pressure drop in reactors calc

4.4 Pressure Drop in Reactors 153

4.4 Pressure Drop in ReactorsPressure drop

phase kineticscalculations

In liquid-phase reactions, the concentration of reactants is insignificantlyaffected by even relatively large changes in the total pressure. Consequently,we can totally ignore the effect of pressure drop on the rate of reaction when

liquid-phase chemical reactors. However, in gas-phase reactions, theconcentration of the reacting species is proportional to the total pressure andconsequently, proper accounting for the effects of pressure drop on the reactionsystem can, in many instances, be a key factor in the success or failure of thereactor operation.

4.4.1 Pressure Drop and the Rate Law

We now focus our attention on accounting for the pressure drop in therate law. For an ideal gas, the concentration of reacting species i is

gas-phase

important

reactions pressuredrop may be very (3-46)

For isothermal operation

18)

We now must determine the ratio as a function of volume V or the cata-lyst to account for pressure drop. We then can combine thetration, rate law, and design equation. However, whenever accounting for theeffects of pressure drop, the differential form the mole balance (designequation) must be used.

for example, the second-order reaction

is being carried out in a packed-bed reactor, the differential form of the molebalance equation in terms of catalyst weight is

When one

differential formsof the

(2-17)gram moles

gram catalyst mindesign equations I--

The law is

From stoichiometry for gas-phase reactions,

154 isothermal Reactor Design Chap. 4

Equationis needed.

Ergun equation

and the rate law can be written as

(4-20)

Note from Equation (4-20) that the larger the pressure drop the smaller P)from frictional losses, the the reaction rate!

Combining Equation with the mole balance (2-17) and assumingisothermal operation (T = gives

Dividing by yields

For isothermal operation (T = the right-hand side is a function of onlyconversion and pressure:

- P ) (4-21)

We now need to relate the pressure drop to the catalyst weight in order todetermine the conversion as a function of catalyst weight.

4.4.2 Flow Through a Packed Bed

The majority of gas-phase reactions are catalyzed by passing the reactantthrough a packed bed of catalyst particles. The used most to calculatepressure drop in a packed porous bed is the Ergun

R.B. Bird, W.E.Stewart,and E.N.Lightfoot, Phenomena (NewYork:Wiley, p. 200.


where P = pressure,

volume of voidtotal bed volume

= porosity =

volume oftotal bed volume

g, = 32.174 lb, . factor)

4.17 X lb,

(recall that for the metric system = 1.0)

= diameter of particle in the bed, ft

= viscosity of gas passing through the bed, h

= length down the packed bed of pipe, ft

= superficial velocity = volumetric flow cross-sectional

p = gas density,

= = superficial mass velocity, s) or h)

area of pipe,

--

In calculating the pressure drop using the Ergun equation, the onlyparameter that varies with pressure on the right-hand side of Equation (4-22)is the gas density, p. We are now going to calculate the pressure drop throughthe bed.

Because the reactor is operated at steady state, the mass flow rate at anypoint down the reactor, is equal to the entering mass flow rate,

equation of continuity),

= m

=

Recalling Equation we have

Equations and (4-23) gives

56 Isothermal Reactor Design Chap. 4

Simplifying yields

I

where

(4-24)

(4-25)

For tubular packed-bed reactors we are more interested in catalyst weightrather than the distance down the reactor. The catalyst weight up to a dis-tance of down the reactor is

W (1

weight of volume of density of

catalyst solids (4-26)

where A, is the cross-sectional area. bulk density of the (massof catalyst per volume of reactor bed), is just the product of the solid density,

, the fraction of solids, - :

Using the relationship between and [Equation we can changeour variables to express the Ergun equation in terms of catalyst weight:

Use this form formultiple reactions

and membranereactors

(4-27)

Further simplification yields

(4-28)

where

(4-29)

Pressure Drop in Reactors 157

Differential formof equation

for the pressuredrop in packed

beds

Two coupledequations to be

solved numerically

Equation will be the one we use when multiple reactions areoccurring or when there is pressure drop in a membrane reactor. However, forsingle reactions in packed-bed reactors it is more convenient to express theErgun equation in terms of the conversion X . Recalling Equation (3-42) for

+ F,, +

and development leading to Equation

-=

where, as before,

Equation (4-28) can now be written as

(1 + EX)- _- - _-2 To PI P,

(3-42)

(4-30)

We note that when is negative the pressure drop will be lesshigher pressure) than that for E = When E is positive, the pressure drop APwill greater than when E = 0.

