open channels
TRANSCRIPT
1
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
2
Design of Open Channels
3
Longitudinal Slopes Gradient
longitudinal direction of highway to facilitate movement of water along roadway
4
Drains
Along ROW Collect surface water
A typical intercepting drain placed in the impervious zone
http://www.big-o.com/constr/hel-cor.htm
5
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
6
Ditch Shape
Trapezoidal – generally preferred considering hydraulics, maintenance, and safety
Source: Fabriform1.com
7
Ditch Shape
V-shaped – less desirable from safety point of view and maintenance
Source: Fabriform1.com
8
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
9
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
10
11
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%
12
Use spillway or chute if Δelev is large
13
Rip Rap for drainage over high slope
14
Riprap (TN Design Manual)
15
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
16
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
17
18
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
19
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
20
Open Channel ExampleQ = 1.486 R2/3*S1/2 n Hydraulic radius, R = a/P a = area, P = wetted perimeter
P
21
Open Channel Example
Flow depth = d Area = 6 feet x d Wetted perimeter = 6 + 2d
6 feet
Flow depth (d)
22
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’
23
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)
24
25
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’
26
Source: FHWA Hydraulic Design Charts
27A cut slope with ditch
28
A fill slope
29
Inlet or drain marker
30
Ditch treatment near a bridgeUS 30 – should pier be protected?
31
A fill slope
32
Where’s the water going to end up?
Hidden Drain
33
34
35
Median drain
36
Design of Culverts
Source: Michigan Design Manual
37
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
38
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)
39
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
40
41
Sometimes … you want a dam … why?
42
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?
43
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
44
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
45
46
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
47
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
48
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’
49
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
50