Evaporation rate is a function of at least two variables: 1) rate of added heat, e.g., how big is your flame or heat source, and 2) surface area of the boiling wort, i.e., directly proportional to the square of the radius of the kettle. The surface area being heated also plays a role, e.g., if you were able to throw a heating coil into the beer itself, that beer is going to cook pretty dang quickly due to all that extra heated surface area.

The typical homebrewer uses a pretty standard burner and a kettle that is maybe 16 to 20 inches in diameter, to make roughly 5 or 6 gallons, yes? And this typical homebrewer will lose roughly 0.8 to 1.0 gallons per hour with this typical setup, so they'll start with 6 or 7 gallons. With a double batch, they'd probably need two kettles, which would require 12 to 14 gallons. But not 16! But if the kettle size is increased from 16 inches to, say, 24 inches, then the boiloff rate with the same rate of added heat will increase by a ratio of 12 squared over 8 squared = 144/64 = 2.25 times the typical guy. So maybe the bigger kettle will lose closer to 2.25 gallons per hour instead of just 1 gallon, due to the larger surface area of the boiling wort alone.

My guess is that your setup is even more robust than this, in one or more ways. To be able to boil 16+ gallons at one time, and still boil off 5 gallons in one hour, you've got to have a hotter heat source, and a larger diameter kettle, yes? Hence your ability to have a really high boiloff rate. If there's any way you can reduce the size of your kettle diameter, or turn down the heat, then you won't need to sparge so much, and your problems may be reduced.

With respect to sparging... yes, you want the volumes to be equal. First runnings = second runnings, and if you do two sparges, then divide in 1/3 so that the first two runnings = third runnings. Personally I only ever do a double sparge when I'm making a beer with an OG more than 1.100.