Batch Operation

1) Analytically (Ap- 2) Graphically (Ch. 2)

4.2 Scale-up of Liquid-Phase Batch

4.2.1 Batch Operation

Sec. 4.2 Scale-Llp of Liquid-Phase Batch Reactor Data to the Design of a CSTR 129

We can solve the equa- tions in the combine step either

1) Analytically (Ap-

Mole balance

Rate law

Second-order, isothermal, liquid-phase batch reaction

130 Isothermal Reactor Design Chap. 4

volume remains constant, we also have V = Vo. Consequently, for constant-vol- ume (V = V,) (e.g., closed metal vessels) batch reactors the mole balance

can be written in terms of concentration.

Generally, when analyzing laboratory experiments it is best to process the data in terns of the measured variable. Since concentration is the measured variable for most liquid-phase reactions, the general mole balance equation applied to reactions in which there is no volume change becomes

We consider the reaction

which is irreversible and second order in A. The rate at which A is being con- sumed i s given by the rate law

- r A = kCi (4-3)

We combine the rate law and the mole balance to obtain

Initially, CA = CAO a t t = 0. If the reaction is carried out isothermally, we can integrate this equation to obtain the reactant concentration at any time t:

cA (4-5)

Sec. 4.2 Scale-up of Liquid-Phase Batch Reactor Data to the Design of a CSTR 131 This time is the time t needed to reduce the reactant concentration in a batch reactor from an iinitial value C,, to some specified value CAS

The total cycle time in any batch operation is considerably longer than the reaction time, tR, as one must account for the time necessary to fill

(9)

and empty (t,) the reactor together with the time necessary to clean the reactor between batches, t,. In sbme cases the reaction time calculated from Equation (4-5) may be only a small fraction of the total cycle time, ti.

t, = ff

+

t,

+

^{fc 4- tR}

Typical cycle times for a batch polymerization process are shown in Table 4- 1.

Batch polymerization reaction times may vary between 5 and 60 h. Clearly, decreasing the reaction time with a 6 0 4 reaction is a critical problem. As the reaction time is reduced, it becomes important to use large lines and pumps to achieve rapid transfers and to utilize efficient sequencing to minimize the cycle time.

TABLE 4-1. TYPICAL CYCLE TIMES FOR A BATCH

POLYMERIZATION PROCESS

-

Activity Time (h)

1. Charge feed to the reactor and agitate, 2. Heat to reaction temperature, f, 3 Carry out reaction, rR

4. Empty and clean reactor, r,

1.5-3.0 1.c2.0 (varies) 0 5-1.0 3.0-6 0 Batch operation

times

Total time excluding reaction

-

It is important to have a grasp of the order of magnitude of batch reaction times, t R , in Table 4-1 to achieve a given conversion, say 90%, for the different values of the specific reaction rate, k. We can obtain these estimates by constd- ering the irreversjble reaction

A - - + B

carried out in a constant-volume batch reactor for a first- and a second-order reaction. We start with a mole balance and then follow our algorithm as shown in Table 4-2.

TABLE 4-2. ALGORITHM TO ESTIMATE REACTION TIMES

- Mole balance

Rate law Firs t-order Second-order

-rA = k C i - r A = kCA

c,

⁼

5

^{= c,, (1 - X )}

Stoichiometry (V = Yo) VO

Combine = k ( 1 - X ) dt

dX -& ⁼ kC*,(1 -

t = X Integrate t = j ; l n - 1 1

1 - x kCAd1 - X )

132 Isothermal Reactor Design Chap. 4

For first-order reactions the reaction time to reach 90% conversion (Le.,

X

= 0.9) in a constant-volume batch reactor scales as

If k = 10-4 S-1,

tR = 2'3 = 23,000 s = 6.4 h

The time necessary to achieve 90% conversion in a batch reactor for an irre- versible first-order reaction in which the specific reaction rate is ^s-l is 6.4 h.

10-4 S - 1

For second-order reactions, we have

tR = =9OOOs = 2.5h 10-3 S-1

Table 4-3 gives the order of magnitude of time to achieve 90% conversion for first- and second-order irreversible batch reactions.