For isothermal operation, Equation (4-30) is only a function of conver-sion pressure:

Recalling Equation

- (4-21)

we see that we have two coupled first-order differential equations, (4-31) andthat be solved simultaneously. A variety of software packages and

numerical integration schemes are for this purpose.

Analytical Solution. If E 0, or if we can neglect ( E X ) with respect i o 11 can obtain an analytical solution to Equation (4-30) for iso-

thermal operation (Le., T isothermal operation with E = 0, Equa-tion becomes

158 Isothermal Reactor Design

Isothermal with

Rearranging gives us

---- -a

Taking inside the derivative, we have

Integrating with P = at W = 0 yields

2

Taking the square root of both sides gives

I IPressure ratio P

only for E 0

where again

Chap. 4

(4-32)

(4-33)

Equation (4-33) can be used to substitute for the pressure in the rate law, inwhich case the mole balance can be written solely as a function of conversionand catalyst weight. The resulting equation can readily be solved either analyt-ically or numerically.

If we wish to express the pressure in terms of reactor length we can useEquation (4-26)to substitute for Win Equation (4-33).Then

(4-34)

Example 4-5 Calculating Pressure Drop in a Packed Bed

Calculate the pressure drop in a 60 ft length of 1 schedule 40 pipe packedwith catalyst pellets in diameter when 104.4 of gas is passingthe bed, The temperature is constant along the length of pipe at 260°C. The voidfraction is 45% and the properties of the gas are similar of at this tem-perature. The entering pressure is 10 atm.

Evaluatingthe pressure drop

parameters

4.4 Pressure Drop in Reactors

At the end of the reactor z L and Equation (4-34) becomes

For -in. schedule 40 pipe, A, 0.01414

104.47383.3-

0.01414 h

For air at and atm,

0.0673 h

= 0.413

From the problem statement,

= 0.0208ft

lb4.17 X

Substituting the values above into Equation (4-25)gives us

159

17383.3 1 -0.45)

-h)0.0208ft

0.01244-(266.9+ 12,920.8)-ft h

1 1 atm144 14.7

164.1 X-x

atmm

0.0775- 25.8-

ft3


W

W

Onlyfor

(E4-5.5)2 0.0775 X 60 ftatm

P = = 2.65 atm

AP = P = = 7.35 atm(E4-5.6)

Reaction with Pressure Drop

Analytical solution: Now that we have expressed pressure as a functionof catalyst weight [Equation we can return to the second-order isother-mal reaction,

to relate conversion and catalyst weight. Recall our mole balance, rate law, and

Mole balance: - (2-17)

Rate law: = (4-19)

Stoichiometry. Gas-phase isothermal reaction with E 0 :

P= - (4-35)

Using Equation (4-33) to substitute for PIP, in terms of the catalyst weight,we obtain

=

( 1 - [ ( I -Combining: - =-

Separating variables: --

Integrating with limits X = 0 when W 0 and substituting for =yields

Solving for conversion gives

Pressure Drop in Reactors

1- { I - -

161

Catalyst weight forsecond-order

reaction inwith AP

The economics

I

Solving for the catalyst weight, we have

(4-36)

We now proceed (Example 4-6) to combine pressure drop with reaction in apacked bed for the case where we will that EX 1in the Ergun equationbut in the rate law in order to obtain an analytical solution. Example 4-7removes this assumption and solves Equations (4-21) and (4-31) numerically.

Example Calculating X in a Reactor with Pressure Drop

7 billion pounds of ethylene oxide were produced in the UnitedStates in 1997.The 1997 selling price was $0.58 a pound, amounting to a commer-cial value of $4.0 billion. Over 60% of the ethylene oxide produced is used to makeethylene glycol. The major end uses of ethylene oxide are antifreeze polyes-ter surfactants and solvents (5%).We want to calculate the catalystweight necessary to achieve 60% conversion when ethylene oxide is to be made bythe vapor-phase catalytic oxidation of ethylene with air.

