the case in which we assume the concentration of B on the sweep gas side of the membrane is essentially zero, C_BS ≈ 0, to obtain

$\begin{matrix} {R_{B} = k_{C} C}_{B} \end{matrix} \begin{matrix} (E6-2.8) \end{matrix}$

where k_C is a transport coefficient. In this example, we shall assume that the resistance to species B out of the membrane is a constant and, consequently, k_C is a constant.
Stoichiometry: Recalling Equation (4-17) for the case of constant temperature and pressure, we have for isothermal operation, T = T₀, and no pressure drop, P = P₀, (i.e., p = 1).

Concentrations

$\begin{matrix} C_{A} = C_{T0} \frac{F_{A}}{F_{T}} & (E6-2.9) \end{matrix}$

$\begin{matrix} C_{B} = C_{T0} \frac{F_{B}}{F_{T}} & (E6-2.10) \end{matrix}$

$\begin{matrix} C_{C} = C_{T0} \frac{F_{C}}{F_{T}} & (E6-2.11) \end{matrix}$

$\begin{matrix} F_{T} = F_{A} + F_{B} + F_{C} & (E6-2.12) \end{matrix}$

$\begin{matrix} X = \frac{F_{A0} - F_{A}}{F_{A0}} & (E6-2.13) \end{matrix}$
Combining and summarizing:

$\begin{matrix} \begin{matrix} \begin{matrix} \frac{d F_{A}}{d V} & = & r_{A} \\ \frac{d F_{B}}{d V} & = & - r_{A} - k_{C} C_{T0} (\frac{F_{B}}{F_{T}}) \\ \frac{d F_{C}}{d V} & = & - r_{A} \\ - r_{A} & = & k C_{T0} [(\frac{F_{A}}{F_{T}}) - \frac{C_{T0}}{K_{C}} (\frac{F_{B}}{F_{T}}) (\frac{F_{C}}{F_{T}})] \\ F_{T} & = & F_{A} + F_{B} + F_{C} \end{matrix} \end{matrix} \end{matrix}$

Summary of equations describing flow and reaction in a membrane reactor
Parameter evaluation:

$C_{T0} = \frac{P_{0}}{R T_{0}} = \frac{830.6 kPa}{[8.314 k Pa \cdot {dm}^{3} / (mol \cdot K)] (500 K)} = 0.2 \frac{mol}{{dm}^{3}}$

$\begin{matrix} k & = & 0.7 {min}^{- 1}, K_{C} = 0.05 mol / {dm}^{3}, k_{C} = 0.2 {min}^{- 1} \\ F_{A0} & = & 10 mol / min \\ F_{B0} & = & F_{C0} = 0 \end{matrix}$
Numerical solution: Equations (E6-2.1)–(E6-2.13) were solved using Poly-math, MATLAB, Wolfram, and Python. The profiles of the molar flow rates are shown below. Table E6-2.1 shows the Polymath programs, and Figure E6-2.1 shows the results of the numerical solution for the entering conditions.

$\begin{matrix} \begin{matrix} \begin{matrix} V = 0 : & F_{A} = F_{A0,} \end{matrix} & F_{B} \end{matrix} = 0, & F_{C} \end{matrix} = 0$

Information on how to obtain and download the Polymath, Wolfram, and Python software can be found on the Web site as well as in Appendix D.

TABLE E6-2.1 POLYMATH PROGRAM

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = ra-kcCto(Fb/Ft) 3 d(FC)/d(V) = -ra
Explicit equations 1 Kc = 0.05 2 Ft = Fa+Fb+Fc 3 k = 0.7 4 Cto = 0.2 5 ra = -kCto((Fa/Ft)-Cto/Kc(Fb/Ft)(Fc/Ft)) 6 kc = 0.2 7 Rate = -ra 8 Rb = kcCto(Fb/Ft) 9 Fao = 10 10 X = (Fao-Fa)/Fao
Calculated values of DEQ variables
	Variable	Initial value	Final value
1	Cto	0.2	0.2
2	Fa	10	3.995179
3	Fao	10.	10.
4	Fb	0	1.832577
5	Fc	0	6.004821
6	Ft	10.	11.83258
7	k	0.7	0.7
8	kc	0.2	0.2
9	kc	0.05	0.05
10	ra	-0.14	-0.0032558
11	Rate	0.14	0.0032558
12	Rb	0	0.006195
13	V	0	500.
14	X	0	0.6004821

A figure shows a cylindrical tube with an inlet and a outlet. — A enters the cylinder through an inlet on the left. A series of small inlets on the top of the cylinder are shown through which B enters. The products A, B, and C exit through an outlet on the right.

A graph representing polymath solutions is shown. — **Figure E6-2.1** Polymath solutions.

The vertical axis represents F subscript j baseline (mol/min) ranging from 0.0 to 10.0 in increments of 2.0. The horizontal axis represents V (decimeter cubed) ranging from 0 to 500 in increments of 100. An upward open and decreasing curve which represents F subscript A is drawn. A downward open and increasing curve which represents F subscript C is drawn. A downward open and decreasing dashed curve which represents F subscript B is drawn.

Figure E6-2.2 Wolfram sliders.

A screenshot of a graph from Wolfram profiles is shown. — **Figure E6-2.3** Wolfram profiles.

The horizontal axis represents the conversion ranging from 0.0 to 1.0 in increments of 0.2. The horizontal axis represents V (decimeter cubed) ranging from 0 to 500 in increments of 100. The conversion starts at the origin, increases to 0.4 for a volume of 100, and shows a gradual increase after that. A point is marked at (400,0.62).

(b) Figures E6-2.2 and E6-2.3 show the Wolfram variables and the conversion profiles respectively. We note that F_B goes through a maximum as a result of the competition between the rate of B being formed from A and the rate of B being removed through the sides of the reactor.

(c) From Figure E6-2.1, we see that the exit molar flow rate of A at 500 dm³ is 4 mol/min, for which the corresponding conversion is

$X = \frac{F_{A0} - F_{A}}{F_{A0}} = \frac{10 - 4}{10} = 0.60$

We note and emphasize that the conversion profile X was calculated after the fact and not used to obtain the molar flow rates.

Analysis: The molar flow rate of A drops rapidly until about 100 dm³, where the reaction approaches equilibrium. At this point, the reaction will only proceed to the right at the rate at which B is removed through the sides of the membrane, as noted by the similar slopes of F_A and F_B in this plot. You will want to use the Wolfram or Python sliders in Living Example Problem, LEP 6-1_A(b) to show that if B is removed rapidly, F_B will become close to zero and the reaction behaves as if it is irreversible, and that if B is removed slowly, F_B and C_B will be large throughout the reactor and the rate of reaction, –r_A, will be small as we approach equilibrium.

Use of Membrane Reactors to Enhance Selectivity. In addition to species leaving through the sides of the membrane reactor, species can also be fed to the reactor through the sides of the membrane. For example, for the reaction

A + B → C + D

species A will be fed only to the entrance, and species B will be fed only through the membrane as shown here.

A diagram for the use of membrane reactors to enhance selectivity is shown. — A reactor is given in which species B (F subscript B0) are distributed evenly throughout the reactor. Species A (F subscript A0) is fed through the entrance of the reactor.

As we will see in Chapter 8, this arrangement is often used to improve selectivity when multiple reactions take place. Here, B is usually fed uniformly through the membrane along the length of the reactor. The balance on B is

$\begin{matrix} \frac{{d F}_{B}}{d V} = r_{B} + R_{B} & (6 -7) \end{matrix}$

where R_B = F_B0/V_t with F_B0, representing the total molar feed rate of B through the sides and V_t the total reactor volume. The feed rate of B can be controlled by controlling the pressure drop across the reactor membrane.⁶ This arrangement will keep the concentration of A high and the concentration of B low to maximize the selectivity given by Equation (E8-2.2) for the reactions given in Section 8.6, for example, Example 8-8.

⁶ The velocity of B through the membrane, U_B, is given by Darcy’s law

U_B = K(P_s – P_r)

where K is the membrane permeability, P_s is the shell-side pressure, and P_r the reactor-side pressure

$F_{B0} = \overset{R_{B}}{\overset{︷}{C_{B0} a U_{B} V_{t}}} = R_{B} V_{t}$

where, as before, a is the membrane surface area per unit volume, C_B0 is the entering concentration of B, and V_t is the total reactor volume.

