The potential energy of 13.5 km^3 of water that is converted to kinetic energy applied at the generator input is:

U = m*g*h

g = 9.8 m/sec^2               acceleration of gravity

h = 690 m                          head (average elevation of lake – 20 m head loss with turbines at 8 m elevation and 12 m head loss due to penstock)

m = 1.35 * 10^13 kg         mass of water in 13.5 km^3 of water 

U = 9.15 * 10^16 kg*m^2/sec^2

1 kg*m^2/sec^2  = 2.778*10^-7 kwh

U = 2.5*10^10 kwh

Dividing by 8,760 hours in a year we get a continuous power at the generator input on a year-round basis:

Power at generators input = 2.5* 10^10 kwh /8760 hrs. = 2.85 GW

Pelton generator efficiency = .92

Generators electrical output = 2.6 GW