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