1 TAME synthesis problem Tert-Amyl Methyl Ether (TAME) is an ...

TAME synthesis problem
Tert-Amyl Methyl Ether (TAME) is an oxygenated additive for green gasolines. Besides its use as an
octane enhancer, it also improves the combustion of gasoline and reduces the CO and HC (and, in a
smaller extent, the NOx) automobile exhaust emissions. Due to the environmental concerns related to
those emissions, this and other ethers (MTBE, ETBE, TAEE) have been lately studied intensively.
TAME is currently catalytically produced in the liquid phase by the reaction of methanol (MeOH) and
the isoamylenes 2-methyl-1-butene (2M1B) and 2-methyl-2-butene (2M2B). There are three
simultaneous equilibrium reactions in the formation and splitting of TAME: the two etherification
reactions and the isomerization between the isoamylenes:
2 M1B + MeOH ⇔ TAME
(1)
2 M 2B
+ MeOH ⇔ TAME
(2)
2M1B ⇔ 2M
2B
(3)
These reactions are to be carried out in a plug flow reactor and a membrane reactor in which MeOH is
fed uniformly through the sides.
For isothermal operation:
a) Plot the concentration profiles for a 10 m 3 PFR.
b) Vary the entering temperature, T0, and plot the exit concentrations as a function of T0.
For a reactor with heat exchange (U = 10 J.m -2 .s -1 .K -1 ):
c) Plot the temperature and concentration profiles for an entering temperature of 353 K
d) Repeat (a) through (c) for a membrane reactor.
NOTE: To simplify the problem, consider the solution density and heat capacity constant (evaluate
them for the entering temperature and the feed mole fractions) and consider the solution as ideal: ai = xi
in the rate expressions.
A new and improved kinetic model, including the non-ideal character of the reacting solution, can be
found in “Number of Actives Sites in TAME Synthesis: Mechanism and Kinetic Modeling”, Manuela
V. Ferreira and José M. Loureiro, Ind. Eng. Chem. Res. 43 (2004) 5156-5165.
1

Additional information for solving the problem
• Thermodynamic equilibrium constants, activity based (Vilarinho Ferreira and Loureiro, 2001)
3
⎛ 5.
0166×
10 ⎞
Keq = exp ⎜
− 10.
839 ⎟
1
(I.1)
⎝ T
⎠
3
⎛ 3.
7264×
10 ⎞
Keq = exp ⎜
− 9.
6367 ⎟
2
(I.2)
⎝ T
⎠
Keq
Keq = (I.3)
1
3
Keq2
with: T in Kelvin
• Kinetic constants for the direct reactions (Kiviranta-Pääkkönen et al., 1998)
k
k
k
1
2
3
3
10 ⎛ 76.
8×
10 ⎞
= 3 . 2870 × 10 exp ⎜
⎜−
⎟
(I.4)
⎝ R T ⎠
3
13 ⎛ 99.
7 × 10 ⎞
= 3 . 9682 × 10 exp ⎜
⎜−
⎟
(I.5)
⎝ R T ⎠
3
10 ⎛ 81.
7 × 10 ⎞
= 7 . 4767 × 10 exp ⎜
⎜−
⎟
(I.6)
⎝ R T ⎠
with: k in mol.
T in Kelvin
−1
kg cat .s -1
R = 8.314 J.mol -1 .K -1
• Adsorption constants for each component, activity based (calculated and adapted from Oktar et
al., 1999)
3
⎛ 4.
6825×
10 ⎞
K1B = exp ⎜
− 10.
157 ⎟ , 1B = 2M1B (I.7)
⎝ T
⎠
3
⎛ 3.
4420 × 10 ⎞
K = exp ⎜
− 6.
5849 ⎟
2B
, 2B = 2M2B (I.8)
⎝ T
⎠
3
⎛1.
0014 × 10 ⎞
K M = exp ⎜
+ 4.
7496 ⎟ , M = MeOH (I.9)
⎝ T
⎠
3
⎛ 2.
3934 × 10 ⎞
KT = exp ⎜
− 3.
5736 ⎟ , T = TAME
⎝ T
⎠
with: T in Kelvin
(I.10)
2

