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Hydraulic Design of Energy Dissipators for Culverts and Channels
Hydraulic Engineering Circular Number 14, Third Edition

Chapter 9: Streambed Level Dissipators

This chapter contains energy dissipators for culvert outlets that are designed to operate at the streambed level and reestablish natural flow conditions downstream from the culvert outlet. They are also intended to drain by gravity when not in operation. The following sections contain limitations, design guidance, and design examples for the following energy dissipators:

  • Colorado State University (CSU) rigid boundary basin
  • Contra Costa basin
  • Hook basin
  • U.S. Bureau of Reclamation (USBR) Type VI impact basin

9.1 CSU Rigid Boundary Basin

The Colorado State University (CSU) rigid boundary basin, illustrated in Figure 9.1, uses staggered rows of roughness elements to initiate a hydraulic jump (Simons, 1970). CSU tested a number of basins with different roughness configurations to determine the average drag coefficient over the roughened portion of the basins. The effects of the roughness elements are reflected in a drag coefficient that was derived empirically for each roughness configuration. The experimental procedure was to measure depths and velocities at each end of the control volume illustrated in Figure 9.2, and compute the basin drag coefficient, CB, from the momentum equation by balancing the forces acting on the volume of fluid.

Figure 9.1. CSU Rigid Boundary Basin

Schematic showing the dimensions of the CSU Rigid Boundary Basin as described in the text.

The CSU test results indicate several design limitations. The height of the roughness elements, h, must be between 0.31 and 0.91 of the approach flow average depth, yA, and, the relative spacing, L/h, between rows of elements, must be either 6 or 12. The latter is not a severe restriction since relative spacing is normally a fixed parameter in a design procedure and other tests (Morris, 1968) have shown that the best range for energy dissipation is from 6 to 12.

Figure 9.2. Definition Sketch for the Momentum Equation

Defines depth and velocity at the culvert outlet, location A (a distance twice the culvert width downstream of the culvert outlet), and location B at the basin exit. The basin length, L sub B, is the distance from the culvert outlet to the basin exit.

The roughness configurations tested and the corresponding test results for CB are shown in Figure 9.3 and Table 9.1, respectively. To design a basin, the designer selects a basin from Figure 9.3 and uses the CB value from Table 9.1 in the following momentum equation to determine the velocity from the basin (VB) if the slope is less than 10%:

(9.1)

ρVoQ + Cp γ (yo2 /2)Wo = CBAFN ρ VA2 /2 + ρVBQ + γ Q2 /(2VB2 WB)

where,

yo = depth at the culvert outlet, m (ft)

Vo = velocity at the culvert outlet, m/s (ft/s)

Wo = culvert width at the culvert outlet, m (ft)

VA = approach velocity at two culvert widths downstream of the culvert outlet, m/s (ft/s)

VB = exit velocity, just downstream of the last row of roughness elements, m/s (ft/s)

WB = basin width, just downstream of the last row of roughness elements, m/s (ft/s)

N = total number of roughness elements in the basin

AF = frontal area of one full roughness element, m2 (ft2)

CB = basin drag coefficient (see Figure 9.3)

Cp = momentum correction coefficient for the pressure at the culvert outlet (see Figure 9.4)

γ = unit weight of water, 9810 N/m3 (62.4 lbs/ft3)

ρ = density of water, 1000 kg/m3 (1.94 slugs/ft3)

The CB values listed are for expansion ratios, WB /Wo, from 4 to 8 based on the configurations tested. CB values developed for the WB/Wo = 4 configuration are also valid for expansion ratios less than 4, but greater than or equal to 2, as long as the same number of roughness elements, N, are placed in the basin. For these smaller expansion ratios, this may require increasing the number of rows, Nr, to achieve the required N shown in Figure 9.3. The elements for all basins are arranged symmetrical about the basin centerline. All basins are flared to the width WB of the corresponding abrupt expansion basin.

Figure 9.3. Roughness Configurations Tested

Defines roughness configurations for five basin width to culvert width ratios: 4, 5, 6, 7, and 8.

Table 9.1. Design Values for Roughness Elements
WB/Wo2 to 45678
W1/Wo0.570.630.60.580.62
Rows (Nr)456456456566
Elements (N)141721151923172227243030
Rectangularh/yAL/hBasin Drag Coefficient, CB
0.9160.320.280.240.320.280.240.310.270.230.260.220.22
0.7160.440.400.370.420.380.350.400.360.330.340.310.29
0.48120.600.550.510.560.510.470.530.480.430.460.390.35
0.37120.680.660.650.650.620.600.620.580.550.540.500.45
Circular0.9160.210.200.480.210.190.170.210.190.170.180.16 
0.7160.290.270.400.270.250.230.250.230.220.220.20 
0.3160.380.360.340.360.340.320.340.320.300.300.28 
0.48120.450.420.250.400.380.360.360.340.320.300.28 
0.37120.520.500.180.480.460.440.440.420.400.380.36 

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 straight-line water surface profile across the jump:

(9.2)

CP γ yo2 Wo /2 + ρVoQ + w (sin θ) = CB AF N ρVA2 /2 +γ Q2 /(2VB2 WB) + ρVB Q

where,

w = weight of water within the basin

Volume = (yo Wo + yA WA) Wo + (0.75LQ/ VB) [(Nr -1) - (WB/Wo - 3) (1 - WA /WB)/2]

Weight = (Volume) γ

θ = arc tan of the channel slope, So

Nr = number of rows of roughness elements

L = longitudinal spacing between rows of elements.

Figure 9.4. Energy and Momentum Coefficients (Simons, 1970)

Figure defines C sub P for rectangular and circular culverts. For rectangular culverts, C sub P equals 0.7. For circular culverts C sub P equals 0.56 for reference discharges less than 4 and decreases linearly to 0.3 as the reference discharge increases to 6. (Reference discharge is Ku times Q divided by D sub o to the five-halves power; Ku equals 1.811 is metric units and 1.0 in customary units.) Other information on this figure is not used in this manual.

The depth yA 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 VA can be computed using Equations 4.1 or 4.2. Where slopes are greater than 10 percent, VA and yA 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)

2Wo So + yA + (0.25) (Q/(WA yA))2 /2g = yo + 0.25(Vo2 /(2g))

where,

WA = Wo [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 + yA + 0.5(VA sinφ)2 /g

where,

FB = necessary freeboard, m (ft)

h = roughness element height, m (ft)

yA = depth approaching first row of roughness elements, m (ft)

g = 9.81 m/s2 (32.2 ft/s2)

φ = 45° (function of yA/h and the Froude number but no relationship has been derived)

φ is believed to be a function of yA/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

Plan and profile view showing location of the splash shield with its leading edge flush with the front edge of the first row of roughness elements. The splash shield has a width of L divided by 2 located a distance h sub 3 above the basin floor. h sub 3 equals 1.5 times y sub c plus h plus h sub 2.

Although the tests were made with abrupt expansions, the configurations recommended for use are the combination flared-abrupt 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, ue, is a function of the longitudinal spacing between rows of elements, L, and the culvert barrel width, Wo:

(9.5)

ue = 4/7 + (10/7)L/Wo

The design procedure for the CSU rigid boundary basin may be summarized as follows:

Step 1. Compute the velocity, Vo, depth, yo, 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 WB /Wo expansion ratio that best matches the site geometry or satisfies other constraints. Choose WB /Wo, number of rows, Nr, number of elements, N, and ratios h/yA and L/h.

