How to Bag It

In the last post I introduced the concept of Brew in a Bag, a method of creating your wort that does not require a lauter tun, but simply uses a mesh bag to mash directly in the brew pot.  (Mesh mash?)   I am going to review how I plan for the brewing of a BIAB batch.

When formulating for a brew day, I always start at the end: how much beer do I want and what gravity do I want it to be?  Batch sizes are usually constrained by the size of the boil kettle.  I have a 10 gallon and a 5 gallon boil kettle.  I am writing this post around the smaller kettle simply because I use it more often, especially for brewing in a bag. My kettle actually has about a 20 liter capacity, which is actually slightly larger than 5 gallons.  Now I cannot expect to boil 20 liters of wort (and keep it in the pot).  The maximum is about 19 liters, and that requires extreme patience during the first quarter of the boil process.  I can expect to boil off about 3.5 liters during a one hour boil.  And then as the wort cools there is another 4% loss in volume.  Add this all together and it means that the largest post boil volume I can expect is about 14.9 liters.  Because I don’t like to fret over the boil pot, let’s just make the final volume 14 liters (about 3.7 gallons).

The next question is gravity.  Let’s start with something simple: 12 °P, 1.048 gravity.  A simple beer of moderate alcoholic strength.  The first calculation to be made is the amount of extract that is required for 14 liters of 12 °P wort:

Extract = (12/100) * (14 * 1.048) = 1.76 kg

The 14*1.048 factor is the mass of the 14 liters of wort, whereas the 12/100 factor is the percent of that mass that is extract.

Next let us make some assumptions about the grain.  In earlier posts we discuss the expected yield of grain, and for ease here we will use a FGDB yield of 80% for this illustration.  Further I will assume that our mash conversion efficiency at 90%, meaning that at most we can expect 72% (90% of  80%) of the grain mass will convert into extract.

So now it is a simple matter to calculate that we need about 2.44 kg of grain (1.76/0.72) in order to get the necessary extract.  Except that is not enough grain.  Why?  Think about what happens in the mash-boil kettle.  If  we add 16.2 liters of water to mash our 2.44 kg of grain, several things will occur.  First we will have about 18 liters in the boil kettle, which is a good working volume. Second we will have a water to grain ratio of about 6.7 liters per kg of grain, which is somewhere north of 3 quarts per pound.  Don’t worry – it works. What will be the gravity of the liquid in the pot?  Well if we have indeed managed to extract 1.76 kg, then it is:

°P = 100*1.76/(1.76+16.2) =  9.8

The problem is that we are now going to pull the bag of grain out of the pot.  Based on my experience, the grain can be expected to absorb around 0.8 kg of water per kg of grain.  Because we know the gravity of the wort in the pot and the amount of water (16.2 kg less 0.8*1.76, which is 14.8 kg of water) we can calculate how much extract is in the pot:

Extract = Mass Water*ºP/(100-ºP) = 14.8*9.8/(100-9.8) = 1.61 kg

What I have found effective is to use an iterative process.  Since our target extract is 1.76, use a new starting grain weight of (1.76/1.61)*2.44 or 2.67 kg of grain. Repeating this process three or four times (Excel helps a lot here) one can get to the required grain weight (2.84 kg, for those keeping score at home).  This represents a brewhouse yield of around 77%, which is not too shabby for such a simple process.

In the next post I will elaborate a bit more on the practical details of this brewing example.

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