Introducing the channel conveyance, KCAR1/2(i.e. the discharge for
slope1), equation (8.9) becomes
dy/dxS 0 {[1 (K 0 /K)^2 ]/(1 Fr^2 )} (8.10)
(equation (8.10) implies that 0 gRSf– equation (8.1) – as well as
0 gR 0 S 0 ). Equation (8.10) can be conveniently used to analyse and
compute various surface profiles in non-uniform flow as, generally, K^2 "yN,
where the exponent N is called the hydraulic exponent. Numerical
methods may also be used to solve equation (8.9) or (8.10) for prismatic
channels, and must be used for non-prismatic channels.
For unsteady flow the treatment of continuity and Bernoulli’s equa-
tion yields (Saint Venant equation)
SfS 0 ∂y/∂x (V/g)(∂V/∂x) (1/g)(∂V/∂t). (8.11)
The first term on the right-hand side of equation (8.11) signifies uniform
flow, and the first three terms signify non-uniform flow. From continuity it
follows that a change of discharge in ∆xmust be accompanied by a change
in depth in ∆t, i.e.
∂Q/∂xB∂y/∂t 0 (8.12)
(for no lateral discharge in ∆x); thus
A∂V/∂xV∂A/∂x∂y/∂t0. (8.13)
The first term on the left-hand side of equation (8.13) represents the prism
storage and the second the wedge storage (Section 8.6).
For a rectangular channel, equation (8.13) becomes
y∂V/∂xV∂y/∂x∂y/∂t0. (8.14)
The solution of the above equations can be achieved only by numeri-
cal techniques applied, for example, to their finite difference form (Cunge,
Holly and Verwey, 1980).
In the case of rapidly varying unsteady flow, a surge is formed which
has a steep front with substantial energy dissipation (analogous to a
moving hydraulic jump). From the momentum and continuity equations
(y 1 andV 1 refer to the section ahead of the surge height ∆ymoving with
celerityc),
c V 1 {g[(A 1 ∆A)∆y/∆A(A 1 ∆A)y 1 /A 1 ]}1/2. (8.15)
For a rectangular section, equation (8.15) converts into
324 RIVER ENGINEERING