Step 3. Determine the flow condition VA and yA at the approach to the roughness element field (two culvert widths downstream).

Calculate VA using Equations 4.1 or 4.2.

For 4 < WB /Wo < 8, read yA from Figure 4.3 and Figure 4.4.

For WB /Wo < 4, compute yA using Equation 9.3.

For slopes > 10 percent, use Equation 9.3 to find both VA and yA

Step 4. For the trial roughness height to depth ratio h/yA and length to height ratio determine dissipator parameters from Figure 9.3:

  1. roughness element height, h
  2. longitudinal spacing between rows of elements, L
  3. width of basin, WB
  4. element width, W1, which equals element spacing
  5. divergence, ue
  6. basin drag, CB
  7. roughness element frontal area, AF = W1 h
  8. Cp from Figure 9.4
  9. Total basin length, LB = 2Wo + LNr. This provides a length downstream of the last row of elements equal to the length between rows, L.

Step 5. Confirm that the trial basin produces an exit velocity, VB, and depth, yB, that matches the downstream conditions. If WB matches the downstream channel width or tailwater controls follow option 1. If WB is less than the channel width follow option 2.

Option 1. Use the downstream depth, yn, or tailwater if higher, to solve Equation 9.1 or Equation 9.2 for the quantity CBAFN. Using the CB, AF, and N values found in steps 2 and 4 compute CBAFN (for basin). The basin value should be greater than or equal to the CBAFN value from the equation. If not, select a new roughness configuration.

Option 2. Use the CB, AF, and N values found in steps 2 and 4 to solve for VB in Equation 9.1 or 9.2. Three solutions for VB 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 VB. If VB does not match the downstream velocity, select a new roughness configuration. If VB is satisfactory, calculate yB. Compare yB to yn. If yB < yn, use the smaller positive root for VB and calculate yB. If tailwater is greater than yB, VB 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 W1/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 site-specific 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:

  • RCB = 2438 x 2438 mm culvert:
  • L = 71.6 m
  • S = 0.02 m/m
  • Q = 39.64 m3/s
  • n = 0.013
  • yC = 2.987 m
  • yn = 1.829 m
  • Downstream natural channel:
  • W = 12.5 m (width)
  • TW = 1.00 m (from downstream control)

Solution

Step 1. Compute the velocity, Vo, depth, yo, and Froude number, Fr, at the culvert outlet

yo = yn = 1.829 m (from HDS 3)

Vo = Vn = 8.87 m/s

Fr = Vo/(g yo)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 WB/Wo expansion ratio which best matches the site geometry or satisfies other constraints. Choose WB /Wo, number of rows, Nr, number of elements, N, and ratios h/yA and L/h.

Channel Width/Culvert Width = 12.5/2.438 = 5.1

Try the following rectangular basin:

WB /Wo = 5 and W1/Wo = 0.63

Nr = 4 and N =15

h/yA = 0.71 and L/h = 6

Step 3. Determine the flow condition VA and yA at the approach to the roughness element field (two culvert widths downstream) = 2Wo or 2(2.438) = 4.876 m.

Calculate VA using Equations 4.1 or 4.2.

VA /Vo = 1.65 - 0.3Fr = 1.65 - 0.3(2.1) = 1.02 from Equation 4.1

VA = 8.870(1.02) = 9.047 m/s

For 4 < WB /Wo < 8, read yA from Figure 4.3 or Figure 4.4.

yA /yo = 0.33 from Figure 4.3 for Fr = 2.1 and L = 2B

yA = 1.829(0.33) = 0.604 m

Step 4. For the trial roughness height to depth ratio h/yA and length to height ratio determine dissipator parameters from Table 9.1:

  1. roughness element height, h = (h/yA)yA = 0.71(0.604) = 0.429 m
  2. spacing between rows of elements, L = (L/h)h = 6(0.429) = 2.574 m
  3. width of basin, WB = (WB /Wo)Wo = 5(2.438) = 12.190 m
  4. element width, W1 = (W1/Wo) Wo = 0.63(2.438) =1.536 m; use 1.524 m
  5. divergence, ue = 4/7+10L/(7Wo) = 4/7 + 10(2.574)/(7(2.438)) = 2.07 use 2
  6. basin drag, CB = 0.42
  7. roughness element frontal area, AF = W1 h = 1.524(0.429) = 0.65 m2
  8. Cp from Figure 9.4 = 0.7
  9. Total basin length, LB = 2Wo + LNr = 2(2.438) + 4(2.574) = 15.172 m

Step 5. Since the width of the basin (WB = 12.190 m) matches the downstream channel width (12.5 m), confirm trial basin using option 1. Use the normal flow conditions (Vn and yn) and solve Equation 9.1 for CBAFN, which will be compared to CBAFN for basin:

Calculate CBAFN from Equation 9.1

yn Downstream = 1.001 m

VB = Q/ (WByn) = 39.64/ [12.190(1.001)] = 39.64/12.178 = 3.255 m/s

ρVoQ + Cp γ Yo2Wo /2 = CBAFN ρ VA2/2 + ρVBQ +γQ2 /(2VB2 WB) (Equation 9.1)

Terms with Vo and yo: 1000(8.870) (39.64) + 0.7(9810) (1.829)2 (2.438)/2 = 379609

Terms with VB: 1000(3.255) (39.64) + 9810(39.64)2 /(2 (3.225)2 (12.190)) = 189820

Term with CBAFN is CBAFN (1000)(9.047)2 /2 = 40924 CBAFN

(379609 - 189820) = 40924 CBAFN

CBAFN = 4.63

Calculate CBAFN for basin based on parameters determined in steps 2 and 4 (N =15, CB = 0.42, AF = 0.65 m2). Using these values CBAFN = 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/yA and length to height ratio determine dissipator parameters from Table 9.1: WB /Wo = 5 that had Nr = 5, N =19, h/yA = 0.71, L/h = 6, and CB = 0.38.

  1. roughness element height, h = (h/yA)yA = 0.71(0.604) = 0.429 m
  2. spacing between rows of elements, L = (L/h)h = 6(0.429) = 2.574 m
  3. width of basin, WB = (WB /Wo)Wo = 5(2.438) = 12.190 m
  4. element width, W1 = (W1/Wo) Wo = 0.63(2.438) =1.536 m; use 1.524 m (5 ft)
  5. divergence, ue = 4/7+10L/(7Wo) = 4/7 + 10(2.574)/(7(2.438)) = 2.07 use 2
  6. basin drag, CB = 0.38
  7. roughness element frontal area, AF = W1 h = 1.524(0.429) = 0.654 m2
  8. Cp from Figure 9.4 = 0.7
  9. Total basin length, LB = 2Wo + LNr = 2(2.438) + 5(2.574) = 17.746 m

Step 5 (2nd iteration). Calculate CBAFN from Equation 9.1.

CBAFN from Equation 9.1 = 4.63 (basin width did not change)

Calculate CBAFN for basin

CBAFN 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.)

