Bought a new keggle, and wondering if there is anyway of calculating the amount that will boil off during a boil.

Figure just fill it with 10 gallons of water and boil for an hour, and take the resultant amount and calculate percentage from there, but not sure if the percentage would be exponential or not in relation to the amount of water in the kettle in the first place.

Here's the loooooooooooooooong answer:

In reality, no matter what container you boil in, the percentage boiloff rate is a farce. What is relatively constant is the boiloff rate in terms of gallons per hour. For most homebrewers on "normal" homebrew systems, this rate is anywhere from 1 to 2 gallons per hour. The variables are surface area of the boiling wort, and boil vigor, i.e., is it just a simmer, or is the wort practically leaping out of the kettle, or somewhere in between. Personally I shoot for leaping. And, the wider the kettle, the faster the boiloff rate. Then in order to get your gallons per hour rate to work in your brewing software, usually you will need to convert that to a percentage, even though the percentage is not actually constant if you boil shorter or longer than an hour, or especially if you change your batch size. For instance, if you normally make 10 gallon batches and lose 10% per hour, you will lose somewhere closer to 20% per hour if you switch to a 5-gallon batch size on the same equipment. Fairly straight-forward mathematics. But if you always boil the same sized batch for the same amount of time using the same boil vigor, then this all becomes moot as you should have a constant percentage boiloff rate.