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Design of Culverts

Design of Culverts. CE154 Hydraulic Design Lectures 8-9. 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.

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Design of Culverts

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  1. Design of Culverts CE154 Hydraulic Design Lectures 8-9 CE154

  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. CE154

  3. Culverts • End treatmentincludes projected, flared, & head and wing walls CE154

  4. CE154

  5. Concrete Box Culvert CE154

  6. Box culvert with fish passage CE154

  7. Corrugated metal horseshoe culvert CE154

  8. Bottomless culvert USF&W CE154

  9. Some culvert, huh? CE154

  10. Culvert or Bridge? CE154

  11. Study materials • Design of Small Dams (DSD) pp. 421–429 (culvert spillway), 582-589 (hydraulic calculation charts) • US Army Drainage Manual (ADM),TM 5-820-4/AFM 88-5, Chapter 4, Appendix B - Hydraulic Design Data for Culverts CE154

  12. Study Objectives • Recognize different culvert flow conditions • Learn the steps to analyze culvert hydraulics • Learn to design culverts CE154

  13. Definition Sketch CE154

  14. Definition Sketch CE154

  15. Relevant technical terms • Critical depthThe depth at which the specific energy (y+v2/2g) of a given flow rate is at a minimum • Soffit or crownThe inside top of the culvert • Invert & thalwegChannel bottom & lowest point of the channel bottom • Headwater The water body at the inlet of a culvert CE154

  16. Relevant technical terms • TailwaterThe water body at the outlet of a culvert • Submerged outletAn outlet is submerged when the tailwater level is higher than the culvert soffit. CE154

  17. Relevant technical terms • Inlet controlOccurs 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. CE154

  18. Relevant technical terms • Outlet controlOccurs 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. CE154

  19. Relevant technical terms • Normal depthOccurs 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 slopeOccurs when the normal depth is less than the critical depth. The flow is called supercritical flow. CE154

  20. Relevant technical terms • Mild slopeOccurs when the normal depth is higher than the critical depth. The flow is called subcritical flow. • Submerged inletAn 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?) CE154

  21. Relevant Technical Terms • FreeboardSafety margin over design water level before overflow occurs (in a unit of length) • Free outletAn outlet condition at which the tailwater level is below the critical depth, whence further lowering of the tailwater will not affect the culvert flow CE154

  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) CE154

  23. Analysis Setting • An existing culvert or bridge (known size) • a river passing underneath • determine water level under certain flood condition or vice versa CE154

  24. Inlet control (1) CE154

  25. Inlet control (2) > CE154

  26. Inlet control (3) – sharp edge inlet CE154

  27. Outlet control (1) CE154

  28. Outlet control (2) CE154

  29. Outlet control (3) CE154

  30. Outlet control (4) CE154

  31. Intermittent control CE154

  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 > 1.2-1.5] • On mild slopes, most likely it’s outlet control CE154

  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 CE154

  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 numberE = specific energyy = depthc = subscript denotes critical flow condition CE154

  35. Weir Flow • Weir flow equation B = culvert widthCw = weir discharge coefficient, aninitial estimate may be 3.0note that this eq. is similar to equations for ogee crest weir, broadcrested weir, sharp crest weir CE154

  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 CE154

  37. Outlet control hydraulics • Energy balance between inlet and outlet CE154

  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 CE154

  39. Outlet control hydraulics • Darcy-Weisbach equation for circular pipesfriction head loss hf = f L/D V2/2gor 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) CE154

  40. Outlet control hydraulics • Manning’s equation to compute friction lossv = (1.49 R2/3 S1/2) / nS = 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 CE154

  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 CE154

  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) CE154

  43. CE154

  44. CE154

  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 CE154

  46. Example (1) CE154

  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. CE154

  48. CE154

  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 CE154

  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. CE154

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