Mann U. Principles of Chemical Reactor Analysis and Design: New Tools for Industrial Chemical Reactor Operations

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Solving (f) for Z

out

¼ 0.130, t ¼ 1.195. Using Eq. 8.1.3 and (e), the reactor

volume is

¼ v

t ¼ 965:2L

c. If the reactor volume is larger than 965.2 L, the extent is larger than 0.130,

and a portion of product C is formed by reaction (a) and a portion by the fol-

lowing reaction:

A(g) þ B(g) ! C(liquid) (g)

Assuming that the volume of product C in the liquid phase is negligible, for

this reaction, D

gas

¼ 22. Using Eq. 2.7.10, the total gas-phase molar ﬂow

rate at the reactor outlet is now

tot

)

gas

¼ (F

tot

)

[1 þ (1)0:130 þ (2)(Z  0:130)] ¼ 1:13 2Z (h)

Using Eq. 8.1.11, the concentration of reactant A is now

¼ C

þ s

1:13  2Z

(i)

Substituting (i) into the rate expression (d) and the latter into (c), for

Z . 0.130 (or t . 1.195), the design equation is

out

¼ t

( y

 Z

out

)( y

 Z

out

)

(1:13 2Z)

t . 1:195 ( j)

Using Eq. 2.6.5, for f

¼ 0.85,

out

¼

out

¼

0:364

1



0:85 ¼ 0:3094 (k)

Substituting Z

out

¼ 0.3094 into ( j), we obtain

t ¼

out

(1:13 2Z)

(0:364  Z

out

)(0:482  Z

out

)

¼ 8:58 (l)

Figure E8.2.1 Reaction operating curve.

330 CONTINUOUS STIRRED-TANK REACTOR

Using Eq. 8.1.3 and (e), the volume of the reactor is

¼ v

t ¼ 6, 932 L

d. The design of a CSTR is described by two design equations: (f) for

0 , Z

out

, 0.130 and ( j) for Z

out

. 0.130. Figure E8.2.1 shows the reaction

curves, and Figure E8.2.2 shows the species curves.

Example 8.3 A biological waste, A, is decomposed in aqueous solution by an

enzymatic reaction:

A ! B þC

The rate expression of the reaction is

r ¼

þ C

An aqueous solution with C

¼ 2 mol/L is fed into a CSTR at a rate of 200 L/

min. For the enzyme type and concentration, k ¼ 0.1 mol/L min and

¼ 4 mol/L.

a. Derive and plot the reaction and species operating curves.

b. Determine the needed reactor volume to achieve 80% conversion.

Solution The stoichiometric coefﬁcients of the reaction are

¼1 s

¼ 1 s

¼ 1 D ¼ 1

We select the inlet stream as the reference stream; hence, Z

¼ 0. Since reactant

A is the only species fed to the reactor, C

¼ C

, y

¼ 1, y

¼ 1, and y

¼ 1.

Figure E8.2.2 Species operating curves.

8.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 331

Using Eq. 8.2.2, the design equation is

out

¼ tr

out



(a)

Using Eq. 8.2.7,

¼ C

(1  Z) (b)

Substituting (b) in the rate expression and the latter in (a), the design equation is

out

¼ k

1  Z

out

) þ (1  Z

out

)



(c)

Applying the procedure described in Section 3.5, the characteristic reaction

time is

¼ 20 min (d)

a. Substituting (d) into (c), for K

¼ 2, the design equation becomes

t ¼

out

(3  Z

out

)

1  Z

out

(e)

Figure E8.3.1 shows the reaction curve. Using Eq. 2.7.8, the species curves are

out

tot

)

¼ 1  Z

out

tot

)

out

tot

)

¼ Z

out

Figure E8.3.2 shows the species curves.

Figure E8.3.1 Reaction operating curve.

332 CONTINUOUS STIRRED-TANK REACTOR

b. Using Eq. 2.6.5, for f

out

¼ 0:8

out

¼

(0)

out

¼

1

(0:80) ¼ 0:80

From the reaction curve (or tabulated data), Z

out

¼ 0.80 is reached at

t ¼ 8.8. Using Eq. 8.1.3 and (d), the needed reactor volume is

¼ 35,200 L

8.2.2 Determination of the Reaction Rate Expression

Since the design equation contains the reaction rate at the outlet conditions, r

out

,we

can easily determine the rate expression from data obtained on a CSTR. Selecting

the inlet stream as the reference stream, Z

¼ 0, we write Eq. 8.2.2 as

out



out

(8:2:12)

Since t

is not known, we use Eq. 8.1.3 to express the design equation in terms of

the reactor volume:

out

tot

)

out

(8:2:13)

Hence, for known (F

tot

)

and V

and for experimentally determined Z

out

, we can

calculate r

out

.Fornth-order reactions,

r ¼ kC

Substituting in Eq. 8.2.13,

out

tot

)

out

Figure E8.3.2 Species operating curves.

