**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

The rate Equation derived from the literature survey for the production of formaldehyde using a silver catalyst is

Formaldehyde and Water are formed in the following reactions:

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

The rate expression may be simplified to -r_{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+ฮต_{A}X_{A})

Finally, the rate expression is

1/-r_{A} = [(1+ฮต_{A}X_{A})/k_{1}RTC_{A}_{O}(1-X_{A})] + k_{2}/k_{1}

Where ฮต_{A} is the fractional change in volume

ฮต_{A} = (moles of products – moles of reactants)/ moles of reactants

= ( 2 + 1.88) โ (1 + 0.5 + 1.88 ) / (1 + 0.5 + 1.88 )

= 0.147

C_{AO} = P_{AO} / RT= 1 atm/RT

Substituting the values ฮตA 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

The design Equation for a 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/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 the Volume of the reactor to be three times 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

The volume of the reactor = 21m^{3}

ฯ d^{2}L/4 = 21

Assume the L/D ratio is 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 /ฯdl

Assume the diameter of the tube = 0.1m

Assume the length of the tube = 4m

Therefore Number of Tubes N_{t} = 25

## Related Search โ Click Here

- MASS TRANSFER DRYING
- What Is Diffusivity And How It Is Related To Mass Transfer And Its Coefficient, How They Are Related?
- Level Measuring Methods | MAGNESIUM AND ALKALINITY
- Hydrorefining Of Crude Benzol Flow Sheet
- Mass Transfer Operation Humidification
- HEAT TRANSFER In Boiling Condition
- HOLDโ UP STUDIES (FLOODING VELOCITY)
- Calculation Of Steam Consumption And Vaporization Capacity Using OPEN PAN EVAPORATOR
- What Is Glass And What Types Of Glasses