FLOOD09052

• County: Routt
• County: Moffat
• Basin: Yampa/White
• Title: Dam Failure Floodplain Information for the Yampa River Downstream of Stagecoach Reservoir
• Date: 12/1/1989
• Prepared For: Moffat County / Routt County
• Prepared By: CWCB
• Floodplain - Doc Type: Flood Documentation Report

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7.0 Colorado DWR Model

The Colorado Division of Water Resources (DWR), also
known as the Office of the State Engineer, has
developed a procedure to prepare dam failure flood
inundation maps below each of the 238 high hazard dams
in the state. This procedure is based on an empirical
dam breach algorithm, a recognized flood routing
technique, and an approximate mapping method.

,

The DWR procedure for dam-failure flood mapping
consists of the following steps:

1. Using an empirical formula derived from
regression analysis of actual dam failures considering .
physical dimensions of the dam and the reservoir as
,well ,as the relative erosiveness of the embankment
soils, compute the peak discharge and base time of a
triangular shaped hydrograph at the dam site due to
failure of the dam.

2. Using the method of successive averages,
route the dam failure hydrograph downstream to typical
trapezoidal shaped valley cross sections by calculating
the depth of flow, average velocity, and travel time.

3. Draw the flood inundation zone on U.S.G.S.
quadrangle maps using flood depths calculated at the
typical sections.

A computer program, called HAZARD2.BAS, has been
created to make the computations in steps 1 and 2 for
dams with with earth embankments. A simplified method
for calculating the peak flow for concrete and masonry
dams was developed by applying the weir equation in the
standard form:

Q = CLh3/.
where C = discharge coefficient
L = length of the weir
h = depth of flow over the weir

Assuming the dam fails abruptly forming a
rectangular breach which has a width of one half the
length of the dam crest (LC) and a depth which is the
full heigth of the dam (H) less the freeboard (FB), and
a discharge coefficient of 3.0, the above equation
becomes:

Q. = {3.0} {(LC)/2} {(1.00)(H-FB)}3/'
= 1.50 (LC) (1I-FB)3/.

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