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FHWA > Engineering > Hydraulics > HEC 14 > Chapter 10 
Hydraulic Design of Energy Dissipators for Culverts and Channels

C_{o} = 1.4  TW/y_{e} < 0.75 
C_{o} = 4.0(TW/y_{e}) 1.6  0.75 < TW/y_{e} < 1.0 
C_{o} = 2.4  1.0 < TW/y_{e} 
A best fit design relationship that minimizes the RMS error when applied to the experimental data was also developed. Equation 10.1 still applies, but the description of the tailwater parameter, C_{o}, is defined in Equation 10.3. The best fit relationship for Equations 10.1 and 10.3 exhibits a RMS error on the experimental data of 0.56.
(10.3)C_{o} = 2.0  TW/y_{e} < 0.75 
C_{o} = 4.0(TW/y_{e}) 1.0  0.75 < TW/y_{e} < 1.0 
C_{o} = 3.0  1.0 < TW/y_{e} 
Use of the envelope design relationship (Equations 10.1 and 10.2) is recommended when the consequences of failure at or near the design flow are severe. Use of the best fit design relationship (Equations 10.1 and 10.3) is recommended when basin failure may easily be addressed as part of routine maintenance. Intermediate risk levels can be adopted by the use of intermediate values of C_{o}.
Frequency tables for both box culvert data and pipe culvert data of relative length of scour hole (L_{s}/h_{s }< 6, 6 < L_{s}/h _{s}< 7, 7 < L_{s}/h_{s }<8 . . . 25 < L_{s}/h_{s }< 30), with relative tailwater depth TW/y_{e} in increments of 0.03 m (0.1 ft) as a third variable, were constructed using data from 346 experimental runs. For box culvert runs L_{s}/h_{s} was less than 10 for 78% of the data and L_{s}/h_{s} was less than 15 for 98% of the data. For pipe culverts, L_{s}/h_{s} was less than 10 for 91% of the data and, L_{s}/h_{s} was less than 15 for all data. A 3:1 flare angle is recommended for the basins walls. This angle will provide a sufficiently wide energy dissipating pool for good basin operation.
Tailwater influenced formation of the scour hole and performance of the dissipator. For tailwater depths less than 0.75 times the brink depth, scour hole dimensions were unaffected by tailwater. Above this the scour hole became longer and narrower. The tailwater parameter defined in Equations 10.2 and 10.3 captures this observation. In addition, under high tailwater conditions, it is appropriate to estimate the attenuation of the flow velocity downstream of the culvert outlet using Figure 10.3. This attenuation can be used to determine the extent of riprap protection required. HEC 11 (Brown and Clyde, 1989) or the method provided in Section 10.3 can be used for sizing riprap.
Figure 10.3. Distribution of Centerline Velocity for Flow from Submerged Outlets
Based on experience with conventional riprap design, the recommended thickness of riprap for the floor and sides of the basin is 2D_{50} or 1.50D_{max}, where D_{max} is the maximum size of rock in the riprap mixture. Thickening of the riprap layer to 3D_{50} or 2D_{max} on the foreslope of the roadway culvert outlet is warranted because of the severity of attack in the area and the necessity for preventing undermining and consequent collapse of the culvert. Figure 10.1 illustrates these riprap details. The mixture of stone used for riprap and need for a filter should meet the specifications described in HEC 11 (Brown and Clyde, 1989).
The design procedure for a riprap basin is as follows:
Step 1. Compute the culvert outlet velocity, V_{o}, and depth, y_{o}.
For subcritical flow (culvert on mild or horizontal slope), use Figure 3.3 or Figure 3.4 to obtain y_{o}/D, then obtain V_{o} by dividing Q by the wetted area associated with y_{o}. D is the height of a box culvert or diameter of a circular culvert.
For supercritical flow (culvert on a steep slope), Vo will be the normal velocity obtained by using the Manning's Equation for appropriate slope, section, and discharge.
Compute the Froude number, Fr, for brink conditions using brink depth for box culverts (y_{e}=y_{o}) and equivalent depth (y_{e} = (A/2)^{1/2}) for nonrectangular sections.
Step 2. Select D_{50} appropriate for locally available riprap. Determine C_{o} from Equation 10.2 or 10.3 and obtain h_{s}/y_{e} from Equation 10.1. Check to see that h_{s}/D_{50}≥ 2 and D_{50}/y_{e}≥ 0.1. If h_{s}/D_{50} or D_{50}/y_{e} is out of this range, try a different riprap size. (Basins sized where h_{s}/D_{50} is greater than, but close to, 2 are often the most economical choice.)
Step 3. Determine the length of the dissipation pool (scour hole), L_{s}, total basin length, L_{B}, and basin width at the basin exit, W_{B}, as shown in Figures 10.1 and 10.2. The walls and apron of the basin should be warped (or transitioned) so that the cross section of the basin at the exit conforms to the cross section of the natural channel. Abrupt transition of surfaces should be avoided to minimize separation zones and resultant eddies.
