Page:EB1911 - Volume 14.djvu/85

Fig. 111.—Scale 20 ft. = 1 in.

Fig. 112.—Scale 80 ft. = 1 in.

Fig. 113.

§ 111. Egg-Shaped Channels or Sewers.—In 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 obtain uniform scouring action, the velocity of flow should be constant or nearly so; a complete uniformity of velocity cannot be obtained with any form of section suitable for sewers, but an approximation 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. 113, the numbers marked on the figure being proportional numbers.

§ 112. Problems on Channels in which the Flow is Steady and at Uniform Velocity.—The general equations given in §§ 96, 98 are

     ζ = α(1 + β/m);

(1)

ζv²/2g = mi;   

(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;

⁠χ = b + 2d √ (n² + 1).

Both Ω and χ are to be minima. Differentiating, and equating to zero.

(db/dd + n) d + b + nd = 0,
db/dd + 2 √ (n² + 1) = 0;

eliminating db/dd,

{n − 2√ (n² + 1)} d + b + nd = 0;
b = 2 {√ (n² + 1) − n} d.

But

Ω / χ = (b + nd) d / {b + 2d √ (n² + 1)}.

Inserting the value of b,

m = Ω/χ = {2d √ (n² + 1) − nd} / {4d √ (n² + 1) − 2nd} = 1/2 d.

That is, with given side 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 m = 1/2d the sides of the channel are tangential to a semicircle drawn on the water line.

Since

Ω / χ = 1/2 d,

therefore

Ω = 1/2 χd.

(1)

Let ABCD be the channel (fig. 115); from E the centre of AD drop perpendiculars EF, EG, EH on the sides.

Let

AB = CD = a; BC = b; EF = EH = c; and EG = d.

Ω = area AEB + BEC + CED,
 = ac + 1/2 bd.
χ  = 2a + b.

Putting these values in (1),

ac + 1/2 bd = (a + 1/2 b) d; and hence c = d.

Fig. 115.

Fig. 116.

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 channel, describe a semicircle on a horizontal line with radius = depth of channel. The bottom will be a horizontal tangent of that semicircle, and the sides tangents drawn at the required side slopes.

The above result may be obtained thus (fig. 116):—

χ = b + 2d / sin β.

(1)

Ω = d (b + d cot β);

Ω/d = b + d cot β;

(2)

Ω/d ² = b/d + cot β.

(3)

From (1) and (2),

χ = Ω / d − d cot β + 2d / sin β.

This will be a minimum for

dχ / dd = Ω / d ² + cot β − 2 / sin β = 0,

or

Ω/d ² = 2 cosec. β − cot β.

(4)

or

d = √ {Ω sin β / (2 − cos β)}.

From (3) and (4),

b/d = 2 (1 − cos β) / sin β = 2 tan 1/2 β.