Saturday, November 3, 2012

Lagering outside?



To properly lager a beer the fermentation temperature needs to be near 50 degrees for the
first two weeks of fermentation.  Many home brewers have dedicated refrigerators, some with temperature controllers, or other more elaborate setups to accomplish this.  A swamp cooler with bottles of ice is also an alternative that some home brewers use.  As was seen in the data collected for my previous post on swamp coolers, without insulation the ice will melt in less than an hour, and the temperature will climb back up to the ambient temperature over the course of about 12 hours.  This is adequate when the goal is to ferment about five degrees or less below the ambient air temperature, but for lagering this might be a stretch.  I don't know about you, but my home isn't kept at 55 degrees in the winter.

The average temperature outside, however, could be at the lagering temperatures for several months throughout the year depending on your location.  Here in Malden, Massachusetts the average temperature is between the lagering temperatures of 48 and 58 for four months.  April and May in the spring, and October and November in the Fall.

So the question is, what will happen when the swamp cooler is brought outside?

Previously the Tou or temperature-coefficient of my system was found to be 0.136 degrees.  That number didn't mean much to me until I thought further about it.  In the equations used before Tou is the amount of energy that the system absorbs from the outside per hour.  This means that 13% of the heat from the outside system is absorbed by the fermenter, and 87% of the heat in the fermenter is contained.  Using this we can model the reaction of the swamp cooler for a non-static temperature using an iterative processes. 

In the graph at the top of this post the blue curve is the outside air temperature.  the red curve is the modeled temperature of the swamp cooler.  Most of the numbers in column F are fairly self explanatory except for perhaps the "hold%" This is the amount to temperature that the swamp cooler "holds" from the previous hour.  Each cell in the wort column is equal to 87% of the previous temperature plus 13% of the outside air temperature.

C3=F$5*C2+B3*(1-F$5)

Will it work?  I really don't know, but it looks like I'll be lagering soon!

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