7 replies to this topic

[Art Montemayor]

Selvan:

Let's review the basics in physical chemistry first:

1) Your methanol, as it exits the overhead condenser is at 10 oC and its corresponding vapor pressure is 54 mm Hg (I get 55.6, but it's essentially the same value) in an "atmospheric system" where the pressure is 760 mm Hg. This establishes your methanol product as a sub-cooled liquid.

2) Why are you worrying about MeOH losses when you have an excellent overhead product temperature that will keep your methanol in the liquid, controlled state?

3) Your presumption that you will lose a proportional MeOH amount (by evaporation?) is not correct as stated. Here, I'm presuming that the overhead product is 100%, "neat" MeOH - although you haven't stated it. I realize you may not have pure overhead MeOH due to an azeotrope, but you have not mentioned this (as well as a lot of other basic data). You have to consider a rate of evaporation after you've successfully condensed the vapor MeOH in order to evaluate it's loss. You haven't considered rates, so you can't proportionate the assumed loss.

Basically, the system you've described (or what I've understood) is one where you are distilling off MeOH vapors and subsequently condensing them in an overhead condenser with a coolant fluid. A minimal of pressure is required for this operation and it is nominally termed "atmospheric" distillation. But in your case - since you don't desire to lose any condensed MeOH due to subsequent vaporization - you actually have a partial vacuum - the 55 mm Hg! Normally, what would be done is that the distillation would be carried out under this induced partial vacuum and the condensed MeOh would be pumped (as a sub-cooled liquid) to an atmospheric storage tank where a Nitrogen gas blanket is kept over the liquid at a pressure of approximately 10-20" of water column. This method practically eliminates any MeOH vaporization losses during the operation.

I believe the main point you are overlooking is that in a distillation unit operation it is the temperature of the overhead product that sets the column pressure (with adjustments for pressure drops through the column, piping, & condenser of course). It's in equilibrium; remember? In other words, the vapor pressure of the essentially pure MeOH at the top of the column is the pressure existing there. Here, I'm also assuming that you have a conventional distillation set-up without any restrictions or backpressure imposed on the column - such as a back-pressure control. Here, there would be no need for such a device or system because it would raise the pressure and operating temperature in the column - something that obviously would have no merit. If you wanted (or had a need) to operate the top of the column at atmospheric pressure, then you would have to maintain the overheads condenser with product liquid at approximately 65 oC – which is close to the MeOH vapor pressure being 760 mmHg. Of course, this would give you a saturated overheads liquid product – and you would really suffer from MeOH losses!

If you employ a brine to further sub-cool the MeOh product, that is fine. However, bear in mind that this would lower the column pressure even further. What I would do (if the main concern is MeOH evaporation losses) is use the brine coolant in a liquid MeOH subcooler exchanger downstream of the overheads condenser, just prior to it being put into the storage tank.

I hope the above helps to explain the system and your concern with MeOH losses.

Art Montemayor
Spring, TX

[Art Montemayor]

Selvan:

I must not be explaining very well the basic features of how distillation works and a distillation column is operated. Allow me to make another attempt.

You seem to believe that you are arbitrarily fixing the distillation column pressure by means other than the vapor pressure that is fixed by the overheads condenser. What means is that? The normal and conventional manner that atmospheric distillation is carried out is that the column pressure – and that in the overheads condenser – is fixed by the temperature at which you maintain on the top tray(s) and in the overheads condenser. This is the temperature of pure, saturated MeOH in your case. We know that MeOH boils at 64.6 °C under atmospheric pressure (called the Normal Boiling Point, NBP). Therefore, if your column is at atmospheric pressure, then the temperature of the liquid and the exiting vapors on the top distillation tray is = 64.6 °C. This is the same vapor that enters the overhead condenser and is condensed there. If you supply sufficient area and cooling fluid to the overheads condenser (which is what we Chem. E.s do), then the rate at which the vapor is generated will either:

1. Be totally condensed at its saturation temperature (64.6 °C); or,

2. Be condensed and subcooled to a lower temperature (if the cooling fluid is cold enough), which also produces a lower product vapor pressure.

