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Methanol Molar Ratio
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Hi! I just came through a study converting a feedstock through two-stage transesterification. During the Acid Esterification process, the authors utilized such equation to obtain methanol(g) to be used:


(g)methanol = FFA*(wt. of oil)*(1/MW of oil)*(molar ratio alcohol: oil)*(32)

where : MW of the oil = 282.5g/mol (as oleic acid)


Is there an explanation to the equation they've used? thank you! Smile
 
Registered: August 12, 2010Reply With QuoteReport This Post
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Who is the author and name of the paper?


" I don't know what I don't know until I know"
1994 GMC 6.5 Tubo 2005 Dodge ram 3500, 3 VW's 2000, 2002, 2005.
 
Location: Manitoba Canada | Registered: March 24, 2009Reply With QuoteReport This Post
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It is local, it hasn't been published yet. I only found it in the thesis archives of our university library. They were able to convert two new edible feedstocks into biodiesel.
 
Registered: August 12, 2010Reply With QuoteReport This Post
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There are a couple of ways to calculate the appropriate amount of methanol to use.

When doing Chemistry, one converts everything to "Molar" or "Normal" quantities. The difference being "Molar" is the quantity of a product. "Normal" is normalized for the active parts.

For example, since Sulfuric Acid (H2SO4) has two hydrogens, then 1 Molar acid is 2 Normal acid. And, you would use 2 moles of NaOH to neutralize 1 mole of H2SO4.

So, you are essentially converting from weight to moles of oil. Then converting to moles of alcohol, and then to grams of oil.

The "32" is the molecular weight of Methanol. (CH3-OH) (C:12 + O:16 + (H:1)*4)

They are "cheating" a bit by assuming your oil consists of 1 Normal oleic acid, rather than 3 Normal Triglycerides. But, that shouldn't really make a difference, except in the final ratio you are using.

So... to calculate moles of oil.

Take weight of oil (in grams) divided by the molecular weight of the oil (Oleic acid) to give Moles (Normal) of oil.

To get grams of methanol, multiply the moles of oil times the molecular weight of oil (32 g/mol).

Typically I think the ratio used was about 6 to 9 moles of methanol per mole of oil (tryglyceride) If it is 6 to 9 moles methanol per mole of triglyceride, then it would be about 1/3 of that per mole of Free Fatty Acid... or about 2 to 3 moles of methanol per mole of Free Fatty Acid. In this case you are using the molecular weight of oleic acid.


I'm not quite sure what your "FFA" value is in the equation.... it seems to be redundant. Unless is is the 3:1 ratio between Free Fatty Acids and Triglycerides (see more notes below for more precise calculations).

Now...
The molecular weight is used for the conversion into moles....

But, technically it is unitless (essentially defined as the units of the weight of a proton+electron, I think)...

And if you use the same units throughout, you can do it in grams, kilos, pounds, tons, troy ounces, whatever you like as long as they are units of mass/weight and not volume.

We can try some sample numbers if you need.

--------------

Once you convert it all back to volume... one gets somewhere close to the 4:1 or 5:1 volume ratio that most people are using.

----------------

Editing Note:
Ratios should be 6 to 9 moles methanol per mole triglyceride or 2 to 3 moles methanol per mole FFA for transesterification. See notes below. I had referred to 20:1 which had been mis-stated from previous information, or perhaps relates acid Esterification which requires more alcohol.

This message has been edited. Last edited by: keelec,
 
Location: Oregon | Registered: October 17, 2007Reply With QuoteReport This Post
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I actually use that equation when I do AE, I first saw it in a Van Gerpen article. I should note that there are better (more efficient) ways if you dig around on this forum.

