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A two step process that always works, but is not fool proof.
Before Acit Pretreatment of High Free Fatty Acid WVO Feedstocks, it is necessary to determine the Acid Value of WVO Equipment needed: Sulpfuric Acid, KOH, phenolphthalein, Methanol, High FFA Feedstock Scale (accurate to 0.01G) Glass containers w/ tight lids Acid Value Measurement: 1) Prepare a beaker by adding ~ 125 solvent. Ideal solvent consist of combination of 100% isopropyl alcohol and toluene 2) To the beaker add ~ 2 ml phenolphthalein 3) Titrate the beaker with 0.1 KOH to first permanent pink 4) Add about 5 grams of sample (weigh to get exact amount) to the beaker 5) Titrate the beaker with 0.1 KOH to first permanent pink 6) Calculate the Acid Value (AV) of the WVO from the following: AV= A x N x 56.1/W Where: A = (Titration in ML of KOH) N = normality of KOH (0.1) W = sample of WVO in Gram 7) % FFA =~~ 1/2 (AV) Procedure: 1) Measure the AV of the feedstock. Remember that FFA% is equal to approximately 1/2 of the AV. 2) Put 1,000G of the feedstock heated to 140 deg. F in ~2L container 3) Measure out in another container an amount of methanol = to 2.25 g of methanol for every gram FFA in the flask. 4) Add an amount of concentrated sulfuric acid equal to 0.05g for every gram of FFA in the ALWAYS ADD THE ACID TO THE METHANOL AND NOT THE OTHER WAY AROUND Procedure (continued): 1) Agitate the methanol/acid mixture until well mixed. 2) Add to the feed stock 3) Agitate at 140 deg. F for one hour 4) Let cool 5) Separate the two layers 6) Measure the AV of the WVO 7) Repeat until the AV value comes close to Virgin Oil Stock:~ 3~4 Then proceed as normal with base transesterification using NOH or KOH amounts determined by titration of the treated WVO. Note: water is created by this process. Water stops the process. Therefore, especially high FFA content WVOs will need to be treated several times to get them from say 90%FFA to AV of 3 or 4. But generally, using 93% sulfuric acid, I can change the AV of WVO from 14 to 3 in one step. Hope that herps. Oko San This message has been edited. Last edited by: Kenito Oko, |
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Oko San
You don't make it clear in your method where you refer to 0.1 KOH that it is meant to be 0.1N KOH which is 5.611g KOH per litre. That is how you do it in the industrial process. In the Tickell method you use a solution with 1g KOH per litre, a 0.1% solution. Most biodiesel makers are using the Tickell method and could easily get this wrong. If using NaOH for the reaction you would use 4g NaOH per litre to make a 0.1N solution for the above method. You would use 1g NaOH per litre of solution for a Tickell titration. This message has been edited. Last edited by: neutral, |
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Girl Mark Fire, neutral, et al
Kento Oko is so sorry for nevel explesing, how deep is his nevel ending glatitude to both of you. It is learry velly hald fol him to contemprate his rife without both of you. Thank you both fol arr you have given me. I am deapry glatefurr. Kenito Oko |
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Girl Mark
I don't know who first suggested using 1g NaOH per litre of water for titration. Possibly it has been thought of many times in many places, because of its obvious convenience, and no one has thought it remarkable enough to comment on. I call it Tickell titration simply because it was in his book that I first saw it used. It is not only convenient but also gets rid of part, but not all, of the error that will occur if the NaOH is not as pure as it is believed to be. I think it makes sense to call it Tickell titration on this forum because most people here saw it first in Tickell's book and know the method. If you know who published the method first I will be only too happy to acknowledge the fact. The magic number to convert mls of Tickell Titration to FFA% is of course 1.3, near enough. TT(mls) / 1.3 = FFA% |
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Girl Mark and neutral
I was intrigued that Kenito Oko San's thread got so little response. As far as I can tell, it is the only solid thing he has contributed here. He probably stole it :-) Fascinating though that so few people have made comment about this appraoch to pretreating fatty acids, but over 2,000 have tuned in to that five star fire storm he started over on the biodiesel politics side. Diff |
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re :Acid Pre-treatment Method,
After running with the Acid method ( 8k ltrs) I regretfully had to cease and return to Base only. Shame as now have to accept lower Yields (and higher By !!). Acid method produced low conversion Bio that was too vulnerable to Low Temps and as its Winter here (Land of Ozz)..not-good. Probable cause being water created during acid stage yep, but without a suitable fix ... buggered at present...grr. Just sharing my experience Cheers |
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Should it not be possible to separate the water by allowing it to settle for a sufficient period in a manner similar to that used to separate water from the original WVO? If this is not workable could some sort of dessicant be used to absorb the water? As a last resort I would imagine you could heat and boil the water off, but you then of course would boil off the methanol as well. I would expect similar results with vacuum distillation. Perhaps a vacuum recovery of methanol followed by a vacuum distillation of the water would be workable.
