Scaling up recipes
TroubleShooting
Glen Alden • Ormond Beach, Florida asks,
I started with 5-gallon (19-L) kits, but after about 30 batches I decided to double my capacity and brew all-grain 10-gallon (38-L) batches. My basic need is to convert a 5-gallon (19-L) recipe to 10-gallons (38 L). In the past I simply doubled the amount of each ingredient. Is this correct?
In general terms, I agree that when recipes are scaled up in volume
that the ingredients are simply not proportioned to the change in
volume. The reason that this method normally does not work is that
there are usually differences between the efficiencies of small and
large brewhouses.
This scaling is often done when a commercial brewer is taking a batch
of brew from their pilot brewery to a commercial size. Even in a small
brewery this may mean scaling a 5-gallon (19-L) batch to a 15-barrel
batch, or a 100-fold increase in volume. This jump in volume is
significant because small differences in material efficiency become
very apparent during this sort of scale change.
It is easy to demonstrate with standard brewhouse calculations that
differences in brewhouse efficiency do affect wort original gravity.
Also, the shape of your brew kettle and the type and size of heater
used for the kettle can affect wort color, hop utilization and
evaporative rate. These factors all are important if you want to make
the same batch of beer twice on different scales.
All recipes published in BYO are based on assumed raw material yields.
Since the actual efficiency of a particular brewhouse setup is fairly
consistent and easy to calculate, those who brew frequently typically
know their brewhouse efficiency and can modify recipes to account for
the difference between an assumed efficiency and actual efficiency. Not
all brewers calculate efficiency the same way, but the numbers are
handy no matter the method. I prefer comparing how much extract I
produce during wort production to the weight of malt used.
For example, let’s assume that I produce 20 liters of 12 °Plato wort
and used 3.4 kg of malt in the process. My extract yield is equal to
(liters wort)*(decimal equivalent of °Plato)*(equivalent specific
gravity to Plato). You can convert Plato to specific gravity using the
following formula:
Specific gravity = {Plato/(258.6-([Plato/258.2]*227.1)}+1
Once you determine that 12 °Plato is equivalent to 1.048, the rest
of the calculation is easy. I have included units below to show how the
units cancel, resulting in kg of extract:
(20 liter)*(0.12 kg extract/kg wort)*(1.048 kg wort/liter) = 2.52 kg extract
The 2.52 kg of extract represents what was extracted from the 3.4
kg of malt during wort production. When 2.52 kg is compared to 3.4 kg
the result shows that 74% of the malt added to the mash ended up as
extract in the wort.
Most homebrew set ups do not have a yield this high and a more typical
number is 68%. So if I gave you one of my recipes and told you that my
brewhouse yield was 74% you could easily adjust the malt bill
accordingly by simply multiplying my malt bill by 1.09 (74 ÷ 68),
assuming your brewhouse has a 68% efficiency.
Scaling up hops is much more difficult as there is no easy method
homebrewers can use to determine how much iso-alpha-acids end up in
their finished beers. When scaling recipes up and down without actually
knowing hop utilization, brewers rely on rules of thumb, tables
containing scaling factors and good old sensory evaluation.
However, if I were scaling a 5-gallon (19-L) batch up to a
10-gallon (38-L) batch I would be very tempted to just double the whole
bloody recipe and forget the math. After all, we’re not blending
something hazardous here, we’re brewing beer. If the result is off,
then you can always tweak it next time.