+A + f B C

Ethylene and oxygenare fed in stoichiometricproportionsto apacked-bed reactor operated isothermally at 260°C.Ethylene is fed at a rate of 0.30 at a pressure of 10 atm. It is proposed to use 10 banks of 1 -in.-diameter schedule 40packed with catalyst with 100 tubes per bank. Consequently,the molar flow rate toeach tube is to be 3 The properties of the reacting fluid are beconsidered identical to those of air at this temperature and pressure. The density ofthe -in.-catalystparticles is 120 and thebed void fraction is 0.45.Therate law is

= lb cat. h

with3

at 260°Clb mol

atm . cat.. hk = 0.0141

Chem., 45, 234 (1953).

The algorithm

We evaluatethecombinestepeither

1) Analytically2) Graphically3) Numerically,or4) Using software

is valid

orE = 0

162 isothermal Reactor Design

Solution

1. mole balance:

2. Rate law:113 213

=

1/3

3. Stoichiometry.Gas-phase, isothermal +

Combiningthe rate law and concentrations:

For stoichiometric feed, :

Chap. 4

(E4-6.1)

(E4-6.2)

(E4-6.4)

(E4-6.5)

(E4-6.7)

(E4-6.8)

where k’ =

between and W when 1 is5. Developing the design equation. For a packed-bed reactor, the relationship

-PO

(4-33)

(E4-6.9)

Combining Equations and (E4-6. we have


Evaluating thedrop

parameters

Separatingvariables to form the integrals yields

Integrating gives us

Solving for we obtain

(E4-6.11)

6. Parameter evaluation per tube divide feed rates by 1000):

Ethylene:

Oxygen:

I inerts

= 3 X

= 1.5 X

lb = 1.08lb

lb = 0.54 lb

0.79 mol= 1.5 X lb X

= 5.64 X = 2.03

Summing: + + 3.65

E 1)

atm

lb mol3 atm 0.63 = 0.0266-cat

k' = = 0.0141

1- [ 1 -0.15)a

For 60% conversion, Equation (E4-6.11) becomes

1 (1-(E4-6.12)

In order to calculate

= -

we need the superficial mass velocity, G.The mass flow rates of each enteringspecies are:

Neglectingpressure drop

results in poordesign (here 53%

vs. 60%

164 Isothermal Reactor Design Chap.

mol-X 28- 30.24

- 17.28

-X 28 56.84

h mol

molh

molh lb mol

0.54

The total mass flow rate is

h104.4-

This is essentially the same superficial mass velocity, temperature, and pres-sure as in Example 4-5. Consequently, we can use the value of calculatedin Example 4-5.

atm0.0775-

- (0.01414

0.0166cat

_-

Substituting into Equation (E4-6.12) yields

lb mol

cat

45.4 lb of catalystper tubeor 45,400 lb of catalyst total

This catalyst weight corresponds to a pressure drop of approximately 5 atm.If we had neglected pressure drop, the result would have been

1

1k

1-0.60.0266

= 35.3 of catalystper tube (neglectingpressure drop) ,

and we would have had insufficient catalyst to achieve the desiredSubstitutingthis catalyst weight 35,300 total) into Equation (E4-6.10)gives a conversion of only 53%.

4.4 Pressure Drop in Reactors

Example 4- Pressure Drop with Reaction-Numerical

Rework Example for the case where volume change is norErgun equation and the two coupled differential equations

conversion and are solved

SolutionI

Rather than rederive everything starting theetry, and pressure drop equations, we will use th4-6 Combining (E4 and

Program examplesPOLYMATH,

MatLab can beloaded from the

CD-ROM (seethe Introduction)

Next. we

.3)

For the reaction conditions in Example 4-6, we theW 0, 0, and y 1.0 and the parameter values cat,

E -0.15, k' 0.0266 cat, and 1.08large number of ordinary differential equation solver software

ODE solvers) are extremely user friendly have become available. We shalluse POLYMATH4 to solve the examples in the text. However, the CD-ROMcontains an example that uses ASPEN, as as all the MATLAB andMATH solution programs to the example With POLYMATH simplyenters Equations (E4-7.3) and (E4-7.4) and the corresponding parameter value intothe computer (Table with the (rather, boundary) conditions theyare solved and displayed as shown Figure E4-7.1,

We note that neglecting E X the Ergun equation in Example-0.09) to obtain an solution resulted in less than a 10% error.

