continuing this from the 'straight acid' thread...

quote:

---

Originally posted by neutral:
You are right - most of the separation is methanol as shown by the fact that it normally floats up, but there will be a trace of glycerol there. If you run the reaction long enough with low methanol the phase will sink.

---

I have been told that this is due to the water. I feel like you're perhaps saying it's due to glycerol since this was in the context of 'how much, if any, glyerol is formed in the acid stage' (or am I misinterpreting?).


methanol management techniques:
There is an industrial process by which they do an interesting round of 'methanol management' (I think Diff hinted at this once).

They do multistage acid-base process such as acid-acid-base, and in between the acid stages, they make sure that one of the times they get a a bottom layer of 'alcohol' to separate instead of a top one (by managing the amount of methanol used for that stage), and let that (sinking) methanol/concentrated water bottom layer separate and remove it for a different form of processing (ie run it through a distillation column or something like that).

What they are removing is a layer of methanol/water/ffa/some oil/sulfuric acid/color bodies .

Here's the trick (supposedly):

If, for the second of the acid stages, you use much less methanol, you may or may not not get as much of a reaction as you'd want, but with little enough methanol used you might see the methanol layer form underneath the biodiesel/oil rather than on top. This phenomenol frequently happens with aleks kac's quantities of methanol (a small layer of what was mistakenly called 'first stage glycerol' by todd and/or other users).

If I understand correctly, you can supposedly predict that this bottom-sinking layer will contain much of the water that was in the reactor.

Since you didn't use much methanol, the water's specific gravity dominates, and the alcohol/water layer goes to the bottom instead of the floating to the top which would have happened if that layer had more methanol than water present (correct me if I'm misunderstanding chemistry somehow here)

You remove that bottom-sinking layer, do a titration on the feedstock, figure out where you're at with ffa reduction, and then keep going with whatever the next stage needed. It has given you a drier feedstock for whichever stage (acid or base) you do next. So if I understand what happens in practice in a plant (I may not because I only did this process in a lab), you're essentially running a 'sacrificial' water-scavenging stage, which is supposedly predictable enough that it can be commercially viable. It of course takes less energy and time to dry out a feedstock by methanol scavenging rather than by distillation of the water from your oil. It seems like an easy way to 'clean up the feedstock and keep going' (and then deal with 'drying' your recovered methanol somehow, which is a smaller volume to heat and deal with, than ferreting the pesky water out of the feedstock itself.



On that same topic, I was talking to a chemist the other day about anhydrous sodium methoxide and we were wondering how it;s made (or rather how it's dewatered).

He waid that silica gel is used for dewatering methanol industrially, get a big mesh size of beads (?), scavenge water from the methanol with them, and then strain out the beads (?). He pointed out that it's probably a NASTY substance at that point and regenerating it would pose some hazards, but... I didn't realize there was something that would absorb water but not methanol.

Mark

quote:

---

neutral wrote:
This is a tricky area. It would be nice to remove the layer after the acid stage, regardless of catalyst type, thus removing water and the acid, but at the same time the layer will take with it some esters which will lower yield. It seems too complex to justify the steps needed to recapture this loss.

---

Not sure how this (recovering any esters in the methanol/water layer) is done but it can be, so it is either not a huge problem (industrial scale, not homebrew) or it does not take so much esters with it as you may think. ( I don't pay attention to yield)

Perhaps the high water content prevents them from mixing into the solvent methanol?

By the way the people working this way are using trap grease and things like that. Yield might not be the biggest issue on their minds, so if there is a loss of yield it might not matter to them as much as it would to a homebrewer who has access to better feedstock.

[This message was edited by girl mark on 17 January 2004 at 04:37 AM.]

Very interesting Girl Mark. Yes, if the methanol content in a second acid stage is kept low the water content will be high and the solubility of FFA and esters in it will be low, perhaps too low to be worth recovery.

One would imagine however that in the previous step with higher methanol content, so that the layer floated, there would be enough ester and FFA to make recovery worthy of consideration.

