Page:EB1911 - Volume 14.djvu/86

Rh ''Proportions of Channels of Maximum Discharge for given Area and Side Slopes. Depth of channel = d; Hydraulic mean depth = d''; Area of section =.

Half the top width is the length of each side slope. The wetted perimeter is the sum of the top and bottom widths.

§ 114. Form of Cross Section of Channel in which the Mean Velocity is Constant with Varying Discharge.—In designing waste channels from canals, and in some other cases, it is desirable that the mean velocity should be restricted within narrow limits with very different volumes of discharge. In channels of trapezoidal form the velocity increases and diminishes with the discharge. Hence when the discharge is large there is danger of erosion, and when it is small of silting or obstruction by weeds. A theoretical form of section for which the mean velocity would be constant can be found, and, although this is not very suitable for practical purposes, it can be more or less approximated to in actual channels.

Let fig. 117 represent the cross section of the channel. From the symmetry of the section, only half the channel need be considered. Let obac be any section suitable for the minimum flow, and let it be required to find the curve beg for the upper part of the channel so that the mean velocity shall be constant. Take o as origin of coordinates, and let de, fg be two levels of the water above ob.

Let ob = b/2; de = y, fg = y + dy, od = x, of = x + dx; eg = ds.

The condition to be satisfied is that

v = c √ (mi)

should be constant, whether the water-level is at ob, de, or fg. Consequently

m = constant = k

for all three sections, and can be found from the section obac. Hence also

$Increment of section⁄Increment of perimeter$ = $ydx⁄ds$ = k.

y2dx2 = k2ds2 = k2(dx2+dy2) and dx = kdy / √ (y2 − k2).

Integrating,

x = k logundefined {y + √ (y2 − k2)} + constant;

and, since y = b/2 when x = 0,

x = k logundefined [{y + √ (y2 − k2)} / { b + √ ( b2 − k2) }].

Assuming values for y, the values of x can be found and the curve drawn.

The figure has been drawn for a channel the minimum section of which is a half hexagon of 4 ft. depth. Hence k = 2; b = 9.2; the rapid flattening of the side slopes is remarkable.

§ 115. In every stream the discharge of which is constant, or may be regarded as constant for the time considered, the velocity at different places depends on the slope of the bed. Except at certain exceptional points the velocity will be greater as the slope of the bed is greater, and, as the velocity and cross section of the stream vary inversely, the section of the stream will be least where the velocity and slope are greatest. If in a stream of tolerably uniform slope an obstruction such as a weir is built, that will cause an alteration of flow similar to that of an alteration of the slope of the bed for a greater or less distance above the weir, and the originally uniform cross section of the stream will become a varied one. In such cases it is often of much practical importance to determine the longitudinal section of the stream.

The cases now considered will be those in which the changes of velocity and cross section are gradual and not abrupt, and in which the only internal work which needs to be taken into account is that due to the friction of the stream bed, as in cases of uniform motion. Further, the motion will be supposed to be steady, the mean velocity at each given cross section remaining constant, though it varies from section to section along the course of the stream.

Let fig. 118 represent a longitudinal section of the stream, A0A1 being the water surface, B0B1 the stream bed. Let A0B0, A1B1 be cross sections normal to the direction of flow. Suppose the mass of water A0B0A1B1 comes in a short time to C0D0C1D1, and let the work done on the mass be equated to its change of kinetic energy during that period. Let l be the length A0A1 of the portion of the stream considered, and z the fall, of surface level in that distance. Let Q be the discharge of the stream per second.

Change of Kinetic Energy.—At the end of the time there are as many particles possessing the same velocities in the space C0D0A1B1 as at the beginning. The change of kinetic energy is therefore the difference of the kinetic energies of A0B0C0D0 and A1B1C1D1.

Let fig. 119 represent the cross section A0B0, and let be a small element of its area at a point where the velocity is v. Let 0 be the whole area of the cross section and u0 the mean velocity for the whole cross section. From the definition of mean velocity we have

u0 = v / 0.

Let v = u0 + w, where w is the difference between the velocity at the small element and the mean velocity. For the whole cross section, w = 0.

The mass of fluid passing through the element of section, in seconds, is (G/g) v, and its kinetic energy is (G/2g) v3. For the whole section, the kinetic energy of the mass A0B0C0D0 passing in seconds is

(G / 2g) v3 = (G/2g) (u03 + 3u02w + 3u02 + w3),

= (G / 2g) {u03 + w2 (3u0 + w)}.

The factor 3u0 + w is equal to 2u0 + v, a quantity necessarily positive. Consequently v3 &gt; 0u03, and consequently the kinetic energy of A0B0C0D0 is greater than

(G / 2g) 0u03 or (G) / 2g) Qu02,

which would be its value if all the particles passing the section had the same velocity u0. Let the kinetic energy be taken at

(G / 2g) 0u03 = (G / 2g) Qu02,

where is a corrective factor, the value of which was estimated by J. B. C. J. Bélanger at 1.1. Its precise value is not of great importance.

In a similar way we should obtain for the kinetic energy of A1B1C1D1 the expression

(G / 2g) 1u13 = (G / 2g) Qu12,

where 1, u1 are the section and mean velocity at A1B1, and where a may be taken to have the same value as before without any important error.

Hence the change of kinetic energy in the whole mass A0B0A1B1 in seconds is

Motive Work of the Weight and Pressures.—Consider a small filament a0a1 which comes in seconds to c0c1. The work done by gravity during that movement is the same as if the portion a0c0 were carried to a1c1. Let dQ be the volume of a0c0 or a1c1, and y0, y1 the depths of a0, a1 from the surface of the stream. Then the volume