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1、1Design of Open Channels and CulvertsCE453 Lecture 25Ref: 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, acc
2、essed March 18, 20062Design of Open Channels3Longitudinal SlopesnGradient longitudinal direction of highway to facilitate movement of water along roadway4DrainsnAlong ROWnCollect surface waterA typical intercepting drain placed in the impervious zone http:/www.big- Channels (Ditches)lDesignlAdequate
3、 capacitylMinimize hazard to trafficlHydraulic efficiencylEase of maintenancelDesirable design (for safety): flat slopes, broad bottom, and liberal rounding6Ditch ShapelTrapezoidal generally preferred considering hydraulics, maintenance, and safetySource: F7Ditch ShapelV-shaped less desirable from s
4、afety point of view and maintenanceSource: F8TermsSteady Flow: rate of discharge does not vary with time (Mannings applies)Uniform: channel properties are constant along length of channelSlopeRoughnessCross-section Water surface is parallel to slope of channelNon-uniform: properties vary9TermsUnstea
5、dy flow: rate of discharge varies with time Critical deptha hydraulic control in design depth of water where flow changes from tranquil to rapid/shootingCritical velocity: velocity corresponding to critical depthCritical slope: slope corresponding to critical depth1011Flow VelocityDepends on lining
6、type Should be high enough to prevent deposit of transported material (sedimentation)For most linings, problem if S 2 fps when full)Should be low enough to prevent erosion (scour)For most types of linings, problem if S 5%12Use spillway or chute if elev is large13Rip Rap for drainage over high slope1
7、4Riprap (TN Design Manual)15Side Ditch/Open Channel Design-BasicsEstimate Q at point of interestSelect ditch cross sectionErosion control?Mannings formula used for designAssume steady flow in a uniform channel16Mannings Formula V = R2/3*S1/2 (metric) V = 1.486 R2/3*S1/2 n nwhere: V = mean velocity (
8、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 = Mannings roughness coefficient1718Side Ditch/Open Channel Design-BasicsQ = VAQ = discharge (ft3/sec, m3/sec)A = area of flow cross section (ft2, m2)FHW
9、A has developed chartsto solve Mannings equation for different cross sections19Open Channel ExamplenRunoff = 340 ft3/sec (Q)nSlope = 1%nMannings # = 0.015nDetermine necessary cross-section to handle estimated runoffnUse rectangular channel 6-feet wide20Open Channel ExampleQ = 1.486 R2/3*S1/2 nnHydra
10、ulic radius, R = a/Pna = area, P = wetted perimeterP21Open Channel ExamplenFlow depth = dnArea = 6 feet x dnWetted perimeter = 6 + 2d6 feetFlow depth (d)22Example (continued)Q = 1.486 a R2/3*S1/2 n340 ft3/sec = 1.486 (6d) (6d)2/3 (0.01)1/2 (6 + 2d) 0.015d 4 feetChannel area needs to be at least 4 x
11、623Example (continued)Find flow velocities.V = 1.486 R2/3*S1/2 nwith 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 0.015If you already know Q, simpler just to do V=Q/A = 340/24 = 14.2)2425Example (continued)Find critical velocities.From chart along critic
12、al curve, vc 13 ft/secCritical slope = 0.007Find critical depth: yc = (q2/g)1/3g = 32.2 ft/sec2q = flow per foot of width = 340 ft3/sec /6 feet = 56.67ft2/secyc = (56.672/32.2)1/3 = 4.64 feet depth of 426Check lining for max depth of flow 2728Source: FHWA Hydraulic Design Charts29A cut slope with di
13、tch30A fill slope31Inlet or drain marker32Ditch treatment near a bridgeUS 30 should pier be protected?33A fill slope34Wheres the water going to end up?Hidden Drain353637Median drain38Design of CulvertsSource: Michigan Design Manual39Culvert Design - Basics lTop of culvert not used as pavement surfac
14、e (unlike bridge), usually less than 20 foot spanl 20 feet use a bridge lThree locations lBottom of Depression (no watercourse) lNatural stream intersection with roadway (Majority) lLocations where side ditch surface drainage must cross roadway 40Hydrologic and Economic ConsiderationsnAlignment and
15、grade of culvert (wrt roadway) are important nSimilar to open channelnDesign flow rate based on storm with acceptable return period (frequency)41Culvert Design Steps lObtain site data lRoadway cross section at culvert locationl(best is at existing channel)lEstablish inlet/outlet elevations, length,
16、and slope of culvert 4243Sometimes you want a dam why?44Culvert Design Steps lDetermine allowable headwater depth (and probable tailwater depth) lduring design flood lcontrol on design size f(topography and nearby land use) lSelect type and size of culvert lExamine need for energy dissipaters lEmerg
17、ency overflow?45Headwater DepthnConstriction due to culvert creates increase in depth of water just upstreamnAllowable/desirable level of headwater upstream usually controls culvert size and inlet geometrynAllowable headwater depth depends on topography and land use in immediate vicinity, as well as
18、 need to protect roadway subgrade46Inlet controlnFlow is controlled by headwater depth and inlet geometrynUsually occurs when slope of culvert is steep and outlet is not submergednSupercritical, high v, low dnMost typicalnFollowing methods ignore velocity head4748Example:Design ElevHW = 230.5 (max)S
19、tream bed at inlet = 224.0Drop = 6.5Peak Flow = 250cfs5x5 boxHW/D = 1.41HW = 1.41x5 = 7.1Need 7.1, have 6.5Drop box 0.6 below stream 223.4 - OK49Outlet controlnWhen flow is governed by combination of headwater depth, entrance geometry, tailwater elevation, and slope, roughness, and length of culvert
20、 nSubcritical flownFrequently occur on flat slopesnConcept is to find the required HW depth to sustain Q flownTail water depth often not known (need a model), so may not be able to estimate for outlet control conditions50Example:Design ElevHW = 230.5Flow = 250cfs5x5 box (D=5)Stream at invert = 22420
21、0 culvertOutlet invert = 224-0.02x200 = 220.0 (note: = 223.4-.017x200)Given tail water depth = 6.5Check critical depth, dc = 4.3 from fig. 17.23Depth to hydraulic grade line = (dc+D)/2 = 4.7 6.5, use 6.551Example (cont.)Design ElevHW = 230.5Flow = 250cfs5x5 boxOutlet invert = 220.0Depth to hydraulic grade line = 6.5Head drop = 3.3 (from chart)220.0+6.5+3.3 = 229.8230.5 OK52