Step 4. Determine the basin exit depth, y_{B} = y_{c}, and exit velocity, V_{B} = V_{c} and compare with the allowable exit velocity, V_{allow}. The allowable exit velocity may be taken as the estimated normal velocity in the tailwater channel or a velocity specified based on stability criteria, whichever is larger. Critical depth at the basin exit may be determined iteratively using Equation 7.14:
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + zy_{c})]^{3}/ (W_{B} + 2zy_{c}) by trial and success to determine y_{B}.
V_{c} = Q/A_{c}
z = basin side slope, z:1 (H:V)
If V_{c}≤ V_{allow}, the basin dimensions developed in step 3 are acceptable. However, it may be possible to reduce the size of the dissipator pool and/or the apron with a larger riprap size. It may also be possible to maintain the dissipator pool, but reduce the flare on the apron to reduce the exit width to better fit the downstream channel. Steps 2 through 4 are repeated to evaluate alternative dissipator designs.
Step 5. Assess need for additional riprap downstream of the dissipator exit. If TW/y_{o}≤0.75, no additional riprap is needed. With high tailwater (TW/y_{o}≥ 0.75), estimate centerline velocity at a series of downstream cross sections using Figure 10.3 to determine the size and extent of additional protection. The riprap design details should be in accordance with specifications in HEC 11 (Brown and Clyde, 1989) or similar highway department specifications.
Two design examples are provided. The first features a box culvert on a steep slope while the second shows a pipe culvert on a mild slope.
Design Example: Riprap Basin (Culvert on a Steep Slope) (SI)
Determine riprap basin dimensions using the envelope design (Equations 10.1 and 10.2) for a 2440 mm by 1830 mm reinforced concrete box (RCB) culvert that is in inlet control with supercritical flow in the culvert. Allowable exit velocity from the riprap basin, V_{allow}, is 2.1 m/s. Riprap is available with a D_{50} of 0.50, 0.55, and 0.75 m. Consider two tailwater conditions: 1) TW = 0.85 m and 2) TW = 1.28 m. Given:
Solution
Step 1. Compute the culvert outlet velocity, V_{o}, depth, y_{o}, and Froude number for brink conditions. For supercritical flow (culvert on a steep slope), V_{o} will be V_{n }
y_{o} = y_{e} = 1.22 m
V_{o} = Q/A = 22.7/ [1.22 (2.44)] = 7.63 m/s
Fr = V_{o} / (9.81y_{e})^{1/2} = 7.63/ [9.81(1.22)]^{1/2} = 2.21
Step 2. Select a trial D_{50} and obtain h_{s}/y_{e} from Equation 10.1. Check to see that h_{s}/D_{50}≥ 2 and D_{50}/y_{e}≥ 0.1.
Try D_{50} = 0.55 m; D_{50}/y_{e} = 0.55/1.22 = 0.45 (≥ 0.1 OK)
Two tailwater elevations are given; use the lowest to determine the basin size that will serve the tailwater range, that is, TW = 0.85 m.
TW/y_{e} = 0.85/1.22 = 0.7, which is less than 0.75. Therefore, from Equation 10.2, C_{o} = 1.4
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 1.55 (1.22) = 1.89 m
h_{S}/D_{50} = 1.89/0.55 = 3.4 and h_{S}/D_{50}≥ 2 is satisfied
Step 3. Size the basin as shown in Figures 10.1 and 10.2.
L_{S} = 10h_{S} = 10(1.89) = 18.9 m
L_{S} min = 3W_{o} = 3(2.44) = 7.3 m, use L_{S} = 18.9 m
L_{B} = 15h_{S} = 15(1.89) = 28.4 m
L_{B} min = 4W_{o} = 4(2.44) = 9.8 m, use L_{B} = 28.4 m
W_{B} = W_{o} + 2(L_{B}/3) = 2.44 + 2(28.4/3) = 21.4 m
Step 4. Determine the basin exit depth, y_{B} = y_{c}, and exit velocity, V_{B} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + zy_{c})]^{3}/ (W_{B} + 2zy_{c})
22.7^{2}/9.81 = 52.5 = [y_{c}(21.4 + 2y_{c})]^{3}/ (21.4 + 4y_{c})
By trial and success, y_{c} = 0.48 m, T_{c} = 23.3 m, A_{c} = 10.7 m^{2}
V_{B} = V_{c} = Q/A_{c} = 22.7/10.7 = 2.1 m/s (acceptable)
The initial trial of riprap (D_{50} = 0.55 m) results in a 28.4 m basin that satisfies all design requirements. Try the next larger riprap size to test if a smaller basin is feasible by repeating steps 2 through 4.
Step 2 (2nd iteration). Select riprap size and compute basin depth.
Try D_{50} = 0.75 m; D_{50}/y_{e} = 0.75/1.22 = 0.61 (≥ 0.1 OK)
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 1.09 (1.22) = 1.34 m
h_{S}/D_{50} = 1.34/0.75 = 1.8 and h_{S}/D_{50}≥ 2 is not satisfied. Although not available, try a riprap size that will yield h_{S}/D_{50} close to, but greater than, 2. (A basin sized for smaller riprap may be lined with larger riprap.) Repeat step 2.
Step 2 (3rd iteration). Select riprap size and compute basin depth.