W1/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)

Sketch showing the dimensions calculated for the SI CSU Rigid Boundary Basin design example

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:

  • RCB = 8 ft x 8 ft culvert
  • L = 235 ft
  • S = 0.02 ft/ft
  • Q = 1400 ft3/s
  • n = 0.013
  • yC= 9.8 ft
  • yn = 6.0 ft
  • Downstream natural channel
  • W = 41 ft (width)
  • TW = 3.3 ft (from downstream control)

Solution

Step 1. Compute the velocity, Vo, depth, yo, and Froude number, Fr, at the culvert outlet

yo = yn = 6.0 ft (from HDS 3)

Vo = Vn = 29.1 ft/s

Fr = Vo/ (g yo)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 WB/Wo expansion ratio which best matches the site geometry or satisfies other constraints. Choose WB /Wo, number of rows, Nr, number of elements, N, and ratios h/yA and L/h.

Channel Width/Culvert Width = 41/8 = 5.13

Try the following rectangular basin:

WB /Wo = 5 and W1/Wo = 0.63

Nr = 4 and N =15

h/yA = 0.71 and L/h = 6

Step 3. Determine the flow condition VA and yA at the approach to the roughness element field (two culvert widths downstream) = 2Wo or 2(8) = 16 ft.

Calculate VA using Equations 4.1 or 4.2.

VA /Vo = 1.65 - 0.3Fr = 1.65 - 0.3(2.1) = 1.02 from Equation 4.1

VA = 29.1(1.02) = 29.7 ft/s

For 4 < WB /Wo < 8, read yA from Figure 4.3 or 4.4.

yA /yo = 0.33 from Figure 4. 3 for Fr = 2.1 and L = 2B

yA = 6.0(0.33) = 1.98 ft

Step 4. For the trial roughness height to depth ratio h/yA and length to height ratio determine dissipator parameters from Table 9.1:

  1. roughness element height, h = (h/yA)yA = 0.71(1.98) = 1.4 ft
  2. spacing between rows of elements, L = (L/h)h = 6(1.4) = 8.4 ft
  3. width of basin, WB = (WB /Wo)Wo = 5(8) = 40 ft
  4. element width, W1 = (W1/Wo) Wo = 0.63(8) = 5.04 ft; use 5 ft
  5. divergence, ue = 4/7+10L/(7Wo) = 4/7 + 10(8.4)/(7(8)) = 2.07 use 2
  6. basin drag, CB = 0.42
  7. roughness element frontal area, AF = W1 h = 5(1.4) = 7 ft2
  8. Cp from Figure 9.4 = 0.7
  9. Total basin length, LB = 2Wo + LNr = 2(8) + 4(8.4) = 49.6 ft

Step 5. Since the width of the basin (WB = 40 ft) matches the downstream channel width (41 ft) confirm trial basin using option 1. Use the normal flow conditions (Vn and yn) and solve Equation 9.1 for CBAFN, which will be compared to CBAFN for basin:

Calculate CBAFN from Equation 9.1

yn Downstream = 3.3 ft

VB = Q/(WByn) = 1400/ [40(3.3)] = 1400/132 = 10.6 ft/s

ρVoQ + Cp γ Yo2Wo /2 = CBAFN ρ VA2/2 + ρVBQ +γQ2 /(2VB2 WB) (Equation 9.1)

Terms with Vo and yo: 1.94(29.1) (1400) + 0.7(62.4) (6)2 (8)/2 = 85,325.5

Terms with VB: 1.94(10.6) (1400) + 62.4(1400)2 / (2 (10.6)2 (40)) = 42,395.9

Term with CBAFN is CBAFN (1.94) (30.6)2 /2 = 908.3(CBAFN)

(85,325.5 - 42,395.9 = 908.3(CBAFN)

CBAFN = 47.3

Calculate CBAFN for basin based on parameters determined in steps 2 and 4 ((Nr = 4, CB = 0.42, AF = 7 ft2). Using these values CBAFN = 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/yA and length to height ratio determine dissipator parameters from Table 9.1: WB /Wo = 5 that had Nr = 5, N =19, h/yA = 0.71, L/h = 6, and CB = 0.38.

  1. roughness element height, h = (h/yA)yA = 0.71(1.98) = 1.4 ft
  2. spacing between rows of elements, L = (L/h)h = 6(1.4) = 8.4 ft
  3. width of basin, WB = (WB /Wo)Wo = 5(8) = 40 ft
  4. element width, W1 = (W1/Wo) Wo = 0.63(8) = 5.04 ft; use 5 ft
  5. divergence, ue = 4/7+10L/(7Wo) = 4/7 + 10(8.4)/(7(8)) = 2.07 use 2
  6. basin drag, CB = 0.38
  7. roughness element frontal area, AF = W1 h = 5(1.4) = 7 ft2
  8. Cp from Figure 9.4 = 0.7
  9. Total basin length, LB = 2Wo + LNr = 2(8) + 5(8.4) = 58 ft

Step 5 (2nd iteration). Calculate CBAFN from Equation 9.1.

CBAFN from Equation 9.1 = 47.3 (basin width did not change)

Calculate CBAFN for basin

CBAFN 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.)

W1/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)

Sketch showing the dimensions calculated for the CU CSU Rigid Boundary Basin design example

9.2 Contra Costa Basin

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 self-cleaning 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 L2 / h2 ratios from 2.5 to 7. The equation is in terms of culvert exit velocity, Vo, and depth, yo, for a circular culvert.

(9.6)

L sub 2 divided by h sub 2 equals 1.2 times Fr squared times (h sub 2 divided by y sub o) to the -1.83 power

where,

yo = outlet depth, m (ft)

Vo = outlet velocity, m/s (ft/s)

Fr = Vo/(g yo)1/2

h2 = height of large baffle, m (ft)

L2 = Length from culvert exit to large baffle, m (ft)

Figure 9.6. Contra Costa Basin

Definition schematic in profile and end views. Dimensions defined in the text.

Equation 9.6 is generalized for other shapes by substituting the equivalent depth of flow, ye, for yo. 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, ye = yn orybrink.

(9.7a)

L sub 2 divided by h sub 2 equals 1.35 times Fr squared times (h sub 2 divided by y sub e) to the -1.83 power

or:

(9.7b)

h sub 2 divided by y sub e equals (1.35 Fr squared divided by (L sub 2 divided by h sub 2)) to the 0.546 power

where,

ye = equivalent depth, (A/2)1/2 , m (ft)

A = outlet flow area, m2 (ft2)

Vo = outlet velocity, m/s (ft/s)

Fr = Vo/ (g ye)1/2

Equation 9.7b is solved by assuming a value of L2/h2 between 2.5 and 7. A trial height of the second baffle,h2, can be determined. If the recommended L2 /h2 = 3.5 value is substituted into Equation 9.7, the design equation becomes Equation 9.8. The value of h2/ye should always be greater than unity.

(9.8)

h sub 2 divided by y sub e equals 0.595 times Fr to the 1.092 power

After determining the values of h2 and L2, the length from the large baffle to the end sill, L3, can be obtained using Equation 9.9.

(9.9)

L sub 3 divided by L sub 2 equals 3.75 times (h sub 2 divided by L sub 2) to the 0.68 power

The height of the small baffle, h1, is half the height of the large baffle, h2. The position of the small baffle is half way between the culvert outlet and the large baffle or L2/2. 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, h3, may vary from 0.06y2 to 0.10y2. After obtaining satisfactory basin dimensions, the approximate maximum water surface depth, y2, without tailwater, can be obtained from Equation 9.10 which is for basins with WB/Wo = 2. The depth y3 is equal to yc for the basin + h3.

