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RE: [new_distillers] Re: My reflux column

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  • Gavin Flett
    Bob........ just super. That s awesome info, thanks so much for taking the time to write all of this. I m re-running the distillate through the still tomorrow.
    Message 1 of 32 , May 2, 2012
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      Bob........ just super. That's awesome info, thanks so much for taking the time to write all of this. I'm re-running the distillate through the still tomorrow. I think my main problem was not letting the column reach equilibrium. But we shall see.

      The steps to calculate the condenser is fantastic. I'll do that for sure if I see I'm not getting the desired result. Also I purchased a digital thermometer so I don't have to climb 8 1/2 feet up just to read it, that'll make things easier. I'll for sure use your checklist to tackle the problem.


      To: new_distillers@yahoogroups.com
      From: bobg542492@...
      Date: Tue, 1 May 2012 14:03:05 -0400
      Subject: Re: [new_distillers] Re: My reflux column

       

      Gavin,
       
      What matters is how much heat your condenser is absorbing.  A low output coolant temperature may simply mean that there isn't enough heat exchange in the condenser to heat up the cooling water; it does not have to mean that there is more than enough heat exchange (but I could mean this as well).  As discussed previously, you must remove the LHV from the column - physics demands this.  If you don't remove it via reflux and its associated heat exchange, the column will continue to heat up until the product is the same proof as the vapors that came off of your boiler, thus guaranteeing that the product removes the LHV, as conservation of energy demands.  If you have some heat removal from your condenser but not enough, you will get some intermediate proof product.  Its all about conservation of energy.
       
      This being said, it is rather complicated to try and compute the design of a condenser system.  You have to know a lot more than I do about the heat conductivity of the tubing and of the heat transfer capacity of the cooling medium.  But you already have a condenser, so you can measure what it is doing, as follows:
       
      1.  Take the temperature of the cooling water before it enters the condenser.
      2.  Take the temperature of the cooling water after it leaves the condenser.
      3.  Measure the coolent flow rate through the condenser (just collect some of the output for a few, carefully timed minutes, and measure how much that is, e.g. in liters, to determine the flow rate of liters/minute).
       
      Now you know the temperature change and the mass flow rate.  You can look up the definition of a unit of heat energy on the Internet, e.g. a Calorie or BTU, defined as a certain
      number of degrees of temperature change over a certain mass.  You can look up the weight (mass) of your coolent (water, I assume) to convert your volume measurement to weight (mass) measurement.  Doing this will ultimately give you the amount of energy that your condenser is actually transferring out per unit of time.  Whatever units you use (joules per minute, BTUs per hour, etc.) can be converted to electrical power units of KW.  You then compare this calculated value of heat transfer (power) of the condenser with your input power and see if the latter exceeds the former.  If it does, you need a bigger condenser, which likely means more tubing - either in length or in diameter, or tubing with a better thermal conduction.  Knowing what your current condeser actually does will at least tell you how much more of the same you need.
       
      This analysis is only approximate, as heat transfer is temperature dependent and the definition of a calorie or BTU is at some specific temperature.  But it ought to be good enough to figure out if the problem is in the condenser or somewhere else.  Note that there is no harm whatsoever in having too much heat transfer capability in your condenser.  If it removes more than the LHV, then it will also remove "sensible heat" of the resulting liquid product/reflux meaning that the condensed liquid will be cooler than its dew point.  That's OK, as you are adjusting the reflux ratio to keep the column in equilibrium anyway, so this only impacts the actual setting of your reflux ratio adjustment.
       
      There are other reasons for low proof of product -- inadequate reflux, too low a HETP, too much vapor volume into the column (vapors ascent too fast reducing the real HETP of the packing), heat loss from the column walls that distorts the necessary temperature gradient up the column, reflux condensing on the column walls and not flowing down through the center of the column and evenly over the packing, clogs in the packing so that vapor/liquid interaction is contrained (reducing HETP), inaccurate temperature reading,... and probably a whole lot more.  So here is a more complete (I'm sure not comprehensive) checklist:
       
