1. CE154 Hydraulic Design Lectures 8-9
Design of Culverts
CE154 Hydraulic Design
Lectures 8-9
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2. Culverts
Definition - A structure used to convey surface runoff through embankments.
It may be a round pipe, rectangular box, arch, ellipse, bottomless, or other shapes.
And it may be made of concrete, steel, corrugated metal, polyethylene, fiberglass, or other materials.
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3. Culverts
End treatment includes projected, flared, & head and wing walls
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5. Concrete Box Culvert
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6. Box culvert with fish passage
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7. Corrugated metal horseshoe culvert
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8. Bottomless culvert USF&W
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9. Some culvert, huh?
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10. Culvert or Bridge?
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11. Study materials
Design of Small Dams (DSD) pp. 421–429 (culvert spillway), (hydraulic calculation charts)
US Army Drainage Manual (ADM),TM /AFM 88-5, Chapter 4, Appendix B - Hydraulic Design Data for Culverts
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12. Study Objectives Recognize different culvert flow conditions
Learn the steps to analyze culvert hydraulics
Learn to design culverts
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13. Definition Sketch
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14. Definition Sketch
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15. Relevant technical terms
Critical depth The depth at which the specific energy (y+v2/2g) of a given flow rate is at a minimum
Soffit or crown The inside top of the culvert
Invert & thalweg Channel bottom & lowest point of the channel bottom
Headwater The water body at the inlet of a culvert
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16. Relevant technical terms
Tailwater The water body at the outlet of a culvert
Submerged outlet An outlet is submerged when the tailwater level is higher than the culvert soffit.
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17. Relevant technical terms
Inlet control Occurs when the culvert barrel can convey more flow than the inlet will accept. The flow is only affected by headwater level, inlet area, inlet edge configuration, and inlet shape. Factors such as roughness of the culvert barrel, length of the culvert, slope and tailwater level have no effect on the flow when a culvert is under inlet control.
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18. Relevant technical terms
Outlet control Occurs when the culvert barrel can not convey more flow than the inlet can accept. The flow is a function of the headwater elevation, inlet area, inlet edge configuration, inlet shape, barrel roughness, barrel shape and area, slope, and tailwater level.
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19. Relevant technical terms
Normal depth Occurs in a channel reach when the flow, velocity and depth stay constant. Under normal flow condition, the channel slope, water surface slope and energy slope are parallel.
Steep slope Occurs when the normal depth is less than the critical depth. The flow is called supercritical flow.
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20. Relevant technical terms
Mild slope Occurs when the normal depth is higher than the critical depth. The flow is called subcritical flow.
Submerged inlet An inlet is submerged when the headwater level is higher than approximately 1.2 times the culvert height D. (Why is it not simply higher than 1.0 times D?)
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21. Relevant Technical Terms
Freeboard Safety margin over design water level before overflow occurs (in a unit of length)
Free outlet An outlet condition at which the tailwater level is below the critical depth, whence further lowering of the tailwater will not affect the culvert flow
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22. Design Setting a river a plan to build a road crossing
need to design the road crossing - given river slope, geometry, & design flood - given desirable roadway elevation - design culvert (unknown size) to pass “Design Flood” with suitable freeboard (design criteria)
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23. Analysis Setting An existing culvert or bridge (known size)
a river passing underneath
determine water level under certain flood condition or vice versa
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24. Inlet control (1)
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25. Inlet control (2)
>
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26. Inlet control (3) – sharp edge inlet
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27. Outlet control (1)
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28. Outlet control (2)
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29. Outlet control (3)
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30. Outlet control (4)
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31. Intermittent control
