Radial and Axial Tubular Reactor - University of Michigan

© 2006 Pearson Education, Inc.All rights reserved.

From Elements of Chemical Reaction Engineering,Fourth Edition, by H. Scott Fogler.

15-1

DVD

15

Radial and AxialTemperature Variations in a

Tubular Reactor

15.1 Diffusion Fundamentals

The first step in our CRE algorithm is the mole balance, which we now needto extend to include the molar flux,

W

A

z

, and diffusional effects. The molarflow rate of A in a given direction, such as the z direction down the length ofa tubular reactor, is just the product of the flux,

W

A

z

(mol/m

2

• s), and thecross-sectional area,

A

c

(m

2

), that is,

F

A

z

=

A

c

W

A

z

In the previous chapters we have only considered plug flow, in which case

Overview. Up to now we neglected gradients in radial concentration, reac-tion rate and temperature in tubular reactors. This chapter introduces theanalysis of reactors with both axial and radial gradients.

• We first discuss diffusion fundamentals in order to develop PartialDifferential Equations (PDEs) for the mole balance.

• We next develop the PDE for the energy balance.• We next solve this set of coupled PDEs numerically.

The solutions to these equations bring us to a secondary goal of thischapter: to introduce the use of computational fluid dynamics (CFD) soft-ware to solve CRE problems. We have chosen to use the software programCOMSOL to solve PDEs for the concentration and temperature profiles.

The Algorithm1. Mole balance2. Rate law3. Stoichiometry4. Combine5. Evaluate

WAzCAvAc

----------�



15-2

Radial and Axial Temperature Variations in a Tubular Reactor Chap. 15

We now will extend this concept to consider diffusion superimposed onthe molar average velocity.

15.1.1 Definitions

Diffusion

is the spontaneous intermingling or mixing of atoms or molecules byrandom thermal motion. It gives rise to motion of the species

relative

tomotion of the mixture. In the absence of other gradients (such as temperature,electric potential, or gravitational potential), molecules of a given specieswithin a single phase will always diffuse from regions of higher concentrationsto regions of lower concentrations. This gradient results in a molar flux of thespecies (e.g., A),

W

A

(moles/area

�

time), in the direction of the concentrationgradient. The flux of A,

W

A

, is relative to a fixed coordinate (e.g., the labbench) and is a vector quantity with typical units of mol/m

2

�

s. In rectangularcoordinates

W

A

�

iW

A

x

�

jW

A

y

�

kW

A

z

(15-1)

where

i

,

j

, and

k

are the coordinate vectors in a Cartesian coordinate system.We now apply the mole balance to species A, which flows and reacts in an ele-ment of volume

D

V

=

D

x

D

y

D

z

, shown in Figure 15-1,

to obtain the variation ofthe molar fluxes in three dimensions.

+

y

z

x

x, y, zFAy|y

FAz|z

FAx|x

z+Δz, y+Δy, z+Δz

z+Δz, y, z+Δz

z, y, z+Δz

Figure 15-1 Mole balance on a cubical element of volume DxDyDz.

FAz WAz x y, FAy WAy x z, FAx WAx z y��

Molar

flow rate

in

Molar

flow rate

out

�

Molar

flow rate

in

Molar

flow rate

out

��

z+Dzz y y+Dy

Mole Balance

x yWAz z x yWAz z z��x zWAy y x zWAy y y��

+��



Sec. 15.1 Diffusion Fundamentals

15-3

Dividing by

D

x

D

y

D

z

and taking the limit as

D

x,

D

y,

and Dz go to zero, weobtain the molar flux balance in rectangular coordinates

(15-2)

The corresponding balance in cylindrical coordinates with no variation in therotation about the z-axis is

(15-3)

We will now evaluate the flux terms WA. We have taken the time to derive themolar flux equations in this form because they are now in a form that is con-sistent with the partial differential equation (PDE) solver COMSOL.

15.1.2 Molar Flux

The molar flux of A, WA, is the result of two contributions: JA, the moleculardiffusion flux relative to the bulk motion of the fluid produced by a concentra-tion gradient, and BA, the flux resulting from the bulk motion of the fluid:

(15-4)

The bulk flow term for species A is the total flux of all molecules relative to afixed coordinate times the mole fraction of A, yA; i.e., BA � yA Wi .

The bulk flow term BA can also be expressed in terms of the concentra-tion of A and the molar average velocity V:

BA � CAV (15-5)

where the molar average velocity is

V � yiVi

Here V is the particle velocity of species i, and yi is the mole fraction of spe-cies i. By particle velocities, we mean the vector-average velocities of millions

Molar

flow rate

in

�

x

Molar

flow rate

out x+ x�

�Rate of

generation�

Rate of

accumulation��

z yWAx x z yWAx x x��rA x y z�� x y z

�CA

�t----------��

�WAx

�x-------------�

�WAy

�y-------------

�WAz

�z------------� rA��

�CA

�t----------�

1r--- �

�r----- rWAr( )�

�WAz

�z------------� rA�

�CA

�t----------�

Total flux �diffusion � bulk

motionWA JA BA��

Â

molm2 s�------------ mol

m3--------- m

s----��

Molar averagevelocity Â



15-4 Radial and Axial Temperature Variations in a Tubular Reactor Chap. 15

of A molecules at a point. For a binary mixture of species A and B, we let VA

and VB be the particle velocities of species A and B, respectively. The flux ofA with respect to a fixed coordinate system (e.g., the lab bench), WA, is justthe product of the concentration of A and the particle velocity of A:

WA � CAVA (15-6)

The molar average velocity for a binary system is

V � yAVA � yBVB (15-7)

The total molar flux of A is given by Equation (15-4). BA can beexpressed either in terms of the concentration of A, in which case

(15-8)

or in terms of the mole fraction of A:

(15-9)

We now need to evaluate the molar flux of A, JA, that is superimposed on themolar average velocity V.

