Biopowered - vegetable oil and biodiesel forum
Biodiesel => Chemistry and process => Topic started by: nathanrobo on March 28, 2013, 07:51:11 PM
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I can't take the credit for this thought as it's Steve's idea... Still I can't remember it being pointed out elsewhere, so I thought it was worth mentioning.
But when we do our mix, how do know
1. that the reaction has stalled
2. why it stalled - in other words was it because the the lye was all used up or because the meth has been used up
Whether the reaction stalled or not, if there is still unused lye present, we're doing a calculation based on all of the idea that it's all used up. So potentially we're overdosing.
Steve's solution is to put a far higher percentage 90% plus of the meth in the first stage with 10 % in the second stage.
Comments...
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Should it not be:
Base catalyst + 20% meth of volume, then 2nd stage base catalyst + 20% meth of unreacted volume?
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I think there's milage in this.
I used to do non titration with NaOH adding 1000G and 25L of meths (200L batch) for the first stage,
Steve suggested I may be stalling due to lack of meth,
I then started using only 800g and had the same amount of dropout in the first stage as I had before.
I now use ASM so didn't take things further.
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I've always advocated using a high percentage of meth on stage one, 75 - 80%, this should produce and excess at the end of the stage, yet only 65 -70ish % for stage two.
That way I'me fairly confident that any stalling of stage one will be due to my meager ration of ASM, 3.75 base, 4 base for stage 2.
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For dosing using the No Titration method the stall in reaction will always be the catalyst being exhausted and used up. The whole idea is to under dose stage 1. When the catalyst has been used in conversion the process stalls
The only time that methanol would come into the equation is if the amount as been scrimped away or the calculations are wrong.
I think we all know, as experienced brewers, that you can only cut back on raw materials so far before an adverse reaction takes place. This is why stage 1 of a No Tit, a deliberate under dose, always shows a fail.
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Jim, so are your meth %s based on the stoich of 14%, or the generally accepted 20%?
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Jim, so are your meth %s based on the stoich of 14%, or the generally accepted 20%?
The trade standard 20%.
If your recovering the methanol why skimp and risk a stall, demeth times don't vary much irrespective of volume being reclaimed.
Using ASM and reclaimed topped up with Virgin I'm only using 10 - 11% virgin per batch.
1800+ltr of bio from a 200ltr drum of meth.
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For dosing using the No Titration method the stall in reaction will always be the catalyst being exhausted and used up. The whole idea is to under dose stage 1. When the catalyst has been used in conversion the process stalls
I thought the idea of the catalyst was to cause the reaction to happen, but is not part of the reaction, so what do you mean by used up? I know most of it ends up in the glyc so does that mean it is neutralised in some way or just locked into it?
Sorry if it's a dum question but if something doesn't add up in my head I prefer to ask an expert rather than lose sleep ;D
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I think your question needs to directed at a chemist. I'm not suitably qualified to give an answer that would, in all probability, raise more questions.
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Strictly speaking, you should be able to reclaim and reuse a true catalysts ie it is not consumed by plays a regenerative role in the reaction.. In our case, you get water and sodium or poatssium salts of ffas and glycerol as by products. So, the base we use is somewhere in between. Niether stoichiometric nor catalytic.
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I think your question needs to directed at a chemist. I'm not suitably qualified to give an answer that would, in all probability, raise more questions.
Are you a politician in your spare time :)
Some things I am reading that may contribute to the debate:
http://www.sciencelearn.org.nz/Contexts/Nanoscience/Science-Ideas-and-Concepts/Chemical-reactions-and-catalysts (http://www.sciencelearn.org.nz/Contexts/Nanoscience/Science-Ideas-and-Concepts/Chemical-reactions-and-catalysts)
Catalysts
Catalysts play an important part in many chemical processes. They increase the rate of reaction, are not consumed by the reaction and are only needed in very small amounts.
There are two main ways that catalysts work.
Adsorption
Particles stick onto the surface of the catalyst (called adsorption) and then move around, so they are more likely to collide and react.
Intermediate compounds
In this process, a catalyst first combines with a chemical to make a new compound. This new compound is unstable, so it breaks down, releasing another new compound and leaving the catalyst in its original form. Many enzymes (special biological catalysts) work in this way. Many industrial chemical processes rely on such catalysts.
