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 Fj are obtained from a mole balance on each species, as given in Table 6-2. For example, for a plug-flow reactor

1-11

image

Must write a mole balance on each species

The generic power law rate law is

Rate Law

3-3

image

To relate concentrations to molar flow rates, recall Equation (4-17), with y = P/P0

Stoichiometry

4-17

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

5-28

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

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Table 6-2. Algorithm for Gas-Phase Reactions

images

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

2. Rates:

Rate Law:

image

Relative Rates:

image

then

image

3. Stoichiometry:

Concentrations:

image

Total molar flow rate: FT = FA + FB + FC + FD + FI

4. Combine:

Appropriate Reactor Mole Balance on Each Species

Rate Law

Concentration for Each Species

Pressure Drop Equation

5. Evaluate:

1. Specify parameter values: kA,CT0,α,β,T0,a,b,c,d

2. Specify entering molar flow rates: FA0, FB0,FC0,FD0,and final volume,Vfinal

6. Use an ODE solver.

Many times we will let the ODE solver replace the combine step.

image

Gas phase

when species A, B, C, D, and inert I are the only ones present. Then

FT = FA + FB + FC + FD + FI

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

6.3 Applications of the Molar Flow Rate Algorithm to Microreactors

Microreactors are emerging as a new technology in CRE. Microreactors 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 20,000 μm (2 cm). The resulting high surface-area-to-volume ratio (ca. 10,000 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 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.

Advantages of microreactors

Figure 6-2 shows (a) a microreactor with heat exchanger and (b) a microplant with reactor, valves, and mixers. Heat, image, is 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

image

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

Figure 6-2. Microreactor (a) and Microplant (b). Courtesy of Ehrfeld, Hessel, and Löwe, Microreactors: New Technology for Modern Chemistry (Weinheim, Germany: Wiley-VCH, 2000).

image

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

1-12

image

or in laminar flow, in which case we will use the segregation model discussed in Chapter DVD13. 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 gas-phase reaction

image

is 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 ten microreactors in parallel. Each microreactor has 100 channels with each channel 0.2 mm square and 250 mm in length.

image

a. Plot and analyze the molar flow rates as a function of volume down the length of the reactor. The volume of each channel is 10–5 dm3.

b. What is the reactor volume necessary to achieve 85% conversion?

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 rate constant is

image

Solution

For one channel,

image

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 where conversion cannot be used as a variable. We first write the reaction in symbolic form and then divide by the stoichiometric coefficient of the limiting reactant, NOCl.

image

image

  1. Mole balances on species A, B, and C:

    E6-1.1

    image

    E6-1.2

    image

    E6-1.3

    image

  2. Rates:

    a. Rate Law

    E6-1.4

    image

    b. Relative rates

    image

  3. Stoichiometry: Gas phase with T = T0 and P = P0, then image Concentration Gas Phase

    4-17

    image

    Applying Equation (4-17) to species A, B, and C, for T = T0, P = P0, the concentrations are

    E6-1.5

    image

  4. Combine: The rate law in terms of molar flow rates is

    image

    combining all

    E6-1.6

    image

    E6-1.7

    image

    E6-1.8

    image

  5. Evaluate:

    image

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).

Figure E6-1.1. Profiles of microreactor molar flow rates.

image

Table E6-1.1. Polymath Program

Information on how to obtain and load the Polymath software can be found in Appendix E.

image

Analysis: This gas phase reaction in a PFR example could just as easily 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 through 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 6×10–6 dm3 down the reactor. Other interesting variables you will want to plot when you load this program from the Living Example Problem file are the total molar flow rate, FT, the concentrations of the reacting species, CA, CB, and CC (for CB and CC you will need to type in two additional equations), and the rates –rA, rB and rC.

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 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 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.

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 of 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.

By having one of the products pass through the membrane, we drive the reaction 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. For example, in the reversible reaction

image

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.

H2 diffuses through the membrane, while C6H6 does not.

image

Figure 6-3. Membrane reactors. (a) Photo of ceramic reactors, (b) cross section of IMRCF, (c) cross section of CRM, (d) schematic of IMRCF for mole balance.

image

Photo courtesy of Coors Ceramics, Golden, Colorado.

