Mass balance in a fixed bed reactor is similar to that of a plugflow reactor (eq. 1.1):
dX  rA

dV FA 0
(1.1)
Recalling dW = B dV, then
dX
 rA

dW B FA 0
(2.1)
For a reaction:
A + bB  cC + dD
Species
A
B
C
D
I
TOTAL
EFFLUENT
Feed Rate
Change within
Reactor
– FA0 X
– b FA0 X
c FA0 X
d FA0 X
Effluent Rate
FA0 = FA0
FA = FA0 (1 – X)
FB0 = B FA0
FB = FA0 (B – bX)
FC0 = C FA0
FC = FA0 (C + cX)
FD0 = D FA0
FD = FA0 (D + dX)
FI0 = I FA0
FI = FI0 I
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:
PV
P0 V0

NTRT NT 0RT0
P0   T   NT 

 V0
V      
 P   T0   NT 0 
(2.2)
Recalling the last row in the previous table:
NT  NT 0   NA 0 X
FT = FT0 +  FA0 X
NT NT 0   NA 0 X

 1   y A0X
NT 0
NT 0
The above equation is further simplified by letting:
change in total number of moles for complete conversion

total number of moles fed to the reactor
NA 0
  d  c  b  1
  y A0
NT 0
(2.3)
Equation (2.2) now becomes:
P0   T 

V      1  X V0
 P   T0 
(2.4)
Kalau reaksinya berlangsung sempurna (X = 1):
NT *  NT 0   NA 0 X  NT 0   NA 0
Perubahan jumlah mol kalau reaksinya berlangsung
sempurna (X = 1):
 NT  * NT 0  NT 0  NA 0  NT 0
 NA 0  d  c  b  1NA 0

d  c  b  1NA 0
NT 0
  y A0
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
FT NT P
CT   
v V RT
(2.5)
At the entrance of the reactor:
FT 0 NT 0 P0
CT0 


v 0 V0 RT0
(2.6)
Taking the ratio of eq. (2.6) to eq. (2.5), we have upon
rearrangement
P0   T   FT 

v        v 0
 P   T0   FT 0 
(2.7)
From the above table:
FT  FT 0  FA 0  X
(2.8)
Substituting eq. (2.8) in eq. (2.7):
P0   T   FT 0  FA 0 X 
P0   T   FA 0 


  v 0      1  X 
v  v 0     
FT 0
FT 0 
 P   T0  
 P   T0  

P0   T   FA 0 
P0   T 


v  v 0      1  X   v 0     1  y A 0 X 
FT 0 
 P   T0  
 P   T0 
P0   T 

v  v 0     1  X 
 P   T0 
(2.9)
We can now express the concentration of species j for a flow
system in terms of conversion:
Fj
Cj  
v
Fj
 FT 0   Fj   P   T0 
        
 FT P0 T   v 0   FT   P0   T 

v0 
 FT 0 P T0 
 Fj   P   T0 
C j  C T 0      
 FT   P0   T 
(2.10)
Substituting F and FT in terms of conversion in eq. (2.10) yields
FA 0  j   jX  P   T0 
   
C j  CT0
FT 0  FA 0 X  P0   T 
 FA 0   j   jX   P   T0 
   
 C T 0  
 FT 0  1  FA 0 FT 0 X  P0   T 
 j   jX  P   T0 
   
 CT0 y A0
1  y A 0 X  P0   T 
 j   jX  P   T0 
   
Cj  CA0
1  X  P0   T 
(2.11)
Energy balance in a fixed bed reactor is similar to that of a
plug-flow reactor (eq. 1.56):
dT Ua Ta  T    rA   HRx  T 

dV
FA 0 X Cp    iCpi 
(1.56)
Recalling dW = B dV, then
dT Ua Ta  T    rA   HRx  T 

dW
B FA 0 X Cp    iCpi 
(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:

dP
G  1    150 1    

 1.75G
 3 
dz
 g c Dp    
Dp

(2.13)
Where
P : 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 h
z
: length down the packed bed of pipe, ft
u : 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)
 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
 (kg/s), is equal
flow rate at any point down the reactor, m
 0 (i.e., equation of
to the entering mass flow rate, m
continuity),
 0 m

m
0 v 0  v
Recalling eq. (2.9), we have
P0   T 

v  v 0     1  X 
 P   T0 
(2.9)
 P   T0   1 
v0
   0   0     

v
 P0   T   1  X 
(2.14)
Combining eqs. (2.13) and (2.14) gives:
  P0   T 
G1    150 1    
dP