1B
r
2B
• Heat of reaction (Vilarinho Ferreira and Loureiro, 2001)
R
∆ H1 = - 41.708 kJ.mol -1
R
∆ H 2 = - 30.981 kJ.mol -1
R
∆ H 3 = - 10.727 kJ.mol -1
• Bulk density and bed porosity
The PFR is filled with a macroreticular strong cation ion-exchange resin in hydrogen form
(Amberlyst 15 Wet, Rohm & Haas).
The bulk density is: ρ = 770 g / L
The bed porosity is: ε = 0.4
b
• Rate equations for the formation of each component (Vilarinho Ferreira and Loureiro, 2001)
=
=
k
1
K
2
( 1 + K a + K a + K a + K a ) ( 1 + K a + K a + K a + K a )
k
2
1B
K
M
K
1B
1B
a
M
2B
a
1B
⎛ 1
⎜
⎝ Keq
2B
1
M
aT
a a
M
M
1B
⎞
−1
⎟
⎠
T
T
+
1B
k
3
1B
K
1B
a
1B
2B
⎛ 1
⎜
⎝ Keq
2B
3
a
a
2B
1B
M
⎞
−1
⎟
⎠
2
( 1 + K a + K a + K a + K a ) ( 1 + K a + K a + K a + K a )
1B
rM r1
B r2
B +
T
M
K
1B
2B
a
M
2B
a
1B
⎛ 1
⎜
⎝ Keq
2B
2
M
aT
a a
M
M
2B
⎞
−1
⎟
⎠
T
T
−
1B
k
3
1B
K
1B
a
1B
2B
⎛ 1
⎜
⎝ Keq
2B
3
a
a
2B
1B
M
M
⎞
−1
⎟
⎠
M
T
T
T
T
(I.11)
(I.12)
= (I.13)
r = − r
(I.14)
M
with: r in mol.
−1
kg cat .s -1
ai stands for the activity of component i in the liquid phase, and since we are considering the
solution as ideal:
a = x
(I.15)
i
i
where xi is the mole fraction of component i in the liquid phase.
We can also define the rate of each reaction:
3

k
K
K
a
a
1
a
⎞
−1⎟
( ) 2
T
1 M 1B
M 1B
⎜ Keq1
aM
a ⎟
1B
1 =
⎝
⎠
(I.16)
1 + K1B
a1B
+ K 2B
a2
B + K M aM
+ KT
aT
r
k
K
K
a
a
⎛
⎜
1
a
⎞
−1⎟
( ) 2
T
2 M 2B
M 1B
⎜ Keq2
aM
a ⎟
2B
2 =
⎝
⎠
(I.17)
1 + K1B
a1B
+ K 2B
a2
B + K M aM
+ KT
aT
r
r
3
=
1 + K
1B
k
a
3
1B
K
1B
a
+ K
1B
2B
2B
⎛
⎜
⎛ 1 a
⎜
⎝ Keq3
a
a + K
• Heat capacity of each component
2B
1B
M
⎞
−1
⎟
⎠
a +
M
K
T
a
T
(I.18)
Cpi 2
3
= ai
+ bi
T + ci
T + di
T
(I.19)
with: Cp in kJ.mol -1 .K -1
T in Kelvin
component i 10 ai 10 4 bi 10 7 ci 10 10 di
MeOH a
2M1B b
2M2B b
TAME c
0.077 1.62 2.06 2.87
1.27 -0.609 5.08 1.69
1.33 -1.48 7.51 -0.882
1.73 2.29 -6.00 20.0
a Zhang and Datta, 1995; b Kitchaiya and Datta, 1995; c Estimated by the Missenard
method (Reid et al., 1987)
To simplify the problem we are going to consider the heat capacity of each component constant
and equal to its value at the entry conditions: T = T0.
The solution heat capacity will also be considered constant and equal to its value at the entry
conditions:
4
∑
i=
1
IN
Cp = Cpi
xi
(I.20)
where
IN
x i is the feed mole fraction of component i.
4

