CULVERT HYDRAULICS
ROBERSON ET AL., WITH ADDITIONS

CULVERTS
A CULVERT IS A CONDUIT PLACED UNDER A FILL, SUCH AS A HIGHWAY EMBANKMENT, OR UNDER A ROAD, TO CONVEY STREAMFLOW.

Culvert at intersection of Campo Creek with Highway 94, San Diego County.

CULVERTS ARE DESIGNED TO PASS THE DESIGN DISCHARGE WITHOUT OVERTOPPING.
THE DESIGN STORM MAY BE A 50-YR STORM.
THE FLOW IN A CULVERT IS A FUNCTION OF:

CROSS-SECTIONAL SIZE AND SHAPE
SLOPE
LENGTH
ROUGHNESS
ENTRANCE AND EXIT DESIGN.

FLOW IN A CULVERT MAY BE FREE SURFACE (OPEN CHANNEL), CLOSED CONDUIT (PIPE FLOW), OR BOTH.
HEADWATER AND TAILWATER ARE MAJOR FACTORS THAT DICTATE WHETHER THE CULVERT FLOWS PARTIALLY OR COMPLETELY FULL.
HEADWATER (HW) IS THE DEPTH OF WATER ABOVE THE INVERT OF THE CULVERT AT THE INLET.
TAILWATER (TW) IS THE DEPTH OF WATER ABOVE THE INVERT OF THE CULVERT AT THE OUTLET.

DESIGN OBJECTIVE: MOST ECONOMICAL DESIGN THAT WILL PASS THE DESIGN DISCHARGE WITHOUT EXCEEDING A SPECIFIED HEADWATER ELEVATION.
DESIGN DEPENDS ON WHETHER THE CULVERT IS UNDER INLET OR OUTLET CONTROL.

INLET CONTROL
ASSUME DIAMETER D, LENGTH L, AND SLOPE S OF CULVERT (SLOPE OF ORIGINAL STREAM) ARE KNOWN.
COMPUTE NORMAL DEPTH y_n AND CRITICAL FLOW DEPTH y_c.
FIND THE TAILWATER DEPTH TW FROM STREAMFLOW RECORDS.
IF TW < D, AND D > y_c > y_n, THE FLOW IS OPEN CHANNEL (FREE SURFACE), SUPERCRITICAL, WITH OUTLET OPEN TO THE ATMOSPHERE.
BECAUSE THE FLOW IS SUPERCRITICAL, THE TAILWATER HAS NO INFLUENCE ON THE UPSTREAM CONDITIONS.
THE HEADWATER IS SOLELY CONTROLLED BY THE CONDITIONS AT THE INLET.
DISCHARGE DEPENDS ONLY ON CONDITIONS AT THE INLET.

OCCURRENCE OF INLET CONTROL
INLET CONTROL OCCURS WHEN THE MAIN PART OF THE CULVERT IS CAPABLE OF CONVEYING MORE DISCHARGE THAT THE INLET WILL ALLOW.
THE CONTROL SECTION UNDER INLET CONTROL IS LOCATED JUST INSIDE THE ENTRANCE OF THE CULVERT.
THE FLOW PASSES THROUGH CRITICAL DEPTH AT THE INLET AND BECOMES SUPERCRITICAL DOWNSTREAM OF THE INLET.
FOR INLET CONTROL, ASUMME THAT THE CULVERT ACTS AS A SLUICE GATE OR AS A WEIR.
THE INLET CONTROL CAPACITY DEPENDS PRIMARILY ON THE GEOMETRY OF THE CULVERT ENTRANCE.
THE INLET CONFIGURATION AND DISCHARGE RATE ARE THE MAIN FACTORS THAT CONTROL THE WATER SURFACE ELEVATION U/S OF THE INLET.
IF INLET IS OPEN TO THE ATMOSPHERE, CONDITIONS ARE LIKE WEIR FLOW.
IF INLET IS SUBMERGED (A COMMON DESIGN SITUATION), CONDITIONS ARE LIKE ORIFICE FLOW.
IF THE HEADWATER IS LESS THAN 1.2D, THE INLET WILL BE UNSUBMERGED.
IF INLET IS OPEN TO THE ATMOSPHERE, BUT THE OUTLET IS SUBMERGED, A HYDRAULIC JUMP WILL FORM INSIDE THE CULVERT.

