Multiple Reactions
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Multiple Reactions. 授課教師:林佳璋. Definitions. There are four basic type of multiple reactions: series, parallel, complex, and independent. These types of multiple reactions can occur by themselves, in pair, or all together. When there is a combination of parallel and series reactions, they

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Multiple reactions

Multiple Reactions

授課教師:林佳璋


Multiple reactions

Definitions

There are four basic type of multiple reactions: series, parallel, complex, and independent. These types of multiple reactions can occur by themselves, in pair, or all together. When there is a combination of parallel and series reactions, they

are often referred to as a complex reactions.

Parallel reactions (competing reaction)

k1

B

A

k2

C

The reactant is consumed by two different reaction pathways to form different products.

Series reactions (consecutive reaction)

k1

The reactant forms an intermediate product, where reacts further to form another product.

k2

A

B

C


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Complex reactions

formation of butadiene from ethanol

Involving a combination of both series and parallel reactions

Independent reactions

cracking of crude oil to form gasoline

Occurring at the same time but neither the products nor reactants react with themselves or one another.


Multiple reactions

Desired and Undesired Reactions

Of particular interest are reactants that are consumed in the formation of a desired product, D, and the formation of an undesired product, U, in a

competing or side reaction.

For a parallel reaction sequence

For a series reaction sequence

We want to minimize the formation of U and maximize the formation of D because the greater the amount of undesired product formed, the greater the cost of separating the undesired product U from

the desired product D (Figure 6-1).

As the cost of a reactor system increases in a attempt to minimize U, the cost of separating species U from D decreases.


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Selectivity tells us how one product is favored over another when we have multiple reactions. We can quantify the formation of D with respect to U by

defining the selectivity and yield of the system.

The instantaneous selectivity of D with respective to U is the ratio of the rate formation of D to the rate of formation of U.

Overall selectivity

for a batch reactor


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Example 6-1

Develop a relationship between SD/U and for a CSTR.

Solution

Consider the instantaneous selectivity for the two parallel reactions just discussed:

Overall selectivity

For a CSTR the overall and instantaneous selectivities are equal.

Carrying out a similar analysis of the series reaction


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instantaneous yield

overall yield

~Instantaneous yield and overall yield are identical

for a CSTR.

~From an economic standpoint, the overall selectivities

and yields are important in determining profits.

However, the rate-based selectivities give insights in

choosing reactors and reaction schemes that will help

maximize the profit.

~There often is a conflict between selectivity and

conversion (yield) because you want to a lot of your

desired product (D) and at the same time minimize

the undesired product (U). However, in many

instances, the greater conversion you achieve, not

only do you make more D, but you also form more U.

for a batch system

for a flow system


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Parallel reactions

rate of disappearance of A is

instantaneous selectivity is

Case 1:1>2

To make this ratio as large as possible, we want to carry out the reaction in a manner that will keep the concentration of reactant A as high as possible during the reaction.

A batch or plug flow reactor should

be used in this case.

CSTR should not be chosen under these

circumstances.

If the reaction is carried out in the gas phase, we should run it without inerts and at high pressures to

keep CA high.

If the reaction is in the liquid phase, the use of diluents

should be kept to a minimum.


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Case 2:2>1

For this ratio rD/rU to be high, the concentration of A should be as low as possible

This low concentration may be accomplished by diluting the feed with inerts and running the reactor at low concentrations of A.

A CSTR should be used because the concentrations of reactants are

maintained at a low level.

A recycle reactor in which the product stream acts as a diluent could be used to maintain the entering concentrations of A at a low

value.


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Because the activation energies of the two reactions in case 1 and 2 are not given, it cannot be determined whether the reaction should be run at high or

low temperatures.

The sensitivity of the instantaneous selectivity to temperature can be determined from the ratio of the specific reaction rates,

A is the frequency factor

E is the activation energy

Case 4:EU>ED

Case 3:ED>EU

SD/U

SD/U

T

T

The reaction system should be carried out at a low temperature to maximize SD/U.

The reaction system should be operated at the highest possible temperature to maximize SD/U.


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Example 6-2

Reactant A decomposes by three simultaneous reactions to form three products, one that is desired, B, and two that are undesired, X and Y. These gas-phase reactions,

along with the appropriate rate laws, are called the Trambouze reactions.

The specific reaction rate are given at 300 K and the activation energies for reactions (1), (2), and (3) are E1=10000 kcal/mol, E2=15000 kcal/mol, and E3=20000 kcal/mol. How and under what conditions (e.g., reactor type(s), temperature, concentrations) should the reaction be carried out to maximize the selectivity of B for an entering

concentration of A of 0.4M and a volumetric flow rate of 2.0 dm3/s.


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maximum

Solution

The selectivity with respect to B is

CA*=?

