5 fixed bed reactor1
TRANSCRIPT
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