6.5 Unsteady-State Operation of Stirred Reactors

In Chapter 5 we discussed the unsteady operation of one type of reactor, the batch reactor. In this section, we discuss two other aspects of unsteady operation: startup of a CSTR and of semibatch reactors. First, the startup of a CSTR is examined to determine the time necessary to reach steady-state operation (see Figure 6-4(a)), and then semibatch reactors are discussed. In each of these cases, we are interested in predicting the concentration and conversion as a function of time. Closed-form analytical solutions to the differential equations arising from the mole balance of these reaction types can be obtained only for zero-and first-order reactions. ODE solvers must be used for other reaction orders.

A diagram of a semi-batch reactor: CSTR startup, semi-batch with cooling, and semi batch with reactive distillation is shown. — **Figure 6-4** Semibatch reactors: **(a)** CSTR startup, **(b)** semibatch with cooling, and **(c)** reactive distillation.

A cylindrical container is given. In the CSTR startup, a cylindrical container is given in which more than half of the container is filled with liquid. A rod like material with two fans at the bottom is fixed in the container. A liquid with concentration C subscript A0 is the input while C subscript A is the concentration for output. In the semi-batch with cooling, a cylindrical container is given in which half of the container is filled with liquid. A rod like material with two fans at the bottom is fixed in the container. A tube is placed on the top of the container in which the reactant B is given as input into the container. Heat is fixed with the container. In the reactive distillation, a cylindrical container is given in which more than half of the container is filled with liquid. A rod like material with two fans at the bottom is fixed in the container. A tube is fixed on the top of the container. The mixture of A and B reacts to form C and is given as a output through the tube fixed on top of the container.

An unsteady-state analysis can be used to determine the startup time for a CSTR (Figure 6-4(a)) and this analysis is given in the Expanded Material for Chapter 6 on the CRE Web site (http://www.umich.edu/~elements/6e/06chap/expanded_ch06_cstr.pdf). Here, we show the time to steady state for a first-order reaction is approximately

$\begin{matrix} t_{s} = \frac{4.6 τ}{1 + τ k} & (6-8) \end{matrix}$

For most first-order systems, the time to reach steady state is three to four space times.

There are two basic types of semibatch operations. In one type, one of the reactants in the reaction

$\begin{matrix} A + B \to C + D & (6-9) \end{matrix}$

(e.g., B) is slowly fed to a reactor containing the other reactant (e.g., A), which has already been charged to a reactor such as that shown in Figure 6-4(b). This type of reactor is generally used when unwanted side reactions occur at high concentrations of B (see Section 8.1) or when the reaction is highly exothermic (Chapter 11). In some reactions, the reactant B is a gas and is bubbled continuously through liquid reactant A. Examples of reactions used in this type of semibatch reactor operation include ammonolysis, chlorination, and hydrolysis. The other type of semibatch reactor is reactive distillation and is shown schematically in Figure 6-4(c). Here, reactants A and B are charged simultaneously and one of the products vaporizes and is withdrawn continuously. Removal of one of the products in this manner (e.g., C) shifts the equilibrium toward the right, increasing the final conversion well above that which would be achieved had C not been removed. In addition, removal of one of the products further concentrates the reactant, thereby producing an increased rate of reaction and decreased processing time. This type of reaction operation is called reactive distillation. Examples of reactions carried out in this type of reactor include acetylation reactions and esterification reactions in which water is removed.

6.6 Semibatch Reactors

6.6.1 Motivation for Using a Semibatch Reactor

One of the best reasons to use semibatch reactors is to enhance selectivity in liquid-phase reactions. For example, consider the following two simultaneous reactions. One reaction produces the desired product D

$\begin{matrix} A + B \overset{k_{D}}{\to} D & (6-10) \end{matrix}$

with the rate law

$\begin{matrix} r_{D} = k_{D} C_{A}^{2} C_{B} & (6-11) \end{matrix}$

and the other reaction produces an undesired product U

$\begin{matrix} A + B \overset{k_{U}}{\to} U & (6-12) \end{matrix}$

with the rate law

$\begin{matrix} r_{U} = k_{U} C_{A} C_{B}^{2} & (6 -13) \end{matrix}$

The instantaneous selectivity of D to U, S_D/U, is the ratio of the rate of formation of the desired product D to the rate of formation of the undesired product U and is given in Equation (6-14), and in Chapter 8 Equation (8-1).

We want S_D/U as large as possible.

$\begin{matrix} S_{D/U} = \frac{r_{D}}{r_{U}} = \frac{k_{D} C_{A}^{2} C_{B}}{k_{U} C_{A} C_{B}^{2}} = \frac{k_{D}}{k_{U}} \frac{C_{A}}{C_{B}} & (6 -14) \end{matrix}$

and guides us in how to produce the most amount of our desired product and the least amount of our undesired product (see Section 8.1). We see from Equation (6-14) that the selectivity, S_D/U can be increased by keeping the concentration of A high and the concentration of B low. This result can be achieved through the use of the semibatch reactor, which is charged with pure A and to which B is fed slowly to A in the vat.

6.6.2 Semibatch Reactor Mole Balances

We will focus our attention primarily on a semibatch reactor with a constant molar feed. A schematic diagram of this semibatch reactor is shown in Figure 6-5. We shall consider the elementary liquid-phase reaction

$\begin{matrix} A + B \to C & (6-15) \end{matrix}$

A diagram for semi-batch reactor is shown. — **Figure 6-5** Semibatch reactor.

A container is given in which most of the container is filled with reactant A. There are two openings on the top of the container in which one of the opening is closed and reactant B is poured through the other opening. A rod like material with a fan at its end is fixed in the container and mixes the reactant A and reactant B. The combination of these two reactants then gives product C.

in which reactant B is slowly added to a well-mixed vat containing reactant A. A mole balance on species A yields

Mole balance on species A

$\begin{matrix} \begin{array}{l} [\underset{in}{Rate}] - [\underset{out}{Rate}] + [\underset{generation}{Rate of}] = [\underset{accumulation}{Rate of}] \\ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \overset{︷}{0} \end{matrix} & - \end{matrix} \end{matrix} & \overset{︷}{0} \end{matrix} \end{matrix} & + \end{matrix} \end{matrix} & \overset{︷}{\begin{matrix} r_{A} \end{matrix}} V (t) \end{matrix} \end{matrix} & = \end{matrix} \end{matrix} & \frac{\overset{︷}{d N_{A}}}{d t} \end{matrix} \end{array} & (6-16) \end{matrix}$

Three variables can be used to formulate and solve semibatch reactor problems: the concentrations, C_j, the number of moles, N_j, of species j respectively and the conversion, X. We will use the subscript “0” to represent feed concentrations, for example, C_B0, and the subscript “i” to represent initial conditions, for example, C_A_i.

We shall use concentration as our preferred variable, leaving the analysis of semibatch reactors using the number of moles, N_j, and conversion X to the CRE Web site Summary Notes (http://www.umich.edu/~elements/6e/06chap/summary.html) and Professional Reference Shelf (http://www.umich.edu/~elements/6e/06chap/prof-prs_cstr.html) for Chapter 6.

Recalling that the number of moles of A, N_A, is just the product of the concentration of A, C_A, and the volume, V, (i.e., (N_A = C_AV)), we can rewrite Equation (6-16) as

$\begin{matrix} r_{A} V = \frac{{d N}_{A}}{d t} = \frac{d (C_{A} V)}{d t} = \frac{{V d C}_{A}}{d t} + C_{A} \frac{d V}{d t} & (6-17) \end{matrix}$

We note that since the reactor is being filled, the volume, V, varies with time. The reactor volume at any time t can be found from an overall mass balance of all species. The mass flow rate into the reactor, ${\dot{m}}_{0}$ , is just the product of the liquid density, ρ₀, and volumetric flow rate υ₀. The mass of liquid inside the reactor, m, is just the product of the liquid density ρ and the volume of liquid V in the reactor, that is m = ρV. There is no mass flow out and no generation of mass.

