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Old 12-06-2012, 11:56 AM   #11
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It will add to the thermal mass that is true. But unless you sit the carboy in a container much bigger than the carboy it won't be a lot. Easy to work out, the increase in thermal mass is the volume of the water left in the container the carboy sits in (not the volume of the container) divided by the volume of the carboy liquid, approx.

However you are changing the air insulation of the carboy to water which is a very poor insulator - yous are effectively increasing the surface area of the total exposure to the room. Not to mention the cooling-by-evapouration effect of the exposed water. And the bigger the volume of water in the container the more the surace area exposure. I suspect on balance it will just make it worse for the carboy.

More effective would be to better insulate the existing thermal mass of the carboy by wrapping it in a dry towel and standing it on some insulation material. It's also easier. Alternatively, belts and braces, sit it in the container and then wrap the external container.

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Old 12-06-2012, 12:02 PM   #12
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You did notice the original question was posed 3.5 years ago, yes?

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Old 12-06-2012, 02:40 PM   #13
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It will add to the thermal mass that is true. But unless you sit the carboy in a container much bigger than the carboy it won't be a lot.
It's more significant than you imply. My 5 gallon better bottle has a roughly 10 inch diameter and 15 inch height. If it's in a bucket with just a 1 inch space around it, filled 2/3 of the way up---10 inches deep---that adds 30% to the water mass. If it's a 2 inch gap, that is now almost 70% more mass.

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However you are changing the air insulation of the carboy to water which is a very poor insulator - yous are effectively increasing the surface area of the total exposure to the room.
Two problems with this. First, the insulating properties of the water is not terribly relevant here. That water is still sitting in poorly heat-conducting air (mostly through thick, fairly insulating plastic), so it's not any better coupled (per unit surface area) to the air temperature than the fermentor was.

Second, dimensionally, the surface area increases with radius squared, while the volume increases with its cube. The bigger you make the container the smaller the ratio, and it's the ratio that determines the rate of temperature change. So even if the enclosing water were perfectly coupled to the contents of your fermentor, you still benefit because you've decreased the surface area to volume ratio.

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Not to mention the cooling-by-evapouration effect of the exposed water.
I'm fairly certain this is negligible. Using this calculator for evaporation rate (http://rpaulsingh.com/problems/Ex10_2.htm), if the tub has a 2 inch gap of exposed water, the cooling effect will be measured in milliKelvin per hour.

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More effective would be to better insulate the existing thermal mass of the carboy by wrapping it in a dry towel and standing it on some insulation material. It's also easier. Alternatively, belts and braces, sit it in the container and then wrap the external container.
Not sure how to compare the either-or effect here and I've done all the calculating I have time for now. But it's not obvious to me that simply insulating it is the best approach. One additional wrinke is that the fermentation inside the fermentor is producing heat of its own; simply insulating will make this effect worse, while a water bath will serve to reduce temperature swings regardless of the source of the heat.
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Old 12-07-2012, 11:50 AM   #14
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just briefly five errors there.
1) The main error. It's not 30% more thermal mass, its 13% more, check your math.
2) Your surrounding water is open to the air so will lose/gain a lot more heat than the enclosed barrel encased in plastic.
3) your surface area to voume ratio relates to a sphere not a cylinder which is a lot less (can't be arsed to look it up). Do you have a spherical barrel and spherical surrounding tub? (If it runs 10" up the 15" of the barrel you will never be able to remove the barrel).
4) if you read the thread (or the title) the OP was asking about conditioning temperature not while fermenting so heat production by the barrel contents will be minimal if any.
5) cooling by evapouration depends on the airflow amongst other things so may not be tiny

I did see this was started 3.5 years ago, I came across it this week by Google and wanted to correct the error for others coming across it the same way.

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Old 12-07-2012, 02:17 PM   #15
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just briefly five errors there.
1) The main error. It's not 30% more thermal mass, its 13% more, check your math.
As far as I can tell, my math is correct.

Fermenter is 10 inches diameter, 5 inches radius, So a 1 inch gap is a 6 inch radius. Area of the annulus is pi*(36 - 25) = 34.5 sq in. If it's 10 inches deep, that is 345 cubic inches. A 5 gallon carboy is 1155 cu in. So the new volume is 1500 cu in. 1500/1155 = 1.3, or 30% greater.

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2) Your surrounding water is open to the air so will lose/gain a lot more heat than the enclosed barrel encased in plastic.
Do you have some method of containing the water in direct contact with the air? Otherwise it's not actually very different. (And, anyway, water is not that great a thermal conductor. It's better than plastic by a factor of 2-3, but it's worse than glass by a similar factor.) The water is also acting as an additional layer of insulation, even if it's not a very good one: you still have the layer of plastic (or whatever your carboy is made of) that the heat needs to pass through.