Try D_{50} = 0.71 m; D_{50}/y_{e} = 0.71/1.22 = 0.58 (≥ 0.1 OK)
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 1.16 (1.22) = 1.42 m
h_{S}/D_{50} = 1.42/0.71 = 2.0 and h_{S}/D_{50}≥ 2 is satisfied.
Step 3 (3rd iteration). Size the basin as shown in Figures 10.1 and 10.2.
L_{S} = 10h_{S} = 10(1.42) = 14.2 m
L_{S} min = 3W_{o} = 3(2.44) = 7.3 m, use L_{S} = 14.2 m
L_{B} = 15h_{S} = 15(1.42) = 21.3 m
L_{B} min = 4W_{o} = 4(2.44) = 9.8 m, use L_{B} = 21.3 m
W_{B} = W_{o} + 2(L_{B}/3) = 2.44 + 2(21.3/3) = 16.6 m
However, since the trial D_{50} is not available, the next larger riprap size (D_{50} = 0.75 m) would be used to line a basin with the given dimensions.
Step 4 (3rd iteration). Determine the basin exit depth, y_{B} = y_{c}, and exit velocity, V_{B} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + zy_{c})]^{3}/ (W_{B} + 2zy_{c})
22.7^{2}/9.81 = 52.5 = [y_{c}(16.6 + 2y_{c})]^{3}/ (16.6 + 4y_{c})
By trial and success, y_{c} = 0.56 m, T_{c} = 18.8 m, A_{c} = 9.9 m^{2}
V_{B} = V_{c} = Q/A_{c} = 22.7/9.9 = 2.3 m/s (greater than 2.1 m/s; not acceptable). If the apron were extended (with a continued flare) such that the total basin length was 28.4 m, the velocity would be reduced to the allowable level.
Two feasible options have been identified. First, a 1.89 m deep, 18.9 m long pool, with a 9.5 m apron using D_{50} = 0.55 m. Second, a 1.42 m deep, 14.2 m long pool, with a 14.2 m apron using D_{50} = 0.75 m. Because the overall length is the same, the first option is likely to be more economical.
Step 5. For the design discharge, determine if TW/y_{o} ≤0.75.
For the first tailwater condition, TW/y_{o} = 0.85/1.22 = 0.70, which satisfies TW/y_{o} ≤ 0.75. No additional riprap needed downstream.
For the second tailwater condition, TW/y_{o} = 1.28/1.22 = 1.05, which does not satisfy TW/y_{o} ≤ 0.75. To determine required riprap, estimate centerline velocity at a series of downstream cross sections using Figure 10.3.
Compute equivalent circular diameter, D_{e}, for brink area:
A = π D_{e}^{2} /4 = (y_{o})(W_{o}) = (1.22)(2.44) = 3.00 m^{2}
D_{e} = [3.00(4)/ π ]^{1/2} = 1.95 m
Rock size can be determined using the procedures in Section 10.3 (Equation 10.6) or other suitable method. The computations are summarized below.
L/D_{e}  L (m)  V_{L}/V_{o} (Figure 10.3)  V_{L} (m/s)  Rock Size D_{50} (m) 

10  19.5  0.59  4.50  0.43 
15  29.3  0.42  3.20  0.22 
20  39.0  0.30  2.29  0.11 
21  41.0  0.28  2.13  0.10 
The calculations above continue until V_{L} ≤ V_{allow}. Riprap should be at least the size shown. As a practical consideration, the channel can be lined with the same size rock used for the basin. Protection must extend at least 41.0 m downstream from the culvert brink, which is 12.6 m beyond the basin exit. Riprap should be installed in accordance with details shown in HEC 11.
Design Example: Riprap Basin (Culvert on a Steep Slope) (CU)
Determine riprap basin dimensions using the envelope design (Equations 10.1 and 10.2) for an 8 ft by 6 ft reinforced concrete box (RCB) culvert that is in inlet control with supercritical flow in the culvert. Allowable exit velocity from the riprap basin, V_{allow}, is 7 ft/s. Riprap is available with a D_{50} of 1.67, 1.83, and 2.5 ft. Consider two tailwater conditions: 1) TW = 2.8 ft and 2) TW = 4.2 ft. Given:
Solution
Step 1. Compute the culvert outlet velocity, V_{o}, depth, y_{o}, and Froude number for brink conditions. For supercritical flow (culvert on a steep slope), V_{o} will be V_{n}.
y_{o} = y_{e} = 4 ft
V_{o} = Q/A = 800/ [4 (8)] = 25 ft/s
Fr = V_{o} / (32.2y_{e})^{1/2} = 25/ [32.2(4)]^{1/2} = 2.2
Step 2. Select a trial D_{50} and obtain h_{s}/y_{e} from Equation 10.1. Check to see that h_{s}/D_{50} ≥ 2 and D_{50}/y_{e} ≥ 0.1.
Try D_{50} = 1.83 ft; D_{50}/y_{e} = 1.83/4 = 0.46 (≥ 0.1 OK)
Two tailwater elevations are given; use the lowest to determine the basin size that will serve the tailwater range, that is, TW = 2.8 ft.
