What is the efficiency?
Okay, so I get that efficiency is calculated based on grain weight and original gravity. Now when a recipe gives a target gravity and you hit it, is there a quick and dirty way to calculate efficiency off of that?
For brewhouse efficiency: yes, but only if the recipe lists the efficiency it's based on. Assuming your batch size (volume in fermenter) is the same as the recipe, then you can do:
(your gravity) / (recipe gravity) * recipe efficiency = your efficiency
If the recipe says 5 gallons at OG 1.050 with 70% efficiency, and you get 5 gallons at 1.055, your efficiency is:
55/50 * 70% = 77%
You can do the same for mash efficiency, if the recipe also lists a preboil volume (and you hit it).
The efficiency, as it is commonly used, is an efficiency based on the laboratory extract of the grain. Meaning how much of the maximum extract did you get out of the grain. Most grains have a laboratory extract of 80 - 82%. This means that 80 - 82% of the grains weight can be converted and extracted by mashing.
I like to calculate the efficiency this way (note the usefulness of metric and the Plato scale ;) ):
Eff = 100% * actual extract weight / maximum extract weight
actual extract weight = Volume * SG * Plato/100%
maximum extract weight = grain weight * lab extract/100%
Example: you used 4 kg of grain to make 20l of a 12Plato wort (1.048 SG) assuming a lab extract of 80%:
actual extract weight = 20 l * 1.048 * 12 Plato / 100% = 2.52kg
max extract = 4kg * 80%/100% = 3.2kg
Efficiency = 2.52 kg / 3.2 kg *100% = 78.7%
There are ways to do this with lb and gal as well. They involve knowing the gravity potential of the grain. The latter is basically another expression of the laboratory extract.
|All times are GMT. The time now is 07:56 PM.|
Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.