Because the concentration changes down the length of a PFR, we cannot operate at this maximum. As a result, a CSTR would be used and operated at this maximum.

The corresponding selectivity at CA* is


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The net of formation of A from reactions (1), (2), and (3) is

The mole balance on a CSTR for this liquid-phase reaction (v=v0) is


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Maximize the selectivity wrt temperature

Run at as high a temperature as possible with existing equipment and watch out for other side reactions that

might occur at high temperatures.

Case 1: If

Run at low temperatures but not so low that a significant conversion is not achieved.

Case 2: If

For the activation energies given above

The selectivity for this combination of activation energies is independent of temperature!


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What is the conversion of A in the CSTR?

If greater than 72% conversion of A is required, then the CSTR operated with a reactor concentration of 0.112 mol/dm3 should be followed by a PFR because the concentration and hence selectivity will decrease continuously from CA* as we move down the PFR to

an exit concentration CAf. Hence the system

would give the highest selectivity while forming more of the desired product B, beyond

what was formed at CA* in a CSTR.


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Optimum CSTR followed by a PFR

The exit concentrations of X, Y, and B can be found from the CSTR mole balances


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If 90% conversion were required then the exit concentration would be CAf=(1-0.9)/(0.4 mol/dm3)=0.04 mol/dm3.

The PFR mole balances for this liquid-phase reaction (v=v0) are

At τ=0, the values of the entering concentrations to the PFR are the exit concentrations from the CSTR.


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maximizing SD/U

The two reactions with recycle shown in (i) and (j) can be used for highly exothermic reactions. Here the recycle stream is cooled and returned to the reactor to dilute and cool the inlet stream thereby avoiding hot spots and runaway reactions. The PFR with recycle is used for gas-phase reactions, and the CSTR is used for liquid-phase reactions.

The last two reactors, (k) and (I), are used for thermodynamically limited reactions where the equilibrium lies far to the left (reactant

side)

A+BC+D

and one of the products must be removed (e.g., C) for the reaction to continue to completion.

The membrane reactor (k) is used for thermodynamically limited gas-phase reactions, while reactive distillation (I) is used for liquid-phase reactions when one of the products has a higher volatility (e.g., C) than

the other species in the reactor.


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Example 6-3

For the parallel reactions

Consider all possible combinations of reaction orders and select the reaction scheme that will maximize SD/U.

Solution

Case I:1>2, 1>2

To maximize the ratio rD/rU, maintain the concentration of both A and B as high as possible. To do this, use

~A tubular reactor (Figure 6.3 (b))

~A batch reactor (Figure 6.3 (c))

~High pressures (if gas phase), and reduce inerts


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Case II:1>2, 1<2

To make SD/U as large as possible, we want to make the concentration of A high and

the concentration of B low. To achieve this result, use

~A semibatch reactor in which B is fed slowly into a large amount of A (Figure 6.3(d))

~A membrane reactor or a tubular reactor with side streams of B continually fed to

the reactor (Figure 6.3(f))

~A series of small CSTRs with A fed only to the first reactor and small amounts of B

fed to each reactor. In this way B is mostly consumed before the CSTR exit stream

flows into the next reactor (Figure 6.3(h))


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Case III:1<2, 1<2

To make SD/U as large as possible, the reaction should be carried out at low concentrations of A and B. To achieve this result, use

~A CSTR (Figure 6.3(a))

~A tubular reactor in which there is a large recycle ratio (Figure 6.3(i))

~A feed diluted with inerts

~Low pressure (if gas pressure)


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Case IV:1<2, 1>2

To make SD/U as large as possible, run the reaction at high concentration of B and low concentration of A. To achieve this result, use

~A semibatch reactor with A slowly fed to a large amount of B (Figure 6.3(e))

~A membrane reactor or a tubular reactor with side streams of A (Figure 6.3(g))

~A series of small CSTRs with fresh A fed to each reactor (Figure 6-3(h))


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Maximizing the Desired Product in Series Reactions

For series of consecutive reactions, the most important variable is time: space-time for a flow reactor and real time for a batch reactor. To illustrate the

important of the time factor, we consider the sequence

B is the desired product

~If the first reaction is slow and the second reaction is fast, it will be extremely

difficult to produce species B.

~If the first reaction (formation of B) is fast and the reaction to form C is slow,

a large yield of B can be achieved.

~However, if the reaction is allowed to proceed for a long time in a batch

reactor, or if the tubular flow reactor is too long, the desired product B will be

converted to the undesired product C.


