**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:

- CH
_{3}OH + ½ O_{2}----> HCHO + H_{2}O

_{A}= k

_{1}p

_{A}/1+k

_{2}p

_{A}

_{}.

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

^{3}.

log

_{10}k

_{1}= 11.43-3810/T

log

_{10}k

_{2}= 10.79-7040/T

Where T is reaction temperature 520 K

But we know that,

C

_{A}= P

_{A}/RT , P

_{A}= C

_{A}RT

And C

_{A}= C

_{A}

_{O}(1-X

_{A}) / (1+ε

_{A}X

_{A})

Finally, the rate expression is

1/-r

_{A}= [(1+ε

_{A}X

_{A})/k

_{1}RTC

_{A}

_{O}(1-X

_{A})] + k

_{2}/k

_{1}

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

_{A}

_{O}= P

_{A}

_{O}/ 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

_{A}

_{O}= dX

_{A}/ -r

_{A}

W / F

_{A}

_{O}= 205000 Kg-sec / K-mole

F

_{A}

_{O}= 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/ft

^{3}

= 100 X 0.453 Kg/0.3048m

^{3}= 1599 kg/m

^{3}

Volume of the catalyst = Weight of the catalyst/Density of the catalyst = 11,491.38 Kg/1599 Kg/m

^{3}

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 = 21m

^{3}

As,

Volume of the reactor = 21m

^{3}

π d

^{2}L/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

^{o}C

Heat Transfer Area AH = Q / U X T

_{ln}= 35.33 m

^{2}

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