Methanol=kg oil*%FFA/100*(ratio of methanol to FFA)*32/282

If you combine everything and use a 20:1 Methanol to FFA ratio it works out to about
Methanol (g/kg)=Acid#*11.42

Keelec I think you are being confused. The molar ratio is methanol:FFA not methanol:Oil so it's based on the mass of FFA. Also he's talking about AE not TE

Longhand
Determine mass of FFA in entire batch of oil
kg oil*%FFA/100=Mass FFA

divide by assumed molar mass of FFA
Mass FFA/282=kMoles FFA

multiply this by molar ratio of methanol to get moles methanol
Moles FFA*(molar ratio)=kMoles Methanol

Convert moles methanol to grams/kilograms methanol
Moles Methanol*32=kg methanol required
 
Registered: February 25, 2010Reply With QuoteReport This Post
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quote:
Originally posted by BWilder:
I actually use that equation when I do AE, I first saw it in a Van Gerpen article. I should note that there are better (more efficient) ways if you dig around on this forum.

Methanol=kg oil*%FFA/100*(ratio of methanol to FFA)*32/282

If you combine everything and use a 20:1 Methanol to FFA ratio it works out to about
Methanol (g/kg)=Acid#*11.42

Keelec I think you are being confused. The molar ratio is methanol:FFA not methanol:Oil so it's based on the mass of FFA. Also he's talking about AE not TE

Longhand
Determine mass of FFA in entire batch of oil
kg oil*%FFA/100=Mass FFA

divide by assumed molar mass of FFA
Mass FFA/282=kMoles FFA

multiply this by molar ratio of methanol to get moles methanol
Moles FFA*(molar ratio)=kMoles Methanol

Convert moles methanol to grams/kilograms methanol
Moles Methanol*32=kg methanol required

I agree, I have seen 20:1 used as the ration of Methanol:FFA for the Acid Esterification reaction not Methanol: oil.
Although I can't quite see why the equation in the OP uses the ratio of Methanol: Oil for AE.
 
Location: East Yorkshire | Registered: January 14, 2006Reply With QuoteReport This Post



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Thank you for your responses Smile Our professor also noticed the error and suggested that methanol:FFA makes more sense than methanol: oil. Is it because the free fatty acids are the ones that will be esterified during the reaction?

BWilder, may I know which Van Gerpen article you found the equation in? Thank you.

Also, would anyone know where Van Gerpen derived the "0.217" in this base transesterification equation:

Methanol (g) = (wt. of oil) x 0.217


Thank you everyone! I am very grateful for the knowledge and insight Smile

This message has been edited. Last edited by: chuckeroo,
 
Registered: August 12, 2010Reply With QuoteReport This Post
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When doing Chemistry, one converts everything to "Molar" or "Normal" quantities. The difference being "Molar" is the quantity of a product. "Normal" is normalized for the active parts.


Molar is (moles/liter)

Normal is (moles/kg)

What's the difference? Normal is temperature INDEPENDENT while molar is temperature DEPENDENT.
 
Location: CO, CA, KS, or FL | Registered: January 17, 2009Reply With QuoteReport This Post
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quote:
Originally posted by chuckeroo:
Keelec, may I know which Van Gerpen article you found the equation in? Thank you.

You provided the equation above. I just broke it down on a piecewise basis.

A basic Chemistry textbook (either organic or inorganic) should help. Organic Chemistry deals with interactions of carbon containing molecules and would be appropriate.

While "Lab Technicians" will be most interested in weight percent (the 0.217) or volume percent, the scientific literature will be using almost exclusively moles.

One of the advantages of using moles is that if you wish to change reagents (such as using Ethanol instead of Methanol), you just substitute the molecular weight of Ethanol (CH3CH2-OH) 46 for that of Methanol 32.

I just use the periodic table to calculate molecular weights.
http://en.wikipedia.org/wiki/P..._%28large_version%29
Just round and add up the numbers at the bottom. Some information like how to calculate molecular weights and moles will just be assumed and won't be in scientific literature.

These equations are occasionally referenced on this BBS, and similar ones elsewhere.
http://biodiesel.infopop.cc/ev...=122103474#122103474
quote:
Also, would anyone know where Van Gerpen derived the "0.217" in this base transesterification equation:
Methanol (g) = (wt. of oil) x 0.217
Take your original equation:
quote:
(g)methanol = FFA*(wt. of oil)*(1/MW of oil)*(molar ratio alcohol: oil)*(32)
Ok, everything in the equation EXCEPT for the weight of the reagents can be treated as constants.

