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

Chapter 8: Stilling Basins

Stilling basins are external energy dissipators placed at the outlet of a culvert, chute, or rundown. These basins are characterized by some combination of chute blocks, baffle blocks, and sills designed to trigger a hydraulic jump in combination with a required tailwater condition. With the required tailwater, velocity leaving a properly designed stilling basin is equal to the velocity in the receiving channel.

Depending on the specific design, they operate over a range of approach flow Froude numbers from 1.7 to 17 as summarized in Table 8.1. This chapter includes the following stilling basins: USBR Type III, USBR Type IV, and SAF. The United States Bureau of Reclamation (USBR) basins were developed based on model studies and evaluation of existing basins (USBR, 1987). The St. Anthony Falls (SAF) stilling basin is based on model studies conducted by the Soil Conservation Service at the St. Anthony Falls Hydraulic Laboratory of the University of Minnesota (Blaisdell, 1959).

Table 8.1. Applicable Froude Number Ranges for Stilling Basins
Stilling BasinMinimum Approach Froude NumberMaximum Approach Froude Number
USBR Type III4.517
USBR Type IV2.54.5
SAF1.717

The selection of a stilling basin depends on several considerations including hydraulic limitations, constructibility, basin size, and cost. The design examples in this chapter all use the identical site conditions to provide a comparison between the size of basins and a free hydraulic jump basin for one case. Table 8.2 summarizes the results of these examples with the incoming Froude number, the required tailwater at the exit of the basin along with basin length and depth. For this example, the SAF stilling basin results in the shortest and shallowest basin. Details of the design procedures and this design example are found in the following sections.

Table 8.2. Example Comparison of Stilling Basin Dimensions
Basin Type1Froude NumberRequired Tailwater3, m (ft)Basin Length,m (ft)Basin Depth,m (ft)
Free jump7.63.1 (10.1)33.7 (109.2)4.8 (15.5)
USBR Type III6.93.0 (9.6)20.6 (67.3)3.8 (12.5)
USBR Type IV28.03.5 (11.2)38.1 (121.8)5.5 (17.4)
SAF6.12.4 (7.9)12.4 (39.7)2.7 (8.6)
1Based on a 3 m by 1.8 m (10 ft by 6 ft) box culvert at a design discharge of 11.8 m3/s (417 ft3/s). All basins have a constant width equal to the culvert width. Detailed description of the example is found in Section 8.1.
2The USBR Type IV approach Froude number is outside of the recommended range, but was included for comparison.
3Required tailwater influences basin depth. Velocity leaving each of these basins is the same and depends on the tailwater channel.

8.1 Expansion And Depression For Stilling Basins

As explained in Chapter 4, the higher the Froude number at the entrance to a basin, the more efficient the hydraulic jump and the shorter the resulting basin. To increase the Froude number as the water flows from the culvert to the basin, an expansion and depression is used as is shown in Figure 8.1. The expansion and depression converts depth, or potential energy, into kinetic energy by allowing the flow to expand, drop, or both. The result is that the depth decreases and the velocity and Froude number increase.

Figure 8.1. Definition Sketch for Stilling Basin

Schematic (Plan and Profile) showing the dimensions of a generalized stilling basin as described in the text. Of primary interest is the location of section 1 at the entrance to the basin and section 2 at the exit of the basin.

The Froude number used to determine jump efficiency and to evaluate the suitability of alternative stilling basins as described in Table 8.1 is defined in Equation 8.1.

(8.1)

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1)

where,

Fr1 = Froude number at the entrance to the basin

V1 = velocity entering the basin, m/s (ft/s)

y1 = depth entering the basin, m (ft)

g = acceleration due to gravity, m/s2 (ft/s2)

To solve for the velocity and depth entering the basin, the energy balance is written from the culvert outlet to the basin. Substituting Q/(y1WB) for V1 and solving for Q results in:

(8.2)

Q equals y sub 1 times W sub B times (2 times g times (z sub o minus z sub 1 plus y sub o minus y sub 1) plus V sub o squared) raised to the one-half

where,

WB = width of the basin, m/s (ft/s)

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

y1 = depth entering the basin, m (ft)

yo = culvert outlet depth, m (ft)

z1 = ground elevation at the basin entrance, m (ft)

zo = ground elevation at the culvert outlet, m (ft)

Equation 8.2 has three unknowns y1, WB, and z1. The depth y1 can be determined by trial and error if WB and z1 are assumed. WB should be limited to the width that a jet would flare naturally in the slope distance L.

(8.3)

W sub B is less than or equal to W sub o plus 2 times L sub T times square root of (S sub T squared plus 1) divided by (3 times Fr sub o)

where,

LT = length of transition from culvert outlet to basin, m (ft)

ST = slope of the transition, m/m (ft/ft)

Fro = outlet Froude number

Since the flow is supercritical, the trial y1 value should start near zero and increase until the design Q is reached. This depth, y1, is used to find the sequent (conjugate) depth, y2, using the hydraulic jump equation:

(8.4)

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1)

where,

y2 = conjugate depth, m (ft)

y1 = depth approaching the jump, m (ft)

C = ratio of tailwater to conjugate depth, TW/y2

Fr1 = approach Froude number

For a free hydraulic jump, C = 1.0. Later sections on the individual stilling basin types provide guidance on the value of C for those basins. For the jump to occur, the value of y2 + z2 must be equal to or less than TW + z3 as shown in Figure 8.1. If z2 + y2 is greater than z3 +TW, the basin must be lowered and the trial and error process repeated until sufficient tailwater exists to force the jump.

In order to perform this check, z3 and the basin lengths must be determined. The length of the transition is calculated from:

(8.5)

L sub T equals (z sub o minus z sub 1) divided by S sub T

where,

LT = length of the transition from the culvert outlet to the bottom of the basin, m (ft)

ST = slope of the transition entering the basin, m/m (ft/ft)

The length of the basin, LB, depends on the type of basin, the entrance flow depth, y1, and the entrance Froude number, Fr1. Figure 8.2 describes these relationships for the free hydraulic jump as well as several USBR stilling basins.

Figure 8.2. Length of Hydraulic Jump on a Horizontal Floor

Graph with L sub B divided by y sub 2 versus Fr sub 1. In increasing order, the graph shows curves for the Type III, Type II, and Free Jump. The Type IV curve is also shown as an extension of the Free Jump curve for low Froude numbers.

