Ice Water For Immersion Chiller

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The Pol

Well-Known Member
Feb 12, 2007
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Anyone circulated say 10 gallons of 32F water through thier immersion chiller, back to the tank, through the ice water again, then back through the immersion chiller? Will the ICE water remain cold enough (10 gallons) to cool (5.5 gallons) of wort?

I always do the final chilling this way. I use the ground water to get the wort to 120-100 and then make the switch to ice water. It works great.
Well, I was hoping to use ONLY ice water... I may have to run a test with boiling water and see how well 10 gallons will chill it without becoming heat saturated.
The Pol said:
Well, I was hoping to use ONLY ice water... I may have to run a test with boiling water and see how well 10 gallons will chill it without becoming heat saturated.
If that's the case then I would not add the initial output back into recirculation until it gets to around 150 or so, I think it will melt the ice to quickly to get the wort to pitching temps. Let us know how that works and how much ice you used, I am sort of interested to see if it works well.
When I get all of my plumbing finished I will do a dry run... If I fill the 10 gallon tank with 32f water and run about 5 gallons of it out, then start to recirculate, it may work. I am trying to keep the system self contained and save water, even if it means stocking up on ice once every 8 weeks.
You can use the initial output of water when its hot for cleanup. I save my output in a 25 gallon trash can and also use it to mix 5 gallons of iodaphor for the carboys, tubing and such, then I recirculate. Gotta save water out here in my town.
For one of my last batches I chilled with only snow, mainly because all my hoses were frozen, and I didn't like that I didn't have warm water for cleaning.

You don't have to go through the effort of using 32*F water for the initial chill. Just use room temp water and save the ice for later. The few more degrees initially won't make a big difference.

You should dry water with dry ice in it. The melting temp of dry ice is significantly lower then water. I use dry ice for this and it works perfect.
just did that last brew day, had no hoses. started out with 20 pounds of ice took another 20 pounds to get down to 68 degrees. normally i use tap water to get down to 100 degrees and 20 pounds of ice to get to 68 degrees with some ice left over for seebee John to use.

My whole epiphany here was to get away from using the garden hose,
I have been successful at getting my entire brewing system with HERMS efficiently mounted on a rolling rack that is 18" deep, 36" wide and 72" tall. That includes my keggle (burner), MLT and HLT, along with my march pump, JC digital temp controller for the HERMS and LP tank.

Ideally I would love to get my 10 gallon cooler filled with 5 gallons water, the rest full of ice, the initial return to the cooler would further melt the ice, the ice absorbing the heat. I will obviously have to make a trial run to see what the heat absorbing capacity of this system would be.

I am also pretty sure I am ordering one of these:

It is called a sentry and I thought it would be great in helping me maintain say 170F in my keggle while I am running my wort through it for HERMS. Better control of the water temp, less work for me, more automation.

I have always used tap water to chill with, works swell, I just want to use less water and be able to brew anywhere, even if I dont have a garden hose available to run the chiller.
5.5 gallon of 212*f wort being cooled by 10 gallon of 32*f water should reach equilibrium at about 90*f.
Wow - a three year old thread. :)

According to my calculations, if you start with 5 gallons of water at 60 ºF with 20 lbs of ice at 0 ºF added to it, and try to cool 5.5 gallons of water (since I don't know the heat capacity of wort vs. water) from 212 ºF, they will reach equilibrium at 77 ºF.

5 gallons water at 60 ºF = 5*8.33 lbs = 41.7 lbs = 18.9 Kg at 15.6 ºC
20 lbs ice at 0 ºF = 9.1 Kg at -18 ºC
5.5 gallons water at 212 ºF = 5.5*8 lbs = 44 lbs = 20 Kg at 100 ºC

specific heat of water = 4.2 KJ/Kg ºC
specific heat of ice = 2.0 KJ/Kg ºC
latent heat of fusion of water = 333 KJ/Kg

heat to raise ice temp to freezing + heat of fusion + heat to raise temp of melted ice and water to x degrees = heat removed when "wort" is cooled to x degrees where x is the equilibrium temperature

(2.0 KJ/Kg ºC)(18 ºC)(9.1 Kg) + (333 KJ/Kg)(9.1 Kg)
+ (4.2 KJ/Kg ºC)(9.1 Kg + 18.9 Kg)(x ºC) =
(4.2 KJ/Kg ºC)(20 Kg)(100 ºC - x ºC)

328 KJ + 3030 KJ + (118 x) KJ = 8400 KJ - (84 x) KJ

202 x = 5042

x = 25

25 ºC = 77 ºF

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