Mass balance in a fixed bed reactor is similar to that of a plug-flow reactor (eq. 1.1):

(2.1)

Recalling dW = B dV, then

0A

A

F

r

dV

dX (1.1)

0AB

A

F

r

dW

dX

For a reaction:

A + bB cC + dD

Species Feed Rate Change within Reactor

Effluent Rate

A FA0 = FA0 – FA0 X FA = FA0 (1 – X)

B FB0 = B FA0 – b FA0 X FB = FA0 (B – bX)

C FC0 = C FA0 c FA0 X FC = FA0 (C + cX)

D FD0 = D FA0 d FA0 X FD = FA0 (D + dX)

I FI0 = I FA0 FI = FI0 I

TOTAL EFFLUENT

FT = FT0 + (d + c – b – 1) FA0 X = FT0 + FA0 X

For the above reaction, (1 + b) mol of reactants give (c + d) mol of product.

In flow systems where this type of reaction occurs, the molar flow rate will be changing as the reaction progresses.

Because only equal numbers of moles occupy equal volumes in the gas phase at the same temperature and pressure, the volumetric flow rate will also change.

Assuming ideal gas behavior:

00T

00

T RTN

VP

RTN

PV

(2.2) 0

0T

T

0

0 VN

N

T

T

P

PV

The above equation is further simplified by letting:

Xy1N

XNN

N

N0A

0T

0A0T

0T

T

Recalling the last row in the previous table:

XNNN 0A0TT

(2.3)

reactorthetofedmolesofnumbertotal

conversioncompleteformolesofnumbertotalinchange

0A

0T

0A yN

N1bcd

Equation (2.2) now becomes:

0

0

0 VX1T

T

P

PV

(2.4)

FT = FT0 + FA0 X

Kalau reaksinya berlangsung sempurna (X = 1):

0A0T0A0TT NNXNN*N

Perubahan jumlah mol kalau reaksinya berlangsungsempurna (X = 1):

0T0A0T0TT NNNN*N

0A0A N1bcdN

0A

0T

0A yN

N1bcd

To derive the concentrations of the species in terms of conversion for a variable-volume flow system, we shall use the relationships for the total concentration.

The total concentration at any point in the reactor is

RT

P

V

N

v

FC TT

T

At the entrance of the reactor:

0

0

0

0T

0

0T0T RT

P

V

N

v

FC (2.6)

(2.5)

Taking the ratio of eq. (2.6) to eq. (2.5), we have upon rearrangement

0

0T

T

0

0 vF

F

T

T

P

Pv

From the above table:

XFFF 0A0TT

(2.7)

Substituting eq. (2.8) in eq. (2.7):

(2.8)

X

F

F1

T

T

P

Pv

F

XFF

T

T

P

Pvv

0T

0A

0

00

0T

0A0T

0

00

X

F

F1

T

T

P

Pvv

0T

0A

0

00

Xy1T

T

P

Pv 0A

0

00

X1T

T

P

Pvv

0

00

(2.9)

We can now express the concentration of species j for a flow system in terms of conversion:

T

T

P

P

F

F

v

F

T

T

P

P

F

Fv

F

v

FC 0

0T

j

0

0T

0

0

0T

T0

jj

j

T

T

P

P

F

FCC 0

0T

j0Tj (2.10)

Substituting F and FT in terms of conversion in eq. (2.10) yields

T

T

P

P

XFF

XFCC 0

00A0T

jj0A

0Tj

T

T

P

P

XFF1

X

F

FC 0

00T0A

jj

0T

0A0T

T

T

P

P

Xy1

XyC 0

00A

jj

0A0T

T

T

P

P

X1

XCC 0

0

jj

0Aj (2.11)

Energy balance in a fixed bed reactor is similar to that of a plug-flow reactor (eq. 1.56):

piip0A

RxAa

CCXF

THrTTUa

dV

dT(1.56)

piip0AB

RxAa

CCXF

THrTTUa

dW

dT

Recalling dW = B dV, then

(2.12)

The majority of gas-phase reactions are catalyzed by passing the reactant through a packed bed of catalyst particles.

We now must determine the ratio P/P0 as a function of volume V or the catalyst weight, W, to account for pressure drop.

We then can combine the concentration, rate law, and design equation. However, whenever accounting for the effects of pressure drop, the differential form of the mole balance (design equation) must be used.