The length of the basin from the floor to the sill is calculated from:

(8.6)

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o)

where,

LS = length of the basin from the bottom of the basin to the basin exit (sill), m (ft)

SS = slope leaving the basin, m/m (ft/ft)

The elevation at the entrance to the tailwater channel is then calculated from:

(8.7)

z sub 3 equals (L sub S times S sub S) plus z sub 1

where,

z3 = elevation of basin at basin exit (sill), m (ft)

Figure 8.1 also illustrates a radius of curvature between the culvert outlet and the transition to the stilling basin. If the transition slope is 0.5V:1H or steeper, use a circular curve at the transition with a radius defined by Equation 8.8 (Meshgin and Moore, 1970). It is also advisable to use the same curved transition going from the transition slope to the stilling basin floor.

(8.8)

r equals y divided by (e raised to (1.5 divided by Fr squared) minus 1)

where,

r = radius of the curved transition, m (ft)

Fr = Froude number

y = depth approaching the curvature, m (ft)

For the curvature between the culvert outlet and the transition, the Froude number and depth are taken at the culvert outlet. For the curvature between the transition and the stilling basin floor, the Froude number and depth are taken as Fr1 and y1.

8.2 General Design Procedure

The design procedure for all of these stilling basins may be summarized in the following steps. Basin specific variations to these steps are discussed in the following sections on each basin.

Step 1. Determine the velocity and depth at the culvert outlet. For the culvert outlet, calculate culvert brink depth, yo, velocity, Vo, and Fro. For subcritical flow, use Figure 3.3 or Figure 3.4. For supercritical flow, use normal depth in the culvert for yo. (See HDS 5 (Normann, et al., 2001) for additional information on culvert brink depths.)

Step 2. Determine the velocity and TW depth in the receiving channel downstream of the basin. Normal depth may be determined using Table B.1 or other appropriate technique.

Step 3. Estimate the conjugate depth for the culvert outlet conditions using Equation 8.4 to determine if a basin is needed. Substitute yo and Fro for y1 and Fr1, respectively. The value of C is dependent, in part, on the type of stilling basin to be designed. However, in this step the occurrence of a free hydraulic jump without a basin is considered so a value of 1.0 is used. Compare y2 and TW. If y2 < TW, there is sufficient tailwater and a jump will form without a basin. The remaining steps are unnecessary.

Step 4. If step 3 indicates a basin is needed (y2 > TW), make a trial estimate of the basin bottom elevation, z1, a basin width, WB, and slopes ST and SS. A slope of 0.5 (0.5V:1H) or 0.33 (0.33V:1H) is satisfactory for both ST and Ss. Confirm that WB is within acceptable limits using Equation 8.3. Determine the velocity and depth conditions entering the basin and calculate the Froude number. Select candidate basins based on this Froude number.

Step 5. Calculate the conjugate depth for the hydraulic conditions entering the basin using Equation 8.4 and determine the basin length and exit elevation. Basin length and exit elevation are computed using Equations 8.5, 8.6, and 8.7 as well as Figure 8.2. Verify that sufficient tailwater exists to force the hydraulic jump. If the tailwater is insufficient go back to step 4. If excess tailwater exists, the designer may either go on to step 6 or return to step 4 and try a shallower (and smaller) basin.

Step 6. Determine the needed radius of curvature for the slope changes entering the basin using Equation 8.8.

Step 7. Size the basin elements for basin types other than a free hydraulic jump basin. The details for this process differ for each basin and are included in the individual basin sections.

Design Example: Stilling Basin with Free Hydraulic Jump (SI)

Find the dimensions for a stilling basin (see Figure 8.1) with a free hydraulic jump providing energy dissipation for a reinforced concrete box culvert. Given:

  • Q = 11.8 m3/s
  • Culvert
  • B = 3.0 m
  • D = 1.8 m
  • n = 0.015
  • So = 0.065 m/m
  • zo = 30.50 m
  • Downstream channel (trapezoidal)
  • B = 3.10 m
  • Z = 1V:2H
  • n = 0.030

Solution

Step 1. Determine the velocity and depth at the culvert outlet. By trial and error using Manning's Equation, the normal depth is calculated as:

Vo = 8.50 m/s, yo = 0.463 m

Fr sub o equals V sub o divided by square root of (g times y sub o) equals 8.50 divided by square root of (9.81 times 0.463) equals 4.0

Since the Froude number is greater than 1.0, the normal depth is supercritical and the normal depth is taken as the brink depth.

Step 2. Determine the velocity and depth (TW) in the receiving channel. By trial and error using Manning's Equation or by using Table B.1:

Vn = 4.84 m/s, yn = TW = 0.574 m

Step 3. Estimate the conjugate depth for the culvert outlet conditions using Equation 8.4. C = 1.0.

y sub 2 equals C times y sub o divided by 2 times (square root of (1 plus 8 times Fr sub o squared) minus 1) equals 1.0 times 0.463 divided by 2 times (square root of (1 plus 8 times 4.0 squared) minus 1) equals 2.4 meters

Since y2 (2.4 m) > TW (0.574 m) a jump will not form and a basin is needed.

Step 4. Since y2 - TW = 2.64 - 0.574 = 2.07 m, try z1 = zo - 2.07 = 28.4 m

Also, choose WB = 3.0 m (no expansion from culvert to basin) and slopes ST = 0.5 and SS = 0.5.

Check WB using Equation 8.3, but first calculate the transition length from Equation 8.5.

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (30.50 minus 28.4) divided by 0.5 equals 4.2 meters

W sub B is less than or equal to W sub o plus 2 times L sub T times square root of (S sub T squared plus 1) divided by (3 times Fr sub o) equals 3.0 plus 2 times 4.2 times square root of (0.5 squared plus 1) divided by (3 times 4.0) equals 3.8 metersWB is OK

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 10.74 m/s, y1 = 0.366 m

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 10.74 divided by square root of (9.81 times 0.366) equals 5.7

Step 5. Calculate the conjugate depth for a free hydraulic jump (C=1) using Equation 8.4.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 1.0 times 0.366 divided by 2 times (square root of (1 plus 8 times 5.7 squared) minus 1) equals 2.77 meters

From Figure 8.2 basin length, LB/y2 = 6.1. Therefore, LB = 6.1(2.77) = 16.9 m.

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (4.2 times (0.5 minus 0.065) minus 16.9 times 0.065) divided by (0.5 plus 0.065) equals 1.29 meters

The elevation at the entrance to the tailwater channel is from Equation 8.7:

z sub 3 equals L sub S times S sub S plus Z sub 1 equals 1.29 times (0.5) plus 28.4 equals 29.05 meters

Since y2 +z2 (2.77+28.4) > z3 + TW (29.05+ 0.574), tailwater is not sufficient to force a jump in the basin. Go back to step 4.

