Page:Lowell Hydraulic Experiments, 4th edition.djvu/65

Let DN = the number of buckets.

Let Dn = the number of guides.

Let DP = the horse-power of the turbine; a horse-power being 550 pounds avoir, raised one foot per second.

Let Dh = the fall acting upon the wheel.

Let DQ = the quantity of water expended by the turbine, in cubic feet per second.

Let DV = the velocity due the fall acting upon the wheel.

Let DV′ = the velocity of the water passing the narrowest sections of the wheel.

Let Dv = the velocity of the interior circumference of the wheel: all the velocities being in feet per second.

Let DC = the coefficient of V, or the ratio of the real velocity of the water passing the narrowest sections of the wheel, to the theoretical velocity due the fall acting upon the wheel.

The unit of length is the English foot.

It is assumed that the useful effect is seventy-five per cent, of the total power of the water expended.

According to rule 1, we have the sum of the widths of the orifices of discharge, equal to D. Then the sum of the areas of all the orifices of discharge, is equal to DH.

By the fundamental law of hydraulics we have

$$V={\sqrt {2gh}}.$$

therefore

$$V'=C{\sqrt {2gh}}.$$

We can find the value of C in the last equation by experiment 30, on the Tremont Turbine. In that wheel we have for the sum of the widths of the orifices of discharge, ${\textstyle 44\times 0.18757=8.25308}$ feet, and the height of the orifices of discharge = 0.9314 feet. Then we have, for the sum of the areas of all the ori- fices of discharge,

$$HD=8.25308\times 0.9314=7.68692\ \mathrm {square\ feet.}$$

By experiment 30, we have

00000Q = 138.1892 cubic feet per second,

00000h = 12.903 feet,

${\textstyle {\sqrt {2g}}}$ = 8.0202 feet,