Flow Through a Packed Bed

5. Parameter evaiuation

4.4 Pressure Drop in Reactors

4.4.2 Flow Through a Packed Bed

The

majority of gas-phase reactions are catalyzed by passing the reactant through a packed bed of catalyst particles. The e q u a ~ o n used most to calculate pressure drop in a packed porous bed is the Ergun equation:2

R. _{B. Bird,} W. E. Stewart, and E. N. Lightfoot, Transport Phenomena (New York:

Wiley, 1960), p. 200.

Sec. 4.4 Pressure Drop in Reactors 155

where P ⁼ pressure, Ib/ft2

volume of void total bed volume

+

^{= porosity =}

l - + , = volume of polid total bed volume

g, = 32.174 lb,

.

ft/s2. lb,(conversion factor)

= 4.17 ^X lo8 lb, *ft/h2 lbf

(recall that for the metric system g , = 1.0) D p = diameter of particle in the bed, ft

p = viscosity of gas passing through the bed, lb,/ft h

z

= length down the packed bed of pipe, ft

u = superficial velocity = volumetric flow ^f cross-sectional p = gas density, Ib/ft3

G = pu = superficial mass velocity, (g/cm2. s) or (lb,/ft2 ⁸ h) area of pipe, ft/h

--

In calculating the pressure drop using the Ergun equation, the only parameter 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 through the bed.

Because the reactor is operated at steady state, the mass flow rate at any point down the reactor, riz (kg/s), is equal to the entering mass flow rate, rizo (i.e., equation of continuity),

m o = m

POVO = P V

Recalling Equation (3-41), we have

(3-41)

(4-23) Combhing Equations (4-22) and (4-23) gives

1 56 Isothermal Reactor Design Chap. 4

Simplifying yields

1 I

where

(4-24)

(4-25) For tubular packed-bed reactors we are more interested in catalyst weight rather than the distance z down the reactor. The catalyst weight up to a dis- tance of z down the reactor is

W

= (1 -+)A,z X Pc

weight of volume of density of

[

^catalyst

]

⁼

[

^solids [ s d i d c a t a l y s ~ (4-26) where A, is the cross-sectional area. Tbe bulk density of the catdyst, p b (mass of catalyst per volume of reactor bed), is just the product of the solid density,

pc , the fraction of solids, (1

-

^$) :

P b = pc (l -

4)

Using the relationship between z and W [Equation (4-26)] we can change our variables to express the Ergun equation in terms of catalyst weight:

Use this form for multiple reactions and membrane reactors

(4-27) dP

Further simplification yields Lety=P/Po

(4-28)

where

(4-29)

Sec. ,4.4 Pressure Drop in Reactors 157

Differential form of Esgun equation for the pressure drop in packed beds

Two coupled equations to be solved numerically

Equation (4-28) will be the one we use when multiple reactions are occurring or when there is pressure drop in a membrane reactor. However, for single reactions in packed-bed reactors it is more convenient to express the Ergun equation in terms of the conversion X . Recalling Equation (3-42) for F7,

F , = F,,

+

F,$X = F,, 1

+

6 X

FTo

i

and the development leading to Equation (3-43),

FT

Fro

-

^{= 1 + E X}

where, as before,

Equation (4-28) can now be written as

( 1

+

^EX)

d P - CY T Po

- _ - - _ -

dW 2 To P I P ,

1

(3-42)

(4-30) We note that when ⁶ is negative the pressure drop AP will be less (Le., higher pressure) than that for ^E = 0. When E is positive, the pressure drop AP will be greater than when E = 0.

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

Recalling Equation (4-21), d X

dW - = F , ( X , P )

(4-31)

(4-21) we see that we have two coupled first-order differential equations, (4-31) and (4-21), that must be solved simultaneously. A variety of software packages and numerical integration schemes are availabIe for this purpose.

Analytical Solution. If E = 0, or if we can neglect (EX ) with respect i o 1 .O

&e., 1 % - E X ) , we can obtain an analytical solution to Equation (4-30) for iso- thermal operation (Le., T = To). For isothermal operation with E = 0, Equa- tion (4-30) becomes

158 Isothermal Reactor Design

Isothermal with

_ _

dP - ^-- aP0

dW 2 ( P / P o )

E = O

Rearranging gives us

---

^{- - a}

2P d(PIP0) Po dW Taking PIPo inside the derivative, we have

- a d(P/Po)’ - -

dW Integrating with P = Po at W = 0 yields

(1

2 = 1 - a W Taking the square root of both sides gives

I I

Pressure 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, in which case the mole balance can be written solely as a function of conversion and 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 z, we can use Equation (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 1/2-in. schedule 40 pipe packed with catalyst pellets 114-h. in diameter when 104.4 lb/h of gas is passing thzough the bed, The temperature is constant along the length of pipe at 260°C. The void fraction is 45% and the properties of the gas are similar to&ose of air at this tem- perature. The entering pressure is 10 atm.