      - is your heat input matched to your column diameter so that vapors flow up the column at the recommended rate?
      - is your heat removal (condenser) adequate to remove the LHV that you put into the column at the product proof that you desire?
      - is the hetp of the packing and height of the packed part of the column sufficient for the proof that you are trying to achieve?
      - is your column well insulated, so that heat loss from the walls does not alter the temperature gradient necessary to achieve the desired distillation?
      - is the vapor flow through the packing even and is the liquid reflux flow equally even so that vapor and liquid exchange heat efficiently?  This isn't easy to observe unless you have a glass still, but the packing must be clean and free of clogging material and scrubbers must NOT be compressed.
      - is your product thermometer measuring the temperature of the vapor only and not of liquid reflux?  (Recall from above that the reflux can be cooler than the dew point; thus you may think that you are getting azeotrope from a thermometer reading and not actually get that proof in the product). 
      - is your thermometer losing heat and giving an incorrect rewading?
      - is your thermometer well calibrated and accurate enough for you to use it to control the reflux?  (This is not a trival question.  You said that your product vapors were at 79 degrees C indicating 87%; azeotrope, 95.6%, is 78.1 degrees C -- less than one degree C difference.  Take care not to confuse the resolution of a digital thermometer with the accuracy of the thermometer.  A cheap digital thermometer may read out on 0.1 degree C increments but may only be accurate to +/- 2 degrees C).
       
      The first two items are what we have been talking about here.  The other items should be reviewed as well.  The point here is that once you understand how the reflux column still works, there are only a limited number of factors that impact the still's ability to provide the desired proof product.
       
      Bob

       
      -----Original Message-----
      From: gavinflett <gavin_flett@...>
      To: new_distillers <new_distillers@yahoogroups.com>
      Sent: Tue, May 1, 2012 10:02 am
      Subject: [new_distillers] Re: My reflux column

       
      Thanks for the great explanation

      So you figure it's my condenser huh? How will I know how much bigger to go? I have used a 1/4" soft copper coil (about 10' worth) and coiled it into a 1.5" X 13" coil. Should I be using 3/8 soft copper coil then? Or can I get a way with just turning the water up? It make sense in theory what you are saying about the input of heat energy. But does having a colder condenser really produce more distillate? The water coming out of the condenser I have now is not warm, it's fairly cool to the touch (maybe about 10C).