TABLE 4-3. BATCH REACTION TIMES First-Order Second-Order Reaction 7ime

k (s-l) kC,o (s-') tR Estimating Reaction

Times

10-4 10-3 Hours

Minutes

10-2 10-1

1 10 Seconds

loo0 10,Ooo Milliseconds

I

^{Example 4-1} Determining k from Batch Data

It is desired to design a CSTR to produce 200 million pounds of e,.ylene glycol per year by hydrolyzing ethylene oxide. However, before the design can be carried out, it is necessary to perform and analyze a batch reactor experiment to determine the specific reaction rate constant. Since the reaction will' be carried out isothermally, the specific reaction rate will need to be determined only at the reaction temperature of the CSTR. At high temperatures there is a significant by-product formation, while at temperatures below 40°C the reaction does not proceed at a significant rate; con- sequently, a temperature of 55°C has been chosen. Since the water is usually present in excess, its concentration may be considered constant during the course of the reaction. The reaction is first-order in ethylene oxide.

CH2-OH

/"\ I

CH,-CH,

+

H 2 0 ^H2s04 > CH,-OH

catalyst

A + B C

Check IO types of homework problems on the CD-ROM for more solved examples usidg this algorithm.

Sec. 4.2 Scale-up of Liquid-Phase Batch Reactor Data to the Design of a CSTR 133

In the laboratory experiment, 500 rnL of a 2 M solution (2 kmol/rn3) olf eth- ylene oxide in water was mixed with 500 mL of water containing 0.9 wt % sulfuric acid, which is a catalyst. The temperature was maintained at 55°C. The concentra- tion of ethylene glycol was recorded as a function of time (Table E4-1.1). From these data, determine the specific reaction rate at 55°C.

TABLE E4- I . 1. CONCENTRATION-TIME DATA

Concentratiori of Ethylene Time (min) Glycol ( k r n ~ H m ~ ) ~

0.0 0.000

0.5 0.145

1

.o

^0.270

1.5 0.376

2.0 0.467

3.0 0.610

4.0 0.7 15

6.0 0.848

10.0 0.957

“ I kmol/rn7 = 1 mol/dm3 = 1 mol/L.

In this example we use the problem-solving algorithm (A through G) that is given in the CD-ROM and on the web “http://www.engin.umich.edu/-problernsolvmg”.

You may wish 10 follow this algorithm in solving the other examples in this chapter and the problems given at the end of the chapter. However, to conserve space it will not be repeated for other example problems.

A. Problem statement. Determine the specific reaction rate, kA.

B. Sketch: rt

[

^A,B,C

-

Batch

C. fdenfib:

C 1. Relevant theories

Rate law: - r A = k,CA d iVA

d t Mole balance:

-

⁼ r,V

134

Following the algorithm

Rate Law

Isothermal Reactor Design Chap. 4

C2. Variables

Dependent: concentrations Independent: time

Knowns: concentratlon of ethylene glycol as a function of time Unknowns:

1. Concentration of ethylene oxide as a function of time 2. Specific reaction rate

3. Reactor volume C3. Knowns and unknowns

C4. Inputs and outputs: reactant fed all at once to a batch reactor

C5. Missing information: None; does not appear that other sources need to be sought.

D. Assumptions and approximations:

Assumptions 1. Well mixed

2. All reactants enter at the same time 3. No side reactions

4. Negligible filling and emptying time

5. Isothermal operation I .*""

Approximations

1. Water in excess so that its concentration is essentially constant.

E. Spec$cation. The problem is neither overspecified nor underspecified.

F. Related material. This problem uses the mole balances developed in Chapter 1 for a batch reactor and the stoichiomeq and rate laws developed in Chapter 3.

G . Use an algorithm. For an isothermal reaction, use the chemical reaction engi-

t

neering algorithm shown in Figure 4-2.

Solution

1. A mole balance on a batch reactor that is well mixed is 1 dNA -

- - - - r A V dt 2. The rate law is

- r A = kC,

(E4- 1.1)

(E4- 1.2) Since water is present in such excess, the coccentration of water at any time t is virtually the same as the initial concentration and the rate law is indepen- dent of the concentration of H20. (CB E CBO.)

3. Stoichiometry. Liquid phase, no volume change, V = Vo (Table E4-1.2):

Sec. 4.2 Scale-up of Liquid-Phase Batch Reactor Data to the Design of a CSTR 135

Stoichiometric table for constant volume

Combining mole balance, rate law, and stoichiometry

t

cc

kfl

t

NA N A

c

_A^{- - = -}

-

v v,

4. Combining the rate law and the mole balance, we have

(E4-1.3) For isothermal operation we can integrate this equation,

using the initial condition that when t = 0, then ^CA = GAo. The initial con- centration of A after mixing the two volumes together is 1.0 kmol/m3 (1 mom).