(9.10)

y sub 2 divided by h sub 2 equals 1.3 times (L sub 2 divided by h sub 2) to the 0.36 power

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, yo, is D/2 or less, the Contra Costa basin is applicable.

Step 2. Compute equivalent depth, ye, and Froude number, Fr.

ye = yo for rectangular culvert

ye = (A/2)1/2 for other shapes

Fr = Vo / (gye)1/2

Step 3. The width of the basin floor, WB, is selected to conform to the natural channel, but must be 1Wo to 3Wo. 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 L2/h2 between 2.5 and 7. If L2/h2 = 3.5, use Equation 9.8 to determine h2. Use Equation 9.7 for other values. Calculate L2 = 3.5 h2. Calculate the first baffle height, h1 = 0.5h2 and position, L1 = 0.5L2.

Step 5. Determine the length from the large baffle to the end sill, L3, 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, y2, using Equation 9.10 which is for WB = 2Wo. Set the end sill height, h3, between 0.06y2 and 0.1y2. If the above dimensions are compatible with the topography at the site, the dimensions are final. If not, a different value of L2 /h2 is selected and the design procedure repeated.

Step 7. Determine the basin exit depth, y3 = yc and exit velocity, V2 = Vc.

Q2/g = (Ac)3/Tc = [yc(WB + yc)]3/ (WB + 2yc) (Substituting for Ac and Tc using the properties of a trapezoid.)

Vc = Q/Ac

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:

  • D = 1.219 m diameter RCP culvert
  • Q = 8.49 m3/s
  • yo = 0.701 m
  • Vo = 12.192 m/s
  • Channel bottom width = 2.438 m

Solution

Step 1. Determine the flow conditions at the outfall of culvert for the design discharge.

yo = 0.701 m is approximately D/2, OK.

Step 2. Compute equivalent depth, ye, and Froude number, Fr.

Using Equations 7.11 and 7.13, flow area in the culvert = 0.696 m2.

ye = (A/2)1/2 = (0.696/2)1/2 = 0.590 m

Fr = Vo / (gye)1/2 = 12.192 / [9.81(0.590)]1/2 = 5.07

Step 3. The width of the basin floor, WB, 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 L2/h2 = 3.5, use Equation 9.8 to determine h2. Calculate L2 = 3.5 h2. Calculate the first baffle height, h1 = 0.5h2 and position, L1 = 0.5L2

h2 / ye = 0.595 Fr 1.092 = 0.595 (5.07)1.092 = 3.5

h2 = ye (h2 / ye) = 0.590 (3.50) = 2.065 m

L2 = 3.5 h2 = 3.5 (2.065) = 7.228 m

h1 = 0.5h2 = 0.5 (2.065) = 1.032 m

L1 = 0.5 L2 = 0.5 (7.228) = 3.614 m

Step 5. Determine the length from the large baffle to the end sill, L3, using Equation 9.9. Repeat the procedure, if necessary, until a dissipator is defined which optimizes the design requirements.

L3 / L2 = 3.75(h2 /L2)0.68 = 3.75(1/3.5)0.68 = 1.6

L3 = (L3 / L2) L2 = 1.6 (7.228) = 11.56 m

Step 6. Estimate the approximate maximum water surface depth without tailwater, y2, using Equation 9.10 which is for WB = 2D. Determine end sill height, h3 = 0.1y2

y2 / h2 = 1.3(L2 /h2)0.36 = 1.3(3.5)0.36 = 2.04

y2 = (y2 / h2) h2 = (2.04) 2.065 = 4.21 m

h3 = 0.1(y2) = 0.1(4.21) = 0.42 m

A summary of physical dimensions is shown in the following table.

Contra costa basin (SI): Summary of physical dimensions
 First BaffleSecond BaffleEnd Sill
Distance from exit (m)3.617.2318.79
Height (m)1.032.070.42

Step 7. Determine the basin exit depth, y3 = yc and exit velocity, V2 = Vc.

Q2/g = (Ac)3/Tc = [yc (WB + yc)]3/ (WB + 2yc)

8.492/9.81 = 7.35 = [yc (2.438 + yc)]3/ (2.438 + 2yc)

By trial and success, yc = 0.938 m, Tc = 4.314 m, Ac = 3.17 m2

Vc = Q/Ac = 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:

  • D = 4 ft diameter RCP culvert
  • Q = 300 ft3/s
  • yo = 2.3 ft
  • Vo = 40 ft/s
  • Channel bottom width = 8 ft

Solution

Step 1. Determine the flow conditions at the outfall of culvert for the design discharge.

yo = 2.3 ft is approximately D/2, OK.

Step 2. Compute equivalent depth, ye, and Froude number, Fr.

Using Equations 7.11 and 7.13, flow area in the culvert = 7.5 ft2.

ye = (A/2)1/2 = (7.5/2)1/2 = 1.94 ft

Fr = Vo / (gye)1/2 = 40 / [32.2(1.94)]1/2 = 5.06

Step 3. The width of the basin floor, WB, 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 L2/h2 = 3.5, use Equation 9.8 to determine h2. Calculate L2 = 3.5 h2. Calculate the first baffle height, h1 = 0.5h2 and position, L1 = 0.5L2

h2 / ye = 0.595 Fr 1.092 = 0.595 (5.06)1.092 = 3.5

h2 = ye (h2 / ye) = 1.94 (3.50) = 6.8 ft

L2 = 3.5 h2 = 3.5 (6.8) = 23.8 ft

h1 = 0.5h2 = 0.5 (6.8) = 3.4 ft

L1 = 0.5 L2 = 0.5 (23.8) = 11.9 ft

Step 5. Determine the length from the large baffle to the end sill, L3, using Equation 9.9. Repeat the procedure, if necessary, until a dissipator is defined which optimizes the design requirements.

L3 / L2 = 3.75(h2 /L2)0.68 = 3.75(1/3.5)0.68 = 1.6

L3 = (L3 / L2) L2 = 1.6 (23.8) = 38.1 ft

Step 6. Estimate the approximate maximum water surface depth without tailwater, y2, using Equation 9.10 which is for WB = 2D. Determine end sill height, h3 = 0.1y2

y2 / h2 = 1.3(L2 /h2)0.36 = 1.3(3.5)0.36 = 2.04

y2 = (y2 / h2) h2 = (2.04) 6.8 = 13.9 ft

h3 = 0.1(y2) = 0.1(13.9) = 1.4 ft

A summary of physical dimensions is shown in the following table.

Contra costa basin (CU): Summary of physical dimensions
 First BaffleSecond BaffleEnd Sill
Distance from exit (ft)11.923.861.9
Height (ft)3.46.81.4

Step 7. Determine the basin exit depth, y3 = yc and exit velocity, V2 = Vc.

Q2/g = (Ac)3/Tc = [yc (WB + yc)]3/ (WB + 2yc)

3002/32.2 = 2795 = [yc (8 + yc)]3/ (8 + 2yc) = 34.063/14.15 = 2792

By trial and success, yc = 3.075 ft, Tc = 14.15 ft, Ac = 34.06 ft2

Vc = Q/Ac = 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.