Overall mass balance

$\begin{matrix} [\begin{matrix} Mass \\ \begin{matrix} \begin{matrix} rate \end{matrix} \\ in \end{matrix} \end{matrix}] & - & [\begin{matrix} Mass \\ \begin{matrix} \begin{matrix} rate \end{matrix} \\ out \end{matrix} \end{matrix}] & + & [\begin{matrix} \begin{matrix} Rate of \\ mass \end{matrix} \\ generation \end{matrix}] & = & [\begin{matrix} \begin{matrix} Rate of \\ mass \end{matrix} \\ accumulation \end{matrix}] \\ \begin{matrix} {\dot{m}}_{0} \end{matrix} & - & 0 & + & 0 & = & \frac{d m}{d t} \\ \begin{matrix} \overset{︷}{ρ_{0} υ_{0}} \end{matrix} & - & \overset{︷}{0} & + & \overset{︷}{0} & = & \overset{︷}{\frac{d (ρ V)}{d t}} & (6-18) \end{matrix}$

For a constant-density system, ρ₀ = ρ, and

$\begin{matrix} \frac{d V}{d t} = υ_{0} & (6-19) \end{matrix}$

with the initial condition V = V₀ at t = 0, integrating for the case of constant volumetric flow rate υ₀ yields

Semibatch reactor volume as a function of time

$\begin{matrix} V = V_{0} + υ_{0} t \end{matrix} \begin{matrix} (6-20) \end{matrix}$

Substituting Equation (6-19) into the right-hand side of Equation (6-17) and rearranging gives us

${{- υ}_{0} C}_{A} + {V r}_{A} = \frac{{V d C}_{A}}{d t}$

The balance on A, seen in Equation (6-17), can be rewritten as

Mole balance on A

$\begin{matrix} \frac{d C_{A}}{d t} = r_{A} - \frac{υ_{0}}{V} C_{A} & (6-21) \end{matrix}$

A mole balance on B that is fed to the reactor at a rate F_B0 is

$\begin{matrix} In & - & Out & + & Generation & = & Accumulation \\ \begin{matrix} \overset{︷}{F_{B0}} \end{matrix} & - & \overset{︷}{0} & + & \overset{︷}{r_{B} V} & = & \overset{︷}{\frac{d N_{B}}{d t}} \end{matrix}$

Rearranging gives

$\begin{matrix} \frac{{d N}_{B}}{d t} = r_{B} V + F_{B0} & (6-22) \end{matrix}$

Substituting for N_B in terms of concentration and reactor volume (N_B = C_BV), differentiating, and then using Equation (6-19) to substitute for (dV/dt), and recalling F_B0 = C_B0 υ₀ the mole balance on B given in Equation (6-22) becomes

$\begin{matrix} \frac{{d N}_{B}}{d t} = \frac{d ({V C}_{B})}{d t} = \frac{d V}{d t} C_{B} + \frac{{V d C}_{B}}{d t} = r_{B} V + F_{B0} = r_{B} V + υ_{0} C_{B0} \end{matrix}$

Rearranging gives

Mole balance on B

$\begin{matrix} \frac{{d C}_{B}}{d t} = r_{B} + \frac{{υ_{}}_{0} (C_{B0} - C_{B})}{V} \end{matrix} \begin{matrix} (6-23) \end{matrix}$

Similarly, for species C we have

\begin{matrix} \frac{{d N}_{C}}{d t} = r_{C} V = {- r}_{A} V & (6-24) \end{matrix}

\begin{matrix} \frac{{d N}_{C}}{d t} = \frac{d (C_{C} V)}{d t} = V \frac{{d C}_{C}}{d t} + C_{C} \frac{d V}{d t} = V \frac{{d C}_{C}}{d t} + υ_{0} C_{C} & (6-25) \end{matrix}

Combining Equations (6-24) and (6-25) and rearranging we obtain

Mole balance on C

\begin{matrix} \frac{d C_{C}}{d t} = r_{C} - \frac{υ_{0} C_{C}}{V} & (6 -26) \end{matrix}

Following the same procedure for species D

Mole balance on D

\begin{matrix} \frac{d C_{D}}{d t} = r_{D} - \frac{υ_{0} D_{D}}{V} & (6 -27) \end{matrix}

At time t = 0, the initial concentrations of B, C, and D in the vat are zero, C_B_i = C_C_i = C_D_i = 0. The concentration of B in the feed is C_B0. If the reaction order is other than zero- or first-order, or if the reaction is nonisothermal, we must use numerical techniques to determine the concentrations and conversion as a function of time. Equations (6-21), (6-23), (6-26), and (6-27) are easily solved with an ODE solver.

Example 6–3 Isothermal Semibatch Reactor with Second-Order Reaction

Cyanogen Bromide

Methylamine

The production of methyl bromide is an irreversible liquid-phase reaction that follows an elementary rate law. The reaction

CNBr + CH₃NH₂ → CH₃Br + NCNH₂

is carried out isothermally in a semibatch reactor. An aqueous solution of methyl amine (B) at a concentration of C_B0 = 0.025 mol/dm³ is to be fed at a volumetric rate of 0.05 dm³/s to an aqueous solution of bromine cyanide (A) contained in a glass-lined reactor.

A diagram for isothermal semi-batch reactor with second order reaction is given. — A container is given in which most of the container is filled with bromine cyanide. There are two openings on the top of the container in which one of the opening is closed and methyl amine is poured through the other opening. A rod like material with a fan at its end is fixed in the container and mixes the bromine cyanide and methyl amine. The combination of these two liquids then give methyl bromide.

The initial volume of liquid in the vat is to be 5 dm³ with a bromine-cyanide concentration of C_A = C_A_i = 0.05 mol/dm³. The specific reaction-rate constant is

k = 2.2dm³/s·mol

Solve for the concentrations of bromine cyanide (A), methyl amine (B), methyl bromide (C), and cyanamide (D), and the rate of reaction as a function of time, and then analyze your results.

Solution

Symbolically, we write the reaction as

A + B → C + D

Mole Balances:

Mole Balance on every species

$\begin{matrix} \begin{matrix} \frac{{d C}_{A}}{d t} = r_{A} - \frac{υ_{0} C_{A}}{V} & [At t = 0, C_{A} = C_{Ai}] \end{matrix} & (6-21) \end{matrix}$

$\begin{matrix} \begin{matrix} \frac{{d C}_{B}}{d t} = \frac{υ_{0} (C_{B0} - C_{B})}{V} + r_{B} & [At t = 0, C_{B} = C_{B i} = 0] \end{matrix} & (6-23) \end{matrix}$

$\begin{matrix} \begin{matrix} \frac{{d C}_{C}}{d t} = r_{C} - \frac{υ_{0} C_{C}}{V} & [At t = 0, C_{C} = 0] \end{matrix} & (6-26) \end{matrix}$

$\begin{matrix} \begin{matrix} \frac{{d C}_{D}}{d t} = r_{D} - \frac{{υ_{0} C}_{D}}{V} & [At t = 0, C_{D} = 0] \end{matrix} & (6 -27) \end{matrix}$
Rates:

Rates: Law Relative
1. Rate Law (Elementary)
  
  $\begin{matrix} {- r}_{A} {= {k C}_{A} C}_{B} & (E6-3.1) \end{matrix}$
2. Relative Rates
  
  $\begin{matrix} {{{{- r}_{A} = - r}_{B} = r}_{C} = r}_{D} & (E6-3.2) \end{matrix}$
Combine:

Before going to the stoichiometry step, normally Step 3, let’s combine the mole balances from Equations (6-21), (6-23), (6-26), and (6-27), the rate law Equation (E6-3.1), and the relative rates Equation (E6-3.2), to arrive at the following forms of the mole balances on A, B, C, and D solely in terms of concentrations

Combined mole balances and rate laws on A, B, C, and D

$\begin{matrix} \frac{{d C}_{A}}{d t} = - {{k C}_{A} C}_{B} - \frac{{υ_{0} C}_{A}}{V} & (E6-3.3) \end{matrix}$

$\begin{matrix} \frac{d C_{B}}{d t} = - k C_{A} C_{B} + \frac{υ_{0} (C_{B0} - C_{B})}{V} & (E6-3.4) \end{matrix}$

$\begin{matrix} \frac{{d C}_{C}}{d t} = {k C}_{A} C_{B} = \frac{υ_{0} C_{C}}{V} & (E6-3.5) \end{matrix}$

$\begin{matrix} \frac{{d C}_{D}}{d t} = {k C}_{A} C_{B} = \frac{υ_{0} C_{D}}{V} & (E6-3.6) \end{matrix}$
Stoichiometry:

The volume of liquid in the reactor at any time t is

$\begin{matrix} V = V_{0} + υ_{0} t & (E6-3.7) \end{matrix}$

These coupled equations are easily solved with an ODE solver such as Polymath.