15.1.3 Fick’s First Law

Our discussion on diffusion will be restricted primarily to binary systems con-taining only species A and B. We now wish to determine how the molar diffu-sive flux of a species (i.e., JA) is related to its concentration gradient. As an aidin the discussion of the transport law used to describe diffusion, recall similarlaws from other transport processes. For example, in conductive heat transferthe constitutive equation relating the heat flux q and the temperature gradientis Fourier’s law:

q � �kt ————T (15-10)

where kt is the thermal conductivity.In rectangular coordinates, the gradient is in the form

———— � i

The one-dimensional form of Equation (15-10) is

qz � �kt (15-11)

In momentum transfer, the constitutive relationship between shear stress, t, andshear rate for simple planar shear flow is given by Newton’s law of viscosity:

moldm2 s�---------------

Ë ¯Á ˜Ê ˆ mol

dm3----------

Ë ¯Á ˜Ê ˆ

dms

------- Ë ¯Á ˜Ê ˆ �

WA � JA � CAV

Binary system ofA and B

WA � JA � yA(WA � WB)

Experimentationwith frog legs led to

Fick’s first law.

Constitutiveequations in heat,

momentum, andmass transfer

��x----- j

�� y ----- k

�� z ----- � �

dTdz------Heat Transfer



Sec. 15.1 Diffusion Fundamentals

15-5

t

� ��

The mass transfer flux law is analogous to the laws for heat and momentumtransport, i.e., for constant total concentration

(15-12)

The general 3-dimensional constitutive equation for

J

A

, the diffusional flux of

A

resulting from a concentration difference, is related to the mole fraction gra-dient by Fick’s first law:

J

A

�

�

cD

AB

————

y

A

(15-13)

where

c

is the total molar concentration (mol/dm

3

),

D

AB

is the diffusivity of Ain B (dm

2

/s), and

y

A

is the mole fraction of A. Combining Equations (15-9)and (15-13), we obtain an expression for the total molar flux of A:

(15-14)

In terms of concentration for constant total concentration

(15-15)

15.1.4 Diffusion and Convective Transport.

When accounting for diffusional effects, the molar flow rate of species A,

F

A

,in a specific direction

z

, is the product of molar flux in that direction,

W

A

z

, andthe cross-sectional area normal to the direction of flow,

A

c

:

In terms of concentration the flux is

The molar flow rate is

(15-15A)

Similar expressions follow for WAx and WAy. Substituting for the flux WAx,WAy, and WAz into Equation (15-2), we obtain

(15-16)

dudz------Momentum Transfer

JAz DABdCA

dz----------��Mass Transfer

Molar flux equation WA = –cDAB————yA + yA(WA + WB)

Molar flux equation WA � �DAB ————CA � CAV

FAz AcWAz�

WAz DABdCA

dz----------� CAUz��

FAz WAz Ac DABdCA

dz----------� CAUz� Ac� �

Flow, diffusion, andreaction.

This form isused in COMSOL

Multiphysics.

DAB�

2CA

�x2------------

�2CA

�y2------------

�2CA

�z2------------� � Ux

�CA

�x---------- Uy

�CA

�y---------- Uz

�CA

�z----------� rA��

�CA

�t----------�




Equation (15-16) is in a user-friendly form directly applicable to the PDEsolver, COMSOL. For one-dimension at steady state, Equation (15-16) reduces to

(15-17)

In order to solve Equation (15-17) we need to specify the boundary conditions.In this chapter we consider some of the simpler boundary conditions, and inChapter CD14 on the CD-ROM we consider the more complicated boundaryconditions, such as the Danckwerts’ boundary conditions.

We will now use this form of the molar flow rate in our mole balance inthe z direction of a tubular flow reactor (see Chapter 1)

(1-11)

However, we first have to discuss the boundary conditions in solving the equa-tion.

15.1.5 Boundary Conditions

The most common boundary conditions are presented in Table 15-1.

TABLE 15-1. TYPES OF BOUNDARY CONDITIONS

1. Specify a concentration at a boundary (e.g., z � 0, CA � CA0). For an instantaneous reactionat a boundary surface, the concentration of the reactants at the boundary is taken to be zero(e.g., CAs � 0). See CD-ROM Chapter CD14 for the more exact and complicated Danckwerts’boundary conditions at z = 0 and z = L.

2. Specify a flux at a boundary.a. No mass transfer to a boundary,

(15-18)

For example, at the wall of a nonreacting pipe,

at r � R

That is, because the diffusivity is finite, the only way the flux can be zero is if the concen-tration gradient is zero.

b. Set the molar flux to the surface equal to the rate of reaction on the surface,

WA(surface) � � (surface) (15-19)

c. Set the molar flux to the boundary equal to convective transport across a boundary layer,

WA(boundary) � kc(CAb � CAs) (15-20)

where kc is the mass transfer coefficient and CAs and CAb are the surface and bulk concen-trations, respectively.

3. Planes of symmetry. When the concentration profile is symmetrical about a plane, the concen-tration gradient is zero at the plane of symmetry. For example, in the case of radial diffusionin a pipe, at the center of the pipe

at r � 0

COMSOL

DABd2CA

dz2------------ Uz

dCA

dz----------� rA� 0�

dFAz

dV----------- rA�

WAR

dCA

dr---------- 0�

rA

dCA

dr---------- 0�



Sec. 15.2 Radial and Axial Variations in a Tubular Reactor 15-7

15.2 Radial and Axial Variations in a Tubular Reactor

In the previous chapters we have assumed that there were no radial variationsin velocity, concentration, temperature, or reaction rate in the tubular andpacked-bed reactors. As a result, the axial profiles could be determined usingan ordinary differential equation (ODE) solver. In this section we will considerthe case where we have both axial and radial variations in the system variables,in which case we will require a partial differential (PDE) solver. A PDE solversuch as COMSOL will allow us to solve tubular reactor problems for both theaxial and radial profiles, as shown here and on the Web module.

We are going to carry out differential mole and energy balances on thedifferential cylindrical annulus shown in Figure 15-2.