Step 4 (2nd iteration). Try z1 = 25.7 m. Maintain WB, ST, and SS.

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (30.50 minus 25.7) divided by 0.5 equals 9.6 meters

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 13.02 m/s, y1 = 0.302 m

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 13.02 divided by square root of (9.81 times 0.302) equals 7.6

Step 5 (2nd iteration). Calculate the conjugate depth for a free hydraulic jump (C=1) using Equation 8.4.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 1.0 times 0.302 divided by 2 times (square root of (1 plus 8 times 7.6 squared) minus 1) equals 3.10 meters

From Figure 8.2 basin length, LB/y2 = 6.1. Therefore, LB = 6.1(3.10) = 18.9 m.

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (9.6 times (0.5 minus 0.065) minus 18.9 times 0.065) divided by (0.5 plus 0.065) equals 5.2 meters

The elevation at the entrance to the tailwater channel is from Equation 8.7:

z sub 3 equals L sub S times S sub S plus Z sub 1 equals 5.2 times (0.5) plus 25.7 equals 28.30 meters

Since y2 +z2 (3.10+25.7) < z3 + TW (28.30+ 0.574), tailwater is sufficient to force a jump in the basin. Continue on to step 6.

Step 6. For the slope change from the outlet to the transition, determine the needed radius of curvature using Equation 8.8 and the results from step 1.

r equals y divided by (e raised to (1.5 divided by Fr squared) minus 1) equals 0.463 divided by (e raised to (1.5 divided by 4.0 squared) minus 1) equals 4.71 meters

Step 7. Size the basin elements. Since this is a free hydraulic jump basin, there are no additional elements and the design is complete. The basin is shown in the following sketch.

Total basin length = 9.6 + 18.9 + 5.2 = 33.7 m

Sketch for Free Hydraulic Jump Stilling Basin Design Example (SI)

Schematic showing the dimensions calculated for the SI design example

Design Example: Stilling Basin with Free Hydraulic Jump (CU)

Find the dimensions for a stilling basin (see Figure 8.1) with a free hydraulic jump providing energy dissipation for a reinforced concrete box culvert. Given:

  • Q = 417 ft3/s
  • Culvert
  • B = 10.0 ft
  • D = 6 ft
  • n = 0.015
  • So= 0.065 ft/ft
  • zo = 100 ft
  • Downstream channel (trapezoidal)
  • B = 10.2 ft
  • Z = 1V:2H
  • n = 0.030

Solution

Step 1. Determine the velocity and depth at the culvert outlet. By trial and error using Manning's Equation, the normal depth is calculated as:

Vo = 27.8 ft/s, yo = 1.50 ft

Fr sub o equals V sub o divided by square root of (g times y sub o) equals 27.8 divided by square root of (32.2 times 1.50) equals 4.0

Since the Froude number is greater than 1.0, the normal depth is supercritical and the normal depth is taken as the brink depth.

Step 2. Determine the velocity and depth (TW) in the receiving channel. By trial and error using Manning's Equation or by using Table B.1:

Vn = 15.9 ft/s, yn = TW = 1.88 ft

Step 3. Estimate the conjugate depth for the culvert outlet conditions using Equation 8.4. C = 1.0.

y sub 2 equals C times y sub o divided by 2 times (square root of (1 plus 8 times Fr sub o squared) minus 1) equals 1.0 times 1.50 divided by 2 times (square root of (1 plus 8 times 4.0 squared) minus 1) equals 7.8 feet

Since y2 (7.8 ft) > TW (1.88 ft) a jump will not form and a basin is needed.

Step 4. Since y2 - TW = 8.55 - 1.88 = 6.67 ft, try z1 = zo -6.67 = 93.3 ft, use 93.

Also, choose WB = 10.0 ft (no expansion from culvert to basin) and slopes ST = 0.5 and SS = 0.5.

Check WB using Equation 8.3, but first calculate the transition length from Equation 8.5.

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (100 minus 93) divided by 0.5 equals 14 feet

W sub B is less than or equal to W sub o plus 2 times L sub T times square root of (S sub T squared plus 1) divided by (3 times Fr sub o) equals 10.0 plus 2 times 14 times square root of (0.5 squared plus 1) divided by (3 times 4.0) equals 12.6 feet; WB is OK

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 35.3 ft/s, y1 = 1.18 ft

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 35.3 divided by square root of (32.2 times 1.18) equals 5.7

Step 5. Calculate the conjugate depth for a free hydraulic jump (C=1) using Equation 8.4.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 1.0 times 1.18 divided by 2 times (square root of (1 plus 8 times 5.7 squared) minus 1) equals 8.94 feet

From Figure 8.2 basin length, LB/y2 = 6.1. Therefore, LB = 6.1(8.94) = 54.5 ft.

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (14 times (0.5 minus 0.065) minus 54.5 times 0.065) divided by (0.5 plus 0.065) equals 4.5 feet

The elevation at the entrance to the tailwater channel is from Equation 8.7:

z sub 3 equals L sub S times S sub S plus Z sub 1 equals 4.5 times (0.5) plus 93.0 equals 95.25 feet

Since y2 +z2 (8.94+93) > z3 + TW (95.25+1.88), tailwater is not sufficient to force a jump in the basin. Go back to step 4.

Step 4 (2nd iteration). Try z1 = 84.5 ft. Maintain WB, ST, and SS.

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (100 minus 84.5) divided by 0.5 equals 31.0 feet

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 42.5 ft/s, y1 = 0.98 ft

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 42.5 divided by square root of (32.2 times 0.98) equals 7.6

Step 5 (2nd iteration). Calculate the conjugate depth for a free hydraulic jump (C=1) using Equation 8.4.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 1.0 times 0.98 divided by 2 times (square root of (1 plus 8 times 7.6 squared) minus 1) equals 10.07 feet

From Figure 8.2 basin length, LB/y2 = 6.1. Therefore, LB = 6.1(10.07) = 61.4 ft.

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (31.0 times (0.5 minus 0.065) minus 61.4 times 0.065) divided by (0.5 plus 0.065) equals 16.8 feet

The elevation at the entrance to the tailwater channel is from Equation 8.7:

z sub 3 equals L sub S times S sub S plus Z sub 1 equals 16.8 times (0.5) plus 84.5 equals 92.90 feet

Since y2 + z2 (10.1 + 84.5) < z3 + TW (92.90 + 1.88), tailwater is sufficient to force a jump in the basin. Continue on to step 6.

Step 6. For the slope change from the outlet to the transition, determine the needed radius of curvature using Equation 8.8 and the results from step 1.

r equals y divided by (e raised to (1.5 divided by Fr squared) minus 1) equals 1.50 divided by (e raised to (1.5 divided by 4.0 squared) minus 1) equals 15.3 feet

Step 7. Size the basin elements. Since this is a free hydraulic jump basin, there are no additional elements and the design is complete. The basin is shown in the following sketch.

Total basin length = 31.0 + 61.4 + 16.8 = 109.2 ft

Sketch for Free Hydraulic Jump Stilling Basin Design Example (CU)

Schematic showing the dimensions calculated for the CU design example

8.3 USBR Type III Stilling Basin

The USBR Type III stilling basin (USBR, 1987) employs chute blocks, baffle blocks, and an end sill as shown in Figure 8.3. The basin action is very stable with a steep jump front and less wave action downstream than with the free hydraulic jump. The position, height, and spacing of the baffle blocks as recommended below should be adhered to carefully. If the baffle blocks are too far upstream, wave action in the basin will result; if too far downstream, a longer basin will be required; if too high, waves can be produced; and, if too low, jump sweep out or rough water may result.

The baffle blocks may be shaped as shown in Figure 8.3 or cubes; both are effective. The corners should not be rounded as this reduces energy dissipation.

The recommended design is limited to the following conditions:

  1. Maximum unit discharge of 18.6 m3/s/m (200 ft3/s/ft).
  2. Velocities up to 18.3 m/s (60 ft/s) entering the basin.
  3. Froude number entering the basin between 4.5 and 17.
  4. Tailwater elevation equal to or greater than full conjugate depth elevation. This provides a 15 to 18 percent factor of safety.
  5. The basin sidewalls should be vertical rather than trapezoidal to insure proper performance of the hydraulic jump.

Figure 8.3. USBR Type III Stilling Basin

Schematic showing the dimensions of the USBR Type III stilling basin as described in the text.

The general design procedure outlined in Section 8.1 applies to the USBR Type III stilling basin. Steps 1 through 4 and step 6 are applied without modification. For step 5, two adaptations to the general design procedure are made:

  1. For computing conjugate depth, C=1.0. (This value is also applicable for the free hydraulic jump.) At a minimum, C=0.85 could be used, but C=1.0 is recommended.
  2. For obtaining the length of the basin, LB, use Figure 8.2 based on the Froude number calculated in step 4.

For step 7, sizing the basin elements (chute blocks, baffle blocks, and an end sill), the following guidance is recommended. The height of the chute blocks, h1, is set equal to y1. If y1 is less than 0.2 m (0.66 ft), then h1 = 0.2 m (0.66 ft).

The number of chute blocks is determined by Equation 8.9 rounded to the nearest integer.

(8.9)

N sub c equals W sub B divided by (2 times y sub 1)

where,

Nc = number of chute blocks

Block width and block spacing are determined by:

(8.10)

W sub 1 equals W sub 2 equals W sub B divided by (2 times N sub c)

where,

W1 = block width, m (ft)

W2 = block spacing, m (ft)

Equations 8.9 and 8.10 will provide Nc blocks and Nc-1 spaces between those blocks. The remaining basin width is divided equally for spaces between the outside blocks and the basin sidewalls. With these equations, the height, width, and spacing of chute blocks should approximately equal the depth of flow entering the basin, y1. The block width and spacing may be reduced as long as W1 continues to equal W2.

The height, width, and spacing of the baffle blocks are shown on Figure 8.3. The height of the baffles is computed from the following equation:

(8.11)

h sub 3 equals y sub 1 times (0.168 Fr sub 1 plus 0.58)

where,

h3 = height of the baffle blocks, m (ft)

The top thickness of the baffle blocks should be set at 0.2h3 with the back slope of the block on a 1:1 slope. The number of baffle blocks is as follows:

(8.12)

N sub B equals W sub B divided by (1.5 times h sub 3)

where,

NB = number of baffle blocks (rounded to an integer)

Baffle width and spacing are determined by:

(8.13)

W sub 3 equals W sub 4 equals W sub B divided by (2 times N sub B)

where,

W3 = baffle width, m (ft)

W4 = baffle spacing, m (ft)

As with the chute blocks, Equations 8.12 and 8.13 will provide NB baffles and NB-1 spaces between those baffles. The remaining basin width is divided equally for spaces between the outside baffles and the basin sidewalls. The width and spacing of the baffles may be reduced for narrow structures provided both are reduced by the same amount. The distance from the downstream face of the chute blocks to the upstream face of the baffle block should be 0.8y2.

The height of the final basin element, the end sill, is given as:

(8.14)

h sub 4 equals y sub 1 times (0.0536 Fr sub 1 plus 1.04)

where,

h4 = height of the end sill, m (ft)

The fore slope of the end sill should be set at 0.5:1 (V:H).

If these recommendations are followed, a short, compact basin with good dissipation action will result. If they cannot be followed closely, a model study is recommended.

Design Example: USBR Type III Stilling Basin (SI)

Design a USBR Type III stilling basin for a reinforced concrete box culvert. Given:

  • Q = 11.8 m3/s
  • Culvert
  • B = 3.0 m
  • D = 1.8 m
  • n = 0.015
  • So = 0.065 m/m
  • zo = 30.50 m
  • Downstream channel (trapezoidal)
  • B = 3.10 m
  • Z = 1V:2H
  • n = 0.030

Solution

The culvert, design discharge, and tailwater channel are the same as considered for the free hydraulic jump stilling basin addressed in the design example in Section 8.1. Steps 1 through 3 of the general design process are identical for this example so they are not repeated here. The tailwater depth from the previous design example is TW=0.574 m.

Step 4. Try z1 = 26.7 m. WB = 3.0 m, ST = 0.5 m/m, and SS = 0.5 m/m. From Equation 8.5:

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (30.5 minus 26.7) divided by 0.5 equals 7.6 meters

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 12.2 m/s, y1 = 0.322 m

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 12.1 divided by square root of (9.81 times 0.322) equals 6.9

Step 5. Calculate the conjugate depth in the basin (C=1) using Equation 8.4.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 1.0 times 0.322 divided by 2 times (square root of (1 plus 8 times 6.9 squared) minus 1) equals 2.98 meters

From Figure 8.2 basin length, LB/y2 = 2.7. Therefore, LB = 2.7(2.98) = 8.0 m.