After the fact, we could also calculate the conversion of A from the concentration of A:

$\begin{matrix} X = \frac{N_{A i} - N_{A}}{N_{A i}} & (E6-3.8) \end{matrix}$

Substituting for N_Ai and N_A

$\begin{matrix} X = \frac{C_{A i} V_{0} - C_{A} V}{{C_{A i} V}_{0}} \end{matrix} \begin{matrix} (E6-3.9) \end{matrix}$

Evaluate:

The initial conditions are t = 0, C_A_i = 0.05 mol/dm³, C_B = C_C, = C_D = 0, and V₀ = 5 dm³.

Equations (E6-3.2)–(E6-3.9) are easily solved with the aid of an ODE solver such as Polymath, Wolfram, Python, or MATLAB (Table E6-3.1).

The concentrations of bromine cyanide (A), methyl amine (B), and methyl bromide (C) are shown as a function of time in Figure E6-3.1, and the rate is shown in Figure E6-3.2.

We note that the concentration of methyl bromide (C), which is identical to the concentration of cynanamide (D) goes through a maximum. The maximum occurs because once all of A has been consumed, then no more C will be formed and the continual flow of B into the reactor will thus dilute the moles of C produced and hence the concentration of C.

TABLE E6-3.1 POLYMATH PROGRAM

Differential equations 1 d(Ca)/d(t) = ra- voCa/V 2 d(Cb)/d(t) = ra+ (Cbo-Cb)vo/V 3 d(Cc)/d(t) = -ra-voCc/V d(Cd)/d(t) = -ra-voCd/V
Explicit equations 1 vo = 0.05 2 Vo = 5 3 V = Vo+vot 4 k = 2.2 5 Cbo = 0.025 6 ra = -kCaCb 7 Cao = 0.05 8 rate = -ra 9 X = (CaoVo-CaV)/(CaoVo)
Calculated values of DEQ variables
	Variable	Initial value	Final value
1	Ca	0.05	7.731E-06
2	Cao	0.05	0.05
3	Cb	0	0.0125077
4	Cbo	0.025	0.025
5	Cc	0	0.0083256
6	Cd	0	0.0083256
7	k	2.2	2.2
8	ra	0	-2.127E-07
9	rate	0	2.127E-07
10	t	0	500.
11	V	5.	30.
12	vo	0.05	0.05
13	Vo	5.	5.
14	X	0	0.9990722

A graph represents the polymath output. — **Figure E6-3.1** Polymath output: concentration–time trajectories.

A graph plots the concentration for different time periods. The vertical axis represents the concentration ranging from 0 to 5 in increments of 1. The horizontal axis represents the time ranging from 0 to 500 in increments of 100. Curves representing concentrations C subscript A, C subscript B, and C subscript C is shown. The curve for C subscript A starts at (0,5) and gradually decreases to 0 concentration at 400 seconds. The curve for C subscript B is represented using dashed lines. It starts at the origin and increases gradually to (500,1). The curve for C subscript C also starts at the origin, increases gradually (200,1.2), and decreases to (500, 0.8).

Why does the concentration of CH₃Br (C) go through a maximum wrt time?

LEP Sliders

Use Wolfram or Python to learn more about this reaction.

A graph of the reaction rate versus the time trajectory is given. — **Figure E6-3.2** Reaction rate–time trajectory.

In the graph, the vertical axis represents the reaction rate ranging from 0.00000 to 0.00020 increments of 0.00005. The horizontal axis represents the time ranging from 0 to 250 in increments of 50. The curve plotted starts at the origin, increases to a maximum value of 0.00017, and decreases gradually to (250, 0.000025).

Takeaway Lesson: Wrong choice of type of reactor.

Analysis: Let’s look at the trends. The concentration of A falls close to zero at about 250 seconds, as does the reaction rate. Consequently, very little C and D are formed after this time, and what has been formed begins to be diluted as B continues to be added to the reactor and stops just before it overflows. Now what do you think of the time to carry out this reaction? It’s about 5 minutes, hardly enough time to turn the valves on and off. Takeaway lesson: While this example showed how to analyze a semibatch reactor, you would not use a semibatch reactor to carry out this reaction at this temperature because the times are too short. Instead, you would use a tubular reactor with B fed through the sides or a number of CSTRs in series with A fed to the first reactor and small amounts of B fed to each of the following reactors. We will discuss this situation further in Chapter 8.

6.6.3 Equilibrium Conversion

For reversible reactions carried out in a semibatch reactor, the maximum attainable conversion (i.e., the equilibrium conversion) will change as the reaction proceeds because more reactant is continuously added to the reactor. This addition shifts the equilibrium continually to the right to form more product.

An outline of what is given in the Summary Notes is on the CRE Web site follows:

At equilibrium

\begin{matrix} K_{C} = \frac{N_{Ce} N_{De}}{N_{Ae} N_{Be}} & (6 -28) \end{matrix}

Using the number of moles in terms of conversion

\begin{matrix} \begin{matrix} N_{Ae} = N_{A0} (1 - X_{e}) & N_{C e} = N_{A0} X_{e} \\ \begin{matrix} N_{De} = N_{A0} X_{e} \end{matrix} & N_{Be} = F_{B0} t - N_{A0} X_{e} \end{matrix} & (6 - 29) \end{matrix}

Substituting

\begin{matrix} K_{C} = \frac{N_{A0} X_{e}^{2}}{({1 - X}_{e}) (F_{B0} t - N_{A0} X_{e})} & (6-30) \end{matrix}

LEP Sliders

Solving for X_e

\begin{matrix} X_{e} = \frac{\begin{matrix} K_{C} (1 + \frac{F_{B0} t}{N_{A0}}) - \sqrt{{[K_{C} (1 + \frac{F_{B0} t}{N_{A0}})]}^{2} - 4 (K_{C} - 1) K_{C} \frac{t F_{B0}}{N_{A0}}} \end{matrix}}{2 (K_{C} - 1)} \end{matrix} \begin{matrix} (6-31) \end{matrix}

Go to the Chapter 6 LEPs and click on Equation (6-31) to learn how the various parameters affect the X_e versus t curve. One notes the equilibrium conversion, X_e, changes with time. Further discussion on this point and calculation of the equilibrium conversion can be found in Professional Reference Shelf R6.1, in the example problem on the CRE Web site.

6.7 And Now… A Word from Our Sponsor–Safety 6 (AWFOS–S6 The BowTie Diagram)

The BowTie diagram is applied to a hazard to help us identify and address both the preventative actions (safeguards) and the contingency actions (mitigating actions). The left-hand side of the Initiating Event shows the potential problems and safeguards, while the right side shows actions that should be taken if the incident were to occur.

The safety analysis of the incident worksheet discussed in AWFOS–S5 can be used to help construct our BowTie diagram. We first brainstorm to pro-actively identify the potential problems and then brainstorm all the things that could cause each potential problem to occur. For each cause we identify an action, (safeguard) we could take to prevent the initiating event from occurring. Next, we brainstorm all the mitigating actions we could take to limit the severity of the incident, should the initiating event occur. There are three outstanding videos that discuss the BowTie Diagram: (1) a 2-minute quick overview (https://www.youtube.com/watch?v=vB0O4TPm2q0), (2) a 5-minute video (https://www.youtube.com/watch?v=P7Z6L7fjsi0) that has all the components, and (3) an in-depth video (https://www.youtube.com/watch?v=VsKgSDbHP3A). Examples of the BowTie diagram for CSB accidents can be found on the safety Web site (http://umich.edu/~safeche/bowtie.html). Figure 6-7 applies the BowTie Diagram in Figure 6-6 to the incident discussed in Chapter 5, the Acetone Drum Explosion.