Evaluating the pressure drop parameters

Sec. 4.4 Pressure Drop in Reactors Soltition

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

For 1; -in. schedule 40 pipe, A, = 0.01414 ft2:

104.4 lb,/h 1bm

= 7383.3

-

0.01414 ft2 h ^* ft2 G =

For air at 260°C and 10 atm,

p = 0.0673 lb,/ft. h po = 0.413 lb,/ft3 From the problem statement,

D, = !in. = 0.0208 ft lb aft Ib,

-

^h2

g , = 4.17 X lo8

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

-

1 59

(E4-5.1)

(4-25)

(E4-5.2)

1

7383.3 lb,/ft2. h( 1 - 0.45)

= l(4.17 X lo8 Ib,*ft/lb,. h2)(04413 lb,/ft3)(0.0208 ft)(O.45)31

(E4-5.3)

+

1.75(7383.3)

-

0.45)(0.0673 lb,/ft

-

^h)

0.0208 ft

1brn 1bf.h

= 0.01244

-

^(266.9

⁺

^12,920.8)

-

^{= 164.}

ft Ib, ft2 h

lb, 1 ft2 1 atm ft3 144 in.2 14.7 lbf/in.2

= 164.1

-

^X

-

^x

atm kPa

ft m

= 0.0775

-

⁼ 25.8

-

'bf

-

ft3

(E4-5.4)

160 Isothermal Reactor Design Chap. 4

a

W

I W

Only for

E = O

(E4-5.5) 2 X 0.0775 a t d f t X 60 ft

10 atm P = 0.265Po = 2.65 atm

AP = P o - P = 10-2.65 = 7.35 atm ^(E4-5.6)

Reaction with Pressure Drop

Analytical solution: Now that we have expressed pressure as a function of catalyst weight [Equation (4-33)] we can return to the second-order isother- mal reaction,

A - B

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

Mole balance: FAO

dW

dX - - -r; (2-17)

Rate law: -r; = k C i (4-19)

Stoichiometry. Gas-phase isothermal reaction with E = 0 :

c,

^{= CA0(l} - X ) Po

-

P ^(4-35)

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

c,

= CA0(l -X ) ( 1 - a W ) ’ D ( 1 - X ) 2 [(I - ff W)”2]2

Combining: - = - dX kC10 dW ^FAO

Separating variables: 7

-

dx - ( l - a W ) d W

-

kCAo (1 -

x)’

Integrating with limits

X

= 0 when W = 0 and substituting for FAo = CAou, yields

Solving for conversion gives

Sec. 4..4 Pressure Drop in Reactors

1 - { I - [(2uoff)/kC,,][X/(1 - X)]}”Z W =

o!

161

(4-37)

Catalyst weight for second-order reaction in PER

with AP

The economics

X =

I

Solving for the catalyst weight, we have

(4-36)

We now proceed (Example 4-6) to combine pressure drop with reaction in a packed bed for the case where we will assunle that EX 4 1 in the Ergun equation but nolt in the rate law in order to obtain an analytical solution. Example 4-7 removes this assumption and solves Equations (4-21 ) and (4-3 1) numerically.

Example 4 6 Calculating X in a Reactor with Pressure Drop

Approxim.*tely 7 billion pounds of ethylene oxide were produced ⁱⁿ the United States 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 make ethylene glycol. The major end uses of ethylene oxide are antifreeze (30%), polyes- ter (30%), surfactants (lo%), and solvents (5%). We want to calculate the catalyst weight necessary to achieve 60% conversion when ethylene oxide is to be made by the vapor-phase catalytic oxidation of ethylene with air.

C2H4

+

^{i 0 2 _ _ _ f} CH2-CH, /O\

A + f B ---+ C

Ethylene and oxygen are fed in stoichiometric proportions to a packed-bed reac- tor operated isothermally at 260°C. Ethylene is fed at a rate of 0.30 Ib mol/s at a pres- sure of 10 atm. It is proposed to use 10 banks of 1 f -in.-diameter schedule 40 tubes packed with catalyst with 100 tubes per bank. Consequently, the molar flow rate to each tube is to be 3 X Ib mol/s. The properties of the reacting fluid are to be considered identical to those of air at this temperature and pressure. The density of the

a

-in.-catalyst particles is 120 lb/ft3 and the bed void fraction is 0.45. The rate law is

113 3 3

-r6, = k P , P, lb molflb cat. h with3

at 260°C lb mol

atm

.

Ib cat.. h k = 0.0141

Ind. Eng. Chem., 45, 234 (1953).

- ^_I--- ^{-- -} -

- -

^~

The algorithm

We can evaluate the combine step either 1) Analytically 2) Graphically 3) Numerically, or 4) Using software

Eq, (4-33) is valid or ^{e x 4 1} oalyif E = 0

162 isothermal Reactor Design

Solution

1. biffereatial mole balance: