open chanal

1. 1
Design of Open Channels and
Culverts

2. 2
Transverse Slopes
 Removes water from pavement
surfaces in shortest amount of time
possible

3. 3
Longitudinal Slopes
 Gradient
longitudinal
direction of
highway to
facilitate
movement of
water along
roadway

4. 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. 5
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

6. 6
Ditch Shape
 Trapezoidal – generally preferred
considering hydraulics, maintenance,
and safety
 V-shaped – less desirable from
safety point of view and
maintenance

7. 7

8. 8
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%

9. 9
Use spillway or chute if Δelev is large

10. 10
Rip Rap for drainage over high slope

11. 11
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

12. 12
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

13. 13
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

14. 14

15. 15
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

16. 16
Open Channel Example
Q = 1.486 R2/3*S1/2
n
 Hydraulic radius, R = a/P
 a = area, P = wetted perimeter
P

17. 17
Open Channel Example
 Flow depth = d
 Area = 6 feet x d
 Wetted perimeter = 6 + 2d
6 feet
Flow depth (d)

18. 18
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’

19. 19
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)

20. 20

21. 21
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’

22. 22
Check lining for max depth of flow …

23. 23

24. 24
Rounded

25. 25
A cut slope with ditch

26. 26
A fill slope

27. 27
Inlet or drain marker

28. 28
Ditch treatment near a bridge
US 30 – should pier be protected?

29. 29
A fill slope

30. 30

31. 31

32. 32
Median drain

33. 33
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

34. 34
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)

35. 35
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

36. 36
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

37. 37
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

38. 38
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

39. 39

40. 40
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

41. 41
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

42. 42

43. 43
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

44. 44