8.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 333

To determine the reaction order, we take the logarithm of both sides and obtain

n ln ( C

out

) ¼ ln

out



(8:2:14)

Modifying the term on the right-hand side, Eq. 8.2.14 reduces to

ln v

out

ðÞ¼n ln (C

out

) þ ln



(8:2:15)

Thus, by operating a CSTR isothermally and measuring C

out

for different values of

, we can plot ln(v

out

) versus ln(C

out

) and determine the reaction order from the

slope of the line. Once the order is known, we can determine the value of the reac-

tion rate constant from the design equation. We repeat this procedure at different

reactor temperatures to determine the activation energy.

To apply Eq. 8.2.15, we have to relate Z

out

to a measurable quantity. Usually, the

species concentrations are measured at the reactor outlet and we use either Eq. 8.2.7

or Eq. 8.2.8. This is illustrated in Example 8.4.

Example 8.4 The gas-phase reaction

2A ! R

is investigated on a CSTR. A stream of species A at 3 atm and 308C (120 mmol/

L) is fed into a 1-L CSTR at different ﬂow rates, and the exit concentration of

reactant A is measured. From the data below, determine the order of the reaction

and the value of the rate constant, k.

(L=min) 0:035 0:18 0:45 1:7

out

(mmol=L) 30 60 80 105

Solution The stoichiometric coefﬁcients of the chemical reaction are

¼2 s

¼ 1 D ¼1

The rate expression is

r ¼ kC

(a)

and we want to determine a and k. We select the inlet stream as the reference

stream; hence, Z

¼ 0, y

¼ 1, and y

¼ 0. To relate the extent to the exit con-

centration, we use Eq. 8.2.8:

out

¼ C

1  2 Z

out

1 þ D Z

out

(b)

334 CONTINUOUS STIRRED-TANK REACTOR

Substituting D ¼ 21 into (b), and rearranging,

out

 C

out

 C

out

(c)

We calculate Z

out

and then (v

out

) for each run. The table below shows the cal-

culated values.

(L=min) 0:035 0:180 0:430 1:65

out

(mmol=L) 30 60 80 105

out

0:429 0:333 0:250 0: 111

out

(L=min) 0:015 0:060 0:108 0: 183

Using Eq. 8.2.18,

ln(v

out

) ¼ a ln C

out

þ ln



(d)

We plot ln(v

out

) versus ln (C

out

), shown in Figure E8.4.1, and the slope is a.

From the ﬁgure,

a ¼

ln(0:17)  ln(0:0155)

ln(100)  ln(30)

¼ 2:06  2

Now that we know that a ¼ 2, the characteristic reaction time is

(e)

Using Eq. 8.2.3 and (e), we calculate the dimensionless space time for each run:

t ¼ Z

out

1  Z

out

1  2Z

out



(f)

Figure E8.4.1 Determination of reaction order.

8.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 335

Using Eq. 8.1.3 and (e), (f) reduces to

out



(g)

The average value of k for the four runs is 0.002 L/mmol min.

8.2.3 Cascade of CSTRs Connected in Series

Consider several CSTRs connected such that the efﬂuent from one reactor is fed

into the next reactor, as shown schematically in Figure 8.4. We select the inlet

stream to the cascade as the reference stream and write Eq. 8.2.3, for each reactor.

We can present the design equation graphically by plotting (r

/r) versus Z,as

shown schematically in Figure 8.5. Here, the dimensionless space time of each

reactor is represented by a rectangle. Note that for a given outlet extent, the total

volume of the cascade is smaller than the volume of a single CSTR. Also note

that when a cascade of numerous small CSTRs is used, the rectangular areas

approach the area under the curve. Hence, the total volume of the cascade con-

verges to the volume of a plug-ﬂow reactor.

Consider a liquid-phase, ﬁrst-order reaction of the form A ! P þ R in an iso-

thermal cascade of CSTRs where only reactant A is fed to the system. Taking the

inlet stream to the cascade as the reference stream and since only reactant A is fed,

¼ 1. Using Eq. 8.2.3, the design equation for the nth CSTR is

out

 Z

1  Z

out

(8:2:16)

where t

is the dimensionless space time of the nth reactor in the cascade,

¼ V

=(v

). For the ﬁrst reactor in the cascade,

Figure 8.4 Cascade of CSTRs.

336 CONTINUOUS STIRRED-TANK REACTOR

1  Z

or Z

1  t

(8:2:17)

For the second reactor in the cascade,

 Z

1  Z

or Z

þ t

(1 þ t

)

(1  t

)(1  t

)

(8:2:18)

Similarly, for the nth reactor in the cascade,

þ t

(1 þ t

) þþt

(1 þ t

) (1 þ t

n1

)

(1  t

)(1  t

) (1  t

)

(8:2:19)

For the special case that each reactor has the same volume, t

¼ t

...