Developed by Professor M. Cutlip of the University of Connecticut. and Professor M.Shacham of Ben Gurion University. Available from the Corporation,Box 7939, Austin, TX 78713.

166 Reactor Design Chap. 4

TABLE POLYMATH SCREEN SHOWING EQUATIONSTYPED

IN AND READY TO BE SOLVED.

Equations Initial Values

15

0266

f = 60= a , w

1

0

TScale:

0.000 oooU

Figure E4-7.1 Reaction rate profile down the PBR.

However, larger errors will result if large values of EX are neglected! By taking intoaccount the change in the volumetric flow rate E = -0.15) in the pressuredrop term, we see that 44.0 lb of catalyst is required per tube as opposed to 45.4lbwhen E was neglected in the analytical solution, Equation (E4-7.4).Why was lesscatalyst required when was not neglected in Equation The thatthe numerical solution accounts for the fact that the pressure drop will be lessbecause E is negative.


Volumetric flowrate increases

with increasingpressure drop

of added

It is also interesting to learn what happens to the volumetric flow rate along the length of the reactor. Recalling Equation

= ---- (3-44)

let f be the ratio of the volumetric flow rate, to the entering volumetric flowrate, at any point down the reactor. For isothermal operation Equation (3-44)becomes

E4-7.2 shows X, y y = andf down the length of the reactor. Wesee that both the conversion and the volumetric flow increase along the length of thereactor while the pressure decreases. For gas-phase reactions with orders greaterthan zero, this decrease in pressure will cause the reaction rate to be less than in thecase of no pressure drop.

4.000

3.200

2.400

1.600

0.800

0.000

W

Figure Output in graphical from POLYMATH

We note from Figure that the catalyst weight necessary to raisethe conversion the last 1%from 65% to 66% (3.5 lb) is 8.5 times more than that (0.41 lb) required to raise the conversion 1% at the reactor's entrance.Also, during the last 5% increase in conversion, the pressure decreases from 3.8 atm to 2.3 atm.

conversion


4.4.3 Spherical Packed-Bed Reactors

When small catalyst pellets are required, the pressure drop can be signif-icant. In 4-6 we saw that significant design flaws can result if pressuredrop is or if steps are not taken to minimize pressure drop. One typeof reactor that minimizes pressure drop and is also inexpensive to build is thespherical reactor, shown in Figure 4-8. In this reactor, called an ultraformer,dehydrogenation reactions such as

paraffin aromatic t

are carried out.

Figure 4-8 Reactor. (Courtesy of Amoco PetroleumProducts.) This reactor one in a series of SI X used by Amoco for reformingpetroleum naphtha. by K. R Sr.

Another advantage of spherical reactors that they are the most eco-nomical shape for high pressures. As a first approximation we will assume thatthe fluid moves down through the reactor in plug Consequently, because


Spherical reactorcatalyst weight

of the increase in cross-sectional area, A,, as the fluid enters the sphere, thesuperficial velocity, G = will decrease. From the Ergun equation[Equation

we that by decreasing G, the pressure drop he reduced significantly,resulting in higher conversions.

Because the cross-sectional area of the reactor is small near the inlet andoutlet, the presence of catalyst there would cause substantial pressure drop;thereby reducing the efficiency of the spherical reactor. To solve this problem,screens to hold the catalyst are placed near the reactor entrance and (Fig-ures 4-9 and 4-10). Were is the location of the screen from the center of the

Feed I

+

Products axis

Figure 4-9 Schematic drawing the inside Figure 4-10 Coordinate system andof a reactor. variables used with a spherical reactor. The

initial and final integration values are slhownas and

reactor. We can use elementary geometry and integral calculus to derive thefollowing expressions for cross-sectional area and catalyst weight as a functionof the variables defined in Figure 4-10:

A,

By using these formulas and the standard pressure drop algorithm, one cana variety of spherical reactor prablems. Note that Equations and


(4-39) make use of L and not L'. Thus, one does not need to adjust these for-mulas to treat spherical reactors that have different amounts of empty space atthe entrance and exit L L ' ) . Only the upper limit of integration needsto be changed, = L + .