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (7.6 times (0.5 minus 0.065) minus 8.0 times 0.065) divided by (0.5 plus 0.065) equals 4.9 meters

The elevation at the entrance to the tailwater channel is from Equation 8.7:

z sub 3 equals L sub S times S sub S plus z sub 1 equals 4.9 times (0.5) plus 26.7 equals 29.15 meters

Since y2 +z2 (2.98+26.7) < z3 + TW (29.15+ 0.574), tailwater is sufficient to force a jump in the basin. If the tailwater had not been sufficient, repeat step 4 with a lower assumption for z1.

Step 6. Determine the needed radius of curvature for the slope changes entering the basin. See the design example Section 8.1 for this step. It is unchanged.

Step 7. Size the basin elements. For the USBR Type III basin, the elements include the chute blocks, baffle blocks, and end sill.

For the chute blocks, the height of the chute blocks, h1=y1=0.322 m (round to 0.32 m). The number of chute blocks is determined by Equation 8.9 and rounded to the nearest integer.

N sub c equals W sub B divided by (2 times y sub 1) equals 3.0 divided by (2 times 0.322) equals 4.6 is approximately equal to 5

Block width and block spacing are determined by Equation 8.10:

W sub 1 equals W sub 2 equals W sub B divided by (2 times N sub c) equals 3.0 divided by (2 times 5) equals 0.30 meters

With 5 blocks at 0.30 m and 4 spaces at 0.30 m, 0.30 m of space remains. This is divided equally for spaces between the outside blocks and the basin sidewalls.

For the baffle blocks, the height of the baffles is computed from Equation 8.11:

h sub 3 equals y sub 1 times (0.168 Fr sub 1 plus 0.58) equals 0.322 times (0.168 times (6.9) plus 0.58) equals 0.56 meters

The number of baffles blocks is calculated from Equation 8.12:

N sub B equals W sub B divided by (1.5 times h sub 3) equals 3.0 divided by (1.5 times 0.56) equals 3.6 is approximately equal to 4

Baffle width and spacing are determined by Equation 8.13:

W sub 3 equals W sub 4 equals W sub B divided by (2 times N sub B) equals 3 divided by (2 times 4) equals 0.38 meters

With 4 baffles at 0.38 m and 3 spaces at 0.38 m, 0.34 m of space remains. This is divided equally for spaces between the outside baffles and the basin sidewalls. The distance from the downstream face of the chute blocks to the upstream face of the baffle block should be 0.8y2=0.8(2.98)=2.38 m.

For the end sill, the height of the end sill is given by Equation 8.14:

h sub 4 equals y sub 1 times (0.0536 Fr sub 1 plus 1.04) equals 0.322 times (0.053 times (6.9) plus 1.04) equals 0.45 meters

Total basin length = 7.6 + 8.0 + 4.9 = 20.5 m. The basin is shown in the following sketch.

Sketch for USBR Type III Stilling Basin Design Example (SI)

Schematic showing the dimensions calculated for the SI design example

Design Example: USBR Type III Stilling Basin (CU)

Design a USBR Type III stilling basin for a reinforced concrete box culvert. Given:

  • Q = 417 ft3/s
  • Culvert
  • B = 10.0 ft
  • D = 6 ft
  • n = 0.015
  • So = 0.065 ft/ft
  • zo = 100 ft
  • Downstream channel (trapezoidal)
  • B = 10.2 ft
  • Z = 1V:2H
  • n = 0.030

Solution

The culvert, design discharge, and tailwater channel are the same as considered for the free hydraulic jump stilling basin addressed in the design example in Section 8.1. Steps 1 through 3 of the general design process are identical for this example so they are not repeated here. The tailwater depth from the previous design example is TW=1.88 ft.

Step 4. Try z1 = 87.5 ft. WB = 10.0 ft, ST = 0.5 ft/ft, and SS = 0.5 ft/ft. From Equation 8.5:

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (100 minus 87.5) divided by 0.5 equals 25.0 feet

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 40.1 ft/s, y1 = 1.04 ft

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 40.1 divided by square root of (32.2 times 1.04) equals 6.9

Step 5. Calculate the conjugate depth in the basin (C=1) using Equation 8.4.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 1.0 times 1.04 divided by 2 times (square root of (1 plus 8 times 6.9 squared) minus 1) equals 9.64 feet

From Figure 8.2 basin length, LB/y2 = 2.7. Therefore, LB = 2.7(9.64) = 26.0 ft.

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (25.0 times (0.5 minus 0.065) minus 26.0 times 0.065) divided by (0.5 plus 0.065) equals 16.3 feet

The elevation at the entrance to the tailwater channel is from Equation 8.7:

z sub 3 equals L sub S times S sub S plus z sub 1 equals 16.3 times (0.5) plus 87.5 equals 95.65 feet

Since y2 +z2 (9.64+87.5) < z3 + TW (95.65+1.88), tailwater is sufficient to force a jump in the basin. If the tailwater had not been sufficient, repeat step 4 with a lower assumption for z1.

Step 6. Determine the needed radius of curvature for the slope changes entering the basin. See the design example Section 8.1 for this step. It is unchanged.

Step 7. Size the basin elements. For the USBR Type III basin, the elements include the chute blocks, baffle blocks, and end sill.

For the chute blocks, the height of the chute blocks, h1=y1=1.04 ft (round to 1.0 ft). The number of chute blocks is determined by Equation 8.9 and rounded to the nearest integer.

N sub c equals W sub B divided by (2 times y sub 1) equals 10 divided by (2 times 1.04) equals 4.8 is approximately equal to 5

Block width and block spacing are determined by Equation 8.10:

W sub 1 equals W sub 2 equals W sub B divided by (2 times N sub c) equals 10 divided by (2 times 5) equals 1.0 feet

With 5 blocks at 1.0 ft and 4 spaces at 1.0 ft, 1.0 ft of space remains. This is divided equally for spaces between the outside blocks and the basin sidewalls.

For the baffle blocks, the height of the baffles is computed from Equation 8.11:

h sub 3 equals y sub 1 times (0.168 Fr sub 1 plus 0.58) equals 1.04 times (0.168 times (6.9) plus 0.58) equals 1.8 feet

The number of baffles blocks is calculated from Equation 8.12:

N sub B equals W sub B divided by (1.5 times h sub 3) equals 10 divided by (1.5 times 1.8) equals 3.7 is approximately equal to 4

Baffle width and spacing are determined by Equation 8.13:

W sub 3 equals W sub 4 equals W sub B divided by (2 times N sub B) equals 10 divided by (2 times 4) equals 1.3 feet

With 4 baffles at 1.3 ft and 3 spaces at 1.3 ft, 0.9 ft of space remains. This is divided equally for spaces between the outside baffles and the basin sidewalls. The distance from the downstream face of the chute blocks to the upstream face of the baffle block should be 0.8y2=0.8(9.64)=7.7 ft.

