[RX-Heven]

Injector setup is 2x720cc primaries and 2x1600cc secondaries.

To run 15% water to fuel at 70% IDC, I calculate this way:

(IDC) x (Total fuel cc's) x (Target % water) =
(.70) x (1440 +3200) x (0.15) = 448 cc's of water

Then, also assuming the water nozzles operate at 70% DC and one 0.9 mm WI nozzle flows 315cc (max), I get:

(15% water to fuel) / [(WI flow rate) x (IDC)] =
(448) / [(315) x (.70)] = 2

So I would need two 0.9 mm nozzles to meet my 15%.

Alternatively,
The water (mixture)-to-fuel percentage is roughly 10% with one .8mm nozzle and one .5 mm nozzle in my setup.
Here is how to find that:
To run 10% water to fuel at 70% IDC (assuming), calculate this way:

(IDC) x (Total fuel injector cc's) x (Target % water) =
(.70) x (1440 +3200) x (0.10) = 325cc's of water

Then, knowing the water nozzles operate at 70% DC (from our assumption of IDC) and one 0.8 mm WI nozzle flows 280cc (max) and one 0.5 mm WI nozzleflows 180cc (max), I get:

(required cc's of water) / [(WI flow rate x IDC)] = 10%
(325) / [(280 x .70) + (180 x .70) ] = 10%