15.2.1 Molar Flux in Radial Coordinates

In order to derive the governing equations, we need to define a couple ofterms. The first is the molar flux of species i, Wi (mol/m2 • s). The molar fluxhas two components, the radial component Wir, and the axial component, Wiz.The molar flow rates are just the product of the molar fluxes and the cross-sec-tional areas normal to their direction of flow. For example, for species i flow-ing in the axial (i.e., z) direction

Fiz = Wiz Acz

where Wiz is the molar flux in the z direction (mol/m2/s), and Acz (m2) is thecross-sectional area of the tubular reactor.

In Chapter CD11 on the CD-ROM we discuss the molar fluxes in somedetail, but for now let us just say they consist of a diffusional component,

, and a convective flow component,

(15-21)

Go to the COMSOL Web site http://www.comsol.com/ecre to see thetutorial and chemical reaction engineering examples.

COMSOLapplication

Figure 15-2 Cylindrical shell of thickness Dr, length Dz, and volume 2prDrDz.

De �Ci/�z( )� UzCi

Wiz De�Ci

�z--------� UzCi��




where is the effective diffusivity (or dispersion coefficient) (m2/s), and isthe axial molar average velocity (m/s). Similarly, the flux in the radial direction is

(15-22)

where Ur (m/s) is the average velocity in the radial direction. For now we willneglect the velocity in the radial direction, i.e., Ur = 0.

15.2.2 Mole Balances on Species A

A mole balance on a cylindrical system volume of length �z and thickness �ras shown in Figure 14-2 gives

Dividing by and taking the limit as Dr and Dz Æ 0

Similarly, for any species i and steady-state conditions,

(15-23)

Using Equations (15-21) and (15-22) to substitute for and in Equation(15-23) and then setting the radial velocity to zero, Ur = 0, we obtain

De Uz

Radial Direction Wir De�Ci

�r--------� UrCi��

Moles of A

in at rË ¯Á ˜Ê ˆ

WArCross-sectional area

normal to radial flux Ë ¯Á ˜Ê ˆ

∑ WAr 2r�z∑� �

Moles of A

in at zË ¯Á ˜Ê ˆ

WAzCross-sectional area

normal to axial flux Ë ¯Á ˜Ê ˆ

∑ WAz 2r�r∑� �

Moles of A

in at rË ¯Á ˜Ê ˆ

Moles of A

out at r �r�( )Ë ¯Á ˜Ê ˆ

� Moles of A

in at zË ¯Á ˜Ê ˆ Moles of A

out at z �z�( )Ë ¯Á ˜Ê ˆ

��

Moles of A

formedË ¯Á ˜Ê ˆ

� Moles of A

AccumulatedË ¯Á ˜Ê ˆ

�

WAr2r�z r WAr 2r�z r �r�� WAz 2r�r z WAz 2r�r z �z�

��

rA2r�r�z��CA 2r�r�z( )

�t------------------------------------- �

2r�r�z

1r---

� rWAr( )�r

-------------------��WAz

�z------------- rA��

�CA

�t----------�

1r---

� rWir( )�r

------------------��Wiz

�z----------- ri�� 0�

Wiz Wir

1r--- �

�r----- De

�Ci

�r--------r�Ë ¯

Ê ˆ��z----- De

�Ci

�z--------� UzCi�� ri� 0�



Sec. 15.3 Energy Flux 15-9

For steady-state conditions and assuming does not vary in the axial direction,

(15-24)

This equation is also discussed further in Chapter CD14.

15.3 Energy Flux

When we applied the first law of thermodynamics to a reactor to relate eithertemperature and conversion or molar flow rates and concentration, we arrived atEquation (11-9). Neglecting the work term we have for steady-state conditions

(15-25)

In terms of the molar fluxes and the cross-sectional area and (q = )

(15-26)

The q term is the heat added to the system and almost always includes a con-duction component of some form. We now define an energy flux vector, e,(J/m2 ◊s), to include both the conduction and convection of energy.

e = Conduction + Convection

(15-27)

where the conduction term q (J/m2 ◊s) is given by Fourier’s law. For axial (z)and radial (r) conduction Fourier’s laws are

and qr = –ke

where is the thermal conductivity (J/m◊s◊K). The energy transfer (flow) isthe vector flux times the cross-sectional area, Ac, normal to the energy flux

Energy flow = e ◊ Ac

15.3.1 Energy Balance

Using the energy flux, e, to carry out an energy balance on our annulus (Figure15-2) with system volume 2prDrDz, we have

Uz

De�

2Ci

�r2----------

De

r------

�Ci

�r-------- De

�2Ci

�z2---------- Uz

�Ci

�z--------� ri� � � 0�

Conduction

Q

Convection

Fi0Hi0 FiHii 1��

i 1�

m

�� 0�

} ¸ Ô Ô ˝ Ô Ô ˛

Q/Ac

Ac q �Wi0Hi0 �WiHi�( )�[ ] 0�

e q �WiHi��

e = energy flux (J/s◊m2)

qz ke�T�z------�� T

�r------

ke

Energy flow in at r( ) er Acr er 2r�z��

Energy flow in at z( ) ez Acz ez 2r�r��




(15-28)

Dividing by 2prDrDz and taking the limit as Dr and Dz Æ 0,

(15-29)

The radial and axial energy fluxes are

Substituting for the energy fluxes, er and ez,

(15-30)

and expanding the convective energy fluxes, ,

Radial: (15-31)

Axial: (15-32)

Substituting Equations (15-31) and (15-32) into Equation (15-30), we obtainupon rearrangement

Recognizing that the term in brackets for steady-state conditions is just the rateof formation of species i, ri, we have

(15-33)

