QUESTIONS

1. What is a culvert?

Culverts are hydraulically short drainage conduits placed in locations where the drainage network intersects the transportation network.

2. What is the typical return period for culvert design?

10 to 50 years.

3. When is a culvert under inlet control?

Culvert flow is under inlet control when the discharge depends only on the conditions at the inlet. Inlet control occurs when the culvert barrel is capable of conveying more discharge than the inlet will allow.

4. When is a culvert under outlet control?

Outlet control occurs when TW > 1.2 D, i.e., when the tailwater exceeds by 20% the culvert diameter. In this case, the culvert barrel will be completely full of water, resembling closed-conduit flow. The headwater may be computed by applying the energy equation from the upstream pool elevation to the downstream pool elevation.

5. List the hydraulic variables affecting culvert flow.

(1) Cross-sectional size and shape, (2) bottom slope, (3) barrel length, (4) roughness, and (5) entrance and exit characteristics.

6. List other considerations in culvert design.

(1) Piping in the embankment surrounding the culvert, (2) local scour at culvert outlet, (3) erosion of fill material near the inlet, (4) clogging with excessive debris, and (5) provision for fish passage.

PROBLEMS

1. Design a circular concrete culvert with the following data:  Q = 300 cfs; inlet invert elevation z1 = 100 ft; tailwater depth y2 = 4 ft; barrel slope So = 0.02; barrel length L = 200 ft; Manning's n = 0.013; roadway shoulder elevation Er = 112 ft; upstream freeboard Fb = 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.

Fig. 8-11  Typical culvert underpass.

The upstream design pool elevation [HW elevation] is = 112 - 2 = 110 ft.

The downstream invert elevation is z2 = 100 - (So L) = 100 - (0.02 × 200) = 96 ft.

The downstream pool elevation [TW elevation] is = 96 + 4 = 100 ft.

First assume outlet control, and apply the energy equation between u/s and d/s. Velocities are zero in the u/s and d/s pools.

110 = 100 + [Ke + KE + fD (L /D)] [V 2/(2g)]

From Chapter 5, Problem 1:

fD = 8 g n 2 / (k 2 R 1/3) = 116.55 n 2 / R 1/3

fD = 116.55 n 2 / R 1/3 = 185.01 n 2 / D 1/3

10 = [0.5 + 1.0 + (185.01 n 2/ D 1/3) (200 / D)] [V 2/(2g)]

10 = [1.5 + (6.253 / D 4/3)] [V 2/(2g)]

[V 2/(2g)] = Q 2/ { [(π/4)D 2] 2 2g }   = (300)2/(0.6168 × 64.34 × D 4) = 2268 / D 4

10 = [1.5 + (6.253/D 4/3)] (2268 / D 4)

Solve by trial and error: D = 4.8 ft. Assume D = 5 ft = 60 in.

From Fig. 8-10, HW/D = 2.6

HW depth = 2.6 × 5.0 = 13.0 ft.

HW elevation = 100 + 13.0 = 113 ft > 110 ft       [Too high!]

Next assume D = 5.5 ft = 66 in.

From Fig. 8-10, HW/D = 1.9

HW depth = 1.9 × 5 = 9.5 ft.

HW elevation = 100 + 9.5 = 109.5 ft < 110 ft       [OK]

Use ONLINECHANNEL 07 to calculate critical depth in the culvert.

yc = 4.766 ft.

Use ONLINECHANNEL 06 to calculate normal depth in the culvert.

yn = 3.171 ft.

The flow is supercritical.

Since TW = 4 > yn = 3.171, there will be a hydraulic jump in the culvert.

Since the outlet is open to the atmosphere, there is inlet control, but with HJ at the outlet.

2. Design a circular concrete culvert with the following data:  Q = 500 cfs; inlet invert elevation z1 = 100 ft; tailwater depth y2 = 4 ft; barrel slope So = 0.01; barrel length L = 200 ft; Manning's n = 0.013; roadway shoulder elevation Er = 115 ft; upstream freeboard Fb = 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.

The upstream design pool elevation [HW elevation] is = 115 - 2 = 113 ft.

The downstream invert elevation is z2 = 100 - (So L) = 100 - (0.01 × 200) = 98 ft.

The downstream pool elevation [TW elevation] is = 98 + 4 = 102 ft.

First assume outlet control, and apply the energy equation between u/s and d/s. Velocities are zero in the u/s and d/s pools.

113 = 102 + [Ke + KE + fD (L /D)] [V 2/(2g)]

From Chapter 5, Problem 1:

fD = 8 g n 2 / (k 2 R 1/3) = 116.55 n 2 / R 1/3

fD = 116.55 n 2 / R 1/3 = 185.01 n 2 / D 1/3

11 = [0.5 + 1.0 + (185.01 n 2 / D 1/3) (200 /D)] [V 2/(2g)]

11 = [1.5 + (6.253 /D 4/3)] [V 2/(2g)]

[V 2/(2g)] = Q 2/ { [(π/4) D 2] 2 2g }   = (500)2/(0.6168 × 64.34 × D 4) = 6279 / D 4

11 = [1.5 + (6.253 /D 4/3)] (6279 / D 4)

Solve by trial and error: D = 5.9 ft. Assume D = 6 ft = 72 in.

From Fig. 8-10, HW/D = 2.7

HW depth = 2.7 × 6 = 16.2 ft.

HW elevation = 100 + 16.2 = 116.2 ft > 113 ft       [Too high!]

Next assume D = 6.5 ft = 78 in.

From Fig. 8-10, HW/D = 2.1

HW depth = 2.1 × 6.5 = 13.6 ft.

HW elevation = 100 + 13.6 = 113.6 ft > 113 ft       [Too high!]

Next assume D = 7 ft = 84 in.

From Fig. 10-8, HW/D = 1.6

HW depth = 1.6 × 7 = 11.2 ft.

HW elevation = 100 + 11.2 = 111.2 ft < 113 ft       [OK]

Use ONLINECHANNEL 07 to calculate critical depth in the culvert.

yc = 5.824 ft.

Use ONLINECHANNEL 06 to calculate normal depth in the culvert.

yn = 4.661 ft.

The flow is supercritical.

Since TW = 4 < yn = 4.661, there will be no hydraulic jump in the culvert.

Since the outlet is open to the atmosphere, there is inlet control.

The same results are obtained using ONLINECULVERT.

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140822 11:00