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

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Substituting (f) through (i) and (e) into (b), (c), and (d), the design equations

become

(0:5  4Z

 4Z

)(0:5  5Z

 3Z

 Z

)

(1 þ Z

þ Z

 Z

)

 1:5

4=3

(0:5  4Z

 4Z

)

2=3

(4Z

 2Z

)

(1 þ Z

þ Z

 Z

)

5=3

(j)



(0:5  4Z

 4Z

)(0:5  5Z

 3Z

 Z

)

(1 þ Z

þ Z

 Z

)

þ 2:5

4=3

(0:5  4Z

 4Z

)

2=3

(4Z

 2Z

)

(1 þ Z

þ Z

 Z

)

5=3

(k)



(0:5  5Z

 3Z

 Z

)(4Z

 2Z

)

(1 þ Z

þ Z

 Z

)

(l)

For the given data, the parameters of the dimensionless design equations are



¼ 0:05



¼ 2

4=3



¼ 0:1

We substitute these values into ( j), (k), and (l), and solve them numerically

for Z

, Z

, and Z

, subject to the initial conditions that at t ¼ 0,

¼ Z

¼ 0. Figure E7.9.1 shows the reaction curves. Now that we

have the reaction curves, we use Eq. 2.7.8 to obtain the species curves:

tot

)

¼ 0:5 4Z

 4Z

(m)

tot

)

¼ 0:5 5Z

 3Z

 Z

(n)

Figure E7.9.1 Reaction operating curves.

280 PLUG-FLOW REACTOR

tot

)

¼ 4Z

 2Z

(o)

tot

)

¼ 6Z

þ 6Z

(p)

tot

)

¼ 2Z

(q)

tot

)

¼ 2Z

(r)

Figure E7.9.2 shows the reaction curves.

b. From the NO curve (or the table of calculated data), the maximum production

is F

/(F

tot

)

¼ 0.1373, and it is achieved at t ¼ 1.36. Using Eq. 7.1.3 and

(e), the required reactor volume is

¼ tv

¼ 10,200 L

At t ¼ 1.36, Z

¼ 0.0388, Z

¼ 0.0232, and Z

¼ 0.0089. Substituting these

values into (m) through (r), the species molar ﬂow rates in the reactor

outlet are

out

¼ 2:42 F

out

¼ 2:19 F

out

¼ 1:32 F

out

¼ 3:57

out

¼ 0:445 F

out

¼ 0:171 mol=min

7.4 NONISOTHERMAL OPERATIONS

The design formulation of nonisothermal plug-ﬂow reactors with multiple reactions

follows the same procedure outlined in the previous section—we write design

Figure E7.9.2 Species operating curves.

7.4 NONISOTHERMAL OPERATIONS 281

equation Eq. 7.1.1 for each independent reaction. Since the temperature varies

along the reactor, we should solve the design equations simultaneously with the

energy balance equation, Eq. 7.1.16.

The energy balance equation contains another variable—the temperature of the

heating (or cooling) ﬂuid, u

, which may also vary along the reactor. (Note that u

is constant only when the ﬂuid either evaporates or condenses.) We distinguish

between two conﬁgurations: co-current ﬂow, shown in Figure 7.5a, and countercur-

rent ﬂow, shown in Figure 7.5b.

To derive an equation for u

, we write the energy balance equation over the heat-

ing ﬂuid in the shell element. For co-current ﬂow,

Q ¼ U(T

 T) dS ¼



(7:4:1)

where

and



are, respectively, the mass ﬂow rate and the heat capacity of the

heating ﬂuid. Using Eq. 5.2.51, dS ¼ (S/V ) dV

, and Eq. 7.4.1 reduces to

¼



U(T

 T)



(7:4:2)

Figure 7.5 Conﬁguration of heating of PFR; (a) co-current, (b) countercurrent.

282 PLUG-FLOW REACTOR

To convert Eq. 7.4.2 to dimensionless form, we use the deﬁnition of the dimension-

less space time, Eq. 7.1.3, and divide both sides by the reference thermal energy,

¼HTN

tot

)





 u)(7:4:3)

where HTN is the heat-transfer number, deﬁned by Eq. 7.1.17. For countercurrent

ﬂow, the energy balance over the heating ﬂuid is

Q ¼ U(T

 T) dS ¼

(7:4:4)

Following a similar procedure to the one we used for co-current ﬂow, Eq. 7.4.4

reduces to

¼ HTN

tot

)





 u)(7:4:5)

Hence, when designing a plug-ﬂow reactor with a heating/cooling ﬂuid whose

temperature varies, we have to solve either Eq. 7.4.3 or 7.4.5 simultaneously

with the design equations and the energy balance equation of the reactor.

