heating strike water with immersion chiller?

killian · 03-15-2010, 05:42 AM

So when I was brewing my last beer I started the water for my immersion chiller (I ran it a little to fast) to cool my wort and when I did the water came out so hot and fast the water sprayed all over and burned my arm.

This dumb*ss mistake got me thinking...

I have seen threads about people using copper coils and propane burners to heat strike water, Why not use the same idea.

I'm thinking that I could put my immersion chiller in a boil pot and bring 2 or 3 gallons of water to a boil then run my strike water through the chiller. Would that bring 10 gallons of water to 170?

I should probably look for some high temp tubing. Would there be any issues trying it with rv hose?

shortyjacobs · 03-15-2010, 03:25 PM

3 gallons of water at boiling will raise 10 gallons of water from 68F to 101F. Why not just heat up the 10 gallons directly??

killian · 03-15-2010, 04:27 PM

whats the equation?

just trying to mash indoors using an electric stove.

shortyjacobs · 03-15-2010, 05:54 PM

Quote:

Originally Posted by killian

whats the equation?

just trying to mash indoors using an electric stove.

Since it's all water, heat capacities are the same. Although you have separate systems, (pot of boiling water and IC inside of it), they can be viewed as a single system, since all temperatures will go towards equilibrium. This means the boiling water will cool as it heats up the room temperature water.

The basic formula for Energy going into thermal change is E=mC(deltaT), where E is energy, m is mass, C is heat capacity, and deltaT is change in temperature.

Put simply, Energy of boiling water + Energy of mash water = Energy of system. Or m1C1(deltaT1)+ m2C2(deltaT2) = m3C3(deltaT3), where 1 is your boiling water, 2 is your mash water, and 3 is the whole kit-n-kaboodle put together. The C1, C2, and C3 terms cancel out because they are the same. Through some funky math, it turns out that you can instead write m1T1+m2T2 = m3T3...math whizes will now recognize that we are just taking a weighted average of the components in the system. Substitute Volume for Mass because all densities are (practically) equal, and you get v1T1+v2T2 = v3T3. So, in your case, you want to know T3, the equilibrium temperature.

(3 gallons * 212°F + 10 gallons * 68°F)/13 gallons (the total volume of the system), (3*212+10*68)/13= 101°F.

Basically, it's no different than dumping 3 gallons of boiling water into 10 gallons of 68 degree water. Beersmith agrees with me when I plug it in this way.

Now what this doesn't account for is you adding heat to the boiling water with your stove. In this case, it's possible to heat all 10 gallons up to whatever temp you want, as long as you are patient enough to wait for it to recirculate long enough to get that extra energy out of your stovetop. But again, I ask, if this is the case, why not just heat your strike water directly, instead of using the IC?

maskednegator · 03-15-2010, 08:18 PM

The real benefit to doing this would be to use your cooling water as strike water for the next batch. That's how the big breweries do it, and it saves a tremendous amount of money on their power bills.

**Bobby_M** · 03-15-2010, 08:19 PM

It would actually be much faster to use a second or even third small pot on the adjacent burners to heat up faster since you're getting more BTU into the total volume.

**springer** · 03-15-2010, 08:32 PM

Quote:

Originally Posted by Bobby_M

It would actually be much faster to use a second or even third small pot on the adjacent burners to heat up faster since you're getting more BTU into the total volume.

This is what I did on my last brew. Needed 8 gallons of sparge water I just used two pots heated up fairly fast .. I only used the stove because it was freezing out and SWMBO wasn't home .... No beer or maple syrup related on the kitchen stove. You think a wort boil over is bad try maple syrup