• Density of each component (Perry and Green, 1997)
C1,
i M i
ρ (I.21)
i
= ⎡
C ⎤
⎢ ⎛ ⎞ 4,
i
⎜ T ⎟ ⎥
⎢1
+ 1 −
⎜ ⎟ ⎥
⎢
⎣
⎝
C3,
i ⎠ ⎥
⎦
C
2,
i
with ρ in g.L -1
T in Kelvin
M in g.mol -1
component i Mi C1,i C2,i C3,i C4,i
MeOH
2M1B
2M2B
TAME
32.042 2.288 0.2685 512.64 0.2453
70.135 0.91619 0.26752 465 0.28164
70.135 0.93322 0.27251 471 0.26031
102.177 * * * *
* as there is no data available for TAME, we will consider its density constant and equal to its
value at 293 K: ρ = 770 g / L
TAME
To simplify the problem we are going to consider the heat capacity of each component constant
and equal to its value at the entry conditions: T = T0.
The solution density will also be considered constant and equal to its value at the entry
conditions:
4
∑
i=
1
IN
ρ = ρ
(I.22)
i xi
5

Starting solving the problem
Writing mass and energy balances
In the figure is a representation of a PFR filled with a catalyst.
Fig. 1:Representaion of a PFR filled with catalyst.
A. Steady state mass balance
For the steady state mass balance of component i, in the volume element of length dz, we can write:
Total Flux IN = Total Flux OUT − What is formed by Chemical Re action
(A.1)
z
z+
dz
If ε is the bed porosity, the area, A, available for the catalyst particles is [(1-ε) A] and the area available
for the fluid is [ε A]. Equation (A.1) becomes:
( ε ) ϕ ( ε A)
ϕ − ρ r Adz
A i,
z = i,
z+
dz b i
(A.2)
where ϕi is the molar flux of component i, ρb is the bulk density and ri is the rate of formation of
component i.
Re-writing equation (A.2):
ϕ
dϕ
dz
i,
z+
dz
− ϕ i,
z ρ b
− ri
= 0
dz ε
ρ
i b
− ri
ε
= 0
IN
Ci
Considering a plug flow with axial dispersion, the molar flux is given by:
z
dz
z+dz
z = 0 z = L
A
OUT
Ci
(A.3)
(A.4)
6

dCi
ϕ i = ui
Ci
− Dax
(A.5)
dz
where ui is the interstitial velocity of the fluid, Ci is the concentration of component i in the fluid and
Dax is the axial dispersion coefficient.
u
i
dC
dz
d
2
i
− Dax
i
2 −
b
ri
dz
C
ρ
ε
= 0
Normalizing some of the variables:
IN
C i is the feed concentration of component i),
ui
C
L
IN
total
dividing by
df
i
dX
df i
dX
u
D
d
2
−
ax
2
IN
Ctotal
i
2 −
b
ri
L
i IN
Ctotal
L
1 d
Pe
f
dX
2
i b
− − τ r
2 IN i
dX ε Ctotal
f
ρ
ε
(A.6)
z
Ci
IN
IN
X = (L is the PFR length), f = ( C IN total ∑C
i where
L
= 0
, equation (A.7) becomes:
ρ
= 0
where Pe is the dimensionless Peclet number given by
ρ b
Grouping the term τ r IN i
ε C
component i becomes:
df
i
dX
total
1 d f
Pe dX
2
i
− − ℜ 2 i
ax
i
Ctotal
= i
ui
L
, and τ is the space-time given by
D
L
.
ui
(A.7)
(A.8)
as a reaction term represented by ℜ i , the steady state mass balance of
= 0
⎛ 1 ⎞
In the limiting case of absence of dispersion ⎜ →0 ⎟ , equation (A.9) becomes:
⎝ Pe ⎠
df i
dX
− ℜi
= 0
(A.9)
(A.10)
7