In both above cases the MeOH product is produced as a liquid to serve as reflux to the column and as final overheads product. The only difference between the two cases is that saturated liquid is prone to high vaporization losses, while the subcooled product is subject to less loss. The colder the product, the less the vaporization loss.

Note that if you subcool the product within the overheads condenser, you also lower the corresponding MeOH vapor pressure. And this vapor pressure is the pressure within the system (at the top tray). This means that if you want to subcool the overheads product, you don’t have to operate the column with 64.6 °C on the top tray. You can operate the column much colder and still have effective distillation. However, you will have to maintain the column at the MeOH corresponding vapor pressure – which is sub-atmospheric or partial vacuum. If you don’t have any problems with 64.6 °C as your top temperature, then there is no reason to operate under a vacuum - which involves more capital expense and operating costs (as well as dealing with non-condensable leakage).

Operating at atmospheric pressure within the column (64.6 °C on the top tray) is very common and the system should not present any MeOH losses since the vapor and produced condensate (whether saturated or subcooled) are kept piped in and in closed atmospheres – such as a reflux drum and subsequent storage tank. It is at the storage tank where the producer faces MeOH potential losses – especially if he is producing a saturated MeOH product and the storage tanks are vented to the atmosphere. If you subcool the MeOH prior to storage, you will diminish storage losses (assuming you have atmospheric vented tanks) proportional to the low temperature. However, as I previously mentioned, you can eliminate 95-99% of vaporization losses with the application of a Nitrogen blanket in your storage tanks – whether the product is saturated or subcooled. And that is why I mentioned this procedure in the first posting.

Your assumptions contain flaws in engineering logic and are leading you to suspect that you will have MeOH vaporization losses in spite of subcooling. For example, your statement:

“…even if we are taking a fresh methanol and start distillation I presume that since it has a 54 - 55 mm Hg vapor pressure at 10 ° C this amount could not be condensed as this will still be in vapor form and will escape to atmosphere”

is totally incorrect and is not logical. If you produce liquid MeOH with a vapor pressure of only 54 mmHg (a partial vacuum) you have no basis to fear that it “will escape to atmosphere”. Rather, the atmosphere (760 mmHg) will be driven towards the low pressure MeOH! This is the vacuum condition that I was trying to explain to you in the first posting. The lower and lower you drive the MeOH vapor pressure (by subcooling), the less and less vaporization losses you will have. You can take this to the ridiculous extreme of subcooling the MeOH until it solidifies. Then you might still have to contend with sublimation losses, but they will be probably very much less than vaporization losses.

I don’t know what your average, ambient temperature in Gujarat is like, but I suspect that a MeOH liquid product of 25 – 29 oC (note that this is subcooled) would be very manageable instead of having to manage a 10 oC product with insulation and other expenses. That’s why I would opt for the Nitrogen blanket if MeOH vaporization losses are a concern.

The answer to your question: “What I am tending to find is that will there be a compulsory loss when u distill unless otherwise we provide better coolant to minimise the vapor loss ?” is NO! and this is explained above.

I hope this rather long response has succeeded in explaining how the distillation column pressure is fixed and controlled as well as to how to reduce and control distillation MeOH losses. I wish you the best of luck.