¼ t

tot

/n, Eq. 8.2.19 reduces to

¼ 1 

[1 þ (t

tot

=n)]

(8:2:20)

In practice, the cascade usually consists of either equal-size reactors or a set of

speciﬁed numbers of CSTRs in which the volume of each is selected such that total

volume of the cascade, for a given outlet conversion, is the smallest. A cascade of

equal-size CSTRs is represented in Figure 8.4 by rectangles whose areas are the

same, and a cascade with the total smallest volume is represented by a given

number of rectangles whose combined area is the smallest. Below, we consider

the performance of a cascade of CSTRs when the reaction rate is given.

Figure 8.5 Graphical presentation of the design equation for a cascade of CSTRs.

8.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 337

Example 8.5 The ﬁrst-order, gas-phase reaction

A ! B þC

is carried out in a cascade of CSTRs. A stream of reactant A is fed into the cas-

cade at a rate of 10 mol/min. The feed concentration is C

¼ 0:04 mol=L, and,

at the reactor temperature, k ¼ 0.1 s

. For outlet conversion of 80%, determine:

a. The volume of each reactor in a series of two equal-size CSTRs.

b. The volume of each reactor in a series of two optimized CSTRs.

c. The volume of each reactor in a series of three equal-size CSTRs.

d. The volume of each reactor in a series of three optimized CSTRs.

e. Compare the results in (a) and (b) to the values obtained in Example 7.1 and

Example 8.1.

Solution The stoichiometric coefﬁcients of the reaction are

¼1 s

¼ 1 s

¼ 1 D ¼ 1

We select the inlet stream to the cascade as the reference stream, and since only

reactant A is fed, C

¼ C

, y

¼ 1, and y

¼ 0. The volumetric ﬂow rate of

the reference stream is

tot

)

¼ 250 L= min

Using Eq. 8.2.8 to express the species concentrations, the reaction rate is

out

¼ kC

1  Z

out

1 þ D Z

out

(a)

Using Eq. 3.5.4, the characteristic reaction time is

¼ 10 s ¼ 0:167 min (b)

Substituting (a) and (b) into Eq. 8.2.3, the design equation of each reactor is

t ¼

out

 Z

)(1 þ Z

out

)

1  Z

out

(c)

In a cascade of two CSTRs, for the ﬁrst reactor,

(1 þ Z

)

1  Z

(d)

For the second reactor, Z

¼ Z

, Z

out

¼ Z

, and (c) reduces to

 Z

)(1 þ Z

)

1  Z

(e)

338 CONTINUOUS STIRRED-TANK REACTOR

a. For a cascade of equal-size reactors, t

¼ t

, and we combine (d) and (e):

)(1 þ Z

)

1  Z

 Z

)(1 þ Z

)

1  Z

(f)

Substituting Z

¼ 0.8, we solve (f) and obtain Z

¼ 0.569. Now that we

know Z

and Z

, we can use either (d) or (e) to determine t and obtain

¼ t

¼ 2.075. We use Eq. 8.1.3 and (b) to calculate the volume of each

CSTR,

¼ tv

¼ 86:46 L (g)

The total volume of the cascade is V

þ V

¼ 172:92 L:

b. To determine Z

for an optimized cascade of two CSTRs, we take the deriva-

tive of t

þ t

with respect to Z

and equate it to zero:

d(t

þ t

)

¼ 0 (h)

Substituting (d) and (e), (h) reduces to

(1 þ 2Z

)(1  Z

) þ Z

(1 þ Z

) 

1 þ Z

1  Z



(1  Z

)

¼ 0 (i)

Substituting Z

¼ 0.8, we solve (i) and obtain Z

¼ 0.5528. Now that we

know the values of Z

and Z

, we use (d) and (e) to calculate the t of each

reactor. We ﬁnd that t

¼ 1.919 and t

¼ 2.225. Using Eq. 8.3.1 and (b),

the volume of the two reactors are V

¼ 79:96 L and V

¼ 92:71 L, and

the total volume of the cascade is 172.67 L.

c. In a cascade of three equal-size CSTRs, we write the design equation for each

reactor. For the ﬁrst reactor,

(1 þ Z

)

1  Z

(j)

For the second reactor,

 Z

)(1 þ Z

)

1  Z

(k)

For the third reactor,

 Z

)(1 þ Z

)

1  Z

(l)

For a cascade of equal-size reactors, t

¼ t

and t

¼ t

. We combine ( j),

(k), and (l) and obtain two equations with two unknowns, Z

and Z

(1 þ Z

)

1  Z

 Z

)(1 þ Z

)

1  Z

(m)

8.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 339