Water temp vs final temp with IC

[theonetrueruss]

I usually pump ice water through my IC once my temp reaches 100F or a little less. I do that until I get down to 65F. I also recirc in my kettle so the temp drops nice and fast. Tomorrow night I'll be brewing and I just measured my tap temp and it is 61F. I figure I could skip the ice and just use the tap water and eventually hit 65F.

I am figuring that it will take a lot longer for 61F water to cool wort to 65F though theoretically possible. Should I give up this idea and just get some ice tomorrow like I would normally do?

I don't suppose anyone has some good knowledge of thermodynamics and could figure the time for a 50' 1/2" copper chiller with recirculating 11G of wort at 1.08OG to chill from 100F to 65F if the water running through the chiller is 61F. Assume a re-circulation rate of 5 GPM. I always sucked at fluids and my math is way out of practice. Though I am tempted to dig back in on this. The problem sounds interesting to me.. and solvable as opposed to the origin of the universe from nothing stuff I am interested in which i lack the brain power to solve)

 

[TheWall]

Hello, we can do some pretty simple thermo calculations to get a truely rough estimates of time. What you should take away from these calculations are the difference in the two times, more of a ratio of time to cool.

First Ice Water

Flow of Ice Water through immersion chiller
Q = U * A * dT

U = Heat Transfer Coefficient (BTU/ft^2-degF-hr) Assumed to be 70, just a solid starting point for a heat transfer coeff value
A = Area Of Chiller (ft^2) (2*pi*r*l)
dT = T2 - T1 = Temperature Difference (degF)

Assuming ice water = 32 degF and a 10 degF approach on cooling. Approach is defined as the "The difference in temperature between the discharge from a cooler (intercooler or aftercooler) and the inlet temperature of the cooling medium; usually air or water"

Q = 70 (BTU/ft^2-hr-F)* 6.54 ft^2 * (55 F - 32 F)
Q = 10,530 BTU/hr = 175 BTU/min

Next - Calculate the heat needed to be removed from a batch of wort assuming 5 gallons of wort cooling from 100 F to 65 F. Knowing Cp = 1 BTU/lb-F and 1 gallon of wort = 8.34 lb.

Q = m * Cp * dT
Q = 5 gal * (8.34 lb/gal) * 1 BTU/lb-F * (100 F - 65 F)
Q = 1460 BTU removed to get 5 gallons of wort down to 65 F

Calculate rough time to complete cooling.

Q_Removable / Q_Avaliable = 1460 BTU / 175 BTU/min
Time = 8.4 minutes

We can do the same w/ the tap water. I am going to assume that the tap water temperature is 60 F and there is no approach (because its not going to make a huge difference). So the heat that can be transfered from your wort chiller is as follows:

Q = 70 * 6.54 ft^2 * (65-60)
Q = 2290 BTU/hr
Q = 39 BTU/min

So time for heat to be removed is as follows:

Q_Removable / Q_Avaliable = 1460 BTU / 39 BTU/min
Time = 38 minutes

So the time ratio (from what I calculated which is based on rough numbers and some pretty crude thermo calcs) is as follows:

Time with Tap / Time with Ice Water = 38 minutes / 9 minutes
Ratio = 4.2 times longer to cool using your standard tap water from 100 F to 65 F

So that is my answer. Obviously the cooling is faster with ice water... and these calculations are very crude making a lot of assumptions and ignoring alot of external variables. I hope this helps! Let me know if you have any more questions.

 

[TheWall]

Also, if you want to get really geeky, you can record your time of cooling. Keep a record of how long it took you to cool your wort usings a standard flow rate. You need to make sure that the flow rate of your cooling water is the same every time you record time of cooling. You should also practice the same techniques such as stirring the wort or not stirring the wort. It would be more accurate if you just put the chiller in and let it sit until you had your wort down to your specified temperature.

Just ideas that will give you a good idea of how long it takes to cool your wort. You can do this with both ice and tap water.