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Yep Dave_D...anything is possible ...!!! but sussing the most energy efficient, most practically efficient,etc etc fix is the prob at present hey..???
cheers |
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Airborne, an intersting whilst unfortunate outcome, I have my 4th batch from acid base sitting nicely, I used it in a trip to places cold and suffered only where the cold is best (Canberra) must be the politicians and the crap they sprout. A bit of diesel mixed in fixed that.
I suggest that the fact I start with a titration of about 3 to 4 may reduce the amount of water produced in the frist phase. I was believe the suphuric is a good drying agent absorbing water, since its not used up in the reaction but merely a catalyst (H2SO4) its doing its job as far as I can see here. Could not be happier with it, and its working out cheaper, 1 litre of 98% acid is cheaper than 3 to 4 Kg of KOH, I use the same heat as the acid phase to do the base, insulated of course. After 24 hours the titration is (last 4 times) 1.5 and does not go lower. Matt |
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maybe you could try a cat litter of some type. i know they absorb water and some clump up possibly making removal easier. since it is relatively cheap it should be worth a shot.
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Those contriversial acusorb beads are supposed to do the job.
Or ask a chemical supply house about molecualr sieves. Cat litter might work but I seem to remember someone saying it all gets gacked up with oil and BD or falls into sludge or something. You need something selective that will absorb water but not oil. A molecular sieve will do that. Zeolite springs to mind as a word I have read in this connection. Ant mathematical elegance -- desired result achieved with minimal complication |
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I finally met a real live user of Accusorb and apparently the product failed miserably, and the company isn't returning his emails. see the 'accusorb' topic in the general forum for more details.
http://biodiesel.infopop.cc/eve/forums?a=tpc&s=447609751&f=419605551&m=553100923 Mark www.localB100.com |
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quote: Check out the topic called 'methanol management' in this forum for more details on settling out water during acid-base process... the link is http://biodiesel.infopop.cc/eve/forums?a=tpc&s=447609751&f=629605551&m=585600516 mark www.localB100.com |
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I haven't been able to find the rate of density increase in free fatty acids. it seems to me that this number is necessary to figure out the amount of grams of free fatty acids present in grams from the titration info. [the value in the equation is in milliliters. I have the density of the FFA at 15 degrees C. Does anyone have this rate of change or is it pretty negligible? Thanks!
hugo. |
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If water and menthenol have different densities couldn't a centrifuge be used to seperate the water from mix after acid reaction and before the base reaction. This way you could use large amounts of menthol and obtain a more complete reaction. I guess the cost of a centrifuge would be the predominant problem, but the unit could also be used to remove water in a pre filter stage and seperate the glycerine from bd after base reaction and water washing.
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pardon my poor spelling (methanol) and tendancy to forget articles |
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Water and methanol are miscible. The two do not separate by gravity, so a centrifuge will not work either.
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I see that this thread has long been dead, but here's hoping. I recently tried this procedure in two small 200ml samples to see what would happen. For one I put in the exact amount of methanol and sulfuric acid specified, and for the other I only used roughly half of the suggested amounts for both (not on purpose, just ran out of room but continued anyway). For the full-fledged reaction, my NaOH titration (after separating the oil from the methanol, etc. layer) dropped from 29.5 to a lovely .7ml. For the other, half-of-suggested, it dropped from 29.5 to 3.5ml. Was I lucky? Does this sometimes happen? This, or close to it, should be low enough to run a base reaction.
I'm asking because if this is reproducible, you could save a lot of methanol and sulfuric while using cheap WVO and greatly reduce the cost. Also, have you thought about reusing some of the methanol layer for the base esterification, again reducing the amount of methanol required? I just ran a small test on this theory, and results are looking good (I have two distinct layers of what appears to be biodiesel and glycerol), but haven't washed and my GC isn't working so I can't say for sure just yet. Waiting to hear back from you, RUBI |
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this method is incorrect. normality is not what you use for that relation. for volumetric titration you go off of molarity instead. always look at the units!
normality is g/litre, molarity is mole/litre. 56.1 is the molecular weight of KOH which is in g/mole. so if you multiply normality by molecular weight you get g2/mole-litre (not what you want). if you multiply molarity by molecular weight it's g KOH, which is what you should get. if you are titrating based on a weight (find from volume and density) instead of volume then that relation is correct. The correct formula is AV = (molarity*molecular weight KOH (or NaOH)*volume titrant dispensed)/weight of oil sample |
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