      --- In new_distillers@yahoogroups.com, Bob Glicksman <bobg542492@...> wrote:
      >
      >
      >
      > Hi Gavin,
      >
      > If you got higher proof when your column was uninsulated, then I would say that this is a clear indication that you don't have enough heat removal capacity in your condenser(s). Remember conservation of energy -- the energy that goes into the still must all come out of the still. However, you can't just radiate heat arbitrarily out the column wall because the fractionating "magic" that causes the ethanol and water to separate requires a very specific temperature gradient to exist up the length of the packing. If the temperature gradient is not maintained exactly throughout the run, your column will fall out of equilibrium and the proof will go down (or the product rate will go down). I never fully answered your question about taking heat off of the column walls, so I will do so below. But the bottom line is that your condensor(s) is probably undersized for the heat that you are puting in.
      >
      > Figuring out the right size for a condensor is a little tough. I forget the metric units, but in English units, 1 BTU raises one pound of water one degree F. Recall that heat transfer can only take place from a higher temperature to a lower temperature. To get azeotrope, the vapor temperature from which you are removing heat in the condenser(s) is the boiling point of the azeotrope which is something like 173 deg F. If you are using cooling water at 73 degrees F to remove heat in the condenser, then at MOST you will be able to raise the water temperature 100 degress F (below boiling), therefore absorb 100 BTU of heat for every pound of water that flows through the condenser. Convert pounds of water to liquid volume and you will know the minimum flow rate needed. If you measure the input and output coolant temperature from your condenser(s) and the coolant flow rate, you can calculate the anount of heat energy that you are removing and compare this with the heat ene rgy that you are puting into the column. I'll bet that you don't have enough removal and need a larger condenser(s).
      >
      > Here is why you must insulate your column. The fractional distallation process upon which your still column works requires a constant heat exchange at various plates running up the column of the still. You don't have actual plates with a packed column, but the principle is the same (you have 'theoretical plates" and the number of theoretical plates that you have depends upon the HETP of the packing material and the packing height). Each plate in the column represents a mixture of ethanol and water that corresponds to a liquid/vapor equilibrium at some specific concentration of ethanol/water in the mixture. The specifics are not arbitrary. They are absolutely dictated by physics and depend only upon the initial concentration of ethanol/water in your beer. If your beer is, say, 10% ethanol and 90% water (by volume), then physics dictates that the vapors resulting from boiling this mixture will contain 55% ethanol/45% water, by volume when condensed back to liqui d. You can't change this -- it is required by the laws of physics. So your boiler, which is the first "plate" will be the the boiling point of a 10% ethanol/90% water mixture, your first theoretical plate in your column must be at the boiling point of a 55% ethanol/45% water liquid mixture, the next plate up must be at whatever the concentration ratio phsycis dictates , etc. You can compute what physics dictates from the LHV of water and ethanol and the molar concentration at each step, or simply find it in a table or graph in a book (sorry, my copy of Nixon and McCaw is loaned out right now so I can't cite you the page and figure number, but it is there for sure). In any event, each plate (real or theoretical) in your column must be maintained at the specific temperature that is the boiling point of the mixture at that plate, as required by the physics of liquid/vapor equilibrium (LVE). If the column is not held precisely in this state of equilibirum, your pr oduction and proof will change. So however you do it, you are required, by the laws of physics, to keep the column in exact equilibrium.
      >
      > Now you can see why simply removing insulation won't work. You have no way of precisely controlling the heat transfer out of the walls of the column. Yes, you will lose heat (which in your case is necessary) but you can't lose it in precisely the way to maintain the temperature gradient up the packing that is required to maintain equilibirum. The temperature and degree of airflow past the column will change and the heat removed will at best be even up the column and more generally be determined by air currents and imprefections in the column wall and will in no way represent what is needed to maintain equilibrium.
      >
      > Now, it is not impossible to do what you were thinking. You could insulate the column really well and put a computer controlled heat exchanger on each plate (if you had real plates) and control the heat removal on each plate to keep that plate at the precise temperature that equilibrium requires. This, in fact, is exactly what the petrochemical industry does. An oil refinery is just a large fractionating column where it is desired to take off product from each plate -- gasoline has the lowest boiling point, followed by diesel fuel, kerosine, light lubricating oil, bunker oil, heavy lubricating oil, etc. The technology to do this on a small ethanol still is not beyond the reach of an individual distiller, but the question is: why would you want to go to all of this troube if you are only interested in separating ethanol from water? Some smart person figured out, a long time ago, that you can maintain still colimn equilibrium via only one measurement and one control -- by sending some condensed product back down the still column (reflux). This is a whole lot easier and cheaper than physically controlling the temperature of each plate and it works with continuous packing as well as with discrete plates. So this is why reflux is required.
