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FHWA > Engineering > Hydraulics > HEC 14 > Chapter 9 
Hydraulic Design of Energy Dissipators for Culverts and Channels

W_{B}/W_{o}  2 to 4  5  6  7  8  
W_{1}/W_{o}  0.57  0.63  0.6  0.58  0.62  
Rows (N_{r})  4  5  6  4  5  6  4  5  6  5  6  6  
Elements (N)  14  17  21  15  19  23  17  22  27  24  30  30  
Rectangular  h/y_{A}  L/h  Basin Drag Coefficient, C_{B}  

0.91  6  0.32  0.28  0.24  0.32  0.28  0.24  0.31  0.27  0.23  0.26  0.22  0.22  
0.71  6  0.44  0.40  0.37  0.42  0.38  0.35  0.40  0.36  0.33  0.34  0.31  0.29  
0.48  12  0.60  0.55  0.51  0.56  0.51  0.47  0.53  0.48  0.43  0.46  0.39  0.35  
0.37  12  0.68  0.66  0.65  0.65  0.62  0.60  0.62  0.58  0.55  0.54  0.50  0.45  
Circular  0.91  6  0.21  0.20  0.48  0.21  0.19  0.17  0.21  0.19  0.17  0.18  0.16  
0.71  6  0.29  0.27  0.40  0.27  0.25  0.23  0.25  0.23  0.22  0.22  0.20  
0.31  6  0.38  0.36  0.34  0.36  0.34  0.32  0.34  0.32  0.30  0.30  0.28  
0.48  12  0.45  0.42  0.25  0.40  0.38  0.36  0.36  0.34  0.32  0.30  0.28  
0.37  12  0.52  0.50  0.18  0.48  0.46  0.44  0.44  0.42  0.40  0.38  0.36 
Equation 9.1 is applicable for basins on less than 10 percent slopes. For basins with greater slopes, the weight of the water within the hydraulic jump must be considered in the expression. Equation 9.2 includes the weight component by assuming a straightline water surface profile across the jump:
(9.2)C_{P} γ y_{o}^{2} W_{o }/2 + ρV_{o}Q + w (sin θ) = C_{B} A_{F} N ρV_{A}^{2} /2 +γ Q^{2} /(2V_{B}^{2} W_{B}) + ρV_{B} Q
where,
w = weight of water within the basin
Volume = (y_{o} W_{o} + y_{A} W_{A}) W_{o} + (0.75LQ/ V_{B}) [(N_{r} 1)  (W_{B}/W_{o}  3) (1  W_{A} /W_{B})/2]
Weight = (Volume) γ
θ = arc tan of the channel slope, S_{o}
N_{r} = number of rows of roughness elements
L = longitudinal spacing between rows of elements.
Figure 9.4. Energy and Momentum Coefficients (Simons, 1970)
The depth y_{A} at the beginning of the roughness elements can be determined from Figure 4.3 and Figure 4.4. These figures are based on slopes less than 10 percent. The velocity V_{A} can be computed using Equations 4.1 or 4.2. Where slopes are greater than 10 percent, V_{A} and y_{A} can be computed using the following energy equation written between the end of the culvert (section o) and two culvert widths downstream (section A).
(9.3)2W_{o} S_{o} + y_{A} + (0.25) (Q/(W_{A} y_{A}))^{2} /2g = y_{o} + 0.25(V_{o}^{2} /(2g))
where,
W_{A} = W_{o} [4/(3Fr) + 1] which is adapted from Equations 4.3 and 4.4
Substantial splashing over the first row of roughness elements will occur if the elements are large and if the approach velocity is high. This problem can be addressed by locating the dissipator partially or totally within the culvert barrel, providing sufficient freeboard in the splash area, or providing some type of splash plate. If feasible both structurally and hydraulically, locating the dissipator partially or totally within the culvert barrel may result in economic, safety, and aesthetic advantages.
The necessary freeboard can be obtained from:
(9.4)FB = h + y_{A} + 0.5(V_{A} sinφ)^{2} /g
where,
FB = necessary freeboard, m (ft)
h = roughness element height, m (ft)
y_{A} = depth approaching first row of roughness elements, m (ft)
g = 9.81 m/s^{2} (32.2 ft/s^{2})
φ = 45° (function of y_{A}/h and the Froude number but no relationship has been derived)
φ is believed to be a function of y_{A}/h and the Froude number, but no relationship has been derived. 45 degrees is a reasonable approximation.
Another solution is a splash shield, which has been investigated in the FHWA Hydraulics Laboratory by J.S. Jones (unpublished research). A splash shield is a plate with a stiffener suspended between the first two rows of roughness elements as shown in Figure 9.5. The height to the plate was selected rather arbitrarily as a function of the critical depth since flow usually passed through critical in the vicinity of the large roughness elements.
Figure 9.5. Splash Shield
Although the tests were made with abrupt expansions, the configurations recommended for use are the combination flaredabrupt expansion basins shown in Figure 9.3 and above. These basins contain the same number of roughness elements as the abrupt expansion basin. The flare divergence, u_{e}, is a function of the longitudinal spacing between rows of elements, L, and the culvert barrel width, W_{o}:
(9.5)u_{e} = 4/7 + (10/7)L/W_{o}
The design procedure for the CSU rigid boundary basin may be summarized as follows:
Step 1. Compute the velocity, V_{o}, depth, y_{o}, and Froude number, Fr, at the culvert outlet or, if the basin is partially or totally located within the culvert barrel, at the beginning of the flared portion of the barrel.
Step 2. Select a trial basin from Table 9.1 based on the W_{B} /W_{o} expansion ratio that best matches the site geometry or satisfies other constraints. Choose W_{B} /W_{o}, number of rows, N_{r}, number of elements, N, and ratios h/y_{A} and L/h.
Step 3. Determine the flow condition V_{A} and y_{A} at the approach to the roughness element field (two culvert widths downstream).
Calculate V_{A} using Equations 4.1 or 4.2.
For 4 < W_{B} /W_{o} < 8, read y_{A} from Figure 4.3 and Figure 4.4.
For W_{B} /W_{o} < 4, compute y_{A} using Equation 9.3.
For slopes > 10 percent, use Equation 9.3 to find both V_{A} and y_{A}
Step 4. For the trial roughness height to depth ratio h/y_{A} and length to height ratio determine dissipator parameters from Figure 9.3:
Step 5. Confirm that the trial basin produces an exit velocity, V_{B}, and depth, y_{B}, that matches the downstream conditions. If W_{B} matches the downstream channel width or tailwater controls follow option 1. If W_{B} is less than the channel width follow option 2.
Option 1. Use the downstream depth, y_{n}, or tailwater if higher, to solve Equation 9.1 or Equation 9.2 for the quantity C_{B}A_{F}N. Using the C_{B}, A_{F}, and N values found in steps 2 and 4 compute C_{B}A_{F}N (for basin). The basin value should be greater than or equal to the C_{B}A_{F}N value from the equation. If not, select a new roughness configuration.
Option 2. Use the C_{B}, A_{F}, and N values found in steps 2 and 4 to solve for V_{B} in Equation 9.1 or 9.2. Three solutions for V_{B} are determined by trial and error: two positive roots and a negative root. The negative root may be discarded. The larger positive root is normally used for V_{B}. If V_{B} does not match the downstream velocity, select a new roughness configuration. If V_{B} is satisfactory, calculate y_{B}. Compare y_{B} to y_{n}. If y_{B} < y_{n}, use the smaller positive root for V_{B} and calculate y_{B}. If tailwater is greater than y_{B}, V_{B} should be calculated using the tailwater depth and the trial basin checked using option 1.
Step 6. Sketch the basin. The basin layout is shown on Figure 9.3. The elements are symmetrical about the basin centerline and the spacing between elements is approximately equal to the element width. In no case, should this spacing be made less than 75 percent of the element width. The W_{1}/h ratio must be between 2 and 8 and at least half the rows of elements should have an element near the wall to prevent high velocity jets from traversing the entire basin length. Alternate rows are staggered so that all streamlines are disrupted.