Hydrogen, species B, flows out through the sides of the reactor as it flows down the reactor with the other products that cannot leave until they exit the reactor.

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, ρb (i.e., W = ρbV). The mole balances on the chemical species that stay within the reactor, namely A and C, are shown in Figure 6-3(d).

1-11

image

image

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

6-1

image

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

Balance on B in the catalytic bed:

6-2

image

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

where RB 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

6-3

image

The rate of transport of B out through the membrane RB is the product of the molar flux of B normal to the membrane, WB (mol/m2/s), and the surface area per unit volume of reactor, a (m2/m3). The molar flux of B, WB in (mol/m2/s) out through the sides of the reactor is the product of the mass transfer coefficient, image, and the concentration driving force across the membrane.

image

6-4

image

Here, image is the overall mass transfer coefficient in m/s and CBS 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 DVD-ROM Chapter 11 and on the Essentials Web site. 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, RB (mol/m3/s), we need to multiply the flux through the membrane, WB (mol/m2·s), by the membrane surface area per volume of reactor, a (m2/m3); that is

6-5

image

The membrane surface area per unit volume of reactor is

image

Letting image and assuming the concentration in the sweep gas is essentially zero (i.e., CBS ≈ 0), we obtain

6-6

image

Rate of B out through the sides.

where the units of kC 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.3 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.

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:

image

and of butane to butene:

image

The dehydrogenation of propane is another reaction that has proven successful with a membrane reactor.4

image

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

image

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., KC = 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 (CT0 = 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, RB, is proportional to the concentration of B (i.e., RB = kCCB).

a. Perform differential mole balances on A, B, and C to arrive at a set of coupled differential equations to solve.

b. Plot and analyze the molar flow rates of each species as a function of reactor volume.

c. Calculate the conversion of A at V = 400 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 image). For the bulk catalyst density of ρb = 1.5 g/cm3 and a 2-cm inside diameter tube containing the catalyst pellets, the specific reaction rate, k, and the transport coefficient, kC, are k = 0.7 min–1 and kC = 0.2 min–1, respectively.

Solution

We shall 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, ρb, (i.e., W = ρbV). First, we shall perform mole balances on the volume element ΔV shown in Figure 6-3(d).

  1. Mole balances:

    Balance on A in the catalytic bed:

    image

    Mole balance on each and every species

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

    E6-2.1

    image

    Balance on B in the catalytic bed:

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

    E6-2.2

    image

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

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

    E6-2.3

    image

  2. Rates:

    Rate Law

    E6-2.4

    image

    Relative Rates

    E6-2.5

    image

    E6-2.6

    image

    E6-2.7

    image

  3. Transport out of the reactor. We apply Equation (6-5) for the case in which the concentration of B of the sweep side is zero, CBS = 0, to obtain

    E6-2.8

    image

    where kC 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, kC is a constant.

  4. Stoichiometry. Recalling Equation (4-17) for the case of constant temperature and pressure, we have for isothermal operation and no pressure drop (T = T0, P = P0),

    Concentrations:

    E6-2.9

    image

    E6-2.10

    image

    E6-2.11

    image

    E6-2.12

    image

  5. Combining and summarizing:

    image

  6. Parameter evaluation:

    image

    Summary of equations describing flow and reaction in a membrane reactor

  7. Numerical solution. Equations (E6-2.1) through (E6-2.11) were solved using Polymath and MATLAB, another ODE solver. The profiles of the molar flow rates are shown here. Table E6-2.1.1 shows the Polymath programs, and Figure E6-2.1 shows the results of the numerical solution for the entering conditions.

    V = 0: FA = FA0, FB = 0, FC = 0

Information on how to obtain and load the Polymath software can be found in Appendix E.

Table E6-2.1. Polymath Program

image

Figure E6-2.1. Polymath solution.

image

We note that FB 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 400 dm3 is 4 mol/min, for which the corresponding conversion is

image

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 FA and FB in this plot. You will want to use Problem 6-2A(b) to show that if B is removed rapidly FB will close to zero and the reaction behaves as if it is irreversible and that if B is removed slowly, FB will be large throughout the reactor and the rate of reaction, –rA, will be small.

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