We adopt the following procedure for setting up the energy balance equation:

1. Select and deﬁne the temperature of the reference stream, T

2. Determine the speciﬁc molar heat capacity of the reference stream,

3. Determine the dimensionless activation energies, g

’s, of all chemical

reactions.

4. Determine the dimensionless heat of reactions, DHR

’s, of the independent

reactions.

5. Determine the correction factor of the heat capacity of the reacting ﬂuid,

CF(Z

, u).

6. Specify the dimensionless heat-transfer number, HTN (using Eq. 7.1.24).

7. Determine (or specify) the inlet temperature, u

8. Determine (or specify) the inlet dimensionless temperature of the heating/

cooling ﬂuid.

9. Solve the design equations simultaneously with the energy balance equation,

and, if necessary, the energy balance equation of the heating/cooling ﬂuid to

obtain Z

’s, u, and u

as functions of the dimensionless space time, t.

The design formulation of nonisothermal plug-ﬂow reactors consists of n

Iþ2

nonlinear ﬁrst-order differential equations. Note that usually the inlet temperature

of the heating/cooling ﬂuid, T

, is known. Hence, the case of co-current

7.4 NONISOTHERMAL OPERATIONS 283

conﬁguration can be readily solved, but the countercurrent conﬁguration is solved

iteratively by trial and error. In the latter, we guess the value of T

out

as an initial

value and solve the formulation equations. We then check if the value of T

the end of the reactor is in agreement with the given value of T

. If not, we

guess another value of T

out

and repeat the calculation.

Example 7.10 The ﬁrst-order, gas-phase reaction

A ! B þC

takes place in a tubular (plug-ﬂow) reactor. The reactor is made of a 20-cm tube,

and it operates at 2 atm. A stream consisting of 80% reactant A and 20% inert I

(by mole) is fed into the reactor at 450 K at the rate of 30 L/min.

a. Determine the needed reactor length for isothermal operation at 450 K.

b. Determine the overall heating (or cooling) load in (a).

c. Determine the local and estimate the average isothermal HTN.

d. Determine the needed reactor length for adiabatic operation.

e. Examine the effect of HTN on the temperature proﬁle along the reactor and

the production of product B.

Data: At 450 K,

k(T

) ¼ 15:0hr

1

) ¼ 8000 cal=mol extent E

=R ¼ 2000 K

¼ 20 cal=mol K

¼ 15 cal=mol K

¼ 12 cal=mol K

¼ 10 cal=mol K

The temperature of the reactor wall is maintained at 470 K.

Solution The stoichiometric coefﬁcients are

¼1 s

¼ 1 s

¼ 0 D ¼ 1

Using Eq. 7.1.1, the design equation is

¼ r



(a)

We select the inlet stream as the reference stream, Z

¼ 0, and T

¼ 450 K.

(0:08206)(450)

¼ 0:0542 mol=L

tot

)

¼ v

¼ 1:626 mol=min

(b)

284 PLUG-FLOW REACTOR

For the given feed composition, y

¼ 0:80, y

¼ 0:20, and y

¼ y

¼ 0. We

use Eq. 7.1.15 to express the species concentrations, and the reaction rate is

r ¼ k(T

g(u1)=u

0:8  Z

(1 þ Z)u

(c)

Using Eq. 3.3.5, the dimensionless activation energy is

g ¼

¼ 4:44

We deﬁne a characteristic reaction time,

k(T

)

¼ 4 min (d)

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

0:8  Z

(1 þ Z)u

g(u1)=u

(e)

Using Eq. 7.1.16, the energy balance equation becomes

CF(Z

, u)

HTN(u

 u)  DHR



(f)

We have to solve (e) and (f) simultaneously. Using Eq. 7.1.20, the speciﬁc molar

heat capacity of the reference stream is

) ¼ y

) þ y

) ¼ 18 cal=mol K (g)

Using Eq. 7.1.18, the dimensionless heat of reaction is

DHR ¼

)

¼ 0:9877 (h)

Using Eq. 5.2.61, the correction factor of the heat capacity of the reacting ﬂuid is

CF(Z

, u) ¼ 1 þ

[

(u)(Z) þ

(u)Z þ

(u)Z] ¼

18 þ 7Z

(i)

7.4 NONISOTHERMAL OPERATIONS 285

Once we solve the design equation, we use Eq. 2.7.8 to determine the species

operating curves:

tot

)

¼ 0:8  Z (j)

tot

)