Most of us have access to used oil which is not too high in FFA so that if acid is used one would hope that only one stage would be needed. Would you recommend in this case that the methanol be kept low in order to ensure it settles to the bottom and is wet enough to avoid much product loss due to solubility?

Does all the water end up in the methanol or is some bound up with (by?) the sulfuric acid? One process uses sodium bicarbonate to neutralize the sulfuric acid which supposedly creates a hydrate and gets the water down to less than 1%. Hence my earlier question about how much water can the sulfuric acid tie up.

quote:

---

Originally posted by neutral:

Would you recommend in this case that the methanol be kept low in order to ensure it settles to the bottom and is wet enough to avoid much product loss due to solubility?

---

I can't possibly recommend anything at this point, just provide some info on what I understand of this process. I simply have not done it enough to make any recommendations.

Is Tilly's alcohol tralles hydrometer a good way to see what's in the methanol layer? Or would it be too 'thrown off' by the amount of ffa and sulfuric in the mixture? or should I do an acid value test on the methanol to see what's in it next time?

Dropout,
I don't have much experience with this and I am not a chemist so I am somewhat taking a shot in the dark here.

I believe that Neutral has said here in another post (please correct me if I'm remembering wrong),
that in the Aleks Kac method, if you do a titration on the oil/ester mixture resulting from the acid stage and you get 1.5 ml (using the homebrew titration that we do), you are actually reading the leftover catalyst (and that is with Kac's very small quantities of catalyst, 1:1000 sulfuric : oil ratio, which is less than the ISU acid-base method and industrial acid-base patents used). It has been my experience with the Aleks Kac method that I have never gone below 1.5 ml on the homebrew titration results after the acid stage. I started to compensate by using appropriately more NaOH.

When doing the Iowa State method (once again: http://www.me.iastate.edu/biodiesel , click on ''technical papers'), I can get a lower reduction in acid number. But note- I've done lots of Aleks Kac acid-base in real life but have only done the ISU method on a 'half liter' kind of scale. So there might be other differences in measurements, etc that are causing the discrepancy.

Doing the weight-based acid value test as described here by Diff in the post "Acid Value", I have seen this method reduce FFA from dozens of percent to .78% (I actually do not have access to such bad grease here at home, the most dramatic case of this that I saw, was in the ISU course, where they proudly presented us with some melted trap grease of 80% ffa to work with. We got it to go to .78% (that's POINT 78) using only one stage of acid esterication. What the water content of the sample was after this I do not know, also it's important to point out that we were working in a beaker (easier reaction conditions than my reactor, perhaps?) and I have never upscaled to anything bigger than a beaker when trying this method since then - because of the stupid reason that I had thousands of gallons of better oil to work with so I had no need of acid-base production since May- and more importantly, I was unsure of my conversion from the weight-based ISU directions to my liters-based reactors (yes, a very stupid reason. It's that 10th grade education level thing again))

Anyway since the percent ffa was so low after this (what does that come out to in ml in homebrew-style titration, Neutral?), I have been assuming that much the sulfuric went to the methanol layer, otherwise it would have turned up on the titration and that titration (acid value, etc) would have seemed higher than ".78 % ffa (in that it would have also read residual sulfuric acid)

I grilled the guy who told me about the methanol management method, about reuse of the recovered methanol. He said that in his pilot plant they reused the sulfuric/methanol layer up to 5 or 6 times (and I didn't get an answer on what they did with it afterwards).

My impression from Dr. Van Gerpen is that he worries about the water content of that methanol building up during such recycling, etc, and was going to do some research on getting the real answer of 'how much is the water a problem' and what methods to use for getting rid of it. I also may well have some of this confused with straightforward flash evaporator methanol recover, though.


Dropout, what's the process with sodium bicarbonate neutralisation?
Mark

quote:

---

Originally posted by neutral:
One would imagine however that in the previous step with higher methanol content, so that the layer floated, there would be enough ester and FFA to make recovery worthy of consideration.