For the end sill, the height of the end sill is given by Equation 8.14:

h sub 4 equals y sub 1 times (0.0536 Fr sub 1 plus 1.04) equals 1.04 times (0.053 times (6.9) plus 1.04) equals 1.5 feet

Total basin length = 25.0 + 26.0 + 16.3 = 67.3 ft. The basin is shown in the following sketch.

Sketch for USBR Type III Stilling Basin Design Example (CU)

Schematic showing the dimensions calculated for the CU design example

8.4 USBR Type IV Stilling Basin

The USBR Type IV stilling basin (USBR, 1987) is intended for use in the Froude number range of 2.5 to 4.5. In this low Froude number range, the jump is not fully developed and downstream wave action may be a problem as discussed in Chapter 4. For the intermittent flow encountered at most highway culverts, wave action is not judged to be a severe limitation. The basin, illustrated in Figure 8.4, employs chute blocks and an end sill.

The recommended design is limited to the following conditions:

  1. The basin sidewalls should be vertical rather than trapezoidal to insure proper performance of the hydraulic jump.
  2. Tailwater elevation should be equal to or greater than 110 percent of the full conjugate depth elevation. The hydraulic jump is very sensitive to tailwater depth at the low Froude numbers for which the basin is applicable. The additional tailwater improves jump performance and reduces wave action.

The general design procedure outlined in Section 8.1 applies to the USBR Type IV basin. Steps 1 through 4 and step 6 are applied without modification. For step 5, two adaptations to the general design procedure are made:

  1. For computing conjugate depth, C = 1.1.
  2. For obtaining the length of the basin, LB, use Figure 8.2 (dashed portion of the free jump curve) based on the Froude number calculated in step 4.

For step 7, sizing the basin elements (chute blocks and an end sill), the following guidance is recommended. The height of the chute blocks, h1, is set equal to 2y1. The top surface of the chute blocks should be sloped downstream at a 5 degree angle.

The number of chute blocks is determined by Equation 8.15a and rounded to the nearest integer.

(8.15a)

N sub c equals W sub B divided by (2.625 times y sub 1)

where,

Nc = number of chute blocks

Block width and block spacing are determined by:

(8.15b)

W sub 1 equals W sub B divided by (3.5 times N sub c)

(8.15c)

W sub 2 equals 2.5 times W sub 1

where,

W1 = block width, m (ft)

W2 = block spacing, m (ft)

Figure 8.4. USBR Type IV Stilling Basin

Schematic showing the dimensions of the USBR Type IV stilling basin as described in the text.

With Equation 8.14, the block width, W1, should be less than or equal to the depth of the incoming flow, y1. Equations 8.14 and 8.15 will provide Nc blocks and Nc-1 spaces between those blocks. The remaining basin width is divided equally for spaces between the outside blocks and the basin sidewalls.

The height of the end sill, is given as:

(8.16)

h sub 4 equals y sub 1 times (0.0536 F sub 1 plus 1.04)

where,

h4 = height of the end sill, m (ft)

The fore slope of the end sill should be set at 0.5:1 (V:H).

Design Example: USBR Type IV Stilling Basin (SI)

Design a USBR Type IV stilling basin for a reinforced concrete box culvert. Given:

  • Q = 11.8 m3/s
  • Culvert
  • B = 3.0 m
  • D = 1.8 m
  • n = 0.015
  • So = 0.065 m/m
  • zo = 30.50 m
  • Downstream channel (trapezoidal)
  • B = 3.10 m
  • Z = 1V:2H
  • n = 0.030

Solution

The culvert, design discharge, and tailwater channel are the same as considered for the free hydraulic jump stilling basin addressed in the design example in Section 8.1. Steps 1 through 3 of the general design process are identical for this example so they are not repeated here. The tailwater depth from the previous design example is TW=0.574 m.

Step 4. Try z1 = 25.00 m. WB=3.0 m, ST=0.5 m/m, and SS=0.5 m/m. From Equation 8.5:

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (30.50 minus 25.75) divided by 0.5 equals 11.0 meters

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 13.55 m/s, y1 = 0.290 m

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 13.55 divided by square root of (9.81 times 0.290) equals 8.0

It should be noted that this Froude number is outside the applicability range for the Type IV basin, therefore the Type IV is not appropriate for this situation. However, we will proceed with the calculations in order to compare basin dimensions with the other basin options.

Step 5. Calculate the conjugate depth in the basin (C=1.1) using Equation 8.4.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 1.1 times 0.303 divided by 2 times (square root of (1 plus 8 times 8.0 squared) minus 1) equals 3.46 meters

From Figure 8.2 basin length, LB/y2 = 6.1. Therefore, LB = 6.1(3.46) = 21.1 m.

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (11.0 times (0.5 minus 0.065) minus 21.1 times 0.065) divided by (0.5 plus 0.065) equals 6.0 meters

The elevation at the entrance to the tailwater channel is from Equation 8.7:

z sub 3 equals L sub S times S sub S plus z sub 1 equals 6.0 times (0.5) plus 25.00 equals 28.00 meters

Since y2 +z2 (3.46+25.00) < z3 + TW (28.00+0.574), tailwater is sufficient to force a jump in the basin. If this had not been the case, repeat step 4 with a lower assumption for z1.

Step 6. Determine the needed radius of curvature for the slope changes entering the basin. See the design example Section 8.1 for this step. It is unchanged.

Step 7. Size the basin elements. For the USBR Type IV basin, the elements include the chute blocks and end sill.

For the chute blocks:

The height of the chute blocks, h1=2y1=2(0.290) = 0.58 m. The number of chute blocks is determined by Equation 8.15a and rounded to the nearest integer.

N sub c equals W sub B divided by (2.625 times y sub 1) equals 3.0 divided by (2.625 times 0.290) equals 3.9 which approximates 4

Block width and block spacing are determined by Equations 8.15b and 8.15c:

W sub 1 equals W sub B divided by (3.5 times N sub c) equals 3.0 divided by (3.5 times 4) equals 0.21 meters

W sub 2 equals 2.5 times W sub 1 equals 2.5 times 0.21 equals 0.53 meters

With 4 blocks at 0.21 m and 3 spaces at 0.53 m, 0.57 m of space remains. This is divided equally for spaces between the outside blocks and the basin sidewalls.