Step 7. Consider erosion protection downstream of the basin. If option 1 (step 5) is applicable, the flow conditions leaving the basin match the downstream conditions and additional riprap downstream of the basin is not required unless sitespecific concern regarding localized turbulence is identified. If, however, option 2 (step 5) is applicable, riprap is likely to be required until flow conditions fully transition to downstream conditions. Chapter 10 contains a section on riprap protection that may be used to size the required riprap.
Design Example: CSU Rigid Boundary Basin (SI)
Design a CSU rigid boundary basin to provide a transition from a RCB culvert to the natural channel. The basin should reduce velocities to approximately the downstream level. Given:
Solution
Step 1. Compute the velocity, V_{o}, depth, y_{o}, and Froude number, Fr, at the culvert outlet
y_{o} = y_{n} = 1.829 m (from HDS 3)
V_{o} = V_{n} = 8.87 m/s
Fr = V_{o}/(g y_{o})^{1/2} = 8.87 / [9.81(1.829)]^{1/2} = 2.1
Step 2. Select a trial basin from Table 9.1 based on the W_{B}/W_{o} expansion ratio which best matches the site geometry or satisfies other constraints. Choose W_{B} /W_{o}, number of rows, N_{r}, number of elements, N, and ratios h/y_{A} and L/h.
Channel Width/Culvert Width = 12.5/2.438 = 5.1
Try the following rectangular basin:
W_{B} /W_{o} = 5 and W_{1}/W_{o} = 0.63
N_{r} = 4 and N =15
h/y_{A} = 0.71 and L/h = 6
Step 3. Determine the flow condition V_{A} and y_{A} at the approach to the roughness element field (two culvert widths downstream) = 2W_{o} or 2(2.438) = 4.876 m.
Calculate V_{A} using Equations 4.1 or 4.2.
V_{A} /V_{o} = 1.65  0.3Fr = 1.65  0.3(2.1) = 1.02 from Equation 4.1
V_{A} = 8.870(1.02) = 9.047 m/s
For 4 < W_{B} /W_{o} < 8, read y_{A} from Figure 4.3 or Figure 4.4.
y_{A} /y_{o} = 0.33 from Figure 4.3 for Fr = 2.1 and L = 2B
y_{A} = 1.829(0.33) = 0.604 m
Step 4. For the trial roughness height to depth ratio h/y_{A} and length to height ratio determine dissipator parameters from Table 9.1:
Step 5. Since the width of the basin (W_{B }= 12.190 m) matches the downstream channel width (12.5 m), confirm trial basin using option 1. Use the normal flow conditions (V_{n} and y_{n}) and solve Equation 9.1 for C_{B}A_{F}N, which will be compared to C_{B}A_{F}N for basin:
Calculate C_{B}A_{F}N from Equation 9.1
y_{n} Downstream = 1.001 m
V_{B} = Q/ (W_{B}y_{n}) = 39.64/ [12.190(1.001)] = 39.64/12.178 = 3.255 m/s
ρV_{o}Q + C_{p} γ Y_{o}^{2}W_{o} /2 = C_{B}A_{F}N ρ V_{A}^{2}/2 + ρV_{B}Q +γQ^{2} /(2V_{B}^{2} W_{B}) (Equation 9.1)
Terms with V_{o} and y_{o}: 1000(8.870) (39.64) + 0.7(9810) (1.829)^{2} (2.438)/2 = 379609
Terms with V_{B}: 1000(3.255) (39.64) + 9810(39.64)^{2} /(2 (3.225)^{2} (12.190)) = 189820
Term with C_{B}A_{F}N is C_{B}A_{F}N (1000)(9.047)^{2} /2 = 40924 C_{B}A_{F}N
(379609  189820) = 40924 C_{B}A_{F}N
C_{B}A_{F}N = 4.63
Calculate C_{B}A_{F}N for basin based on parameters determined in steps 2 and 4 (N =15, C_{B} = 0.42, A_{F} = 0.65 m^{2}). Using these values C_{B}A_{F}N = 4.12. Since 4.12 is less than the 4.40 calculated from Equation 9.1, try a basin with more resistance (5 rows).
Step 4 (2nd iteration). For the trial roughness height to depth ratio h/y_{A} and length to height ratio determine dissipator parameters from Table 9.1: W_{B} /W_{o} = 5 that had N_{r} = 5, N =19, h/y_{A} = 0.71, L/h = 6, and C_{B} = 0.38.
Step 5 (2nd iteration). Calculate C_{B}A_{F}N from Equation 9.1.
C_{B}A_{F}N from Equation 9.1 = 4.63 (basin width did not change)
Calculate C_{B}A_{F}N for basin
C_{B}A_{F}N for basin = 0.38(0.654)(19) = 4.72 > 4.63 which is OK
Step 6. Sketch the basin and distribute roughness elements. (See sketch below. All dimensions shown in meters.)
W_{1}/h = 1.524/0.429 = 3.55 which is between the target range of 2 to 8.
Step 7. Since the design matches the downstream conditions, minimum riprap will be required. See Chapter 10 for guidance on riprap placement.
Sketch for the CSU Rigid Boundary Basin Design Example (SI)
Design Example: CSU Rigid Boundary Basin (CU)
Design a CSU rigid boundary basin to provide a transition from a RCB culvert to the natural channel. The basin should reduce velocities to approximately the downstream level. Given:
Solution
Step 1. Compute the velocity, V_{o}, depth, y_{o}, and Froude number, Fr, at the culvert outlet
y_{o} = y_{n} = 6.0 ft (from HDS 3)
V_{o} = V_{n} = 29.1 ft/s
Fr = V_{o}/ (g y_{o})^{1/2} = 29.1 / [32.2(6.0)]^{1/2} = 2.1
Step 2. Select a trial basin from Table 9.1 based on the W_{B}/W_{o} expansion ratio which best matches the site geometry or satisfies other constraints. Choose W_{B} /W_{o}, number of rows, N_{r}, number of elements, N, and ratios h/y_{A} and L/h.
Channel Width/Culvert Width = 41/8 = 5.13
Try the following rectangular basin:
W_{B} /W_{o} = 5 and W_{1}/W_{o} = 0.63
N_{r} = 4 and N =15
h/y_{A} = 0.71 and L/h = 6
Step 3. Determine the flow condition V_{A} and y_{A} at the approach to the roughness element field (two culvert widths downstream) = 2W_{o} or 2(8) = 16 ft.
Calculate V_{A} using Equations 4.1 or 4.2.
V_{A} /V_{o} = 1.65  0.3Fr = 1.65  0.3(2.1) = 1.02 from Equation 4.1
V_{A} = 29.1(1.02) = 29.7 ft/s
For 4 < W_{B} /W_{o} < 8, read y_{A} from Figure 4.3 or 4.4.
y_{A} /y_{o} = 0.33 from Figure 4. 3 for Fr = 2.1 and L = 2B
y_{A} = 6.0(0.33) = 1.98 ft
Step 4. For the trial roughness height to depth ratio h/y_{A} and length to height ratio determine dissipator parameters from Table 9.1:
Step 5. Since the width of the basin (W_{B }= 40 ft) matches the downstream channel width (41 ft) confirm trial basin using option 1. Use the normal flow conditions (V_{n} and y_{n}) and solve Equation 9.1 for C_{B}A_{F}N, which will be compared to C_{B}A_{F}N for basin:
Calculate C_{B}A_{F}N from Equation 9.1
y_{n} Downstream = 3.3 ft
V_{B} = Q/(W_{B}y_{n}) = 1400/ [40(3.3)] = 1400/132 = 10.6 ft/s
ρV_{o}Q + C_{p} γ Y_{o}^{2}W_{o} /2 = C_{B}A_{F}N ρ V_{A}^{2}/2 + ρV_{B}Q +γQ^{2} /(2V_{B}^{2} W_{B}) (Equation 9.1)
Terms with V_{o} and y_{o}: 1.94(29.1) (1400) + 0.7(62.4) (6)^{2} (8)/2 = 85,325.5
Terms with V_{B}: 1.94(10.6) (1400) + 62.4(1400)^{2} / (2 (10.6)^{2} (40)) = 42,395.9
Term with C_{B}A_{F}N is C_{B}A_{F}N (1.94) (30.6)^{2} /2 = 908.3(C_{B}A_{F}N)
(85,325.5  42,395.9 = 908.3(C_{B}A_{F}N)
C_{B}A_{F}N = 47.3
Calculate C_{B}A_{F}N for basin based on parameters determined in steps 2 and 4 ((N_{r} = 4, C_{B} = 0.42, A_{F} = 7 ft^{2}). Using these values C_{B}A_{F}N = 44.1. Since 44.1 is less than the 47.3 calculated from Equation 9.1, try a basin with more resistance (5 rows).
