Today I did my 90 Shilling Ale in traditional fashion. Pulling a gallon of the first runnings and boiling them down for 45 minutes. From there I tossed in my hops and took the rest of my first runnings. I then batch sparged and collected enough wort that I ended up with a little over 5 gallons of wort. Now here is what I am wondering about. I only used 9 pounds of Marris Otter. That was my entire grain bill. Yet I ended up with an OG of 1.058. Why? I know I concentrated that wort, but I then replaced all the water I boiled out of it. How does that work?
I wonder if you got a reading where the wort wasn’t mixed well. 9 lbs ought to be closer to 1045-ish, depending on your efficiency. I don’t know when/how you take your readings. But I finally started taking a small sample at flameout when the wort is mixed thoroughly, cooling it in the freezer, and using my refractometer for it. I think it’s more consistent. I used to take a sample after chilling to pitching temps, but you get temp stratification more easily then.
I expected around 1.050
I took the reading after I got the wort chilled to 65F. I stir vigorously during cooling to get it to cool faster so it was well mixed.
I think I have this figured out due to input on another forum. I extracted 1 gallon and did a 45 minute boil on it while the rest of the wort stayed in the mash. So it had an extra 45 minutes for conversion. Then I used more water in my sparge to make up for the water I boiled off of the first gallon which would have washed more sugars out of the grain than I normally do. I am really curious about how this beer is going to taste.
yup, you got almost 90% efficiency. not out of the question. if your temps were solid you should be fine.
Yep, that explains it.