Page:EB1911 - Volume 14.djvu/86

Proportions of Channels of Maximum Discharge for given Area and Side Slopes.
Depth of channel = d; Hydraulic mean depth = 1/2d;

Area of section = Ω.

	Inclination  of Sides to  Horizon.	Ratio of  Side Slopes.	Area of  Section Ω.	Bottom  Width.	Top width =  twice length of each Side Slope.
Semicircle	..	..	1.571d 2	0	2d
Semi-hexagon	60°   0′	3   : 5	1.732d 2	1.155d	2.310d
Semi-square	90°   0′	0   : 1	2d 2	2d	2d
	75°  58′	1   : 4	1.812d 2	1.562d	2.062d
	63°  26′	1   : 2	1.736d 2	1.236d	2.236d
	53°   8′	3   : 4	1.750d 2	d	2.500d
	45°   0′	1   : 1	1.828d 2	0.828d	2.828d
	38°  40′	11/4 : 1	1.952d 2	0.702d	3.202d
	33°  42′	11/2 : 1	2.106d 2	0.606d	3.606d
	29°  44′	13/4 : 1	2.282d 2	0.532d	4.032d
	26°  34′	2  : 1	2.472d 2	0.472d	4.472d
	23°  58′	21/4 : 1	2.674d 2	0.424d	4.924d
	21°  48′	21/2 : 1	2.885d 2	0.385d	5.385d
	19°  58′	23/4 : 1	3.104d 2	0.354d	5.854d
	18°  26′	3   : 1	3.325d 2	0.325d	6.325d

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.

Fig. 117.

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 log_ε {y + √ (y² − k²)} + constant;

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

x = k log_ε [{y + √ (y² − k²)} / {1/2 b + √ (1/4 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.

Steady Motion of Water in Open Channels of Varying Cross Section and Slope

§ 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.

Fig. 118.

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 θ 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.

Fig. 119.

Change of Kinetic Energy.—At the end of the time θ 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 ω be a small element of its area at a point where the velocity is v. Let Ω₀ 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 = Σ ωv / Ω₀.

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

The mass of fluid passing through the element of section ω, in θ seconds, is (G/g) ωvθ, and its kinetic energy is (G/2g) ωv³θ. For the whole section, the kinetic energy of the mass A₀B₀C₀D₀ passing in θ seconds is

(Gθ / 2g) Σωv³ = (Gθ/2g) Σω (u₀³ + 3u₀²w + 3u₀² + w³),
= (Gθ / 2g) {u₀³Ω + Σωw² (3u₀ + w)}.

The factor 3u₀ + w is equal to 2u₀ + v, a quantity necessarily positive. Consequently Σωv³ > Ω₀u₀³, and consequently the kinetic energy of A₀B₀C₀D₀ is greater than

(Gθ / 2g) Ω₀u₀³ or (Gθ) / 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

α (Gθ / 2g) Ω₀u₀³ = α (Gθ / 2g) Qu₀²,

where α is a corrective factor, the value of which was estimated by J. B. C. J. Bélanger at 1.1.^[1] Its precise value is not of great importance.

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

α (Gθ / 2g) Ω₁u₁³ = α (Gθ / 2g) Qu₁²,

where Ω₁, 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 θ seconds is

α (Gθ / 2g)Q(u12−u02).

(1)

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

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1. Boussinesq has shown that this mode of determining the corrective factor α is not satisfactory.