Step 4 (2nd iteration). For the trial roughness height to depth ratio h/y_{A} and length to height ratio determine dissipator parameters from Table 9.1: W_{B} /W_{o} = 5 that had N_{r} = 5, N =19, h/y_{A} = 0.71, L/h = 6, and C_{B} = 0.38.
Step 5 (2nd iteration). Calculate C_{B}A_{F}N from Equation 9.1.
C_{B}A_{F}N from Equation 9.1 = 47.3 (basin width did not change)
Calculate C_{B}A_{F}N for basin
C_{B}A_{F}N for basin = 0.38(7) (19) = 50.5 > 47.3 which is OK
Step 6. Sketch basin and distribute roughness elements. (See following figure). All dimensions shown in feet.)
W_{1}/h = 5/1.4 = 3.57 which is between the target range of 2 to 8.
Step 7. Since the design matches the downstream conditions, minimum riprap will be required. See Chapter 10 for guidance on riprap placement.
Sketch for the CSU Rigid Boundary Basin Design Example (CU)
The Contra Costa energy dissipator (Keim, 1962) was developed at the University of California, Berkeley, in conjunction with Contra Costa County, California. It is intended for use primarily in urban areas with defined tailwater channels. A sketch of the dissipator is shown in Figure 9.6.
The dissipator was developed to be selfcleaning with minimum maintenance requirements. It is best suited to small and medium size culverts of any cross section where the depth of flow at the outlet is less than or equal to half the culvert height, but is applicable over a wide range of culvert sizes and operating conditions as noted in Table 1.1. The flow leaving the dissipator will be at minimum energy when operating without tailwater. When tailwater is present, the performance will improve. Field experience with this dissipator has been limited. Designers should not extrapolate parameter values in this guidance beyond the ranges cited for the model tests.
Equation 9.6 was obtained by testing model Contra Costa dissipators that had L_{2 }/ h_{2} ratios from 2.5 to 7. The equation is in terms of culvert exit velocity, V_{o}, and depth, y_{o}, for a circular culvert.
(9.6)where,
y_{o} = outlet depth, m (ft)
V_{o} = outlet velocity, m/s (ft/s)
Fr = V_{o}/(g y_{o})^{1/2}
h_{2} = height of large baffle, m (ft)
L_{2} = Length from culvert exit to large baffle, m (ft)
Figure 9.6. Contra Costa Basin
Equation 9.6 is generalized for other shapes by substituting the equivalent depth of flow, y_{e}, for y_{o}. Equivalent depth is found by converting the area of flow at the culvert outfall to an equivalent rectangular cross section with a width equal to twice the depth of flow. For box culverts, y_{e} = y_{n} ory_{brink}.
(9.7a)or:
(9.7b)where,
y_{e} = equivalent depth, (A/2)^{1/2 }, m (ft)
A = outlet flow area, m^{2} (ft^{2})
V_{o} = outlet velocity, m/s (ft/s)
Fr = V_{o}/ (g y_{e})^{1/2}
Equation 9.7b is solved by assuming a value of L_{2}/h_{2} between 2.5 and 7. A trial height of the second baffle,h_{2}, can be determined. If the recommended L_{2} /h_{2} = 3.5 value is substituted into Equation 9.7, the design equation becomes Equation 9.8. The value of h_{2}/y_{e} should always be greater than unity.
(9.8)After determining the values of h_{2} and L_{2}, the length from the large baffle to the end sill, L_{3}, can be obtained using Equation 9.9.
(9.9)The height of the small baffle, h_{1}, is half the height of the large baffle, h_{2}. The position of the small baffle is half way between the culvert outlet and the large baffle or L_{2}/2. Side slopes of the trapezoidal basin for all experimental runs were 1:1 (V:H). The width of basin, W, may vary from one to three times the width of the culvert. The floor of the basin should be essentially level. The height of the end sill, h_{3}, may vary from 0.06y_{2} to 0.10y_{2}. After obtaining satisfactory basin dimensions, the approximate maximum water surface depth, y_{2}, without tailwater, can be obtained from Equation 9.10 which is for basins with W_{B}/W_{o} = 2. The depth y_{3} is equal to y_{c} for the basin + h_{3}.
(9.10)The following steps outline the design procedure for the design of the Contra Costa basin:
Step 1. Determine the flow conditions at the outfall of the culvert for the design discharge. If the depth of flow at the outlet, y_{o}, is D/2 or less, the Contra Costa basin is applicable.
Step 2. Compute equivalent depth, y_{e}, and Froude number, Fr.
y_{e} = y_{o } for rectangular culvert
y_{e} = (A/2)^{1/2} for other shapes
Fr = V_{o }/ (gy_{e})^{1/2}
Step 3. The width of the basin floor, W_{B}, is selected to conform to the natural channel, but must be 1W_{o} to 3W_{o}. If there is no defined channel, the width should be no greater than 3 times the culvert width. The basin side slopes should be 1:1.
Step 4. Assume a value of L_{2}/h_{2} between 2.5 and 7. If L_{2}/h_{2} = 3.5, use Equation 9.8 to determine h_{2}. Use Equation 9.7 for other values. Calculate L_{2} = 3.5 h_{2}. Calculate the first baffle height, h_{1} = 0.5h_{2} and position, L_{1} = 0.5L_{2}.
Step 5. Determine the length from the large baffle to the end sill, L_{3}, using Equation 9.9. Repeat the procedure, if necessary, until a dissipator is defined which optimizes the design requirements.
Step 6. Estimate the approximate maximum water surface depth without tailwater, y_{2}, using Equation 9.10 which is for W_{B} = 2W_{o}. Set the end sill height, h_{3}, between 0.06y_{2} and 0.1y_{2}. If the above dimensions are compatible with the topography at the site, the dimensions are final. If not, a different value of L_{2} /h_{2} is selected and the design procedure repeated.
Step 7. Determine the basin exit depth, y_{3} = y_{c} and exit velocity, V_{2} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c}(W_{B} + y_{c})]^{3}/ (W_{B} + 2y_{c}) (Substituting for A_{c} and T_{c} using the properties of a trapezoid.)
V_{c} = Q/A_{c}
Step 8. Riprap may be necessary downstream especially for the low tailwater cases. See Chapter 10 for design recommendations. Freeboard to prevent overtopping and a cutoff wall to prevent undermining of the basin also should be considered.
Design Example: Contra Costa Basin (SI)
Determine the design dimensions for a Contra Costa basin. Given:
Solution
Step 1. Determine the flow conditions at the outfall of culvert for the design discharge.
y_{o} = 0.701 m is approximately D/2, OK.
Step 2. Compute equivalent depth, y_{e}, and Froude number, Fr.
Using Equations 7.11 and 7.13, flow area in the culvert = 0.696 m^{2}.
y_{e} = (A/2)^{1/2} = (0.696/2)^{1/2} = 0.590 m
Fr = V_{o} / (gy_{e})^{1/2} = 12.192 / [9.81(0.590)]^{1/2} = 5.07
Step 3. The width of the basin floor, W_{B}, is selected to conform to the natural channel. The basin side slopes should be 1:1 (V:H).