An excellent representation of the BowTie method applied to prevent/mitigate an elderly person falling is demonstrated in the link (http://www.patientsafetybowties.com/knowledge-base/6-the-bowtie-method-Topevent). Further Reading on the BowTie Diagram:

A bow tie diagram of the hazards of Initiating events, is shown. It depicts two potential problems and their Preventative action on the left, and also two consequences and their Mitigating actions on the right. — **Figure 6-6** BowTie diagram.

A bow tie diagram of the acetone accident is shown. — **Figure 6-7** BowTie diagram of acetone accident.

The threat is performing hot work on drum that held flammable substances. The preventive action is as follows: Clean drum as instructed by manufacturer and Test for chemical residue in drum. The consequence is Fatality from combustion. The mitigating action is as follows: Bolt down the drum and Wearing fire-resistant and protective clothing.

The Bowtie Method. (2018). Retrieved from Patient Safety BowTies: http://www.patientsafetybowties.com/knowledge-base/6-the-bowtie-method#Top%20event.

D. Hatch, P. McCulloch, and I. Travers, “Visual HAZOP,” The Chemical Engineer, (917), 27–32 (2017, November).

B. K. Vaughen, and K. Bloch, “Use the bow tie diagram to help reduce process safety risks,” Chemical Engineering Progress, 30–36 (2016, December).

The CRE Algorithm

Closure. Chapters 5 and 6 present the heart of chemical reaction engineering for isothermal reactors. After completing these chapters, the reader should be able to apply the algorithm building blocks

to any of the reactors discussed in this chapter: BR, CSTR, PFR, PBR, membrane reactor, and semibatch reactor. The reader should be able to account for pressure drop and describe the effects of the system variables. The reader should also be able to use either conversions (Chapter 5) or concentration and molar flow rates (Chapter 6) to solve chemical reaction engineering problems.

Summary

Solution Algorithm—Measures Other Than Conversion

Gas-Phase PFR: When using measures other than conversion for reactor design, the mole balances are written for each species in the reacting mixture:

Mole balances on each and every species

$\begin{matrix} \frac{{d F}_{A}}{d V} = r_{A}, \frac{{d F}_{B}}{d V} = r_{B}, \frac{{d F}_{C}}{d V} = r_{C}, \frac{{d F}_{D}}{d V} = r_{D} & (S6-1) \end{matrix}$

The mole balances are then coupled through their relative rates of reaction. If

Rate Law

$\begin{matrix} {- r}_{A} = {k C}_{A}^{α} C_{B}^{β} & (S6-2) \end{matrix}$

for aA + bB → cC + dD, then

Relative Rates

$\begin{matrix} r_{B} = \frac{b}{a} r_{A}, r_{C} = - \frac{C}{a} r_{A}, r_{D} = - {\frac{d}{a} r}_{A} & (S6-3) \end{matrix}$

Concentration can also be expressed in terms of the number of moles (batch) and in terms of molar flow rates (PFR, CSTR, PBR).

$\begin{matrix} \begin{matrix} G a s : & C_{A} = C_{T0} \frac{F_{A}}{F_{T}} \frac{P}{P_{0}} \frac{T_{0}}{T} \end{matrix} = C_{T0} \frac{F_{A}}{F_{T}} \frac{T_{0}}{T} p & (S6-4) \end{matrix}$

$\begin{matrix} C_{B} = C_{T0} \frac{F_{B}}{F_{T}} \frac{T_{0}}{T} p & (S6-5) \end{matrix}$

Stoichiometry

$\begin{matrix} p = \frac{P}{P_{0}} \end{matrix}$

$\begin{matrix} \begin{matrix} \begin{matrix} F_{T} = F_{A} + F_{B} + F_{C} + F_{D} + F_{1} \end{matrix} & (S6-6) \end{matrix} \end{matrix}$

$\begin{matrix} \frac{d p}{d W} = \frac{- α}{2 p} (\frac{F_{T}}{F_{T0}}) (\frac{T}{T_{0}}) & (S6-7) \end{matrix}$

$\begin{matrix} \begin{matrix} L i q u i d : & C_{A} = \frac{F_{A}}{υ_{0}} \end{matrix} & (S6-8) \end{matrix}$
Membrane Reactors: The mole balances for the reaction

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

when reactant A and product C do not diffuse out the membrane

Mole Balance

$\begin{matrix} \frac{{d F}_{A}}{d V} = r_{A}, \frac{{d F}_{B}}{d V} = r_{B} - R_{B}, and \frac{{d F}_{C}}{d V} = r_{C} & (S6-9) \end{matrix}$

with

$\begin{matrix} R_{B} = k_{C} C_{B} & (S6-10) \end{matrix}$

Transport Law

and k_c is the overall mass transfer coefficient.
Semibatch Reactors: Reactant B is fed continuously to a vat initially containing only A

A + B \vec{\leftarrow} C + D

Mole Balances

\begin{matrix} \frac{{d C}_{A}}{d t} = r_{A} - \frac{υ_{0}}{V} C_{A} \end{matrix} \begin{matrix} (S6-11) \end{matrix}

\begin{matrix} \frac{{d C}_{B}}{d t} = r_{B} + \frac{υ_{0} (C_{B0} - C_{B})}{V} \end{matrix} \begin{matrix} (S6-12) \end{matrix}

\begin{matrix} \frac{{d C}_{C}}{d t} = r_{C} - \frac{υ_{0} C_{C}}{V} \end{matrix} \begin{matrix} (S6-13) \end{matrix}

\begin{matrix} \frac{{d C}_{D}}{d t} = r_{D} - \frac{υ_{0} C_{D}}{V} \end{matrix} \begin{matrix} (S6-14) \end{matrix}

\begin{matrix} \begin{matrix} V o l u m e : & V = V_{0} + υ_{0} \end{matrix} t & (S6-15) \end{matrix}

Rate Law

\begin{matrix} {- r}_{A} = k [C_{A} C_{B} - \frac{C_{C} C_{D}}{K_{e}}] \end{matrix} \begin{matrix} (S6-16) \end{matrix}

ODE Solver Algorithm

When using an ordinary differential equation (ODE) solver such as Polymath, MATLAB, Wolfram, or Python it is usually easier to leave the mole balances, rate laws, and concentrations as separate equations, rather than combining them into a single equation as we did to obtain an analytical solution. Writing the equations separately leaves it to the computer to combine them and produce a solution. The formulations for a gas-phase packed-bed reactor with pressure drop and a liquid-phase semibatch reactor are given below for two elementary reactions carried out isothermally.

LEP 6-4 Algorithm for a Gas-Phase Reaction	LEP 6-5 Algorithm for a Liquid-Phase Reaction
A + B → 3C *Packed-Bed Reactor*	A + B ⇄ 2C *Semibatch Reactor*
$\begin{matrix} \frac{d F_{A}}{d W} = r_{A}^{'} \\ \frac{d F_{B}}{d W} = r_{B}^{'} \\ \frac{d F_{C}}{d W} = r_{C}^{'} \\ r_{A}^{'} = - k C_{A} C_{B} \end{matrix}$	$\begin{matrix} \frac{d C_{A}}{d t} = r_{A} - \frac{υ_{0} C_{A}}{V} \\ \frac{d C_{B}}{d t} = r_{A} + \frac{υ_{0} (C_{B0} - C_{B})}{V} \\ \frac{d C_{C}}{d t} = - 2 r_{A} - \frac{υ_{0} C_{C}}{V} \end{matrix}$
$\begin{matrix} r_{B}^{'} = r_{A}^{'} \\ r_{C}^{'} = 3 (r_{A}^{'}) \\ C_{A} = C_{T 0} \frac{F_{A}}{F_{T}} p \\ C_{B} = C_{T 0} \frac{F_{B}}{F_{T}} p \\ C_{C} = C_{T 0} \frac{F_{C}}{F_{T}} p \\ \frac{d p}{d W} = - \frac{α}{2 p} \frac{F_{T}}{F_{T0}} \end{matrix}$ $\begin{matrix} \begin{matrix} F_{T0} = 30, F_{A0} = 15, C_{T0} = 0.02, C_{A0} = 0.01, \\ C_{B0} = 0.01, k = 5000, α = 0.009 \\ W_{final} = 80 \end{matrix} \end{matrix}$	$\begin{matrix} r_{A} = - k [C_{A} C_{B} - \frac{C_{C}^{2}}{K_{C}}] \\ V = V_{0} + υ_{0} t \\ X = \frac{V_{0} C_{A i} - V C_{A}}{V_{0} C_{A i}} \end{matrix}$ $\begin{matrix} \begin{matrix} k = 0.15, K_{C} = 16.0, V_{0} = 10.0 \\ ν_{0} = 0.1, C_{B0} = 0.1, C_{A i} = 0.04 \\ t_{final} = 200 \end{matrix} \end{matrix}$

Use Wolfram to learn more about this reactor and reaction.