---

Neutral,
Do you have much data on this? maybe a simple yield measurement maybe could do the trick and tell you what you're leaving. Of course I have no idea if all the methanol separates or if some of it stays in the oil/ester layer. If some of it stays in the oil lyer it would seem that it'd throw off any yield test results.

from the sodium sulphate thread:

quote:

---

Originally posted by Airborne:
I have made zillions of lts of BCT Bio here and my feedstock titrates routinely between 2.5 and 3.5,..

I have also tried both Alex's and the ISU method exactly with Flop results !!!

---

I am also under the impression that acid esterification gives much more dramatic ffa reductions with HORRIBLE feedstock- getting 80% ffa trap grease going down to a few percent- and slightly less 'dramatic' reductions with lower-ffa feedstock.

In the ISU course example I just gave (80% FFA reduced to under 1% FFA in one step), the instructors were mildly surprised that it was this dramatic- they had expected to see it drop to a few percent and to then need another acid step. So this reduction to under 1% in one step was not completely normal results, so don't judge your own results by this example. I believe they said that it's more normal to get a huge reduction at first and then less of one if you have to do a subsequent acid step.


The ISU method involves varying both the methanol and the catalyst depending on the %ffa (which becomes difficult for a homebrewer without methanol recovery to justify- hence Aleks' one-size-fits-all approach for homebrew situations)

I think I've heard that there's some evidence that below a few percent ffa, there could be a problem of getting the methanol molecules to actually contact the remaining ffa molecules dispersed throughout the oil and ester- that that ISU formula doesn't provide quite enough methanol for low ffa feedstock to work as well as for high-ffa oil. I hope I got the information right, again I've done no experiments to prove or disprove this theory in my own experience.

So it might actually need an adaptation to deal with lowish (??) levels of ffa such as Airborne's (which homebrewers are handling with base-only process just fine in general. Though I found it useful at times to use Aleks Kac's method for reducing "3.5 ml titration" oil even though it's not the worst feedstock you might face).


mark

Yep lets cut to the chase....PHWOAR...Brain ache though....

Doesn't the ISU method use high molar ratios of methanol to overcome this? Unless we can build a simple way of separating the methanol and the wvo and then pulling out the water, it won't be a backyard process.

Girl Mark, the process I was referring to can be found at http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=/netahtml/search-bool.html&r=5&f=G&l=50&co1=AND&d=ptxt&s1=biodiesel&OS=biodiesel&RS=biodiesel

It does use higher molar ratios (than ALeks Kac method). I was just saying that there was one explanation for the less-dramatic reductions in ffa at lower ffa feedstock, which is that (possibly) the formula doesn't apply as written across all levels of ffa. But it works well enough, better than aleks's for instance.
yes, its not a completely backyard process.

Mark

Girl Mark

I referred above to small amounts of glycerol being produced in the acid stage. You agreed that under some conditions the methanol/water/acid phase sank and ascribed this to the domination of water when little methanol was used.

If you consider a bad oil, say 16% FFA, and use a low amount of methanol, say 8%, the water that could be produced would amount to 1%, if all the FFA converted, which it cannot do. So there cannot be more than about 10% water in the aqueous layer.

As the density of methanol is about 0.79 the addition of 10% water will only move the density of the mixture to about 0.81, well below the density of the oil. It seems therefore a significant amount of glycerol must be liberated in the acid stage if it gets to the point that this layer sinks.

It is curious that the US patent referred to above claims the use of sodium carbonate to neutralize the acid used in the acid stage but does not mention it in the examples. It seems to me that the idea has merit as neutralization with NaOH, while it produces the dessicant sodium sulphate, also produces water, while the carbonate does not.

The patent is also curious in that it does not go beyond 100g sample size in its examples, does not mention weight-for-weight yield, does not mention the FFA content of the product. It seems to confuse conversion with yield. Perhaps it is a bit of a rush job. The use of THF certainly appears to speed up the process although some of the extra speed would be due to using more methanol and more sulphuric acid than typical.

Does the sodium carbonate absorb or "tie up" the water and then drop to the bottom of the tank as a solid? Is there an easy way to calculate the water created by the sulphiric acid and the amount "cleaned up" by the sodium carbonate?

quote:

---

Originally posted by neutral:
It is curious that the US patent referred to above claims the use of sodium carbonate to neutralize the acid used in the acid stage but does not mention it in the examples. It seems to me that the idea has merit as neutralization with NaOH, while it produces the dessicant sodium sulphate, also produces water, while the carbonate does not.

