Misplaced_Canuck

Recipes 

5 gallons of beer @ 212 = 660gal-degF
5 gallons of ice water @32 = 160 gal-degF

660 + 160 = 820 / 10 gal = 82F.

[This is in a perfect world].

You'll need more than 5 gallons. Probably 2x 5-gallomn buckets.

MC

__________________

SGFBeerBuzz

Recipes 

Ouch. Math hurts me tiny brain.

__________________
Beer Buzz Radio
Listen live every Sunday at 4 pm
KBFL 99.9 FM 1060 AM or streaming at RadioSpringfield

Watch the show on our webcam at TAGsgf.com

"Beer, if drunk in moderation, softens the temper, cheers the spirit, and promotes health."

Lennie

Recipes 

You don't need a big pump to pump through 75' of 1/4" hose as long as its not going vertically any great distance. Plus you want the water to move fairly slowly to get the most heat out per unit of water. A small fountain pump will work. The only thing you have to watch, is the temp of the water. A little pump won't take real hot water. I'd fill your sink and pump this through the system, then switch to the ice water after you've brought the temp down.

They do make a rubber sleeve kind of thing that will slip over a faucet without being screwed in or anything, it doesn't have much back-pressure so it ought to work unless your faucet has a large head on it. Maybe something like this:

http://www.amazon.com/04410201-06-Shampoo-Spray-Shower-vinyl/dp/B001H2OPS8/ref=sr_1_157?ie=UTF8&qid=1330725519&sr=8-157

DeafSmith

Recipes 

It's actually more complicated than that.

I worked out some numbers to cool 5 gallons from 212 to an equilibrium temperature using various amounts of water and ice. Assumptions are:
heat capacity of wort is almost the same as water (very nearly true)
water at boiling weighs 8.0 lbs per gallon
water at room temp. weighs 8.33 lbs per gallon
ice has a latent heat of fusion of 143.9 BTU per lb
ice has a heat capacity just about 1/2 that of water
Assume we start with W gallons of water at 65º F and I lbs of ice at 0º F

BTU to cool 5 gallons of water from 212º F to equilibrium temp. X =
5(8.0)(212-X) = 40(212-X)

BTU to heat I lbs of ice from 0º F to 32º F = (1/2)(I)(32) = 16I

BTU to melt I lbs of ice = 143.9 I

BTU to heat the water from I lbs of melted ice to equil. temp. X =
(I)(X-32)

BTU to heat W gallons of water from 65º F to equilibrium temp. X =
(W)(8.33)(X-65)



therefore:
40(212-X) = 16I + 143.9I + I(X-32) + 8.33W(X-65)
or
X = [(40)(212)+(8.33)(65)W - 128I]/(8.33W+40+I)

W (gallons)......I (lbs)......Equilibrium temp. X (º F)
----------------------------------------------------
......5...............10.................108.1
....10...............10...................94.6
....15...............10...................87.6
....20...............10...................83.2
......5...............20...................84.9
....10...............20...................79.1
....15...............20...................75.9
....20...............20...................73.9
......5...............30...................65.8
....10...............30...................65.6
....15...............30...................65.5
....20...............30...................65.4

So you can see how hard it is to cool 5 gallons of 212º wort to pitching temp.

DeafSmith

Recipes 

How much pressure the pump needs to develop depends on your objective. The coil has a lot of resistance to flow and needs a lot of pressure to get a large volume of water through it per unit time. If your objective is to cool using the minimum amount of water through the chiller, then you want the water to go through slowly and don't need much pressure from the pump. On the other hand, if your objective is to cool quickly and not to conserve water, then you want the maximum flow you can get and the more pressure the pump puts out (lift capacity), the better.

Lennie

Recipes 

My experience with ICs is that low flow early on is adequate to get a significant cooling to below 140F. I'm guessing you'd be OK with less than a gallon a minute. After that, things slow but its not critical at that point. I wouldnt bother with ice until the wort in under 100F. 30lb of ice is probably less efficient that 10gal of water, from a price perspective.

brbecker87

Recipes 

Thanks for all the help guys.

I can't remember who said it, but they mentioned that a March pump wouldn't work so well. I was thinking that if I buy a march pump to run the water through IC for now (with the intention of buying a plate chiller eventually?), but I get the impression that it might just be best to get a small to medium submersible pump for the price and just use lots of water/ice.

Just for my information. How does a plate chiller compare to an IC?

Copbrew133

Recipes 

I had an issue with cooling quickly as living here in Florida our water supply is usually around 80 degrees. I solved the problem cheaply through a closed cooling system. I got an old styrofoam Omaha Steak cooler with two holes on the end to run a half inch outlet from the pond pump, which came from Lowes for like $30, to my 25 foot copper immersion chiller. the warm water returns to the cooler through the other hole in the end and the warm water flows over the top of the ice water. The pump sits in the bottom of the old cooler and pumps cold water just fine through a total of about 40 feet of tubing and copper. It cools boiling wort down to pitch temps in about 15 minutes. When I used to use an ice bath and tap water it took about an hour. I do still use an ice bath in a circular pond liner. It speeds things up a bit. I also get free ice so I understand the ice thing could get a little extra pricy.

adixon3

Recipes 

Have you tried taking your shower head off and replacing it with a water hose? or if it doesn't fit cut the hose and attach it with a hose clamp.

brbecker87

Recipes 

I haven't tried that shower head idea, but my wife has already warned me that I'm not to have a hose running through the house, especially since we have new carpet...