The Polymath, MATLAB, Wolfram, and Python solutions to the above equations are given on the CRE Web site in Chapter 6 (http://www.umich.edu/~elements/6e/06chap/live.html).

CRE Web Site Materials

(http://umich.edu/~elements/6e/06chap/obj.html#/)

A figure represents the useful links and evaluation of the CRE website. — In the CRE website, the list of useful links include living example problems, Extra help, Additional Materials, Professional Reference Shelf, and Computer simulation problem statements. Evaluation of CRE website materials include self tests and i greater than clicker questions.

Interactive Computer Games (http://umich.edu/~elements/6e/icm/index.html)

A screenshot of the user interface of Tic- Tac interactive game is shown. The interactive game involves the concepts of reactors and chemical reactions. — Tic-Tac Interactive Game
(*http://umich.edu/~elements/6e/icm/tictac.html*)

A screenshot of a webpage displaying information regarding "Wetlands and Phytoremediation". — Wetlands
(*http://umich.edu/~elements/6e/web_mod/wetlands/index.htm*)

Aerosol Reactors (http://www.umich.edu/~elements/6e/web_mod/aerosol/index.htm)

Aerosol reactors are used to synthesize nano-size particles. Owing to their size, shape, and high specific surface area, nanoparticles can be used in a number of applications such as pigments in cosmetics, membranes, photocatalytic reactors, catalysts and ceramics, and catalytic reactors.

We use the production of aluminum particles as an example of an aerosol plug-flow reactor (APFR) operation. A stream of argon gas saturated with Al vapor is cooled.

Aerosol reactor and temperature profile.

As the gas is cooled, it becomes supersaturated, leading to the nucleation of particles. This nucleation is a result of molecules colliding and agglomerating until a critical nucleus size is reached and a particle is formed. As these particles move down the reactor, the supersaturated gas molecules condense on the particles, causing them to grow in size and then to flocculate. In the development on the CRE Web site in the Web modules category, we will model the formation and growth of aluminum nanoparticles in an AFPR.

Questions, Simulations, and Problems

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

A = • B = ▪ C = ♦ D = ♦♦

An illustration depicts three books tied together with a rope.

In each of the following 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. You may wish to refer to W. Strunk and E. B. White, The Elements of Style, 4th ed. New York: Macmillan, 2000 and Joseph M. Williams, Style: Ten Lessons in Clarity & Grace, 6th ed. Glenview, IL: Scott, Foresman, 1999, to enhance the quality of your sentences. See the preface 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

Q6-1_A QBR (Question Before Reading). What are the steps in the CRE algorithm that must be added when conversion is not used as a variable and which step is usually the most difficult to implement?

Q6-2_A i>clicker. Go to the Web site (http://www.umich.edu/~elements/6e/06chap/iclicker_ch6_q1.html) and view at least 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: explain why the question should not be on the next exam. In either case, explain your reasoning.

Q6-3_A Read through all the problems at the end of this chapter. Make up and solve an original problem based on the material in this chapter. (a) Use real data and reactions for further instructions. (b) Make up a reaction and data. (c) Use an example from everyday life (e.g., cooking spaghetti; see Question Q5-4_A).

Q6-4_A How would you modify Table 6-2 for

A constant-volume gas-phase reaction, and
A variable-volume gas-phase reaction?

Q6-5_B What are the similarities and differences between the BowTie Diagram, the Safety Analysis of the Incident, and the Swiss Cheese Model?

Q6-6_A Robert the Worrier: What if... you were asked to explore the example problems in this chapter to learn the effects of varying the different parameters? This sensitivity analysis can be carried out by downloading the examples from the CRE Web site. For each of the example problems you investigate, write a paragraph describing your findings.

Q6-7_A Go to the LearnChemE screencasts link for Chapter 6 (http://www.learncheme.com/screencasts/kinetics-reactor-design).

View one or more of the screencast 5- to 6-minute videos and write a two sentence evaluation.
In the membrane reactor, what are the benefits of removing hydrogen?
View two other LearnChemE videos for this chapter and list two things that should be improved upon and two things that were well done.

Computer Simulations and Experiments

P6-1_B AWFOS–S6 List two similarities and two differences between the Safety Analysis of the Incident Algorithm and the BowTie Diagram.

Example 6-1: Gas-Phase Reaction in a Microreactor

Wolfram and Python
1. Use Wolfram and/or Python to Compare what happens when T and E_A are each set at their maximum and minimum values (e.g., E_Amax, T_min, E_Amax, T_max), and then move the other slider and write a set of conclusions.
  
  Polymath
2. Compare Figure E6-1.1 profiles with those for a reversible reaction with K_C = 0.02 mol/dm³ and describe the differences in the profiles.
3. How would your profiles change for the case of an irreversible reaction with pressure drop when α_p = 99 × 10³ dm^–3 for each tube?
Example 6-2: Membrane Reactor

Wolfram and Python

A Stop and Smell the Roses Simulation. Vary the sliders to develop an intuitive feel of membrane reactors.
1. Starting with the default settings (all in proper default units) on the LEP simulation (i.e. , K_C = 0.05, k = 0.7, C_T0 = 0.2, and k_C = 0.2), vary each parameter individually and describe what you find. Note and explain any maximum or minimum values of your plots down the length (i.e., volume = 500 dm³) of your reactor. Hint: Go to the extremes of the range of variables.
2. Repeat (i) but set K_C at its maximum value and then vary k and k_C, and describe what you find.
3. Write a set of conclusions from your experiments in (i) and (ii).
  
  Polymath
4. Vary ratios of parameters such as (k/k_C) and (kτC_A0/K_C) (note: τ = 400 min) and write a paragraph describing what you find. What ratio of parameters has the greatest effect on the conversion X = (F_A0 – F_A)/F_A0? (Trial and Error Solution)
5. Include pressure drop with α = 0.002 dm^–3 and compare the conversion profiles for the two cases.
6. Write a summary paragraph of all the trends and your results.
7. Make up a question/problem on membrane reactors with a solution in which Wolfram must be used to obtain the answer. Hint: Apply to this problem one or more of the six ideas discussed in Table P-4 in the Complete Preface-Introduction on the Web site (http://www.umich.edu/~elements/6e/toc/Preface-Complete.pdf). Also comment on what types of questions you would ask when using Wolfram.
Example 6-2: Membrane Reactor Again

Polymath

Rework part (b) for the case when the reaction produces 3 mole of hydrogen

C₆H₁₂ ⇄ 3H₂ + C₆H₆

You will need to make minor modifications to the LEP Polymath code on the Web site. All the parameters remain the same except the equilibrium constant, which is K_C = 0.001 ${(\frac{mol}{{dm}^{3}})}^{3}$ .
Example 6-3: Isothermal Semibatch Reactor with a Second-Order Reaction

Wolfram and Python
1. Describe what happens as C_B0, υ₀, and V₀ are varied one at a time between their maximum and minimum values. Explain why the variations from the base case look the way they do.
2. After using the slider to vary the parameter, write a set of conclusions.
  
  Polymath
3. Redo this example problem assuming the reaction is reversible with K_C = 0.1. Modify the Polymath code and compare with the irreversible case. (Only a couple of changes in the Poly-math program are necessary.)
Example 6-4: Algorithm for a Gas-Phase Reactor with Pressure Drop

Wolfram and Python
1. Determine which slider has the greatest effect on conversion.
2. The entering pressure in reactor was set at 10 atm. Which parameter(s) cause(s) the exit pressure drop to atmospheric pressure, i.e., p = 0.1? Give the values of those parameters.
3. After using the sliders to vary the parameters, write a set of conclusions.
Example 6-5: Algorithm for a Semibatch Liquid-Phase Reactor

Wolfram and Python
1. Why is the conversion almost negligible below 20 minutes for the values of the initial settings?
2. Describe what happens when you decrease K_C and increase k at the same time. How does this change affect the maximum concentration of B?
3. What slider variables have the greatest effect on conversion?

