The mods have reviewed, discussed and addressed the recent issues and borderline comments referenced in this thread. No need to continue mentioning them and trying to chum the waters to keep this kind of thing going.
Yes, this topic has been discussed before. Looking back through old posts I found where I had been involved in the discussion on the exact same topic on this forum in 2006. Nine years later still no progress. If you think it is redundant then don't subscribe to this thread or respond in the discussion. If you want to be involved with this thread, then don't make snide comments.
Now turn those keg boilers 90 degrees please!
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Lazarus Long
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Re: Now turn those keg boilers 90 degrees please!
Now back on topic or begone. 
I use a pot still.Sometimes with a thumper
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shadylane
- Master of Distillation
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Re: Now turn those keg boilers 90 degrees please!
Made two runs. One with the boiler vertical and the other with it leaning at a 45' angle.
I couldn't lay the korny keg on it's side because of the heating element placement.
There was 3-1/2 gallons in the keg and the measurements were done after the keg had been boiling for an hour.
Both runs distilled a quart of water in 25 minutes.
I couldn't lay the korny keg on it's side because of the heating element placement.
There was 3-1/2 gallons in the keg and the measurements were done after the keg had been boiling for an hour.
Both runs distilled a quart of water in 25 minutes.
Re: Now turn those keg boilers 90 degrees please!
I don't know much about the vapor bed disturbance due to boiling turbulence, so for all of the following please take it with a grain of salt. I wanted to introduce some real numbers to what we are talking about here.
All of the following calculations are based on a regular 15.5 gallon Sanke Keg. For simplicity i ignored the slight doming at the top and bottom of the keg in the math except for 1 value which I'll point out below.
If you charge at 83% ish you end up pretty close to 13 gallons. At 13 gallons you get the following values:
Vertical Keg:
Surface Area ±198 sqin
Head Space Volume ±572 cuin
Head Space Distance ±3.5 in This value I added domed height of keg too.
Horizontal Keg:
Surface Area ±240 sqin
Head Space Volume ±572 cuin
Head Space Distance ±3.5 in
This might seem really surprising considering the conviction in the debate in this thread. I know I was surprised. Its just trigonometry though. As you fill a horizontal cylinder past halfway the surface area of the liquid decreases. The rate at which this decrease in surface area occurs increases as the vessel level rises. Surface area is equal to (1/2r^2)*(2arccos(m/r)-sin(2arccos(m/r)) where r=radius of the vessel & m=filled height. That's a lot of math to find out that the surface area is only 20% different.
Many of you have already covered that watts in = watts out so i don't see how there is a difference in rate due to surface area anyway. The part I don't understand is the gas bed stuff. I don't know if this is helpful or not. As many of you said there is a pretty big difference in the complexity of construction between a vertical and horizontal still, so it may not even be an option for lots of folks, but there's the number either way.
All of the following calculations are based on a regular 15.5 gallon Sanke Keg. For simplicity i ignored the slight doming at the top and bottom of the keg in the math except for 1 value which I'll point out below.
If you charge at 83% ish you end up pretty close to 13 gallons. At 13 gallons you get the following values:
Vertical Keg:
Surface Area ±198 sqin
Head Space Volume ±572 cuin
Head Space Distance ±3.5 in This value I added domed height of keg too.
Horizontal Keg:
Surface Area ±240 sqin
Head Space Volume ±572 cuin
Head Space Distance ±3.5 in
This might seem really surprising considering the conviction in the debate in this thread. I know I was surprised. Its just trigonometry though. As you fill a horizontal cylinder past halfway the surface area of the liquid decreases. The rate at which this decrease in surface area occurs increases as the vessel level rises. Surface area is equal to (1/2r^2)*(2arccos(m/r)-sin(2arccos(m/r)) where r=radius of the vessel & m=filled height. That's a lot of math to find out that the surface area is only 20% different.
Many of you have already covered that watts in = watts out so i don't see how there is a difference in rate due to surface area anyway. The part I don't understand is the gas bed stuff. I don't know if this is helpful or not. As many of you said there is a pretty big difference in the complexity of construction between a vertical and horizontal still, so it may not even be an option for lots of folks, but there's the number either way.
