Biopowered - vegetable oil and biodiesel forum
Biodiesel => Chemistry and process => Topic started by: MattC on October 11, 2012, 03:48:40 PM
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Hi guys
I'm about to start my first batch using Sodium Methylate (picking up from Nigel on Saturday) and I've been trying to work through the calculations. The Wiki page gives JRL's simplified formula but I wanted to check a couple of things:
1. Is this based on 30% or 25% methylate - or is it an approximation that gives a rough answer for both?
2. Is the titration value based on standard 0.1% NaOH titration solution or does it assume titration using a solution made up using the methylate, and if so, at what concentration?
I found this calculator http://www.trianglebiofuels.com/calc.html (http://www.trianglebiofuels.com/calc.html) that claims to calculate the required volume of methylate, but it seems to produce roughly twice the quantity suggested by Jim's formula. Any idea why this might be the case?
Matt
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Hi Matt, does this thread help at all?
http://www.biopowered.co.uk/forum/index.php/topic,278.msg3789.html#msg3789
I have been working on a Methylate calculator for the tools section based on CountryPaul's formula (which factors in concentration - the 30*100 bit) a couple of posts below that one, but not completed it yet.
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Thanks. I'll give it a read and let you know. VOD's been down today so I couldn't look it up on there.
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The figures on my wiki page work very well for 28% Sodium Methylate which is what I use.
If you want to be sure that you are not using more methylate than you need, just drop the base figure slightly (ie maybe try 3.75 as opposed to 4.0)
As for titration fluid, I have no idea - I have not done such a thing since Vince went public with the no titration two stage process.
Two stage no titration lends itself so well to Methylate use because it results in much lower methylate consumption than a single stage titrated process - and at the best part of £2 per litre, you dont want to use any more than you have to.
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OK. So, if I understand it right, the formula:
(((titration +base) x litres)/40*54)/30*100)/0.95
... assumes you have titrated with 0.1% NaOH solution and uses these constants:
MW of NaOH = 40
MW NaOCH3 = 54
Methylate by weight = 30%
SG of Methylate solution = 0.95
Is that right?
Putting those together you get a multiplication factor of 4.74, rather than Jim's factor of 5, which might partially explain he and others have been able to reduce the base at little.
Next question: How would the formula change if you made up titration solution using methylate?
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28% methylate would give a factor of 5.1, which would sound about right.
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I went with a base of 4grams for both stages at the weekend with the 30%ASM I'm peddling and got a slightly hazy 5/45 at the end. Some say this is down the meth deficiency rather than methylate. I normally use a base of 4 and then 5 with NaOH.
i used just 3.5 lts of methylate for 150lts of oil.
Nige