8.9 radial and axial variations in a tubular reactorelements/344/probsets/ps23/4thed-ch08p501.pdf8.9...

4th EditionChapter 8

4thEd/Ch8p501Section8.9.doc

Page 501A

8.9 Radial and Axial Variations in a Tubular ReactorIn the previous sections we have assumed that there were no radial

variations in velocity, concentration, temperature or reaction rate in the tubularand packed 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 inwhich case will require a partial differential (PDE) solver. A PDE solver such asFEMLab, will allow us to solve tubular reactor problems for both the axial andradial profiles.

Dr

ez, Fiz

ez, Fiz

er (Wir)

!Figure 8-25 Cylindrical shell of thickness Dr, length Dz and volume 2prDrDz

Molar FluxIn order to derive the governing equations we need to add a couple of

definitions. The first is the molar flux of species “i”, Wi mol/m2•s. The molarflux has 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 crosssectional areas normal to their direction of flow Acz. E.g., for species i flowing inthe z direction

Fiz = Wiz Aczwhere Wiz is the molar flux in the z direction (mol/m

2/s), and Acz (m2) is the

cross sectional area of the tubular reactor.In chapter 11 we discuss the molar fluxes in some detail, but for now let us

just say they consist of a diffusional component,

†

-De∂Ci∂z

Ê

Ë Á

ˆ

¯ ˜ , and a convective

flow component,

†

UzCi( )

†

Wiz = -De∂Ci∂z

+ UzCi (8-83)

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

B


†

Wir = -De∂Ci∂r

+ UrCi (8-84)

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. A mole balance on acylindrical system volume of length Dz and thickness Dr as shown in Figure 8-25gives

Mole Balance on Species A

†

Moles of A In at r

Ê Ë Á ˆ

¯ ˜ = WAr •

cross sectional areanormal to radial flux

Ê Ë Á ˆ

¯ ˜ = WAr •2prDz

†

Moles of AIn at z

Ê Ë Á ˆ

¯ ˜ = WAz •

cross sectional areanormal to axial flux

Ê Ë Á ˆ

¯ ˜ = WAz •2prDr

†

Moles of AIn at r

Ê Ë Á ˆ

¯ ˜ -

Moles of Aout at r + Dr( )

Ê Ë Á

ˆ ¯ ˜ + Moles of AIn at z

Ê Ë Á ˆ

¯ ˜ -

Moles of Aout at z + Dz( )

Ê Ë Á

ˆ ¯ ˜ + Moles of Aformed

Ê Ë Á ˆ

¯ ˜ = Moles of AAccumulated

Ê Ë Á ˆ

¯ ˜

WAr 2prDz r - WAr 2prDz r +Dr + WAr 2prDr z - WA2prDr z+Dz + rA2prDrDz =∂CA 2prDrDz( )

∂tDividing by 2prDrDz and taking the limit as Dr and Dz Æ 0

†

-1r

∂ rWAr( )∂r

-∂WAz

∂z+ rA =

∂CA∂t

Similarly for any species i and steady state conditions

†

-1r

∂ rWir( )∂r

-∂Wiz

∂z+ ri = 0 (8-85)

Using Equation (8-83) and (8-84) to substitute for Wiz and Wir in Equation(8-85) and then setting Ur!=!0 we obtain

†

-1r

∂∂r

-De∂Ci∂r

rÊ

Ë Á

ˆ

¯ ˜

È

Î Í

˘

˚ ˙ -

∂∂z

-De∂Ci∂z

+ UzCiÈ

Î Í ˘

˚ ˙ + ri = 0

For steady state conditions and assuming Uz does not vary in the axial direction.

†

De∂2Ci∂r 2

+Der

∂Ci∂r

+ De∂2Ci∂z2

- Uz∂Ci∂z

+ ri = 0 (8-86)

Energy FluxWhen we applied the first law of thermodynamics to a reactor to relate

either temperature and conversion or molar flow rates and concentration, wearrived at Equation (8-9). Neglecting the work term we have for steady stateconditions

C


†

˙ Q

Conduction}

+i=1

n

Â Fi0Hi0 -i=1Â FiHi

Convection6 7 4 4 8 4 4

= 0 (8-87)

In terms of the molar fluxes and the corresponding cross sectional area

†

Ac˙ Q

Ac+ ÂWi0Hi0 - ÂWiHi( )

È

Î Í

˘

˚ ˙ = 0 (8-88)

The

†

˙ Q term is the heat added to the system most always includes a conductioncomponent of some form. We now define an energy flux, e, (J/m2•s), to includeboth the conduction and convection of energy.