Set W = 2.438 m (channel bottom width). 1 ≤ W/D ≤ 3 OK
Step 4. Assume L_{2}/h_{2} = 3.5, use Equation 9.8 to determine h_{2}. Calculate L_{2} = 3.5 h_{2}. Calculate the first baffle height, h_{1} = 0.5h_{2} and position, L_{1} = 0.5L_{2}
h_{2} / y_{e} = 0.595 Fr ^{1.092} = 0.595 (5.07)^{1.092} = 3.5
h_{2} = y_{e} (h_{2} / y_{e}) = 0.590 (3.50) = 2.065 m
L_{2} = 3.5 h_{2} = 3.5 (2.065) = 7.228 m
h_{1} = 0.5h_{2} = 0.5 (2.065) = 1.032 m
L_{1} = 0.5 L_{2} = 0.5 (7.228) = 3.614 m
Step 5. Determine the length from the large baffle to the end sill, L_{3}, using Equation 9.9. Repeat the procedure, if necessary, until a dissipator is defined which optimizes the design requirements.
L_{3} / L_{2} = 3.75(h_{2} /L_{2})^{0.68} = 3.75(1/3.5)^{0.68} = 1.6
L_{3} = (L_{3} / L_{2}) L_{2} = 1.6 (7.228) = 11.56 m
Step 6. Estimate the approximate maximum water surface depth without tailwater, y_{2}, using Equation 9.10 which is for W_{B} = 2D. Determine end sill height, h_{3} = 0.1y_{2}
y_{2} / h_{2} = 1.3(L_{2} /h_{2})^{0.36} = 1.3(3.5)^{0.36} = 2.04
y_{2} = (y_{2} / h_{2}) h_{2} = (2.04) 2.065 = 4.21 m
h_{3} = 0.1(y_{2}) = 0.1(4.21) = 0.42 m
A summary of physical dimensions is shown in the following table.
First Baffle  Second Baffle  End Sill  

Distance from exit (m)  3.61  7.23  18.79 
Height (m)  1.03  2.07  0.42 
Step 7. Determine the basin exit depth, y_{3} = y_{c} and exit velocity, V_{2} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c }(W_{B} + y_{c})]^{3}/ (W_{B} + 2y_{c})
8.49^{2}/9.81 = 7.35 = [y_{c }(2.438 + y_{c})]^{3}/ (2.438 + 2y_{c})
By trial and success, y_{c} = 0.938 m, T_{c} = 4.314 m, A_{c} = 3.17 m^{2}
V_{c} = Q/A_{c} = 8.49/3.17 = 2.68 m/s
Step 8. Riprap may be necessary downstream especially for the low tailwater cases. See Chapter 10 for design recommendations. Freeboard to prevent overtopping and a cutoff wall to prevent undermining of the basin also should be considered.
Design Example: Contra Costa Basin (CU)
Determine the design dimensions for a Contra Costa basin. Given:
Solution
Step 1. Determine the flow conditions at the outfall of culvert for the design discharge.
y_{o} = 2.3 ft is approximately D/2, OK.
Step 2. Compute equivalent depth, y_{e}, and Froude number, Fr.
Using Equations 7.11 and 7.13, flow area in the culvert = 7.5 ft^{2}.
y_{e} = (A/2)^{1/2} = (7.5/2)^{1/2} = 1.94 ft
Fr = V_{o} / (gy_{e})^{1/2} = 40 / [32.2(1.94)]^{1/2} = 5.06
Step 3. The width of the basin floor, W_{B}, is selected to conform to the natural channel. The basin side slopes should be 1:1 (V:H).
Set W = 8 ft (channel bottom width). 1 ≤ W/D ≤ 3 OK
Step 4. Assume L_{2}/h_{2} = 3.5, use Equation 9.8 to determine h_{2}. Calculate L_{2} = 3.5 h_{2}. Calculate the first baffle height, h_{1} = 0.5h_{2} and position, L_{1} = 0.5L_{2}
h_{2} / y_{e} = 0.595 Fr ^{1.092} = 0.595 (5.06)^{1.092} = 3.5
h_{2} = y_{e} (h_{2} / y_{e}) = 1.94 (3.50) = 6.8 ft
L_{2} = 3.5 h_{2} = 3.5 (6.8) = 23.8 ft
h_{1} = 0.5h_{2} = 0.5 (6.8) = 3.4 ft
L_{1} = 0.5 L_{2} = 0.5 (23.8) = 11.9 ft
Step 5. Determine the length from the large baffle to the end sill, L_{3}, using Equation 9.9. Repeat the procedure, if necessary, until a dissipator is defined which optimizes the design requirements.
L_{3} / L_{2} = 3.75(h_{2} /L_{2})^{0.68} = 3.75(1/3.5)^{0.68} = 1.6
L_{3} = (L_{3} / L_{2}) L_{2} = 1.6 (23.8) = 38.1 ft
Step 6. Estimate the approximate maximum water surface depth without tailwater, y_{2}, using Equation 9.10 which is for W_{B} = 2D. Determine end sill height, h_{3} = 0.1y_{2}
y_{2} / h_{2} = 1.3(L_{2} /h_{2})^{0.36} = 1.3(3.5)^{0.36} = 2.04
y_{2} = (y_{2} / h_{2}) h_{2} = (2.04) 6.8 = 13.9 ft
h_{3} = 0.1(y_{2}) = 0.1(13.9) = 1.4 ft
A summary of physical dimensions is shown in the following table.
First Baffle  Second Baffle  End Sill  

Distance from exit (ft)  11.9  23.8  61.9 
Height (ft)  3.4  6.8  1.4 
Step 7. Determine the basin exit depth, y_{3} = y_{c} and exit velocity, V_{2} = V_{c}.
Q^{2}/g = (A_{c})^{3}/T_{c} = [y_{c }(W_{B} + y_{c})]^{3}/ (W_{B} + 2y_{c})
300^{2}/32.2 = 2795 = [y_{c }(8 + y_{c})]^{3}/ (8 + 2y_{c}) = 34.06^{3}/14.15 = 2792
By trial and success, y_{c} = 3.075 ft, T_{c} = 14.15 ft, A_{c} = 34.06 ft^{2}
V_{c} = Q/A_{c} = 300/34.06 = 8.8 ft/s
Step 8. Riprap may be necessary downstream especially for the low tailwater cases. See Chapter 10 for design recommendations. Freeboard to prevent overtopping and a cutoff wall to prevent undermining of the basin also should be considered.
The Hook basin was developed at the University of California in cooperation with the California Division of Highways and the Bureau of Public Roads (MacDonald, 1967). The basin was originally developed for large arch culverts with low tailwater, but can be used with box or circular conduits. The dissipator can be used for culvert outlet Froude numbers from 1.8 to 3.0. Two hydraulic model studies were conducted: (1) a basin with wingwalls warped from vertical at the culvert outlet to side slopes of 1:1.5 (V:H) at the end sill and a tapered basin floor which is discussed in Section 9.3.1 and (2) a trapezoidal channel of uniform cross section which is discussed which is discussed in Section 9.3.2.
The hook basin with warped wingwalls is shown in Figure 9.7. The design procedure is deterministic except for selecting the width of the hooks. Judgment is necessary in choosing this dimension to insure that the width is sufficient for effective energy dissipation, but not so great that flow passage between the hooks is inadequate. A ratio of W_{4}/W_{o} = 0.16, which was the minimum tested, is recommended. Each design should be checked to see that the spacing between hooks is 1.5 to 2.5 times the hook width.
The height of wingwalls, h_{6}, should be at least twice the flow depth at the culvert exit or 2y_{e}. This height is based on the highest water surface elevations observed in the basin during the study. Therefore, setting h_{6} = 2y_{e} does not provide freeboard to contain splashing. Depending on the site conditions, the designer should provide for additional freeboard.
Figure 9.7. Hook Basin with Warped Wingwalls
The best range of design dimensions that were tested are indicated in the design procedure. In most cases, the ratio that will produce the smallest dimension was used. The recommended hook configuration is shown in Figure 9.8. The recommended dimensions are:
Figure 9.8. Hook for Warped Wingwall Basin
A flare angle, α, of 5.7 degrees per side (tan α = 0.10) is the optimum value for Fr > 2.45. Increasing the length beyond L_{B} = 3W_{o} does not improve basin performance. The effectiveness of the dissipator falls off rapidly with increasing Froude number regardless of hook width, for flare angle exceeding 5.7 degrees. The exit velocity of the dissipator, V_{B}, is estimated from Figure 9.9. The higher the velocity ratio, V_{o}/V_{B}, the more effective the basin is in dissipating energy and distributing the flow downstream.
