What you are actually looking for here is not the amount of evaporation you will be working with it is the amount of vaporization. This is a technical but important distinction the overwhelming amount of water loss is from the boil not from evaporation. Now this can be calculated but unfortunately I won’t be able to give you a magic bullet for this there are simply to many variables such as ambient temp the thermal conductivity of your pot how windy it is that day and many more besides but this equation will get you most of the way there.
The Three variables in that equation are BTU which is the BTUs produces by your burner in an hour, EF the efficiency of transfer of those BTUs into your pot and T the time in hours of the boil.
Now how to determine your efficiency. I have made a formula that should let us do that with one empirical measurement. Now you will want to do this in conditions similar to your planed boil day or ideally you could do it during brew day and then add additional water to get you to your pre boil volume.
Where BTU is the BTU per hour output of your burner ST is the starting temp in F of your water mL is the volume of the water you are heating in milliliters and T is time in hours that it takes for your water to reach a boil.
This should be done while stirring if not constantly then often and the final time should be counted when your water just starts to boil.
As an example let us say I bring 5 gallons of water to a boil over this burner rated 210000 BTU/h.
Let us say that this water just finished mashing and is at 160 F and that it took one hour to reach a boil.
So our efficiency calculation would look like this.
(210000*1)/(((100-((160/1.8)-32))*18927*4.2)/1055)/100 = EF =.646
So we would put that EF value in the first calculation.
Now we plan on doing a 90-minute boil so we set T equal to 1.5.
So in this example we would need to add 25 gallons of liquid to the pot bringing total volume at boil time to 30 gal. Suffice it to say my example got a bit away from me.
Now all of this math assumes you are using a constant heat source in this instance we would have that burner running at full heat the entire time if you reduced the heat at all you would need to adjust the calculations accordingly by reducing the BTU value in the last equation.