Example 4-8 Dehydrogenation Reactions in a Spherical Reactor

Reforming reactors are used to increase the octane number of petroleum. In areforming process 20,000 barrels of petroleum are to be processed per day. The cor-responding mass and molar feed rates are 44 and 440 molls, respectively. In thereformer, dehydrogenationreactions such as

paraffin olefin

occur. The reactionis first-orderin paraffin.Assumethat pure paraffin enters the reac-tor at a pressure of 2000 and a corresponding concentration of 0.32Comparethe and conversion when this reaction is carried out in a tubu-lar packed bed 2.4 in diameter and 25 m in length with that of a sphericalpackedbed 6 m in diameter. The catalyst weight is the same in each reactor, 173,870 kg.

k'

Additional information:

= 0.032

= 0.02

= L' = 27

= 0.4

.= 2.6

Solution

We begin by performing mole balance over the cylindrical core of thickness Azshown in Figure

Figure E4-8.1 Spherical reactor.


Followingthe algorithm

(E4-8.2)

The equations inboxes are the key

equations usedin the ODE solver

program

I . Mole balance:

In -out generation = 0

Dividing by and taking the limit as 0 yields

In terms of conversion

(E4-8.1)

(E4-8.3)

I

x (1+ 1- = 1 (E4-8.4)

where

P

that ( y with a subscript) represents the mole fraction and y alone representsthe pressure ratio,

The variation in the dimensionless pressure, y, is given by incorporating thevariable y in Equation (4-24):

The units of for this problem are

For a reactor

=

(E4-8.7)

(E4-8.8)

(E4-8.10)

172 Isothermal Reactor Design Chap.

A comparisonbetween reactors

ParameterRecall that = 1 for metric units.

-0.4)

X

0.02

(E4-8.11)

= + (25,630 X (0.01

I

The last term in,brackets converts s) to Recalling other param-eters, m 44 L = 27 dm, R = 30 dm, and = 2.6

Table E4-8.1 shows the POLYMATH input used to solve the above equations.The MATLAB program is given as a living example problem on the CD-ROM.

TABLE POLYMATH PROGRAM

Initial ValuesEquations

01

.02

rhoca

1-phi)

10.015-phi)

= 0, 54

For the spherical reactor, the conversion and the pressure at the exit are

X 0.81 = 1980

If similar calculations are performed for the tubular packed-bed reactor (PBR), onefinds that for the same catalyst weight the conversion and pressure at the exit are

X = 0.71 = 308

Figure E4-8.2 shows how conversion, and pressure, vary withcatalyst weight in each reactor. Here and represent the tubular reactor and

4.4 Pressure Drop Reactors 173

IKEY:

y2

0.800

0.400

n. 200

oao0. 1.200 1.600

Figure Pressure and conversion for: tubular PBR; 2, spherical PBR.

and represent the reactor In addition to the higher thespherical reactor has the economic benefit of reducing the and compres-sion cost because of higher at the exit

Because pressure drop in the spherical reactor is very small, onecould increase the reactant flow rate significantly and still maintain adequatepressure at the exit. In fact, Amoco uses a reactor with similar specifications toprocess 60,000 barrels of petroleum naphtha per day.

Pressure Drop in Pipes

Wormally, the pressure drop for gases flowing through pipes withoutpacking can be neglected. For flow in pipes, the pressure drop along the lengthof the pipe is given by

du(4-40)

where D pipe diameter, cm

= average velocity of gas,

= Fanning friction factor

G =4

The friction factor is a function of the Reynolds number and pipe roughness.The mass velocity G is constant along the length of the pipe. Replacing with

and combining with Equation for the case of constant T andEquation (4-40) becomes

P

��

pressure drop in reactors calc

Documents