Figure 9.9. Velocity Ratio for Hook Basin With Warped Wingwalls
Depending on final velocity and soil conditions, some scour can be expected downstream of the basin. The designer should, where necessary, provide riprap protection in this area. Chapter 10 contains design guidance for riprap. Where large debris is expected, armor plating the upstream face of the hooks with steel is recommended.
The recommended design procedure for a Hook basin with warped wingwalls is as follows:
Step 1. Compute the culvert outlet velocity, V_{o}, equivalent depth, y_{e}, and Froude number, Fr = V_{o}/(gy_{e})^{1/2}. If 1.8 < Fr < 3.0, proceed with design.
Step 2. Compute the downstream channel velocity, V_{n}, and depth, y_{n}.
Step 3. Select width of the basin at the basin exit, W_{B} (W_{B} = W_{6}), and compute L_{B}. W_{6} should be approximately equal to the channel width, if the downstream channel is defined.
L_{B} = (W_{6}  W_{o}) / (2tan<span α), use α = 5.7° (tan = 0.10)
Step 4. Compute the position and spacingof the hooks (see Figure 9.7):
Step 5. Compute hook dimensions (see Figure 9.8):
Step 6. Compute the end sill and wingwall dimensions (see Figure 9.7):
Step 7. Find V_{o}/V_{B} from Figure 9.9 and compute basin exit velocity, V_{B}. Compare V_{B} with V_{n} from step 2. If V_{B} is unacceptable, adjust basin length. Assess scour potential downstream based on soil condition and outlet velocity. If riprap is needed, see Chapter 10.
Step 8. Where large debris is expected, the upstream face of the hooks should be armored with steel.
Design Example: Hook Basin With Warped Wingwalls (SI)
Determine dimensions for a Hook basin with warped wingwalls (see Figure 9.7) for a long concrete semicircular arch culvert that is 3.658 m wide and 3.658 m from the floor to the crown. Given:
The downstream channel has a trapezoidal shape with the following properties:
Solution
Step 1. Compute the culvert outlet velocity, V_{o}, equivalent depth, y_{e}, and Froude number, Fr, V_{o}/(gy_{e})^{1/2}. V_{o }= 11.43 m/s and y_{e} = 1.829 m were given.
Fr_{o} = V_{o} / (gy_{e})^{1/2} = 11.43/ (9.81 x 1.829)^{1/2} = 2.70
Since 1.8 < 2.70 < 3, proceed to step 2.
Step 2. Compute the downstream channel velocity, V_{n}, and depth, y_{n}. V_{n} = 5.27 m/s and y_{n} = 1.676 m were given.
Step 3. Select W_{6} and compute L_{B}.
Use W_{6} = W_{c} = 6.096 m and tan α = 0.10.
L_{B} = (W_{6}  W_{o}) / (2tan α) = (6.096  3.658)/ [2(0.10)] = 12.19 m or 3.3W_{o}
Step 4. Compute the position and spacing of the hooks (see Figure 9.7):
Step 5. Compute hook dimensions (see Figure 9.8):
Step 6. Compute the end sill and wingwall dimensions (see Figure 9.7):
Step 7. Find V_{o}/V_{B} from Figure 9.9 and compute V_{B}. Compare with V_{n} from step 2. Assess scour potential downstream based on soil condition and outlet velocity. If riprap is needed, see Chapter 10.
With Fr = 2.7 and L_{B} = 3.3 W_{o}, V_{o} /V_{B} will be less than 1.9 making V_{B } ≈ 11.43/1.9 = 6.016 m/s. This is somewhat higher than the normal velocity in the downstream channel indicating riprap protection may be desirable. See Chapter 10.
The dissipator design is shown on the sketch below. (All dimensions are in meters.)
Step 8. Since no large debris is expected at this site, the hook face will not be armored with steel.
Sketch for the Hook Basin with Warped Wingwalls Design Example (SI)
Design Example: Hook Basin With Warped Wingwalls (CU)
Determine dimensions for a Hook basin with warped wingwalls (see Figure 9.7) for a long concrete semicircular arch culvert that is 12 ft wide and 12 ft from the floor to the crown. Given:
The downstream channel has a trapezoidal shape with the following properties:
Solution
Step 1. Compute the culvert outlet velocity, V_{o}, equivalent depth, y_{e}, and Froude number, Fr V_{o}/(gy_{e})^{1/2}. V_{o} = 37.5 ft/s and y_{e} = 6 ft were given.
Fr_{o} = V_{o} / (gy_{e})^{1/2} = 37.5/ (32.2 x 6)^{1/2} = 2.7
Since 1.8 < 2.7 < 3, proceed with step 2.
Step 2. Compute the downstream channel velocity, V_{n}, and depth, y_{n}. V_{n} = 16.1 ft/s and y_{n} = 5.5 ft were given.
Step 3. Select W_{6} and compute L_{B}.
Use W_{6} = W_{c} = 20 and tan<span α = 0.10.
L_{B} = (W_{6}  W_{o}) / (2tan α) = (20  12)/ [2(0.10)] = 40 ft or 3.3W_{o}
Step 4. Compute the position and spacing of the hooks (see Figure 9.7):
Step 5. Compute hook dimensions (see Figure 9.8):
Step 6. Compute the end sill and wingwall dimensions (see Figure 9.7):
Step 7. Find V_{o}/V_{B} from Figure 9.9 and compute V_{B}. Compare with V_{n} from step 2. Assess scour potential downstream based on soil condition and outlet velocity. If riprap is needed, see Chapter 10.
With Fr = 2.7 and L_{B} = 3.3 W_{o}, V_{o} /V_{B} will be less than 1.9 making V_{B } ≈ 37.5/1.9 = 19.7 ft/s. This is somewhat higher than the normal velocity in the downstream channel (16.1 ft/s) indicating riprap protection may be desirable. See Chapter 10.
The dissipator design is shown on the sketch below. (All dimensions are in feet.)
Step 8. Since no large debris is expected at this site, the hook face will not be armored with steel.
Sketch for the Hook Basin with Warped Wingwalls Design Example (CU)
The Hook basin with a uniform trapezoidal channel with end sill is shown in Figure 9.10. The hooks and end sill are closer to the outfall of the culvert than the hooks and end sill with warped wingwalls. The research report (MacDonald, 1967) presents several charts depicting the effect of various variables on the performance of the dissipator. These charts show that for a given discharge condition widening the basin produces some reduction in the velocity downstream, and flattening the side slopes improves the performance of the dissipator for values of the Froude number up to 3.0.
Figure 9.10. Hook Basin with Uniform Trapezoidal Channel
The best range of design dimensions that were tested are indicated in the design procedure. In most cases, the ratio that will produce the smallest dimension was used. The recommended hook configuration is shown in Figure 9.11. The height dimensions are different than those used for the hook energy dissipator with warped wingwalls. The recommended dimensions are:
Figure 9.11. Hook for Uniform Trapezoidal Channel Basin
Figure 9.12. Velocity Ratio for Hook Basin With Uniform Trapezoidal Channel
The design procedure for a Hook basin with a uniform trapezoidal channel is as follows:
Step 1. Compute the culvert outlet velocity, V_{o}, equivalent depth, y_{e} = (A/2)^{1/2}, and Froude number, Fr = V_{o}/(gy_{e})^{1/2}.
Culvert width, W_{o} = width of rectangular culvert or W_{o }= 2y_{e} for circular and other shapes.
If 1.8 < Fr < 3.0, proceed with design.
Step 2. Compute the downstream channel velocity, V_{n}, and depth, y_{n}.
Step 3. Select a basin width, W_{B} (W_{6} = W_{B}), side slope, and length, L_{B}. W_{6} should be approximately equal to the channel width if the downstream channel is defined.
W_{6} = W_{o} to 2W_{o}
Basin side slope can be from 1:1.5 to 1:2 (V:H).