TW/y_{e} = 2.8/4 = 0.7, which is less than 0.75. From Equation 10.2, C_{o} = 1.4
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 1.50 (4) = 6.0 ft
h_{S}/D_{50} = 6.0/1.83 = 3.3 and h_{S}/D_{50}≥ 2 is satisfied
Step 3. Size the basin as shown in Figures 10.1 and 10.2.
L_{S} = 10h_{S} = 10(6.0) = 60 ft
L_{S} min = 3W_{o} = 3(8) = 24 ft, use L_{S} = 60 ft
L_{B} = 15h_{S} = 15(6.0) = 90 ft
L_{B} min = 4W_{o} = 4(8) = 32 ft, use L_{B} = 90 ft
W_{B} = W_{o} + 2(L_{B}/3) = 8 + 2(90/3) = 68 ft
Step 4. Determine the basin exit depth, y_{B} = y_{c}, and exit velocity, V_{B} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + zy_{c})]^{3}/ (W_{B} + 2zy_{c})
800^{2}/32.2 = 19,876 = [y_{c}(68 + 2y_{c})]^{3}/ (68 + 4y_{c})
By trial and success, y_{c} = 1.60 ft, T_{c} = 74.4 ft, A_{c} = 113.9 ft^{2}
V_{B} = V_{c} = Q/A_{c} = 800/113.9 = 7.0 ft/s (acceptable)
The initial trial of riprap (D_{50} = 1.83 ft) results in a 90 ft basin that satisfies all design requirements. Try the next larger riprap size to test if a smaller basin is feasible by repeating steps 2 through 4.
Step 2 (2nd iteration). Select riprap size and compute basin depth.
Try D_{50} = 2.5 ft; D_{50}/y_{e} = 2.5/4 = 0.63 (≥ 0.1 OK)
From Equation 10.1,
h_{s} = (h_{s} /y_{e})y_{e} = 1.04 (4) = 4.2 ft
h_{s}/D_{50} = 4.2/2.5 = 1.7 and h_{S}/D_{50}≥ 2 is not satisfied. Although not available, try a riprap size that will yield h_{s}/D_{50} close to, but greater than, 2. (A basin sized for smaller riprap may be lined with larger riprap.) Repeat step 2.
Step 2 (3rd iteration). Select riprap size and compute basin depth.
Try D_{50} = 2.3 ft; D_{50}/y_{e} = 2.3/4 = 0.58 (≥ 0.1 OK)
From Equation 10.1,
h_{s} = (h_{s} /y_{e})y_{e} = 1.15 (4) = 4.6 ft
h_{s}/D_{50} = 4.6/2.3 = 2.0 and h_{s}/D_{50}≥ 2 is satisfied.
Step 3 (3rd iteration). Size the basin as shown in Figures 10.1 and 10.2.
L_{S} = 10h_{S} = 10(4.6) = 46 ft
L_{S} min = 3W_{o} = 3(8) = 24 ft, use L_{S} = 46 ft
L_{B} = 15h_{S} = 15(4.6) = 69 ft
L_{B} min = 4W_{o} = 4(8) = 32 ft, use L_{B} = 69 ft
W_{B} = W_{o} + 2(L_{B}/3) = 8 + 2(69/3) = 54 ft
However, since the trial D_{50} is not available, the next larger riprap size (D_{50} = 2.5 ft) would be used to line a basin with the given dimensions.
Step 4 (3rd iteration). Determine the basin exit depth, y_{B} = y_{c}, and exit velocity, V_{B} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + zy_{c})]^{3}/ (W_{B} + 2zy_{c})
800^{2}/32.2 = 19,876 = [y_{c}(54 + 2y_{c})]^{3}/ (54 + 4y_{c})
By trial and success, y_{c} = 1.85 ft, T_{c} = 61.4 ft, A_{c} = 106.9 ft^{2}
V_{B} = V_{c} = Q/A_{c} = 800/106.9 = 7.5 ft/s (not acceptable). If the apron were extended (with a continued flare) such that the total basin length was 90 ft, the velocity would be reduced to the allowable level.
Two feasible options have been identified. First, a 6ftdeep, 60ftlong pool, with a 30ftapron using D_{50} = 1.83 ft. Second, a 4.6ftdeep, 46ftlong pool, with a 44ftapron using D_{50} = 2.5 ft. Because the overall length is the same, the first option is likely to be more economical.
Step 5. For the design discharge, determine if TW/y_{o} ≤ 0.75.
For the first tailwater condition, TW/y_{o} = 2.8/4.0 = 0.70, which satisfies TW/y_{o} ≤ 0.75. No additional riprap needed downstream.
For the second tailwater condition, TW/y_{o} = 4.2/4.0 = 1.05, which does not satisfy TW/y_{o} ≤ 0.75. To determine required riprap, estimate centerline velocity at a series of downstream cross sections using Figure 10.3.
Compute equivalent circular diameter, D_{e}, for brink area:
A = π D_{e}^{2} /4 = (y_{o})(W_{o}) = (4)(8) = 32 ft^{2}
D_{e} = [32(4)/ π ]^{1/2} = 6.4 ft
Rock size can be determined using the procedures in Section 10.3 (Equation 10.6) or other suitable method. The computations are summarized below.
