OPEN-CHANNEL HYDRAULICS: LECTURE 054 - COMPUTATION OF UNIFORM FLOW

1. CONVEYANCE
1.01
The depth of uniform flow is the normal depth.
1.02
Thus, uniform or equilibrium flow is also referred to as the normal flow.
1.03
The discharge under normal flow is:

            1
Q = v A =
A R^2/3 S^1/2
            n

Eq. 1

1.04
For some applications, it is useful to define the conveyance:

      1
K =
A R^2/3
      n

Eq. 2

1.05
Thus, in terms of conveyance, the discharge is:

Q = K S^1/2

Eq. 3

1.06
In terms of discharge, the conveyance is:

       Q
K =

      S^1/2

Eq. 4

1.07
The conveyance has information on friction and cross-sectional shape, and is independent of the energy slope.

2. COMPUTATION
2.01
Under uniform or normal flow, the discharge is:

      1
Q =
A R^2/3 S^1/2
      n

Eq. 5

2.02
Since the hydraulic radius is:

      A
R =

      P

Eq. 6

2.03
Thus, the normal flow equation can be expressed as:

  Q n

P^2/3 - A^5/3 = 0
  S^1/2

Eq. 7

2.04
In a typical prismatic channel, the input hydraulic variables are:
1. Discharge Q,
2. Bottom width b,
3. Side slope z: z horizontal to 1 vertical
4. Bottom slope S, and
5. Manning's n.

Fig. 01

Definition sketch for normal flow computation.

2.05
For b different than zero, and z different than zero, the channel is trapezoidal.
2.06
For z = 0, the channel is rectangular.
2.07
For b = 0, the channel is triangular.
2.08
The wetted perimeter is:

P = b + 2y (1 + z²)^1/2

Eq. 8

2.09
where y is the flow depth.
2.10
The flow area is:

A = (b + zy)y

Eq. 9

2.11
Substituting the wetted perimeter and flow area into the normal flow equation:

Q n

[b + 2y_n(1 + z²)^1/2]^2/3 - [(b + zy_n)y_n]^5/3 = 0
S^1/2
Eq. 10

2.12
where y_n is the normal depth.

2.13
Given the input variables Q, n, S, b, and z, this equation is solved for the normal depth by iteration.

Fig. 01

Definition sketch for normal flow computation.

Narrator: Victor M. Ponce
Music: Fernando Oñate
Editor: Flor Pérez

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