design of culverts -...
TRANSCRIPT
-
CE154
Design of Culverts
CE154 Hydraulic Design
Lectures 8-9
Fall 2009 1
-
CE154
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.
Fall 2009 2
-
Fall 2009 CE154 3
Culverts
End treatment includes projected, flared, & head and wing walls
-
CE154
Fall 2009 4
-
CE154
Concrete Box Culvert
Fall 2009 5
-
CE154
Box culvert with fish passage
Fall 2009 6
-
CE154
Corrugated metal horseshoe culvert
Fall 2009 7
-
CE154
Bottomless culvert USF&W
Fall 2009 8
-
CE154
Some culvert, huh?
Fall 2009 9
-
CE154
Culvert or Bridge?
Fall 2009 10
-
CE154
Study materials
Design of Small Dams (DSD) pp. 421429 (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
Fall 2009 11
-
CE154
Study Objectives
Recognize different culvert flow conditions
Learn the steps to analyze culvert hydraulics
Learn to design culverts
Fall 2009 12
-
Fall 2009 CE154 13
Definition Sketch
-
CE154
Definition Sketch
Fall 2009 14
-
CE154
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
Fall 2009 15
-
CE154
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.
Fall 2009 16
-
CE154
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.
Fall 2009 17
-
CE154
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.
Fall 2009 18
-
CE154
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.
Fall 2009 19
-
CE154
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?)
Fall 2009 20
-
CE154
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
Fall 2009 21
-
CE154
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)
Fall 2009 22
-
CE154
Analysis Setting
An existing culvert or bridge (known size)
a river passing underneath
determine water level under certain flood condition or vice versa
Fall 2009 23
-
CE154
Inlet control (1)
Fall 2009 24
-
CE154
Inlet control (2)
Fall 2009 25
>
-
CE154
Inlet control (3) sharp edge inlet
Fall 2009 26
-
CE154
Outlet control (1)
Fall 2009 27
-
CE154
Outlet control (2)
Fall 2009 28
-
CE154
Outlet control (3)
Fall 2009 29
-
CE154
Outlet control (4)
Fall 2009 30
-
CE154
Intermittent control
Fall 2009 31
-
CE154
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 its outlet control
Fall 2009 32
-
CE154
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
Fall 2009 33
-
CE154
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
Fall 2009 34
-
Fall 2009 CE154 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
-
Fall 2009 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
Outlet control hydraulics
Energy balance between inlet and outlet
tcoefficienlossfriction
tofromtypicallytcoefficienlossexit
elevationtailwaterTW
ADMBDSDponcoeflossentrance
gTWLSoH
k
k
k
Vkkk
f
ex
en
fexen
=
=
=
=
++=+
0.11.0
12&426..
2)(
2
Fall 2009 37
-
CE154
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
Fall 2009 38
-
CE154
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)
Fall 2009 39
-
CE154
Outlet control hydraulics
Mannings 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
Fall 2009 40
-
CE154
Design Procedure
1. Establish design criteria - Q, HWmax, and other design data L, S, TW, etc.
2. Determine trial size (e.g., A=Q/10)
3. Assume inlet control, compute HW -unsubmerged, weir flow eq. -submerged, orifice flow eq.
4. Assume outlet control, compute HW
5. Compare results of 3 & 4. The higher HW governs.
6. Try a different size until the design criteria are met
Fall 2009 41
-
CE154
Example (1)
A circular corrugated metal pipe culvert, 10 in diameter, 50 long, square edge with headwall, on slope of 0.02, Mannings 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)
Fall 2009 42
-
CE154
Fall 2009 43
-
CE154
Fall 2009 44
-
CE154
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
Fall 2009 45
-
CE154
Example (1)
Fall 2009 46
-
CE154
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.
Fall 2009 47
-
CE154 Fall 2009 48
-
CE154
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
Fall 2009 49
-
CE154
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.
Fall 2009 50
-
CE154
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
Fall 2009 51
-
CE154
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
Fall 2009 52
-
CE154
Example (3)
Calculate friction loss coefficient R = A/P = D/4 = 2.5 n = 0.024 Kf = 29.1 n2 L / R4/3 = 1.48
Eq. (32) on p. 425 shows that H/D + L/D So 0.5 = 0.0252(kex + ken + kf)(Q/D5/2)2 H/10 + 30*0.02 0.5 = 0.0252 (1+0.5+1.48)(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
Fall 2009 53
-
CE154
Fall 2009 54
-
CE154
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
Fall 2009 55
-
CE154
Example (4)
1. Select a trial culvert pipe size Assuming culvert flow velocity V = 10 fps A = Q/V = 800/10 = 80 ft2 D = sqrt(804/) = 10.1 ft Say D = 10 ft
Fall 2009 56
-
CE154
Example (4)
2. Assuming inlet control: - using rounded inlet to reduce headloss - Q/D5/2 = 2.53 - 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)
Fall 2009 57
-
CE154
Example (4)
3. 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
Fall 2009 58
-
CE154
Example (4)
The normal flow condition is: - from Table B-3 again, A/D2 = 0.492 - 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 = 1.7 - 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.
Fall 2009 59
-
CE154
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/R1.333 = 0.24
Fall 2009 60
-
CE154
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 = 5.94 + 8 0.01100 = 12.94
HW/D = 12.94/10 = 1.3 4. Compare the headwater depth for inlet
and outlet conditions, select the higher value for design.
Fall 2009 61
-
CE154
Example (4)
The governing headwater depth is 13 ft 5. 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
Fall 2009 62
-
Culvert failure modes along forest
roads in northern CA
Fall 2009 CE154 63
-
CE154
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
Fall 2009 64