design of open channels and culverts
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Design of Open Channels and Culverts. Transverse Slopes. Removes water from pavement surfaces in shortest amount of time possible. Longitudinal Slopes. Gradient longitudinal direction of highway to facilitate movement of water along roadway. Drains. Along ROW Collect surface water.

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Presentation Transcript
transverse slopes
Transverse Slopes
  • Removes water from pavement surfaces in shortest amount of time possible
longitudinal slopes
Longitudinal Slopes
  • Gradient longitudinal direction of highway to facilitate movement of water along roadway
drains
Drains
  • 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
Drainage Channels (Ditches)
  • Design
    • Adequate capacity
    • Minimum hazard to traffic
    • Hydraulic efficiency
    • Ease of maintenance
  • Desirable design (for safety): flat slopes, broad bottom, and liberal rounding
ditch shape
Ditch Shape
  • Trapezoidal – generally preferred considering hydraulics, maintenance, and safety
  • V-shaped – less desirable from safety point of view and maintenance
flow velocity
Flow Velocity
  • Depends on lining type
  • Typically 1 to 5% slopes used
  • Should be high enough to prevent deposit of transported material (sedimentation)
    • For most linings, problem if S < 1%
  • Should be low enough to prevent erosion (scour)
    • For most types of linings, problem if S > 5%
side ditch open channel design basics
Side Ditch/Open Channel Design-Basics
  • 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
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,f)

S = slope of channel

n = Manning’s roughness coefficient

side ditch open channel design basics1
Side Ditch/Open Channel Design-Basics

Q = VA

Q = discharge (ft3/sec, m3/sec)

A = area of flow cross section (ft2, m2)

FHWA Hydraulic Design Charts

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

open channel example
Open Channel Example
  • 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 example1
Open Channel Example

Q = 1.486 R2/3*S1/2

n

  • Hydraulic radius, R = a/P
  • a = area, P = wetted perimeter

P

open channel example2
Open Channel Example
  • Flow depth = d
  • Area = 6 feet x d
  • Wetted perimeter = 6 + 2d

Flow depth (d)

6 feet

example continued
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 continued1
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 continued2
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’

slide28

Ditch treatment near a bridge

US 30 – should pier be protected?

culvert design basics
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
hydrologic and economic considerations
Hydrologic and Economic Considerations
  • Alignment and grade of culvert (with respect to roadway) are important
  • Similar to open channel
  • Design flow rate based on storm with acceptable return period (frequency)
culvert design steps
Culvert Design Steps
  • Obtain site data and roadway cross section at culvert crossing location (with approximation of stream elevation) – best is natural stream location, alignment, and slope (may be expensive though)
  • Establish inlet/outlet elevations, length, and slope of culvert
culvert design steps1
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
headwater depth
Headwater Depth
  • Constriction due to culvert creates increase in depth of water just upstream
  • Allowable level of headwater upstream usually controls culvert size and inlet geometry
  • Allowable headwater depth depends on topography and land use in immediate vicinity
types of culvert flow
Types of culvert flow
  • Type of flow depends on total energy available between inlet and outlet
  • 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
slide40

Ans.

Example:

Design ElevHW = 230.5

Stream bed at inlet = 224.0

Drop = 6.5’

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 - OK

types of culvert flow1
Types of culvert flow
  • 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
slide43

Example:

Design ElevHW = 230.5

Flow = 250cfs

5x5 box

Stream elev at inlet = 240

200’ culvert

Outlet invert = 240-0.02x200

= 220.0’

Given tail water depth = 6.5’

Check critical depth = 4.3’

from fig. 17.23 (next page)

Depth to hydraulic grade line =

(dc+D)/2 = 4.7 < 6.5, use 6.5’

Head drop = 3.3’ (from chart)

220.0+6.5+3.3 = 229.8’<230.5 OK

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