calculations help

[BetterSense]

I thought I understood this by now. Tell me where I am going wrong.

I want 19l of 12% beer. Suppose I get 70% efficiency. Then I need 19*0.12/0.70=3.25kg of grain.

So that's what I bought, and I am mashing now, but every online calculator says that I am only going to get 1.037 rather than 1.048, and that I should have bought like 25% more grain. That's not a small difference. Clearly I am having a GCF (Gross Conceptual Failure). What is it?

 

[LLBeanJ]

12%(?) aside, let's just work on gravity points. I work in imperial units, so I'll leave it to you to do the conversion to SI.

1.048 post-boil would be 48 gravity points x 5 gallons = 240 points needed.
Most grain is going to yield about 36 points/gallon @ 100% efficiency.
240 / 36 = 6.67 lbs / .70 = 9.52 lbs of grain needed at 70% efficiency.

Edit: S/B 1.048 post-boil, not pre-boil. Fixed.

 

[EJay]

I thought I understood this by now. Tell me where I am going wrong.

I want 19l of 12% beer. Suppose I get 70% efficiency. Then I need 19*0.12/0.70=3.25kg of grain.

So that's what I bought, and I am mashing now, but every online calculator says that I am only going to get 1.037 rather than 1.048, and that I should have bought like 25% more grain. That's not a small difference. Clearly I am having a GCF (Gross Conceptual Failure). What is it?

You calculation 19*.012 basically equals liters (2.28) of 100% ethanol in the 19L. so far so good.
Your dividing by efficiency is basically converting it to liters of 70% alcohol. (3.25 liters). This is NOT kg of grain.

You'll need to go a bit further to figure out points per kg of grain and how much alcohol that produces. Efficiency is not the weight of alcohol divided by weight of grain. Its the weight of extracted sugar vs. the theoretical maximum available in the grain.

I usually run about 3.25 kg of grain in my 2.5g batches of 6%

 

[BetterSense]

Alcohol is totally irrelevant at this stage. I am only trying to hit original gravity of 12 degrees which is roughly 12% sugar ~1.048.

 

[ajdelange]

You want 19L of 12° wort. The specific gravity of 12 ° wort is 1.048 so 19L of it are going to weigh 1.048*0.998203*19 = 19.8762 kg. Twelve percent of this is 1.048*0.998203*19*.12 = 2.38515 kg of extract required. If you achieve 70% efficiency in the mash then you will need 1.048*0.998203*19*.12/0.7 = 3.40735 kg grain - pretty close to what you got with the difference being that you didn't account for the increased density of the wort ( 1.048*0.998203 = 1.04612 gram/cc) before applying the 12% - an error of 4.6%. But in essence you are spot on.

Now the calculators used by home brewers don't do things in the common sense way you approached the problem. The first thing they do is assume that specific gravity and extract content are linear and use 'points' to calculate extract amount. The errors that this introduces are small and in terms of the accuracy to which we measure grain mass and water not, therefore, significant. The other thing they do is consider the efficiency as a loss relative to what a Congress mash for the grain produces. Thus if the Congress mash for a particular grain indicates 70% efficiency (ratio of extract to weight of grain) and you get 65% efficiency they call this 100*65/70 = 92.8571% efficient, not 65% efficient. I expect that your discrepancy may depend on that.

 

[BetterSense]

You want 19L of 12° wort. The specific gravity of 12 ° wort is 1.048 so 19L of it are going to weigh 1.048*0.998203*19 = 19.8762 kg. Twelve percent of this is 1.048*0.998203*19*.12 = 2.38515 kg of extract required. If you achieve 70% efficiency in the mash then you will need 1.048*0.998203*19*.12/0.7 = 3.40735 kg grain - pretty close to what you got with the difference being that you didn't account for the increased density of the wort ( 1.048*0.998203 = 1.04612 gram/cc) before applying the 12% - an error of 4.6%. But in essence you are spot on.

Now the calculators used by home brewers don't do things in the common sense way you approached the problem. The first thing they do is assume that specific gravity and extract content are linear and use 'points' to calculate extract amount. The errors that this introduces are small and in terms of the accuracy to which we measure grain mass and water not, therefore, significant. The other thing they do is consider the efficiency as a loss relative to what a Congress mash for the grain produces. Thus if the Congress mash for a particular grain indicates 70% efficiency (ratio of extract to weight of grain) and you get 65% efficiency they call this 100*65/70 = 92.8571% efficient, not 65% efficient. I expect that your discrepancy may depend on that.

This is brilliant. Thank you for explaining everything. I was taught by a German and usually follow recipes, but this time I was winging it. My main problem is that I looked up some stuff that the 70% was a typical number for efficiency, but that was deviant American "efficiency". I think I need to be using a number closer to 50% when calculating it the proper way. And start with the proper wort density.


I swear there is a secret society out there responsible for seeding these absurd roundabout ways of reckoning. "Points per pound per gallon"? Dear lord my head already hurts.

 

[EJay]

Alcohol is totally irrelevant at this stage. I am only trying to hit original gravity of 12 degrees which is roughly 12% sugar ~1.048.

sorry, when you said 12% I thought you meant 12% ABV.