2[pi]r_o d_o = 2[pi]r_i d_i = Q/0.125 [root](2gH).
If we take r_o, so that the axial velocity of discharge from the
central orifices of the wheel is equal to u_o, we get
r_o = 0.3984 [root](Q/[root]H),
d_o = r_o.
If, to obtain considerable steadying action of the centrifugal head,
r_i = 2r_o, then d_i = ½d_o.
_Speed of the Wheel._--Let V_i = 0.66 [root](2gH), or the speed due to
half the fall nearly. Then the number of rotations of the turbine per
second is
N = V_i/2[pi]r_i = 1.0579 [root](H[root]H/Q);
also
V_o = V_i r_o/r_i = 0.33 [root](2gH).
_Angle of Vanes with Outlet Surface._
Tan[phi] = u_o/V_o = 0.125/0.33 = .3788;
[phi] = 21º nearly.
If this value is revised for the vane thickness it will ordinarily
become about 25º.
_Velocity with which the Water enters the Wheel._--The head producing
the velocity is
H - (V_i²/2g) (1 + u_o²/V_i²) + u_i²/2g
= H {1 - .4356 (1 + 0.0358) + .0156}
= 0.5646H.