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The answer to this lies in the sensible heat of the liquid. Sensible heat is energy added to a substance which does not change its phase. Latent heat is heat that changes its phase. Obviously the hotter a liquid is, the easier it is to get it to evaporate.


Yes but it cools down when you do and the amount it cools down by is 1100kJ per kg evaporated and this is the latent heat of vapourisation which I have always understood to be a constant.

If the graph were of the energy you had to add to evaporate a kg of methanol that was already at that temp then it would make more sense but not be the latent heat of vapourisation but something else.

Even then you wold expect the graph to terminate at 165degC not 240degC.

The fact is that at any temp you have to add heat to keep the liquid at that temp while some of it evaporates and the amount you have to add is 1100kJ per kg evaporated. Otherwise the liquid cools down.

Producer "As you can tell, I am really fishing here." LOL yes as it would then need some kind of mix ratio to be involved as well. And although the temp would change as it does with vac or pressure I believe the energy would stay the same at 1100kJ per kg.


Whatever though. We are in danger of hijacking a really valuble thread with a side issue of how methanex presents data. So unless anyone really knows for sure they have a definitve answer I would suggest we either leave it for now or anyone wanting to carry on starts a thread and points us to it. I'll come if you do but am not going to start one. I can stay confused and secretly believe Mehanex are doing something wrong whilst still knowing it must be my lack of understanding really.


mathematical elegance -- desired result achieved with minimal complication
 
Location: Manchester UK | Registered: June 03, 2003Reply With QuoteReport This Post
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I've been using vacuum distillation to recover methanol from biodiesel for one year. It has been my experience that the batch will lose 10-15 degrees F of temp during the distillation process. I generally start at a temp of 140-150 degrees and then add vacuum. The methanol really doesn't start to show up much in the clear tube at the end of the condenser until I hit 20"Hg on the gauge. The distillation process takes about 45 minutes, and I generally recover 1 gallon on 50 gallon batches using 20% methanol.

I've tried adding heat to balance the heat loss, but haven't done any quantitative testing to see if it makes any difference. Just eyeballing it, I can't see a difference.
 
Registered: March 10, 2007Reply With QuoteReport This Post
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Oh, I am using a standard 80 gallon water heater and black pipe. The black pipe is not insulated, nor is any portion of the condenser piping.
 
Registered: March 10, 2007Reply With QuoteReport This Post
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Graham: for what I've seen, the foaming trap on top of the reactor could also be calculated to act as a reflux column. There's a lot of information on how to design a reflux column on this site:

http://homedistiller.org/refluxdesign.htm

My question is: would it be advisable to use a reflux column with this processor? Would it improve the methanol purity?

Since either way we'd have the foaming trap, it'd seem like a good idea...


************************

"When you don't think what you say, you say what you think" Jacinto Benavente.

"Wars not make one great" Yoda.

"A pessimist is a well informed optimist"

WWVhaCwgSSdtIGEgZ2Vlay4gU08gV0hBVD8=
 
Location: Miami, Florida. | Registered: April 06, 2008Reply With QuoteReport This Post
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Jon H,

We also have used vacuum distillation over the last 9 months. 250 liter tank with 18" of vac. but it takes significantly longer than 45 min. for the same 2% removal rate...

What's the horse power of your lab vac.pump.?
What kind of vac. pump (diaphragm, rotary, etc.)?
Finally, is it located before or after the condenser and does it draw the vac. on the receiver tank or the reactor/80 gal. water heater tank?

All this can be applied to the circulating liquid vac. in the push pull set up.

Thanks,

GCG
 
Location: Michigan | Registered: May 08, 2007Reply With QuoteReport This Post
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quote:
What's the horse power of your lab vac.pump.?


Graham

I was wondering that myself: how big of a pump do i need to generate enough vacuum to make this worth while? On first thought, i had considered the HF pump from my appleseed paired to a home-beaten venturi, but now i'm reconsidering if that is going to be underpowered.

I'm no expert in matching venturis to pumps, in fact i've never used a venturi before, but is it feasable to use this low budget pump/venturi combo? I can intuit that an underpowered vacuum system would merely draw less of a vacuum in the processor necessitating "normal" distillation temperatures, so is there a baseline minimum that is needed for this to be effective in terms of returns on investment?
 
