_Work expended in Overcoming the Friction of the Stream Bed._--Let
A´B´, A´´B´´ be two cross sections at distances s and s + ds from
A0B0. Between these sections the velocity may be treated as uniform,
because by hypothesis the changes of velocity from section to section
are gradual. Hence, to this short length of stream the equation for
uniform motion is applicable. But in that case the work in overcoming
the friction of the stream bed between A´B´ and A´´B´´ is
GQ[theta][zeta](u²/2g)([chi]/[Omega]) ds,
where u, [chi], [Omega] are the mean velocity, wetted perimeter, and
section at A´B´. Hence the whole work lost in friction from A0B0 to
A1B1 will be
_
/ l
GQ[theta] | [zeta](u²/2g)([chi]/[Omega]) ds. (3)
_/ 0
Equating the work given in (2) and (3) to the change of kinetic energy
given in (1),
[alpha](GQ[theta]/2g)(u1² - u0²)
_
/ l
= GQz[theta] - GQ[theta] | [zeta](u²/2g)([chi]/[Omega]) ds;
_/ 0
_
/ l
.: z = [alpha](u1² - u0²)/2g + | [zeta](u²/2g)([chi]/[Omega]) ds.
_/ 0
[Illustration: FIG. 120.]
§ 116. _Fundamental Differential Equation of Steady Varied
Motion._--Suppose the equation just found to be applied to an
indefinitely short length ds of the stream, limited by the end
sections ab, a1b1, taken for simplicity normal to the stream bed (fig.
120). For that short length of stream the fall of surface level, or
difference of level of a and a1, may be written dz. Also, if we write
u for u0, and u + du for u1, the term (u0² - u1²)/2g becomes udu/g.
Hence the equation applicable to an indefinitely short length of the
stream is
dz = udu/g + ([chi]/[Omega])[zeta](u²/2g) ds. (1)
From this equation some general conclusions may be arrived at as to
the form of the longitudinal section of the stream, but, as the
investigation is somewhat complicated, it is convenient to simplify it
by restricting the conditions of the problem.
_Modification of the Formula for the Restricted Case of a Stream
flowing in a Prismatic Stream Bed of Constant Slope._--Let i be the
constant slope of the bed. Draw ad parallel to the bed, and ac
horizontal. Then dz is sensibly equal to a´c. The depths of the
stream, h and h + dh, are sensibly equal to ab and a´b´, and therefore
dh = a´d. Also cd is the fall of the bed in the distance ds, and is
equal to ids. Hence
dz = a´c = cd - a´d = i ds - dh. (2)