L/D_{e}  L (ft)  V_{L}/V_{o} (Figure 10.3)  V_{L} (ft/s)  Rock Size D_{50} (ft) 

10  64  0.59  14.7  1.42 
15  96  0.42  10.5  0.72 
20  128  0.30  7.5  0.37 
21  135  0.28  7.0  0.32 
The calculations above continue until V_{L} ≤ V_{allow}. Riprap should be at least the size shown. As a practical consideration, the channel can be lined with the same size rock used for the basin. Protection must extend at least 135 ft downstream from the culvert brink, which is 45 ft beyond the basin exit. Riprap should be installed in accordance with details shown in HEC 11.
Design Example: Riprap Basin (Culvert on a Mild Slope) (SI)
Determine riprap basin dimensions using the envelope design (Equations 10.1 and 10.2) for a pipe culvert that is in outlet control with subcritical flow in the culvert. Allowable exit velocity from the riprap basin, V_{allow}, is 2.1 m/s. Riprap is available with a D_{50} of 0.125, 0.150, and 0.250 m. Given:
Solution
Step 1. Compute the culvert outlet velocity, V_{o}, and depth, y_{o}.
For subcritical flow (culvert on mild slope), use Figure 3.4 to obtain y_{o}/D, then calculate V_{o} by dividing Q by the wetted area for y_{o}.
K_{u} Q/D^{2.5} = 1.81 (3.82)/1.83^{2.5} = 1.53
TW/D = 0.61/1.83 = 0.33
From Figure 3.4, y_{o}/D = 0.45
y_{o} = (y_{o}/D)D = 0.45(1.83) = 0.823 m (brink depth)
From Table B.2, for y_{o} /D = 0.45, the brink area ratio A/D^{2} = 0.343
A = (A/D^{2})D^{2} = 0.343(1.83)^{2 }= 1.15 m^{2}
V_{o} = Q/A = 3.82/1.15 = 3.32 m/s
y_{e} = (A/2)^{1/2} = (1.15/2)^{1/2} = 0.76 m
Fr = V_{o} / [9.81(y_{e})]^{1/2} = 3.32/ [9.81(0.76)]^{1/2} = 1.22
Step 2. Select a trial D_{50} and obtain h_{s}/y_{e} from Equation 10.1. Check to see that h_{s}/D_{50}≥ 2 and D_{50}/y_{e}≥ 0.1.
Try D_{50} = 0.15 m; D_{50}/y_{e} = 0.15/0.76 = 0.20 (≥ 0.1 OK)
TW/y_{e} = 0.61/0.76 = 0.80. Therefore, from Equation 10.2,
C_{o} = 4.0(TW/y_{e}) 1.6 = 4.0(0.80) 1.6 = 1.61
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 0.933 (0.76) = 0.71 m
h_{S}/D_{50} = 0.71/0.15 = 4.7 and h_{S}/D_{50}≥ 2 is satisfied
Step 3. Size the basin as shown in Figures 10.1 and 10.2.
L_{S} = 10h_{S} = 10(0.71) = 7.1 m
L_{S} min = 3W_{o} = 3(1.83) = 5.5 m, use L_{S} = 7.1 m
L_{B} = 15h_{S} = 15(0.71) = 10.7 m
L_{B} min = 4W_{o} = 4(1.83) = 7.3 m, use L_{B} = 10.7 m
W_{B} = W_{o} + 2(L_{B}/3) = 1.83 + 2(10.7/3) = 9.0 m
Step 4. Determine the basin exit depth, y_{B} = y_{c} and exit velocity, V_{B} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + zy_{c})]^{3}/ (W_{B} + 2zy_{c})
3.82^{2}/9.81 = 1.49 = [y_{c}(9.0 + 2y_{c})]^{3}/ (9.0 + 4y_{c})
By trial and success, y_{c} = 0.26 m, T_{c} =10.0 m, A_{c} = 2.48 m^{2}
V_{c} = Q/A_{c} = 3.82/2.48 = 1.5 m/s (acceptable)
The initial trial of riprap (D_{50} = 0.15 m) results in a 10.7 m basin that satisfies all design requirements. Try the next larger riprap size to test if a smaller basin is feasible by repeating steps 2 through 4.
Step 2 (2nd iteration). Select a trial D_{50} and obtain h_{s}/y_{e} from Equation 10.1.
Try D_{50} = 0.25 m; D_{50}/y_{e} = 0.25/0.76 = 0.33 (≥ 0.1 OK)
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 0.320 (0.76) = 0.24 m
h_{S}/D_{50} = 0.24/0.25 = 0.96 and h_{S}/D_{50}≥ 2 is not satisfied. Although not available, try a riprap size that will yield h_{S}/D_{50} close to, but greater than 2. (A basin sized for smaller riprap may be lined with larger riprap.) Repeat step 2.
Step 2 (3rd iteration). Select a trial D_{50} and obtain h_{s}/y_{e} from Equation 10.1.