The equation used most to calculate pressure drop in a packed porous bed is the Ergun equation:

G75.1

D

11501

Dg

G

dz

dP

p

3

pc

WhereP : pressure, lb/ft2

: porosity (volume of void / total bed volume)Dp : diameter of particle in the bed, ft : viscosity of gas passing through the bed, lbm/ft hz : length down the packed bed of pipe, ftu : superficial velocity = volumetric flow : cross sectional

area of pipe, ft/h : gas density, lb/ft3

G : u = superficial mass velocity, (g/cm2 s) or (lbm/ft2 h)

(2.13)

In calculating the pressure drop using the Ergun equation, the only parameter that varies with pressure on the right-hand side of eq. (2.13) is the gas density, .

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, (kg/s), is equal to the entering mass flow rate, (i.e., equation of continuity),

m

0m

mm0

vv00

Recalling eq. (2.9), we have

X1

1

T

T

P

P

v

v 0

0

00

0(2.14)

Combining eqs. (2.13) and (2.14) gives:

X1

T

T

P

PG75.1

D

1150

Dg

1G

dz

dP

0

0

p

3

pc0

X1T

T

P

Pvv

0

00

(2.9)

Simplifying yields:

X1T

T

P

P

dz

dP

0

00

G75.1

D

1150

Dg

1G

p

3

pc0

0

where:

(2.16)

(2.15)

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 distance of z down the reactor is:

catalystsolid

ofdensity

solids

ofvolume

catalyst

ofweight

CczA1W

where Ac is the cross-sectional area. The bulk density of the catalyst, B (mass of catalyst per volume of reactor bed), is just the product of the solid density, C , the fraction of solids, (1 – ) :

1CB

(2.17)

(2.18)

Using the relationship between z and W [eq. (2.17)] we can change our equation (2.15) into:

X1

T

T

P

P

1AdW

dP

0

0

Cc

0

(2.19)

Further simplification yields:

where:

X1T

T

PP

P

2dW

dP

00

0

0Cc

0

P1A

2

(2.21)

(2.20)

X1T

T

P

P

dz

dP

0

00

(2.15)

We note that:

when is negative, the pressure drop P will be less than that for = 0.

When is positive, the pressure drop P will be greater than when = 0.

(2.1)0AB

A

F

r

dW

dX

piip0AB

RxAa

CCXF

THrTTUa

dW

dT (2.12)

X1T

T

PP

P

2dW

dP

00

0

(2.20)

B

AA

r'r

Calculate the pressure drop in a 60 ft length of 1 1/2-in. schedule 40 pipe packed with catalyst pellets 1/4-in. in diameter when 104.4 lb/h of gas is passing through 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 those of air at this temperature. The entering pressure is 10 atm.

EXAMPLE 2.1

SOLUTION

X1T

T

PP

P

2dW

dP

00

0

0TT 0

P2

P

PP

P

2dW

dP 20

0

0

dWdPP

P22

0

W

0

P

P2

0

dWPdPP

2

0

WPP

1 P

P

2

2

00

WPPP

1 2

0

2

2

0

W1P

P2

0

2

W1P

P2

0

21

0

W1P

P

21

0

W1P

P

0Cc

0

P1A

2

CczA1W

CC

0Cc

0 zA1P1A

2W

0

0

P

z2W

21

0

0

0 P

z21

P

P

0Cc

0

P1A

2

G75.1

D

1150

Dg

1G

p

3

pc0

0

2

m

2

m

ft.h

lb3.7383

ft0141.0

hlb4.104G

For 1½-in. schedule 40 pipe, A = 0.01414 ft2

For air at 260C and 10 atm:

= 0.0673 lbm/ft.h

0 = 0.413 lbm/ft3

Dp = ¼ in = 0.0208 ft

2

f

m8

c h.lb

ft.lb1017.4g

3.738375.10208.0

0673.045.01150

45.00208.01017.4413.0

45.013.7383380

2

f

2

2

3

f

inlb7.14

atm1

in144

ft1

ft

lb1.164

m

kPa8.25

ft

atm0775.0

265.0

10

600775.021

P

z21

P

P2121

0

0

0

atm65.2P265.0P 0

atm35.7PPP 0

Ethylene oxide is made by the vapor-phase catalytic oxidation ofethylene with air:

OC2H4 + ½ O2 CH2 CH2

We want to calculate the catalyst weight necessary to achieve 60%conversion. Ethylene and oxygen are fed in stoichiometricproportions to a packed-bed reactor operated isothermally at260°C. Ethylene is fed at a rate of 0.30 Ib mol/s at a pressure of 10atm. It is proposed to use 10 banks of 1½-in.-diameter schedule 40tubes packed with catalyst with 100 tubes per bank.Consequently, the molar flow rate to each tube is to be 3 10–4 Ibmol/s. The properties of the reacting fluid are to be consideredidentical to those of air at this temperature and pressure. Thedensity of the a ¼-in.-catalyst particles is 120 lb/ft3 and the bedvoid fraction is 0.45

