Doug is partially right. I never even considered the weight of the soluble sugars that are removed from the mash tun.
But I'm not so sure of Doug's calculation that 60% of the weight is removed, leaving only 40%.
So here are some calculations with the following assumptions: 10 lbs of grain,a 5 gallon batch, 6.5 gallons of pre- boil 1.040 wort.
6.5 gallons of water x 8.34 lbs/gallon= 54.21 lbs
54.21 lbs x 1.040=56.38 or a difference of 2.17 lbs.
So using the above assumptions, if the spent grain was dried out, it would weigh 7.83 lbs? (I'm including the question mark, because I've had another 16 hr day with 5 hrs of sleep, so I figured I messed this up somewhere)
So the spent grain would be 78% of its original weight? (using the above brewhouse assumptions) I'm throwing this out as a question, not making a boilerplate assertion.
This analysis makes the common error of assuming 6.5 gal of wort contains 6.5 gal of water, which it does not. °Plato is defined as the weight % of extract (mostly sugar) in the wort, thus the weight % of water in the wort is 100° - °Plato. So, to work the example in the quoted post:
6.5 gal of wort @ 1.040 SG @ 68°F/20°C = 6.5 gal * 8.33 gal/lb * 1.040 = 56.311 lb
1.040 SG = 9.994 °P, so
Extract Wt in BK = 56.311 lb * 0.09994 = 5.628 lb, and
Water Wt in BK = 56.311 lb * (1 - 0.09995) = 50.683 lb, so
Water Vol in BK = 50.683 / 8.33 = 6.0844 gal, and
Extract Vol in BK = 6.5 gal - 6.084 = 0.4156 gal
Calculation of the dried spent grain weight is complicated by the fact that about 4% of the original grain weight is from absorbed water, and the residual dried material will contain unconverted grain material and dried extract (from the absorbed wort.) Plus, some of the dried extract weight is due to water that chemically combined with the starch during saccharification (i.e. didn't come from the original dried grain weight.) I have a spreadsheet that takes all of these factors into account (you can find a link to it
here.) The calculated dried spent grain material weight is 2.817 lb, and the dried retained extract weight is 1.417 lb, so the total dried spent grain mass would weigh 4.234 lb. This is 42.3% of the original grain weight. In this specific example, the conversion efficiency was 91.6%. If instead we had 98% efficiency, the dried spent grain would weigh 2.34 lb + 1.54 lb = 3.88 lb, or 38.8% of the original grain weight. So, my previous ~40% of original grain weight estimate was right on.
I do all my calculations with volumes corrected to 68°F/20°C since that is the temp that was used to measure the °Plato to SG data. Water at 68°F/20°C has a density of 8.33 lb/gal. According to the Bettin & Spieweck polynomial, water has a density of 8.34 lb/gal at 55.7°F/13.2°C.
But, as noted by Augie, the dried spent grain weight directly relevant to the OP's question, as OP is interested in the wet spent grain weight.
Brew on