Recalling

, , ,

Energy Flow

in at rË ¯Á ˜Ê ˆ Energy Flow

out at r �r�Ë ¯Á ˜Ê ˆ Energy Flow

in at zË ¯Á ˜Ê ˆ Energy Flow

out at z �z�Ë ¯Á ˜Ê ˆ

��Accumulation

of Energy in

Volume 2r�r�z( )Ë ¯Á ˜Á ˜Á ˜Ê ˆ

�

er2r�z( ) r er2r�z( ) r �r�ez2r�r z ez2r�r z �z�

�� 0�

1r---

� rer( )�r

--------------��ez

�z-------� 0�

er qr �Wir Hi��

ez qz �Wiz Hi��

1r---

� r qr �Wir Hi�[ ][ ]�r

--------------------------------------------�� qz �Wiz Hi�[ ]

�z-------------------------------------� 0�

�Wi Hi

1r--- �

�r----- r�Wir Hi( ) 1

r---�Hi

� rWir( )�r

------------------�Wir�Hi

�r---------------------��

Neglect

� �Wiz Hi( )�z

-------------------------- �Hi�Wiz

�z----------- �Wiz

�Hi

�z--------��

1r---�

� rqr( )�r

--------------�qz

�z------- �Hi

1r---

� rWir( )�r

------------------�Wiz

�z-----------�Ë ¯

Ê ˆ� �Wiz�Hi

�z--------�� 0�

¸ Ô Ô Ô ˝ Ô Ô Ô ˛

ri

1r---�

��r----- rqr( ) �qz

�z------- �Hiri� �Wiz

�Hi

�z--------�� 0�

qr ke�T�r------�� qz ke

�T�z------��

�Hi

�z-------- CPi

�T�z------�



Sec. 15.4 Some Initial Approximations 15-11

and

we have the energy balance in the form

(15-34)

15.4 Some Initial Approximations

Assumption 1. Neglect the axial diffusive term [i.e., ] wrt the con-

vective term in the expression involving heat capacities

With this assumption, Equation (15-34) becomes

(15-35)

For laminar flow, the velocity profile is

(15-36)

where U0 is the average velocity inside the reactor and R is the tabular reactor radius.

Assumption 2. Assume that the sum is constant.The energy balance now becomes

(15-37)

Equation (15-36) is the form we will use in the COMSOL problem that fol-lows. In many instances, the term is just the product of the solution den-sity and the heat capacity of the solution (J/kg • K).

15.5 Heat Exchange Fluid Balance

We also recall that a balance on the coolant gives the variation of coolant tem-perature with axial distance where Uht is the overall heat transfer coefficientand R is the reactor wall radius.

ri �i rA�( )�

�riHi ��iHi rA�( ) �HRx rA��

ke

r---- �

�r----- r�T

�r---------Ë ¯

Ê ˆ ke��

2T

�z2-------- �HRxrA �WizCPi

( )�T�z------�� 0�

-ÊËÁ

ˆ¯

DC

zei∂

∂

�CPiWiz �CPi

0 UzCi�( ) �CPiCiUz� �

ke

r---- �

�r----- r�T

�r---------

Ë ¯Á ˜Ê ˆ

ke�

2T

�z2-------- �HRxrA Uz�CPi

Ci( )�T�z------�� 0�

Uz 2U0 1 rR---Ë ¯

Ê ˆ2

��

Energy balancewith radial andaxial gradients

CPm�CPi

Ci CA0� iCPi� �

ke�

2T

�z2--------

ke

r---- �

�r----- r�T

�r------

Ë ¯Á ˜Ê ˆ

�HRxrA UzCPm

�T�z------�� 0�

CPm




(15-38)

15.6 Boundary and Initial Conditions

A. Initial conditions if other than steady statet = 0, Ci = 0, T = T0, for z > 0 all r (A)

B. Boundary conditions

1) Radial(a) At r = 0, we have symmetry and .

(b) At the tube wall r = R, the temperature flux to the wall onthe reaction side equals the convective flux out of thereactor into the shell side of the heat exchanger.

(B)

(c) There is no mass flow through the tube walls

at r = R. (C)

2) Axial(a) At the entrance to the reactor z = 0,

T = T0 and Ci = Ci0 (D)

(b) At the exit of the reactor z = L,

and (E)

The following examples will solve the preceding equations using COM-SOL. For the exothermic reaction with cooling, the expected profiles are

Example 15-1 Radial Effects in a Tubular Reactor

This example will highlight the radial effects in a tubular reactor, which up untilnow have been neglected to simplify the calculations. Now, the effects of parameterssuch as inlet temperature and flow rate will be studied using the software programCOMSOL. Follow the step-by-step procedure in the Web Module and on the CD-ROM.

mcCPc

�Ta

�z-------- Uht2R T R,z( ) Ta�[ ]�

�T �r� 0� �Ci �r� 0�

ke�T�r-------�

RUht T R,z( ) Ta�( )�

�Ci �r� 0�

�T�z------ 0�

�Ci

�z-------- 0�

Tr=0

z

T0

FA0

mc Ta0•

mc Ta0•



Sec. 15.6 Boundary and Initial Conditions 15-13

We continue Example 12-3, which discussed the reaction of propylene oxide(A) with water (B) to form propylene glycol (C). The hydrolysis of propylene oxidetakes place readily at room temperature when catalyzed by sulfuric acid.

This exothermic reaction is approximated as a first-order reaction, given that thereaction takes place in an excess of water.

The CSTR in Example 12-3 has been replaced by a tubular reactor 1.0 m inlength and 0.2 m in diameter. The feed to the reactor consists of two streams thatare mixed just before entering the tubular reactor. One stream is an equivolumetricmixture of propylene oxide and methanol, and the other stream is water containing0.1 wt % sulfuric acid. The water is fed at a volumetric rate 2.5 times larger thanthe propylene oxide–methanol feed. The molar flow rate of propylene oxide fed tothe tubular reactor is 0.1 mol/s.

There is an immediate temperature rise upon mixing the two feed streamscaused by the heat of mixing. In these calculations, this temperature rise is alreadyaccounted for, and the inlet temperature of both streams is set to T0 = 312 K.

