CHAPTER 8: CULVERT HYDRAULICS 
8.1 CULVERTS
Culverts are hydraulically short drainage conduits placed in locations where the drainage network intersects
the transportation network (roads, railroad tracks). Culverts differ from bridges
in that they are much smaller; thus, there are many more
culverts than bridges. Culverts are designed to operate under gradually
varied flow; therefore, the principles
of Chapter 7 are applicable.
Culverts are designed to pass the design
discharge without overtopping of the superstructure.
The design discharge is derived from the
design storm,
which is based on hydrologic considerations.
The return period
(the reciprocal of the frequency) typically varies from
10 to 50 years.
The longer the return period, the
greater the design discharge and, therefore,
the larger the required culvert size (Fig. 81).
Fig. 81 Large culvert at Yogurt Canyon, U.S.Mexico
international border, California.


The flow in a culvert is a function of:
Crosssectional size and shape,
Bottom slope,
Barrel length,
Roughness, and
Entrance and exit characteristics.
The flow in a culvert may be either (a) completely freesurface (openchannel flow), (b) completely
closedconduit (pipe flow), or
(c) partially freesurface and closedconduit flow. Headwater (HW)
is the depth of water above the culvert invert at the inlet.
Tailwater (TW) is the depth of water above the culvert invert at the outlet.
The design headwater and tailwater elevations are major
factors in determining whether the culvert flows partially full or completely full.
The design objective is to find the most economical design (i.e., the smallest culvert size)
that will pass the design discharge without exceeding a specified headwater elevation (Fig. 82). The design depends
on whether the culvert flow is under: (a) inlet control, or (b) outlet control.
Fig. 82 Large culvert crossing a
railroad embankment,
Tecate,
Baja California, Mexico.


8.2 INLET CONTROL
Culvert flow is under inlet control when the discharge depends only on the conditions at the inlet.
Assume a circular culvert of diameter D, length L, slope S,
headwater depth HW, and tailwater depth TW. Calculate the normal depth y_{n} and the critical flow depth y_{c} .
Then, the following conditions hold:
If y_{n} <
y_{c} , the flow in the culvert barrel is supercritical
and the tailwater has no influence on the upstream conditions (Fig. 83). Therefore, the headwater is solely controlled
by the conditions at the inlet.
Fig. 83 Culvert flow under supercritical conditions, with inlet submerged and outlet
unsubmerged.


If TW > y_{n}, a hydraulic
jump may form at or near the culvert outlet (Fig. 84).
Fig. 84 Culvert flow under supercritical conditions, with
inlet unsubmerged and outlet
submerged due to high tailwater.


Occurrence of inlet control
Inlet control occurs when the culvert barrel is capable of conveying more discharge than the inlet will allow.
The control section is located just inside the entrance of the culvert. The flow passes through critical depth at the control
section and becomes supercritical downstream of the inlet.
Under inlet control, the culvert acts as an orifice or weir.
If the inlet is submerged (a common design situation), the flow condition resembles that of an orifice.
If the inlet is unsubmerged (i.e., open to the atmosphere), the flow condition resembles that of a weir.
[If HW < 1.2 D, the inlet will be unsubmerged]. If the inlet is unsubmerged but the outlet is submerged,
a hydraulic jump will form inside the culvert barrel.
8.3 OUTLET CONTROL
Outlet control occurs when TW > 1.2 D, i.e., for high tailwater. In this case, the culvert barrel will be completely
full of water, resembling closedconduit flow. The headwater may be computed by applying the energy equation from the upstream (u/s) pool
elevation to the downstream (d/s) pool elevation. The headwater is directly controlled by the tailwater elevation and the frictional characteristics of the culvert barrel.
Occurrence of outlet control
Outlet control occurs when inlet and outlet are submerged. Outlet control also occurs when the culvert slope is mild
(subcritical flow) and both the headwater and tailwater are less than the culvert diameter (HW < D; TW < D).
In this case, the best approach is to calculate the watersurface profile.
Figure 85 shows a schematic portrayal of flow rate as a function of
headwater energy under inlet and outlet control (U.S. Army Corps of Engineers, 2014).
Fig. 85 Discharge
as a function of headwater energy under inlet and outlet control (U.S. Army Corps of Engineers, 2014).


8.4 CULVERT DESIGN
The following steps are followed in culvert design:
Assemble design data
Discharge,
Tailwater elevation, and
Slope of culvert barrel.
Choose the culvert characteristics
Crosssectional shape (circular, square, rectangular, arch),
Dimensions (diameter, if circular),
Barrel length,
Kind of material (Figs. 86 and 87) (concrete,
corrugated steel,
corrugated aluminum, stone masonry), and
Type of entrance (squareedged or rounded).
Fig. 86 A set of
two culverts made with corrugated steel.