 1.75G     1  X 
3 
dz
 0 g c Dp  
Dp
  P   T0 
Simplifying yields:
dP
P0   T 

  0     1  X 
dz
 P   T0 
(2.15)
where:

G1    150 1    
0 
 1.75G
3 
 0 g c Dp  
Dp

(2.16)
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:
 weight of   volume of  density of 
catalyst   solids
  solid catalyst 

 
 

W
 1   A c z 
C
(2.17)
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 – ) :
B  C 1  
(2.18)
Using the relationship between z and W [eq. (2.17)] we can
change our equation (2.15) into:
dP
P0   T 

  0     1  X 
dz
 P   T0 
dP
0
P0   T 


    1  X 
dW
A c 1     C  P   T0 
(2.15)
(2.19)
Further simplification yields:
dP
  P0   T 
   1  X 
  
dW
2  P P0   T0 
(2.20)
where:
2 0

A c  C 1   P0
(2.21)
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.
dX
 rA

dW B FA 0
 rA
 r' A 
B
(2.1)
dT Ua Ta  T    rA   HRx  T 

dW
B FA 0 X Cp    iCpi 
(2.12)
dP
  P0   T 
   1  X 
  
dW
2  P P0   T0 
(2.20)
EXAMPLE 2.1
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.
SOLUTION
dP
  P0   T 
   1  X 
  
dW
2  P P0   T0 
T  T0
0
 P02
dP
  P0 
  
  
dW
2  P P0 
2P
2P
dP    dW
2
P0
W
2 P
PdP     dW
2 
P0 P
0
0
 1  2P
 2  P P   W
 P0 
P2
 1   W
2
P0
2
P
   1  W
 P0 
0
1 2 2
 2  P  P0    W
 P0 
P
12


 1  W
P0
P
 1  W 1 2
P0
2 0

A c  C 1   P0
W  1   A c z  C
P  2 0 z 

  1 
P0 
P0 
12
2 0
1    A C z C
W 
A c  C 1   P0
2 0 z
W 
P0
2 0

A c  C 1   P0
G1   
0 
 0 g c Dp  3
150 1    

 1.75G

Dp


For 1½-in. schedule 40 pipe, A = 0.01414 ft2
104.4 lbm h
lbm
G
 7383.3
2
0.0141 ft
h. ft2
For air at 260C and 10 atm:
 = 0.0673 lbm/ft.h
0 = 0.413 lbm/ft3
Dp = ¼ in = 0.0208 ft
lbm . ft
g c  4.17  10
lb f .h2
8
7383.3 1  0.45 
0  
8
3 
0.413  4.17  10 0.0208 0.45


150 1  0.45 0.0673   1.757383.3 


0.0208
1 atm
lb f 1 ft2
 164.1 3 

2
ft 144 in 14.7 lb f in2
atm
kPa
 0.0775
 25.8
ft
m
P  2 0 z 

  1 
P0 
P0 
12
 1 

P  0.265P0  2.65 atm
P  P0  P  7.35 atm
12



2 0.0775 60 
10

 0.265
EXAMPLE 2.2
Ethylene oxide is made by the vapor-phase catalytic oxidation of
ethylene with air:
O
C2H4 + ½ O2  CH2  CH2
We want to calculate the catalyst weight necessary to achieve 60%
conversion. Ethylene and oxygen are fed in stoichiometric
proportions to a packed-bed reactor operated isothermally at
260°C. Ethylene is fed at a rate of 0.30 Ib mol/s at a pressure of 10
atm. It is proposed to use 10 banks of 1½-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  10–4 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 ¼-in.-catalyst particles is 120 lb/ft3 and the bed
void fraction is 0.45
The rate law is:
 rA'  k PA1 3 PB2 3 lbmol lb cat .h
lb mol
k  0.0141
at 260C
atm.lb cat .h
SOLUTION
Reaction:
A+½BC
Mole balance:
dX  rA'

dW FA 0
Rate law :
 rA'  k PA1 3 PB2 3  k C ART 1 3 CBRT 2 3  kRTC1A 3CB2 3
Stoichiometry: gas phase, isothermal
P0   T 

v  v 0     1  X 
 P   T0 
(2.9)
 j   jX  P   T0 
   
Cj  CA0
1  X  P0   T 
(2.11)
For an isothermal gas phase reaction:
P0 

v  v 0 1  X   
P
 j   jX  P 
 
Cj  CA0
1  X  P0 
Ethylene and air are fed to the reactor where ethylene and
oxygen are in stoichiometric proportions:
A  1
A   1
FB 0
B 
 0.5
FA 0
B   0.5
FI0 79 21FB 0
I 