Try D_{50} = 0.205 m; D_{50}/y_{e} = 0.205/0.76 = 0.27 (≥ 0.1 OK)
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 0.545 (0.76) = 0.41 m
h_{S}/D_{50} = 0.41/0.205 = 2.0 and h_{S}/D_{50}≥ 2 is satisfied. Continue to step 3.
Step 3 (3rd iteration). Size the basin as shown in Figures 10.1 and 10.2.
L_{S} = 10h_{S} = 10(0.41) = 4.1 m
L_{S} min = 3W_{o} = 3(1.83) = 5.5 m, use L_{S} = 5.5 m
L_{B} = 15h_{S} = 15(0.41) = 6.2 m
L_{B} min = 4W_{o} = 4(1.83) = 7.3 m, use L_{B} = 7.3 m
W_{B} = W_{o} + 2(L_{B}/3) = 1.83 + 2(7.3/3) = 6.7 m
However, since the trial D_{50} is not available, the next larger riprap size (D_{50} = 0.25 m) would be used to line a basin with the given dimensions.
Step 4 (3rd iteration). Determine the basin exit depth, y_{B} = y_{c} and exit velocity, V_{B} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + zy_{c})]^{3}/ (W_{B} + 2zy_{c})
3.82^{2}/9.81 = 1.49 = [y_{c}(6.7 + 2y_{c})]^{3}/ (6.7 + 4y_{c})
By trial and success, y_{c} = 0.31 m, T_{c} =7.94 m, A_{c} = 2.28 m^{2}
V_{c} = Q/A_{c} = 3.82/2.28 = 1.7 m/s (acceptable)
Two feasible options have been identified. First, a 0.71 m deep, 7.1 m long pool, with an 3.6 m apron using D_{50} = 0.15 m. Second, a 0.41 m deep, 5.5 m long pool, with a 1.8 m apron using D_{50} = 0.25 m. The choice between these two options will likely depend on the available space and the cost of riprap.
Step 5. For the design discharge, determine if TW/y_{o} ≤ 0.75
TW/y_{o} = 0.61/0.823 = 0.74, which satisfies TW/y_{o} ≤ 0.75. No additional riprap needed.
Design Example: Riprap Basin (Culvert on a Mild Slope) (CU)
Determine riprap basin dimensions using the envelope design (Equations 10.1 and 10.2) for a pipe culvert that is in outlet control with subcritical flow in the culvert. Allowable exit velocity from the riprap basin, V_{allow}, is 7.0 ft/s. Riprap is available with a D_{50} of 0.42, 0.50, and 0.83 ft. Given:
Solution
Step 1. Compute the culvert outlet velocity, V_{o}, depth, y_{o} and Froude number.
For subcritical flow (culvert on mild slope), use Figure 3.4 to obtain y_{o}/D, then calculate V_{o} by dividing Q by the wetted area for y_{o}.
K_{u}Q/D^{2.5} = 1.0(135)/6^{2.5} = 1.53
TW/D = 2.0/6 = 0.33
From Figure 3.4, y_{o}/D = 0.45
y_{o} = (y_{o}/D)D = 0.45(6) = 2.7 ft (brink depth)
From Table B.2 for y_{o}/D = 0.45, the brink area ratio A/D^{2} = 0.343
A = (A/D^{2})D^{2} = 0.343(6)^{2 }= 12.35 ft^{2}
V_{o} = Q/A = 135/12.35 = 10.9 ft/s
y_{e} = (A/2)^{1/2} = (12.35/2)^{1/2} = 2.48 ft
Fr = V_{o} / [32.2(y_{e})]^{1/2} = 10.9/ [32.2(2.48)]^{1/2} = 1.22
Step 2. Select a trial D_{50} and obtain h_{s}/y_{e} from Equation 10.1. Check to see that h_{s}/D_{50}≥ 2 and D_{50}/y_{e}≥ 0.1.
Try D_{50} = 0.5 ft; D_{50}/y_{e} = 0.5/2.48 = 0.20 (≥ 0.1 OK)
TW/y_{e} = 2.0/2.48 = 0.806. Therefore, from Equation 10.2,
C_{o} = 4.0(TW/y_{e}) 1.6 = 4.0(0.806) 1.6 = 1.62
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 0.923 (2.48) = 2.3 ft
h_{S}/D_{50} = 2.3/0.5 = 4.6 and h_{S}/D_{50}≥ 2 is satisfied
Step 3. Size the basin as shown in Figures 10.1 and 10.2.
L_{S} = 10h_{S} = 10(2.3) = 23 ft
L_{S} min = 3W_{o} = 3(6) = 18 ft, use L_{S} = 23 ft
L_{B} = 15h_{S} = 15(2.3) = 34.5 ft
L_{B} min = 4W_{o} = 4(6) = 24 ft, use L_{B} = 34.5 ft
W_{B} = W_{o} + 2(L_{B}/3) = 6 + 2(34.5/3) = 29 ft
Step 4. Determine the basin exit depth, y_{B} = y_{c} and exit velocity, V_{B} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + zy_{c})]^{3}/ (W_{B} + 2zy_{c})
135^{2}/32.2 = 566 = [y_{c}(29 + 2y_{c})]^{3}/ (29 + 4y_{c})
By trial and success, y_{c} = 0.86 ft, T_{c} =32.4 ft, A_{c} = 26.4 ft^{2}
V_{c} = Q/A_{c} = 135/26.4 = 5.1 ft/s (acceptable)
The initial trial of riprap (D_{50} = 0.5 ft) results in a 34.5 ft basin that satisfies all design requirements. Try the next larger riprap size to test if a smaller basin is feasible by repeating steps 2 through 4.