The reaction rate law is

–rA = kCA (E15-1.1)

with

k = Ae–E/RT (E15-1.2)

where E = 75,362 J/mol and A = 16.96 ¥ 1012 h–1, which can also be put in the form

k(T) = k0(T0) exp (E15-1.3)

with k0 = 1.28 h–1 at 300 K. The thermal conductivity ke of the reaction mixture andthe diffusivity De are 0.599 W/m/K and 10–9 m2/s, respectively, and are assumed tobe constant throughout the reactor. In the case where there is heat exchange betweenthe reactor and its surroundings, the overall heat-transfer coefficient is 1300 W/m2/Kand the temperature of the cooling jacket temperature is assumed to be constant andis set to 273 K. The other property data are shown in Table E15-1.1.

TABLE 15-1.1. PHYSICAL PROPERTY DATA

Propylene Oxide Methanol Water Propylene Glycol

Molar weight (g/mol) 58.095 32.042 18 76.095

Density (kg/m3) 830 791.3 1000 1040

A B ææÆ C �

ER--- 1

T0

----- 1T---�Ë ¯

Ê ˆ



15-14

Radial and Axial Temperature Variations in a Tubular Reactor Chap. 15

Solution

Mole Balances:

Recalling Equation (14-24) and applying it to species A

(E15-1.4)

Rate Law:

(E15-1.5)

Stoichiometry:

The conversion along a streamline (

r

) at a distance

z

from the reac-tor entrance

X

(

r,

z

) = 1 –

C

A

(

r,

z

)/

C

A0

(E15-1.6)

The overall conversion at a given distance

z

from the reactor entrance is

(E15-1.7)

The mean concentration at any distance

z

from the reactor entrance is

(E15-1.8)

For plug flow, the velocity profile is

U

z

=

U

0

(E15-1.9)

The laminar flow velocity profile is

(E15-1.10)

Recalling the Energy Balance

(15-37)

Heat capacity (J/mol • K) 146.54 81.095 75.36 192.59

Heat of formation (J/mol) –154911.6 238,400 –286098 –525676

T

ABLE

15-1.1.

P

HYSICAL

P

ROPERTY

D

ATA

(

CONTINUED

)

Propylene Oxide Methanol Water Propylene Glycol

De�

2CA

�r2------------ 1

r---De

�CA

�r---------- De

�2CA

�z2------------ Uz

�CA

�z----------� rA� � � 0�

rA� k T1( ) exp ER--- 1

T1

----- 1T---�Ë ¯

Ê ˆ CA�

X z( ) 12 CA r z,( )Uzr rd

0

R

�FA0

----------------------------------------------��

CA z( )2 CA r z,( )Uzr rd

0

R

�R2U0

----------------------------------------------�

Uz 2U0 1 rR---Ë ¯

Ê ˆ2

��

ke�

2T

�z2--------

ke

r---- �

�r----- r

�T�r------

Ë ¯Á ˜Ê ˆ

�H �RxrA UzCPm

�T�z------ 0=��




Assumptions

1. is zero.2. Neglect axial diffusion/dispersion flux with regard to convective flux when

summing the products of the heat capacities and their respective fluxes.3. Steady state.

Cooling jacket

(15-38)

Boundary conditions

At r = 0, then and (E15-1.11)

At r = R, then and (E15-1.12)

At z = 0, then Ci = Ci0 and T = T0 (E15-1.13)

These equations were solved using COMSOL for a number of cases including adi-abatic and non-adiabatic plug flow and laminar flow; they were also solved with andwithout axial and radial dispersion. A detailed accounting on how to change theparameter values in the COMSOL program can be found in the COMSOL Instruc-tions section on the Web in screen shots similar to Figure E15-1.1. Figure E15-1.2gives the data set in SI units used for the COMSOL example.

Color surfaces are used to show the concentration and temperature profiles, similarto the black and white figures shown in Figure E15-1.2. Read through the COMSOLWeb module entitled “Radial and Axial Temperature Gradients.” One notes inFigure E15-2.1 that the conversion is lower near the wall because of the cooler fluidtemperature. These same profiles can be found in color on the Web and CD-ROM inthe Web Modules.

Ur

mCPC

�Ta

�z-------- 2RUht T R,z( ) Ta�( )�

�Ci

�r-------- 0� �T

�r------ 0�

�Ci

�r-------- 0� ke

�T�r------� Uht T R,z( ) Ta�( )�

Define expression

Figure E15-1.1 COMSOL screen shot of data set for a second order reaction.




Analysis: One notes the maximum and minimum in these profiles. Near the wall,the temperature of the mixture is lower because of the cold wall temperature. Con-sequently, the rate will be lower, and thus the conversion will be lower. However,right next to the wall, the velocity through the reactor is almost zero so the reactantsspend a long time in the reactor; therefore, a greater conversion is achieved, as notedby the upturn right next to the wall.

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

Temperature Surface

Radial Location (m)

Axi

al L

ocat

ion

(m)

-0.2 -0.1 0 0.1 0.2

350

340

330

320

310

300

290

280

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

Conversion Surface

Radial Location (m)

Axi

al L

ocat

ion

(m)

-0.2 -0.1 0 0.1 0.2

Tem

pera

ture

(K

)

Radial Temperature Profiles

Radial Location (m)

0 0.02 0.04 0.06 0.08 0.1

350

340

330

320

310

300

290

280

Outlet

Half Axial Location

Inlet

Con

vers

ion

Radial Conversion Profiles

Radial Location (m)

0 0.02 0.04 0.06 0.08 0.1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

10.8

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0Average Outlet Conversion=90.3%

Average Outlet Conversion=90.3%

Outlet

Half Axial Location

Inlet

Figure E15-1.2 (a) Temperature surface, (b) temperature surface profiles,(c) conversion surface, and (d) radial conversion profile.

