Turbogator
Member
I recently brewed a batch of American amber ale and I went through the procedure to calculate the mash efficiency. But the number I came up with seems crazy high and I was hoping someone could check me and comment on the reliability of my calculation.
This is the resource I used to determine the calculation procedure. https://blog.eckraus.com/calculating-improving-mash-efficiency
The grain bill was 10 lbs. pale at 1.036 potential, 2.5 lbs. Vienna malt at 1.034 , 1.75 Crystal L60 at 1.036 and 1 lbs. of toasted malt at 1.033. I calculated the total points to be 541. These potential numbers all came out of beer smith.
I mashed it at 151 degrees for 60 minutes and sparged in two batches at 165 degrees. After the mash and sparge I collected 8.5 gallons of wort in my kettle
I calculated the total potential of the wort as ((541/8.5)/1000)+1 = 1.0636.
I measured the gravity of the wort I collected with both a hygrometer and a refractometer. The gravity was 1.050. Now 1.050 / 1.0636 =.9872 or 98.72 % is what the calculation calls for.
This seems crazy high. Am I doing this calculation wrong? Am I missing something?
This is the resource I used to determine the calculation procedure. https://blog.eckraus.com/calculating-improving-mash-efficiency
The grain bill was 10 lbs. pale at 1.036 potential, 2.5 lbs. Vienna malt at 1.034 , 1.75 Crystal L60 at 1.036 and 1 lbs. of toasted malt at 1.033. I calculated the total points to be 541. These potential numbers all came out of beer smith.
I mashed it at 151 degrees for 60 minutes and sparged in two batches at 165 degrees. After the mash and sparge I collected 8.5 gallons of wort in my kettle
I calculated the total potential of the wort as ((541/8.5)/1000)+1 = 1.0636.
I measured the gravity of the wort I collected with both a hygrometer and a refractometer. The gravity was 1.050. Now 1.050 / 1.0636 =.9872 or 98.72 % is what the calculation calls for.
This seems crazy high. Am I doing this calculation wrong? Am I missing something?