Art Montemayor
Spring, TX

[Art Montemayor]

Selvan:

A lot of confusion is taking place on this subject due to the terms employed and the basic, pure science that is applied instead of basic, practical engineering. Let me cite some examples, based on what you have stated:

1. You say “what I am worried is the Saturated MeOH that is going at 10 ° C from the secondary condenser”.
My comment is: There is no “saturated” MeOH in the secondary condenser unless it is at an absolute pressure of approximately 55 mmHg – a goodly partial vacuum. Here I am assuming that the system is enclosed –as all distillation systems are. If the system is enclosed, then the vapor above the liquid MeOH is in equilibrium with the parent liquid and its pressure will be the liquid’s vapor pressure, 55 mmHg. This is a scientific fact and well-proven in the actual, industrial distillation process itself. Once this sub-cooled liquid MeOH is introduced into the storage tank(s), it can be also enclosed or vented to the atmosphere. Here, you fail to state how you will store it, so I’m assuming you will have your tanks vented to the atmosphere. This is why, and where, you will experience your vaporization losses. If you maintain your entire system enclosed – including the storage tanks – then your vaporization losses are nil or zero. In this, your instance, where you would be storing 10 oC MeOH, the pressure in the storage tanks would be 55 mmHg. But this is a vacuum, as proven above; so what would have to be done is to introduce a non-condensable gas to lend a positive partial pressure in keeping with Dalton’s Law of partial pressures. In other words, Nitrogen gas would be employed as a blanket gas to keep a positive pressure in the storage tanks (approximately 10 – 25 inches of water column pressure). Conventional cone-roofed storage tanks are incapable of sustaining a reasonable vacuum without collapsing and that is why a positive pressure must be administered. Otherwise, if you are unable to do this with your tanks, then you will have to vent them to the atmosphere – and this is where the losses start to appear! Note that the moment you impose a pressure greater than 55 mmHg on the 10 oC MeOH, it is no longer “saturated”; it is sub-cooled at that point.

2. You seem to believe that you will suffer vaporization losses from even your sub-cooled MeOH. While it is basically true that even at 10 oC the MeOH exhibits a vapor pressure, I maintain that it is of no consequence if you maintain a totally enclosed system. And if you cannot maintain an enclosed system, the lowering of the vapor pressure by sub-cooling decreases the vaporization losses to a level that is so small as to make it immeasurable. I cannot justify the existence of a 5% MeOH vaporization loss – even in a situation where the overheads condenser is producing 25 oC liquid MeOH product (& reflux) and storing at the same temperature (25 oC). By enclosing the storage tanks (& subjecting them to the vapor pressure of MeOH at 25 oC) you virtually eliminate all the potential vaporization losses.

3. Everything in the Cosmos, to my knowledge, exhibits a vapor pressure – albeit some substances’ vapor pressure, such as steel, are hard to measure. But the fact remains: all liquids and solids are trying to revert to the gaseous phase unless restrained by external forces. You will never reduce the vapor pressure of a liquid (by sub-cooling) down to a level where you will have absolute zero vaporization losses. It is the rate of these losses that is substantially reduced when the liquid is cooled that is important – not the fact that there still exists a vapor pressure to account for. Allow me to give you a practical example of what I mean: your cooling water at 30 oC (86 oF) has a vapor pressure of 31.9 mmHg. This is already getting close to the MeOH vapor pressure at 10 oC. If you put a liter of this water in an open tray that has 400 cm2 of surface area and also put a similar liter of the 10 oC MeOH into an Erlenmeyer flask, I will wager that if you leave both exposed to the open air that the water will evaporate sooner than the MeOH. What I’m trying to demonstrate is that it is the RATE of evaporation, not merely the vapor pressure that determines the losses. The larger evaporation surface area promotes faster losses – even though the vapor pressure is less. But both liquids will suffer evaporation. However, if I cleverly put a rubber stopper on the Erlenmeyer flask and insulate it so it retains its 10 oC temperature, I dare say I will reduce the MeOH losses down to nil, although I will have to reinforce the flask so it can withstand the developed partial vacuum of 55 mmHg that exists within its interior for reasons stated above