Registered: April 01, 2005Reply With QuoteReport This Post



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but it takes significantly longer than 45 min. for the same 2% removal rate...


What is your headspace? This will make a difference to your removal rate. It should be as small as possible.


mathematical elegance -- desired result achieved with minimal complication
 
Location: Manchester UK | Registered: June 03, 2003Reply With QuoteReport This Post
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Thanks Ant,

Ya head space could be an issue in one of our set-ups but in the other unit I believe its because the return circulation flow is being injected just below the surface of the volume.

So the head space is right but we aren't liberating enough methanol since the freshly heated bio goes directly into the fluid rather than spraying into the head space.

It has a side injection point which comes in at a 90 deg. angle. So it sprays in to the tank along the wall causing a swirl effect.

This swirl causes the biodiesel to rise up the sides of the tank causing other issues as well (I didn't build these tanks; they were a gift so to speak) - anyway we're going to be working on it over the next couple days so if you or anyone else has some suggestions - lay them on me...

Thanks again,

GCG
 
Location: Michigan | Registered: May 08, 2007Reply With QuoteReport This Post
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Project Purity,

Thank you for the 3D rendering. Plumbing is not my strong suit and I have a hard time picturing how to layout the 2D stuff in 3D. However, I have a question. There are 2 crosses in your diagram 1. on the pump and 2. below the sight tube. I am referring to number 2. What does that section lead to as it is hidden behind the pump.

Doug
 
Location: Los Angeles | Registered: March 25, 2008Reply With QuoteReport This Post
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On the rendering, the cross at the pump inlet handles suction from the bottom of the main tank (for the pull phase) and suction from the top of the reactor (for the push phase). The cross at the outlet of the reactor serves to direct the flow side of the pump to either the top inlet through the clear tubing (for the pull stage), and to the bottom inlet of the main tank (for the push stage).

That clear tubing isn't just a site tube. Its in the main plubing during the main processing (pull) stage when the temps aren't above 130F, but during the distillation (push)phase, all of the plumbing being used is hard pipe and the clear line will be totally isolated.

Graham,
Is there any problem with having that clear line there?

Some better pictures (updated ones) are here:
http://www.biodieselpictures.com/viewtopic.php?t=128

On a side note, maybe it will be easier to understand next week when i replace each of those black iron crosses with a couple tees... 1 1/4" black iron crosses are very hard to find and expensive when you do find them (if anyone has a link to a piece that size that's less expensive please let me know)
 
Registered: April 01, 2005Reply With QuoteReport This Post
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Ant, Berny and Producer

At http://www.thermexcel.com/english/tables/vap_eau.htm the table indicates declining LHV versus temperature (and pressure) for water too.

At http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/phase.html#c6 under body temperature there is a statement "...the binding energy of the water molecules is greater at that lower temperature, and it therefore takes more energy to break them apart into the gaseous state." which probably holds true for methanol too.

As GL states, the relationship between mass, pressure, temperature and volume is complex and multi-dimensional. LHV is looking at a slice of this relationship. Don't forget that LHV is the amount of energy needed to change a liquid to a vapour without a change in temperature.

But it is surprising that high up in the Andes where pressure and temperature are lower it does appear to take more energy from a gas cooker to boil a certain mass of water than at sea level, even though the boiling point is lower. Probably you wouldn't notice the difference, but you would notice that your tea isn't well brewed and your food undercooked.

I think where we are getting confused is that we expect reduced pressure somehow to "suck" water or methanol from liquid into vapour. LHV tables are theoretical to a degree - if you have 1kg of fluid at a certain temperature (and maybe in a fixed volume) on the very verge of boiling, how much energy do you have to add to change it all to vapour without changing its temperature? IE, to move from one equilibrium to another. It's not dealing with a dynamic system such as a bio reactor or refrig plant where a particular pressure is being maintained or replenished.

To further confuse things, have a look at NASA's animated gas lab. http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/frglab2.html However I don't seem to be able to get this to demonstrate the LHV relationship, probably because it's just dealing with gases.