.5. Integrating yields

In ⁼ kt

L A

(E4-1.4) The concentration of ethylene oxide at any time t is

The concentration of ethylene glycol at any time ^t can be obtained from the reaction stoichiometry:

A + B ^__$ C

N C = N A O X = NAo- NA For liquid-phase reactions V = V,,

Evaluating the specific reaction rate from batch reactor concentration-time data

136 Isothermal Reactor Design Chap. 4

1

Rearranging and taking the logarithm of both sides yields

(E4- 1.73

Ve see that a plot of ln[(CAo - c C ) / c A o ] as a function o f t will be a straight line with slope - k. Calculating the quantity (CAo - Cc)/CAo (Table E4-1.3) and then plot-

'rABLE FA-1.3

O.OO0 1.000

0.0 0.145 0.855

0.5 1

.o

0.270 0.730

1.5 0.376 0.624

0.467 0.533

2.0

0.610 0.390

3 .O

0.715 0.285

4.0

0.848 0.152

10.0 6.0 0.957 0.043

ting (CAo - Cc)/CAo versus t on semilogarithmic paper is shown in Figure E4-1.1.

The slope of this plot is also equal to - k . Using the decade method (see Appendix D) between (CAo - Cc)/CAo = 0.6 (t ;= 1.55 min) and (CA0 - C C ) / ~ A O = 0.06 ( t = 8.95 min) to evaluate the slope

0.02

1

Mole balance

Sec. 4.2 Scale-Up of Liquid-Phase Batch Reactor Data to the Design of a CSTR 137

2‘3 = 0.311 min-I In10 -

k = _ _ _

1% - t , 8.95 - I .55 ^(E4-1.8)

the rate law becomes

(E% 1.9) Thiis rate law can now be used in the design of an industrial CSTR.

4.2.2 Design of CSTRs

In Chapter 2 we derived the following design equation for a CSTR:

(2- 13) which gives the volume V necessary to achieve a conversion X. When the vol- umetrrc flow rate does not change with reaction, (Le., u = u,) we can write

or in terms of the space time,

For a first-order irreversible reaction, the rate law is

Rate law -rA = kCA

We can combine the rate law and mole balance to give Combine

Solving for the effluent concentration of A, CA, we obtain

(4-6)

(4-7)

(4-8) For the case we are considering, there is no volume change during the course of the reaction, so we can use Equation (3-29),

Relationship between space time and conversion ^for a

to Combine with Equation (4-8) to give first-order liquid-

phase reaction

x =

- tk

1 + t k (4-9)

1 38 Isothermal Rqactor Design Chap. 4

For a first-order reactioq the product zk is often referred to as the reaction The Damkohler is a dimensionless number that can give us a quick esti- mate of the degree of conversion that can be achieved in continuous-flow reac- tors. The Damkohler number is the ratio of the rate of reaction of A to the rate of convective transport of A at the entrance to the reactor. For first- and sec- ond-order irreversible reactions the Damkijhler numbers are

Da =

-

Damkohler number.

FAO

and

respectively. It is important to know what values of the Dardcohler number, Da, give high and low conversion in continuous-flow reactors. A value of Da = 0.1 or less will usually give less than 10% conversion and a value of Da = 10.0 or greater will usually give greater than 90% conversion.

0.1 <Da< 10

CSTRs in Series. A first-order reaction with no volume change (v = u , ) is to be carried out in two CSTRs placed in series (Figure 4-3). The effluent con-

'

^cA2

C A I

CAO X I

X 2

Figure 4-3 Two CSTRs in series.

centration of A from reactor 1 is

From a mole balance on reactor

2,

- FA2 - - vO(cAl -

vz

⁼

-'A2 k 2 C A 2

Solving for CA2, the concentration exiting the second reactor, we get

Sec. 4.2 Scale-up of Liquid-Phase Batch Reactor Data to the Design of a CSTR 139

If instead of two GSTRs in series we had n equal-sized CSTRs connected in series (7, = z2 = = z, = z ) operating at the s a n e temperature ( k , == k , =

...

^{z z} k, = k ) , the concentration leaving the last reactor would be

The conversion for these n tank reactors in series would be Conversion as a

function of the number of tanks in series

CSTRs in series

I

(4-10)

(4-1 1) A plot of the conversion as a function of the number of reactors in series for a first-order reaction is shown in Figure 4-4 for various values of the Damkohler

0 1 2 3 4 5 6 7 8 9 1 0 1 1 I 2 1 3

Number of tonks, n

Figure 4-4 Conversion as a function of the number of tanks in series for different Damkohler numbers for a first-order reaction.

number zk. Observe from Figure 4-4 that when the product of the space time and the specific reaction rate is relatively large, say, Da ² 1, approximately 90% conversion is achieved in two or three reactors; thus the cost of adding subsequent reactors might not be justified. When the product zk is small, Da

-

^(3.1, the conyersion continues to increase significantly with each reactor added.