9.3 Hook Basin

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.

9.3.1 Hook Basin with Warped Wingwalls

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 W4/Wo = 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, h6, should be at least twice the flow depth at the culvert exit or 2ye. This height is based on the highest water surface elevations observed in the basin during the study. Therefore, setting h6 = 2ye 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

Definition sketch with plan, profile, and end views. Dimensions defined in the text.

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:

  1. h3 = ye
  2. h2= 1.28h1
  3. h1 = ye/1.4
  4. β = 135°
  5. r = 0.4h1

Figure 9.8. Hook for Warped Wingwall Basin

Definition sketch showing dimensions. r is the radius of the curved underside of the hook, Beta is the angle from the horizontal to the tip of the hook using the center point of the radius as the reference point. The following vertical distances measured from the basin floor are provided: h sub 1: to the center point of the radius, h sub 2: to the hook tip, and h sub 3: to the highest point of the curved underside of the hook.

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 LB = 3Wo 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, VB, is estimated from Figure 9.9. The higher the velocity ratio, Vo/VB, 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

Graph provides velocity ratio (V sub o divided by V sub B) as a function of Froude number at the culvert exit. Curves for L sub B equal to 3.0, 4.5, and 6.0 times W sub o are shown.

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, Vo, equivalent depth, ye, and Froude number, Fr = Vo/(gye)1/2. If 1.8 < Fr < 3.0, proceed with design.

Step 2. Compute the downstream channel velocity, Vn, and depth, yn.

Step 3. Select width of the basin at the basin exit, WB (WB = W6), and compute LB. W6 should be approximately equal to the channel width, if the downstream channel is defined.

LB = (W6 - Wo) / (2tan<span α), use α = 5.7° (tan = 0.10)

Step 4. Compute the position and spacingof the hooks (see Figure 9.7):

  1. distance to first hooks, L1 = 0.75 LB
    (allowable range: 0.75 < L1/LB < 0.80)
  2. width at first hooks, W1 = 2L1(tan α) + Wo
  3. distance between first (row A) hooks, W2 = 0.66 W 1
    (allowable range: 0.66 < W2/W1 < 0.70)
  4. distance to second (row B) hook, L2 = 0.83 LB
    (allowable range: 0.83 < L2/LB < 0.89)
  5. width of hooks, W4 = 0.16Wo
  6. lateral spacing between A and B hook, W3 = (W2 - W4)/2
    If spacing does not satisfy 1.5 < W3/W4 < 2.5, adjust W4.

Step 5. Compute hook dimensions (see Figure 9.8):

  1. height to center of radius, h1 = ye /1.4
  2. height to point, h2 = 1.28h1
  3. height to top of radius, h3 = ye
  4. angle of radius, β = 135°
  5. radius, r = 0.4h1

Step 6. Compute the end sill and wingwall dimensions (see Figure 9.7):

  1. height of end sill, h4 = 0.67ye
  2. width of slot in end sill, W5 = 0.33W6
  3. height to top of warped wingwall, h6 = 2ye minimum
  4. height to top of end sill, h5 = 0.94h6

Step 7. Find Vo/VB from Figure 9.9 and compute basin exit velocity, VB. Compare VB with Vn from step 2. If VB 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:

  • So = 0.020 m/m
  • n = 0.012
  • Q = 76.41 m3/s
  • ye = 1.829 m
  • Vo = 11.43 m/s

The downstream channel has a trapezoidal shape with the following properties:

  • Wc = 6.096 m
  • So = 0.020 m/m
  • Z = 1.5
  • n = 0.030
  • Vn= 5.27 m/s
  • yn= 1.676 m

Solution

Step 1. Compute the culvert outlet velocity, Vo, equivalent depth, ye, and Froude number, Fr, Vo/(gye)1/2. Vo = 11.43 m/s and ye = 1.829 m were given.

Fro = Vo / (gye)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, Vn, and depth, yn. Vn = 5.27 m/s and yn = 1.676 m were given.

Step 3. Select W6 and compute LB.

Use W6 = Wc = 6.096 m and tan α = 0.10.

LB = (W6 - Wo) / (2tan α) = (6.096 - 3.658)/ [2(0.10)] = 12.19 m or 3.3Wo

Step 4. Compute the position and spacing of the hooks (see Figure 9.7):

  1. distance to first hooks, L1 = 0.75 LB = 0.75(12.19) = 9.143 m
  2. width at first hooks, W1 = 2L1(tanα) + Wo = 2(9.143)(0.1) + 3.658 = 5.487 m
  3. distance between first (row A) hooks, W2 = 0.66 W 1 = 0.66(5.487) = 3.621 m
  4. distance to second (row B) hook, L2 = 0.83 LB = 0.83(12.19) = 10.118 m
  5. width of hooks, W4 = 0.16Wo = 0.16(3.658) = 0.585 m
  6. lateral spacing between A and B hook, W3
    W3 = (W2 - W4)/2 = (3.621 - 0.585) /2 = 1.518 m
    W3/W4 = 1.518/0.585 = 2.6, which does not satisfy 1.5 < W3/W4 < 2.5
    Adjust W4 = W3/2.5 = 1.518/2.5 = 0.607 m

Step 5. Compute hook dimensions (see Figure 9.8):

  1. height to center of radius, h1 = ye /1.4 = 1.829/1.4 = 1.306 m
  2. height to point, h2 = 1.28h1 = 1.28(1.306) = 1.672 m
  3. height to top of radius, h3 = ye = 1.829 m
  4. angle of radius, β = 135°
  5. radius, r = 0.4h1 = 0.4(1.306) = 0.522 m

Step 6. Compute the end sill and wingwall dimensions (see Figure 9.7):

  1. height of end sill, h4 = 0.67ye = 0.67(1.829) = 1.225 m
  2. width of slot in end sill, W5 = 0.33W6 = 0.33(6.096) = 2.012 m
  3. height to top of warped wingwall, h6 = 2ye = 2 (1.829) = 3.658 m
  4. height to top of end sill, h5 = 0.94h6 = 0.94 (2ye) = 3.439 m

Step 7. Find Vo/VB from Figure 9.9 and compute VB. Compare with Vn 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 LB = 3.3 Wo, Vo /VB will be less than 1.9 making VB ≈ 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)

Sketch showing the dimensions calculated for the SI Hook Basin with Warped Wingwalls design example

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:

  • So = 0.020 ft/ft
  • n = 0.012
  • Q = 2700 ft3/s
  • ye = 6 ft
  • Vo = 37.5 ft/s

The downstream channel has a trapezoidal shape with the following properties:

  • Wc = 20 ft
  • So = 0.020 ft/ft
  • Z = 1.5
  • n = 0.030
  • Vn = 16.1 ft/s
  • yn = 5.5 ft

Solution

Step 1. Compute the culvert outlet velocity, Vo, equivalent depth, ye, and Froude number, Fr Vo/(gye)1/2. Vo = 37.5 ft/s and ye = 6 ft were given.

Fro = Vo / (gye)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, Vn, and depth, yn. Vn = 16.1 ft/s and yn = 5.5 ft were given.

Step 3. Select W6 and compute LB.

Use W6 = Wc = 20 and tan<span α = 0.10.

