I am trying to figure out if we can get by with a small chiller (e.g. 5 ton or less) and a big tank of cold water. Our water rates are quite high and we live in a village where everyone is very concerned about water consumption, so using a one pass system where we dump hot water down the drain won't really fly. Can someone out there check my numbers, since I'm really not sure if any of my assumptions make sense?
For mash cooling:
Assume we need to cool 300 gallons of mash (assume same thermal properties and density as water) from 150F down to 90F or so for pitching. This should use about 150,120 Btu (300 gal * 8.34 lb/gal*60F). If we use cold tap water (at 50F) and let it heat up on average to about (100F), we would need an empty tank big enough to hold 360 gallons of hot water, which we could use for mashing, cleaning, etc. Alternatively, we could use a small chiller to cool the water down overnight and start in the morning with 360 gallons of water at 50F.
For condensing our stripping still:
Assume we are producing low wine with 60% alcohol, starting with 300 gallon of wash at 10%. This should yield a maximum of 50 gallons at 60%ABV (20 gallons of water and 30 gallons of alcohol). The energy to condense is 8100 Btu per gallon for water and about 2400 Btu per gallon of ethanol. Therefore, condensing the vapour should take about 234,000 Btu. We'll need a little extra cooling to bring the temperature of the condensate down from about 180 to 70F, this should take about 45,870 Btu (50 gallons * 8.34 lb/gal * 110F). Our total cooling requirement is, therefore about 280,000 Btu. This would require about 668 gallons of water (assuming it starts around 50 and we let it warm up to 100F).
If we cool the 668 gallons of 100 F water overnight (e.g. 12 hours) back down to 50F, we should need about 23,213 Btu/h or roughly 2 tons.
Obviously, we would need multiple tanks and/or do some careful scheduling of our mashing and stripping. Does any of this make sense?
Thanks in advance for your feedback!