Formaldehyde Fixed Bed Reactor

Fixed bed reactor design:

The 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 -r_A = k₁p_A/1+k₂p_A.

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,
C_A = P_A/RT , P_A = C_ART
And C_A = C_AO (1-X_A) / (1+ε_AX_A)

Finally, the rate expression is
1/-r_A = [(1+ε_AX_A)/k₁RTC_AO(1-X_A)] + k₂/k₁
Where ε_A is the fractional change in volume
ε_A = (moles of products - moles of reactants)/ moles of reactants
     = ( 2 + 1.88) – (1 + 0.5 + 1.88 ) / (1 + 0.5 + 1.88 )

     = 0.147
C_AO = P_AO / RT= 1 atm/RT
Substituting the values εA and CAo in the final rate expression and plotting the graph between 1/-r_A Vs X_A.



X_A 1/-r_A

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

From the graph the area under the curve for X_A = 0.9 = 205000 Kg-sec / K-mole
Design Equation for fixed bed reactor is W / F_AO = dX_A / -r_A
W / F_AO = 205000 Kg-sec / K-mole
F_AO = 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

The 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 (V_r = 3V_c)
V_r = Volume of the reactor
V_c = 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/m^{2 o}K
Tln = (149.6-60)-(343-100)/ln{(149.6-60)/(343-100)}= 153.35 ^oC
Heat Transfer Area AH = Q / U X T_ln = 35.33 m²
Number of Tubes present in the reactor N_t = AH /πdl
Assume diameter of the tube = 0.1m
Assume length of the tube = 4m
Therefore Number of Tubes N_t = 25