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Design of Open Channels and Culverts

CE453 Lecture 25

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

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Design of Open Channels

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Longitudinal Slopes Gradient

longitudinal direction of highway to facilitate movement of water along roadway

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Drains

Along ROW Collect surface water

A typical intercepting drain placed in the impervious zone

http://www.big-o.com/constr/hel-cor.htm

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Drainage Channels (Ditches)

Design Adequate capacity Minimize hazard to traffic Hydraulic efficiency Ease of maintenance

Desirable design (for safety): flat slopes, broad bottom, and liberal rounding

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Ditch Shape

Trapezoidal – generally preferred considering hydraulics, maintenance, and safety

Source: Fabriform1.com

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Ditch Shape

V-shaped – less desirable from safety point of view and maintenance

Source: Fabriform1.com

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Terms

Steady Flow: rate of discharge does not vary with time (Manning’s applies)

Uniform: channel properties are constant along length of channel

SlopeRoughnessCross-section

Water surface is parallel to slope of channel

Non-uniform: properties vary

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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 depthCritical slope: slope corresponding to critical depth

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Flow Velocity

• Depends on lining type • Should be high enough to prevent deposit

of transported material (sedimentation)• For most linings, problem if S < 1% (generally

velocity should be > 2 fps when full)• Should be low enough to prevent erosion

(scour)• For most types of linings, problem if S > 5%

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Use spillway or chute if Δelev is large

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Rip Rap for drainage over high slope

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Riprap (TN Design Manual)

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Side Ditch/Open Channel Design-Basics

• Estimate Q at point of interest• Select ditch cross section• Erosion control?• Manning’s formula for design• Assume steady flow in a uniform

channel

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Manning’s Formula

V = R2/3*S1/2 (metric) V = 1.486 R2/3*S1/2 n 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,ft)S = slope of channeln = Manning’s roughness coefficient

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Side Ditch/Open Channel Design-Basics

Q = VAQ = discharge (ft3/sec, m3/sec)A = area of flow cross section (ft2, m2)

FHWA has developed chartsto solve Manning’s equation for different cross sections

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Open Channel Example

Runoff = 340 ft3/sec (Q) Slope = 1% Manning’s n = 0.015 Determine necessary cross-section

to handle estimated runoff Use rectangular channel 6-feet

wide

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Open Channel ExampleQ = 1.486 R2/3*S1/2 n Hydraulic radius, R = a/P a = area, P = wetted perimeter

P

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Open Channel Example

Flow depth = d Area = 6 feet x d Wetted perimeter = 6 + 2d

6 feet

Flow depth (d)

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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’

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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)

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Example (continued)

Find critical velocities.From chart along critical curve, vc 13 ft/secCritical 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’

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Source: FHWA Hydraulic Design Charts

27A cut slope with ditch

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A fill slope

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Inlet or drain marker

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Ditch treatment near a bridgeUS 30 – should pier be protected?

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A fill slope

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Where’s the water going to end up?

Hidden Drain

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Median drain

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Design of Culverts

Source: Michigan Design Manual

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Culvert Design - Basics

Top of culvert not used as pavement surface (unlike bridge), usually less than 20 foot span

> 20 feet use a bridge Three locations

Bottom of depression (no watercourse) Natural stream intersection with roadway

(majority) Locations where side ditch surface

drainage must cross roadway

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Hydrologic and Economic Considerations

Alignment and grade of culvert (wrt roadway) are important

Similar to open channel

Design flow rate based on storm with acceptable return period (frequency)

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Culvert Design Steps

Obtain site data Roadway cross section at culvert

location (best is at existing channel)

Establish inlet/outlet elevations, length, and slope of culvert

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Sometimes … you want a dam … why?

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Culvert Design Steps

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 Emergency overflow?

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Headwater Depth 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, as well as need to protect roadway subgrade

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Inlet control

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

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Example:Design ElevHW = 230.5 (max)Stream bed at inlet = 224.0Drop = 6.5’Peak Flow = 250cfs5x5 box

HW/D = 1.41HW = 1.41x5 = 7.1’Need 7.1’, have 6.5’

Drop box 0.6’ below stream @223.4’ - OK

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Outlet control

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

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Example:Design ElevHW = 230.5Flow = 250cfs5x5 box (D=5)Stream at invert = 224200’ culvertOutlet invert = 224-0.02x200 = 220.0’ (note: = 223.4-.017x200)Given tail water depth = 6.5’Check critical depth, dc = 4.3’ from fig. 17.23Depth to hydraulic grade line = (dc+D)/2 = 4.7 < 6.5, use 6.5’

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Example (cont.)Design ElevHW = 230.5Flow = 250cfs5x5 boxOutlet 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

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