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little help please- acid stage problem
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Oil titrated at 21 NAOH so I heated up to 131 and added 2mL/liter of sulfuric and 12 liters of methanol... mixed and settled and reheated, mixed and settled... I drained off about 10 liters of bottom stuff and it's VERY cloudy... Titration dropepd to 8 but the oil looks cloudy. And with a titration of 8 I'm hesitant to move forward.

What should I do next?
 
Registered: February 06, 2008Reply With QuoteReport This Post
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sound like another acid stage to me..

use about 0.3ml/liter and 10liters of methanol.

8 is too high for NaOH..way too high

-dkenny


'84 bluebird school bus, DD8.2L turbo( 4/2011, the bus tranny has died..Frown 8.23.11 bus driven to scrap yard Frown )
2006 Jeep Liberty CRD Smile - the wife's
99 dodge 2500 5.9l 24v..-mine Smile
everything run B100 when its warm enough Smile
 
Location: RTP, North Carolina | Registered: December 15, 2004Reply With QuoteReport This Post
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What % of the total oil amount is your 12L methanol?
 
Location: central virginia | Registered: March 13, 2008Reply With QuoteReport This Post
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quote:
Originally posted by Double D:
What % of the total oil amount is your 12L methanol?


was 12% 100 L batch
 
Registered: February 06, 2008Reply With QuoteReport This Post
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You had enough chems in there to get tht tittration down lower. I would have expected around 4ml NaOH before draining.
How long did you process?
Constant mixing?
Constant temp?

Anything else?

Thanks Ski.
 
Location: central virginia | Registered: March 13, 2008Reply With QuoteReport This Post
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I used the same method I did for the 2 previous batches.. heat to 131 and add chemicals, mix for 30 mins to an hour and then let it sit... then 12 hours later heat back up to 131 and mix another 30-60 mins... then let sit for 12 hours before draining.
It is ia sealed propane hot water heater, so the temp never really drops too far, but no- it's not CONSTANT heat and CONSSTANT mixing... though this method has worked for the last 2 batches... dropped from 20 to 5 last time and from 18 to 4 before that...

This batch also seemed to "pop" during the second heating... which I don't remember before... I assumed it was more water so I drained off mroe than I thought I should... but still, the cloudyness and lack of titration dropping is different with this batch... and it's from the same drum of oil too. Maybe I'll heat and circulate again and then see if it changed before using any more checmicals.
 
Registered: February 06, 2008Reply With QuoteReport This Post



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OK so I added more heat and turned the pump on and the thing is popping like crazy.. sounds like it does when you add the acid to the methanol... this is much worse than before... what is that sound? sounds like something boiling on the bottom of the reactor?

UPDATE: I assumed it was water.. so I did a hot pan test and it was crakling like 4th fo july fireworks.. there must be a TON of water in this oil... I'm going to reheat to 160 and settle and drain until i get back to dryer oil.

lesson learned.. test to make sure you drain enough water off after heat n settle drying!!
 
Registered: February 06, 2008Reply With QuoteReport This Post
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might not be water, but methanol..

-dkenny


'84 bluebird school bus, DD8.2L turbo( 4/2011, the bus tranny has died..Frown 8.23.11 bus driven to scrap yard Frown )
2006 Jeep Liberty CRD Smile - the wife's
99 dodge 2500 5.9l 24v..-mine Smile
everything run B100 when its warm enough Smile
 
Location: RTP, North Carolina | Registered: December 15, 2004Reply With QuoteReport This Post
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Ski,
Thanks for reposting your process. I have a hard time remembering everyones individual process and don't like to make assumptions. I'll likely remember it now.

I agree with dkenny that the boiling could be methanol. From the sounds of it, you may not have tested the water% before processing, just heat and settle? If so, H2O would definately be suspect. All is not lost though, just more chems. In draining the 10L you would have drained out most of the methanol that was left in the processor.
Running another acid stage is a good idea since you ended at 8ml NaOH.
Since you managed a FFA drop of 16ml NaOH from 21ml NaOH, the oil must not have been terribly wet. I'm confident just adding additional chems (sulfuric) would have dropped it further.

Lets look at the math.
Starting T 21ml NaOH
Ending T 8ml NaOH
Sulfuric used 2ml per L
Methanol used 120ml per L (12%)

The T 21 tells us how many mols FFA that are present at the start. 21X.025 = .525mol FFA per Liter.
I'm using .025 since the 1g/L titrant is a .025 molar solution.

The amount of sulfuric that shows up in the titration is roughly 1.57ml titrant per ml Sulfuric. Since you used 2ml H2SO4 that would be 2x1.57= 3.14ml
So I would take your starting titration and add 3.14 to it for the sulfuric added, 21+3.14= 24.14ml then subtract the ending titration to get the titration drop due to FFA esterification.
24.14-8= 16.14ml.