      >
      > Bob
      >
      >
      >
      >
      >
      > -----Original Message-----
      > From: Gavin Flett <gavin_flett@...>
      > To: new_distillers <new_distillers@yahoogroups.com>
      > Sent: Sun, Apr 29, 2012 11:24 pm
      > Subject: RE: [new_distillers] My reflux column
      >
      >
      >
      >
      >
      > Hi Bob, thanks for the attention on this. Answers are below each question in red
      >
      >
      >
      > To: new_distillers@yahoogroups.com
      > From: bobg542492@...
      > Date: Mon, 30 Apr 2012 02:02:10 -0400
      > Subject: Re: [new_distillers] My reflux column
      >
      >
      >
      >
      >
      > Here are a few suggestions, Gavin:
      >
      > 1. Check that everything is as it should be. Is your column well insulated? Is the packing clean and the vors running freely through it? Is the still running at atmospheric pressure (no pressure buildup, e.g. due to clogged packing)? Yes the column is well insulated. But I got my 93% when it wasn't insulated. After I put up the insulation, the reflux seems to stop and the ABV drops into the 80's.
      >
      > 2. Make sure that your temperature probe is measuring only vapor and only above the packing. Yes
      >
      > 3. Is temperature your sole detemrination of proof or have you checked this with a hydrometer? Yes If the latter, do you compenate the hydometer reading for the temperature? It's read at the correct temp of 20C
      >
      > 4. Have you checked to make sure that your temperature probe is reading accurately? No Try testing it in boiling water and make sure to adjust the boiling point for altitude if you are not near sea level. I'm at sea level
      >
      > 5. Make sure (as best that you can determine) that the vapors ascend through the packing evenly and that reflux runs down the center of the column and not down th walls of the column. How am I supposed to do that? I can't see inside the still. I have the column packed with copper mesh, I have no idea if it's allowing the vapours to travel through evenly. I can breath through the column, which I read was a way to determine the packing density.
      >
      > 6. After clearing the above, do what Nixon and McCaw recommend and run at 100% reflux ratio (all product gets poured back down the center of the column). Wait for the column to reestablish equilibrium and see what your proof/temperature is. If the proof is not at azeotrope, then you either (a) have too much heat going in and the vapors are ascending through the packing too fast, reducing its real HETP, or (c) you do not have enough cooling capacity in your condensors and you can't get get the heat back out even with your reduced heat input, or (c) your packing HETP is too low. If you get to this step without a resolution, please post the details of your still and lets run through some numbers: I managed to get at the most 93% ABV, but only for a short while, after that I had to increase the power to the still just so it wouldn't take 3 days to distill everything. My still charge was about 20 L at 40%. So far it's been going for 15 hours and it's still reading 80.5-82 C (which I think is slightly inaccurate as the still ran steady at about 77C for about 8 hours).
      >
      > - boiler heat (what you are running at).
      > - proof of the beer (and how did you determine this?).
      > - column height and packing material.
      > - column diameter.
      > - still head configuration and reflux configuration (CM, LM, VM)
      > - how many condensers and where are they located
      > - what are you running through the condensors (water, beer, other) and at what temperature. Do you have a way to measure coolant flow rate through the condensors and what is it?
      >
      > There isn't a whole lot more to this. A basic batch reflux column still that is well insulated from heat loss only has a few relevant variables to work with:
      > - proof of the beer
      > - heat input, which in conjunction with column diameter determines the vapor flow rate up the column (which can't be too fast or your HETP will be a lot lower than you think it is).
      > - heat output, which MUST balnace heat input and if you can't remove enough heat via the condensors, will limit the proof that you can get.
      > - column height and HETP of the packing, which, in conjunction with reflux ratio and heat removal, determines the proof of the product.
      > - measurement technique (which, if in error, will mislead you).
      >
      > KR,
      > Bob
      >
      >
      >
      >
      > -----Original Message-----
      > From: Gavin Flett <gavin_flett@...>
      > To: new_distillers <new_distillers@yahoogroups.com>
      > Sent: Sat, Apr 28, 2012 8:46 pm
      > Subject: RE: [new_distillers] My reflux column
      >
      >
      >
      >
      >
      > Lowering the power did not seem to make a difference. I am now collecting the distillate at about 87%. Maybe it's just near the end of the run, but the temp still reads 79C.
      >
      >
      >
      > To: new_distillers@yahoogroups.com
      > From: bobg542492@...
      > Date: Sat, 28 Apr 2012 23:20:24 -0400
      > Subject: Re: [new_distillers] My reflux column
      >
      >
      >
      >
      >
      > Please let us know how turning down the heat turned out. As to your question about reflux, the answer depends upon what type of reflux management system you use. Collant management (CM) systems generally require a greater flow rate of cooling water (or whatever you use for coolant). Cooling the water will help, but you will achieve greater results by upping the flow rate. Liquid Management (LM) and Vapor Management(VM) systems have a value that controls how much liquid or vapor is taken off as product; the restis reflux. So you take less product to increase the reflux ratio.