Web Modules

Web Module on Wetlands on the CRE Web site. Download the Polymath program and vary a number of parameters such as rainfall, evaporation rate, atrazine concentration, and liquid flow rate, and write a paragraph describing what you find. This topic is a hot Ch.E. research area.
Web Module on Aerosol Reactors on the CRE Web site. Download the Polymath program and (1) vary the parameters, such as cooling rate and flow rate, and describe their effect on each of the regimes: nucleation, growth, and flocculation. Write a paragraph describing what you find. (2) It is proposed to replace the carrier gas by helium.
1. Compare your plots (He vs. Ar) of the number of Al particles as a function of time. Explain the shape of the plots.
2. How does the final value of d_p compare with that when the carrier gas was argon? Explain.
3. Compare the time at which the rate of nucleation reaches a peak in the two cases (carrier gas = Ar and He). Discuss the comparison.
Data for a He molecule: mass = 6.64 × 10^–27 kg, volume = 1.33 × 10^–29 m³, surface area = 2.72 × 10^–19 m², bulk density = 0.164 kg/m³, at a temperature of 25°C and pressure of 1 atm.

Interactive Computer Games

P6-2_B Download the Interactive Computer Games (ICG) from the CRE Web site (http://www.umich.edu/~elements/6e/icm/tictac.html). Play the game and then record your performance number, which indicates your mastery of the material. Your instructor has the key to decode your performance number. Knowledge of all sections is necessary to pit your wit against the computer adversary in playing a game of Tic-Tac-Toe.

Performance number: __________________

Problems

P6-3_C OEQ (Old Exam Question). The second-order liquid-phase reaction

C₆H₅COCH₂Br + C₆H₅N → C₆H₅COHCH₂NC₅H₅Br

is carried out in a batch reactor at 35°C. The specific reaction-rate constant is 0.0445 dm³/mol/min. Reactor 1 is charged with 1000 dm³, where the concentration of each reactant after mixing is 2M.

What is the conversion after 10, 50, and 100 minutes?

Now, consider the case when, after filling reactor 1, the drain at the bottom of reactor 1 is left open and it drains into reactor 2, mounted below it, at a volumetric rate of 10 dm³/min.
What will be the conversion and concentration of each species in reactor 1 after 10, 50, and 80 minutes in the reactor that is being drained? (Ans: At t = 10 min then X = 0.47)
What is the conversion and concentration of each species in reactor 2 that is filling up with the liquid from reactor 1 after 10 and after 50 minutes? (Ans: At t = 50 min then X = 0.82)
At the end of 50 minutes, the contents of the two reactors are added together. What is the overall conversion after mixing?
Apply one or more of the six ideas in Preface Table P-4, page xxvii, to this problem.

P6-4_B The elementary gas-phase reaction

(CH₃)₃COOC(CH₃)₃ → C₂H₆ + 2CH₃COCH₃

A → B + 2C

is carried out isothermally at 400 K in a flow reactor with no pressure drop. The specific reaction rate at 50°C is 10^–4 min^–1 (from pericosity data) and the activation energy is 85 kJ/mol. Pure di-tert-butyl peroxide enters the reactor at 10 atm and 127°C and a molar flow rate of 2.5 mol/min, that is, F_A = 2.5 mol/min.

Use the algorithm for molar flow rates to formulate and solve the problem. Plot F_A, F_B, F_C, and then X as a function of plug-flow reactor volume and space time to achieve 90% conversion.
Calculate the plug-flow volume and space time for a CSTR for 90% conversion.

P6-5_B For the reaction and data in Problem P6-4_B, we now consider the case when the reaction is reversible with K_C = 0.025 dm⁶/mol² and the reaction is carried out at 300 K in a membrane reactor where C₂H₆ is diffusing out. The membrane transport coefficient is k_C = 0.08 s^–1.

What is the equilibrium conversion and what is the exit conversion in a conventional PFR? (Ans: X_eq = 0.52, X = 0.47)
Plot and analyze the conversion and molar flow rates in the membrane reactor as a function of reactor volume up to the point where 80% conversion of di-tert-butyl peroxide is achieved. Note any maxima in the flow rates.
Apply the ideas in Tables P-2 and P-4 in the Complete Preface-Introduction on the Web site (http://www.umich.edu/~elements/6e/toc/Preface-Complete.pdf) to generate new questions to be added to this problem–three each for both critical and creative thinking.

P6-6_C OEQ (Old Exam Question). (Membrane reactor) The first-order, gas-phase, reversible reaction

A \vec{\leftarrow} B + 2 C

is taking place in a membrane reactor. Pure A enters the reactor, and B diffuses out through the membrane. Unfortunately, a small amount of the reactant A also diffuses through the membrane.

Plot and analyze the flow rates of A, B, and C and the conversion X down the reactor, as well as the flow rates of A and B through the membrane.
Next, compare the conversion profiles in a conventional PFR with those of a membrane reactor from part (a). What generalizations can you make?
Would the conversion of A be greater or smaller if C were diffusing out instead of B?
Discuss qualitatively how your curves would change if the temperature were increased significantly or decreased significantly for an exothermic reaction. Repeat the discussion for an endothermic reaction.

Additional information:

k = 10 min^–1
K_C = 0.01 mol²/dm⁶
k_CA = 1 min^–1
k_CB = 40 min^–1
F_A0 = 100 mol/min
υ₀ = 100 dm³/min
V_reactor = 20 dm³

P6-7_B Fuel Cells Rationale. With the focus on alternative clean-energy sources, we are moving toward an increased use of fuel cells to operate appliances ranging from computers to automobiles. For example, the hydrogen/oxygen fuel cell produces clean energy as the products are water and electricity, which may lead to a hydrogen-based economy instead of a petroleum-based economy.

A large component in the processing train for fuel cells is the water-gas shift membrane reactor. (M. Gummala, N. Gupla, B. Olsomer, and Z. Dardas, Paper 103c, 2003, AIChE National Meeting, New Orleans, LA.)

CO + H_{2} O ⇄ {CO}_{2} + H_{2}

An illustration of a fuel cell is shown. — A fuel cell is shown in the form of a rectangle with layers of Anode, Cathode, 2 Catalysts and Electrolyte present in it. The 2 catalysts are connected by a load. The input of Anode is H2O and the input of Cathode is CO. The output of Cathode is hydrogen molecule and carbon dioxide.

Here, CO and water are fed to the membrane reactor containing the catalyst. Hydrogen can diffuse out the sides of the membrane, while CO, H₂O, and CO₂ cannot. Based on the following information, plot the concentrations and molar flow rates of each of the reacting species down the length of the membrane reactor. Assume the following: The volumetric feed is 10 dm³/min at 10 atm, and the equimolar feed of CO and water vapor with C_T0 = 0.4 mol/dm³. The equilibrium constant is K_e = 1.44, with k = 1.37 dm⁶/mol kg-cat · min, bulk density, ρ = 1000 kg/m³, and the mass transfer coefficient k_H₂ = 0.1 dm³/kg-cat · min. Hint: First calculate the entering molar flow rate of CO and then relate F_A and X.

What is the membrane reactor catalyst weight necessary to achieve 85% conversion of CO?
Sophia wants you to compare the MR with a conventional PFR. What will you tell her?
For that same membrane reactor catalyst weight, Nicolas wants to know what would be the conversion of CO if the feed rate were doubled?

P6-8_C OEQ (Old Exam Question). The production of ethylene glycol from ethylene chlorohydrin and sodium bicarbonate

CH₂OHCH₂Cl + NaHCO₃ → (CH₂OH)₂ + NaCl + CO₂↑

is carried out in a semibatch reactor. A 1.5-molar solution of ethylene chlorohydrin is fed at a rate of 0.1 mol/min to 1500 dm³ of a 0.75-molar solution of sodium bicarbonate. The reaction is elementary and carried out isothermally at 30°C where the specific reaction rate is 5.1 dm³/mol/h. Higher temperatures produce unwanted side reactions. The reactor can hold a maximum of 2500 dm³ of liquid. Assume constant density.