EXAMPLE 2.2

SOLUTION

The rate law is:

h.catlblbmolPPkr 32B

31A

'A

C260ath.catlb.atm

mollb0141.0k

Mole balance:

0A

'

A

F

r

dW

dX

Reaction: A + ½ B C

Rate law :

32B

31A

32B

31A

32B

31A

'A CkRTCRTCRTCkPPkr

Stoichiometry: gas phase, isothermal

P

PX1vv 0

0

X1T

T

P

Pvv

0

00

(2.9)

T

T

P

P

X1

XCC 0

0

jj

0Aj (2.11)

For an isothermal gas phase reaction:

0

jj

0Aj P

P

X1

XCC

1A

5.0F

F

0A

0BB

Ethylene and air are fed to the reactor where ethylene and oxygen are in stoichiometric proportions:

88.1

F

F2179

F

F

0A

0B

0A

0II

1A

5.0B

1C

148.05.021/795.01

115.01

N

Ny

0T

0A0A

0

jj

0Aj P

P

X1

XCC

00A

0

AA0AA P

P

X1

X1C

P

P

X1

XCC

00A

0

BB0AB P

P

X1

X5.05.0C

P

P

X1

XCC

00A

0

CC0AC P

P

X1

XC

P

P

X1

XCC

32B

31A

'A CkRTCr

32

0

0A

31

0

0A'A P

P

X1

X5.05.0C

P

P

X1

X1CkRTr

323231

0

0A'A X15.0X1

P

P

X1

kRTCr

X1P

P

X1

kRTC63.0r

0

0A'A

X1

X1W1kRTC63.0r 21

0A'A

21'

A W1X1

X1'kr

where k’ = 0.63 kRTCA0 = 0.63 kPA0

0

jj

0Aj P

P

X1

XCC

0A

'

A

F

r

dW

dX

21

0A

W1X1

X1'k

F

1

dW

dX

W

0

21X

0

0A dWW1dXX1

X1

'k

F

230A W113

2X

X1

1ln1

'k

F

32

0A XX1

1ln1'k2F311

W (a)

w

0

23W

0

21 W13

2dWW1

2323 013

2W1

3

2

3

2W1

3

2 23

23W113

2

Parameter evaluation per tube (i.e., divide feed rates by 1000):

Ethylene FA0 = 3 10-4 lbmol/s = 1.08 lbmol/h

Oxygen FB0 = 1.5 10-4 lbmol/s = 0.54 lbmol/h

Inert (N2)

= 5.64 10-4 lbmol/s = 2.03 lbmol/h

Jumlah FT0 = FA0 + FB0 + FI = 3.65 lbmol/h

2

24I Omol21

Nmol79slbmol105.1F

30.065.3

08.1

F

Fy

0T

0A0A

15.015.013.0y 0A

atm0.3atm103.0PyP 00A0A

atm3h.catlb.atm

lbmol0141.063.0kP63.0'k 0A

catlb.h

lbmol0266.0

For 60% conversion, eq. (a) becomes

32

0A XX1

1ln1'k2F311

W

320A 'kF303.111

W

h

lb24.30

lbmol

lb28

h

lbmol08.1m 0A

h

lb28.17

lbmol

lb32

h

lbmol54.0m 0B

h

lb84.56

lbmol

lb28

h

lbmol03.2m 0I

h

lb4.104m 0T

22c

0T

ft.h

lb3.7383

ft01414.0

hlb4.104

A

mG

G75.1

D

1150

Dg

1G

p

3

pc0

0

For air at 260C and 10 atm:

= 0.0673 lbm/ft.h

0 = 0.413 lbm/ft3

Dp = ¼ in = 0.0208 ft

2

f

m8

c h.lb

ft.lb1017.4g

1045.0112001414.0

0775.02

P1A

2

0Cc

0

2

f

2

2

3

f

inlb7.14

atm1

in144

ft1

ft

lb1.164

ft

atm0775.0

3.738375.10208.0

0673.045.01150

45.00208.01017.4413.0

45.013.7383380

catlb

0166.0

320A 'kF303.111

W

catlb

0166.0

h.catlb

lbmol0266.0

h

lbmol08.1

catlb

0166.0303.1

11

32

= 45.4 lb of catalyst per tube

= 45,400 lb of catalyst total

21

0

W1P

P

496.0catlb4.45catlb

0166.01

21

P = 0.496 P0 = 4.96 atm

P = P0 – P = 10 – 4.96 = 5.04 atm