Ascertain the prevailing type of control (inlet or outlet), based on (a) headwater elevation,
(b) tailwater elevation, (c) diameter, and (d) slope.
If inlet control prevails, calculate the required
headwater elevation to pass the design discharge.
If outlet control prevails, calculate the required headwater elevation using (a)
the energy equation or (b) watersurface profile computations.
If the calculated headwater elevation is greater than allowed, choose a largersize
culvert and repeat the process.
In some cases, it may not be possible to determine the type of control a priori.
In this case, both calculations are advised. The design type of control is that which results in the greatest headwater elevation.
Other design considerations in culvert design
Piping in the embankment surrounding the culvert,
Local scour at culvert outlet,
Erosion of fill material near the inlet,
Clogging with excessive debris, and
Provision for fish passage.
Fig. 87 A highway underpass featurig a
rectangular stonemasonry culvert.


Design Example
Design a culvert for the following conditions:
Design discharge: Q = 200 cfs.
Return period: T = 25 yr.
Barrel length: L = 200 ft.
Bottom slope: S_{o} = 0.01.
Culvert material: Concrete.
Manning's n = 0.013.
Inlet invert elevation: z_{1} = 100 ft.
Roadway shoulder elevation: E_{s} = 110 ft.
Tailwater depth above outlet invert: TW =
y_{2} = 3.5 ft.
Freeboard: F_{b} = 2 ft.
Solution
The design elevation for the upstream pool is: E_{s}  F_{b} = 100  2 = 108 ft.
Assume a circular concrete pipe, with square edge with headwalls.
Assume outlet control.
Assume that the hydraulic grade line (HGL) is at the elevation of the downstream pool.
Calculate the outlet invert elevation: z_{2} = z_{1} 
(S_{o} L) = 100  (0.01 × 200) = 98 ft.
Calculate the downstream pool elevation: z_{2} + y_{2} = 98 + 3.5 = 101.5 ft.
Set up the energy balance (Fig. 88):
V_{1}^{2} V_{2}^{2}
z_{1} + y_{1} + ^{_____} = z_{2} + y_{2} + ^{_____} + ∑h_{L}
2g 2g
 (81) 
Fig. 88 Energy balance in culvert flow.


Assume V_{1} = 0, i.e., the velocity is zero in the upstream pool.
Assume V_{2} = 0, i.e., the velocity dissipates to zero in the downstream pool.
The head loss ∑h_{L} is equal to the sum of entrance losses (with loss coefficient K_{e}), exit losses (with loss coefficient K_{E}), and barrel losses.
Using the DarcyWeisbach equation, the head loss is:
V^{ 2}
∑h_{L} = [ (K_{e} + K_{E} +
f (L / D ) ] ^{_____}
2g
 (82) 
From Table 81, assume
K_{e} = 0.5 and K_{E} = 1 (Roberson et al., 1985).
Table 81 Loss coefficients in pipe entrance, contraction, and expansion.

Description 
Sketch 
Additional data 
Loss coefficient K 
Pipe entrance h_{L} = K_{e} [V^{ 2}/(2g)] 

r /d 
K_{e} 
0.0 
0.50 
0.1 
0.12 
> 0.2 
0.03 


1.0 
Contraction h_{L} = K_{C} [V^{ 2}/(2g)] 

D_{2} /D_{1} 
K_{C} θ = 60° 
K_{C} θ = 180° 
0.0 
0.08  0.50 
0.2 
0.08  0.49 
0.4 
0.07  0.42 
0.6 
0.06  0.32 
0.8 
0.05  0.18 
0.9 
0.04  0.10 
Expansion h_{L} = K_{E} [V^{ 2}/(2g)] 