Multiple reactions

Example 6-4

The oxidation of ethanol to form acetaldehyde is carried out on a catalyst of 4 wt % Cu-2 wt % Cr on Al2O3. Unfortunately, acetaldehyde is also oxidized on this catalyst to form carbon dioxide. The reaction is carried out in a threefold excess of oxygen and in dilute concentration (ca. 0.1% ethanol, 1% O2, and 98.9% N2). Consequently, the volume change with the reaction can be neglected. Determine the concentration of

acetaldehyde as a function of space-time,

The reactions are irreversible and first order in ethanol and acetaldehyde, respectively.

Solution

O2 in excess


Multiple reactions

Mole balance on A:

Mole balance on B:

integrating factor


Multiple reactions

Mole balance on C:

Optimum yield


Multiple reactions

The yield has been defined as

shown as a function of conversion in Figure E6-4.2

Another technique is often used to follow the progress for two reactions in series. The concentrations of A, B, and C are plotted as a singular point at different space time (e.g., 1’, 2’) on a triangular diagram (see Figure 6.-4). The vertices correspond to

pure A, B, and C.

For (k1/k2)>>1, a large quantity of B can be obtained.

For (k1/k2)<<1, very little B can be obtained.


Multiple reactions

Algorithm for Solution of Complex Reactions

~In complex reactions systems consisting of combination of parallel and

series reactions, the availability of software packing (ODE solvers) makes

it much easier to solve problems using moles Nj or molar flow rates Fj

rather than conversion.

~For liquid systems, concentration is usually the preferred variable used in

the mole balance equations. The resulting coupled differential mole

balance equations can be easily solved using an ODE solver.

~For gas systems, the molar flow rates are usually the preferred variable in

the mole balance equation.


Multiple reactions

Mole Balances

Table 6-1 gives the forms of the mole balance equation we shall use for complex reactions where rA and rB are the net rates of formation of A and B.


Multiple reactions

We find the net rate of reaction for each species in terms of the concentration of the reacting species in order to combine them with their respective mole

balances.


Multiple reactions

Net Rates of Reaction

Consider the following reactions


Multiple reactions

Example 6-5

Consider the following set of reactions:

Write the rate law for each species in each reaction and then write the net rates of formation of NO, O2, and N2.

Solution

Net Rates of Reaction


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Relative Rates of Reaction


Multiple reactions

Net rate NO

Net rate N2

Net rate O2


Multiple reactions

Stoichiometry: Concentration

for liquid-phase reactions, v=v0

for gas-phase reactions

T=T0, P=P0

where fn represents the functional dependence on concentration of the net rate of formation


Multiple reactions

Multiple Reactions in a PFR/PBR

We now insert rate laws written in terms of molar flow rates into the mole balances. After performing this operation for each species, we arrive at a coupled set of first-order ordinary differential equations to be solved for the molar flow rates as a

function of reactor volume.

In liquid-phase reactions, incorporating and solving for total molar flow rate is not necessary

at each step along the solution pathway because there is no volume change with reaction.

for gas-phase reactions

Combing mole balance, rate laws, and stoichiometry

T=T0, P=P0

solved simultaneously with a numerical package or by writing an ODE solver.

V=?


Multiple reactions

Example 6-6

Consider again the reaction in Example 6-5. Write the mole balances on a PFR in terms of molar flow rates for each species.

Solution

T=T0, P=P0

total molar flow rate of all the gases is


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mole balance on NO

mole balance on NH3

mole balance on H2O


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mole balance on N2

mole balance on O2

mole balance on NO2


Multiple reactions

ODE solver algorithm is shown in Table E6-6.1


Multiple reactions

Table 6-2 shows the equation for species j and reaction i that are to be combined when we have q

reactions and n species.


Multiple reactions

Example 6-7

The production of m-xylene by the hydrodealkylation of mesitylene over a Houdry Detrol

catalyst involves the following reactions:

m-xylene can also undergo hydrodealkylation to form toluene:

The second reaction is undesirable, because m-xylene sells for a higher price than toluene ($1.32/lb vs. $0.30/lb). Thus we see that there is a significant incentive to maximize the

production of m-xylene.

The hydrodealkylation of mesitylene is to be carried out isothermally at 1500R and 35 atm in a packed-bed reactor in which the feed is 66.7 mol% hydrogen and 33.3 mol% mesitylene. The volumetric feed rate is 476 ft3/h and the reactor volume (i.e., V=W/b) is

238 ft3.


Multiple reactions

The rate laws for reactions 1 and 2 are, respectively

where the subscripts are: M=mesitylene, X=m-xylene, T=toluene, Me=methane, and

H=hydrogen (H2).

At 1500R, the specific reaction rates are

The bulk density of the catalyst has been included in the specific reaction rate (i.e, k1=k1’b).