B. Steady state energy balance
For the steady state energy balance in the volume element of length dz, considering no dispersion, we
can write:
heat produced by
Total Flux IN + = Total Flux OUT
z
z+
dz
Chemical Re action
M reactions
h
R
h
( ε A) ϕ z + ρ b ∑(
∆H
j rj
) Adz
= ( ε A)
ϕ z+
dz + Alat
U ( T −Tw
)
j=
1
where ϕ h is the heat flux,
R
j
heat losses through
+
reactor walls
(B.1)
(B.2)
∆ H is the heat of reaction j, rj is the rate of reaction j, Alat is the lateral area
of the volume element, U is the overall heat-transfer coefficient, T is the reactor temperature and Tw is
the reactor wall temperature.
Re-writing equation (B.2):
M reactions
b ∑
j=
1
R Alat
( ∆H
r ) + U ( T −T
) = 0
h h
ϕ z+
dz −ϕ
x ε − ρ
j j
w
(B.3)
dz
A dz
The heat flux can be given by:
h
ϕ = ρ CpT
(B.4)
u i
where ρ is the solution density and Cp is the solution heat capacity.
If R0 is the reactor radius, the lateral area of the volume element of length dz is:
= 2π R dz
(B.5)
A lat 0
and its sectional area is:
A = π R
(B.6)
2
0
leading to:
Alat 2
= (B.7)
Adz
R
0
Substituting into equation (B.3):
8

− b ∑
j=
1
R 2
( ∆H
r ) + U ( T −T
) = 0
M reactions
dT
ε ui ρ Cp ρ
j j
w
(B.8)
dz
R
Normalizing the space variable:
rearranging equation (B.8):
M reactions
b
− ∑ ρ Cp j=
1
0
z
X = , remembering the space-time definition
⎛ ⎞
⎜τ
= L ⎟ and
L
⎝ ui
⎠
R 2U
τ
( ∆H
r ) + ( T −T
) = 0
dT ρ τ
ε j j
w
(B.9)
dX
ρ Cp R
The term
2U
τ
ρ Cp R
0
0
is dimensionless and is referred as NTU (number of heat transfer units).
Finally, the steady state energy balance becomes:
M reactions
b
− ∑ ρ Cp j=
1
R ( ∆H
r ) + NTU ( T −T
) = 0
dT ρ τ
ε j j
w
(B.10)
dX
C. Boundary Conditions
In the absence of dispersion, we only need one boundary condition:
IN ⎧ Ci
⎪ fi
=
X = 0 ⇒
IN
⎨ C
(C.1)
total
⎪
⎩ T = T0
where T0 is the initial reactor temperature.
9

Algorithm to solve the problem
We have a non-linear system of Ordinary Differential Equations (ODE) to solve:
⎧ 1B
⎪ − ℜ1B
= 0
⎪
2B
⎪ − ℜ2
B = 0
⎪
⎪ M
⎨ − ℜ M = 0
⎪
⎪
dfT
− ℜT
= 0
⎪ dX
3
⎪ dT b
⎪ − ∑
⎩ dX Cp j=
dX
df
dX
df
dX
df
ρ τ
ε
ρ
1
R ( ∆H
r ) + NTU ( T −T
)
j
j
w
= 0
The program we developed uses subroutine DDASSL (Brenan et al., 1989) to solve this system. This
code solves a system of differential/algebraic equations of the form delta(t, y, yprime)=0, with
delta(i) = yprime(i) – y(i), using the backward differentiation formulas of orders one trough five. t is
the current value of the independent variable (in our case t = X), y is the array that contains the solution
components at t (in our case we have: y(i) = f(i), i=1,4 and y(5) = T) and yprime is the array that
contains the derivatives of the solution components at t.
The program solves the system from t to tout and it is easy to continue the solution to get results at
additional tout. In our case, we are going to get results at different values of X, between 0 and 1.
This problem is rather complex because most of the other variables depend on yi: the kinetic, adsorption
and thermodynamic equilibrium “constants” depend on T (y5), the reaction rate and the components rate
of formation depend on yi (fi (i=1,4) and T), as the activity coefficients.
In Figure 2 is the algorithm to solve our problem.
(1)
10