LB = (W6 - Wo) / (2tan α) = (20 - 12)/ [2(0.10)] = 40 ft or 3.3Wo

Step 4. Compute the position and spacing of the hooks (see Figure 9.7):

  1. distance to first hooks, L1 = 0.75 LB = 0.75(40) = 30 ft
  2. width at first hooks, W1 = 2L1(tanα) + Wo = 2(30)(0.1) + 12 = 18 ft
  3. distance between first (row A) hooks, W2 = 0.66 W 1 = 0.66(18) = 11.9 ft, use 12 ft
  4. distance to second (row B) hook, L2 = 0.83 LB = 0.83(40) = 33.2 ft, use 33 ft
  5. width of hooks, W4 = 0.16Wo = 0.16(12) = 1.92 ft, use 2 ft
  6. lateral spacing between A and B hook, W3
    W3 = (W2 - W4)/2 = (12 - 2) /2 = 5 ft
    W3/W4 = 5/2 = 2.5, which satisfies 1.5 < W3/W4 < 2.5

Step 5. Compute hook dimensions (see Figure 9.8):

  1. height to center of radius, h1 = ye /1.4 = 6/1.4 = 4.3 ft
  2. height to point, h2 = 1.28h1 = 1.28(4.3) = 5.5 ft
  3. height to top of radius, h3 = ye = 6 ft
  4. angle of radius, β = 135°
  5. radius, r = 0.4h1 = 0.4(4.3) = 1.72 ft

Step 6. Compute the end sill and wingwall dimensions (see Figure 9.7):

  1. height of end sill, h4 = 0.67ye = 0.67(6) = 4 ft
  2. width of slot in end sill, W5 = 0.33W6 = 0.33(20) = 6.6 ft
  3. height to top of warped wingwall, h6 = 2ye = 2 (6) = 12 ft
  4. height to top of end sill, h5 = 0.94h6 = 0.94 (2ye) = 11.3 ft

Step 7. Find Vo/VB from Figure 9.9 and compute VB. Compare with Vn 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 LB = 3.3 Wo, Vo /VB will be less than 1.9 making VB ≈ 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)

Sketch showing the dimensions calculated for the CU Hook Basin with Warped Wingwalls design example

9.3.2 Hook Basin with Uniform Trapezoidal Channel

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

Definition sketch with plan, profile, and end views. Dimensions defined in the text.

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:

  1. h1 = 0.78ye
  2. h2 = ye
  3. h3 = 1.4h1
  4. β = 135°
  5. r = 0.4h1

Figure 9.11. Hook for Uniform Trapezoidal Channel Basin

Definition sketch showing dimensions. r is the radius of the curved underside of the hook, Beta is the angle from the horizontal to the tip of the hook using the center point of the radius as the reference point. The following vertical distances measured from the basin floor are provided: h sub 1: to the center point of the radius, h sub 2: to the hook tip, and h sub 3: to the highest point of the curved underside of the hook.

Figure 9.12. Velocity Ratio for Hook Basin With Uniform Trapezoidal Channel

Graph provides velocity ratio (V sub o divided by V sub B) as a function of Froude number at the culvert exit. Curves for W sub 6 equal to 1.0 and 2.0 times W sub o are shown. Other variables shown are explained in the text.

The design procedure for a Hook basin with a uniform trapezoidal channel is as follows:

Step 1. Compute the culvert outlet velocity, Vo, equivalent depth, ye = (A/2)1/2, and Froude number, Fr = Vo/(gye)1/2.

Culvert width, Wo = width of rectangular culvert or Wo = 2ye for circular and other shapes.

If 1.8 < Fr < 3.0, proceed with design.

Step 2. Compute the downstream channel velocity, Vn, and depth, yn.

Step 3. Select a basin width, WB (W6 = WB), side slope, and length, LB. W6 should be approximately equal to the channel width if the downstream channel is defined.

W6 = Wo to 2Wo

Basin side slope can be from 1:1.5 to 1:2 (V:H).

LB = 3.0Wo

Step 4. Compute the position and spacing of the hoooks (see Figure 9.10):

  1. distance to first hooks, L1 = 1.25 Wo
  2. width at first hooks, W1 = Wo
  3. distance between first hooks, W2 = 0.65 Wo
  4. distance to second hook, L2 = 2.085 Wo
  5. width of hooks, W4 = 0.16Wo
  6. spacing between first and second hook, W3 = (W2 - W4)/2
    If spacing does not satisfy W3/W4≥ 1.0, adjust W6.

Step 5. Compute hook dimensions (see Figure 9.11):

  1. height to center of radius, h1 = 0.78ye
  2. height to point, h2 = ye
  3. height to top of radius, h3 = 1.4h1
  4. angle of radius, β = 135°
  5. radius, r = 0.4h1

Step 6. Compute the end sill and wingwall dimensions (see Figure 9.10):

  1. height of end sill, h4 = 0.67ye
  2. width of slot in end sill, W5 = 0.33WB
  3. height to top of side slope, h6
    h6 = 3.33 ye for 1:1.5 side slopes
    h6 = 2.69ye for 1:2 side slopes
  4. height to top of end sill, h5 = 0.70h6

Step 7. Find Vo/VB from Figure 9.12 and calculate VB. Compare with Vn from step 2. If VB is unacceptable, adjust W6, 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:

  • So = 0.020 m/m
  • n= 0.012
  • Q = 76.41 m3/s
  • ye = 1.829 m
  • Vo =11.43 m/s

The downstream channel has a trapezoidal shape with the following properties:

  • Wc = 6.096 m
  • So = 0.020 m/m
  • Z = 1.5
  • n = 0.030
  • Vn = 5.27 m/s
  • yn = 1.676 m

Solution

Step 1. Compute the culvert outlet velocity, Vo, equivalent depth, ye, and Froude number, Fr = Vo/(gye)1/2. Vo = 11.43 m/s and ye = 1.829 m were given.

Fro = Vo / (gye)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, Vn, and depth, yn. Vn = 5.273 m/s and yn = 1.676 m were given.

Step 3. Select a basin width, W6, side slope, and length, LB. W6 should be approximately equal to the channel width, if the downstream channel is defined.

W6 = Wc = 6.096 m, which is 6.096/3.658 = 1.67 Wo

Basin side slope will be 1:1.5 (V:H)

LB = 3.0Wo = 3.0(3.658) = 10.974 m

Step 4. Compute the position and spacing of the hooks (see Figure 9.10):

  1. distance to first hooks, L1 = 1.25 Wo = 1.25 (3.658) = 4.573 m
  2. width at first hooks, W1 = Wo = 3.658 m
  3. distance between first hooks, W2 = 0.65 Wo = 0.65 (3.658) = 2.377 m
  4. distance to second hook, L2 = 2.085 Wo = 2.085 (3.658) = 7.627 m
  5. width of hooks, W4 = 0.16Wo = 0.16(3.658) = 0.585 m
  6. spacing between first and second hook, W3
    W3 = (W2 - W4)/2 = (2.377 - 0.585)/2 = 0.896 m
    W3 /W4 = 0.896/0.585 = 1.5, which satisfies W3/W4≥ 1.0.