The rate of disappearance of A in the nth reactor is

c.40

- r A n = kCAn = k

-

( 1

+

^zk)"

140 Isothermal Reactor ^Design Chap. 4

CSTRs in Parallel. We now consider the case in which equal-sized reactors are placed in parallel rather than in series, and the feed is distributed equally among each of the reactors (Figure 4-5). A balance on any reactor, say i, gives

Figure 4-5 CSTRs in parallel.

the individual reactor volume

(4-12)

Since the reactors are of equal size, operate at the same temperature, and have identical feed rates, the conversion will be the same for each reactor:

x,

⁼

x

^{2 --}

...

⁼

x,

=

x

as will be the rate of reaction in each reactor

The volume of each individual reactor,

Vi,

is related to the total volume, V; of all the reactors by the equation

V Vi = - n

A similar relationship exists for the total molar flow rate:

Sec:. 4.2 Scale-up of Liquid-Phase Batch Reactor Data to the Design of a CSTR 141

Substituting these values into Equation (4-12) yields

Conversion for tanks in parallel

or

(4- 13) This result shows that the conversion achieved in any one of the reactors in parallel is identical to what would be achieved if the reactant were fed in one stream to one large reactor of volume V!

A Second-Order Reaction in a

CSTR.

For a second-order liquid-phase reaction being carried out in a CSTR, the combination of the rate law and the design equation yields

For constant density u = t i o , FAOX = uo(CAo - C,) , then

Using our definition of conversion, we have

We solve Equation (4-15) for the conversion X

Conversion for a second-order liquid-phase reaction in a CSTR

-

-

(1 ^3- 2Da) -

../-

2Da

(4-14)

(4- 15)

(4- 16)

The minus sign must be chosen in the quadratic equation because X can- not be greater than 1. Conversion is plotted as a function of the Damkcjhler parameter, zkC,,,, in Figure 4-6. Observe from this figure that at high conver-

142 Isothermal Reacto- Design Chap. 4

I

.o

T ":"I

0.4

I I I 1 I I

1 I

I I

I I

I I

I I

I I

I I I

1

I I

I 1 I I I I I I P I 1 I I I J

0.1 0.2 0.4 0.6 1.0 2 4 6 IO 20 40 60

r k C A O

Figure 4-6 Conversion as a function of the Damkohler number (.rkCA0) for a second order reaction in a CSTR.

sions (say 70%) a 10-fold increase in the reactor volume (or increase in the specific reaction rate by raising the temperature) will increase the conversion only to 85%. This is a consequence of the fact that the CSTR operates under the condition of the lowest value of the reactant concentration (i.e., the-exit

oncentration), and consequently the smallest value of the rate of reaction.

Uses and economics

I I .b- 1

Example 4-2 Producing 200 Million Pounds per Year in a CSTR

Close to 5.2 billion pounds of ethylene glycol were produced in 1995, which ranked it the twenty-sixth most produced chemical in the nation that year on a total pound basis. About one-half of the ethylene glycol is used for antifreeze while the other half is used in the manufacture of polyesters. In the polyester category, 88% was used for fibers and 12% for the manufacture of bottles and films. The 1997 selling price for ethylene glycol was $0.38 per pound.

It is desired to produce 200 million pounds per year of ethylene glycol. The reactor is to be operated isothermally. A 1 lb mol/ft3 solution of ethylene oxide in water is fed to the reactor together with an equal volumetric solution of water con- taining 0.9 wt % of the catalyst &SO4. If 80% conversion is to be achieved, deter- mine the necessary reactor volume. How many 800-gal reactors would be required if they are arranged in parallel? What is the corresponding conversion? How many 800-gal reactors would be required if they are arranged in series? What is the corre- sponding conversion? The specific reaction rate constant is 0.311 min-', as deter- mined in Example 4- 1 ,

Suludiun

Assumption: Ethylene glycol is the only reaction product formed.

CH,-OH CH,-CH,

/"\ +

H,O H2so' > CH,-OH

I

catalyst C

A + B

Sec. 4.2 Scale-up of Liquid-Phase Batch Reactor Data to the Design of a CSTR 143 The specified production rate in lb mo!/min is

Ib 1 yr 1 day 1 h llbmol ^- 6.137 Ib

--

^mol

yr 365 days 24h 60mm 621b min

F c = 2 x l o * - x - x - - x---,x-- From the reaction stoichiometry

FC = FAOX

we find the required molar flow rate of ethylene oxide to be

We can now calculate the reactor volume using the following equations:

Figure E4-2.1 Single CSTR.