Plot the concentration of hydrogen, mesitylene, and xylene as a function of space time. Calculate the space time where the production of xylene is a maximum (i.e., opt)


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Solution

1.Mole balances:

2.Rate laws and net rates:


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3.Stoichiometry:

T=T0, P=P0


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4.Combining:

If we know CM, CH, and CX, then CMe, and CT can be calculated from the reaction

stoichiometry.

5.Parameter evaluation:


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These equations can be solved analytically.


Multiple reactions

Multiple Reactions in a CSTR

~If there is no volume change with reaction, we use concentration, Cj, as variables.

~If the reactions are gas phase and there is volume change, we use molar flow rates,

Fj, as variables.

The total molar flow for n species is


Multiple reactions

Example 6-8

For the multiple reactions and conditions described in Example 6-7, calculate the conversion of hydrogen and mesitylene along with the exiting concentrations of mesitylene, hydrogen,

and xylene in a CSTR.

Solution


Multiple reactions

The moles of hydrogen consumed in reaction 1 are equal to the moles of mesitylene

consumed. Therefore, the conversion of hydrogen in reaction 1 is

The conversion of hydrogen in reaction 2 is just the overall conversion minus the

conversion in reaction 1; that is,

The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for

=0.5 is

The overall selectivity of xylene relative to toluene is


Multiple reactions

Membrane Reactor to Improve Selectivity in Multiple Reactions

~In addition to using membrane reactors to remove a reaction product in order to

shift the equilibrium toward completion, we can use membrane reactor to

increase selectivity in multiple reactions.

~This increase can be achieved by injecting one of the reactants along the length of

the reactor.

~It is particularly effective in partial oxidation of hydrocarbons, chlorination,

ethoxylation, hydrogenation, nitration, and sulfunation reactions.

By keeping one of the reactants at a low concentration, we can enhance selectivity. By feeding a reactant through the sides of a membrane reactor, we can keep

its concentration low.


Multiple reactions

Example 6-9

The reactions

take place in the gas phase. The overall selectivities, , are to be compared for a membrane reactor (MR) and a conventional PFR. First, we use the instantaneous selectivity to determine

which species should be fed through the membrane

We see that to maximize SD/U we need to keep the concentration of A high and the concentration of B low; therefore, we feed B through the membrane. The molar flow rate of A entering the reactor is 4 mol/s and that of B entering through the membrane is 4 mol/s as

shown in Figure E6-9.1 For the PFR, B enters along with A.

The reactor volume is 50 dm3 and the entering total concentration is 0.8 mol/dm3. Plot the molar flow rates and the overall selectivity, , as a function of reactor volume for both the MR

and PFR.


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Solution

Mole Balances for both the PFR and the MR

Net Rates and Rate Laws


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Transport Law

The volumetric flow rate through the membrane is given by Darcy’s Law:

K is the membrane permeability (m/skPa) and Ps (kPa) and Pt (kPa) are the shell and tube side pressures, and At is the membrane surface area (m2).

The molar flow rate of B per unit volume of reactor is

Stoichiometry

no pressure drop

isothermal operation

Combining

Polymath

Fi vs. V

SD/U vs. V


Multiple reactions

We can easily modify the program, Table E6-9.1, for the PFR simply by setting RB equal to zero (RB=0) and the initial condition for B to be 4.0.

Notes that the enormous enhancement in selectivity the MR has over the PFR.


Multiple reactions

Example 6-10

The following gas-phase reactions take place simultaneously on a metal oxide-supported catalyst:

Writing these equations in terms of symbols yields

Determine the concentrations as a function of position (i.e., volume) in a PFR.

Addition information: Feed rate = 10 dm3/min; volume of reactor = 10 dm3; and CA0 =

CB0 = 1.0 mol/dm3, CT0 = 2.0 mol/dm3


Multiple reactions

Solution

Mole Balances

Stoichiometry

no pressure drop

concentration

relative rates

isothermal operation


Multiple reactions

Notes that there is a maximum in the concentration of NO (i.e., C) at approximately 1.5 dm3.


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Closure

~After completing this section, the student should be able to describe the

different types of multiple reactions (series, parallel, complex, and independent)

and to select a reaction system that maximizes the selectivity.

~The student should be able to write down and use the algorithm for solving

CRE problems with multiple reactions.

~The student should also be able to point out the major differences in the CRE

algorithm for the multiple reactions from that for the single reactions, and then

discuss why care must be taken when writing the rate law and stoichiometric

steps to account for the rate laws for each reaction, the relative rates, and the

net rates of reaction.

~Finally, the student should feel a sense of accomplishment by knowing they

have now reached a level they can solve realistic CRE problems with complex

kinetics.


Multiple reactions

Homework-4

Due 2007.11.26

In Int. J. Chem. Kinet., 35, 555 (2003), does the model overpredict the orthodiethyl benzene concentration? Please give a reason.


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