Paul
 
Location: New Zealand | Registered: August 15, 2006Reply With QuoteReport This Post
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Thanks Paulus, for the info.
Now, here's something odd...
If you look at the table in your first link, the LHV for water actually RISES with the pressure decline. Eek
Am I reading it wrong?
The way I see it, that'd suggest that, for a given temperature, the LHV would be HIGHER at a lower pressure, which seems ridiculous to me. For example: let's suppose we have water at 50*C at 1 atm. vs the same water at the same temperature at 0.1 atm. The water at 1 atm. would need substantially more energy to turn to vapor than the water at 0.1 atm., which already IS vapor.
Where am I messing up?


************************

"When you don't think what you say, you say what you think" Jacinto Benavente.

"Wars not make one great" Yoda.

"A pessimist is a well informed optimist"

WWVhaCwgSSdtIGEgZ2Vlay4gU08gV0hBVD8=
 
Location: Miami, Florida. | Registered: April 06, 2008Reply With QuoteReport This Post



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...the LHV for water actually RISES with the pressure decline.?
Correct, and it was the same with the methanol table.

quote:
...for a given temperature, the LHV would be HIGHER at a lower pressure
At a certain pressure, you can also only have a specific temperature where LHV applies.

quote:
...than the water at 0.1 atm., which already IS vapor.
Not if the temperature is lower than 45.83*C because the water will be liquid.

quote:
Let's suppose we have water at 50*C at 1 atm. vs the same water at the same temperature at 0.1 atm. The water at 1 atm. would need substantially more energy to turn to vapor than the water at 0.1 atm., which already IS vapor.


At 1 atm the concept of LHV only comes into play when the water is at 100*C. Adding 2257.92 kJ to 1kg of 50*C water will increase its sensible heat (and the kinetic energy of the molecules) but will obviously not result in its complete vaporization. Why? Because at atmospheric, water will only boil at 100*C. See http://www.thermexcel.com/english/tables/eau_atm.htm . I saved this table as an Excel spreadsheet, and calculate that to raise the temperature of this water from 50 to 100*C would need 213.9kJ. Subtract this from 2257.92kJ leaves only 2044.02 kJ to vaporize all the water - not enough.

If that 1kg water was at 0.1 atm, it would boil off at 45.83*C and absorb 2392.94 kJ in so doing. If the energy was applied heat, the temperature would stay at 45.83*C. But you are quoting 50*C, so you have raised the temperature of the vapour (steam) by 4.17*C beyond vaporization. This is called superheating. Turning to http://www.thermexcel.com/english/tables/vap_eau.htm, it takes 1.8927 kJ to raise 1kg of steam by 1*C x 4.17 = about another 7.9 kJ.

So you're partly correct but it appears that if the kinetic energy is in the molecules in the first place (the temperature is higher) than less energy is required to get them to launch out into space.

Think of a porous pottery evaporation-style wine cooler. Will it cool better on a hot day than a cold day?

Notice how much less energy is needed to heat the 1kg of steam by 1*C than to heat 1kg of water. About 40% less. Of course, in the 0.1 atm example, pressure has risen too. Why? Because at 0.1 atm, saturated steam can only exist at 45.83*C. To raise its temperature beyond vaporization, pressure must also go up. A closed vessel is implied.

Also note that vastly more energy is needed to change the state of the water to steam, than to heat water and to heat steam.

Clear? Lol. Of course, I could be wrong...

Anyway perhaps this explains why heating as well as circulating is a benefit in GL venturi reactors. If it were the other way around, we'd be cooling the bio instead of heating it during methanol recovery.

Paul
 
Location: New Zealand | Registered: August 15, 2006Reply With QuoteReport This Post
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Hmmm. I think I get it now. So, the concept of LHV only applies at the boiling point for a given pressure. It's basically the "last little push" to force the change of state. Kinda like the extra catalyst to force the conversion, so to speak.

quote:
Clear? Lol. Of course, I could be wrong...


Well, if you're wrong, you're a lot less wrong than me... Big Grin

Anyways, I appreciate you taking the time to explain it to me.


************************

"When you don't think what you say, you say what you think" Jacinto Benavente.