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32. Key Approaches
Critical flow does not occur on mild slopes, except under certain special, temporary condition [such as inlet control (3)]
Critical flow always occurs at the inlet of a steep slope, except when the inlet is deeply submerged [H/D > ]
On mild slopes, most likely it’s outlet control
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33. Approaches
For unsubmerged inlet control, - for culvert on steep slope, use critical flow condition to determine the discharge - for culvert on mild slope, use weir equation to compute flow
For submerged inlet control, use orifice flow equation to compute discharge
For outlet control, perform energy balance between inlet and outlet
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34. Critical Flow Condition
yc = (q2/g)1/3
Fr = vc/(gyc)1/2 = 1
vc = (gyc)1/2
Ec = yc + vc2/2g = 3/2 yc
q = unit discharge = Q/width (for non-circular conduit; for circular pipe use table to find critical condition) Fr = Froude number E = specific energy y = depth c = subscript denotes critical flow condition
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35. Weir Flow Weir flow equation
B = culvert width Cw = weir discharge coefficient, an initial estimate may be 3.0 note that this eq. is similar to equations for ogee crest weir, broadcrested weir, sharp crest weir
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36. Orifice Flow Inlet control with submerged inlet,
Cd = orifice discharge coefficient, an initial estimate  0.60
b = culvert height
HW-b/2 = average head over the culvert
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37. Outlet control hydraulics
Energy balance between inlet and outlet
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38. Outlet control hydraulics
Entrance loss coefficient on p.B-12 of ADM and p. 426 & 454 of Design of Small Dam
Exit loss coefficient: as a function of area change from the culvert (a1) to downstream channel (a2) Kex = (1- a1/a2)2 = 1 for outlet into reservoir
Friction loss coefficient may be computed using Darcy-Weisbach or Manning equation
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39. Outlet control hydraulics
Darcy-Weisbach equation for circular pipes friction head loss hf = f L/D V2/2g or for non-circular channels, using hydraulic radius R=A/P=D/4 to replace D: hf = f L/(4R) V2/2g kf = f L/(4R)
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40. Outlet control hydraulics
Manning’s equation to compute friction loss v = (1.49 R2/3 S1/2) / n S = v2 n2 / (2.22 R4/3) hf = SL = v2/2g (29.1 n2L/R4/3) kf = 29.1 n2L/R4/3 - see Eq. on p. B-1
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41. Design Procedure
Establish design criteria - Q, HWmax, and other design data – L, S, TW, etc.
Determine trial size (e.g., A=Q/10)
Assume inlet control, compute HW -unsubmerged, weir flow eq. -submerged, orifice flow eq.
Assume outlet control, compute HW
Compare results of 3 & 4. The higher HW governs.
Try a different size until the design criteria are met
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42. Example (1)
A circular corrugated metal pipe culvert, 10’ in diameter, 50’ long, square edge with headwall, on slope of 0.02, Manning’s n=0.024, is to convey flood flow of 725 cfs. Tailwater is at the center of the culvert outlet. Determine the culvert flow condition.
Assuming first if the slope is steep, inlet control. If mild, outlet control.
Determine if the slope is steep or mild by comparing normal and critical flow depth, e.g. tables from Design of Small Dams (DSD)
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45. Example (1)
Q = 725, n = 0.024, D = 10 ft, S = 0.02
Qn/(D8/3S1/2) = 0.265
Table B-3, it corresponds to d/D = 0.541, or the normal depth dn = 5.41 ft
Q/D2.5 = 2.293
From Table B-2, find d/D = 0.648, or the critical depth dc = 6.48 ft
dc > dn, so the 0.02 slope is steep  inlet control
Critical flow occurring at the culvert entrance
Use Figure 9-68 (or Figure B-8 of DSD p.585) for circular culverts on steep slope to determine headwater depth
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46. Example (1)
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47. Example (1)
For Q/D2.5 = 2.293, and square edge inlet, Curve A on figure 9-68 shows
H/D = 1.0
The headwater is at the culvert soffit level, and it drops to 6.48 ft at the inlet and continues to drop to 5.41 ft to flow through the culvert, before dropping to 5 ft at the outlet.
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49. Example (2)
Concrete pipe (n=0.015) culvert 10 ft in diameter, 0.02 slope, square edge, vertical headwall, Q = 1550 cfs, tailwater at pipe center at outlet. Determine the culvert flow condition.
Q/D2.5 = 1550/(10)^2.5 = 4.90
Qn/(D8/3S1/2) = 0.35
dn determined from Table B-3, d/D=0.65
dc determined from Table B-2, d/D = 0.913
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50. Example (2)
The culvert will run open-channel, same as in Example (1) and the water level drops to the pipe center level at the outlet.
To compute headwater level, Figure 9-68 shows that H/D = 2.15
The culvert entrance will be submerged, with water level dropping to dc = 9.13 ft at the inlet and continues dropping to dn = 6.5 ft for the bulk length of the pipe.
