Hi, I'm new to homebrewing and am mostly trying to figure it out on my own. There's so much differing information online, and most of it is imperial units, which makes it harder for me to follow. I've already made two BIAB batches, an IPA and an APA, both of which turned out surprisingly well. However, I'm having trouble figuring out how to make calculations about how much water to use at each stage, so some advice would be great! I have a 14 L fermentor and a 16 L kettle (a regular pot with no dead volume). Using a bag to mash 3.5 kg grains, I want to end up with 14 L in my fermentor. In my first two batches I made the mistake of covering the kettle during boiling, and my more experienced friend could taste a little DMO, though not loads, so next time I will leave uncovered during boiling and cooling. I don't yet know the evaporation rate from the kettle if leaving uncovered. Because the kettle is only slightly bigger than the fermentor, and because the grains absorb some water, I would like to know how much water to use at each stage. I was thinking that one way to solve this would be to batch sparge the mashed grains. My brewhouse efficiency last time was around 58%, so sparging might also help improve that slightly. So would something like the following work? • Prepare 13 L strike water at 68°C • Mash 3.5 kg grains in 13 L water at 66°C • Batch sparge grains in 4 L water at 76°C • Boil the two batches together • If the cooled volume is under 14 L and the OG is high enough to do so, top up with water to 14 L I'm largely just guessing here, because I'm having trouble figuring out the calculations because the information online varies so much, so your help would really be appreciated!