Re: Now turn those keg boilers 90 degrees please!
I am a power engineer. Think of evaporation as a change in state of energy. The energy put into the boiler is still there but in the form of steam. That energy can still be used but of course there is a lost of energy with each use and change of state.LWTCS wrote: +100%
Two 15 gallon, identical kegs with two equal beer charges of say,,,,,10%, both heated with 5500 watts applied ( one keg upright and one keg lying on its side) will render the same volume at the same rate.
The squat kettle shapes and horizontal position of pill shaped kettles has much more to do with (not including perceived quality of separation) keeping construction and long term maintenance costs within the total physical plant reduced.
Buildings with taller ceiling height are just more expensive to construct. Keeping the entire distillation apparatus closer to the ground is easier and safer to maintain over the service life of the equipment.
The only thing that can be done is too collect the energy more efficiently at the boiler. Odin suggests laying the boiler on its side. The idea is that the surface of the point of contact is where the change of state occurs but that is wrong. The change of state happens where the heat energy is applied. The surface that is being heated.
The best way to increase the efficiency is to increase the heating surface. That is why fire tube boilers are universally used in large applications for boiling water.
The gains to be had are tiny in systems that are 25 gallons or smaller. Makes my head hurt. I don't like pain in a hobby...
Re: Now turn those keg boilers 90 degrees please!
When I read this thread, discarding of the rubbish that is not already removed, I think two issues have been intertwined.
1) What is the best (flavourwise, energywise or else) design: wide and low or narrow and high? On this point Odin states that he got some efficiency gain with wide and low over narrow and high. (by the way: 4 times as wide, not 4 times wider!!)
2) Will the turning of a keg reach the same gain? This has not been tested, apart from a test by shadylane.
What seems to be forgotten is that a 90 degr turned keg is a cilinder on its side.
When I think of my own 10-l Daalderop-boiler, about 28 cm wide and 60 cm high, used as a potstill, I am curious about the changes in behaviour during a run. I always run at a constant power. Now it runs steadily, slowing down in an almost constant manner. The fluid in the kettle evaporates slowly into a slowly growing headspace. The surface being constant all the time.
I did not do any calculation, but when I would turn it 90 degr, it would react very differently. The amount of vapour per second would be the same (assuming complete isolation), but the force of this boiling would vary: very forceful at 90 % filling to quiet at 50 % filling to forceful at 10 % filling. Also the amount of reflux due to cooling in the headspace would vary in a non-constant manner.
I wonder if this would have any influence on the making of cuts.
This reminds me of the distilling with submarine stills, made of wood panels and iron sheets. These were wide and low, as opposed to the copper kettles.
1) What is the best (flavourwise, energywise or else) design: wide and low or narrow and high? On this point Odin states that he got some efficiency gain with wide and low over narrow and high. (by the way: 4 times as wide, not 4 times wider!!)
2) Will the turning of a keg reach the same gain? This has not been tested, apart from a test by shadylane.
What seems to be forgotten is that a 90 degr turned keg is a cilinder on its side.
When I think of my own 10-l Daalderop-boiler, about 28 cm wide and 60 cm high, used as a potstill, I am curious about the changes in behaviour during a run. I always run at a constant power. Now it runs steadily, slowing down in an almost constant manner. The fluid in the kettle evaporates slowly into a slowly growing headspace. The surface being constant all the time.
I did not do any calculation, but when I would turn it 90 degr, it would react very differently. The amount of vapour per second would be the same (assuming complete isolation), but the force of this boiling would vary: very forceful at 90 % filling to quiet at 50 % filling to forceful at 10 % filling. Also the amount of reflux due to cooling in the headspace would vary in a non-constant manner.
I wonder if this would have any influence on the making of cuts.
This reminds me of the distilling with submarine stills, made of wood panels and iron sheets. These were wide and low, as opposed to the copper kettles.
Re: Now turn those keg boilers 90 degrees please!
Ketelkantel
Like NatiMatt I did some math.
