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Old 04-04-2013, 09:08 PM   #21
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For the mechanically and household-handy-impaired (me) is that half inch or 1 inch copper? The price is literally double between the two.

I.e. would this work:

Lowes copper tubing
Mine are 3/8ths inch. My five gallon setup used a 25' length, when I went to ten gallon batches I needed to double that lenth.

These days, for just a little more money than raw materials, immersion chillers are readily available and ready to go like this one:




http://www.amazon.com/Super-Efficien...ersion+chiller
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Old 04-04-2013, 09:47 PM   #22
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Mine are 3/8ths inch. My five gallon setup used a 25' length, when I went to ten gallon batches I needed to double that lenth.

These days, for just a little more money than raw materials, immersion chillers are readily available and ready to go like this one:


http://www.amazon.com/Super-Efficien...ersion+chiller

Wow, thanks a ton! It's literally only pennies more when all is said and done! Funny, one of the reasons I want to move from 1 gallon to 5 gallon (aside from just moving up) is because of your Blonde Ale!!
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Old 05-22-2013, 09:56 PM   #23
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Originally Posted by BierMuncher View Post
Mine are 3/8ths inch. My five gallon setup used a 25' length, when I went to ten gallon batches I needed to double that lenth.

These days, for just a little more money than raw materials, immersion chillers are readily available and ready to go like this one:




http://www.amazon.com/Super-Efficien...ersion+chiller
Is there a length of tubing, or diameter of tubing, where its efficiency begins to drop off?

I like to build and MacGuyver things, so it's not necessarily about saving money, but if I grabbed a 1/4" set of tubing, and created a 50' coil. Would this be twice as efficient as a 1/2" tube with a 25' coil? I am not a scientist of any kind so this is all just junk in my brain. The way I think is, same amount of coolant water, smaller tube diameter, longer tube, increase flow pressure (and flow rate, or am I confusing my laws here), would this create a net effect of more cooling quicker?

Or is it the exact opposite, because theres less water in any given section of the tubing, and its traveling faster through the tubing because of the decreased diameter, it has a lower thermal mass, so it wouldn't be able to pull as much heat out of the wort? Am I completely insane and should've paid more attention in physics class?
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Old 05-23-2013, 03:28 AM   #24
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Id wait but that's just me. as far as lifting it, that's 25lb minimum empty keggle.... one gallon of water weighs roughly eight lbs, plus a big hunk of oddly shaped metal... u do the math. and considering its gonna be hot as hell, that's a big risk. If uve never used a keggle before either I think u'll be surprised just how hot even the top of the keggle will get.

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Old 05-24-2013, 02:40 AM   #25
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Originally Posted by ArkotRamathorn View Post
Is there a length of tubing, or diameter of tubing, where its efficiency begins to drop off?

I like to build and MacGuyver things, so it's not necessarily about saving money, but if I grabbed a 1/4" set of tubing, and created a 50' coil. Would this be twice as efficient as a 1/2" tube with a 25' coil? I am not a scientist of any kind so this is all just junk in my brain. The way I think is, same amount of coolant water, smaller tube diameter, longer tube, increase flow pressure (and flow rate, or am I confusing my laws here), would this create a net effect of more cooling quicker?

Or is it the exact opposite, because theres less water in any given section of the tubing, and its traveling faster through the tubing because of the decreased diameter, it has a lower thermal mass, so it wouldn't be able to pull as much heat out of the wort? Am I completely insane and should've paid more attention in physics class?
Well, I never did that well in Fluid Mechanics, but I can say most things with heat exchangers relate to surface area...

Surface Area of tubing = PI * diameter * length in feet * 12in/ft

So, 50' of 1/4" = 3.14*.25*50*12 = 471 sq. in.

25' of 1/2" = 3.14*.5*25*12 = 471 sq. in.

When you double the diameter, but halve the length you end up at the same place. Both coils have the same surface area.

However, they do have different volumes...

Volume of tubing = PI * radius squared * length in feet * 12 in/ft

Volume of 50' of 1/4" = 3.14*.125*.125*50*12 = 29.43 cu. in.

Volume of 25' of 1/2" = 3.14*.25*.25*25*12 = 58.88 cu. in.

Maybe some of the factors you mentioned might put one slightly ahead of the other, but i bet there wouldn't be much of a difference between the two setups due to the surface area being the same.

P.S. Awesome screen name. Do you have any liter-a-cola?
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Old 05-24-2013, 09:59 PM   #26
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Well, I never did that well in Fluid Mechanics, but I can say most things with heat exchangers relate to surface area...

Surface Area of tubing = PI * diameter * length in feet * 12in/ft

So, 50' of 1/4" = 3.14*.25*50*12 = 471 sq. in.

25' of 1/2" = 3.14*.5*25*12 = 471 sq. in.

When you double the diameter, but halve the length you end up at the same place. Both coils have the same surface area.

However, they do have different volumes...

Volume of tubing = PI * radius squared * length in feet * 12 in/ft

Volume of 50' of 1/4" = 3.14*.125*.125*50*12 = 29.43 cu. in.

Volume of 25' of 1/2" = 3.14*.25*.25*25*12 = 58.88 cu. in.

Maybe some of the factors you mentioned might put one slightly ahead of the other, but i bet there wouldn't be much of a difference between the two setups due to the surface area being the same.

P.S. Awesome screen name. Do you have any liter-a-cola?
Oh duh, surface area... So basically I would have to try a length of coil up to the point where the coolant water reached equilibrium with the wort and has pulled as much energy as possible from the wort.

What's a liter-a-cola?

I'm super jealous of my younger brother since he got to meet Steve Lemme and Kevin Heffernan during a bacherlor party (I believe they went to their stand up tour).
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Old 05-28-2013, 02:25 PM   #27
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Wait.... A hot Keggle is not an easy thing to move and not worth the risk.

DPB

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Old 05-28-2013, 03:18 PM   #28
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I've carried a hot 10 gal batch in a keggle down a few steps and across a long hallway to chill. It's doable, but not something I'd recommend or really want to do again.

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Old 05-29-2013, 01:03 AM   #29
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I chill my 5 gal. batches in three ice baths to increase the surface area and lighten the load a bit. About 1.75 gal. at a time. The first 2 I get to about 90 degrees and the last one as cold as possible. Sometimes I have to wait a bit before pitching. I also use a whisk to oxygenate the wort.

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