†

e = conduction + convection

†

e = q + ÂWiHi (8-89)

where the conduction term q

†

(kJ m 2 •s) is given by Fourier’s law. For axial andradial conduction Fourier’s laws are

†

qz = -ke∂T∂z

and

†

q r = -k e∂T∂r

where ke is the thermal conductivity (J/m•s•K). The energy transfer (flow) is theflux times the cross sectional area, Ac, normal to the energy flux

Energy flow = e • AcEnergy BalanceUsing the energy flux, e, to carry out an energy balance on our annulus (Figure8-25) with system volume 2prDrDz we have

†

Energy Flow In at r( ) = er Acr = er ⋅ 2prDz

Energy Flow In at z( ) = ezAcz = ez ⋅ 2prDr

†

Energy Flowin at r

Ê Ë Á ˆ

¯ ˜ - Energy Flowout at r + Dr

Ê Ë Á ˆ

¯ ˜ + Energy Flowin at z

Ê Ë Á ˆ

¯ ˜ - Energy Flowout at z + Dz

Ê Ë Á ˆ

¯ ˜ =

Accumulationof Energy in

Volume 2prDrDz

Ê

Ë Á Á

ˆ

¯ ˜ ˜

er 2prDz( ) r - er 2prDz( ) r +Dr + ez2prDr z - ez2prDr z+Dz = 0 (8-90)Dividing by 2prDrDz and taking the limit as Dr and Dz Æ 0

†

-1r

∂ re r( )∂r

-∂ez∂z

= 0 (8-91)

D


The radial and axial energy fluxes are

†

er = q r + ÂWiHi

ez = q z + ÂWizHiSubstituting for the energy fluxes, er and ez

†

-1r

∂ r q r + ÂWirHi[ ][ ]∂r

-∂ q z + ÂWizHi[ ]

∂z= 0 (8-92)

and expanding the convective energy fluxes,

†

ÂWiHi ,Neglect

radial:

†

1r

∂∂r

r ÂHiWir( ) = ÂHi1r

∂ rWir( )∂r

+ÂWir

r∂rHi

∂r(8-93)

axial:

†

∂ ÂWizHi( )∂z

= ÂHi∂Wiz

∂z+ ÂWiz

∂Hir∂z

(8-94)

Substituting Equations (8-93) and (8-94) into Equation (8-92), we obtain uponrearrangement

†

-1r

∂ rq r( )∂r

-∂q z∂z

- ÂHi1r

∂ rWir( )∂r

+∂Wiz

∂zÊ

Ë Á

ˆ

¯ ˜

ri6 7 4 4 4 8 4 4 4

- ÂWiz∂Hi∂z

= 0

Recognizing the term in brackets is related to Eqn (8-85) for steady stateconditions and is just the rate of formation of species i, ri. We have

†

-1r

∂∂r

rq r( ) -∂q z∂z

- ÂHiri - ÂWiz∂Hi∂z

= 0 (8-95)Recalling

†

q r = -k e∂T∂r

, q z = -k e∂T∂z

, ∂Hi∂z

= CPi∂T∂z

,

and

†

ri = ui -rA( )

†

Â riHi = Â u iHi -rA( ) = -DHRx rAwe have the energy in the form

†

k er

∂∂r

r∂T∂r

Ê

Ë Á

ˆ

¯ ˜

Ê

Ë Á

ˆ

¯ ˜ + k e

∂2T∂z2

+ DHRx rA - ÂWizCPi( )∂T∂z

= 0 (8-96)

Some initial approximationsAssumption 1.Neglect the diffusive term wrt the convective term in the expression involvingheat capacities

E


†

ÂCPi Wiz = ÂCPi 0 + UzCi( ) = ÂCPi CiUzwith this assumption Equation (8-96) becomes

†

k er

∂∂r

r∂T∂r

Ê

Ë Á

ˆ

¯ ˜ + k e

∂2T∂z2

+ DHRx rA - Uz ÂCPi Ci( )∂T∂z

= 0 (8-97)

Assumption 2.Assume that the sum

†

CPm = ÂCPiCi = CA0 ÂQiCPi is constant. The energy balancenow becomes

†

k e∂2T∂z2

+k er

∂∂r

r ∂T∂r

Ê

Ë Á

ˆ

¯ ˜ + DHRx rA - UzCPm

∂T∂z

= 0 (8-98)

Equation (8-98) is the form we will use in our FEMLab problem. In manyinstances the term

†

CPm is just the product of the solution density and the heatcapacity of the solution (kJ/kg•K).

