(2)
Q = Ωv.
(3)
Problem I.—Given the transverse section of stream and discharge, to find the slope. From the dimensions of the section find Ω and m; from (1) find ζ, from (3) find v, and lastly from (2) find i.
Problem II.—Given the transverse section and slope, to find the discharge. Find v from (2), then Q from (3).
Problem III.—Given the discharge and slope, and either the breadth, depth, or general form of the section of the channel, to determine its remaining dimensions. This must generally be solved by approximations. A breadth or depth or both are chosen, and the discharge calculated. If this is greater than the given discharge, the dimensions are reduced and the discharge recalculated.
| Fig. 114. |
Since m lies generally between the limits m = d and m = 1⁄2d, where d is the depth of the stream, and since, moreover, the velocity varies as √ (m) so that an error in the value of m leads only to a much less error in the value of the velocity calculated from it, we may proceed thus. Assume a value for m, and calculate v from it. Let v1 be this first approximation to v. Then Q/v1 is a first approximation to Ω, say Ω1. With this value of Ω design the section of the channel; calculate a second value for m; calculate from it a second value of v, and from that a second value for Ω. Repeat the process till the successive values of m approximately coincide.
§ 113. Problem IV. Most Economical Form of Channel for given Side Slopes.—Suppose the channel is to be trapezoidal in section (fig. 114), and that the sides are to have a given slope. Let the longitudinal slope of the stream be given, and also the mean velocity. An infinite number of channels could be found satisfying the foregoing conditions. To render the problem determinate, let it be remembered that, since for a given discharge Ω∞ √χ, other things being the same, the amount of excavation will be least for that channel which has the least wetted perimeter. Let d be the depth and b the bottom width of the channel, and let the sides slope n horizontal to 1 vertical (fig. 114), then
Ω = (b + nd) d;