Step 2 (2nd iteration). Select a trial D_{50} and obtain h_{s}/y_{e} from Equation 10.1.
Try D_{50} = 0.83 ft; D_{50}/y_{e} = 0.83/2.48 = 0.33 (≥ 0.1 OK)
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 0.311 (2.48) = 0.8 ft
h_{S}/D_{50} = 0.8/0.83 = 0.96 and h_{S}/D_{50}≥ 2 is not satisfied. Although not available, try a riprap size that will yield h_{S}/D_{50} close to, but greater than 2. (A basin sized for smaller riprap may be lined with larger riprap.) Repeat step 2.
Step 2 (3rd iteration). Select a trial D_{50} and obtain h_{s}/y_{e} from Equation 10.1.
Try D_{50} = 0.65 ft; D_{50}/y_{e} = 0.65/2.48 = 0.26 (≥ 0.1 OK)
From Equation 10.1,
h_{S} = (h_{S} /y_{e})y_{e} = 0.581 (2.48) = 1.4 ft
h_{S}/D_{50} = 1.4/0.65 = 2.15 and h_{S}/D_{50}≥ 2 is satisfied. Continue to step 3.
Step 3 (3rd iteration). Size the basin as shown in Figures 10.1 and 10.2.
L_{S} = 10h_{S} = 10(1.4) = 14 ft
L_{S} min = 3W_{o} = 3(6) = 18 ft, use L_{S} = 18 ft
L_{B} = 15h_{S} = 15(1.4) = 21 ft
L_{B} min = 4W_{o} = 4(6) = 24 ft, use L_{B} = 24 ft
W_{B} = W_{o} + 2(L_{B}/3) = 6 + 2(24/3) = 22 ft
However, since the trial D_{50} is not available, the next larger riprap size (D_{50} = 0.83 ft) would be used to line a basin with the given dimensions.
Step 4 (3rd iteration). Determine the basin exit depth, y_{B} = y_{c} and exit velocity, V_{B} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + zy_{c})]^{3}/ (W_{B} + 2zy_{c})
135^{2}/32.2 = 566 = [y_{c}(22 + 2y_{c})]^{3}/ (22 + 4y_{c})
By trial and success, y_{c} = 1.02 ft, T_{c} =26.1 ft, A_{c} = 24.5 ft^{2}
V_{c} = Q/A_{c} = 135/24.5 = 5.5 ft/s (acceptable)
Two feasible options have been identified. First, a 2.3ftdeep, 23ftlong pool, with an 11.5ftapron using D_{50} = 0.5 ft. Second, a 1.4ftdeep, 18ftlong pool, with a 6ftapron using D_{50} = 0.83 ft. The choice between these two options will likely depend on the available space and the cost of riprap.
Step 5. For the design discharge, determine if TW/y_{o} ≤0.75
TW/y_{o} = 2.0/2.7 = 0.74, which satisfies TW/y_{o } ≤ 0.75. No additional riprap needed.
The most commonly used device for outlet protection, primarily for culverts 1500 mm (60 in) or smaller, is a riprap apron. An example schematic of an apron taken from the Federal Lands Division of the Federal Highway Administration is shown in Figure 10.4.
Figure 10.4. Placed Riprap at Culverts (Central Federal Lands Highway Division)
They are constructed of riprap or grouted riprap at a zero grade for a distance that is often related to the outlet pipe diameter. These aprons do not dissipate significant energy except through increased roughness for a short distance. However, they do serve to spread the flow helping to transition to the natural drainage way or to sheet flow where no natural drainage way exists. However, if they are too short, or otherwise ineffective, they simply move the location of potential erosion downstream. The key design elements of the riprap apron are the riprap size as well as the length, width, and depth of the apron.
Several relationships have been proposed for riprap sizing for culvert aprons and several of these are discussed in greater detail in Appendix D. The independent variables in these relationships include one or more of the following variables: outlet velocity, rock specific gravity, pipe dimension (e.g. diameter), outlet Froude number, and tailwater. The following equation (Fletcher and Grace, 1972) is recommended for circular culverts:
(10.4)where,
D_{50} = riprap size, m (ft)
Q = design discharge, m^{3}/s (ft^{3}/s)
D = culvert diameter (circular), m (ft)
TW = tailwater depth, m (ft)
g = acceleration due to gravity, 9.81 m/s^{2} (32.2 ft/s^{2})
Tailwater depth for Equation 10.4 should be limited to between 0.4D and 1.0D. If tailwater is unknown, use 0.4D.
