τ
=
∫XAf
CAO
1/RT + k2 CA
0
dXA
k1 CA
Now: CA
CAO (1-XA)
=
(1 + έAXA) HCHO + H2O
CH3OH + 0.5O2 Here έA =
VxA = 1 - VxA = 0 VxA= 0
=
2 - 1.5
=
0.333
1.5
Now, total no. of moles entering.
n
=
408.4135 kmol/hr
PVo
=
nRT
Vo
=
408.4135x8.314x519.3 182.6
Vo
=
9656.6 m hr
VO
3
=
1hr 60 min
nRT P
Vo
3
=
160.9 m /min
Now: F methanol
=
121.78kgmol hr
1hr 60 min
F methanol
=
2.02 kgmol /min
54