So, pull the weight of the reagents out of the equation and you get:

(g)methanol = (wt. of oil) * [(1/MW of oil)*(molar ratio alcohol: oil)*(32)]
Treat everything in the square brackets [] as a constant...

MW of oil from above: 282.5g/mol (as oleic acid)
Molar Ration: 6?

So we get:

(1/282.5)*2*32 = 0.227

Hmmm.... There I go believing what I read Razz I'll have to stop doing that Razz

This looks like about a 2:1 ratio Moles Methanol to Moles FFA. And, NOT a 20:1 ratio...
Now that makes a lot more sense!!!!

Of course I'm off just slightly, so let's recalculate with a triglyceride.

But, you are using oil... assuming to be oil made out of pure Oleic Acid + Glycerol.

Oleic Acid: C18H34O2
Molecular Weight: 12*18 + 34 + 16*2 = 282
Note, due to rounding, you had a Molecular weight for oleic acid of 282.5, I was using 282.

Glycerin: C3H5(OH)3
Molecular Weight: 12*3 + 16*3 + 1*8 = 92

Uhhh.... I think I'm counting 3 extra oxygens and hydrogens (counting them twice, once in the FFA, and once in glycerin).

So, let's put them all together to make a tryglyceride, and one gets:
282*3 + 92 - (17*3) = 887 for the tryglyceride.

Let's try that in the equation above:

(1/887)*6*32 = 0.216 WHEW!!!!!!!!!!!!

And, one derives the equation:

So the constant you're using for the Molar Ratio is 6.

Here is an interesting article...
http://journeytoforever.org/bi...brary/macromodx.html

Essentially... if you use more methanol (9:1 moles Methanol to Oil) instead of (6:1 moles Methanol to Oil) then you get more complete, faster conversions.

Remember, we have simplified this to use exclusively one type of FFA, Oleic Acid. Your typical oils will have a mix of FFAs. Some with different numbers of carbons (we had 18 for Oleic Acid). And different numbers of double bonds (saturation)... Oleic acid has one double bond, so it is "short" 2 hydrogens in the middle.



Because of the variability, you may never know the exact composition of your oil.

(am I using Fatty Acid and Free Fatty Acid interchangeably??? I suppose I should be more careful whether I'm talking about a "Free" acid or the Fatty Acid component that is bound with an ester bond to something else).

-------------

Oops, miscalculated the Molecular weight of Glycerol... fixed.
That dropped me to 0.21646, I think... still within rounding errors of the given value 0.217.

This message has been edited. Last edited by: keelec,
 
Location: Oregon | Registered: October 17, 2007Reply With QuoteReport This Post
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So after hunting through my papers I realized why I couldn't find it. It's actually one of the commons papers. In the AE process you do actually get some transesterification but very little and the main focus is to convert as much of the free fatty acids as possible before hitting the transesterification stage which would just convert them all to soaps anyway. So the equation is based on the amount of FFA not total oil since that is what you are trying to react.

http://seniordesign.engr.uidah...hfryfuel/paper_2.pdf

For the transesterification reaction I actually have no idea where that number came from. Stoiciometric is about 0.10 by weight and I generally hear 1.6x stoiciometric in literature if I remember correctly. 0.217 seems quite high although not unheard of in a lab environment, the more methanol you add the easier it is to get conversion. I assume this is still from the same unpublished paper?
 
Registered: February 25, 2010Reply With QuoteReport This Post
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Wow keelec! Thank you so much for all that effort!! Whew. I'm still trying to absorb all of that Smile


Thanks too BWilder Smile the 0.217 equation is from Van Gerpen's paper entitled Biodiesel Production Technologies. It's on pdf and can easily be downloaded when typed in google. Maybe you can see it too for yourself Smile

Thanks everyone
 
Registered: August 12, 2010Reply With QuoteReport This Post
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Lets be careful not to get mixed up between what was originally posted as the Esterification reaction and then halfway down the OP swithces to asking about the factor used in the Transesterification reaction.

