FLOOD09845

• County: Statewide
• Community: State of Colorado
• Basin: Statewide
• Title: Floodplain Management Workshop
• Date: 2/10/1993
• Prepared For: Greeley
• Prepared By: CWCB
• Floodplain - Doc Type: Educational/Technical/Reference Information

Substituting (2) into (I) we obtain:

Q = 1.486(1)- ) [k]~ sj
n y b+2y ,

(3)

To isolate b or y in this equation so it can be solvable quickly becomes a mathematical nightmare, The
math only gets worse if the channel is trapezoidal, circular, triangulay, or other commonly used shapes, To
help alleviate the headaches of solving equations manually every time one wishes to size a hydranlic struc-
ture, various tables, nomographs and charts have been developed for design aids, Several of these aids are
included in this handout and shonld prove useful.

1. OPEN CHANNELS

An open channel can be of nearly any configuration desired, however the most commonly used design falls
into the trapezoidal category, In this case we will dwell on Olis type of open channel.

Table 2 presents values of K which can be used to detemliue the rEquired bottom width of a trapezoidal
channel with identical bank slopes, Substituting the parameters of a trapezoidal channel into equation (I)
and breaking it down into a usable form results in equation (4);

2 I
QK,S'
=; y 0

(4)

K is a catch-all variable that encompasses the parameters of the channel. For this instance, K is identified
by the equation;

K=

[ IJ ~13
1.486 z + ~

[I ] 2'3
~+ 2V(z2 + I) ,

(5)

In equation (5), z is the channel sideslope, ely, and x h; the (kpth:width ratio, ylb,

Solving equation (4) for K:

-~
K- &'3S1n
y ,

(6)

Knowing the parameters for the right side of equation (6), you can solve for K and find the corresponding
bottom width of the channel utilizing tablll I. For example, a flow of 80 cfs in a grass lined channel with
equal 3;1 side slopes, n = 0.028, desired depth of flow y = 2 ft., and the slope S, =, 0,0042 ft/ft:

80.0.028
K = 2&'30.0042In = 5.443

Look in table I in the column headed 3 for a 3:1 side slope and locate 5.443, The number falls between
5.401 and 5,461. Moving to the left of the table we find Ole corresponding ylb values of 0.47 and 0.48,
Interpolating between these values yields a ylb value of 0.477, Now solve for b:

Pag~ 4
(,reenhonu & O'Mora, lnc.