 1.88
FA 0
FA 0
C  1
NA 0
1
   y A0  
 1  0.5  1
  0.148
NT 0
1  0.5  79 / 210.5
 j   jX  P 
 
Cj  CA0
1  X  P0 
 A  AX  P 
1X  P 
   C A 0
 
CA  CA0
1  X  P0 
1  X  P0 
 B  B X  P 
0.5  0.5X  P 
   C A 0
 
CB  C A 0
1  X  P0 
1  X  P0 
C  CX  P 
X P
   C A 0
 
CC  C A 0
1  X  P0 
1  X  P0 
 rA'  kRTC1A 3CB2 3
13
 rA'
 C A 0 1  X   P  
  
 kRT 
 1  X  P0  
 rA'
kRTC A 0  P 
 1  X 1 3 0.5 2 3 1  X 2 3

1  X  P0 
 rA'
kRTC A 0  P 
 1  X 
 0.63
1  X  P0 
 rA'
1X 
 0.63kRTC A 0 1  W  

 1  X 
1X 
12



 r  k' 
1


W

 1  X 
'
A
 C A 0 0.5  0.5X   P  
  

1  X
 P0  

23
 j   jX  P 
 
Cj  CA0
1  X  P0 
1 2
where k’ = 0.63 kRTCA0 = 0.63 kPA0
dX  rA'

dW FA 0
dX 1  1  X 
12


 k' 
1


W

dW FA 0  1  X 
W
FA 0 X  1  X 
12


dX

1


W
dW



k' 0  1  X 
0

FA 0 
1
2


1   ln
 X 
1  1  W 3 2
 3
k' 
1X

1 





1  1  3FA 0 2k' 1   ln
  X  
1X



W

23
(a)
w
2
12
32




1


W
dW


1


W

3
0
0
W
2
2
32
  1  W   1  0 3 2
3
3
2
2
32
  1  W  
3
3

2

1  1  W 3 2
3

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
79 mol N2
4
FI  1.5  10 lbmol s 
21 mol O2
Inert (N2)
= 5.64  10-4 lbmol/s = 2.03 lbmol/h
Jumlah
FT0 = FA0 + FB0 + FI = 3.65 lbmol/h
FA 0 1.08
y A0 

 0.30
FT 0 3.65
  y A 0   0.31  0.5  1   0.15
PA 0  y A 0 P0  0.3 10 atm  3.0 atm
lbmol
k'  0.63 kPA 0  0.63  0.0141
 3 atm
atm.lb cat.h
lbmol
 0.0266
h.lb cat
For 60% conversion, eq. (a) becomes

1  1  3FA 0

W
1 






2k'  1   ln
  X  
1X



1  1  1.303FA 0 k' 2 3
W

23
lbmol
lb
lb
 A 0  1.08
m
 28
 30.24
h
lbmol
h
lbmol
lb
lb
 B 0  0.54
m
 32
 17.28
h
lbmol
h
lbmol
lb
lb
 I0  2.03
m
 28
 56.84
h
lbmol
h
lb
 T 0  104.4
m
h
 T 0 104.4 lb h
m
lb
G

2  7383.3
A c 0.01414 ft
h.ft2
For air at 260C and 10 atm:
 = 0.0673 lbm/ft.h
0 = 0.413 lbm/ft3
Dp = ¼ in = 0.0208 ft
lbm . ft
g c  4.17  10
lb f .h2
8
G1   
0 
 0 g c Dp  3
150 1    

 1.75G

Dp


7383.3 1  0.45 
0  
8
3 
0.413  4.17  10 0.0208 0.45


150 1  0.45 0.0673   1.757383.3 


0.0208
1 atm
lb f 1 ft2
atm
 164.1 3 

2
2  0.0775
ft 144 in 14.7 lb f in
ft
2 0.0775
2 0


A c C 1   P0 0.01414 120 1  0.45 10 
0.0166

lb cat
1  1  1.303FA 0 k' 2 3
W


 0.0166  
lbmol  
 1.303 lb cat   1.08 h  



1  1 
lbmol


0.0266


lb
cat
.
h



0.0166
lb cat
= 45.4 lb of catalyst per tube
= 45,400 lb of catalyst total
23
P
 1  W 1 2
P0
  0.0166 

 45.4 lb cat 
 1  
  lb cat 

12
P = 0.496 P0 = 4.96 atm
P = P0 – P = 10 – 4.96 = 5.04 atm
 0.496