Whenever the flow is supercritical in the culvert, the culvert diameter is adjusted as follows:
(10.5)where,
D' = adjusted culvert rise, m (ft)
y_{n} = normal (supercritical) depth in the culvert, m (ft)
Equation 10.4 assumes that the rock specific gravity is 2.65. If the actual specific gravity differs significantly from this value, the D_{50} should be adjusted inversely to specific gravity.
The designer should calculate D_{50} using Equation 10.4 and compare with available riprap classes. A project or design standard can be developed such as the example from the Federal Highway Administration Federal Lands Highway Division (FHWA, 2003) shown in Table 10.1 (first two columns). The class of riprap to be specified is that which has a D_{50} greater than or equal to the required size. For projects with several riprap aprons, it is often cost effective to use fewer riprap classes to simplify acquiring and installing the riprap at multiple locations. In such a case, the designer must evaluate the tradeoffs between over sizing riprap at some locations in order to reduce the number of classes required on a project.
Class  D_{50} (mm)  D_{50} (in)  Apron Length^{1}  Apron Depth 

1  125  5  4D  3.5D_{50} 
2  150  6  4D  3.3D_{50} 
3  250  10  5D  2.4D_{50} 
4  350  14  6D  2.2D_{50} 
5  500  20  7D  2.0D_{50} 
6  550  22  8D  2.0D_{50} 
^{1}D is the culvert rise. 
The apron dimensions must also be specified. Table 10.1 provides guidance on the apron length and depth. Apron length is given as a function of the culvert rise and the riprap size. Apron depth ranges from 3.5D_{50} for the smallest riprap to a limit of 2.0D_{50} for the larger riprap sizes. The final dimension, width, may be determined using the 1:3 flare shown in Figure 10.4 and should conform to the dimensions of the downstream channel. A filter blanket should also be provided as described in HEC 11 (Brown and Clyde, 1989).
For tailwater conditions above the acceptable range for Equation 10.4 (TW > 1.0D), Figure 10.3 should be used to determine the velocity downstream of the culvert. The guidance in Section 10.3 may be used for sizing the riprap. The apron length is determined based on the allowable velocity and the location at which it occurs based on Figure 10.3.
Over their service life, riprap aprons experience a wide variety of flow and tailwater conditions. In addition, the relations summarized in Table 10.1 do not fully account for the many variables in culvert design. To ensure continued satisfactory operation, maintenance personnel should inspect them after major flood events. If repeated severe damage occurs, the location may be a candidate for extending the apron or another type of energy dissipator.
Design Example: Riprap Apron (SI)
Design a riprap apron for the following CMP installation. Available riprap classes are provided in Table 10.1. Given:
Solution
Step 1. Calculate D_{50} from Equation 10.4. First verify that tailwater is within range.
TW/D = 0.5/1.5 = 0.33. This is less than 0.4D, therefore,
use TW = 0.4D = 0.4(1.5) = 0.6 m
Step 2. Determine riprap class. From Table 10.1, riprap class 2 (D_{50} = 0.15 m) is required.
Step 3. Estimate apron dimensions.
From Table 10.1 for riprap class 2,
Length, L = 4D = 4(1.5) = 6 m
Depth = 3.3D_{50} = 3.3 (0.15) = 0.50 m
Width (at apron end) = 3D + (2/3)L = 3(1.5) + (2/3)(6) = 8.5 m
Design Example: Riprap Apron (CU)
Design a riprap apron for the following CMP installation. Available riprap classes are provided in Table 10.1. Given:
Solution
Step 1. Calculate D_{50} from Equation 10.4. First verify that tailwater is within range.
TW/D = 1.6/5.0 = 0.32. This is less than 0.4D, therefore,
use TW = 0.4D = 0.4(5) = 2.0 ft
Step 2. Determine riprap class. From Table 10.1, riprap class 2 (D_{50} = 6 in) is required.
Step 3. Estimate apron dimensions.
From Table 10.1 for riprap class 2,
Length, L = 4D = 4(5) = 20 ft
Depth = 3.3D_{50} = 3.3 (6) = 1.65 ft
Width (at apron end) = 3D + (2/3)L = 3(5) + (2/3)(20) = 28.3 ft
Some energy dissipators provide exit conditions, velocity and depth, near critical. This flow condition rapidly adjusts to the downstream or natural channel regime; however, critical velocity may be sufficient to cause erosion problems requiring protection adjacent to the energy dissipator. Equation 10.6 provides the riprap size recommended for use downstream of energy dissipators. This relationship is from Searcy (1967) and is the same equation used in HEC 11 (Brown and Clyde, 1989) for riprap protection around bridge piers.
(10.6)where,
D_{50} = median rock size, m (ft)
V = velocity at the exit of the dissipator, m/s (ft/s)
S = riprap specific gravity
The length of protection can be judged based on the magnitude of the exit velocity compared with the natural channel velocity. The greater this difference, the longer will be the length required for the exit flow to adjust to the natural channel condition. A filter blanket should also be provided as described in HEC 11 (Brown and Clyde, 1989).
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Cynthia Nurmi
Resource Center (Atlanta)
4045623908
cynthia.nurmi@dot.gov