I’ve been reading up on mash efficiency lately and I just can’t seem to find a good way to predict mash efficiency when you are getting started… trial and error is what I’ve found to be the way to go…
Why care about efficiency?…
Well… if you’ve got a few batches under your belt, efficiency doesn’t matter much because you know your numbers and can plan your brews around it…
… But, someone who is just getting started will try to brew a recipe designed with 80% efficiency and being new may get 60% efficiency and next thing you know their beer is nothing like it’s supposed to taste…
Why?
Well efficiency is a measure of how much sugar you extract from your grains… so if you have ten pounds of 2-row malt you should expect to get a 1.057 OG beer at 80% efficiency, but if your efficiency drops to 60%, then your beer will now be a 1.043 OG beer… (assuming a five gallon batch of course)
That’s a smaller beer…. there is less sugar, and you can’t use the same amount of hops and expect the same sweet to bitter balance… your ABV % changes and the beer is just not the same…
If you know your efficiency is 60%, then you can simply use 13.2 pounds of 2-row instead of 10 pounds and you will get a 1.057 OG beer…
But my question remains… how do you help a new brewer predict efficiency when getting started?
There are too many variables… mash thickness, equipment used, mash temperature, sparge method used, etc.
… and these are not even predictable…
Typically you can expect higher efficiency when you fly sparge (continuous sparge) versus batch sparge or no sparge… but I’ve seen brewers get like 70 or even 80% efficiency with no sparge and some get 60% with fly sparge!!!
That’s like opposite… WTF?… I mean Oh Crap!
Anyways… here’s how to calculate your efficiency and what to do if you screw it up the first time… (I know I did)…
Take the example above… 2-row malt has a potential extract of 36 points (1.036) per pound per gallon. So if you have 10 pounds in 5 gallons then you get 72 points (72 = 36 * 10 / 5)…
That means that when you mash and collect your wort your OG would be 1.072 if you were to be 100% efficient… but as is the case with most brewers 80% efficiency is like the upper end and 60% is the lower end (specially if you, like me, drink beer while brewing… and it’s before noon… j/k)…
80% efficiency means you get 80% of the 72 points per pound per gallon which is 57.6 or 1.057 OG… 60% of 72 is 43 or 1.043 OG…
So if you are trying to brew a 1.057 OG beer, but you only have a 1.043 OG wort, then you are going to have to boil off water during the boil and brew a smaller batch…
In this scenario, you have 43 points of sugar in 5 gallons or 215 points total (215 = 43 x 5)… If you want a 1.057 beer then you need to divide 215 by 57 and that’s the size of your batch which is 3.77 gallons (215/57 = 3.77)…
… but you have to boil off the water, not scoop it out or siphon wort because you’ll be removing sugar as well…
If the opposite happens… meaning you get higher efficiency and therefore a higher OG than expected, then you do the same math as above and you’ll find that you will need to add more water to dilute your wort so your OG drops to whatever your goal is. This means you will end up fermenting a bigger batch… although if you have a 5 gallon carboy and can’t ferment more than 5 gallons, then you still dilute and just siphon off only 5 gallons into your fermentor and whatever is left pour it into a growler or jar to ferment or experiment with…
So, there you have it… efficiency learned by trial and error…
All right enough learning,
Time to go Brew Beer And Drink It,
Jorge
