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.
Use ONLINECHANNEL 06
to calculate normal depth and
ONLINECHANNEL 07
to calculate critical depth in the culvert.
Fig. 811 Typical culvert underpass.


The upstream design pool elevation
[HW elevation] is = 112  2 = 110 ft.
The downstream invert elevation is z_{2} =
100  (S_{o} 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 + [K_{e}
+ K_{E} + f_{D} (L /D)]
[V^{ 2}/(2g)]
From Chapter 5, Problem 1:
f_{D} = 8 g n^{ 2} / (k^{ 2} R^{ 1/3}) = 116.55 n^{ 2} / R^{ 1/3}
f_{D} = 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. 810, 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. 810, 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.
y_{c} = 4.766 ft.
Use ONLINECHANNEL 06
to calculate normal depth in the culvert.
y_{n} = 3.171 ft.
The flow is supercritical.
Since TW = 4 > y_{n} = 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.
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.
The upstream design pool elevation [HW elevation] is = 115  2 = 113 ft.
The downstream invert elevation is z_{2} = 100  (S_{o} 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 + [K_{e} + K_{E} + f_{D} (L /D)]
[V^{ 2}/(2g)]
f_{D} = 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. 810, 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. 810, 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. 108, 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.
y_{c} = 5.824 ft.
Use ONLINECHANNEL 06
to calculate normal depth in the culvert.
y_{n} = 4.661 ft.
The flow is supercritical.
Since TW = 4 < y_{n} = 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.