What we do in industry when operating atmospheric distillation such as that of MeOH, is that we try to eliminate as much of the potential MeOH vaporization and leakage losses as possible while running the process at a level that is safe, economically and operationally sound. To do this, we base our MeOH recovery on a contained storage system or on a Nitrogen-blanketed system. We can foresee no losses of any MeOH within the distillation column, the overheads condenser, the reflux drum, and return. All these components are inherently enclosed and subject to the operating pressure that we maintain – the vapor pressure of the liquid MeOH inside. If we want to operate the column at atmospheric pressure, we maintain the top tray liquid content at approximately 65 oC – which corresponds to the 14.7 psia vapor pressure of MeOH. We further condense the overheads vapors to a saturated liquid (at 65 oC), return the required reflux back to the column’s top tray and remove the product saturated liquid (also at 65 oC) while taking it to a sub-cooler where we lower its temperature in order to lower its vapor pressure and reduce its tendency to evaporate. On a practical basis, the colder we subcool, the more stringent the insulation requirements to maintain the colder product in storage and possibly in transport. On an even more practical basis, MeOH has no value in being transported as a liquid colder than ambient – in fact, if it is colder than the atmospheric dew point it fosters and accelerates corrosion of the steel storage and transport equipment due to moisture condensation on the steel surfaces. This is definitely not desirable – especially since the cost of refrigeration is essentially not recoverable and causes further damage. Ultimately, more often than not, MeOH is stored at ambient temperatures. Some manufacturers still ignorantly try to store the ambient product in steel storage tanks that are ill-suited for even 5 inches of W.C. pressure – much less the necessary 10-25 inches of W.C. This is because they have traditionally taken the “Scrooge” attitude in maintaining a minimum of investment in storage and transport facilities – as has been the custom in times prior to the existing stringent environmental laws and regulations as well as when competitive pricing was less fierce. Today this is not the case and zero losses of valuable MeOH are the rule – not the exception. This is why more and more producers are revamping and re-engineering storage facilities while employing nitrogen blanketing where needed to comply with environmental constraints on emissions and their need to reduce critical product losses.

Please pardon my lengthy explanation and comments, but you stated you want to know the exact reason and as you can see, the subjects of vaporization losses and associated vapor pressures are very important subjects that touch or affect a variety of decisions and design parameters within the chemical processing industry. Your case is an example of such an application and I believe that others may also be affected by similar circumstances such as you are currently facing. I hope I complied with your request and have succeeded in accurately describing what I believe to be an answer to your questions and concerns.

Art Montemayor
Spring, TX

[siretb]

Hi Selvan. Much has already been said by the good posts of our fellow Art. I fully agree with him.
1) Your basic statement of your very first post is not correct: The losses are not directly linked to the temperature in a conventional distillation.
At a given pressure, the top tray will have an equilibrium temperature (64.6C at atmospheric for almost pure methanol). You will start to have significant losses only if you have a vector carrying the methanol to the outside. What you want to do, is to operate in a confined (enclosed system). I fully agree with Art that your basic option should be atmospheric + some subcooling + maintaining the tanks under nitrogen blanket at a low enough temperature (say <30°C)
2) One may subcool the reflux. And the idea of a second downstream subcooled condenser is good.
3) For Methanol, I see no real reason of operating the column at a reduced pressure. If you want to go for subcooling, you may very well subcool your reflux at 10°C, you'll still get the top tray at 64.6°C.
4) You will have losses if you feed to your system non condensable components. They will need to escape, carrying with them the partial pressure at their escape temperature. If you continuously feed non condensable, then you may have to cold-trap the vapor methanol. In a plant that uses a nitrogen blanket in its tanks, anywhere in the plant, an option is to use the enthalpy of vaporization of the nitrogen to generate the "cold" (so named cryo-condensation) . After all, you'll need to evaporate the nitrogen, why not make good use of this vaporization? But that's only an option, costly in terms of investment.
If you do not feed non condensables, just maintain a close system
5) The figures you've given of 80-85% recovery for methanol look awfully low to me. This kind of figure cannot be tolerated. If what you describe is a genuine distillation setup, check the design of your condensers. Have been able to match the heat duty at condenser with throughput rate?

Bernard SIRET