Step 5. Compute hook dimenstions (See Figure 9.11):

  1. height to center of radius, h1 = 0.78ye = 1.427 m
  2. height to point, h2 = ye = 1.829 m
  3. height to top of radius, h3 = 1.4h1 = 1.4(1.427) = 1.998 m
  4. angle of radius, β = 135°
  5. radius, r = 0.4h1 = 0.4(1.427) = 0.571 m

Step 6. Compute the end sill and wingwall dimensions (see Figure 9.10):

  1. height of end sill, h4 = 0.67ye = 0.67(1.829) = 1.225 m
  2. width of slot in end sill, W5 = 0.33WB = 0.33(6.096) = 2.012 m
  3. height to top of side slope, h6 for 1:1.5 (V:H) side slopes
    h6 = 3.33 ye = 3.33(1.829) = 6.091 m
  4. height to top of end sill, h5 = 0.70h6 = 0.70(6.091) = 4.264 m

Step 7. Find Vo/VB from Figure 9.12 and compute VB. Compare with Vn 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 W6/Wo = 1.67, Vo/VB ≈ 2.0 making VB ≈ 11.43/2 = 5.72 m/s which is slightly higher than the normal channel velocity, Vn = 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.

Design Example Results (SI)
FeatureElementSymbolWarped Wingwall (m)Trapezoidal (m)
BasinLengthLB12.1910.974
WidthWB6.0966.096
First HooksLengthL19.1434.573
SpacingW23.6212.377
Second HookLengthL210.1187.682
SpacingW31.5180.896
End WallHeighth41.2251.225
SlotW52.0122.012
Toph53.4394.264
HooksHeighth31.8291.998
WidthW40.6070.585

Sketch for the Hook Basin with a Uniform Trapezoidal Channel (SI)

Sketch showing the dimensions calculated for the SI Hook Basin with Uniform Trapezoidal Channel design example

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:

  • So = 0.020 ft/ft
  • n = 0.012
  • Q = 2700 ft3/s
  • ye = 6 ft
  • Vo = 37.5 ft/s

The downstream channel has a trapezoidal shape with the following properties:

  • Wc = 20 ft
  • So = 0.020 ft/ft
  • Z = 1.5
  • n = 0.030
  • Vn = 16.1 ft/s
  • yn = 5.5 ft

Solution

Step 1. Compute the culvert outlet velocity, Vo, equivalent depth, ye, and Froude number, Fr = Vo/(gye)1/2. Vo = 37.5 ft/s and ye = 6 ft were given.

Fro = Vo / (gye)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, Vn, and depth, yn. Vn = 16.1 ft/s and yn = 5.5 ft were given.

Step 3. Select a basin width, W6, side slope, and length, LB. W6 should be approximately equal to the channel width, if the downstream channel is defined.

W6 = Wc = 20 ft, which is 20/12 = 1.67 Wo

Basin side slope will be 1:1.5 (V:H)

LB = 3.0Wo = 3.0(12) = 36 ft

Step 4. Compute the position and spacing of the hooks (see Figure 9.10):

  1. distance to first hooks, L1 = 1.25 Wo = 1.25 (12) = 15 ft
  2. width at first hooks, W1 = Wo = 12 ft
  3. distance between first hooks, W2 = 0.65 Wo = 0.65 (12) = 7.8 ft
  4. distance to second hook, L2 = 2.085 Wo = 2.085 (12) = 25 ft
  5. width of hooks, W4 = 0.16Wo = 0.16(12) = 1.92 ft (use 2 ft)
  6. spacing between first and second hook, W3
    W3 = (W2 - W4)/2 = (7.8 - 2)/2 = 2.9 ft W3 /W4 = 2.9/2 = 1.45, which satisfies W3/W4≥ 1.0.

Step 5. Compute hook dimensions (see Figure 9.11):

  1. height to center of radius, h1 = 0.78ye = 0.78 (6) = 4.7 ft
  2. height to point, h2 = ye = 6 ft
  3. height to top of radius, h3 = 1.4h1 = 1.4(4.7) = 6.6 ft
  4. angle of radius, β = 135°
  5. radius, r = 0.4h1 = 0.4(4.7) = 1.9 ft

Step 6. Compute the end sill and wingwall dimensions (see Figure 9.10):

  1. height of end sill, h4 = 0.67ye = 0.67(6) = 4 ft
  2. width of slot in end sill, W5 = 0.33WB = 0.33(20) = 6.6 ft
  3. height to top of side slope, h6 for 1:1.5 (V:H) side slopes
    h6 = 3.33 ye = 3.33(6) = 20 ft
  4. height to top of end sill, h5 = 0.70h6 = 0.70(20) = 14 ft

Step 7. Find Vo/VB from Figure 9.12 and compute VB. Compare with Vn 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 W6/Wo = 1.67, Vo/VB ≈ 2.0 making VB ≈ 37.5/2 = 18.8 ft/s which is slightly higher than the normal channel velocity, Vn = 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.

Design Example Results (SI)
FeatureElementSymbolWarped Wingwall (ft)Trapezoidal (ft)
BasinLengthLB3640
WidthWB2020
First HooksLengthL11530
SpacingW27.812
Second HookLengthL22533
SpacingW32.95
End WallHeighth444
SlotW56.66.6
Toph51411.3
HooksHeighth36.66
WidthW422

Sketch for Hook Basin with a Uniform Trapezoidal Channel (CU)

Sketch showing the dimensions calculated for the CU Hook Basin with Uniform Trapezoidal Channel design example

9.4 USBR Type VI Impact Basin

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 box-like 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

Definition sketch with plan and profile views. Dimensions defined in the text.

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

Shows curve to determine H sub o divided by W sub B as a function of the Froude number.

Once the basin width, WB, 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 WB 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, ye = (A/2)1/2, entering the dissipator from a pipe or irregular-shaped 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 h3 + (h2/2) which is half of the baffle.

The basin floor should be constructed horizontally and will operate effectively with entrance conduits on slopes up to 15o (27%). For entrance conduits with slopes greater than 15o, 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 one-sixth 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.

