Let

H = the available fall after deducting loss of head in pipes and

channels from the gross fall;

Q = the supply of water in cubic feet per second; and

[eta] = the efficiency of the turbine.

The work done per second is [eta]GQH, and the horse-power of the

turbine is h.p. = [eta]GQH/550. If [eta] is taken at 0.75, an

allowance will be made for the frictional losses in the turbine, the

leakage and the friction of the turbine shaft. Then h.p. = 0.085QH.

The velocity of flow through the turbine (uncorrected for the space

occupied by the vanes and guide-blades) may be taken

u_i = u_i = 0.125 [root](2gH),

in which case about (1/64)th of the energy of the fall is carried away

by the water discharged.

Encyclopaedia Britannica, 11th Edition, "Husband" to "Hydrolysis" Volume 14, Slice 1 · 第 186 / 207 题