v = c [root](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 y dx
---------------------- = ---- = k
Increment of perimeter ds
y²dx² = k²ds² = k²(dx² + dy²) and dx = k dy/[root](y² - k²).
Integrating,
x = k log_[epsilon] {y + [root](y² - k²)} + constant;
and, since y = b/2 when x = 0,
x = k log_[epsilon] [{y + [root](y² - k²)}/{½b + [root](¼b² - 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.
[Illustration: 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 [theta] 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.
[Illustration: FIG. 119.]
_Change of Kinetic Energy._--At the end of the time [theta] 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 [omega] be a
small element of its area at a point where the velocity is v. Let
[Omega]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 = [Sigma][omega]v/[Omega]0.
Let v = u0 + w, where w is the difference between the velocity at the
small element [omega] and the mean velocity. For the whole cross
section, [Sigma][omega]w = 0.
The mass of fluid passing through the element of section [omega], in
[theta] seconds, is (G/g)[omega]v[theta], and its kinetic energy is
(G/2g)[omega]v³[theta]. For the whole section, the kinetic energy of
the mass A0B0C0D0 passing in [theta] seconds is
(G[theta]/2g)[Sigma][omega]v³
= (G[theta]/2g)[Sigma][omega](u0³ + 3u0²w + 3u0² + w³),
= (G[theta]/2g){u0³[Omega] + [Sigma][omega]w²(3u0 + w)}.
The factor 3u0 + w is equal to 2u0 + v, a quantity necessarily
positive. Consequently [Sigma][omega]v³ > [Omega]0u0³, and
consequently the kinetic energy of A0B0C0D0 is greater than
(G[theta]/2g)[Omega]0u0³ or (G[theta])/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
[alpha](G[theta]/2g)[Omega]0u0³ = [alpha](G[theta]/2g)Qu0²,
where [alpha] is a corrective factor, the value of which was estimated
by J. B. C. J. Bélanger at 1.1.[6] Its precise value is not of great
importance.
In a similar way we should obtain for the kinetic energy of A1B1C1D1
the expression
[alpha](G[theta]/2g)[Omega]1u1³ = [alpha](G[theta]/2g)Qu1²,
where [Omega]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
[theta] seconds is
[alpha](G[theta]/2g) Q (u1² - u0²). (1)
_Motive Work of the Weight and Pressures._--Consider a small filament
a0a1 which comes in [theta] 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[theta] be the volume of a0c0 or a1c1, and y0, y1 the
depths of a0, a1 from the surface of the stream. Then the volume
dQ[theta] or GdQ[theta] pounds falls through a vertical height z + y1
- y0, and the work done by gravity is
G dQ[theta](z + y1 - y0).
Putting p_a for atmospheric pressure, the whole pressure per unit of
area at a0 is Gy0 + p_a, and that at a1 is - (Gy1 + p_a). The work of
these pressures is
G(y0 + p_a/G - y1 - p_a/G) dQ[theta] = G(y0 - y1) dQ[theta].
Adding this to the work of gravity, the whole work is GzdQ[theta]; or,
for the whole cross section,