Pelton wheel's theoretical speed for max efficiency

[Michael V]

Homework Statement

A Pelton wheel is driven by five identical jets. The runner has a diameter of 3,8 m, fixed in a horizontal position. The head from reservoir level to nozzles is 350 m and the efficiency of power transmission through the pipeline and nozzles is 90%. The relative velocity decreases by 8% as the water traverses the bucket surfaces which deflect the jet at an angle of 162°. The coefficient of velocity for the jets is 0,97 (Cv) and the discharge is 25920 m[itex]^{3}[/itex]/hour.
Calculate the following:
a) The velocity of water entering the buckets
b) The theoretical wheel speed in r/min for maximum efficiency (u = 0,5 V)

Homework Equations

For a: [itex]V = Cv\sqrt{2g×h}[/itex] where h = Height×90%

For b: [itex]U = \frac{\pi×D×N}{60}[/itex]

The Attempt at a Solution

a) [itex]V = Cv\sqrt{2g×h} = 0,97×\sqrt{2(9,81)×315} = 76,256 m/s[/itex]

b) [itex]U = 0.5V = \frac{\pi×D×N}{60}[/itex]
I'm confused as to whether I must use V from (a) or if I must calc a new V where h = H for max efficiency?

 

[LightHero]

If the former, U = 0.5×76,256 = 38,128 m/sN = \frac{U×60}{\pi×D} = \frac{38,128×60}{3,14×3,8} = 645,2 r/min