Also from the article on using Acetone as a co-solvent
http://pubs.rsc.org/en/Content/ArticleLanding/2011/GC/c1gc15049a (http://pubs.rsc.org/en/Content/ArticleLanding/2011/GC/c1gc15049a)
we have concluded that the retardation of FAME formation after the
formation of GL can be explained as follows: in the case without
solvent,
GL cannot be dissolved in the
oil or FAME, but methanol and KOH catalyst dissolve well in
GL. Therefore, FAME formation is retarded after the formation
of GL due to the dissolution of the important reactant methanol
and the catalyst into the GL phase, which easily precipitates
and is excluded from the reactant solution. On the other hand,
FAME formation is not retarded and does not require excess
methanol in the presence of IPA, which forms a homogeneous
solution that includes GL,
So while everything is still mixed using IPA the reaction can continue, but that of course would mean it would never separate, it just shows the separation is the factor that stops the reaction or a lack of methanol?
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I think your question needs to directed at a chemist. I'm not suitably qualified to give an answer that would, in all probability, raise more questions.
Are you a politician in your spare time :)
LOL ;)
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I think your question needs to be directed at a chemist. I'm not suitably qualified to give an answer that would, in all probability, raise more questions.
Are you a politician in your spare time :)
No I'm not. As you can tell by my response I wouldn't attempt to comment on a question I don't understand....especially when it is very specific.
However, there are some out there who will.....and they often come unstuck as they can't back their "knowledge" up. ;)
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I thought the idea of the catalyst was to cause the reaction to happen, but is not part of the reaction, so what do you mean by used up? I know most of it ends up in the glyc so does that mean it is neutralised in some way or just locked into it?
Sorry if it's a dum question but if something doesn't add up in my head I prefer to ask an expert rather than lose sleep ;D
I'll have a go at explaining in my laymans terms and from info read.
Sodium/Potassium Hydroxide water and Vegoil is the formula for Soap, in our conversion the first reaction is Soap 'cause you can never get bone dry wvo. Mixing either catalyst with Methanol produces water.
When titrating the value in g/ltr is used up in the soap reaction when neutralising the FFA, this produces water so more catalyst is used up.
Soap and Glycerol in the mix will retard or stall the reaction, this is where the titless method comes into its own.
We add what we consider necessary in catalyst/methanol to deliberately stall the reaction once the FFAs have been converted to soap and a good majority of the glycerol has been stripped from the Trigyleride molecule and remove the by product.
This leaves a relatively clean miix to finish, the fact that there are no FFAs and very little glyc we only need a reduced percentage of methanol and catalyst to move forward to completion.
Any glycerol drawn off will contain any catalyst not used in the production of soap, it can be reclaimed and used again BUT it's not a simple process and for the home brewer and not worth the effort and cost.
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I thought the idea of the catalyst was to cause the reaction to happen, but is not part of the reaction, so what do you mean by used up? I know most of it ends up in the glyc so does that mean it is neutralised in some way or just locked into it?
Sorry if it's a dum question but if something doesn't add up in my head I prefer to ask an expert rather than lose sleep ;D
I'll have a go at explaining in my laymans terms and from info read.... "Any glycerol drawn off will contain any catalyst not used in the production of soap, it can be reclaimed and used again BUT it's not a simple process and for the home brewer and not worth the effort and cost."
Mate does that mean that there's not any significant amount of excess catalyst left over in the bio for the 2nd stage? In other words, it doesn't matter why the reaction stalled, or if indeed it did. Once the glyc is removed, the only thing that matters is the calculation based on the 10/90??
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For dosing using the No Titration method the stall in reaction will always be the catalyst being exhausted and used up. The whole idea is to under dose stage 1. When the catalyst has been used in conversion the process stalls
I thought the idea of the catalyst was to cause the reaction to happen, but is not part of the reaction, so what do you mean by used up? I know most of it ends up in the glyc so does that mean it is neutralised in some way or just locked into it?
Sorry if it's a dum question but if something doesn't add up in my head I prefer to ask an expert rather than lose sleep ;D
Have a read of this thread -
http://www.vegetableoildiesel.co.uk/forum/viewthread.php?tid=35594#pid402702
The issue is that whilst the 'catalyst' catalyses one reaction, where, as you say, its a 'catalyst' and, by definition, is not 'used up', that is not the only reaction taking place. I tried to explain the two other main ones, with diagrams from the web, in the thread linked to above.
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ps.
you will see that, in the acid/base reaction -
(http://www.all-about-ph.com/images/acid-alkaline-reaction.jpg)
water is produced. Which then, of course, feeds the sanctification reaction again. Double whammy on 'catalyst' removal!
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I copy and paste this explanation that chemist Neutral from infopop posted.
"Any school girl knows that a catalyst by definition does not take part in the
chemical reaction. A catalyst enhances the reaction but is not "used" up in the reaction."
So far so good. What are the facts? We know that in the biodiesel making reaction we must put in a particular amount of NaOH to get the reaction to go close to completion, even with a new oil with virtually zero FFA. If the NaOH were just behaving as a true catalyst putting in half the NaOH would result in the reaction taking twice as long but still going just as far. Clearly this is not what happens.
