Course Number: PE Exam Review - Civil Engineering Instructor: Russell Briggs, PELecture Number:04 Water Resources & Environmental 02.3

© North Carolina State University, All Rights Reserved

Lecture Notes

P.E. Review

Water Resources

& Environmental

Culverts

Culverts

Example 6.1 (A.M.)

• For a culvert in inlet control, which statement is false?

• a) Whether the pipe has a headwall or not is an important

consideration.

• b) If a pipe is in inlet control, there is no limit to the amount

of water that may be “stacked” behind the pipe.

• c) Inlet control can be true for any type of pipe (i.e.

concrete, corrugated metal, clay)

• d) The downstream water elevation can be above the outlet

of the pipe.

Culverts

Example 6.2 (P.M. Trans, WRE)

A culvert crossing for a road must pass the 100-year flood

without overtopping. The distance from the invert of the

culvert to the minimum elevation of the road is 12 feet. The

estimate of the 100-year flood is 350 cfs, and the culvert will

be approximately 115’ long. The downstream receiving

channel flows at a depth of 3.0 feet in the 100-year flood.

The number and diameter of the new concrete pipe

culvert(s) that should be installed is:

a) 1-48” b) 2-24” c) 1-60” d) 1-72”

Inlet or outlet control?

Downstream channel depth of 3.0, so all but the 2-24” are

inlet control Try inlet control first, then.

Culverts

Example 6.2

Length = 115’

Diameter = ?

Flow = 350 cfs

HW = 12’ TW = 3’

Try 60” pipe

Project through 350 cfs and find HW/D is approximately 3.3

Thus HW is 3.3 X 60 / 12, or 16.5 feet. This is higher than

the allowable HW of 12’, so reject and try a bigger pipe

Try 72” pipe

Project through 350 cfs and find HW/D is approximately 1.7

Thus HW is 1.7 X 72 / 12, or 10.2 feet. Acceptable. (<12’)

Culverts

Example 6.3 (P.M. Trans, WRE)

Due to repeated flooding, a new pipe will be installed parallel to an existing box culvert to accommodate the 100-year flow, which is 1500 cfs.

Existing Box: 8’ Wide X 10’ High Inv. Elevation at entrance: 361.53 Inlet control assured New culvert: Concrete, Entrance flush with headwall Invert elevation at entrance is 361.53 Inlet control assured Allowable water surface elevation is 375.5 and both culverts

have wingwalls (45 degrees). The size of the new pipe to be installed so the 100-year flood

does not exceed the allowable water surface elevation is:

a) 24” b) 96” c) 72” d) 48”

Culverts

Example 6.3

Inv. Elev. = 361.5

w.s.e. = 375.5

8X10 Box Culvert

Flow = ? cfs

HW = 14’

First, determine the

flow through the box

culvert before the

road overtops.

HW = 14; D, or height = 10

so HW/D = 14/10 = 1.4

Connect HW/D of 1.4 &

Height of Box = 10’

Read Q/B – 135 cfs/ft

Therefore,

Qculvert = 135 cfs/ft X B

= 135 cfs/ft X 8 ft

= 1080 cfs

Culverts

Example 6.3

Inv. Elev. = 361.5

w.s.e. = 375.5

Diameter = ? inches

Flow = 1500 - 1080 cfs

HW = 14’

The new concrete pipe must

transmit the difference between the

100-year discharge and the amount

the existing box culvert can convey.

Qpipe = 1500 cfs – 1080 cfs = 420 cfs

Try 48” Pipe, HW/D …

Try 72” Pipe,

HW/D = 2.2; HW = 2.2 X 72/12;

HW = 13.2; ok (allowable HW = 14’)

Choose 72” Pipe.

Culverts

Example 6.4 (P.M. Trans, WRE)

At a particular stream crossing, your estimate of the 100-

year flood is 140 cfs. The municipality requires headwalls

for the end treatment of all pipe 36” in diameter and greater.

The invert of the stream at the inlet end of the culvert is 325.

What should be the minimum road elevation if you install

dual 36” reinforced concrete pipes as the culvert system?

a) 331.0

b) 334.3

c) 337.0

d) 343.0

Culverts

Example 6.4

Elev. = ?

Inv. Elev. = 325

w.s.e. = ?

Diameter = 36”

Flow = 140 cfs

HW = ?

Set a point for diameter at 36”

Set a point for discharge of 70 cfs

Connect those two points, extend and read HW/D of 2.0

Thus HW is 2.0 X 36 / 12, or 6.0 feet.