Read data from file
T0, Tw IN
C i
Q (operation flow)
Input constants values
ε, ρb, U, R 0,
V (reactor volume)
∆ H
R
j
Calculate constants
ε V
τ =
Q
C C
IN
total = ∑
i
IN
i
IN
0 Ci
f i = IN
Ctotal
Initialize the variables
t = X = 0
yi (i=1,4) = f
y5 = T 0
0
i
Solve system of ODEs
Call
subroutine DASSL
New values
of
yi
Cp
ρ
NTU
tout = X =
X + ∆X
No
tout = X = 1 ?
Fig. 2: Algorithm to solve the problem.
Yes
STOP
11

Some results
For both cases, isothermal and non-isothermal, the feed concentrations were the same:
C
C
C
C
IN
2M
1B
IN
2M
2B
IN
MeOH
IN
TAME
=
=
=
= 0
3.
33
3.
33
6.
66
mol / L
mol / L
mol / L
These concentrations lead to a feed mole ratio MeOH/isoamylenes, RM/IA, of 1.0.
a) Figures R.1, R.2 and R.3 represent the concentration profiles for a 10 m 3 isothermal PFR, operating
at 323 K, 343 K and 363 K, and a flow of 40 L/min.
C (mol/L)
7
6
5
4
3
2
1
0
Isothermal: T = 323 K
RM/IA = 1, Q = 40 L/min
0 0.2 0.4 0.6 0.8 1
X
2M1B 2M2B MeOH TAME
Fig. R.1: Concentration profiles for a 10 m 3 isothermal PFR operating at 323 K.
12

C (mol/L)
Fig. R.2: Concentration profiles for a 10 m 3 isothermal PFR operating at 343 K.
C (mol/L)
7
6
5
4
3
2
1
0
7
6
5
4
3
2
1
0
Isothermal: T = 343 K
RM/IA = 1, Q = 40 L/min
0 0.2 0.4 0.6 0.8 1
X
2M1B 2M2B MeOH TAME
Isothermal: T = 363 K
RM/IA = 1, Q = 40 L/min
0 0.2 0.4 0.6 0.8 1
X
2M1B 2M2B MeOH TAME
Fig. R.3: Concentration profiles for a 10 m 3 isothermal PFR operating at 363 K
As the temperature increases, the reactions are faster, favoring TAME production, but the chemical
equilibrium is moved to the opposite direction: for an operating temperature of 363 K, the equilibrium
concentration of TAME is reached faster than for 343 K, but its value is lower.
13

) Figure R.4 represents the exit concentrations as a function of the entering temperature, T0, for a 10
m 3 isothermal PFR operating with a flow of 40 L/min.
C OUT (mol/L)
5
4
3
2
1
0
Isothermal operation
RM/IA = 1.0, Q = 40 L/min
323 333 343 353 363
T0 (K)
2M1B 2M2B MeOH TAME
Fig. R.4: Exit concentrations as function of T0, for a 10 m 3 isothermal PFR
Figure R.4 shows that there is an optimum operating temperature around 333 K that leads to a
maximum exit concentration for TAME. It is due to the fact that we have a system with competition
between kinetics and equilibrium, as we are going to see for the non-isothermal PFR
c) We chose a reactor diameter of 1 m (for a volume of 10 m 3 it leads to a reactor length of 12.7 m
approximately) and for the wall temperature, we decided to use room temperature: 298 K. The entering
temperature is 353 K.
It is important to notice that the catalyst used in TAME production is a macroreticular strong cation ionexchange
resin in hydrogen form (Amberlyst 15 Wet, Rohm & Haas) that has a maximum operating
temperature of 393 K.
Figure R.5 represents the concentration (a) and temperature (b) profiles for an operating flow of 200
L/min.
14