D_{1} /D_{2} 
K_{E} θ = 10° 
K_{E} θ = 180° 
0.0 
 1.00 
0.2 
0.13  0.92 
0.4 
0.11  0.72 
0.6 
0.06  0.42 
0.8 
0.03  0.16 
The relation between DarcyWeisbach friction factor f and Manning's n is:
8 g n^{ 2}
f = ^{__________}
k^{ 2}R^{ 1/3}
 (83) 
in which k = 1 in SI units, and k = 1.486 in U.S. Customary units.
In U.S. Customary units, with k = 1.486, and g = 32.17 ft/s^{2}:
116.55 n^{ 2}
f = ^{____________}
R^{ 1/3}
 (84) 
For a circular pipe: R = D / 4.
Therefore:
185.01 n^{ 2}
f = ^{____________}
D^{ 1/3}
 (85) 
From Eq. 81,
the energy balance reduces to:
z_{1} + y_{1} = z_{2} + y_{2} + ∑h_{L}
 (86) 
108 = 101.5 + ∑h_{L}
 (87) 
The head loss equation (Eq. 82) is repeated here for convenience:
V^{ 2}
∑h_{L} = [ (K_{e} + K_{E} +
f (L / D ) ] ^{_____}
2g
 (82) 
Replacing Eq. 85 in Eq. 82:
V^{ 2}
∑h_{L} = [ 0.5 + 1.0 +
(185.01 n^{ 2} L / D^{ 4/3} ) ] ^{_____}
2g
 (88) 
Combining Eqs. 87 and 88:
V^{ 2}
6.5 = [ 1.5 +
(6.253 / D^{ 4/3} ) ] ^{_____}
2g
 (89) 
The flow velocity is: V = Q / A. Therefore: V = 200 / A = 200 / [ (π/4) D^{ 2} ]
The velocity head is: V^{ 2} / (2g) = { 200^{2} / [ (π/4)^{2} D^{ 4} ] } / (2g) = 1008 / D^{ 4}
Replacing the velocity head in Eq. 89:
1008
6.5 = [ 1.5 +
(6.253 / D^{ 4/3} ) ] ^{_______}
D^{ 4}
 (810) 
Solving Eq. 810 by iteration: D = 4.38 ft. For design purposes, assume the next larger size: D = 4.5 ft.
With Q = 200 cfs, D = 4.5 ft = 54 in, enter Fig. 89 to find the ratio of headwater depth to diameter HW/D = 2.2, for the case of
square edge with headwalls (1).
Fig. 89 Headwater depth for concrete culverts with inlet control.


The headwater depth is: HW = (HW/D) × D = 2.2 × 4.5 = 9.9 ft.
The upstream pool elevation is: 100 + 9.9 = 109.9 ft. This upstream pool elevation is greater than 108 ft; therefore, it is too large.
The chosen D = 4.5 ft is too small. Try the next size: D = 5 ft.
With Q = 200 cfs, D = 5.0 ft = 60 in, enter Fig. 87 to find the ratio of headwater depth to diameter HW/D = 1.6, for the case of
square edge with headwalls (1).
The headwater depth is: HW = HW/D × D = 1.6 × 5.0 = 8.0 ft.
The upstream pool elevation is: 100 + 8.0 = 108.0 ft. This upstream pool elevation is the same as the design elevation; therefore, the design is now OK.
Calculate the normal depth using ONLINE CHANNEL 06: y_{n} = 3.284 ft.
Calculate the critical depth using ONLINE CHANNEL 07: y_{c} = 4.037 ft.
Since y_{n} < y_{c}, the flow is supercritical.
Since TW = 3.5 > y_{n} = 3.284, there will be a small hydraulic jump at or near the outlet.
Since the flow is supercritical for most of the culvert length, it is concluded that inlet control prevails.
The design diameter is: D = 5 ft = 60 in. ANSWER.

QUESTIONS
What is a culvert?
What is the typical return period for culvert design?
When is a culvert under inlet control?
When is a culvert under outlet control?
List the hydraulic variables affecting culvert flow.
List other considerations in culvert design.
PROBLEMS
Design a circular concrete culvert with the following data: Q = 300 cfs;
inlet invert elevation z_{1} = 100 ft;
tailwater depth y_{2} = 4 ft;
barrel slope S_{o} = 0.02;
barrel length L = 200 ft;
Manning's n = 0.013;
roadway shoulder elevation E_{r} = 112 ft;
upstream freeboard F_{b} = 2 ft.
The entrance type is square edge with headwalls (Fig. 810).
Use ONLINECHANNEL 06
to calculate normal depth and
ONLINECHANNEL 07
to calculate critical depth in the culvert.
Fig. 810 Typical culvert underpass.


Design a circular concrete culvert with the following data: Q = 500 cfs; inlet invert elevation z_{1} = 100 ft;
tailwater depth y_{2} = 4 ft;
barrel slope S_{o} = 0.01;
barrel length L = 200 ft;
Manning's n = 0.013;
roadway shoulder elevation E_{r} = 115 ft;
upstream freeboard F_{b} = 2 ft.
The entrance type is square edge with headwalls. Use ONLINECHANNEL 06
to calculate normal depth and ONLINECHANNEL 07
to calculate critical depth in the culvert.
Verify the culvert design using ONLINECULVERT.
REFERENCES
Chow, V. T. 1959. Openchannel Hydraulics. McGraw Hill, New York.
U.S. Army Corps of Engineers. 2014. HECRAS River Analysis System. Version 4.1, Hydrologic Engineering
Center, Davis, California.
http://openchannelhydraulics.sdsu.edu 

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