The only possible explanation for what is observed is that the NaOH is doing at least two things: acting as a catalyst for the desired reaction and also acting as a reagent in another reaction. What can that other reaction be?
Let's look at the desired reaction first. The NaOH in methanol produces methoxide ions. The triglyceride is attacked by the plentiful methoxide ion which attaches to the carbon at the business end of the fatty acid chain. The electrons then rearrange and the glycerol portion falls off leaving the methoxyl group in place, thus producing the methyl ester. The methanol is consumed but the NaOH is not. This is a true catalytic reaction.
Now consider alternative reactions. If, and only if, water is present the following reaction can occur: the NaOH generates hydroxyl ions from the water which compete with the methoxide ions, so that occasionaly one succeedes in attaching to the end carbon. If this happens the glycerol will still fall off but, instead of the methyl ester being made, a free fatty acid will result. This immediately reacts with the NaOH present to form soap, and here's the rub - in doing so a molecule of water is formed, replacing the molecule which was consumed in making the hydroxyl ion. So we see that NaOH is consumed but the water is not. If you like the fine points of symantics you might say that the NaOH was acting as a catalyst for its own consumption. It is simpler to say that the NaOH is a reagent rather than a catalyst.
The important point to note here is that the water participates in the production of soap but is not consumed. It is therefore a catalyst in the unwanted side reaction. This shows the importance of having the oil reasonably dry if a high yield of ester is to be obtained and to minimize the amount of NaOH needed to drive the reaction close to completion. However even if the oil is bone dry the unwanted side reaction will still occur as water is produced when NaOH is dissolved in alcohol. The only way this can be prevented is by using sodium metal or sodium methoxide instead of NaOH, both expensive.
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I copy and paste this explanation that chemist Neutral from infopop posted.
"Any school girl knows that a catalyst by definition does not take part in the
chemical reaction. A catalyst enhances the reaction but is not "used" up in the reaction."
So far so good. What are the facts? We know that in the biodiesel making reaction we must put in a particular amount of NaOH to get the reaction to go close to completion, even with a new oil with virtually zero FFA. If the NaOH were just behaving as a true catalyst putting in half the NaOH would result in the reaction taking twice as long but still going just as far. Clearly this is not what happens.
The only possible explanation for what is observed is that the NaOH is doing at least two things: acting as a catalyst for the desired reaction and also acting as a reagent in another reaction. What can that other reaction be?
Let's look at the desired reaction first. The NaOH in methanol produces methoxide ions. The triglyceride is attacked by the plentiful methoxide ion which attaches to the carbon at the business end of the fatty acid chain. The electrons then rearrange and the glycerol portion falls off leaving the methoxyl group in place, thus producing the methyl ester. The methanol is consumed but the NaOH is not. This is a true catalytic reaction.
Now consider alternative reactions. If, and only if, water is present the following reaction can occur: the NaOH generates hydroxyl ions from the water which compete with the methoxide ions, so that occasionaly one succeedes in attaching to the end carbon. If this happens the glycerol will still fall off but, instead of the methyl ester being made, a free fatty acid will result. This immediately reacts with the NaOH present to form soap, and here's the rub - in doing so a molecule of water is formed, replacing the molecule which was consumed in making the hydroxyl ion. So we see that NaOH is consumed but the water is not. If you like the fine points of symantics you might say that the NaOH was acting as a catalyst for its own consumption. It is simpler to say that the NaOH is a reagent rather than a catalyst.
The important point to note here is that the water participates in the production of soap but is not consumed. It is therefore a catalyst in the unwanted side reaction. This shows the importance of having the oil reasonably dry if a high yield of ester is to be obtained and to minimize the amount of NaOH needed to drive the reaction close to completion. However even if the oil is bone dry the unwanted side reaction will still occur as water is produced when NaOH is dissolved in alcohol. The only way this can be prevented is by using sodium metal or sodium methoxide instead of NaOH, both expensive.
If understood correctly the NaOH or KOH is consumed during the reaction. The question remains outstanding... can a situation arrise whereby the reaction stalls or stops because of insufficient meth, where there is still NaOH or KOH present. Doing a drop out test in this scenario could mean that your 10/90 test results in too much caustic being used in the 2nd reaction.
On a slightly side point - any reason why AHM would be used without the down sides of ASM?
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can a situation arise whereby the reaction stalls or stops because of insufficient meth, where there is still NaOH or KOH present.
Only if you have absolutely no concept of what you're doing. Sadly a far to common situation in Home brewing.
The second stage wouldn't really be affected by excess catalyst, there's no FFAs or water present (apart from that generated in the small amount of methoxide mixed for stage 2).
ASM + methanol does not create water.