(a)
(b)
C (mol/L)
T (K)
7
6
5
4
3
2
1
0
390
380
370
360
350
Non-Isothernal: To = 353 K, Tw = 298 K
RM/IA = 1.0, Q = 200 L/min
0 0.2 0.4 0.6 0.8 1
X
2M1B 2M2B MeOH TAME
0 0.2 0.4 0.6 0.8 1
X
Fig. R.5: Non-isothermal PFR: (a) concentration profiles; (b) temperature profile.
Figure R.5(a) shows the competition between the three reactions: first, 2M1B and 2M2B react with
MeOH to produce TAME and the reactants concentrations decrease and TAME concentration
increases; but then, although MeOH and TAME concentrations are almost constant, 2M1B is still
decreasing and 2M2B starts to increase: the third reaction, the isomerization, is now more evident.
The temperature profile (Fig. R.5(b)) shows that the reactor seams to be almost adiabatic (no heat losses
through the reactor walls) since the reactor temperature is always increasing. This close to adiabatic
behavior was expected since the reactor diameter is rather large: 1 m. But the temperature reaches a
value that is very close to the catalyst limit: remember that its maximum operating temperature is 393 K
and the reactor is reaching 385 K.
15

To see what is the maximum temperature that the reactor reaches, we can make it adiabatic setting the
overall heat-transfer coefficient equal to zero: U = 0. In Figure R.6 are the results for the adiabatic
reactor.
(a)
C (mol/L)
(b)
T (K)
7
6
5
4
3
2
1
0
390
380
370
360
350
Adiabatic operation: To = 353 K
RM/IA = 1.0, Q = 200 L/min
0 0.2 0.4 0.6 0.8 1
X
2M1B 2M2B MeOH TAME
0 0.2 0.4 0.6 0.8 1
X
Fig. R.6: Adiabatic PFR: (a) concentration profiles; (b) temperature profile.
The maximum temperature reached by the reactor is 388 K. Comparing Figures R.5 and R.6 it is easy to
see that the non-isothermal 1m diameter PFR can be considered adiabatic: the concentration and
temperature profiles are almost the same.
To improve the reactor production, i.e., to reach higher exit concentrations of TAME, we can use a
multi-tubular reactor. Lets think of a reactor composed of 4000 tubes (each one considered as a PFR)
16

with a diameter of 1’’ each. To have a total reactor volume of 10 m 3 , each tube has a length of 5 m. In
order to compare the results of the multi-tubular reactor with the ones obtained with the 1 m diameter
PFR (Fig. R.5), we have to choose similar operating conditions: to have a total operating flow of 200
L/min, the equivalent flow in each tube is 0.05 L/min; since the reactor is now really cooled, we will
choose the “best” temperature for the cooling fluid (333 K, as seen with the isothermal behavior runs).
Figure R.7 shows the results for one of this tubes operating in the above conditions.
(a)
C (mol/L)
(b)
T (K)
7
6
5
4
3
2
1
0
360
358
356
354
352
350
Non-Isothermal Multi-tubular reactor
To = 353 K, Tw = 333 K, Qtotal = 200 L/min
0 0.2 0.4 0.6 0.8 1
X
2M1B 2M2B MeOH TAME
0 0.2 0.4 0.6 0.8 1
X
Fig. R.7: Multi-tubular reactor: (a) concentration profiles; (b) temperature profile.
The maximum temperature reached is 358 K - in this case there are no problems with the catalyst - and
the exit concentration of TAME is higher: 2.603 mol/L against 1.862 mol/L for the PFR in Figure R.5,
which represents an increase of 28.5 %.
17

References
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