I think we agree the first equation posted for Acid Esterification probably has an error in that it uses Methanol:Oil ration when it should use Methanol:FFA.

Yes, 6:1 was always the starting point for molar ration for Methanol:Oil for Base Transesterification whereas 20:1 was the molar ration Methanol:FFA for Acid Esterification.

Both have been published in papers by Jon van Gerpen over the years.
 
Location: East Yorkshire | Registered: January 14, 2006Reply With QuoteReport This Post



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Oops, I miscalculated the molecular weight of glycerol above... fixed.

I assume different people use slightly different values...

The 0.217 is essentially 2:1 stoichiometric (see first calculations above, or 1/3 of second set). That would be assuming 100% triglycerides, and ignoring additional elements in the oil (as well as ignoring varying molecular weights of constituents).

And, thus the idea is to push the equation forward to completion faster, and more completely, then recover the excess methanol in a later step.

The article I referenced above suggested going as high as 3:1 stoichiometric for better, faster, more complete conversion.

Now...
If you take:
Density Canola Oil: 0.917 g/cc
Density Methanol: 0.7918 g/cc

So..

If you take your equation:

Methanol (g) = (wt. of oil) x 0.217

You get:

Methanol cc = (Oil cc) * 0.217 * [ (0.7918 g/cc methanol) / (0.917 g/cc oil) ]
And you get by volume:

Methanol volume = 0.187 * Oil Volume. And, is actually independent of volume units as long as you are consistent.

Ignoring, of course, the thermal expansion.

Now, that is interesting... so it leads to a "nice round number"... and essentially identical to the 1l to 250ml ratio in the Tilly Dr. Pepper discussion.

http://biodiesel.infopop.cc/ev...19605551/m/857600061

And, your magic number: 0.217 wt methanol:wt oil ratio is related to:
0.187 volume methanol:volume Oil ratio.
6:1 moles Methanol:moles Triglyceride Oil
2:1 moles Methanol:moles Fatty acids (stoichiometric ratio).
 
Location: Oregon | Registered: October 17, 2007Reply With QuoteReport This Post
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quote:
Originally posted by keelec:
And, your magic number: 0.217 wt methanol:wt oil ratio is related to:
0.187 volume methanol:volume Oil ratio.
6:1 moles Methanol:moles Triglyceride Oil
2:1 moles Methanol:moles Fatty acids (stoichiometric ratio).

Are you sure the 2:1 is correct. I have never seen 2:1 used for Methanol:Fatty Acid ratio but I have seen 20:1 Methanol:Fatty Acid ratio.

Maybe I am missing something in all of you workings above.
 
Location: East Yorkshire | Registered: January 14, 2006Reply With QuoteReport This Post
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Keelec is mis-interpreting the equation above. I think he is just confused between fatty acid and FREE fatty acid so he is still trying to base the molar ratio on the total oil. The equation is based on FREE fatty acid not fatty acid. Again the first equation was for AE.

Example would probably be easiest.

say we have 25kg of oil with an FFA level of 3.5%

If I use a 20:1 molar ratio the equation for methanol then becomes

Methanol=25*3.5/100/282*20*32=1.99kg methanol

I agree with the transesterification one so far though.
Edit: Well most of it, how he derived it is probably exactly how the paper was using it.
 
Registered: February 25, 2010Reply With QuoteReport This Post
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quote:
Originally posted by BWilder:
Keelec is mis-interpreting the equation above. I think he is just confused between fatty acid and FREE fatty acid so he is still trying to base the molar ratio on the total oil. The equation is based on FREE fatty acid not fatty acid. Again the first equation was for AE.

Example would probably be easiest.

say we have 25kg of oil with an FFA level of 3.5%

If I use a 20:1 molar ratio the equation for methanol then becomes

Methanol=25*3.5/100/282*20*32=1.99kg methanol
Yes, that is my understanding also.
 
Location: East Yorkshire | Registered: January 14, 2006Reply With QuoteReport This Post
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