(a) (b)

(c) (d)




Example 15-2 Radial Effects in a Packed-Bed Reactor1

One of the common methods to produce phthalic anhydride is from the partial oxi-dation of o-xylene. This simplified reaction for the formation of phthalic anhydridefrom o-xylene oxidation will be represented by

In terms of symbols

A + 3B Æ C + 3D

Phthalic anhydride is primarily used in plasticizers and in resins used to make boathulls, hot tubs, and synthetic marble surfaces. In the partial oxidation of o-xylenereaction, there are several byproducts as well as products of combustion that areformed if the reactor is not optimized. It is desirable to avoid high temperatures inthis reactor based on temperature limits on the materials (reactor walls and catalyst)as well as reducing the formation of byproducts. As you would expect, high temper-atures will result in the combustion of both the reactant and product, resulting in theformation of CO2 and H2O. Because of the high operating temperatures, this reactoris cooled using molten salt (sodium nitrite-potassium nitrate). For this COMSOLECRE problem we will only use approximate reaction kinetics for the overall reac-tion, as well as making an assumption of constant overall flowrate. This assumptionmust be employed based on the form of the equation given in the COMSOL ECREpackage, which uses a constant velocity through the reactor. For safety considerations,the o-xylene is diluted to a mole fraction of 0.012 and fed to the reactor at 610K.

It is desired to make 76 metric tons of phthalic anhydride per year in apacked-bed reactor 1 m in length with a 0.1 m radius. Other parameters are given inTable E15-2.1.

The reaction follows the following rate law based on partial pressures

Additional Information

1 Example problem by Professor Robert Hesketh, Department of Chemical Engineer-ing, Rowan University, Glassboro, New Jersey.

TABLE E15-2.1 COMSOL ECRE CONSTANTS

E 112971 J/mol

A 8.922 mol/m3 • Pa2 • s

R 8.314 J/mol•K

+ O2 + 3H2O

O-xylene Phthalic Anhydride

O

O

O

Me

Me

C H O C H O H O8 10 2 8 4 3 23 3+ Æ +

- ¢ = ¢r k P PA A O2




Construct a model of this reactor and investigate the radial and axial temperatureprofiles.

(a) Run your model using values of thermal conductivity from 0.1 to 1000 J/(mK s).

(b) Compare this model to the plug flow model for mass and heat transfer. Dis-cuss the overall effect of the value of effective thermal conductivity of the bedon the resulting temperature profile and outlet conversion obtained.

(c) Give a rationale as to why industry does not use tubes of 0.1m in diameter,but instead uses a smaller tube diameter of 0.0254 m.

Preliminary CalculationsTo produce 76 metric ton/y during 350 d/y operation

For 79% conversion

T0 610 K

FA0 0.02149 mol/s

FO2_0 .37155 mol/s

Ra 0.1 m

dHrx -1125924 J/mol

ke 0.78 J/(m s K)

Cp 1150 J/(kg K)

Uht 156 J/(s K m2)

Ta 620 K

FT 1.79078 mol/s

P 1.01325e5 Pa

p 3.141592654

TABLE E15-2.1 COMSOL ECRE CONSTANTS (CONTINUED)

F76T

350d

1d

24h

1h

3600s

1000kg

T

1mol

148g0.17 mol sC = ¥ ¥ * =

FF

XAC

00 17

0 790 215= = =.

.. mol s

FF

F F F F F FTA

A B N A B B00

0 0 0 0 00 012

79

212= = + + = + +

.

FF F

mol sBT A

00 0

10 790 21

0 372= -

+=.

.

.

FT = 1 79. mol s

A mC = ( ) =p 0 1 0 03142. .

v030 0896= . m s




SolutionIn COMSOL we will use a 2D axis-symmetric model in the r and z directions. Thisreaction takes place in a packed-bed reactor, where we will assume that the flow isplug flow with no axial or radial diffusion.

Assumptions:1. Ur = 0 (i.e., assume negligible velocity in the radial direction)2. Uz = constant3. Neglect axial diffusion/dispersion flux with respect to convective flux4. Assume constant density and that mean heat capacity of the gas can be

approximated by the values of air5. Steady-state

Mole Balance:Starting with Equation (15-24)

(15-24)

Neglecting diffusion in the axial and radial directions gives the plug flow equation:

(E15-2.1)

Note that the velocity Uz is constant.

Rate Law:

(E15-2.2)

To convert the rate to per reactor volume we multiply by the bulk density of the cat-alyst bed, rB

(E15-2.3)

For the COMSOL model, the partial pressures need to be converted to concentrationusing the ideal gas law Ci = Pi /RT

(E15-2.4)

with

(E15-2.5)

Stoichiometry:Because of the dilute reactants, e = 0. We also will neglect changes in volumetricflow rate with temperature and pressure

v = v0

DC

r

D

r

C

rD

C

zU

C

zre

i e ie

iz

ii

∂∂

∂∂

∂∂

∂∂

2

2

2

2 0+ + - + =

- + =UC

zrz

ii

∂∂

0

- ¢ = ¢ÊËÁ

ˆ¯

r k P PA A O2

mol

kgcat • Pa • s2

- = - ¢ = ¢ =r r k P P kP PA A b b A O A Or r2 2

- = ( )r k RT C CA A O2

2

k Aem Pa

=∑ ∑

ÊË

ˆ¯

- E

RT mol

s3 2




(E15-2.6)

Energy Balance:Starting with the energy balance from Equation 15-37:

(15-37)

The conductivity used in the energy balance will be an effective conductivityof the bed. The conduction in the axial direction is neglected, resulting in the fol-lowing equation:

( E15-2.7)

The boundary conditions for these PDEs will be:

(A) (E15-2.8)

(B) (E15-2.9)

(C) (E15-2.10)

(D) (E15-2.11)

Findings:

CF

CF

CF

AA

OO

CC= = =

v v v0 0 02

2, ,

kT

z

k

r rr

T

rH r U C

T

zee

Rx A z Pm

∂∂

∂∂

∂∂

∂∂

2

2 0+ ÊËÁ

ˆ¯

+ - =D

k

r rr

T

rH r U C

T

ze

Rx A z Pm

∂∂

∂∂

∂∂

ÊËÁ

ˆ¯

+ - =D 0

- = ( ) -( ) = £ £kT

rU T z T r R ze

Rht a

∂∂

R at and m, 0 1

∂ ∂ = ∂ ∂ = =C r T r ri / and at0 0 0

T T C C z r Ri i= = = £ £0 0 0 0 and at and

∂ ∂ = ∂ ∂ = = £ £T z C z z r Ri0 0 1 0 0and at m and.