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3) your surface area to voume ratio relates to a sphere not a cylinder which is a lot less (can't be arsed to look it up). Do you have a spherical barrel and spherical surrounding tub? (If it runs 10" up the 15" of the barrel you will never be able to remove the barrel).
No, my ratio is approximately correct for any shape you can imagine. The exact factors vary, but the principle is that the larger your container, the lower the ratio of surface area to volume.

Again using the 1 inch gap (and ignoring thickness of the containing vessel), the surface area of the 10 inches height I'm assuming is 2*pi*(6 in) * (10 in) = 377 sq in. For the carboy itself, that's 2*pi*(5in)*(10in) = 314 sq in, so the increase is 20% in this section. That's not exactly the ratio since I'm ignoring the surface area above the enclosing vessel (which is unchanged, so will reduce the increase ratio) and the small additional area that's in contact with the floor (which will increase it). Since we increased the volume by 30% and the area by about 20%, we have a smaller net effect due to the surface area exposure.


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4) if you read the thread (or the title) the OP was asking about conditioning temperature not while fermenting so heat production by the barrel contents will be minimal if any.
True, but it's a possible reason to prefer a water bath in general. In this case, if you need to do something, I think the best solution is what you pointed out---do both.

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5) cooling by evapouration depends on the airflow amongst other things so may not be tiny
Yes, it depends on a few things. I assumed a 1 m/s constant wind, which would be pretty high for in the house, so I'm pretty confident it's a conservative estimate. I'd have had to underestimate by a factor of a few hundred to a thousand to affect the conclusion.
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Old 12-10-2012, 04:55 PM   #16
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I don't really have time or inclination to correct you math, I pointed out it was wrong I would think you could have worked out your error with a little effort. Are you still at school? Area of a 12" diameter circle is 12pi. Area of a 10 inch diameter circle is 10pi. The area of the gap is 12pi - 10pi, i.e. 2pi. Volume of the gap at 10 inches high is 2pi*10, i.e. 20pi. Volume of a 10 inch cylinder at 15 inches high is 15*10pi, i.e. 150pi. The volume of the gap divided by the volumne of the carbouy is 20pi/150pi which is 13.33%. The increase in volume is 13.33%. Anyone just sense checking your claim ought to be able to tell without any math that it would never be 30%. Just visualise it.

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Old 12-29-2012, 06:54 PM   #17
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I was surfing the forum and decided to read this thread, because my son and I are exploring various ideas for fermentation temp control.

I apologize for jumping into this fray and taking sides, but I think some apologies might be in order: from dbdb if he's off-base or from me if I've misunderstood something from this disagreement.

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I don't really have time or inclination to correct you math, I pointed out it was wrong I would think you could have worked out your error with a little effort. Are you still at school?
I no longer have my college calculus book handy, but I believe the formula you've given is for the circumference of a circle, not the area.

1. The circumference of a circle is pi*diameter or 2*pi*radius.
2. The area of a circle is pi R squared or pi*radius squared.

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Area of a 12" diameter circle is 12pi. Area of a 10 inch diameter circle is 10pi.
No, dbdb, the circumference of a 12" diameter circle is 12pi; that of a 10" diameter circle is 10pi. The area of a 12" diameter circle is pi*6squared: 3.1416*36=113 sq in. The area of a 10" diameter circle is pi*5squared: 3.1415*25 = 78.5 sq in.

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The area of the gap is 12pi - 10pi, i.e. 2pi.
No, you've simply given how much bigger around one circle is than the other, appr 6.25" in this case. The gap is simply 1" all the way around(the 10" circle sitting inside the 12" circle).

The difference in the areas is (pi * 6squared) - (pi * 5 squared) or 113-78.5=34.5 sq in. or 30% less than the larger area or 44% larger than the smaller area. Please note that, when comparing these areas, you can't simply plug the difference into the formula "pi*(6-5)squared" as you did when you were unwittingly computing circumferences instead of areas.

From your formula, I get 37.7" circumference for the larger circle (12pi) minus 31.42" circumference of the smaller circle (10pi) for a difference of 6.29 inches or 16.7% fewer linear inches than the larger circle or 20% larger than the smaller circle.

This difference in circumference (gap as you say) is meaningless by itself in this case.

If you applied your calculations to farm land, you've simply calculated how many more linear feet of fence you'd need or save to go around a pasture that was 1000ft vs 1200ft on a side. Your formula won't tell you how much fertilizer or seed you'd need to cover said fields. For square fields, 20% more linear feet of fence, but 44% more volume of seed. The calculations for round fields are a bit more complicated.

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Volume of the gap at 10 inches high is 2pi*10, i.e. 20pi.
No, this just gives you the surface area of one side of a tube that's 2" in diameter x 10" long. This is not the volume of the gap between a 10" long 10" diameter tube set inside a 10" long 12" diameter tube.

Even if you use the correct formula for area of a circle, you cannot compute volume of the gap simply by plugging the difference in diameter into the formula. You have to compute the volumes first, then subtract them.