OUTLET CONTROL
IF TW > 1.2D, THE CULVERT WILL BE COMPLETELY FULL OF WATER.
THE HEADWATER CAN BE COMPUTED BY APPLYING THE ENERGY EQUATION FROM THE U/S POOL TO THE D/S POOL.
THE HEADWATER IS DIRECTLY CONTROLLED BY THE WATER SURFACE ELEVATION AT THE OUTLET.
DISCHARGE DEPENDS ON THE ENTIRE CULVERT AND CONDITIONS AT THE OUTLET.
OCCURRENCE OF OUTLET CONTROL
OUTLET CONTROL OCCURS WHEN INLET AND OUTLET ARE SUBMERGED.
OUTLET CONTROL ALSO OCCURS WHEN THE PIPE HAS A MILD SLOPE (SUBCRITICAL FLOW) AND BOTH TW AND HW ARE LESS THAN D.
IN THIS CASE IT IS BEST TO CALCULATE THE W.S.P.

CULVERT DESIGN
STEPS:

ASSEMBLE INITIAL DESIGN DATA

DESIGN DISCHARGE
TAILWATER ELEVATION
SLOPE OF CULVERT

MAKE INITIAL CHOICE OF CULVERT

LENGTH
CROSS-SECTIONAL SHAPE (CIRCULAR, SQUARE, RECTANGULAR, ARCH)
SIZE (D IF CIRCULAR)
KIND OF MATERIAL (CONCRETE, CORRUGATE STEEL)
TYPE OF ENTRANCE (SQUARE EDGED OR ROUNDED)

TRY TO ASCERTAIN THE TYPE OF CONTROL (INLET OR OUTLET), BASED ON HW, TW, D AND SLOPE.
IF INLET CONTROL PREVAILS, CALCULATE THE REQUIRED WATER SURFACE ELEVATION IN THE HEADWATER TO PASS THE DESIGN DISCHARGE.
IF OUTLET CONTROL PREVAILS, CALCULATE THE REQUIRED WATER SURFACE ELEVATION IN THE UPSTREAM POOL BY THE ENERGY EQUATION OR W.S.P. COMPUTATIONS.
IF WATER SURFACE ELEVATION IN THE HEADWATER IS GREATER THAN ALLOWED, CHOOSE A LARGER SIZE CULVERT AND REPEAT THE PROCESS.

FOR SOME DATA IT MAY BE HARD TO DETERMINE A PRIORI THE TYPE OF CONTROL.
IN THIS CASE MAKE BOTH CALCULATIONS.
THE ONE YIELDING THE GREATEST WATER SURFACE ELEVATION U/S WILL BE THE CONTROLLING ONE.
OTHER DESIGN FACTORS:

PIPING IN EMBANKMENT.
LOCAL SCOUR AT OUTLET.
EROSION OF FILL MATERIAL NEAR INLET (METAL CULVERTS ARE PARTICULARLY SUSCEPTIBLE).
FISH PASSAGE.
CLOGGING BY DEBRIS.

HEC-RAS CULVERT HYDRAULICS

In general, the higher upstream energy "controls" and determines the type of flow in the culvert for a given flow rate and tailwater condition.
For outlet control, the required upstream energy is computed by performing an energy balance from the downstream section to the upstream section.
The HEC-RAS culvert routines consider entrance losses, friction losses in the culvert barrel, and exit losses at the outlet in computing the outlet control headwater of the culvert.
During the computations, if the inlet control answer comes out higher than the outlet control answer, the program will perform some additional computations to evaluate if the inlet control answer can actually persist through the culvert without pressurizing the culvert barrel.
The assumption of inlet control is that the flow passes through critical depth near the culvert inlet and transitions into supercritical flow.
If the flow persists as low flow through the length of the culvert barrel, then inlet control is assumed to be valid.
If the flow goes through a hydraulic jump inside the barrel, and fully develops the entire area of the culvert, it is assumed that this condition will cause the pipe to pressurize over the entire length of the culvert barrel and thus act more like an orifice type of flow.
If this occurs, then the outlet control answer (under the assumption of a full flowing barrel) is used instead of the inlet control answer.

