Page:Lowell Hydraulic Experiments, 4th edition.djvu/71

 96. The trace adopted by the author, for the corresponding guides, is as follows.

The number $$n$$ having been determined, divide the circle, in which the extremities of the guides are found, into $$n$$ equal parts, $$vw$$, $$wx$$, etc.

Put $$\omega'$$ for the width between two adjoining guides,

and $$t'$$ for the thickness of the metal forming the guides.

We have by rule 4, $\omega' = {d \over n}$.

With $$w$$ as a centre, and the radius $$\omega' + t'$$, draw the arc $$yz$$; and with $$x$$ as a centre, and the radius $$2(\omega' + t')$$, draw the arc $$a'b'$$. Through $$v$$ draw the portion of a circle $$vc'$$, touching the arcs $$yz$$ and $$a'b'$$ ; this will be the curve for the essential part of the guide. The remainder of the guide, $$c'd'$$, should be drawn tangent to the curve $$c'v$$, a convenient radius is one that would cause the curve $$c'd'$$, if continued, to pass through the centre $$O$$. This part of the guide might be dispensed with, except that it affords great support to the part $$c'v$$, and thus permits the use of much thinner iron than would be necessary, if the guide terminated at $$c'$$, or near it.

97. Collecting together the foregoing formulas for proportioning turbines, which, it is understood, are to be limited to falls not exceeding forty feet, and to diameters not less than two feet; we have

for the horse-power, $$P = 0.0425 D^2 h \sqrt{h} ;$$ for the diameter, $$D = 4.85 \sqrt{{P}\over{h\sqrt{h}}};$$ for the quantity of water discharged per second, $$Q = 0.5 D^2 \sqrt{h};$$ for the velocity of the interior circumference of the wheel, when the fall is not very variable, $$v = 0.56 \sqrt{2gh},$$ or, for the height of the orifices of discharge, $$H = 0.10 D;$$