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

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Using Eq. 7.1.3, the needed reactor volume is

¼ tv

¼ 100:8L (f)

c. If the variation in the volumetric ﬂow rate along the reactor is not considered

(D ¼ 0), the design equation is

¼ 1  Z (g)

We solve (g) by separating the variables and integrating. The solution is

Z(t) ¼ 1 e

t

For Z

out

¼ 0.8

t ¼

0:8

1  Z

dZ ¼ 1:61 (h)

Hence, the calculated reactor volume is

¼ tv

¼ 67:18 L (i)

Note that by ignoring the effect of the volume expansion, we specify a reactor

volume that is only 68% of the required volume.

d. For a reactor with a volume of 67.1 L, t ¼ 1.61. Using the proper design

equation, (e), the outlet extent, Z

out

, is determined by

out

1 þ Z

1  Z

dZ ¼ 1:61 ( j)

The solution is Z

out

¼ f

out

¼ 0:680. Hence, if the wrongly speciﬁed reactor

volume is used, only 68% conversion is obtained.

e. To attain 80% conversion on the 67.1-L reactor, the feed ﬂow rate should be

reduced. From the design equation (e), t ¼ 2.42. Using Eq. 7.1.3 to deter-

mine the feed rate,

¼ 166:3L=min (k)

tot

)

¼ v

¼ 6:65 mol=min

Hence, when the wrongly speciﬁed reactor volume is used, the feed ﬂow

rate should be lowered by about 34% to maintain the speciﬁed conversion

level of 80%.

250 PLUG-FLOW REACTOR

Example 7.2 The second-order, gas-phase chemical reaction

2A ! B

is carried out in a 1000 L isothermal plug-ﬂow reactor. A feed stream consisting

of 80% A and 20% inert (% mole) is fed to the reactor at a molar ﬂow rate of

125 mol/min. The concentration of reactant A in the feed stream is

¼ 0:05 mol=L. The molar fraction of reactant A at the outlet of the reactor

is 0.16. We want to modify the reactor such that it provides 90% conversion of

A. What is the required additional volume?

Solution The stoichiometric coefﬁcients of the chemical reaction are

¼2 s

¼ 1 s

¼ 0 D ¼1

In this case, the reaction rate constant is not provided, and we have to determine

it from the operating data. First, we derive the design equation for this case. We

select the inlet stream to the system as the reference stream. Hence,

tot

)

¼ (F

tot

)

, y

¼ 0:8, y

¼ 0:2. The reference concentration is

¼ 0:0625 mol=L

and the volumetric ﬂow rate of the reference stream is

tot

)

¼ 2000 L=min

For a gas-phase reaction, the species concentration is given by Eq. 7.2.6, and the

rate expression is

r ¼ kC

þ s

1 þ D Z



(a)

Using Eq. 3.5.4, for a second-order reaction, the characteristic reaction time, t

,is

(b)

Substituting (a) and (b) into Eq. 7.2.2, the design equation is

0:8  2Z

1  Z



(c)

7.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 251

Separating the variables, the design equation for the reactor is

t ¼

out

1  Z

0:8  2Z



dZ (d)

where t is the dimensionless space time of the reactor, deﬁned by Eq. 7.1.3. To

determine Z

out

, we use Eq. 2.7.8 and Eq. 2.7.10 to express the molar fraction of

reactant A at the reactor outlet:

out

;

out

tot

out

0:8  2Z

out

1  Z

out

¼ 0:16 (e)

Solving (e), Z

out

¼ 0.348. Substituting this value into (d), we obtain t ¼ 2.2.

Using Eq. 7.1.3, the characteristic reaction time is

¼ 0:2278 min ( f)

Using (b), the reaction rate constant is

k ¼

¼ 70:39 L =mol min

Now we design a reactor to provide 90% conversion. Using Eq. 2.6.5,

out

¼

out

¼

0:8

2

(0:9) ¼ 0:36

Substituting in (d),

new

0:36

1  Z

0:8  2Z



dZ ¼ 2:8 (g)

The required reactor volume is

new

¼ v

t ¼ (2000 L=min)(0:2278 min)(2:8) ¼ 1275 L (h)

An additional 275 L should be added to the reactor.