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Volume of a 10 inch cylinder at 15 inches high is 15*10pi, i.e. 150pi. The volume of the gap divided by the volume of the carboy is 20pi/150pi which is 13.33%. The increase in volume is 13.33%.
No, your results do not give volumes. What you have computed is the surface area (of one side) of a tube that is 10 inches in diameter and 15" long. The rest of your statement makes no sense to me, because it appears that you're starting your calculations with the wrong formula - you're trying to jump from circumference to volume (1D or linear to 3D) instead of from area (2D) to volume (3D)

Also, when comparing circumferences and surface areas of tubes of different diameters, it's OK to simply compute the surface area of the difference between the two. The differences are directly proportional and the formulas don't contain parentheses. IOW, pi*diameter1*height - pi*diameter2*height is the same thing as pi*(diameter1-diameter2)*height.

OTOH, when comparing the area of circles and volumes of round things, the proportions are exponential, so you can't simply plug the difference between the radii into the formula. This difference, or gap, is a volume. You have to compute the two volumes first, then subtract them.

For example, when comparing the areas of different circles (and, hence, volumes of tubes of various radii), pi*radius1 squared - pi*radius2 squared is not the same as pi*(radius1-radius2)squared. Neither is it the same when you multiply these by height. IOW, the difference between the area of 2 circles of 6" and 5" (your 12" and 10" circles) is the area of the 12" circle minus the area of the 10" circle. You can't simply plug the difference between the diameters or radii into your area formula as you can with the formula for circumference.

1. The outside surface area of a tube is circumference*height and is expressed in square units (square inches, square feet, square whatevers).

2. The volume of a tube is radius squared*pi and expressed in cubic units.

Using the correct formula for area of a circle given earlier in this message, the volume of the 10" high 12" diameter container would be 113 sq in * 10 in high = 1130 cu in. The volume of the 10" high 10" diameter container would be 78.5 sq in * 10 in = 785 cu in. The difference in volume (the volume of the gap/space between) would be 1130 cu in - 785 cu in = 345 cu in. IOW, 30.5% less volume compared with the larger vessel or 44% greater volume compared with the smaller vessel.

I ran a few other calculations on paper to double check my math and yours.

1. When comparing the differences in circumferences of circles, you can subtract the diameters before multiplying by pi. IOW, (pi*diameter #1) - (pi*diameter #2) is the same thing as pi*(diameter #1 = diameter #2).

2. When comparing the differences in areas, you have to compute the individual areas first, then subtract them. For example, the volume of the 7" space between a 5" radius tube inserted into a 12" radius tube is a lot greater than the volume of a a 7" radius tube alone.

If you double the diameter of a circle, you also double the circumference as well as the surface area of one surface of a tube of this diameter.

If you double the radius of a circle, you increase the area by quite a bit more than double.

I think all this is much simpler when dealing with rectangles in 2D and 3D. Using circles to make tubes and spheres makes the calculations more complicated.

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Originally Posted by dbdb View Post
Anyone just sense checking your claim ought to be able to tell without any math that it would never be 30%. Just visualise it.
Visualizing differences in areas of 2D objects and, to an even greater extent, differences in volumes of 3D objects can be deceiving. A rather small difference in one dimension of a vessel can cause a surprising increase in volume.

Dbdb, surely there's a way to explain your stance without being uncharitable or insulting. There are problems in your calculations (area vs volume), so you should assume a posture that's a little more humble. I think you need to recheck your formulas, redo the math, then see if you're still correct.

I will publicly apologize to dbdb ahead of time, though, if I've totally missed something here and criticized his calculations undeservedly. My wife says I tend to do that when I butt in to a conversation that's already in progress.

Respectfully,
Keith

Last edited by kzimmer0817; 12-29-2012 at 09:03 PM. Reason: Corrected a term.
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Old 12-29-2012, 07:57 PM   #18
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I don't really have time or inclination to correct you math, I pointed out it was wrong I would think you could have worked out your error with a little effort. Are you still at school?
No, I'm no longer at school. I wrapped up the PhD at Caltech last year.

I'm pretty confident in my numbers, and as kz pointed out, your calculations are for circumference and area, not area and volume.

He's also correct that visualizing volumes is rather difficult, but as a general rule, most of the volume of a solid (sphere, cylinder, or something approximating that) is found at large radii. This is because the additional volume is roughly proportional to the surface area of a sphere at that radius (exactly proportional in the case of a sphere), and that grows as r^2. So the shell at r=2 has 4x the contribution as that at r=1, r=3 has 9x, etc. That is why a relatively small increase in radius adds such a large fractional volume.

Anyway, after all this, I actually do agree with the idea that insulation is probably preferable for the situation being discussed (however many posts/years ago that was). If you're just trying to keep the fermentor at the average temperature of the room, slinging gallons of water around is annoying enough that it'd have to be far more effective to be worthwhile. But the idea that the water bath is increasing the coupling to the environment doesn't wash (pun regretfully intended).
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