"Wars not make one great" Yoda.

"A pessimist is a well informed optimist"

WWVhaCwgSSdtIGEgZ2Vlay4gU08gV0hBVD8=
 
Location: Miami, Florida. | Registered: April 06, 2008Reply With QuoteReport This Post
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Hi folks

That's right, the key condition for LHV is that there is NO change in temperature.

It is the amount of heat energy needed to change from liquid to vapour WITHOUT changing temperature.

Or, is the amount of heat energy given off when going from vapour to liquid, without changing temperature.

It isn't dealing with temperature change, only state change.

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

Why do you need to add more heat energy to vaporize low pressure liquid?

Seems odd at first, but it is quite simple.

Paulus got it with...
quote:
... it appears that if the kinetic energy is in the molecules in the first place (the temperature is higher) than less energy is required to get them to launch out into space.


Water at 100C is 'hot' !!
Its molecules are in a high state of kinetic agitation, and as such can more easily escape the water-water molecular bond forces, if the only other force acting against them is 1 atmosphere of pressure.

Water at 81.35C is 'cooler' !!
Its molecules are in a lower state of kinetic agitation, so the water-water molecular bond forces have more success to keep the molecules held together as a liquid phase. Although the atmospheric pressure may be 1/2 that of boiling water at 100C, this extra 'suck force' (less pressure, in fact) does not entirely overcome the basic molecular attraction, so more heat energy must be added per volume of water at 81.25C at 0.5 Atm, compared to water at 100C, 1.0 Atm, to cause the water molecules to break free as steam.


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Location: UK | Registered: December 04, 2005Reply With QuoteReport This Post
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Project

Thanks! My brain must have been broken that day! That third picture on your link made all the difference! Obvious when I saw it. Both ts are connected to the pump - inlet and outlet!

Thanks
Doug
 
Location: Los Angeles | Registered: March 25, 2008Reply With QuoteReport This Post
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Originally posted by GCG:
Jon H,

We also have used vacuum distillation over the last 9 months. 250 liter tank with 18" of vac. but it takes significantly longer than 45 min. for the same 2% removal rate...

What's the horse power of your lab vac.pump.?
What kind of vac. pump (diaphragm, rotary, etc.)?
Finally, is it located before or after the condenser and does it draw the vac. on the receiver tank or the reactor/80 gal. water heater tank?

All this can be applied to the circulating liquid vac. in the push pull set up.

Thanks,

GCG


I'll have to check, but I think its 1/3hp. I think it is a rotary pump, it uses an oil lubricant. I have it after the condenser, and the fluid trap.
 
Registered: March 10, 2007Reply With QuoteReport This Post
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quote:
Why do you need to add more heat energy to vaporize low pressure liquid?

Seems odd at first, but it is quite simple.

Paulus got it with...

quote:
... it appears that if the kinetic energy is in the molecules in the first place (the temperature is higher) than less energy is required to get them to launch out into space.



Water at 100C is 'hot' !!
Its molecules are in a high state of kinetic agitation, and as such can more easily escape the water-water molecular bond forces, if the only other force acting against them is 1 atmosphere of pressure.

Water at 81.35C is 'cooler' !!
Its molecules are in a lower state of kinetic agitation, so the water-water molecular bond forces have more success to keep the molecules held together as a liquid phase. Although the atmospheric pressure may be 1/2 that of boiling water at 100C, this extra 'suck force' (less pressure, in fact) does not entirely overcome the basic molecular attraction, so more heat energy must be added per volume of water at 81.25C at 0.5 Atm, compared to water at 100C, 1.0 Atm, to cause the water molecules to break free as steam.


Thanks Graham. Now I finally got it.

I hate to bother you, but a few posts ago I posted a question, and then the discussion got sidetracked, so I guess you missed it. I'm gonna quote it here, so you don't have to sift through all the posts:

quote:
Graham: for what I've seen, the foaming trap on top of the reactor could also be calculated to act as a reflux column. There's a lot of information on how to design a reflux column on this site:

http://homedistiller.org/refluxdesign.htm

My question is: would it be advisable to use a reflux column with this processor? Would it improve the methanol purity?

