timmystank
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How do you tell what your efficiency is when doing stovetop PM?
How to tell efficiency?
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COLObrewer
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Too much math, use beersmith or other brewing software.
Take the weight of any LME, and multiply by 36
Add to that (the weight of any DME multiplied by 45)
Add to that (the weight of any cane sugar used multiplied by 46)
Add to that (the weight of any corn sugar used multiplied by 42)
This gives you the gravity units contributed by the extracts and sugars.
Now divide this number by the volume of the beer in gallons, which gives you the gravity units per gallon from extracts and sugars.
Take the OG of your beer, subtract 1, multiply the difference by 1000, and subtract from this the calculated gravity units per gallon.
This gives you the gravity units per gallon contributed by the mash. Call it "x"
Now look up the extract potential in ppppg (points per pound per gallon) for each of the grains you used in the mash. If this is expressed as 1.0xx, subtract the 1, then multiply by 1000. (i.e. 1.036 = 36) If the ppppg is expressed as an integer (whole number) usually between 32 - 38, then use the number as is.
For each grain, multiply the pounds of grain by the ppppg, and sum the results, then divide the total by the volume of the beer in gallons. This gives you the theoretical extract per gallon contributed by the mash assuming 100% efficiency. Call it "y"
Divide "x" by "y", and then multiply by 100. This will be the brewhouse efficiency of your mash as a percentage.
But it's much easier to buy Promash or other brewing software.
-a.
Add to that (the weight of any DME multiplied by 45)
Add to that (the weight of any cane sugar used multiplied by 46)
Add to that (the weight of any corn sugar used multiplied by 42)
This gives you the gravity units contributed by the extracts and sugars.
Now divide this number by the volume of the beer in gallons, which gives you the gravity units per gallon from extracts and sugars.
Take the OG of your beer, subtract 1, multiply the difference by 1000, and subtract from this the calculated gravity units per gallon.
This gives you the gravity units per gallon contributed by the mash. Call it "x"
Now look up the extract potential in ppppg (points per pound per gallon) for each of the grains you used in the mash. If this is expressed as 1.0xx, subtract the 1, then multiply by 1000. (i.e. 1.036 = 36) If the ppppg is expressed as an integer (whole number) usually between 32 - 38, then use the number as is.
For each grain, multiply the pounds of grain by the ppppg, and sum the results, then divide the total by the volume of the beer in gallons. This gives you the theoretical extract per gallon contributed by the mash assuming 100% efficiency. Call it "y"
Divide "x" by "y", and then multiply by 100. This will be the brewhouse efficiency of your mash as a percentage.
But it's much easier to buy Promash or other brewing software.
-a.
COLObrewer
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Told you
jkarp
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user 30639
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My head just exploded trying to figure that out :cross:Take the weight of any LME, and multiply by 36
Add to that (the weight of any DME multiplied by 45)
Add to that (the weight of any cane sugar used multiplied by 46)
Add to that (the weight of any corn sugar used multiplied by 42)
This gives you the gravity units contributed by the extracts and sugars.
Now divide this number by the volume of the beer in gallons, which gives you the gravity units per gallon from extracts and sugars.
Take the OG of your beer, subtract 1, multiply the difference by 1000, and subtract from this the calculated gravity units per gallon.
This gives you the gravity units per gallon contributed by the mash. Call it "x"
Now look up the extract potential in ppppg (points per pound per gallon) for each of the grains you used in the mash. If this is expressed as 1.0xx, subtract the 1, then multiply by 1000. (i.e. 1.036 = 36) If the ppppg is expressed as an integer (whole number) usually between 32 - 38, then use the number as is.
For each grain, multiply the pounds of grain by the ppppg, and sum the results, then divide the total by the volume of the beer in gallons. This gives you the theoretical extract per gallon contributed by the mash assuming 100% efficiency. Call it "y"
Divide "x" by "y", and then multiply by 100. This will be the brewhouse efficiency of your mash as a percentage.
But it's much easier to buy Promash or other brewing software.
-a.
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