Culvert Design.ppt

CE154

Design of Culverts

CE154 Hydraulic DesignLectures 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 treatmentincludes projected, flared, & head and wing walls

CE154Fall 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. 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

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 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

Fall 2009 15

CE154


• TailwaterThe water body at the outlet of a culvert

• Submerged outletAn outlet is submerged when the tailwater level is higher than the culvert soffit.

Fall 2009 16

CE154


• 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.

Fall 2009 17

CE154


• 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.

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 slopeOccurs when the normal depth is less than the critical depth. The flow is called supercritical flow.

Fall 2009 19

CE154


• 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?)

Fall 2009 20

CE154

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

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 it’s 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 numberE = specific energyy = depthc = subscript denotes critical flow condition

Fall 2009 34

Fall 2009 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

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

kk

k

Vkkk

f

ex

en

fexen

0.11.0

12&426..

2)(

2

Fall 2009 37

CE154


• 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


• 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


• 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-1Fall 2009 40

CE154

Design Procedure1. 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 HW5. Compare results of 3 & 4. The higher HW

governs.6. Try a different size until the design criteria

are metFall 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, 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)

Fall 2009 42

CE154Fall 2009 43

CE154Fall 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

CE154Fall 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 = 1.0Fall 2009 52

CE154

Example (3)

• Calculate friction loss coefficient – R = A/P = D/4 = 2.5n = 0.024Kf = 29.1 n2 L / R4/3 = 1.48

• Eq. (32) on p. 425 shows thatH/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

CE154Fall 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 sizeAssuming culvert flow velocity V = 10 fpsA = 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=0.31dn=0.6DThe 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 + TWHL = (Ken + Kex + Kf)V2/2gKen = 0.2 for rounded edge with

headwallKex = 1.0 being conservative since

not all the velocity head is lost (draw profile)

Kf = 29n2L/R1.333

= 0.24Fall 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 = 1.34.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 ft5.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 inletmaximum 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

Documents

Culvert Design.ppt