Since either way we'd have the foaming trap, it'd seem like a good idea...


What do you think?


************************

"When you don't think what you say, you say what you think" Jacinto Benavente.

"Wars not make one great" Yoda.

"A pessimist is a well informed optimist"

WWVhaCwgSSdtIGEgZ2Vlay4gU08gV0hBVD8=
 
Location: Miami, Florida. | Registered: April 06, 2008Reply With QuoteReport This Post



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Graham, thanks for the endorsement. I've been trying to put a little thought into how the venturi reactor works, and this is where it gets complicated.

I don't believe the point of the venturi arrangement is to lower the pressure in the reactor to such a degree that methanol would boil off at a lower temperature than would be indicated by the air pressure in the reactor at that time.

Rather it appears the venturi purpose is to act as a motor to drive air and methanol vapour past the condenser. Pressure overall can't be lowered in the reactor because the air is reintroduced via the oil circulation.

The methanex document page 13, 3.7 table of methanol vapour pressure vs temperature predicts that at 1 atm (which we have) at about 65 deg C, the MeOH vapour pressure will overcome atmospheric pressure and commence the change of state that we seek. In accordance with the principle of LHV it will absorb heat from the surrounding fluid, cooling it if further heat energy is not introduced externally.

But this is not a pure solution of MeOH. It's a mixture containing some water, and mostly methyl ester. In accordance with the article on distillation http://en.wikipedia.org/wiki/Distillation the boiling (vapour) point of the solution rises depending on the proportion of each, and so it is necessary to run up the temperature until much of the MeOH has been released.

So far so good. If the vapour is almost all MeOH then it will still be at about 65*C as it boils off (LHV principle at work.) In the earlier thread on the eco reactor, you said this vapour is at saturation. But in the flow out of the reactor and through the venturi is a lot of air, mostly nitrogen.

So - question - is it not saturated in the sense of "the flow is exclusively MeOH at maximum concentration" but saturated in the sense that the MeOH part of the flow is at its minimum temperature before it starts to condense into a liquid?

Again refer to LHV. As the flow goes through the condenser MeOH will give up part of its heat - but not immediately change temperature - in order to condense. As liquid MeOH builds up on the walls of the condenser, it will cool below its dew point, becoming supercooled.

Three questions -

What affect do water vapour and air in the flow have on the dew point of the MeOH? (in the same way that boiling points of mixtures are different to their respective, individual boiling points.)

And, I guess accordingly, does much water evaporate/boil off from the bio during MeOH recovery using this setup? I believe it must as the temperature inside the reactor approaches 100*C.

And what overall degree of purity MeOH are you getting by distilling off the MeOH in this way?

Further question - on the aspect of dewatering WVO with the venturi reactor, in my case, the pump on my Murphy reactor is only rated to just over 90*C. I would pick from our newly-established understanding of LHV that to thoroughly dewater a temperature in excess of 100*C is necessary, or else suffer extended processing time as we wait for the water to just evaporate and not boil.

Phew! Sorry for being long-winded. Hopefully this will increase understanding why and how the venturi reactor works. It certainly has mine, as I have been writing this.

Paul
 
Location: New Zealand | Registered: August 15, 2006Reply With QuoteReport This Post
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Hi bernyjb

Ah, sorry, missed that. Yes, I guess it could be acting as a simple reflux column, although that isn't the reason for it.

I don't think reflux is essential - the purity of methanol I am getting is fine, because I add a lot of new methanol for the next reaction.

About 6 litres recovered methanol and 12 litres new methanol for an 85 litre batch.

Using dry, low titrating oil and no 5% prewash, the methanol is fine to use. Reflux can make the distillation process take much longer.

The trap is not essential if you can keep the rate of boiling low, but I occasionally had situations where it foamed up and came through the condenser. With a big boil-over the trap can be swamped with foam and let foam through, so I'm still looking for improvements there.


Rover 75 + Skoda Fabia on B100
http://www.graham-laming.com

Bicycle on G100 12,000 miles p.a. ( http://uk.virginmoneygiving.com/GrahamLaming )
 
Location: UK | Registered: December 04, 2005Reply With QuoteReport This Post
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