600

620

640

660

680

700

720

740

Tem

pera

ture

(K

)

Temperature (K)

z0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

123

4

5

Figure E15-2.1: Surface Plot

Figure E15-2.2: Temperature Profile in reactor tube for ke=0.78 J/(s m K). Positions: for ke=0.78 J/(s m K) 1) r=0.1 m Wall,2) r=0.08m, 3) r=0.06m, 3) r=0.04m, 4) r=0.02m, 5) r=0.00m center.




Results:A surface plot of the temperatures in the reactor tube is shown in Figure E15-2.1which shows a noticeable difference in temperatures of the fluid at the end of thereactor starting at about z = 0.7 m. For this value of thermal conductivity, the wallremains “cold” at approximately the temperature of the molten salt of 720K. Thiscan also be seen in the cross-section plot that is shown in Figure E15-2.2. Each lineon this plot represents a radial position within the reactor. In Figure E15-2.3 a crosssection plot showing the temperature profile as a function of radial position and axialdistance in the reactor is presented. Each line in this figure represents an axial posi-tion in the reactor. In this plot, the difference between the wall temperature and thetemperature within the reactor is clearly shown. The plug flow condition, in whichthe temperature profile has been eliminated in the radial direction, is shown in Fig-ure E15-2.4. This result was obtained with a thermal conductivity of 1000 J/(s m K).

600

620

640

660

680

700

720

740

Tem

pera

ture

(K

)

Temperature (K)

r0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

123

4

5

Figure E15-2.3: Temperature Profile in reactor tube for ke=0.78 J/(s m K). Positions: 1) z=0.0 m, 2) r=0.2m, 3) z=0.4m, 3) z=0.6m, 4) z=0.8m, 5) r=1.0 m This plot shows a large temperature variation in the radial direction.

610

615

620

625

630

635

640

645

650

655

660

Tem

pera

ture

(K

)

Temperature (K)

r0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

123

4

5

Figure E15-2.4: Temperature Profile in reactor tube for ke=1000 J/(s m K). Positions: 1) z=0.0 m, 2) r=0.2m, 3) z=0.4m, 3) z=0.6m, 4) z=0.8m, 5) r=1.0 m This plot shows that there is no temperature variation with respect to radial position.




Table E15-2.2 gives a summary of the output from the COMSOL program.

Analysis: From Table E15-2.2 we observe that the removal of heat is effective inlowering the temperature within the reactor. The advantage of the molten salt cool-ing is that it will help to prevent a runaway reaction in the reactor tubes. As theradial thermal conductivity is increased, the overall rate of heat transfer from thefluid to the reactor walls increases. As you can see from the table, the conversionalso decreases.

S U M M A R Y

To obtain the axial or radial temperature and concentration gradients, the fol-lowing coupled partial differential equations were solved using COMSOL:

(S15-1)

and

(S15-2)

C D - R O M M A T E R I A L

• Learning Resources1. Summary Notes2. Web Module COMSOL Radial and Axial Gradients

TABLE E15-2.2 SUMMARY OF THE RESULTS OF THE EFFECT OF THERMAL CONDUCTIVITY ON T AND X

ke (J/(s m K) 0.1 0.78 10 100 1000Plug Flow Model

T*Ac outlet avg (K m^2)

22.3 21.9 21.0 20.7 20.7

T outlet average (K) 711 697 668 659 658 658

X*Ac 0.00804 0.00726 0.00569 0.00524 0.00519

X outlet average 0.256 0.231 0.181 0.167 0.165 0.165

De�

2Ci

�r2----------

De

r------

�Ci

�r-------- De

�2Ci

�z2---------- Uz

�Ci

�z--------� ri 0=� � �

ke�

2T

�z2--------

ke

r---- �

�r----- r�T

�r------Ë ¯

Ê ˆ �HRxrA UzCPm

�T�z------ 0=��



Chap. 15 CD-ROM Material 15-23

P15-1C In a diving-chamber experiment, a human subject breathed a mixture of O2

and He while small areas of his skin were exposed to nitrogen gas. Afterawhile the exposed areas became blotchy, with small blisters forming on theskin. Model the skin as consisting of two adjacent layers, one of thickness �1

and the other of �2. If counterdiffusion of He out through the skin occurs atthe same time as N2 diffuses into the skin, at what point in the skin layers isthe sum of the partial pressures a maximum? If the saturation partial pressurefor the sum of the gases is 101 kPa, can the blisters be a result of the sum ofthe gas partial pressures exceeding the saturation partial pressure and the gascoming out of the solution (i.e., the skin)?

Before answering any of these questions, derive the concentration pro-files for N2 and He in the skin layers.

Diffusivity of He and N2 in the inner skin layer� 5 � 10�7 cm2/s and 1.5 � 10�7 cm2/s, respectively

Diffusivity of He and N2 in the outer skin layer� 10�5 cm2/s and 3.3 � 10�4 cm2/s, respectively

P15-2B The decomposition of cyclohexane to benzene and hydrogen is mass trans-fer–limited at high temperatures. The reaction is carried out in a 5-cm-ID pipe20 m in length packed with cylindrical pellets 0.5 cm in diameter and 0.5 cmin length. The pellets are coated with the catalyst only on the outside. The bedporosity is 40%. The entering volumetric flow rate is 60 dm3/min.(a) Calculate the number of pipes necessary to achieve 99.9% conversion of

cyclohexane from an entering gas stream of 5% cyclohexane and 95% H2

at 2 atm and 500�C.(b) Plot conversion as a function of length.(c) How much would your answer change if the pellet diameter and length

were each cut in half?(d) How would your answer to part (a) change if the feed were pure cyclo-

hexane?(e) What do you believe is the point of this problem?

