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quote:
Originally posted by RickDaTech:
Mark,

I'm not going to try to get you to change anything especially how you dispose of glycerin. Keep doing however you feel is appropriate.

Repeatability is at the core of the scientific method. You have done all you can. Now it is up to others to verify your method is repeatable.

Ricky


Rick,

I'm not sure why you think I need to change anything, particularly the way I dispose of my glycerin. If you think I'm making up the fact that I've worked in the waste management industry for years, and have a detailed and comprehensive understanding of waste disposal laws and practices, make a few phone calls yourself. Ask the Water Corporation, or whoever the relevant WWTP authority in your area is, whether it's Ok to dump highly alkaline, methanolic glycerin down your sink.

As for repeatability of my method, anyone that understood the chemistry of my method would see that its foolproof nature is inevitable. As it happens there are many people using it, and I've not had a single report of a failure. Many of them, however, are frightened to post on either this site or the Australian site as it can be a very hostile site for those that challenge the status quo. I however, am happy to engage with people that have genuine questions or comments.

But as I've said before, my next batch will be a public demonstration
 
Registered: September 28, 2017Reply With QuoteReport This Post
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Originally posted by WesleyB:
RickDaTech; I used new corn oil from a grocery store. I did not want to complicate matters so I used new oil. I was researching if room temperature tranesterification proceeds at a reasonable rate. Used vegetable oil can be wet. It didn't make sense to use dried methoxide with potentially wet oil. I titrated the new corn oil once at about 0.6, it was not zero.


Wesley,

I'm not sure I understand your logic. The methoxide solution made by any method other than mine contains water:

MeOH + KOH = KMeO + H2O

Why, then, would a bit of water in the oil be a problem with my method? I've said it before, and I'll say it again, I'm astonished at how many people worry about water in their oil, and then introduce water into their mixture from the methoxide solution..

But in any case, as I've said, I've used this method with some very poor quality oil. I once picked up about 1600L of oil in 20L drums from a bloke in Capel. It had been sitting outside for years. Due to the silly design of 20L oil drums, most of them had free phase water in there. In some cases the drum was almost entirely water. When I used to pinch WVO from the back of cafes in my trailer, in many cases the drums had free phase water in there.

As long as the WVO that is used, even if it comes from a drum with free phase water in there, is clear, as opposed to murky and translucent (because of the presence of water micelles as the dispersed phase)there'll be no problem.

Never done a titration of WVO
Never heated my mixture
Never had a problem
 
Registered: September 28, 2017Reply With QuoteReport This Post
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Your equation where methanol reacts with potassium oxide to form methoxide and water needs an arrow pointing both ways, forwards and reverse. An equals sign is misleading since the reverse reaction occurs setting up an equilibrium. Another equation is CaO + H2O yields Ca(OH)2. But removing the water from calcium hydroxide takes a lot of energy. This pushes the methoxide forming reaction to the right by removing available water for the reverse methoxide plus water yields methanol plus OH- ion that accompanies K+ ion.
 
Location: Texas | Registered: April 27, 2011Reply With QuoteReport This Post
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Hi Mark,

quote:
Originally posted by Mark;
I have found that even using poor-quality WVO, as long as the oil is transparent (as opposed to translucent or opaque), it's good to go. And that's even from drums that have free phase water in them. As long as you (obviously) don't decant either the water, or the murky phase above the water, but only the clear oil above that, the level of water it contains is low enough so as to not interfere with the process.
That is interesting.
So you think that having a bit of water in the oil is OK, It is the water produced when mixing KOH with Methanol that will cause a problem.
Just out of curiosity, if the water in the oil was actally interfering with the reaction, what would this interference look like? In other words, how would we know there was a water problem?

