Design of Open Channels and Culverts

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

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

Drains Along ROW Collect surface water A typical intercepting drain placed in the impervious zone http://www.big-o.com/constr/hel-cor.htm

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

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

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%

Use spillway or chute if Δelev is large

Rip Rap for drainage over high slope

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

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

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

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

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

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

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’

Example (continued) If you already know Q, simpler just to do 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)

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’

Check lining for max depth of flow …

Rounded

A cut slope with ditch

A fill slope

Inlet or drain marker

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

A fill slope

Median drain

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

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)

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

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

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

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

Drop box 0.6’ below stream - OK 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

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

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