---

Originally posted by Dropout:
"Does the sodium carbonate absorb or "tie up" the water and then drop to the bottom of the tank as a solid?"

No, it would react with the sulphuric to give off carbon dioxide and make sodium sulphate. It is the sodium sulphate that is the dessicant as it likes to form a hydrate. I don't think potassium carbonate does this but I might be wrong. I was certainly wrong above saying that sodium carbonate will not produce water when it neutralizes sulphuric. It does, but only half as much as when NaOH is used. Sodium sulphate is a bit soluble in glycerol but most will slowly settle out.

"Is there an easy way to calculate the water created by the sulphiric acid and the amount "cleaned up" by the sodium carbonate?"

Yes, 98 grams of sulphuric will produce 18 grams of water with sodium carbonate and 36 grams of water with NaOH. There is a hydrate with 10 molecules of water to one of sodium sulphate so potentially all of the water could be bound. Whether there is enough energy in the formation of this hydrate to go all the way to 10 molecules of water under these conditions I do not know. Some other kind of chemist might know.

[This message was edited by neutral on 20 January 2004 at 01:20 PM.]

Am I correct in understanding that the hydrate stays in solution with the wvo/methanol but that the water is not available to interfere with the base reaction?

If so, how is the hydrate washed out of the finished exters?

quote:

---

Originally posted by Dropout:
Am I correct in understanding that the hydrate stays in solution with the wvo/methanol but that the water is not available to interfere with the base reaction?

If so, how is the hydrate washed out of the finished exters?

---

No, the dissolved hydrate will be virtually entirely in the methanol/water/glycerol phase. The amount formed will exceed its solubility however and there will be a fine powder formed which will slowly settle. Most of any which remains unsettled in the esters after the acid stage will be collected in the glycerol of the base stage. Any which still remains unsettled will be easily washed out. This is exactly the same as what happens when NaOH is used as the neutralzer.

Now it's time for some experimenting.

How about sulfuric acid catalyst winding up partly in that phase? I can't remember it's density.

Mark

quote:

---

Originally posted by neutral:
Girl Mark

I referred above to small amounts of glycerol being produced in the acid stage. You agreed that under some conditions the methanol/water/acid phase sank and ascribed this to the domination of water when little methanol was used.

If you consider a bad oil, say 16% FFA, and use a low amount of methanol, say 8%, the water that could be produced would amount to 1%, if all the FFA converted, which it cannot do. So there cannot be more than about 10% water in the aqueous layer.

As the density of methanol is about 0.79 the addition of 10% water will only move the density of the mixture to about 0.81, well below the density of the oil. It seems therefore a significant amount of glycerol must be liberated in the acid stage if it gets to the point that this layer sinks.

---

Good point Girl Mark - sulphuric acid is very dense, about 1.8, so it could have a significant effect, depending on the amount used. In a more typical situation, say 5% FFA, using 0.25% sulphuric acid the density of this layer would move about 3% toward that of sulphuric acid, barely enough to notice. If you can get the layer to sink in that situation then glycerol is the main component doing it. In the case of bad oil, as in the sample above with 16% FFA, if using a proportionate amount of sulphuric, its contribution would be substantial.

The simplest way to answer the question would be to have a gc analysis done on the oil after the acid stage. From the amount of ester produced, after allowing for the ester from the FFA, and for mono and diglycerides, one could then calculate how much glycerol had been released. It would of course be no use measuring the free glycerol by gc of the oil at this stage because most of it would not be there - it would have settled out.

Girl Mark
I just read this, posted above, "the most dramatic case of this that I saw, was in the ISU course, where they proudly presented us with some melted trap grease of 80% ffa to work with. We got it to go to .78% (that's POINT 78) using only one stage of acid esterication".
That was my trap grease you were working with!
Weird that someting like that gives me such a great degree of pleasure.
Diff
Looks like crap but smells like feces.