Minimum Elevation, then, is 6.0 feet + 325 = 331.0

(1/2 of 140 cfs)

Culverts

Example 6.5 (P.M. Trans, WRE)

Determine the size of pipe culvert required for a crossing

for the following situation:

Max. upstream water surface elevation 325

Upstream invert 315

Downstream invert 314

Downstream water surface elevation 320

Length of concrete culvert 100 feet

50-year discharge 85 cfs

a) 1 – 36” pipe b) 1 – 5’ X 5’ Box culvert

c) 1 – 18” pipe d) 1 – 24” pipe

Note also that 3 choices are pipe culverts; 1 is box culvert. Try pipe culverts first, since have 3 opportunities for correct answer from that chart

Note that the downstream wse – downstream invert is six feet; therefore, all pipes are in outlet control.

Culverts

Example 6.5

Inv. Elev. = 314

Inv. Elev. = 315

w.s.e. = 320

w.s.e. = 325

Length = 100’

Diameter = ?

Flow = 85 cfs

First, determine driving head

as 5’

(upstream w.s.e. –

downstream w.s.e.)

Connect 85 cfs with the

head (H) of 5’

Second, connect the point on

the length line of 100 with the

point on the turning line. Use k

= 0.5.

Read the diameter of 36”

Culverts

Example 6.6 (P.M. Trans, WRE)

What is the upstream water surface elevation for a 5’ X 5’ box

culvert that is 200’ long with a downstream invert of 250.0, a

downstream water surface elevation of 259.0, and an

upstream invert of 253.0. The flow of interest for the culvert is

300 cubic feet per second.

a) 257.0

b) 262.0

c) 264.0

d) 268.0

Use Chart 8.

Note that the downstream wse – downstream invert is nine feet; therefore, a 5’ high box culvert will be in outlet control.

Culverts

Example 6.6

Inv. Elev. = 250 Inv. Elev. = 253

w.s.e. = 259

w.s.e. = ?

Length = 200’

5’X5’ Box Culvert

Flow = 300 cfs

Use ke = 0.5 (no other

information given) & set a

point on the length scale at

200. Connect this point to

the 5X5 box culvert on the

area scale. This sets a point

on the turning line.

Second, connect the point on

the turning line & the

discharge of 300 cfs; project

through to the head scale.

Read H ≈ 4.8’

Upstream w.s.e =

downstream w.s.e. + 4.8’ =

259 + 4.8 = 263.8

Culverts

Example 6.7 (A.M.)

– Outlet control in a culvert generally implies that:

– a) the culvert is undersized.

– b) the culvert is aged and in need of replacement.

– c) the inlet condition of the culvert is not the only

– determining factor in upstream water surface elevation.

– d) the outlet of the culvert should be modified to be more

– efficient.

Culverts

Example 6.8 (P.M. Trans, WRE)

At a particular site, you have been asked to do a study of an

existing pair of 48” corrugated metal pipe culverts (CMP)

under a road. At this site, the drainage area is 123 acres

and is a mixture of urban and rural character. You have

developed the following frequency-discharge relationship for

the culvert in question:

– Frequency Discharge

– 10-year 190 cfs

– 25-year 220 cfs

– 50-year 280 cfs

– 100-year 320 cfs

Culverts

Example 6.8 (cont.)

– The road overtopping elevation is 140 and the length of the

culvert is 120. The culvert projects from the fill on both the

upstream and downstream sides. The culvert will operate in

outlet control for everything equal to or greater than the 5-year

event, with an approximate tail water elevation of 132. The

slope of the culvert is 1.0%, with the upstream invert elevation

of 130. Which is the most frequent event listed that will not

overtop the road?

Culverts

Example 6.8

Inv. Elev. = 250 Inv. Elev. = 130

w.s.e. = 132

w.s.e. = 140

Length = 120’

Diameter = 48”

Flow = ? cfs

Outlet control.

H = (up-down) stream w.s.e.

= 140 – 132 =8’

Determine capacity of a single

48” CMP

Connect diameter = 48”

& length = 120’

Set a point on turning line.

Use that point & H = 8’; extend

to discharge & read 145 cfs.

2 pipes,

therefore 2X145 cfs = 290 cfs

Culverts

Example 6.8

– Frequency Discharge

– 10-year 190 cfs

– 25-year 220 cfs

– 50-year 280 cfs

– 100-year 320 cfs

Capacity = 290 cfs, therefore

50-year is the maximum flow

that is listed which will not

overtop the road.