Next I want to find out how much water was created so I need to know how many moles of FFA were esterified since each mole of FFA that is esterified creates one mol of water.

16.14x.025= .404moles water produced.
To find out the volume of water you need to convert mols to grams then to ml.
The molar mass of water is 18g per mol.
.404x18=7.27g H2O
The SG of H2O is 1g/ml so that tells us we have 7.27ml H2O that were produced per liter of oil.

Now we need to find out how much methanol was potentially consumed. For every mol of FFA esterified, it takes 1 mol of methanol.
We know that there were .404mol FFA esterified so that tells us that potentially .404mols methanol were consumed.
The Molar mass of methanol is 32.05g per mol. So we would multiply the molar mass of methanol by the number of mols consumed to get the number of grams of methanol consumed.
.404x32.05= 12.9g methanol.
The SG of methanol is .792g per ml.
12.9/.792= 16.3ml methanol consumed. That leaves you with 103.7ml methanol and 7.72ml water and 2ml sulfuric combined in your processor (per liter of oil). Most of this will be at the bottom of your processor but not all.
Multiplying all this by the batch size of 100L gives us a total of 10.37L methanol, .772L water, and .2L of sulfuric that is left over.
Combined that would give you a total of up to 11.34L of left over chems. It's actually less due to hydroscopy of the chems and molecule size but never the less pretty close.

Hope this helps.
 
Location: central virginia | Registered: March 13, 2008Reply With QuoteReport This Post
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WHOA... Double D... i had 2 glasses of wine with dinner... and I can't even begin to comprehend what you've posted. I hope it's the wine... I am very interested in the "numbers" as I am a mathematical person in general.. so I need to disect your math big time.. AND I WILL!
The whole mols thing has never made sense to me though (I'm an architect, not a scientist)... so I'll read you post in depth tomorrow and see what I can derive from it. What is a mol and how does it relate to something I learned in chemistray class in 8th grade? I just got schooled BIG TIME.. and I LOVE IT!

have no idea what to do next.. but am excited to try it!! At least dkenny's answer made sense to me! HA!
 
Registered: February 06, 2008Reply With QuoteReport This Post
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OK so I think I understand all the math in there... though I'm still not sure what a mol actually is?

Anyway- based on all that info, I added more heat and circulated and then drained a little more off the bottom... the titration ended up at 5 and it passed the 3/27 test, so i moved on to methanol recovery. it'll go in the wash tank later today or tonight... and I guess that'll really be the tell all of how well I did.

Meanwhile, the next batch has started, and seems to be going more smoothly... SO FAR! I'm using heaty longer and more circulation in the acid stage... hopefully that makes it all run more smoothly!
 
Registered: February 06, 2008Reply With QuoteReport This Post
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Ski2liv,

a mol is short for molar.
if you take the atomic numbers for the elements in a chemical and use this number to represent grams.. this is moler weight of this chemical..

take H2SO4

H - atomic # =1 -> 1gram but we have 2..so total is 2 grams
S - atomic # =16 -> 16gram
O - atomic # = 8 -> 8 grams but we have 4..so total is 32..

add all this up and you get a molar weight
this is case, if my math is correct, is 50

so if we wanted a 1 molar solution we'd take 50grams of pure sulfuric acid and add this to 1 liter of water


got it?

edit->update...if you got it..well..forget it..
oops on my part..right idea wrong set of numbers..its not the atomic number but the atomic mass for an element..
H - atomic mass is 1 but there are 2 so 1 * 2 =2
S - atomic mass is 32
O - atomic mass is 16 but there are 4 so 4 * 16 = 64
the total ... 98 grams per mol..not 50..
sorry about the confusion, but is been 26 yrs since collage chemistry..
Double D thanks Smile

-dkenny

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


'84 bluebird school bus, DD8.2L turbo( 4/2011, the bus tranny has died..Frown 8.23.11 bus driven to scrap yard Frown )
2006 Jeep Liberty CRD Smile - the wife's
99 dodge 2500 5.9l 24v..-mine Smile
everything run B100 when its warm enough Smile
 
Location: RTP, North Carolina | Registered: December 15, 2004Reply With QuoteReport This Post



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Interesting.. I've never come across molar weights before. Prety cool. Bsaically it's the only accurate way to figure out chemical conversions and % concentrations in solutions then?

And to think.. i was drawing houses for fun and writing the equations for science on the inside lid of my graphing calculator!! live and learn!
 
Registered: February 06, 2008Reply With QuoteReport This Post
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