Example 7.3 The elementary, gas-phase reaction

A þ B ! C

252 PLUG-FLOW REACTOR

is carried out in an isothermal plug-ﬂow reactor operated at 2 atm and 1708C. At

this temperature, the reaction rate constant is k ¼ 90 L/(mol min), and the vapor

pressure of the product, C, is 0.3 atm. The reactor is fed with two gas streams: the

ﬁrst one consists of 80% A, 10% B, 10% inert (I), and is at 2.5 atm and 1508C;

the second consists of 80% B, 20% I, and is at 3 atm and 1808C. The ﬁrst stream

is fed at a rate of 100 mol/min and the second at a rate of 120 mol/min.

Determine:

a. The conversion of reactant A when C begins to condense.

b. The reactor volume where C starts to condense.

c. The reactor volume needed for 85% conversion of A.

Solution In the ﬁrst section of the reactor, all the species are gaseous, and the

reaction is

A(g) þ B(g) ! C(g)

and its stoichiometric coefﬁcients are

¼1 s

¼ 1 D ¼1

First, we have to select a reference stream. Since the two streams are mixed at the

reactor inlet, we select a ﬁctitious reference stream that consists of the molar ﬂow

rates of the two streams, and is at 2 atm and 1708C (the reactor operating con-

ditions). Hence,

¼ 0:8F

¼ 80 mol=min

¼ 0:1F

þ 0: 8F

¼ 106 mol=min

¼ 0:1F

þ 0: 2F

¼ 34 mol=min

tot

)

¼ F

þ F

¼ 220 mol=min

The composition of the reference stream is y

¼ 0:364, y

¼ 0:482, and

¼ 0:154. Assuming ideal-gas behavior, the concentration of the reference

stream is

2 atm

(0:08206 L atm=mol K)(443 K)

¼ 0:055 mol=L

Using Eq. 7.1.14, the volumetric ﬂow rate of the reference stream is

tot

)

¼ 4000 L=min

7.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 253

a. Species C starts to condense when its partial pressure in the reactor

is 0.3 atm. At a given point in the reactor, the partial pressure of product

Cis

¼ y

P ¼

tot

Using Eqs. 2.7.8 and 2.7.10,

þ s

1 þ D Z

P ¼

1  Z

(2 atm) ¼ 0:3 atm (a)

Solving (a), Z ¼ 0.130, and using Eq. 2.6.5, the conversion is

¼

Z ¼ 0:358 (b)

b. To determine the reactor volume for Z ¼ 0.130, we use Eq. 7.2.2,

¼ r



(c)

Since the reaction is elementary, r ¼ kC

, and, using Eq. 7.2.6,

r ¼ kC

( y

 Z)( y

 Z)

(1 þ D Z)

(d)

Using Eq. 3.5.4, for a second-order reaction, the characteristic reaction

time is

¼ 0:202 min (e)

Substituting (d) and (e) into (c), the design equation reduces to

(0:364 Z)(0:482  Z)

(1  Z)

(f)

254 PLUG-FLOW REACTOR

To determine the dimensionless space time for Z ¼ 0.130, we solve (f) by

separation of variables,

t ¼

0:130

(1  Z)

(0:364  Z)(0:482  Z)

dZ ¼ 0:9306 (g)

Using Eq. 7.1.3, the volume of the reactor for Z ¼ 0.130 is

¼ tv

¼ (0:9306)(4000 L=min)(0:202 min) ¼ 751:9L

c. Any product C generated after dimensionless extent of 0.130 is reached

cannot be in the vapor phase, as shown in Figure E7.3.1. Hence, the follow-

ing reaction takes place downstream in the reactor:

A(g) þ B(g) ! C(liq)

For this reaction, D

gas

¼ 22. For Z . 0.130, a portion of product C is in the

gas phase and a portion in the liquid phase (with negligible volume). The

total molar ﬂow rate of the gas-phase is now (for Z . 0.130)

tot

)

gas

¼ (F

tot

)

[1 þ (1)0:130 þ (2)(Z  0:130)]

¼ (F

tot

)

(1:13  2Z) (h)

Using Eq. 7.2.6, for Z . 0.130, the concentrations of the two reactants are

¼ C

0:364  Z

1:13  2Z

¼ C

0:482  Z

1:13  2Z

(i)

Substituting (i) and (d) into (c), for Z . 0.130, the design equation is

(0:364 Z)(0:482  Z)

(1:13  2Z)

for Z . 0:13 ( j)

Figure E7.3.1 Zones of product C.