L_{B} = 3.0W_{o}
Step 4. Compute the position and spacing of the hoooks (see Figure 9.10):
Step 5. Compute hook dimensions (see Figure 9.11):
Step 6. Compute the end sill and wingwall dimensions (see Figure 9.10):
Step 7. Find V_{o}/V_{B} from Figure 9.12 and calculate V_{B}. Compare with V_{n} from step 2. If V_{B} is unacceptable, adjust W_{6}, if feasible. Assess scour potential downstream based on soil condition and outlet velocity. If riprap is needed, see Chapter 10.
Step 8. Where large debris is expected, the upstream face of the hooks should be armored with steel.
Design Example: Hook Basin with a Uniform Trapezoidal Channel (SI)
Determine dimensions for a Hook basin with a uniform trapezoidal channel for a long concrete semicircular arch culvert that is 3.658 m wide and 3.658 m from the floor to the crown. Given:
The downstream channel has a trapezoidal shape with the following properties:
Solution
Step 1. Compute the culvert outlet velocity, V_{o}, equivalent depth, y_{e}, and Froude number, Fr = V_{o}/(gy_{e})^{1/2}. V_{o} = 11.43 m/s and y_{e} = 1.829 m were given.
Fr_{o} = V_{o} / (gy_{e})^{1/2} = 11.43/ (9.81 x 1.829)^{1/2} = 2.70
Since 1.8 < 2.70 < 3, proceed to step 2.
Step 2. Compute the downstream channel velocity, V_{n}, and depth, y_{n}. V_{n} = 5.273 m/s and y_{n} = 1.676 m were given.
Step 3. Select a basin width, W_{6}, side slope, and length, L_{B}. W_{6} should be approximately equal to the channel width, if the downstream channel is defined.
W_{6} = W_{c} = 6.096 m, which is 6.096/3.658 = 1.67 W_{o}
Basin side slope will be 1:1.5 (V:H)
L_{B} = 3.0W_{o} = 3.0(3.658) = 10.974 m
Step 4. Compute the position and spacing of the hooks (see Figure 9.10):
Step 5. Compute hook dimenstions (See Figure 9.11):
Step 6. Compute the end sill and wingwall dimensions (see Figure 9.10):
Step 7. Find V_{o}/V_{B} from Figure 9.12 and compute V_{B}. Compare with V_{n} from step 2. Assess scour potential downstream based on soil condition and outlet velocity. If riprap is needed, see Chapter 10.
From Figure 9.12 with a Froude number of 2.70 and W_{6}/W_{o }= 1.67, V_{o}/V_{B} ≈ 2.0 making V_{B } ≈ 11.43/2 = 5.72 m/s which is slightly higher than the normal channel velocity, V_{n} = 5.273 m/s indicating minimum riprap protection will be necessary. A sketch of this dissipator is shown in the sketch on the next page. (All dimensions are shown in meters.)
Step 8. Where large debris is expected the upstream face of the hooks should be armored. Since no large debris is expected, the hook face will not be armored.
The design example dimensions for both the warped wingwall and the trapezoidal basins are shown in the following table.
Feature  Element  Symbol  Warped Wingwall (m)  Trapezoidal (m) 

Basin  Length  L_{B}  12.19  10.974 
Width  W_{B}  6.096  6.096  
First Hooks  Length  L_{1}  9.143  4.573 
Spacing  W_{2}  3.621  2.377  
Second Hook  Length  L_{2}  10.118  7.682 
Spacing  W_{3}  1.518  0.896  
End Wall  Height  h_{4}  1.225  1.225 
Slot  W_{5}  2.012  2.012  
Top  h_{5}  3.439  4.264  
Hooks  Height  h_{3}  1.829  1.998 
Width  W_{4}  0.607  0.585 
Sketch for the Hook Basin with a Uniform Trapezoidal Channel (SI)
Design Example: Hook Basin with a Uniform Trapezoidal Channel (CU)
Determine dimensions for a Hook basin with a uniform trapezoidal channel for a long concrete semicircular arch culvert that is 12 ft wide and 12 ft from the floor to the crown. Given:
The downstream channel has a trapezoidal shape with the following properties:
Solution
Step 1. Compute the culvert outlet velocity, V_{o}, equivalent depth, y_{e}, and Froude number, Fr = V_{o}/(gy_{e})^{1/2}. V_{o} = 37.5 ft/s and y_{e} = 6 ft were given.
Fr_{o} = V_{o} / (gy_{e})^{1/2} = 37.5/ (32.2 x 6)^{1/2} = 2.7
Since 1.8 < 2.7 < 3, proceed with step 2.
Step 2. Compute the downstream channel velocity, V_{n}, and depth, y_{n}. V_{n} = 16.1 ft/s and y_{n} = 5.5 ft were given.
Step 3. Select a basin width, W_{6}, side slope, and length, L_{B}. W_{6} should be approximately equal to the channel width, if the downstream channel is defined.
W_{6} = W_{c} = 20 ft, which is 20/12 = 1.67 W_{o}
Basin side slope will be 1:1.5 (V:H)
L_{B} = 3.0W_{o} = 3.0(12) = 36 ft
Step 4. Compute the position and spacing of the hooks (see Figure 9.10):
Step 5. Compute hook dimensions (see Figure 9.11):
Step 6. Compute the end sill and wingwall dimensions (see Figure 9.10):
Step 7. Find V_{o}/V_{B} from Figure 9.12 and compute V_{B}. Compare with V_{n} from step 2. Assess scour potential downstream based on soil condition and outlet velocity. If riprap is needed, see Chapter 10.
From Figure 9.12 with a Froude number of 2.70 and W_{6}/W_{o }= 1.67, V_{o}/V_{B} ≈ 2.0 making V_{B } ≈ 37.5/2 = 18.8 ft/s which is slightly higher than the normal channel velocity, V_{n} = 16.1 ft/s indicating minimum riprap protection will be necessary. A sketch of this dissipator is shown on the next page. (All dimensions are shown in feet.)
Step 8. Where large debris is expected the upstream face of the hooks should be armored. Since no large debris is expected, the hook face will not be armored.
The design example dimensions for both the warped wingwall and the trapezoidal basins are shown in the following table.
Feature  Element  Symbol  Warped Wingwall (ft)  Trapezoidal (ft) 

Basin  Length  L_{B}  36  40 
Width  W_{B}  20  20  
First Hooks  Length  L_{1}  15  30 
Spacing  W_{2}  7.8  12  
Second Hook  Length  L_{2}  25  33 
Spacing  W_{3}  2.9  5  
End Wall  Height  h_{4}  4  4 
Slot  W_{5}  6.6  6.6  
Top  h_{5}  14  11.3  
Hooks  Height  h_{3}  6.6  6 
Width  W_{4}  2  2 
Sketch for Hook Basin with a Uniform Trapezoidal Channel (CU)
The U.S. Bureau of Reclamation (USBR) Type VI impact basin was developed at the USBR Laboratory (ASCE, 1957). The dissipator is contained in a relatively small boxlike structure that requires no tailwater for successful performance. Although the emphasis in this manual is on its use at culvert outlets, the structure may also be used in open channels.
The shape of the basin has evolved from extensive tests, but these were limited in range by the practical size of field structures required. With the many combinations of discharge, velocity, and depth possible for the incoming flow, it became apparent that some device was needed which would be equally effective over the entire range. The vertical hanging baffle, shown in Figure 9.13, proved to be this device. Energy dissipation is initiated by flow striking the vertical hanging baffle and being deflected upstream by the horizontal portion of the baffle and by the floor, creating horizontal eddies.
Notches in the baffle are provided to aid in cleaning the basin after prolonged periods of low or no flow. If the basin is full of sediment, the notches provide concentrated jets of water for cleaning. The basin is designed to carry the full discharge over the top of the baffle if the space beneath the baffle becomes completely clogged. Although this performance is not good, it is acceptable for short periods of time.
Figure 9.13. USBR Type VI Impact Basin
The design information is presented as a dimensionless curve in Figure 9.14. This curve incorporates the original information contained in ASCE (1957) and the results of additional experimentation performed by the Department of Public Works, City of Los Angeles. The curve shows the relationship of the Froude number to the ratio of the energy entering the dissipator to the width of dissipator required. The Los Angeles tests indicate that limited extrapolation of this curve is permissible.
