Page:Lowell Hydraulic Experiments, 4th edition.djvu/60

t is the time occupied by a particle of water in passing from the section A D to the section B C, and it will evidently be the time required for the discharge of the volume AH. We find t by the proportion

$${Q'':1::AH:t}={{AH} \over {Q''}}.$$

If the wheel was at rest, a particle of water at A would arrive at B in the time t, but the wheel is moving with the angular velocity ω, therefore the point B, in the time t, will have advanced to E, and

$${BE}={R'\omega t}={{R'\omega AH} \over {Q''}},$$

consequently, a particle of water at A, instead of being at B, at the end of the time t, will have arrived, by some path, at the point E. In this manner, by taking successive values of R′, sufficiently near to each other, the entire path of a particle of water, from its entrance into the wheel, up to the moment of its discharge, may be traced; and as, by the hypothesis, all the particles at the same distance from the axis move with the same velocity, and in the same relative direction, the path of the entire stream, from its entrance into the wheel to its discharge, will be determined.

In experiment 30, we have the total quantity discharged by the wheel equal to 138.1892 cubic feet per second; as the wheel has forty-four apertures,

$${Q''}={138.1892 \over 44}={3.14066\ \mathrm {cubic\ feet\ per\ second.} }$$

The velocity of the interior circumference of the wheel was 18.0474 feet per second, and the interior radius of the wheel being 3.375 feet, we have

$$\omega ={18.0474 \over 3.375}={5.3474\ \mathrm {feet\ per\ second,} }$$

consequently,

$${BE}={{5.3474R'AH} \over {3.14066}}={1.7026R'AH.}$$

84. The successive steps in the calculation for the entire path, are given in table III.

The arcs of circles F G, H I, etc. are drawn on a plan of the buckets, figure 2, plate VI., with the radii contained in the first column.

Column 2 contains the entire areas of these circles.