      >
      > I suggest that you get a copy of Nixon and McCaw from The Amphora Society website. The book is excellent and after you read and digest it all, it will save you countless hours of experimentation and wrong turns.
      >
      >
      >
      > -----Original Message-----
      > From: Gavin Flett <gavin_flett@...>
      > To: new_distillers <new_distillers@yahoogroups.com>
      > Sent: Sat, Apr 28, 2012 6:30 pm
      > Subject: RE: [new_distillers] My reflux column
      >
      >
      >
      >
      >
      > You definitely talk good sense mister.
      >
      >
      > Yes, I did think that more heat would equal more product coming out. And I did know that it has to be counteracted with more reflux, I was just unsure of how that could be done (do I pump more water through, thus making the water colder). I am still unsure of why insulating the column is better, wouldn't more reflux happen if the column is air cooled all the way up (vapours are cooling, condensing and revapourizing more times)?
      >
      >
      > I turned down the heat to the point where the output is around a KW, hopefully this will produce a purer spirit.
      >
      >
      >
      > To: new_distillers@yahoogroups.com
      > From: bobg542492@...
      > Date: Sat, 28 Apr 2012 20:39:43 -0400
      > Subject: Re: [new_distillers] My reflux column
      >
      >
      >
      >
      >
      > This is just a guess, Gavin, but I'd say that you are puting in too much heat, and/or not providing enough reflux and/or enough heat removal in your condenser. Distillation is driven by basic physics -- whatever heat you put in must be taken out. Follow this analysis along, if you will:
      >
      > The latent heat of vaporization (LHV) is the heat energy that you must put into a liquid at its boiling pint in order to vaporize it. Conversely, in order to liquify a vapor at the dew point, you need to remove the LHV. When you have a mixture of two substances in a liquid, each substance retains its individual physical and chemical characteristics, including its LHV. The LHV is an intrinsic physical property of any substance and it turns out that the LHV of ethanol, on a volume basis (not on a mole basis) is considerably lower than the LHV for a similar volume of water. Thus, if you boil off a beer that is 10% ethanol and 90% water, you get a vapor that is about 55% ethanol and 45% water (a point on the co-called "VLE" curve that you see all over the place when discussing distillation).
      >
      > Now, your packed column performs some fractionating "magic" such that the vapor at the top of the column should be azepropic ethanol - i.e. 96% ethanol and 4% water. But given the information above, it should be obvious that if you vaporize off a quart of beer and want to get out a quart of azeotropic ethanol, the energy that you remove will be less than the energy that you put in. Since the still column does not otherwise lose energy (it is well insulated) and does not have a means to store energy, then this can't work - conservation of energy says it can't work. So what will happen if you try and remove all of the vapor at the top of the column as product is that the proof MUST go down to the same 55% ethanol 45% water that you put in (for this example). This is why you MUST HAVE REFLUX. The only way to get a product at or near azeotrope is to send a significant amount of condensed product back down the column so that it is recycled enough times to remov e all of the heat that physics says must be removed.
      >
      > This is all covered well in Nixon and McCaw "The Compleat Distiller". The more heat that you put into your still, the more heat that your condensor must remove, so if the condensor design is inadequate, you can't get high proof even with 100% reflux ratio. If you have enough heat removal capacity in your condensor, you must still set your reflux ratio high enough remove the LHV using the azeotrope. You require a minimum reflux ratio if you have enough packing, but your HETP may be too low, in which case you can compensate, but only by increasing the reflux ratio. If 100% reflux (no product takeoff) won't get the proof up high enough (as indicated by the temperature of the vapors at the top of the column) than you either must reduce the heat input to the still or increase the heat removal (bigger or faster condensor).
      >
      > Lastly, Nixon and McCaw also make that point that too much heat can cause the vapor to travel up the column too fast, negating the fractionating "magic" that happens in the column packing.
      >
      > A 2" diameter column will be driven quite well by a 1 KW heater. 1.6 KW is much more than you need. Too much heat is BAD -- it does not make distillation faster, as you might intuitively assume. It only requires you to have more packing HETP, higher reflux ratio, and more condenser capacity or else the proof will go down, as you have already found out the hard way.
      >
      > Bob
      >
      >
      >
      > -----Original Message-----
      > From: gavinflett <gavin_flett@...>
      > To: new_distillers <new_distillers@yahoogroups.com>
      > Sent: Sat, Apr 28, 2012 4:35 pm
      > Subject: [new_distillers] My reflux column
      >
      >
      >
      >
      >
      > Ok guys, I need to get some reflux column experts to weigh in on this one. I built an offset reflux head. It's about 6' tall and 2" thick. I'm using an electric hotplate that puts out up to 1600W. I'm also doing this outside and it'a bit chilly at 11C, but I have the whole thing insulated including the column.
      >
      > Now.... i'm only getting about 90% out this thing right now. It's been going for 8 hours now and have only collected about 3 or 4L. The most I got out this thing was 93%.
      >
      > My question is, why can't I get 96% (or whatever the maximum is) out of this thing? Why is it coming out so slow, and why isn't it more potent?
      >