Coolant BalanceWe also recall that a balance on the coolant gives the variation of coolanttemperature with axial distance where Uht is the overall heat transfer coefficient

†

˙ m cCPc∂Ta∂z

= Uht 2pR TR z( ) - Ta[ ] (8-99)

For laminar flow, the velocity profile is

†

Uz = 2Uo 1-rR

Ê

Ë Á

ˆ

¯ ˜

2Ê

Ë Á Á

ˆ

¯ ˜ ˜ (8-100)

where Uo is the average velocityBoundary and initial conditions

A. Initial conditions if other than steady state

†

t = 0 Ci = 0 , T = T0 , for z > 0 all rB. Boundary condition

(1) Radial

(a) At r = 0, we have symmetry

†

∂T∂r

= 0 and ∂Ci∂r

= 0

(b) At the tube wall r = R, the temperature flux to the wall on thereaction side equals the convective flux out of the reactor into theshell side of the heat exchanger.

†

-ke∂T∂r R

= U TR z( ) - Ta( )

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

†

∂Ci∂r

= 0 @ r = R

Energybalance with

radial andaxial

gradients

F


(d) At the entrance to the reactor z = 0

†

T = T0 and

†

Ci = Ci0(e) at the exit of the reactor z = L

†

∂T∂z

= 0 and ∂Ci∂z

= 0

The following example will solve the above equations using FEMLab. Forthe exothermic reaction with cooling, the expected profiles are

Example 8-12: Radial Effects In Tubular ReactorThis example will highlight the radial effects in a tubular reactor which up untilnow have been neglected to simplify the calculations. Now, the effects ofparameters such as inlet temperature and flow rate will be studied using a thesoftware program FEMLab.

We continue Example 8-8 which discussed the reaction of propylene oxide(A) reacts with water (B) to form propylene glycol (C). The hydrolysis ofpropylene oxide takes place readily at room temperature when catalyzed bysulfuric acid.

A + B Æ CThis exothermic reaction is approximated as a 1st order reaction given that thereaction takes place in an excess of water.

The CSTR from Example 8-8 has been replaced by a tubular reactor 1.6meter in length and 0.2 m in diameter.

The feed to the reactor consists of two streams: One stream is anequivolumetric mixture of propylene oxide and methanol and the other stream iswater containing 0.1 wt % sulfuric acid. The water is fed at a volumetric rate 2.5times larger than the propylene oxide-methanol feed. The molar flow rate ofpropylene oxide fed to the 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 isalready accounted for and the inlet temperature of both streams is set to 312 K.

The reaction rate law is

G


–rA!=!k!CAwith

k=Ae–E/RT

Where E = 75362 J/mol and A = 16.96!x!1012 s–1 which can also be put inthe form

†

k T( ) = k T0( )expER

1T0

-1T

Ê

Ë Á

ˆ

¯ ˜

È

Î Í

˘

˚ ˙

With k0!=!1.28 s–1 at 300 K. The thermal conductivity ke of the reaction mixture

and the diffusivity De are 0.599 (W/mK) and 10–9 m2/s respectively, and

assumed to be constant throughout the reactor. In the case where there is a heatexchange between the reactor and its surroundings, the overall heat-transfercoefficient is 1300 (W/ m2K) and the temperature of the cooling jacket isassumed to be constant and is set to 273K.

Physical dataPropylene oxide Methanol Water Propylene glycol

Molar weight (g/mol) 58.095 32.042 18 76.095Density (kg/m3) 830 791.3 1000 1040Heat capacity (J/mol•K) 146.54 81.095 75.36 192.59Heat of formation (J/mol) -154911.6 -286098 -525676

SolutionMole Balances. Recalling Equation 8-86 and applying it to species A

A:

†

De∂2CA∂r 2

+1r

De∂CA∂r

+ De∂2CA∂z2

- Uz∂CA∂z

+ rA = 0 (E8-12.1)

Rate law

†

-rA = k T1( )expER

1T1

-1T

Ê

Ë Á Á

ˆ

¯ ˜ ˜

È

Î Í Í

˘

˚ ˙ ˙

•CA (E8-12.2)