Table 9.2 (SI). USBR Type VI Impact Basin Dimensions (m) (AASHTO, 1999)
WBh1h2h3H4LL1L2
1.00.790.380.170.431.400.590.79
1.51.160.570.250.622.000.881.16
2.01.540.750.330.832.681.141.54
2.51.930.940.421.043.331.431.93
3.02.301.120.501.254.021.722.30
3.52.681.320.581.464.652.002.68
4.03.121.510.671.675.332.283.08
4.53.461.680.751.886.002.563.46
5.03.821.870.832.086.522.843.82
5.54.192.030.912.297.293.124.19
6.04.602.251.002.507.983.424.60
Table 9.2 (SI). USBR Type VI Impact Basin Dimensions (m) (AASHTO, 1999) (continued)
WBW1W2t1t2t3t4t5
1.00.080.260.150.150.150.150.08
1.50.130.420.150.150.150.150.08
2.00.150.550.150.150.150.150.08
2.50.180.680.160.180.180.160.08
3.00.220.830.200.200.220.200.08
3.50.260.910.200.230.230.210.10
4.00.300.910.200.280.250.250.10
4.50.360.910.200.300.300.300.13
5.00.390.910.220.310.300.300.15
5.50.410.910.220.330.330.330.18
6.00.450.910.250.360.350.350.19
Table 9.2 (CU). USBR Type VI Impact Basin Dimensions (m) (AASHTO, 1999)
WBh1h2h3h4LL1L2
4.3.081.500.671.675.422.333.08
5.3.831.920.832.086.672.923.83
6.4.582.251.002.508.003.424.58
7.5.422.581.172.929.424.005.42
8.6.173.001.333.3310.674.586.17
9.6.923.421.503.7512.005.176.92
10.7.583.751.674.1713.425.757.67
11.8.424.171.834.5814.586.338.42
12.9.174.502.005.0016.006.839.17
13.10.174.922.175.4217.337.4210.00
14.10.755.252.335.8318.678.0010.75
15.11.505.582.506.2520.008.5011.50
16.12.256.002.676.6721.339.0812.25
17.13.006.332.837.0821.509.6713.00
18.13.756.673.007.5023.9210.2513.75
19.14.587.083.177.9225.3310.8314.58
20.15.337.503.338.3326.5811.4215.33
Table 9.2 (CU). USBR Type VI Impact Basin Dimensions (m) (AASHTO, 1999) (continued)
WBW1W2t1t2t3t4t5
4.0.331.080.500.500.500.500.25
5.0.421.420.500.500.500.500.25
6.0.501.670.500.500.500.500.25
7.0.501.920.500.500.500.500.25
8.0.582.170.500.580.580.500.25
9.0.672.500.580.580.670.580.25
10.0.752.750.670.670.750.670.25
11.0.833.000.670.750.750.670.33
12.0.923.000.670.830.830.750.33
13.1.003.000.670.920.830.830.33
14.1.083.000.671.000.920.920.42
15.1.173.000.671.001.001.000.42
16.1.253.000.751.001.001.000.50
17.1.333.000.751.081.001.000.50
18.1.333.000.751.081.081.080.58
19.1.423.000.831.171.081.080.58
20.1.503.000.831.171.171.170.67

Figure 9.15. Energy Loss of USBR Type VI Impact Basin versus Hydraulic Jump

Compares energy loss (H sub L divided by H sub o) as a function of Froude number by showing curves for the impact basin and a hydraulic jump on a horizontal floor. The former always has a higher energy loss.

For erosion reduction and better basin operation, use the alternative end sill and 45o 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 m3/s (400 ft3/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, Vo and check against design limits. Compute the flow area at the end of the approach pipe, A. Compute equivalent depth, ye = (A/2)1/2.

Step 2. Compute the Froude number, Fr, and the energy at the end of the pipe, Ho.

Step 3. Determine Ho/WB from Figure 9.14. Calculate the required width of basin, WB.

WB = Ho / (Ho/ WB)

Step 4. Obtain the remaining dimensions of the USBR Type VI impact basin from Table 9.2 using WB obtained from step 3.

Step 5. Determine exit velocity, VB = V2, 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.)

HB = Q/(WBVB) + VB2/(2g) = Ho(1- HL/Ho)

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:

  • D = 1.219 m (pipe diameter and rectangular channel width)
  • So = 0.15 m/m
  • Q = 8.5 m3/s (pipe)
  • Q = 10.6 m3/s (channel)
  • n = 0.015
  • Vo = 12.192 m/s
  • yo = 0.701 m (for both pipe and channel)

Solution

First design dissipator for the pipe conduit.

Step 1. Determine the maximum discharge, Q, and velocity, Vo. Compute the flow area at the end of the approach pipe, A. Compute equivalent depth, ye = (A/2)1/2.

Since Q is less than 11.3 m3/s and Vo less than 15.2 m/s, the dissipator can be tried at this site.

A = Q/Vo = 8.5/12.192 = 0.697 m2

ye = (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, Ho.

Fr = Vo /(gye)1/2 = 12.192/ [9.81(0.590)]1/2 = 5.07

Ho = ye + Vo2 /(2g) = 0.590 + (12.192)2 /19.62 = 8.166 m

Step 3. Determine Ho/ WB from Figure 9.14. Calculate the required width of basin, WB.

WB = Ho / (Ho/ WB) = 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 WB = 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, VB = V2, 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.)

HB = Q/(WBVB) + VB2/(2g) = Ho(1- HL/Ho)

HB = 8.5/(5.0VB) + VB2/19.62 = 8.166(1 -0.67)

HB = 1.7/VB + VB2/19.62 = 2.695

No tailwater exists so the supercritical solution is chosen. By trial and error, VB = 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. WB = 5.5 m for this case.

USBR Type VI Impact Basin Example Results (SI)
Approach ChannelCircular PipeRectangular Channel
Depth of flow yo (m)0.7010.701
Area of flow, A (m2)0.6970.855
Velocity, Vo (m/s)12.19214.419
Equivalent depth, ye (m)0.5900.701
Velocity Head, Vo2/(2g) (m)7.5767.861
Ho = ye + Vo2/2g (m)8.1668.562
Fr = Vo/(gye)0.55.074.74
Ho/ WB from Figure 9.141.681.55
Width of basin, WB (m)5.05.5
HL/Ho from Figure 9.1567%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:

  • D = 4 ft (pipe diameter and rectangular channel width)
  • So = 0.15 ft/ft
  • Q = 300 ft3/s (pipe)
  • Q = 375 ft3/s (channel)
  • n = 0.015
  • Vo = 40 ft/s
  • yo = 2.3 ft (for both pipe and channel)

Solution

First design dissipator for the pipe conduit.

Step 1. Determine the maximum discharge, Q, and velocity, Vo. Compute the flow area at the end of the approach pipe, A. Compute equivalent depth, ye = (A/2)1/2.

Since Q is less than 400 ft3/s and Vo less than 50 ft/s, the dissipator can be tried at this site.

A = Q/Vo = 300/40 = 7.5 ft2

ye = (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, Ho.

Fr = Vo /(gye)1/2 = 40/ [32.2(1.94)]1/2 = 5.06

Ho = ye + Vo2 /(2g) = 1.94 + (40)2 /64.4 = 26.8 ft

Step 3. Determine Ho/ WB from Figure 9.14. Calculate the required width of basin, WB.

WB = Ho / (Ho/ WB) = 26.8 /1.68 = 16 ft

Step 4. Obtain the remaining dimensions of the USBR Type VI impact basin from Table 9.2 using WB = 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, VB = V2, 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.)

HB = Q/(WBVB) + VB2/(2g) = Ho(1- HL/Ho)

HB = 300/(16VB) + VB2/64.4 = 26.8(1 -0.67)

HB = 18.75/VB + VB2/64.4 = 8.84

No tailwater exists so the supercritical solution is chosen. By trial and error, VB = 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. WB = 18 ft for this case.

USBR Type VI Impact Basin Example Results (CU)
Approach ChannelCircular PipeRectangular Channel
Depth of flow yo (ft)2.32.3
Area of flow, A (ft2)7.59.2
Velocity, Vo (ft/s)4040.9
Equivalent depth, ye (ft)1.92.3
Velocity Head, Vo2/(2g) (ft)24.926
Ho = ye + Vo2/2g (ft)26.828.3
Fr = Vo/(gye)0.55.064.75
Ho/ WB from Figure 9.141.681.55
Width of basin, WB (ft)1618
HL/Ho from Figure 9.1567%65%
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Resource Center (Atlanta)
404-562-3908
cynthia.nurmi@dot.gov

Updated: 04/07/2011
 

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