Suppose we have a cilindrical boiler with a width of 20 cm, so a radius of 10, standing upright. Independent of the degree of filling and the height of the boiler, the surface of the fluid in it will always be the same: πr2 .
When we lay the boiler on its side, the surface is dependent on the degree of filling, as can be seen in this cross section.
The level of the fluid is h. Its length is dependent on the angle φ: h = 2 * r * sin( ⅟₂ φ)
The surface of the remaining part of the circle is: O = {(360-φ)/360} * π * r2
The surface of the triangle formed by h, r and r is: D = sin( ⅟₂ φ) * cos( ⅟₂ φ) * r2
And the surface of the cross section of the fluid is the sum of O and D.
Although the real volume of the filling is dependent of the height of the cilinder, the degree of filling is only dependent of φ. Or rather: φ is the result of a degree of filling.
The resulting surface of the fluid is a rectangle, with one side depending on φ and the other on the height of the cilinder. Given a fixed height, the surface will be at its maximum when the boiler is half filled and so φ is 180°.
Now we can find the surfaces of the fluids in layed down kegs dependent on the degree of filling and dependent on the height of the kegs. I took 3 examples: 1 is as high as it is wide, so r = 10 and H = 20. The second is twice its width and the third is 3 times its width.
The y-axis is the percentage of filling, the x-axis is the surface of the fluid in square whatevers.
The 10 % at the top and at the bottom are hardly relevant, as normally one does not fill his boiler to that levels.
An interesting point is the half-filled square keg, that is 20 wide and 20 high: when it is laid on its side we realise the squaring of the circle! At last we, distillers, solved a millennia old mathemathical problem!!
I was surprised to see how small the gain is for this square keg: when half filled the surface of fluid in a laid down keg is only 27 % larger than in an upright keg.
For higher kegs the surface is more or less proportional to the proportion of the keg: when the keg is 2 times as high as wide, the surface of the fluid will be 2 times as large when laid down compared with upright.
But then I would like to see some opinions of chemical technical engineers on heat and fluid and vapour.
And now for a sphere?
Like NatiMatt I did some math.
Suppose we have a cilindrical boiler with a width of 20 cm, so a radius of 10, standing upright. Independent of the degree of filling and the height of the boiler, the surface of the fluid in it will always be the same: πr2 .
When we lay the boiler on its side, the surface is dependent on the degree of filling, as can be seen in this cross section.
The level of the fluid is h. Its length is dependent on the angle φ: h = 2 * r * sin( ⅟₂ φ)
The surface of the remaining part of the circle is: O = {(360-φ)/360} * π * r2
The surface of the triangle formed by h, r and r is: D = sin( ⅟₂ φ) * cos( ⅟₂ φ) * r2
And the surface of the cross section of the fluid is the sum of O and D.
Although the real volume of the filling is dependent of the height of the cilinder, the degree of filling is only dependent of φ. Or rather: φ is the result of a degree of filling.
The resulting surface of the fluid is a rectangle, with one side depending on φ and the other on the height of the cilinder. Given a fixed height, the surface will be at its maximum when the boiler is half filled and so φ is 180°.
Now we can find the surfaces of the fluids in layed down kegs dependent on the degree of filling and dependent on the height of the kegs. I took 3 examples: 1 is as high as it is wide, so r = 10 and H = 20. The second is twice its width and the third is 3 times its width.
The y-axis is the percentage of filling, the x-axis is the surface of the fluid in square whatevers.
The 10 % at the top and at the bottom are hardly relevant, as normally one does not fill his boiler to that levels.
An interesting point is the half-filled square keg, that is 20 wide and 20 high: when it is laid on its side we realise the squaring of the circle! At last we, distillers, solved a millennia old mathemathical problem!!
I was surprised to see how small the gain is for this square keg: when half filled the surface of fluid in a laid down keg is only 27 % larger than in an upright keg.
For higher kegs the surface is more or less proportional to the proportion of the keg: when the keg is 2 times as high as wide, the surface of the fluid will be 2 times as large when laid down compared with upright.
But then I would like to see some opinions of chemical technical engineers on heat and fluid and vapour.
And now for a sphere?