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?
Glen Alden
Ormond Beach, Florida

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.