Figure 9.14. Design Curve for USBR Type VI Impact Basin
Once the basin width, W_{B}, has been determined, many of the other dimensions shown in Figure 9.13 follow according to Table 9.2. To use Table 9.2, round the value of W_{B} to the nearest entry in the table to determine the other dimensions. Interpolation is not necessary.
In calculating the energy and the Froude number, the equivalent depth of flow, y_{e} = (A/2)^{1/2}, entering the dissipator from a pipe or irregularshaped conduit must be computed. In other words, the cross section flow area in the pipe is converted into an equivalent rectangular cross section in which the width is twice the depth of flow. The conduit preceding the dissipator can be open, closed, or of any cross section.
The effectiveness of the basin is best illustrated by comparing the energy losses within the structure to those in a natural hydraulic jump, Figure 9.15. The energy loss was computed based on depth and velocity measurements made in the approach pipe and also in the downstream channel with no tailwater. Compared with the natural hydraulic jump, the USBR Type VI impact basin shows a greater capacity for dissipating energy.
Although tailwater is not necessary for successful operation, a moderate depth of tailwater will improve the performance. For best performance set the basin so that maximum tailwater does not exceed h_{3} + (h_{2}/2) which is half of the baffle.
The basin floor should be constructed horizontally and will operate effectively with entrance conduits on slopes up to 15^{o} (27%). For entrance conduits with slopes greater than 15^{o}, a horizontal conduit section of at least four conduit widths long should be provided immediately upstream of the dissipator. Experience has shown that, even for conduits with slopes less than 15 degrees, it is more efficient when the horizontal section of pipe recommended for steeper slopes is used. In every case, the proper position of the entrance invert, as shown in Figure 9.13, should be maintained.
If a horizontal section of pipe is provided before the dissipator, the conduit should be analyzed to determine if a hydraulic jump would form in the conduit. When a hydraulic jump is expected and the pipe outlet is flowing full, a vent about onesixth the pipe diameter should be installed at a convenient location upstream from the jump.
To provide structural support to the hanging baffle, a short support should be placed under the center of the baffle wall. This support will also provide an additional energy dissipating barrier to the flow.
W_{B}  h_{1}  h_{2}  h_{3}  H_{4}  L  L_{1}  L_{2} 

1.0  0.79  0.38  0.17  0.43  1.40  0.59  0.79 
1.5  1.16  0.57  0.25  0.62  2.00  0.88  1.16 
2.0  1.54  0.75  0.33  0.83  2.68  1.14  1.54 
2.5  1.93  0.94  0.42  1.04  3.33  1.43  1.93 
3.0  2.30  1.12  0.50  1.25  4.02  1.72  2.30 
3.5  2.68  1.32  0.58  1.46  4.65  2.00  2.68 
4.0  3.12  1.51  0.67  1.67  5.33  2.28  3.08 
4.5  3.46  1.68  0.75  1.88  6.00  2.56  3.46 
5.0  3.82  1.87  0.83  2.08  6.52  2.84  3.82 
5.5  4.19  2.03  0.91  2.29  7.29  3.12  4.19 
6.0  4.60  2.25  1.00  2.50  7.98  3.42  4.60 
W_{B}  W_{1}  W_{2}  t_{1}  t_{2}  t_{3}  t_{4}  t_{5} 

1.0  0.08  0.26  0.15  0.15  0.15  0.15  0.08 
1.5  0.13  0.42  0.15  0.15  0.15  0.15  0.08 
2.0  0.15  0.55  0.15  0.15  0.15  0.15  0.08 
2.5  0.18  0.68  0.16  0.18  0.18  0.16  0.08 
3.0  0.22  0.83  0.20  0.20  0.22  0.20  0.08 
3.5  0.26  0.91  0.20  0.23  0.23  0.21  0.10 
4.0  0.30  0.91  0.20  0.28  0.25  0.25  0.10 
4.5  0.36  0.91  0.20  0.30  0.30  0.30  0.13 
5.0  0.39  0.91  0.22  0.31  0.30  0.30  0.15 
5.5  0.41  0.91  0.22  0.33  0.33  0.33  0.18 
6.0  0.45  0.91  0.25  0.36  0.35  0.35  0.19 
W_{B}  h_{1}  h_{2}  h_{3}  h_{4}  L  L_{1}  L_{2} 

4.  3.08  1.50  0.67  1.67  5.42  2.33  3.08 
5.  3.83  1.92  0.83  2.08  6.67  2.92  3.83 
6.  4.58  2.25  1.00  2.50  8.00  3.42  4.58 
7.  5.42  2.58  1.17  2.92  9.42  4.00  5.42 
8.  6.17  3.00  1.33  3.33  10.67  4.58  6.17 
9.  6.92  3.42  1.50  3.75  12.00  5.17  6.92 
10.  7.58  3.75  1.67  4.17  13.42  5.75  7.67 
11.  8.42  4.17  1.83  4.58  14.58  6.33  8.42 
12.  9.17  4.50  2.00  5.00  16.00  6.83  9.17 
13.  10.17  4.92  2.17  5.42  17.33  7.42  10.00 
14.  10.75  5.25  2.33  5.83  18.67  8.00  10.75 
15.  11.50  5.58  2.50  6.25  20.00  8.50  11.50 
16.  12.25  6.00  2.67  6.67  21.33  9.08  12.25 
17.  13.00  6.33  2.83  7.08  21.50  9.67  13.00 
18.  13.75  6.67  3.00  7.50  23.92  10.25  13.75 
19.  14.58  7.08  3.17  7.92  25.33  10.83  14.58 
20.  15.33  7.50  3.33  8.33  26.58  11.42  15.33 
W_{B}  W_{1}  W_{2}  t_{1}  t_{2}  t_{3}  t_{4}  t_{5} 

4.  0.33  1.08  0.50  0.50  0.50  0.50  0.25 
5.  0.42  1.42  0.50  0.50  0.50  0.50  0.25 
6.  0.50  1.67  0.50  0.50  0.50  0.50  0.25 
7.  0.50  1.92  0.50  0.50  0.50  0.50  0.25 
8.  0.58  2.17  0.50  0.58  0.58  0.50  0.25 
9.  0.67  2.50  0.58  0.58  0.67  0.58  0.25 
10.  0.75  2.75  0.67  0.67  0.75  0.67  0.25 
11.  0.83  3.00  0.67  0.75  0.75  0.67  0.33 
12.  0.92  3.00  0.67  0.83  0.83  0.75  0.33 
13.  1.00  3.00  0.67  0.92  0.83  0.83  0.33 
14.  1.08  3.00  0.67  1.00  0.92  0.92  0.42 
15.  1.17  3.00  0.67  1.00  1.00  1.00  0.42 
16.  1.25  3.00  0.75  1.00  1.00  1.00  0.50 
17.  1.33  3.00  0.75  1.08  1.00  1.00  0.50 
18.  1.33  3.00  0.75  1.08  1.08  1.08  0.58 
19.  1.42  3.00  0.83  1.17  1.08  1.08  0.58 
20.  1.50  3.00  0.83  1.17  1.17  1.17  0.67 
Figure 9.15. Energy Loss of USBR Type VI Impact Basin versus Hydraulic Jump
For erosion reduction and better basin operation, use the alternative end sill and 45^{o }wingwall design as shown in Figure 9.13. The sill should be set as low as possible to prevent degradation downstream. For best performance, the downstream channel should be at the same elevation as the top of the sill. A slot should be placed in the end sill to provide for drainage during periods of low flow. Although the basin is depressed, the slot allows water to drain into the surrounding soil.
For protection against undermining, a cutoff wall should be added at the end of the basin. Its depth will depend on the type of soil present. Riprap should be placed downstream of the basin for a length of at least four conduit widths. For riprap size recommendations see Chapter 10.
The Los Angeles experiments simulated discharges up to 11.3 m^{3}/s (400 ft^{3}/s) and entrance velocities as high as 15.2 m/s (50 ft/s). Therefore, use of the basin is limited to installations within these parameters. Velocities up to 15.2 m/s (50 ft/s) can be used without subjecting the structure to damage from cavitation forces. Some structures already constructed have exceeded these thresholds suggesting there may be some design flexibility. For larger installations where discharge is separable, two or more structures may be placed side by side. The USBR Type VI is not recommended where debris or ice buildup may cause substantial clogging.
