The VL-Ferm Part 1

Posted by Jon Van Lew on June 18, 2014
picture of the fermenter being tested

The problem of tiny apartment space and controlled brewing is a major problem for this poorly paid grad student in an expensive part of town. Having naturally constant temperatures for two weeks seems like it wouldn’t be a problem for LA, but I always run into gigantic weather and temperature swings whenever I throw the wort into the fermenter. My approach thus far has followed along these steps:

  1. deal with it.

Which, while beautiful in its simplicity, doesn’t always yield optimum deliciousness in my beer. Then I saw a kickstarter for the BrewJacket which is meant to specifically address problems like mine. So I totally thought I could steal it. Without having thought too hard about it before, I had thought low-power Peltier coolers wouldn’t be enough to chill 5 gallons of wort, but the BrewJacket must have been using something similar given the footprint of the design. So I thought I would compare some of the stats given in the BrewJacket kickstarter to see how much power they must be removing with their device. On their page, it said

Immersion can bring your beer down to 35º F below ambient in a matter of days and hold it there for as long as it is plugged into the wall.

For something to take that long to bring a bucket of beer down to some temperature, it can’t be pulling out too many Watts. Let’s just assume that the BrewJacket part of the Immersion chiller is providing a completely adiabatic boundary to the walls of the bucket (i.e. the only path heat can leave/enter is through the chiller). So every Joule of energy you pull from the system works to simply drop the temperature. In this system we can then simply do a little volumetric energy balance,

\[\begin{equation} \rho c_p V \cfrac{\mathrm{d}T}{\mathrm{dt}} = Q_\text{out} \end{equation}\]

Beer is predominately water, so we’ll use its material properties:

\[\begin{eqnarray} \rho = 1000 ~\text{kg/m$^3$}\\\ c_p = 4179 ~\text{J/kg-K}\\\ v = 5 ~\text{gal} * 0.00379\cfrac{\text{m$^3$}}{\text{gal}} = 0.0189 \text{m$^3$} \end{eqnarray}\]

Now let’s just say that the beer started at, maybe, 80º F and then fell down to 35º F in 3 days; so dT = 25 K and dt = $2.59\times 10^{5}$ s. This lets us find $Q_\text{out}$

\[\begin{eqnarray} Q_\text{out} & = & 1000 ~\text{kg/m$^3$} \times 4179 ~\text{J/kg-K} \times 0.0189 \text{m$^3$} \times 25~\text{K} / 2.59e5~\text{s}\\\ Q_\text{out} & = & 7 ~\text{J/s} \end{eqnarray}\]

Dude. 7 W? That’s easy! But since real fermenter walls won’t be adiabatic, I’ll throw on a huge design margin. Let’s say we need 20 W of cooling. I think even cheap-o peltier thermoelectrics can pull 20 W. Since I had precisely such cheap-o peltier, I kinda hacked together some hardware with a flat bar of stainless, an old cpu heat sink, and a peltier sandwiched in the middle. My arduino + a relay got a 20 V power supply onto the thermoelectric.

What you see in that image at the top is the first trial run of the whole assembly. I must admit it didn’t work that well at first. There was too much stainless steel rod exposed and interacting with the room’s air. Some insulation wrapped around that would force more energy interaction with the liquid it’s immersed in.

Anyway. I’ve dubbed this monstrosity the VL-Ferm


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