For the sill:

The height of the end sill, is given by Equation 8.16:

h sub 4 equals y sub 1 times (0.0536 Fr sub 1 plus 1.04) equals 0.290 times(0.0536 times (8.0) plus 1.04) equals 0.43 meters

Total basin length = 11.0 + 21.1 +6.0 = 38.1 m. The basin is shown in the following sketch.

Sketch for USBR Type IV Stilling Basin Design Example (SI)

Schematic showing the dimensions calculated for the SI design example

Design Example: USBR Type IV Stilling Basin (CU)

Design a USBR Type IV stilling basin for a reinforced concrete box culvert. Given:

  • Q = 417 ft3/s
  • Culvert
  • B = 10 ft
  • D = 6 ft
  • n = 0.015
  • So = 0.065 ft/ft
  • zo = 100.0 ft
  • Downstream channel (trapezoidal)
  • B = 10.2 ft
  • Z = 1V:2H
  • n = 0.030

Solution

The culvert, design discharge, and tailwater channel are the same as considered for the free hydraulic jump stilling basin addressed in the design example in Section 8.1. Steps 1 through 3 of the general design process are identical for this example so they are not repeated here. The tailwater depth from the previous design example is TW=1.88 ft.

Step 4. Try z1 = 82.6 ft. WB=10.0 ft, ST=0.5 ft/ft, and SS=0.5 ft/ft. From Equation 8.5:

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (100 minus 82.6) divided by 0.5 equals 34.8 feet

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 43.9 ft/s, y1 = 0.95 ft

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 43.9 divided by square root of (32.2 times 0.95) equals 7.9

It should bFr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 43.9 divided by square root of (32.2 times 0.95) equals 7.9e noted that this Froude number is outside the applicability range for the Type IV basin, therefore the Type IV is not appropriate for this situation. However, we will proceed with the calculations in order to compare basin dimensions with the other basin options.

Step 5. Calculate the conjugate depth in the basin (C=1.1) using Equation 8.4.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 1.1 times 0.95 divided by 2 times (square root of (1 plus 8 times 7.9 squared) minus 1) equals 11.15 feet

From Figure 8.2 basin length, LB/y2 = 6.1. Therefore, LB = 6.1(11.15) = 68.0 ft.

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (34.8 times (0.5 minus 0.065) minus 68.0 times 0.065) divided by (0.5 plus 0.065) equals 19.0 feet

The elevation at the entrance to the tailwater channel is from Equation 8.7:

z sub 3 equals L sub S times S sub S plus z sub 1 equals 19.0 times (0.5) plus 82.60 equals 92.10 feet

Since y2 +z2 (11.15+82.60) < z3 + TW (92.10+1.88), tailwater is sufficient to force a jump in the basin. If this had not been the case, repeat step 4 with a lower assumption for z1.

Step 6. Determine the needed radius of curvature for the slope changes entering the basin. See the design example Section 8.1 for this step. It is unchanged.

Step 7. Size the basin elements. For the USBR Type IV basin, the elements include the chute blocks and end sill.

For the chute blocks:

The height of the chute blocks, h1=2y1=2(0.95) = 1.9 ft. The number of chute blocks is determined by Equation 8.15a and rounded to the nearest integer.

N sub c equals W sub B divided by (2.625 times y sub 1) equals 10.0 divided by (2.625 times 0.95) equals 4.01 which approximates 4

Block width and block spacing are determined by Equations 8.15b and 8.15c:

W sub 1 equals W sub B divided by (3.5 times N sub c) equals 10.0 divided by (3.5 times 4) equals 0.7 feet

W sub 2 equals 2.5 times W sub 1 equals 2.5 times 0.7 equals 1.8 feet

With 4 blocks at 0.7 ft and 3 spaces at 1.8 ft, 1.8 ft of space remains. This is divided equally for spaces between the outside blocks and the basin sidewalls.

For the sill:

The height of the end sill, is given by Equation 8.16:

h sub 4 equals y sub 1 times (0.0536 Fr sub 1 plus 1.04) equals 0.95 times (0.0536 times (7.9) plus 1.04) equals 1.4 feet

Total basin length = 34.8 + 68.0 + 19.0 = 121.8 ft. The basin is shown in the following sketch.

Sketch for USBR Type IV Stilling Basin Design Example (CU)

Schematic showing the dimensions calculated for the CU design example

8.5 SAF Stilling Basin

The Saint Anthony Falls (SAF) stilling basin, shown in Figure 8.5, provides chute blocks, baffle blocks, and an end sill that allows the basin to be shorter than a free hydraulic jump basin. It is recommended for use at small structures such as spillways, outlet works, and canals where the Froude number at the dissipator entrance is between 1.7 and 17. The reduction in basin length achieved through the use of appurtenances is about 80 percent of the free hydraulic jump length. The SAF stilling basin provides an economical method of dissipating energy and preventing stream bed erosion.

Figure 8.5. SAF Stilling Basin (Blaisdell, 1959)

Schematic showing the dimensions of the SAF stilling basin as described in the text.

The general design procedure outlined in Section 8.1 applies to the SAF stilling basin. Steps 1 through 3 and step 6 are applied without modification. As part of step 4, the designer selects a basin width, WB. For box culverts, WB must equal the culvert width, Wo. For circular culverts, the basin width is taken as the larger of the culvert diameter and the value calculated according to the following equation:

(8.17)

W sub B equals 1.7 times D sub o times Q divided by (g to the 0.5 power times (D sub o to the 2.5 power))

where,

WB = basin width, m (ft)

Q = design discharge, m3/s (ft3/s)

Do = culvert diameter, m (ft)

The basin can be flared to fit an existing channel as indicated on Figure 8.5. The sidewall flare dimension z should not be greater than 0.5, i.e., 0.5:1, 0.33:1, or flatter.

For step 5, two adaptations to the general design procedure are made. First, for computing conjugate depth, C is a function of Froude number as given by the following set of equations. Depending on the Froude number, C ranges from 0.64 to 1.08 implying that the SAF basin may operate with less tailwater than the USBR basins, though tailwater is still required.

(8.18a)
C equals 1.1 minus Fr sub 1 squared divided by 120when 1.7 < Fr1 < 5.5
(8.18b)
C equals 0.85when 5.5 < Fr1 < 11
(8.18c)
C equals 1.1 minus Fr sub 1 squared divided by 800when 11 < Fr1 < 17

The second adaptation is the determination of the basin length, LB, using Equation 8.19.

(8.19)

L sub B equals 4.5 times y sub 2 divided by C divided by Fr sub 1 to the 0.76 power

For step 7, sizing the basin elements (chute blocks, baffle blocks, and an end sill), the following guidance is recommended. The height of the chute blocks, h1, is set equal to y1.

