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

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