Oh, by the way, you said that you produce about 22% by- product (what you call glycerol) when you make biodiesel.
When I make biodiesel, I make it in a translucent plastic drum with a mark at the full line for oil (130 litres) on the side of the drum.
This year I used about 13% methanol for my reaction (16 litres). I noted that when I drained the by- product (That which you call glycerol). As per normal, after the by- product had been drained, the level of biodiesel was above the full oil mark. that would be about 10%- 12% by- product by volume of the total contents of the reactor.
Over the years, I have used varying amounts of methanol. I have never produced by- product that drained so the remaining biodiesel was below the full oil mark.

This message has been edited. Last edited by: Tilly,






 
Location: ลึก ประเทศอินเดีย | Registered: March 03, 2001Reply With QuoteReport This Post
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WesleyB

Thank you for filing in some of the details.

Rick
 
Location: Cowboy Country | Registered: December 06, 2004Reply With QuoteReport This Post
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Hi everyone,

I just noticed that this "status Quo" changing procedure wastes 15% of the methanol you buy.
The recipe is to add 20% (200ml) methanol for each litre of oil being reacted into the cement powder for drying.
You only recover 17% (170ml) of the methanol from the drying procedure. That means 3% (30ml) or 15% of the methanol you dry is going into the waste stream along with the cement without ever being used in the reaction.

That means that for every 200 litre drum of methanol you buy, you will throw 30 litres of the methanol straight into the rubbish
Here in paradise where a 200 litre drum of methanol costs, at the minimum $250, that is throwing away at least $37 worth of methanol from each 200 litre drum of methanol you buy.

I can not see many people doing that any time soon.






 
Location: ลึก ประเทศอินเดีย | Registered: March 03, 2001Reply With QuoteReport This Post



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quote:
Originally posted by WesleyB:
Your equation where methanol reacts with potassium oxide to form methoxide and water needs an arrow pointing both ways, forwards and reverse. An equals sign is misleading since the reverse reaction occurs setting up an equilibrium. Another equation is CaO + H2O yields Ca(OH)2. But removing the water from calcium hydroxide takes a lot of energy. This pushes the methoxide forming reaction to the right by removing available water for the reverse methoxide plus water yields methanol plus OH- ion that accompanies K+ ion.


Yes of course double arrows are the correct symbol. If you can find a keyboard or a font on this website that has them I'll gladly insert them.

And I'm sorry I don't understand your comment about the Ca(OH)2 (hydrated lime). At no stage have I suggested removing water from the Ca(OH)2. The reactions are (apologies for absence of correct arrows):

Reaction 1: MeOH + KOH = KMeO + H2O (an equilibrium)
Reaction 2: CaO + H2O = Ca(OH)2 (an equilibrium, but lying very much to the right)
__________________________________________________________________
Reaction 3 (1 + 2): MeOH + KOH + CaO = KMeO + Ca(OH)2 (an equilibrium lying very much to the right)

As the dehydration reaction (2) lies very much to the right, it will pull reaction 3 to the right. Therefore, effectively, giving a high conc of catalyst and no water.

As far as dehydration agents goes, we of course normally have to consider the position of the equilibrium.

For example, Na2SO4 and MgSO4 are both very effective drying agents but with different properties.

Na2SO4 has a very high capacity - Na2SO4 + 10H2O = Na2SO4.10H2O

MgSO4 has a much lower capacity - MgSO4 + 2.H2O = MgSO4.2H2O

So at first glance you would say that the sodium salt is the better dehydrating agent. But there is another factor - the position of the equilibrium, which on the Mg salt lies more to the right. So a popular approach is to use the Na salt to sop up most of the water, and then use the Mg salt to pull the last bit of water out to ensure complete dryness.

With the CaO, however, it just doesn't adsorb water (like the sulphates above) it actually forms a new crystal structure with the Ca(OH)2, resulting in an equilibrium lying so far to the right that, effectively, you probably wouldn't even call it an equilibrium. That's why it's the standard compound to use in organic labs to dehydrate alcohols - even more effective than MgSO4 or CaCl2 (drierite)
 
Registered: September 28, 2017Reply With QuoteReport This Post
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Location: New Zealand | Registered: August 15, 2006Reply With QuoteReport This Post
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