Design of Open Channels and
Culverts
CE453 Lecture 26
Ref: Chapter 17 of your text and HYDRAULIC DESIGN OF
HIGHWAY CULVERTS, Hydraulic Design Series Number 5, Federal
Highway Administration, Publication No. FHWA-NHI-01-020,
September 2001; available at
http://www.cflhd.gov/design/hyd/hds5_03r.pdf, accessed March 18,
2006
Design of
Open Channels
Longitudinal Slopes
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Gradient
longitudinal
direction of
highway to
facilitate
movement of
water along
roadway
Drains
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Along ROW
Collect surface water
A Typical intercepting
drain placed in the
impervious zone
http://www.big-o.com/constr/hel-cor.htm
Drainage Channels (Ditches)
Design
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Adequate capacity
Minimize hazard to traffic
Hydraulic efficiency
Ease of maintenance
Desirable design (for safety):
flat slopes, broad bottom, and
liberal rounding
Ditch Shape
Source: Fabriform1.com
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Trapezoidal – generally preferred
considering hydraulics,
maintenance, and safety
V-shaped – less desirable from
safety point of view and
maintenance
Terms
Steady Flow: rate of discharge does not
vary with time (Manning’s applies)
Uniform: channel properties are constant
along length of channel
Slope
Roughness
Cross-section
Water surface is parallel to slope of
channel
Non-uniform: properties vary
Terms
Unsteady flow: rate of discharge varies
with time
Critical depth
a hydraulic control in design
depth of water where flow changes from
tranquil to rapid/shooting
Critical velocity: velocity corresponding to
critical depth
Critical slope: slope corresponding to
critical depth
Flow Velocity
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Depends on lining type
Should be high enough to prevent deposit
of transported material (sedimentation)
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For most linings, problem if S < 1%
Should be low enough to prevent erosion
(scour)
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For most types of linings, problem if S > 5%
Use spillway or chute if Δelev is large
Rip Rap for drainage over high slope
Riprap (TN Design Manual)
Side Ditch/Open Channel Design-Basics
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Find expected Q at point of interest
(see previous lecture)
Select a cross section for the slope,
and any erosion control needed
Manning’s formula used for design
Assume steady flow in a uniform
channel
Manning’s Formula
V = R2/3*S1/2
n
(metric)
V = 1.486 R2/3*S1/2
n
where:
V = mean velocity (m/sec or ft/sec)
R = hydraulic radius (m, ft) = area of the cross section
of flow (m2, ft2) divided by wetted perimeter (m,f)
S = slope of channel
n = Manning’s roughness coefficient
Side Ditch/Open Channel
Design-Basics
Q = VA
Q = discharge (ft3/sec, m3/sec)
A = area of flow cross section (ft2,
m2)
FHWA has developed charts
to solve Manning’s equation
for different cross sections
Open Channel Example
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Runoff = 340 ft3/sec (Q)
Slope = 1%
Manning’s # = 0.015
Determine necessary cross-section to
handle estimated runoff
Use rectangular channel 6-feet wide
Open Channel Example
Q = 1.486 R2/3*S1/2
n
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Hydraulic radius, R = a/P
a = area, P = wetted perimeter
P
Open Channel Example
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Flow depth = d
Area = 6 feet x d
Wetted perimeter = 6 + 2d
Flow depth (d)
6 feet
Example (continued)
Q = 1.486 a R2/3*S1/2
n
340 ft3/sec = 1.486 (6d) (6d)2/3 (0.01)1/2
(6 + 2d)
0.015
d  4 feet
Channel area needs to be at least 4’ x 6’
Example (continued)
Find flow velocities.
V = 1.486 R2/3*S1/2
n
with R = a/P =
6 ft x 4 ft = 1.714
2(4ft) + 6ft
so, V = 1.486(1.714)2/3 (0.01)1/2 = 14.2 ft/sec
0.015
If you already know Q, simpler just to do
V=Q/A = 340/24 = 14.2)
Example (continued)
Find critical velocities.
From chart along critical curve, vc  13 ft/sec
Critical slope = 0.007
Find critical depth: yc = (q2/g)1/3
g = 32.2 ft/sec2
q = flow per foot of width
= 340 ft3/sec /6 feet = 56.67ft2/sec
yc = (56.672/32.2)1/3 = 4.64 feet > depth of 4’
Check lining for max depth of flow …
Rounded
A cut slope with ditch
A fill slope
Inlet or drain marker
Ditch treatment near a bridge
US 30 – should pier be protected?
A fill slope
Hidden Drain
Where’s the water going to end up?
Median drain
Design of Culverts
Source: Michigan Design Manual
Culvert Design - Basics
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Top of culvert not used as pavement
surface (unlike bridge), usually less than
20 foot span
> 20 feet use a bridge
Three locations
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Bottom of Depression (no watercourse)
Natural stream intersection with roadway
(Majority)
Locations where side ditch surface drainage
must cross roadway
Hydrologic and
Economic
Considerations
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Alignment and grade
of culvert (wrt
roadway) are
important
Similar to open
channel
Design flow rate
based on storm with
acceptable return
period (frequency)
Culvert Design Steps
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Obtain site data and roadway cross
section at culvert crossing location
(with approximation of stream
elevation) – best is natural stream
location and slope (may be expensive
though)
Establish inlet/outlet elevations,
length, and slope of culvert
Sometimes … you want a dam … why?
Culvert Design Steps
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Determine allowable headwater depth
(and probable tailwater depth) during
design flood – control on design size –
f(topography and nearby land use)
Select type and size of culvert
Examine need for energy dissipaters
Headwater Depth
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Constriction due to culvert creates
increase in depth of water just upstream
Allowable/desirable level of headwater
upstream usually controls culvert size and
inlet geometry
Allowable headwater depth depends on
topography and land use in immediate
vicinity
Inlet control
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Flow is controlled by
headwater depth and inlet
geometry
Usually occurs when slope
of culvert is steep and
outlet is not submerged
Supercritical, high v, low
d
Most typical
Following methods ignore
velocity head
Example:
Design ElevHW = 230.5 (max)
Stream bed at inlet = 224.0
Drop = 6.5’
Peak Flow = 250cfs
5x5 box
HW/D = 1.41
HW = 1.41x5 = 7.1’
Need 7.1’, have 6.5’
Drop box 0.6’ below stream
@223.4’ - OK
Outlet
control
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When flow is governed by
combination of headwater
depth, entrance geometry,
tailwater elevation, and slope,
roughness, and length of
culvert
Subcritical flow
Frequently occur on flat slopes
Concept is to find the required
HW depth to sustain Q flow
Tail water depth often not
known (need a model), so may
not be able to estimate for
outlet control conditions
Example:
Design ElevHW = 230.5
Flow = 250cfs
5x5 box (D=5)
Stream at invert = 224
200’ culvert
Outlet invert = 224-0.02x200
= 220.0’ (note: = 223.4.017x200)
Given tail water depth = 6.5’
Check critical depth, dc = 4.3’
from fig. 17.23
Depth to hydraulic grade line =
(dc+D)/2 = 4.7 < 6.5, use 6.5’
Example (cont.)
Design ElevHW = 230.5
Flow = 250cfs
5x5 box
Outlet invert = 220.0’
Depth to hydraulic grade
line = 6.5’
Head drop = 3.3’ (from
chart)
220.0+6.5+3.3 =
229.8’<230.5 OK