I make all my beer on my electric stove. Mostly 3gal batches with the occasional 4+.
Efficiency is largely about the bound water in grain (0.125qt/lb) as a percentage of the total water added. If you want 3gal of concentrated wort from 10lb of grain, you could add 4.25gal of water in a no-sparge method. We'll assume 90% conversion (I don't typically see 100%) of a 36ppg malt mix, so you'd have 10x36x0.9= 324points of sugar in your 4.25gal of water. Since the grain holds 1.25gal, you'd get 3.0/4.25=70.6% of that sugar in your 3gal. That'd be 3gal of wort at 229pts in 3gal or 76ppg. Boil that then dilute in the fermentor to 5gal gal and you'd have a 1.046 OG beer.
Keep in mind that the apparent absorbtion (.125gal
/lb or 0.5qt/lb) also includes volume contributed by the dissolved extract. This makes the true starting volume larger, and the efficiency somewhat lower. Extract contributes 0.63 L/kg (.303 qt/lb) of volume. So in the example above the 10 lbs of grain is 36/46.214 = 77.9% extract by mass (10% less if you consider 90% conversion), *10 lbs = 7.8 lbs of extract, *.303qt/lb = 2.36 qts volume. The efficiency then becomes 3/(4.25+2.36/4)=62%, leaving you with 3 gallons of 1.067 wort or 5 gallons of 1.040 wort, per your method above.
A two runoff sparge would be somewhat more efficient, which can be important when dealing with higher gravity beers.
same as above but split the water so that there are two equal runoffs:
strike water: 11 qt
absorbtion: 10*0.5 = 5 qt
runoff 1: 11-5=6 qt
sparge water: 6 qt
runoff 2: 6 qt
e1=vout/vtotal = 6/(11+2.36) = 44.9%
v_left = 11+2.36-6 = 7.36 qt
e2=vout/vtotal = 6/(11+2.36-6+6) = 44.9%
etotal=e1+(1-e1)*e2 = 44.9%+(1-44.9%)*44.9%=69.6%.
total collected: 6 qt + 6 qt = 3 gal
gravity: .696*324/3 = 75.2 points in 3 gal, or 45 points in 5 gal.
As far as how much a difference this is vs. a full volume boil, you could get about 88% efficiency with a two sparge collecting 7 gallons preboil, which would yield 5 gallons of 1.063 wort at the end of the boil for the same 10 lbs of grain. Or to put it another way, you can get the same 1.045 wort by using 7 lbs instead of 10. Not a great example, however, since you'd probably want to back off on the efficiency for quality reasons at this point.