Chemical Engineering 140
Midterm Examination #2 Solution
1)
F1
F2
Dry air
0.21 O
2
0.79 N
2
MEOH/H
2
O
0.5 MEOH
0.5 H
2
O
Recycle F3
Recycle to feed ratio = 0.2
1.0 MEOH
AF4
Water
Methanol
O
2
N
2
MEOH/H
2
O
0.5 MEOH
0.5 H
2
O
MEOH
220
o
C
3.5 atm
MEOH
HCHO
DME
DMM
H2O
H2
PMM
F5
F4
F1
F2
Dry air
0.21 O
2
0.79 N
2
MEOH/H
2
O
0.5 MEOH
0.5 H
2
O
Recycle F3
Recycle to feed ratio = 0.2
1.0 MEOH
AF4
Water
Methanol
O
2
N
2
MEOH/H
2
O
0.5 MEOH
0.5 H
2
O
MEOH
220
o
C
3.5 atm
MEOH
HCHO
DME
DMM
H2O
H2
PMM
F5
F4
2a)
Reaction (1): CH3OH + 0.5 O2 HCHO + H2O
Basis: F1= 100 mol
HCHO = 10 mol
Dry air introduced into the MEOH converter = F2 = 2*F1=200 mol.
O2 in = 200 mol (0.21)= 42 mol
10 mol of HCHO formed, 0.5 (10) = 5 mol of O2 reacted.
2b)
F3/F1 = 0.2 F3 = 20 mol
F4 = F1+F3 = 100+20 = 120 mol, F4 Composition: 70 mol MEOH, 50 mol H2O.