Plot and analyze the conversion, reaction rate, concentration of reactants and products, and number of moles of glycol formed as a function of time.
Suppose you could vary the flow rate between 0.01 and 200 mol/min. What flow rate and holding time would you choose to make the greatest number of moles of ethylene glycol in 24 hours, keeping in mind the downtimes for cleaning, filling, and so on, shown in Table 5-3?
Suppose the ethylene chlorohydrin is fed at a rate of 0.15 mol/min until the reactor is full and then shut in. Plot the conversion as a function of time.
Discuss what you learned from this problem and what you believe to be the point of this problem.

P6-9_C The following elementary reaction is to be carried out in the liquid phase

{{{{{{{{Na OH + CH}_{3} COOC}_{2} H}_{5} \to CH}_{3} COO}^{-} Na}^{+} + C}_{2} H}_{5} OH

The initial concentrations are 0.2 M in NaOH and 0.25 M in CH₃COOC₂H₅ with k = 5.2 × 10^–5 dm³/mol·s at 20°C with E = 42,810 J/mol. Design a set of operating conditions (e.g., υ₀, T, . . .) to produce 200 mol/day of ethanol in a semibatch reactor and not operate above 37°C and below a concentration of NaOH of 0.02 molar.⁷ The semibatch reactor you have available is 1.5 m in diameter and 2.5 m tall. The reactor down time is (t_c + t_e + t_f) = 3 h.

⁷ Manual of Chemical Engineering Laboratory, University of Nancy, Nancy, France, 1994 ([email protected]).

P6-10_B Go to Professor Herz’s Reactor Lab Web site at www.reactorlab.net. From the menu at the top of the page, select Download and then click on the English version link. Provide the required information and then download, install, and open the software. Select Division D2, Lab L2 and there the labeled PFR of The Reactor Lab concerning a packed-bed reactor (labeled PFR) in which a gas with the physical properties of air flows over spherical catalyst pellets. Perform experiments here to get a feeling for how pressure drop varies with input parameters such as reactor diameter, pellet diameter, gas-flow rate, and temperature. In order to get significant pressure drop, you may need to change some of the input values substantially from those shown when you enter the lab. If you get a notice that you can’t get the desired flow, then you need to increase the inlet pressure.

P6-11_B OEQ (Old Exam Question). Pure butanol is to be fed into a semibatch reactor containing pure ethyl acetate to produce butyl acetate and ethanol. The reaction

CH₃COOC₂H₅+C₄H₉OH ⇄ CH₃COOC₄H₉+C₂H₅OH

is elementary and reversible. The reaction is carried out isothermally at 300 K. At this temperature, the equilibrium constant is 1.08 and the specific reaction rate is 9 × 10^–5 dm³/mol·s. Initially, there is 200 dm³ of ethyl acetate in the vat, and butanol is fed at a volumetric rate of 0.05 dm³/s. The feed and initial concentrations of butanol and ethyl acetate are 10.93 and 7.72 mol/dm³, respectively.

Plot and analyze the equilibrium conversion of ethyl acetate as a function of time.
Plot and analyze the conversion of ethyl acetate, the rate of reaction, and the concentration of butanol as a function of time.
Rework part (b), assuming that ethanol evaporates (reactive distillation) as soon as it forms. (This is a graduate level question.)
Use Polymath or some other ODE solver to learn the sensitivity of conversion to various combinations of parameters (e.g., vary F_B0, N_A0, υ₀).
Apply one or more of the six ideas in Preface Table P-4, page xxvii, to this problem.
Write a question that requires critical thinking and then explain why your question requires critical thinking. Hint: See Preface Section G.

P6-12_C Use the reaction data in Problem P6-11_B and the molar flow rate algorithm to carry out the following problems:

Calculate the CSTR volume to achieve 80% of the equilibrium conversion for an equal molar feed and for a volumetric feed of 0.05 dm³/s.
Safety. Now consider the case where we want to shut down the reactor by feeding water at a volu-metric rate of 0.05 dm³/s. How long will it take to reduce the rate to 1% of the rate in the CSTR under the conditions of part (a)?

P6-13_C OEQ (Old Exam Question). An isothermal reversible reaction $A ⇄ B$ is carried out in an aqueous solution. The reaction is first-order in both directions. The forward rate constant is 0.4 h^–1 and the equilibrium constant is 4.0. The feed to the plant contains 100 kg/m³ of A and enters at the rate of 12 m³/h. Reactor effluents pass to a separator, where B is completely recovered. The reactor is a stirred tank of volume 60 m³. A fraction, f₁, of the unreacted effluent is recycled as a solution containing 100 kg/m³ of A and the remainder is discarded. Product B is worth $2 per kilogram and operating costs are $50 per cubic meter of solution entering the separator. What value of f maximizes the operational profit of the plant? What fraction A fed to the plant is converted at the optimum? Source: H. . Shankar, IIT Mumbai.

Supplementary Reading

MAXWELL ANTHONY, presidential inauguration address, “The economic future of Jofostan and the chemical reaction industry and one’s ability to deal with multiple reactions.” Riça, Jofostan, January 1, 2022.

GARRISON KEILLOR and TIM RUSSELL, Dusty and Lefty: The Lives of the Cowboys (Audio CD). St. Paul, MN: High-bridge Audio, 2006.

G. F. FROMENT and K. B. BISCHOFF, Chemical Reactor Analysis and Design, 2nd ed. New York: Wiley, 1990.

Recent information on reactor design can usually be found in the following journals: Chemical Engineering Science, Chemical Engineering Communications, Industrial and Engineering Chemistry Research, Canadian Journal of Chemical Engineering, Jofostan Journal of Chemical Engineering, AIChE Journal, and Chemical Engineering Progress.

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Differential equations 1 d(Fa)/d(V) = rb 2 d(Fb)/d(V) = rb 3 d(FC)/d(V) = rc
Explicit equations 1 T = 698 2 Cto = 1641/8.314/T 3 E = 24000 4 Ft = Fa+Fb+Fc 5 Ca = CtoFa/Ft 6 k = 0.29exp(E/1.987(1/500-1/T)) 7 Fao = 0.0000226 8 vo = Fao/Cto 9 Tau = V/vo 10 ra = -kCa^2 11 X = 1-Fa/Fao 12 rb = -ra 13 rc = -ra/2
Calculated values of DEQ variables
	Variable	Initial value	Final value
1	Ca	0.2827764	0.0307406
2	Cto	0.2827764	0.2827764
3	E	2.4E+04	2.4E+04
4	Fa	2.26E-05	3.495E-06
5	Fao	2.26E-05	2.26E-05
6	Fb	0	1.91E-05
7	Fc	0	9.552E-06
8	Ft	2.26E-05	3.215E-05
9	k	274.4284	274.4284
10	ra	-21.94397	-0.2593304
11	rateA	21.94397	0.2593304
12	rb	21.94397	0.2593304
13	rc	10.97199	0.1296652
14	T	698.	698.
15	Tau	0	0.1251223
16	V	0	1.0E-05
17	vo	7.992E-05	7.992E-05
18	X	0	0.8453416

Table of Contents for 6. Isothermal Reactor Design: Moles and Molar Flow Rates

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6. Isothermal Reactor Design: Moles and Molar Flow Rates

6.1 The Moles and Molar Flow Rate Balance Algorithms

6.2 Mole Balances on CSTRs, PFRs, PBRs, and Batch Reactors

6.2.1 Liquid Phase

6.2.2 Gas Phase

6.3 Application of the PFR Molar Flow Rate Algorithm to a Microreactor

6.4 Membrane Reactors

6.5 Unsteady-State Operation of Stirred Reactors

6.6 Semibatch Reactors

6.6.1 Motivation for Using a Semibatch Reactor

6.6.2 Semibatch Reactor Mole Balances

6.6.3 Equilibrium Conversion

6.7 And Now… A Word from Our Sponsor–Safety 6 (AWFOS–S6 The BowTie Diagram)

Summary

ODE Solver Algorithm

CRE Web Site Materials

Questions, Simulations, and Problems

Questions

Computer Simulations and Experiments

Web Modules

Interactive Computer Games

Problems

Supplementary Reading

Table of Contents for
6. Isothermal Reactor Design: Moles and Molar Flow Rates