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Maintaining Raw Material Ratios In Formaldehyde Manufacturing


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#1 Aldehydes

Aldehydes

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Posted 02 June 2018 - 03:24 AM

I have only recently joined the Formaldehyde industry and am working on the designs of silver catalyst process. IN the process it is necessary to maintain a certain ratio of methanol, water (steam) and air (Oxygen) in the feed to the reactor. 

In one process variant, the methanol is vapourized (through indirect steam heating) with simultaneous feeding of the air. The Methanol to Air ratio in the vapours is kept as 1:1.8. In this process the temperature of the vapourizer is below the boiling point of methanol (approximately 46 deg C). I want to understand how is sufficient vapourization being achieved at this temperature. I found from literature that the Vapour pressure of Methanol at 46 deg C is about 50 kPa and since the liquid in the vapourizer stabilises at about 88 to 90 % by weight Methanol, the partial pressure of the Methanol in the vapour should be about 40 kPa. Similarly the water partial pressure in the vapour would be about 2 kPa. The total pressure in the vapourizer is 2000 mm WC and therefore the partial pressure of the air is about 78kPa. In this process the balance water requirement is added as steam to the air-methanol mixture.

In the above process the raw material ratios are maintained.

However, a side cut is available from the subsequent Formaldehyde absorption stage and this side cut can provide all the water requirement of the reaction. The side cut also contains Methanol and some formaldehyde. In order to avoid a separate vapourizer for the side cut, we added it to the methanol vapourizer. This resulted in the temperature in the Vapourizer increasing to about 76 deg C, However, it was no longer possible to feed all the required air. Instead of 1.8:1 air to methanol, we could only achieve a maximum of 1.3. Beyond this as the air is increased, the levels and the temperature in the vapourizer starts falling and the level cannot be maintained with additional quantities of water mixed with the methanol.

Why does this happen? The Air being at 63 deg C could be having a cooling effect on the water-methanol mixture in the vapourizer. That could explain the fall in temperature but I am unable to explain the fall in the level which seems to be caused by an increased evaporation of the methanol and water as air is increased. 

Why would this be happening. I have tried analysing using the methanol water Vapour Liquid Equilibrium curve but cannot find an answer.

Please help. 

Thanks

Alok Kapoor






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