Formaldehyde Fixed Bed Reactor

Fixed bed reactor design:

Catalyst used in fixed bed : SILVER

Reactants to Fixed bed Reactor: Methanol, Air
Products from Fixed bed Reactor: Methanol, Formaldehyde, Water, Oxygen, Nitrogen

Rate Equation derived from the literature survey for the production of formaldehyde using silver catalyst is
Formaldehyde and Water are formed in the following reactions:
  • CH3OH + ½ O2----> HCHO + H2O
The rate expression may be simplified to -rA = k1pA/1+k2pA.

Where p is a partial pressure in atm,and A refers to methanol . The Catalyst bulk density is 100lb/ft3 .

log10k1 = 11.43-3810/T
log10k2 = 10.79-7040/T
Where T is reaction temperature 520 K



But we know that,
CA = PA/RT , PA = CART
And CA = CAO (1-XA) / (1+εAXA)

Finally the rate expression is
1/-rA = [(1+εAXA)/k1RTCAO(1-XA)] + k2/k1
Where εA is the fractional change in volume
εA = (moles of products - moles of reactans)/ moles of reactants
     = ( 2 + 1.88) – (1 + 0.5 + 1.88 ) / (1 + 0.5 + 1.88 )

     = 0.147
CAO = PAO / RT= 1 atm/RT
Substituting the values εA and CAo in the final rate expression and plotting the graph between 1/-rA Vs XA.



XA 1/-rA

0--- 81632.7
0.1--- 92036.3
0.2--- 105041
0.3--- 121761
0.4--- 144054
0.5--- 175265
0.6--- 222082
0.7--- 300109
0.8--- 456163
0.9--- 924327

Conversion to rate of reaction graph for formaldehyde formation




From the graph the area under the curve for XA = 0.9 = 205000 Kg-sec / K-mole
Design Equation for fixed bed reactor is W / FAO = dXA / -rA
W / FAO = 205000 Kg-sec / K-mole
FAO = molar feed rate of methanol
= 201.80 K-mol/hr
= 0.0506 K-mol/sec
Weight of the Catalyst = 205000 X 0.0506 =11,491.38 Kg

But from the bulk density of silver catalyst = 100lb/ft3
= 100 X 0.453 Kg/0.3048m3= 1599 kg/m3
Volume of the catalyst = Weight of the catalyst/Density of the catalyst = 11,491.38 Kg/1599 Kg/m3

Volume of the catalyst = 7.186 m3
Assume Volume of the reactor to be three times of the catalyst volume, since gas flow into the reactor (Vr = 3Vc)
Vr = Volume of the reactor
Vc = volume of the catalyst
Therefore Volume of the reactor = 21m3

As,
Volume of the reactor = 21m3
π d2L/4 = 21

Assume L/D ratio as 2
Diameter of the Reactor = 2.4m
Length of the Reactor = 5m
Total Heat evolved in the reactor Q = 896043 Kcal/hr
U = Overall Heat Transfer Coefficient = 900 W/m2 oK
Tln = (149.6-60)-(343-100)/ln{(149.6-60)/(343-100)}= 153.35 oC
Heat Transfer Area AH = Q / U X Tln = 35.33 m2
Number of Tubes present in the reactor Nt = AH /πdl
Assume diameter of the tube = 0.1m
Assume length of the tube = 4m
Therefore Number of Tubes Nt = 25