The recommended design procedure for the USBR Type VI impact basin is as follows:
Step 1. Determine the maximum discharge, Q, and velocity, V_{o} and check against design limits. Compute the flow area at the end of the approach pipe, A. Compute equivalent depth, y_{e} = (A/2)^{1/2}.
Step 2. Compute the Froude number, Fr, and the energy at the end of the pipe, H_{o}.
Step 3. Determine H_{o}/W_{B} from Figure 9.14. Calculate the required width of basin, W_{B}.
W_{B} = H_{o} / (H_{o}/ W_{B})
Step 4. Obtain the remaining dimensions of the USBR Type VI impact basin from Table 9.2 using W_{B} obtained from step 3.
Step 5. Determine exit velocity, V_{B} = V_{2}, by trial and error using an energy balance between the culvert exit and the basin exit. Determine if this velocity is acceptable and whether or not riprap protection is needed downstream (see Chapter 10.)
H_{B} = Q/(W_{B}V_{B}) + V_{B}^{2}/(2g) = H_{o}(1 H_{L}/H_{o})
This equation is a cubic equation yielding 3 solutions, two positive and one negative. The negative solution is discarded. The two positive roots yield a subcritical and supercritical solution. Where low or no tailwater exists, the supercritical solution is taken. Where sufficient tailwater exists, the subcritical solution is taken.
Design Example: USBR Type VI Impact Basin (SI)
Determine the USBR Type VI impact basin dimensions for use at the outlet of a concrete pipe. Compare the design with a dissipator at the end of a rectangular concrete channel. Given:
Solution
First design dissipator for the pipe conduit.
Step 1. Determine the maximum discharge, Q, and velocity, V_{o}. Compute the flow area at the end of the approach pipe, A. Compute equivalent depth, y_{e} = (A/2)^{1/2}.
Since Q is less than 11.3 m^{3}/s and V_{o} less than 15.2 m/s, the dissipator can be tried at this site.
A = Q/V_{o} = 8.5/12.192 = 0.697 m^{2}
y_{e} = (A/2)^{1/2} = (0.697/2)^{1/2} = 0.590 m
Step 2. Compute the Froude number, Fr, and the energy at the end of the pipe, H_{o}.
Fr = V_{o} /(gy_{e})^{1/2} = 12.192/ [9.81(0.590)]^{1/2} = 5.07
H_{o} = y_{e} + V_{o}^{2} /(2g) = 0.590 + (12.192)^{2} /19.62 = 8.166 m
Step 3. Determine H_{o}/ W_{B} from Figure 9.14. Calculate the required width of basin, W_{B}.
W_{B} = H_{o} / (H_{o}/ W_{B}) = 8.166 /1.68 = 4.86 m
Step 4. Obtain the remaining dimensions of the USBR Type VI impact basin from Table 9.2 using W_{B} = 5.0 m obtained from step 3. (The basin width is taken to the nearest 0.5 m.) Results are summarized in the following table.
Step 5. Determine exit velocity, V_{B} = V_{2}, by trial and error using an energy balance between the culvert exit and the basin exit. Determine if this velocity is acceptable and whether or not riprap protection is needed downstream (see Chapter 10.)
H_{B} = Q/(W_{B}V_{B}) + V_{B}^{2}/(2g) = H_{o}(1 H_{L}/H_{o})
H_{B} = 8.5/(5.0V_{B}) + V_{B}^{2}/19.62 = 8.166(1 0.67)
H_{B} = 1.7/V_{B} + V_{B}^{2}/19.62 = 2.695
No tailwater exists so the supercritical solution is chosen. By trial and error, V_{B} = 6.9 m/s, therefore velocity has been reduced from 12.2 m/s to 6.9 m/s.
Compare the design for the circular pipe with a second USBR Type VI impact basin at the end of a long rectangular concrete channel. The computations and comparison with the pipe are tabulated below. W_{B} = 5.5 m for this case.
Approach Channel  Circular Pipe  Rectangular Channel 

Depth of flow y_{o} (m)  0.701  0.701 
Area of flow, A (m^{2})  0.697  0.855 
Velocity, V_{o} (m/s)  12.192  14.419 
Equivalent depth, y_{e} (m)  0.590  0.701 
Velocity Head, V_{o}^{2}/(2g) (m)  7.576  7.861 
H_{o} = y_{e} + V_{o}^{2}/2g (m)  8.166  8.562 
Fr = V_{o}/(gy_{e})0.5  5.07  4.74 
H_{o}/ W_{B} from Figure 9.14  1.68  1.55 
Width of basin, W_{B} (m)  5.0  5.5 
H_{L}/H_{o} from Figure 9.15  67%  65% 
Design Example: USBR Type VI Impact Basin (CU)
Determine the USBR Type VI impact basin dimensions for use at the outlet of a concrete pipe. Compare the design with a dissipator at the end of a rectangular concrete channel. Given:
Solution
First design dissipator for the pipe conduit.
Step 1. Determine the maximum discharge, Q, and velocity, V_{o}. Compute the flow area at the end of the approach pipe, A. Compute equivalent depth, y_{e} = (A/2)^{1/2}.
Since Q is less than 400 ft^{3}/s and V_{o} less than 50 ft/s, the dissipator can be tried at this site.
A = Q/V_{o} = 300/40 = 7.5 ft^{2}
y_{e} = (A/2)^{1/2} = (7.5/2)^{1/2} = 1.94 ft
Step 2. Compute the Froude number, Fr, and the energy at the end of the pipe, H_{o}.
Fr = V_{o} /(gy_{e})^{1/2} = 40/ [32.2(1.94)]^{1/2} = 5.06
H_{o} = y_{e} + V_{o}^{2} /(2g) = 1.94 + (40)^{2} /64.4 = 26.8 ft
Step 3. Determine H_{o}/ W_{B} from Figure 9.14. Calculate the required width of basin, W_{B}.
W_{B} = H_{o} / (H_{o}/ W_{B}) = 26.8 /1.68 = 16 ft
Step 4. Obtain the remaining dimensions of the USBR Type VI impact basin from Table 9.2 using W_{B} = 16 ft obtained from step 3. (The basin width is taken to the nearest 1 ft.) Results are summarized in the following table.
Step 5. Determine exit velocity, V_{B} = V_{2}, by trial and error using an energy balance between the culvert exit and the basin exit. Determine if this velocity is acceptable and whether or not riprap protection is needed downstream (see Chapter 10.)
H_{B} = Q/(W_{B}V_{B}) + V_{B}^{2}/(2g) = H_{o}(1 H_{L}/H_{o})
H_{B} = 300/(16V_{B}) + V_{B}^{2}/64.4 = 26.8(1 0.67)
H_{B} = 18.75/V_{B} + V_{B}^{2}/64.4 = 8.84
No tailwater exists so the supercritical solution is chosen. By trial and error, V_{B} = 22.7 ft/s, therefore velocity has been reduced from 40 ft/s to 22.7 ft/s.
Compare the design for the circular pipe with a second USBR Type VI impact basin at the end of a long rectangular concrete channel. The computations and comparison with the pipe are tabulated below. W_{B} = 18 ft for this case.
Approach Channel  Circular Pipe  Rectangular Channel 

Depth of flow y_{o} (ft)  2.3  2.3 
Area of flow, A (ft^{2})  7.5  9.2 
Velocity, V_{o} (ft/s)  40  40.9 
Equivalent depth, y_{e} (ft)  1.9  2.3 
Velocity Head, V_{o}^{2}/(2g) (ft)  24.9  26 
H_{o} = y_{e} + V_{o}^{2}/2g (ft)  26.8  28.3 
Fr = V_{o}/(gy_{e})0.5  5.06  4.75 
H_{o}/ W_{B} from Figure 9.14  1.68  1.55 
Width of basin, W_{B} (ft)  16  18 
H_{L}/H_{o} from Figure 9.15  67%  65% 
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