EXAMPLE
A CONCRETE CULVERT IS TO BE DESIGNED FOR THE 25-YR FLOOD.
THE DESIGN DISCHARGE IS: Q = 200 CFS.
THE INLET INVERT ELEVATION IS: z₁ = 100 FT.
MANNING'S n = 0.013
THE NATURAL STREAM BED SLOPE IS: S₀ = 0.01.
THE TAILWATER DEPTH ABOVE OUTLET INVERT IS: y₂ = 3.5 FT.
THE CULVERT LENGTH IS: L = 200 FT.
THE ROADWAY SHOULDER ELEVATION IS 110 FT.

SOLUTION
ASSUME A 2 FT FREEBOARD BETWEEN U/S DESIGN W.S. ELEVATION IN ROAD SHOULDER.
THEREFORE, THE DESIGN ELEVATION FOR U/S POOL IS: 110 - 2 = 108 FT.
ASSUME A CIRCULAR CONCRETE PIPE, SQUARE EDGE WITH HEADWALLS.
FIRST ASSUME OUTLET CONTROL.
ASSUME THAT THE H.G.L. IS AT THE ELEVATION OF THE D/S POOL.
CALCULATE OUTLET INVERT ELEVATION: z₂ = 100 - (0.01 × 200) = 100 - 2 = 98 FT.
CALCULATE D/S POOL ELEVATION: 98 + 3.5 = 101.5 FT.
SET UP ENERGY EQUATION: z₁ + y₁ + V₁²/(2g) = z₂ + y₂ + V₂²/(2g) + ∑h_L

ASSUME V₁ = 0 [VELOCITY IS ZERO IN THE U/S POOL]
ASSUME V₂ = 0 [VELOCITY DISSIPATES TO ZERO IN THE D/S POOL]
∑h_L = [K_e + K_E + f(L/D)] V²/(2g)
ASSUME K_e = 0.5; K_E = 1. [TABLE 5-3]
f = 8 g n² / (k² R^1/3)
where k = 1.486
f = 116.5 n² / R^1/3
R = D / 4
f = 184.93 n² / D^1/3
ENERGY EQUATION: 108 = 101.5 + [0.5 + 1.0 + (184.93 n² L / D^1/3) ] V²/(2g)
6.5 = [1.5 + (6.251 / D^4/3) ] V²/(2g)
V = Q/A = 200/A = 200/[(π/4)D²]
V²/(2g) = { 200²/[(π/4)²D⁴] } / (2g)
6.5 = 200² [1.5 + (6.251 / D^4/3) ] (1008 / D⁴)
SOLVE BY TRIAL AND ERROR: D = 4.38 FT.
CHOOSE NEXT LARGER SIZE: D = 4.5 FT.
WITH Q = 200, AND D = 4.5 FT = 54 IN, ENTER FIG. 7-5 TO FIND RATIO HEADWATER DEPTH OVER DEPTH HW/D = 2.2

HW DEPTH = 2.2 × 4.5 = 9.9 FT.
U/S POOL ELEVATION = 100 + 9.9 = 109.9 LARGER THAN 108. TOO LARGE.
THE CHOSEN D = 4.5 FT IS TOO SMALL.
TRY D = 5 FT.
WITH Q = 200, AND D = 5 FT = 60 IN, ENTER FIG. 7-5 TO FIND RATIO HEADWATER DEPTH OVER DEPTH HW/D = 1.6
HW DEPTH = 1.6 × 5 = 8 FT.
U/S POOL ELEVATION = 100 + 8 = 108 SAME AS 108. OK.
CHECK CRITICAL DEPTH WITH ONLINECHANNEL 07.

THUS: y_c = 4.037 FT.
CHECK NORMAL DEPTH WITH ONLINECHANNEL 06.
THUS: y_n = 3.28 FT.
BECAUSE y_n < y_c, THE FLOW IS SUPERCRITICAL.
BECAUSE TW > y_n, THERE WILL BE A SMALL HYDRAULIC JUMP NEAR THE OUTLET.
BECAUSE THE FLOW IS SUPERCRITICAL IN THE PIPE, WE CAN CONCLUDE THAT FOR THE MOST PART THERE IS INLET CONTROL.
DESIGN DIAMETER: D = 5 FT = 60 IN.

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