P15-3B Assume the minimum respiration rate of a chipmunk is 1.5 micromoles ofO2/min. The corresponding volumetric rate of gas intake is 0.05 dm3/min atSTP.(a) What is the deepest a chipmunk can burrow a 3-cm diameter hole

beneath the surface in Ann Arbor, Michigan? DAB = 1.8 ¥ 10–5 m/s(b) In Boulder, Colorado?

External Skin Boundary

Partial Pressure

Internal Skin Boundary

Partial Pressure

N2 101 kPa 0He 0 81 kPa

�1 20 �m Stratum corneum�2 80 �m Epidermis




(c) How would your answers to (a) and (b) change in the dead of winterwhen T = 0˚F?

(d) Critique and extend this problem (e.g., CO2 poisoning).P15-4 Instructions: If you have not installed COMSOL Multiphysics, request a trial

version of the software from the Web site http://www.comsol.com/ecre. Fromthis page you can also download model and documentation files required forthe COMSOL exercises outlined in this chapter.

Load COMSOL Multiphysics and follow the installation instructions.Double-click on the COMSOL Multiphysics icon on your desktop. In theModel Library, select model denoted “4-Non-Isothermal Reactor II” and clickthe Dynamic Help icon on the main toolbar. This opens a Help window in theCOMSOL desktop presenting the detailed documentation of the specificmodel. Scroll the side bar to review the model equations. Move further downthe documentation to find the step-by-step instructions that guide you throughthe model-building process. The steps also detail how to solve the model andanalyze the results. If you want bypass the model setup process, return to theModel Library dialog and press OK to open a saved model file.(a) Why is the concentration of A near the wall lower than the concentration

near the center? (b) Where in the reactor do you find the maximum and minimum reaction

rates? Why? Instructions: Select the “2D Plot Group 1 > Surface 1” nodeto access the surface plot dialog. Type “–rA” (replace “cA”) in the“Expression” edit field to plot the absolute rate of consumption of A(moles m–3 s–1).

(c) Increase the activation energy of the reaction by 5%. How do the concen-tration profiles change? Decrease? Instructions: Select the “Global Defi-nitions > Parameters” node. Multiply the value of “E” in the constantslist by 1.05 (just type “*1.05” behind the existing value to increase ormultiply by 0.95 to decrease). Go to the “Study 1” node and select“Compute.”

(d) Change the activation energy back to the original value. Instructions:Select the “Global Definitions > Parameters” node. Remove the factor“0.95” in the constants list.

(e) Increase the thermal conductivity, ke, by a factor of 10 and explain howthis change affects the temperature profiles. At what radial position doyou find the highest conversion? Instructions: Multiply the value of “ke”in the constants lists by 10. Go to the “Study 1” node and select “Compute.”

(f) Increase the coolant flow rate by a factor of 10 and explain how thischange affects the conversion.

(g) In two or three sentences, describe your findings when you varied theparameters (for all parts).

(h) What would be your recommendation to maximize the average outletconversion?

(i) Review Figure E15-1.2 and explain why the temperature profile goesthrough a maximum and why the conversion profile goes through a max-imum and a minimum.

(j) See other problems in the Web Module.



Chap. 15 Supplementary Reading 15-25

P15-5 If you have not installed COMSOL Multiphysics, request a trial version of thesoftware from the Web site http://www.comsol.com/ecre and follow the instal-lation instructions as outlined in P15-4.(a) Before running the program, sketch the radial temperature profile down a

PFR for (1) an exothermic reaction for a PFR with a cooling jacket and(2) an endothermic reaction for a PFR with a heating jacket.

(b) Run COMSOL Multiphysics and compare with your results in (a). Dou-ble-click on the COMSOL Multiphysics icon on your desktop. In theModel Library, select the model denoted “3-Non-Isothermal I” and pressOK. You can use this model to compare your results in (1) and (2),above. Click the Dynamic Help icon on the main toolbar to review theinstructions for this model and other models in COMSOL. Change thevelocity profile from laminar parabolic to plug flow. Select the “Model 1> Definitions > Variables 1” node. Change the expression for uz (thevelocity) to “u0” (replace the expression “2*u0*(1–(r/Ra)2)”, whichdescribes the parabolic velocity profile). You can now continue to varythe input data and change the exothermic reaction to an endothermic one.(Hint: Select the “Global Definitions > Parameters” node to access thelist of constants. Do not forget “Ta0” the jacket temperature at the end ofthe list.) To run a simulation, go to the “Study 1” node and select “Com-pute.” Write a paragraph describing your findings.

(c) The thermal conductivity in the reactor, denoted “ke” in Figure E15-1.1,is the molecular thermal conductivity for the solution. In a plug flowreactor, the flow is turbulent. In such a reactor, the apparent thermal con-ductivity is substantially larger than the molecular thermal conductivityof the fluid. Vary the value of the thermal conductivity “ke” to learn itsinfluence on the temperature and concentration profile in the reactor.

(d) In turbulent flow, the apparent diffusivity is substantially larger than themolecular diffusivity. Increase the molecular diffusivity in the PFR toreflect turbulent conditions and study the influence on the temperatureand concentration profiles. Here you can go to the extremes. Find some-thing interesting to turn in to your instructor. See other problems in theWeb Module.

(e) See other problems in the Web Module.

S U P P L E M E N T A R Y R E A D I N G

1. Partial differential equations describing axial and radial variations in temperatureand concentration in chemical reactors are developed in

BURGESS, THORNTON W., The Adventures of Prickly Porky. New York: DoverPublications, Inc., 1916.

BUTT, JOHN B., Reation Kinetics and Reactor Design, Second Edition, Revisedand Expanded. New York: Marcel Dekker, Inc., 1999.

FROMENT, G. F., AND K. B. BISCHOFF, Chemical Reactor Analysis and Design,2nd ed. New York: Wiley, 1990.

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Radial and Axial Tubular Reactor - University of Michigan