I am a little confused about this topic, perhaps somebody could help shed some light. I've been doing mini-mashes for the last year, and have finally decided to venture into all grain brewing. I have been doing some research on how to calculate mash/sparge water volumes, as well as how to account for water loss due grain absorption, evaporation, trub loss, etc. Where I'm getting confused is when people talk about loss of wort in the "dead space" of their mash tun. When I was doing mini-mashes, I would simply "dump" whatever wort was remaining in the bottom of my cooler right into my kettle. To me, that seems like a simple solution to leaving a half gallon or so of wort behind. Am I missing something here?
If your tun has a false bottom the "dead space" is the area that requires you to fill with water because it is below the grains.
Right. Whether you use a stainless braid or a domed false bottom or whatever your set up is, you will have a small amount that cannot be pushed out of the ball valve and due to the large amount of now wet grains on top, dumping it out doesn't make sense. It's easy to measure and plan accordingly before hand. All you need to do is put some water in above your false bottom and drain it. Then you measure what is left behind and add that to your calculations.
Why not simply just pick up the mashtun and pour out the remaining wort left trapped in the bottom? This is whats confusing me. I always line my cooler with a grain bag when mashing/sparging. When I'm all done, I simply lift the bag out of the cooler and actually dump whatever wort is remaining in my "dead space", leaving nothing behind. Now I'm thinking most/some people simply dump their grains directly into their mash tuns, therefore leaving wort behind as a result of not being able to separate the grains from the remaining wort? Is this a correct assumption?
Some mash in a converted keg and don't have a 15.5 gallon strainer bag on hand (me). Your procedure will work fine for your system.
