2 replies to this topic

[blooma]

Hi all,

I`m working on a biodiesel production on a small scale.I have a 6 m^3 tank and i`m mixing methanol and KOH/NaOH in it at 55 deg C to get methoxide needed as part of the production.

I need to design that tank and take into consideration all safety regulation. the methanol can evaporate during that mixing so i thought about installing a reflux system that can keep the ambient pressure but forbid the methanol from leaving the tank.

My problem is : what is the evaporation rate of the methanol ? i need it to calculate the amount of cooling fluid (water) for the reflux so i can close the mass/energy balances.

methanol BP is ~68 deg C at 1 atm

thanks
Blooma

Faculty of chem. eng
Technion
Israel

[Art Montemayor]

blooma:

Your query involves a RATE. The vapor pressure of methanol - like that of most any other liquid - is dependent on the temperature of the liquid.

Therefore, to find out how fast (or at what rate) you are vaporizing methanol, you must know (or furnish) the temperature of the methanol liquid. We all know at what temperature methanol boils at under atmospheric pressure. What we have to know is what temperature of methanol liquid you have and at what rate are you heating the methanol at that temperature.

You must furnish a heat rate in order to heat the methanol. If you have a batch process, then you have unsteady state heat transfer - something that is a little harder to figure out than steady state heat transfer. Is your KOH/NaOH mixture causing an exothermic reaction with the methanol? You don't mention this. You seem to expect a heat input into the methanol, but where is the heat coming from - and at what rate?

[blooma]

Dear Mr. Montemayor

Thank you for your guidance and for your quick answers. It took me a few weeks but now i think that i have all the information i need in order to continue this project.

I have calculated the enthalpy of the process and made one experiment with NaOH:
It`s two different reactions - the first one is the dissolution of the NaOH and the second one is the methoxidation (methanol+soda caustic=methoxide).
I've calculated the enthalpy differences from their heat of formation and got a delta H of ~5700 kcal for the quantities i`m using.
The proposed reaction time is 25 minutes so assuming linear heat emission the heat rate should be ~230 kcal/min.

two important comments -
* I`m not heating the vessel during the reaction
* The temperature differences calculated should be 60 deg C and it`s actually more like 30 deg C

My question is : how much methanol evaporates during this reaction?

Thank you very much
blooma