¼ Z (k)

tot

)

¼ Z (m)

a. For isothermal operation, u ¼ 1, and (e) reduces to

0:8  Z

1 þ Z

(n)

We solve (n) analytically. Using Eq. 2.6.5, for 60% conversion, the extent is

Z ¼

¼ 0:48

From the solution, Z ¼ 0.48 is reached at t ¼ 1.17. Using (d) and Eq. 7.1.3,

the required reactor volume is

¼ v

¼ 140:4 L (o)

The length of the reactor is 4.46 m.

b. For isothermal operation, du/dt ¼ 0. Combining Eqs. 7.1.16 and 7.1.19, the

local dimensionless heat-transfer rate along the reactor is

tot

)



)



(p)

Integrating (p), the total heating load is

Q ¼ (F

tot

)

out

¼ 6:24 kcal=min

c. Using 7.1.22, the local isothermal HTN is

HTN

iso

(t) ¼

 1

DHR

(q)

286 PLUG-FLOW REACTOR

where u

¼ 470/450 ¼ 1.044, and dZ/dt is given by (n). Figure E7.10.1

shows the local isothermal HTN. Using Eq. 7.1.23, the average isothermal

HTN for 0  t  1.17 is 8.78.

d. For adiabatic operation, HTN ¼ 0, and using (i), the energy balance equation,

(f), reduces to

¼

18 þ 7Z



DHR

(r)

We solve (e) and (r) simultaneously, subject to the initial conditions that at

t ¼ 0, Z ¼ 0 and u ¼ 1. Figure E7.10.2 shows the reaction curve and the

temperature curve, and Figure E7.10.3 shows the species curves determined

by ( j) through (m). From the calculated data, an extent of Z ¼ 0.48 is reached

at t ¼ 6.96, and, at that space time, u ¼ 0.565. Using (g) and Eq. 7.1.3, the

required reactor volume is

¼ v

¼ 835 L

The length of the reactor is 26.5 m. The exit temperature is

T ¼ T

u ¼ 295 K

Figure E7.10.1 The local isothermal HTN.

Figure E7.10.2 Reaction and temperature curves—adiabatic operation.

7.4 NONISOTHERMAL OPERATIONS 287

e. To examine the effect of HTN on the reactor operation, we solve (e) and (f)

for different values of HTN. Figure E7.10.4 shows the effect on the reactor

temperature, Figure E7.10.5 shows the effect of HTN on the reaction

extent (and from (k), on the production rate of product B). Note that when

HTN !1, the temperature of the reactor is the same as the wall temperature

(470 K, u ¼ 1.025).

Figure E7.10.3 Species curves—adiabatic operation.

Figure E7.10.4 Effect of HTN on the temperature proﬁle along the reactor.

Figure E7.10.5 Effect of HTN on the reaction extent.

288 PLUG-FLOW REACTOR

Example 7.11 The gas-phase elementary chemical reactions

Reaction 1: A þ B ! V

Reaction 2: V þ B ! W

take place in a tubular reactor (plug ﬂow). A gas mixture of 40% A, 40% B, and

20% I (inert) is fed to the reactor at a rate of 400 L/min. The inlet temperature is

1508C, and the reactor operates at a constant pressure of 2 atm. It is desirable to

maximize the production of product V. The temperature of the cooling ﬂuid is

1308C.

a. Determine the reactor volume for maximum production of product V for iso-

thermal operation.

b. Determine the heating/cooling load for isothermal operation.

c. Determine the local and average HTN for isothermal operation.

d. Determine the reactor volume for maximum production of product V for

adiabatic operation.

e. Examine the effect of HTN on the temperature proﬁle along the reactor, the

two reactions, and the production of product V.

Data: At 1508C, k

¼ 0:2L=mol s

1

¼ 0:2L=mol s

1

¼12,000 cal=mol extent DH

¼9000 cal=mol extent

¼ 4950 cal=mol E

¼ 7682 cal=mol

¼ 16 cal=mol K

¼ 8 cal=mol K

¼ 20 cal=mol K

¼ 26 cal=mol K

¼ 10 cal=mol K

Solution The purpose of this example is to illustrate the effect of heat transfer

on the production rate and product distribution. The two chemical reactions are

independent, and there is no dependent reaction. The stoichiometric coefﬁcients

of the reactions are

¼1 s

¼ 1 s

¼ 0 D ¼1

¼ 0 s

¼1 s

¼ 1 D ¼1

We write Eq. 7.1.1 for each reaction:

¼ r



(a)

¼ r



(b)

7.4 NONISOTHERMAL OPERATIONS 289