StoichiometryThe conversion along a streamline (r) at a distance z

X(r,z) = 1 – CA(r,z)/CA0 (E8-12.3)The overall conversion is

†

X (z) =1-2p CA0

RÚ (r,z)UzrdrFA0

(E8-12.4)

H


The mean concentration at any distance z

†

C A (z) =2p CA0

RÚ (r,z)rdrpR2

(E8-12.5)

For plug flow the velocity profile is

Uz = U0 (E8-12.6)The laminar flow velocity profile is

†

Uz = 2Uo 1 -rR

Ê

Ë Á

ˆ

¯ ˜

2Ê

Ë

Á Á Á

ˆ

¯

˜ ˜ ˜

(E8-12.7)

Recalling the Energy Balance

†

k e∂2T∂z2

+k er

∂∂r

r ∂T∂r

Ê

Ë Á

ˆ

¯ ˜ + DHRx rA - UzCPm

∂T∂z

= 0 (8-98)

Assumptions

1) Ur is zero.2) Neglect axial diffusion/dispersion flux w.r.t. convective flux when

summing the heat capacity times their fluxes.3) Steady State

Cooling jacket

†

˙ m CpidTadz

= Uht TR (z) - Ta( ) (8-99)

Boundary conditions

At

†

r = 0 then

†

∂Ci∂r

= 0 and

†

∂T∂r

= 0 (E8-12.8)

At

†

r = R then

†

∂Ci∂r

= 0 and

†

-k e∂T∂r

= Uht TR (z) - Ta( ) (E8-12.9)

At

†

z = 0 then

†

Ci = Ci0 and

†

T = T0 (E8-12.10)

These equations were solved using FEMLab for a number of cases includingadiabatic and non-adiabatic plug flow and laminar flow and with and withoutaxial and radial dispersion. A detailed accounting on how to change theparameter values in the FEMLab program can be found in the!FEMLab!Instructions!! section on the web. Color surfaces are used to show theconcentration and temperature profiles. Use the FEMLab Program on theCDROM to develop temperature concentration profiles similar to the onesshown below. Read through the FEMLab web module, “radial and axialtemperature gradients” before you run the program.

I


Define expressionFigure E8-12.1 FEMLab Screen Shot

(a) (b)

(c) (d)

Figure E8-12.2 Temperature surface (a) and profile (b) conversion surface (c)and profile (d)

J


One notes 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 CDROMin the web modules. One notes the maximum and minimum in these profiles.Near the wall the temperature of the mixture is lower because of the cold walltemperature. Consequently, the rate will be lower and thus the conversion willbe lower. However, right next to the wall the velocity through the reactor isalmost zero so the reactants spend a long time in the reactor and therefore agreater conversion is achieved as noted by the upturn right next to the wall.

8.10 The Practical SideScaling up exothermic chemical reactions can be very tricky. Tables 1 and

2 give reactions that have resulted in accidents and causes, respectively.†

!!!!!!

More information is given in the Summary Notes and Professional ReferenceShelf on the web. The use of the ARSST to detect potential problems will bediscussed in Chapter 9.

ClosureVirtually all reactions that are carried out in industry involve heat effects.

This chapter provides the basis to design reactors that operate at steady state andinvolve heat effects. To model these reactors we simply add another step to ouralgorithm, this step is the energy balance. Here it is important to understand howthe energy balance was applied to each reaction type so you will be able todescribe what would happen if you changed some of the operating conditions,e.g., T0. The living example problems (especially T3) and the ICM module willyou achieve a high level of understanding. Another major goal after studyingthis chapter, is to be able to design reactors that have multiple reactions takingplace under nonisotherm conditions. Try working problem 8-30 to be sure you † Ref. Singh, J. “Safe Scale Up of Exothermic Reactions,” Chemical Engineering, p92, May

1997 and Venugopal, B. “Avoiding Runaway Reactions,” Chemical Engineering, p54,June 2002.

K


have achieved this goal. An industrial example that provides a number ofpractical details is included as an appendix to this chapter. The last example ofthe chapter considers a tubular reactor that has both axial and radial gradients.As with the other living example problems, one should vary a number of theoperating parameters to get a feel of how the reactor behaves.

8.9 radial and axial variations in a tubular reactorelements/344/probsets/ps23/4thed-ch08p501.pdf8.9...

Documents