Page:Encyclopædia Britannica, Ninth Edition, v. 12.djvu/511

495 HYDRAULTCS.] HYDROMECHANICS 495 former being the section of a Navigation Canal and the latter the section of an Irrigation Canal. 93. Channels of Circular Section. The following short table facilitates calculations of the discharge with different depths of water in the channel. Let r be the radius of the channel section ; then for a depth of water = KT, the hydraulic mean radius is fj.r and the area of section of the waterway vr -, where K, /x, and v have the following values : Depth of water K _ terms of radius ) 01 05 10 15 20 25 30 35 40 45 50 55 60 65 70 75 8 85 9 95 1-0 Hydraulic mean depth &amp;gt;^ in terms of radius ...1 OOGG8_ 0321 0023 0963 1278 1574 1852 2142 242 269 293 320 343 365 387 408 429 449 466 484 500 Waterway in terms ofl_ v _ square of radius f 00189 0211 0598 10C7 1051 228 294 370 450 532 614 709 795 885 979 1-070 1-17.5 1-276 1-371 1-470 1-571 94. Egg-SJuiped Channels or Sewers. hi sewers for discharging storm water and house drainage the volume of flow is extremely variable ; and there is a great liability for deposits to be left when the flow is small, which are not removed during the short periods when the flow is large. The sewer in consequence becomes choked. To FIG. 107. Scale 80 feet=l inch. obtain uniform scouring action, the velocity of flow should be con stant or nearly so ; a complete uniformity of velocity cannot be obtained with any form of section suitable for sewers, but an approxi mation to uniform velocity is obtained by making the sewers of oval section. Various forms of oval have been suggested, the simplest being one in which the radius of the crown is double the radius of the invert, and the greatest width is two -thirds the height. The section, of such a sewer is shown in fig. 108, the numbers marked on the figure being pro portional numbers. 95. Problems on Channels in which the Flow is Steady and at Uniform Velocity. The general equations given in 88, 90 are 108. = /! + JL 771 mi (2); &quot;U Q = (3). Problem I. Given the transverse section of stream and discharge, to find the slope. From the dimensions of the section find fl and m ; from (1) find &amp;lt;T, 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 thebreadth, depth, or general form of the section of the channel, to determine its remaining dimensions. This must generally be solved by approxi mations. A breadth or depth or both are chosen, and the discharge calculated. If this is greater than the given discharge, the dimen sions arc reduced and the discharge recalculated. Since m lies generally between the limits m = d and m= d, where d is the deptli^of the stream, and since, moreover, the velocity varies as VTO so that an error in the E / value of m leads ; only to a much less error in the value ( of the velocity cal culated from it, we may proceed thus. Assume a vnliifi fnr &amp;gt; A . &quot; / ... ; /D / H in, and calculate v J} G C from it. Let t-j be -., .- this first approxi mation to v. Then -2L is a first approximation to fl, say flj. &quot;V With j this value of fl design the section of the channel ; calculate a second value for m ; calculate from it a second value of r, and from that a second value for fl. Repeat the process till the successive values of m approximately coincide. 96. Problem IV. Most Economical Form of Channel for given Side Slopes. Suppose the channel is to be trapezoidal in section (fig. 109), 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 condi tions. To render the pro blem determinate, let it be remembered that, since for a given discharge floe %/%, other things being the same, the amount of ex cavation 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 horizon tal to 1 vertical (fig. 110), then n=(b + nd}d Both fi and x are to be minima. Differentiating, and equating to zero, (lb_ dd ad eliminating, ad But Inserting the value of b, + I)d + b + nd=0 , X 4d/n* + l-2nd 2 That is, with given slide slopes, the section is least for a given discharge when the hydraulic mean depth is half the actual depth. A simple construction gives the form of the channel which fulfils this condition, for it can be shown that when in = -=- the sides of the channel are tangential to a semi-circle drawn on the water line. Since = |. f therefore Let ABCD be the channel (fig. 109) ; from E the centre of AD drop perpendiculars EF, EG, EH on the sides. Let AB=CD=; BC = &; EF-EH-c; and EG-d. Q = area AEB + BEC + CED , = ac + ^ bd. Putting these values in (1), ac + ^bd=(a + b)d ; and hence c**=d. That is EF EG, EH are all equal, hence a semicircle struck from E with radius equal to the depth of the stream will pass through F and H and be tangential to the sides of the channel. To draw the chan nel, describe a semi circle on a horizontal line with radius = depth T ,. of channel. The bot tom will be a horizontal tangent of that semicircle, and the sices tangents drawn at the required side slopes. The above result may be obtained thus (fig. Ill) :