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51. Example (3)
Same condition as in Example (1), with corrugated pipe 10 ft diameter, S=0.02, L=300 ft, tailwater level at pipe center, Q=2000 cfs. Determine flow condition.
Q/D2.5 = 2000/(10)^2.5 = 6.32
Qn/(D8/3S1/2) = 2000*0.024/(65.4) = 0.73
Critical depth at 9.65 ft, practically full flow
Normal depth shows full flow – since data is out of range of table  outlet control
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52. Example (3)
Calculate entrance loss coefficient – square edge flush with vertical headwall (p.426) Ken = 0.5
Calculate exit loss coefficient – tailwater at pipe centerline, outlet channel is not supported, full exit velocity head is lost Kex = 1.0
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53. Example (3)
Calculate friction loss coefficient – R = A/P = D/4 = 2.5 n = Kf = 29.1 n2 L / R4/3 = 1.48
Eq. (32) on p. 425 shows that H/D + L/D So – 0.5 = (kex + ken + kf)(Q/D5/2)2 H/ *0.02 – 0.5 = ( )(6.32)2
H = 29 ft
Check using Figure B-10 of Design of Small Dams or Figure B-13 of Reader – graphical solution shows H=32’
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55. Example (4)
Design a culvert for the following condition: - Design Flow Q = 800 cfs - culvert length L = 100 ft - Allowable headwater depth HW = 15 ft - Concrete pipe culvert - Slope S = 0.01 (1.0%) - Tailwater level (TW) at 0.8D above invert at outlet
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56. Example (4)
Select a trial culvert pipe size Assuming culvert flow velocity V = 10 fps A = Q/V = 800/10 = 80 ft2 D = sqrt(804/) = 10.1 ft Say D = 10 ft
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57. Example (4)
Assuming inlet control: - using rounded inlet to reduce headloss - Q/D5/2 = From Figure 9-68 of DSD, H/D = 1
This is a conservative design. Reasonably H/D could be designed as high as 1.2 to maintain un-submerged inlet condition.
Check by using Figure B-7 of DSD. The rounded inlet is similar to groove inlet (see Table B-1 of ADM)
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58. Example (4)
Assuming outlet control: - First determine the outlet flow condition. From Table B-2 of DSD, at Q=800 cfs, Q/D5/2=2.53, the critical depth dc=0.682D. Hence, TW=0.8D is above the critical level. The normal flow is determined from Table B-3 of DSD. Use n=0.018 for aged concrete. Qn/D8/3S1/2=0.31 dn=0.6D The normal flow depth is 6.0 ft in the culvert
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59. Example (4)
The normal flow condition is: - from Table B-3 again, A/D2 = An = 49.2 ft2 - Vn = Q/An = 16.3 fps - R = hydraulic radius = 0.278D = 2.78 ft - Fr = Froude number = V/(gR)1/2 = This shows that flow is supercritical in the culvert. It transitions to the tailwater depth at the outlet (S3 or jump). TW flow may be supercritical or subcritical, depending on the downstream slope.
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60. Example (4)
To compute the headloss of the outlet-control condition: HW + SoL = HL + TW HL = (Ken + Kex + Kf)V2/2g Ken = 0.2 for rounded edge with headwall Kex = 1.0 being conservative since not all the velocity head is lost (draw profile) Kf = 29n2L/R = 0.24
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61. Example (4) HL = (1 + 0.24 + 0.2) V2/2g = 1.44 (16.3)2/64.4 = 5.94 ft
The energy balance equation becomes HW = HL + TW – SoL = – 0.01100 = 12.94
HW/D = 12.94/10 = 1.3
Compare the headwater depth for inlet and outlet conditions, select the higher value for design.
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62. Example (4) The governing headwater depth is 13 ft
This is less than the maximum of 15 ft of the allowable headwater depth. Hence, it is acceptable. The culvert size may be reduced slightly to reduce cost and still meets design criteria. Hence, use 10 ft diameter concrete pipe rounded edge at inlet maximum headwater depth 13 ft
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63. Culvert failure modes along forest roads in northern CA
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64. Design Considerations
Flared ends improve efficiency
Use culverts as wide as stream width
Use same gradient as stream channel
Use same alignment as stream channel
Single large culvert is better for debris passage than several small ones
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