ui = ui = 0.125 √2gH,
in which case about 1⁄64th of the energy of the fall is carried away by the water discharged.
The areas of the outlet and inlet surface of the wheel are then
2πrodo = 2πridi = Q / 0.125 √ (2gH).
If we take ro, so that the axial velocity of discharge from the central orifices of the wheel is equal to uo, we get
| ro = 0.3984 √ (Q/√H), do = ro. |
If, to obtain considerable steadying action of the centrifugal head, ri = 2ro, then di = 1⁄2do.
Speed of the Wheel.—Let Vi = 0.66 √2gH, or the speed due to half the fall nearly. Then the number of rotations of the turbine per second is
N = Vi / 2πri = 1.0579 √ (H √ H/Q);
also