    • Bob Glicksman
      Gavin, If you are trying to get to the azeotrope starting with 10% abv beer, the minimum reflux ratio that you need is 2.4:1. This means that for every 1
      Message 32 of 32 , May 8, 2012
      • 0 Attachment
        Gavin,
         
        If you are trying to get to the azeotrope starting with 10% abv beer, the minimum reflux ratio that you need is 2.4:1.  This means that for every 1 liter of product that you take off, you need to send 2.4 liters (or about 70% of the liquified vapor) back down your column.  Theoretically, to achieve this minimum you need an infinite number of plates; in practice, you can get close enough to azeotrope that you can't measure the difference with around 25 theoretical plates.  To a limited degree, you can reduce the number of plates and still get the same proof by increasing the reflux ratio.  At 100% reflux, you can get aeotrope with about 7 plates from a 10% beer (but of course 100% reflux gives you no product and is only useful as a science fair project).  Nixon and McCaw advocate short columns and around 90% reflux (10 liters down the column for even one that you take off). 
         
        This is all dictated by the laws of physics.  The only ways around this are:
        - increase the proof of your beer (decreases reflux for azeotropic product)
        - reduce your product purity requirement (reduce the required proof of the product)
         
        You can also use these same figures and increase your product production rate by increasing the boiler power, but only if you correspondingly increase your condenser heat removal capacity (to maintain the reflux ratio needed for your HETP), and making sure that your column diameter is sufficient to maintain vapor flow up the column at a resasonable rate (12 - 20 in/sec; calculated as I described in an earlier posting).
         
        Note that when you start your batch still off with 10% abv, it won't stay this way.  As product comes off, the %abv of the beer must reduce and therefore you need to monitor still head temperature (as a proxy for proof) throughout your run and as you see the temperature increase, you must increase your reflux rate to get back to azeptrope (or whatever proof you desire).  At some point, your reflux ratio will be so high that you decide that it isn't worth it anymore and you will stop the run and use the low wines on a subsequent run.
         
        I cannot emphasize enough that the "correct" way to get your still set up and going the way that you want it is to start with 100% reflux and get the column up into equilibirium.  This could easilt take a couple of hours.  If you can't achieve equlibrium with 100% reflux, you aren't removing enough heat or don't have enough HETP or have some other issue such as fugitive heat loss from the column, reflux running down the column walls, or plugged packing, etc.  If your beer was only ethanol and water, then at equilibrium the still heat temperature would be around 172.5 degrees F which is the boiling point of the azeotrope.  You would then SLOWLY start to take off product, carefully watching the thermometer, until you saw it rise to a new, higher temperature at equilibrium, and then you would slightly increase the reflux ratio back to azeotrope and keep watching and adjusting, for reasons stated earlier.  Of course you have other stuff in your beer than ethanol and water and some of it has a lower BP than the azeotrope, so your initial equilibrium temperature will be a little lower depending upon how much methanol and other stuff is in your beer.  As a spirit distiller, you probably want to very slowly bleed off this stuff (foreshots) and discard it, then start collecting the heads and the main run and stop at the tails.  I'm more interested in fuel ethanol and can afford to take this other stuff off as product, as long as I see that it is small (my engine won't get a headache :-).
         
        Get you column into equilibrium and keep it in equlibrium at all times!!!  This is what reflux distillation is all about.  Patience is the watchword here.  If you don't have the patience to do this, get yourself a high quality continuous still and then once you get it up into equilibirium and establish the required minimum reflux ratio, you can just leave it there without further adjustment (which doesn't mean unattended -- you are making a highly flammable fuel, regardless of what you actually intend to do with it later :-).
         
        Bob


        -----Original Message-----
        From: Gavin Flett <gavin_flett@...>
        To: new_distillers <new_distillers@yahoogroups.com>
        Sent: Mon, May 7, 2012 10:48 pm
        Subject: RE: [new_distillers] Re: My reflux column

        Just an update on the run Bob. I have tried it with 450W, 600W and 900W of 
        power, I have tried a 2:1, 1:1 and 1/2:1 reflex ratio. Everything that I do 
        achieves 92% ABV, so I am thinking that it is my packing that is preventing 
        further purity. Or, maybe it's my alcoholometer that is faulty....
        
        
        
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