7.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 255

Using Eq. 2.6.5, for f

¼ 0.85,

Z ¼

¼ 0:3094

To determine the dimensionless space time of the reactor section

where product C is formed as a liquid, we integrate ( j) between 0.130 and

0.3094:

t ¼

0:3094

0:130

(1:13  2Z)

(0:364  Z)(0:482 Z)

dZ ¼ 2:52 (k)

Using Eq. 7.1.3, the volume of the reactor section where condensed C is

formed is

¼ tv

¼ 2036 L

The total volume of the reactor is V

¼ 751.9 þ 2036 ¼ 2788 L.

d. The reactor operation is described by two design equations: (f) for

0  Z  0.13 and (k) for Z . 0.13. Figure E7.3.2 shows the reaction operat-

ing curve. Using Eq. 2.7.8, the species curves are

(t)

tot

)

¼ y

 Z(t)

(t)

tot

)

¼ y

 Z(t)

(t)

tot

)

¼ Z(t)

Figure E7.3.3 shows the species curves.

Figure E7.3.2 Reaction operating curve.

256 PLUG-FLOW REACTOR

Example 7.4 Design a packed-bed reactor for the gas-phase, heterogeneous

catalytic cracking reaction

A ! B þC

The reactor should produce 20 metric ton per day of product B at 75% conver-

sion of A. A process stream at 3878C and 2 atm, consisting of 90% A and 10%

inert (I), is available in the plant. Based on the kinetic data below:

a. Determine the feed ﬂow rate to the reactor.

b. Derive the design equation and plot the reaction and species curves.

c. Determine the volume of the reactor.

d. Determine the mass of the catalyst in the bed.

Data: Molecular mass of product B is 28 g/mol

Bulk density of the catalyst bed is 1.30 kg/L

The mass-based reaction rate expression is

1 þ KC

mol (g catalyst)

1

where k

¼ 0.192 cm

/g-cat s, and K ¼ 60 L/mol.

Solution This example sho ws the use of the design equation when the ra te

expression is given on the basis of mass, and it is not a po wer function of the species

concentrations. The s toichiometric coefﬁcients of the chemical reaction ar e

¼1 s

¼ 1 s

¼ 0 D ¼ 1

Figure E7.3.3 Species operating curves.

7.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 257

W e select the feed str eam as the reference s tream; hence, y

¼ 0:9, y

¼ 0:1, and

¼ y

¼ 0. The reference concentration is

2atm

(0:08206 L atm/mol K)(660 K)

¼ 0:0369 mol=L

a. To determine the feed ﬂow rate, we use Eq. 2.7.8 to express the production

rate of product B in terms of (F

tot

)

. Selecting the inlet stream as the reference

stream,

out

¼ (F

tot

)

( y

þ s

out

) (a)

Using Eq. 2.6.5,

out

¼

out

¼ 0:675

and the molar ﬂow rate of product B at the reactor outlet is

out

¼ 8:267 mol=s

Hence, using (a), the molar ﬂow rate of the reference stream is

tot

)

out

¼ 12:25 mol=s

The volumetric ﬂow rate of the reference stream is

tot

)

¼ 332:0L=s (b)

b. Using Eq. 3.2.3, the volume-based rate expression is

r ¼ r

bed

1 þ KC

(c)

where k ¼ r

bed

¼ 0.250 s

is the volume-based reaction rate constant. We

use Eq. 7.2.6 to express the species concentrations:

¼ C

0:9  Z

1 þ Z

(d)

258 PLUG-FLOW REACTOR

Substituting (c) and (d) in Eq. 7.2.2, the design equation becomes

¼ kC

0:9  Z

1 þ Z

1 þ KC

0:9  Z

1 þ Z



(e)

We deﬁne the characteristic reaction time by

¼ 4s (f)

and the design equation reduces to

0:9  Z

1 þ Z þ KC

(0:9 Z)

(g)

Substituting the numerical values of K and C

, (g) becomes

0:9  Z

2:99  1:21Z

(h)

We solve (h) numerically subject to the initial condition that at t ¼ 0, Z ¼ 0.

The reaction curve is shown in Figure E7.4.1. Using Eq. 2.7.8, the species

curves are

(t)

tot

)

¼ y

 Z(t)

(t)

tot

)

(t)

tot

)

¼ Z(t)

Figure E7.4.2 shows the species curves.

Figure E7.4.1 Reaction operating curve.

7.2 ISOTHERMAL OPERATIONS WITH SINGLE REACTIONS 259