The number of chute blocks is determined by Equation 8.20 rounded to the nearest integer.

(8.20)

N sub c equals W sub B divided by (1.5 times y sub 1)

where,

Nc = number of chute blocks

Block width and block spacing are determined by:

(8.21)

W sub 1 equals W sub 2 equals W sub B divided by (2 times N sub c)

where,

W1 = block width, m (ft)

W2 = block spacing, m (ft)

Equations 8.20 and 8.21 will provide Nc blocks and Nc spaces between those blocks. A half block is placed at the basin wall so there is no space at the wall.

The height, width, and spacing of the baffle blocks are shown Figure 8.5. The height of the baffles, h3, is set equal to the entering flow depth, y1.

The width and spacing of the baffle blocks must account for any basin flare. If the basin is flared as shown in Figure 8.5, the width of the basin at the baffle row is computed according to the following:

(8.22)

W sub B2 equals W sub B plus 2 times z times L sub B divided by 3

where,

WB2 = basin width at the baffle row, m (ft)

LB = basin length, m (ft)

z = basin flare, z:1 as defined in Figure 8.5 (z=0.0 for no flare)

The top thickness of the baffle blocks should be set at 0.2h3 with the back slope of the block on a 1:1 slope. The number of baffles blocks is as follows:

(8.23)

N sub B equals W sub B2 divided by (1.5 times y sub 1)

where,

NB = number of baffle blocks (rounded to an integer)

Baffle width and spacing are determined by:

(8.24)

W sub 3 equals W sub 4 equals W sub B2 divided by (2 times N sub B)

where,

W3 = baffle width, m (ft)

W4 = baffle spacing, m (ft)

Equations 8.23 and 8.24 will provide NB baffles and NB-1 spaces between those baffles. The remaining basin width is divided equally for spaces between the outside baffles and the basin sidewalls. No baffle block should be placed closer to the sidewall than 3y1/8. Verify that the percentage of WB2 obstructed by baffles is between 40 and 55 percent. The distance from the downstream face of the chute blocks to the upstream face of the baffle block should be LB/3.

The height of the final basin element, the end sill, is given as:

(8.25)

h sub 4 equals 0.07 times y sub 2 divided by C

where,

h4 = height of the end sill, m (ft)

The fore slope of the end sill should be set at 0.5:1 (V:H). If the basin is flared the length of sill (width of the basin at the sill) is:

(8.26)

W sub B3 equals W sub B plus 2 times z times L sub B

where,

WB3 = basin width at the sill, m (ft)

LB = basin length, m (ft)

z = basin flare, z:1 as defined in Figure 8.5 (z=0.0 for no flare)

Wingwalls should be equal in height and length to the stilling basin sidewalls. The top of the wingwall should have a 1H:1V slope. Flaring wingwalls are preferred to perpendicular or parallel wingwalls. The best overall conditions are obtained if the triangular wingwalls are located at an angle of 45° to the outlet centerline.

The stilling basin sidewalls may be parallel (rectangular stilling basin) or diverge as an extension of the transition sidewalls (flared stilling basin). The height of the sidewall above the floor of the basin is given by:

(8.27)

h sub 5 is greater than or equal to y sub 2 times (1 plus 1 divided by (3 times C))

where,

h5 = height of the sidewall, m (ft)

A cut-off wall should be used at the end of the stilling basin to prevent undermining. The depth of the cut-off wall must be greater than the maximum depth of anticipated erosion at the end of the stilling basin.

Design Example: SAF Stilling Basin (SI)

Design a SAF stilling basin with no flare for a reinforced concrete box culvert. Given:

  • Q = 11.8 m3/s
  • Culvert
  • B = 3.0 m
  • D = 1.8 m
  • n = 0.015
  • So = 0.065 m/m
  • zo = 30.50 m
  • Downstream channel (trapezoidal)
  • B = 3.10 m
  • Z = 1V:2H
  • n = 0.030

Solution

The culvert, design discharge, and tailwater channel are the same as considered for the free hydraulic jump stilling basin addressed in the design example in Section 8.1. Steps 1 through 3 of the general design process are identical for this example so they are not repeated here. The tailwater depth from the previous design example is TW=0.574 m.

Step 4. Try z1 = 27.80 m. WB=3.0 m (no flare), ST=0.5 m/m, and SS=0.5 m/m. From Equation 8.5:

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (30.50 minus 27.80) divided by 0.5 equals 5.4 meters

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 11.29 m/s, y1 = 0.348 m

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 11.29 divided by square root of (9.81 times 0.348) equals 6.1

Step 5. Calculate the conjugate depth in the basin using Equation 8.4. First estimate C using Equation 8.18. For the calculated Froude number, C=0.85.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 0.85 times 0.348 divided by 2 times (square root of (1 plus 8 times 6.1 squared) minus 1) equals 2.41 meters

From Equation 8.19 basin length is calculated:

L sub B equals 4.5 times y sub 2 divided by C divided by Fr sub 1 to the 0.76 power equals 4.5 times 2.41 divided by 0.85 divided by 6.1 to the 0.76 power equals 3.2 meters

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (5.4 times (0.5 minus 0.065) minus 3.2 times 0.065) divided by (0.5 plus 0.065) equals 3.8 meters

The elevation at the entrance to the tailwater channel is from Equation 8.7:

Z3 = LSSS + Z1 = 3.8(0.5) + 27.80 =29.70 m

Since y2 +z2 (2.41+27.80) < z3 + TW (29.70+0.574), tailwater is sufficient to force a jump in the basin. If tailwater had not been sufficient, repeat step 4 with a lower assumption for z1.

Step 6. Determine the needed radius of curvature for the slope changes entering the basin. See the design example Section 8.1 for this step. It is unchanged.

Step 7. Size the basin elements. For the SAF basin, the elements include the chute blocks, baffle blocks, and an end sill.

For the chute blocks:

The height of the chute blocks, h1=y1=0.348 (round to 0.35 m).

The number of chute blocks is determined by Equation 8.20:

N sub c equals W sub B divided by (1.5 times y sub 1) equals 3.0 divided by (1.5 times 0.348) equals 5.7 which approximates 6

Block width and block spacing are determined by Equation 8.21:

W sub 1 equals W sub 2 equals W sub B divided by (2 times N sub c) equals 3.0 divided by (2 times 6) equals 0.25 meters

A half block is placed at each basin wall so there is no space at the wall.

For the baffle blocks:

The height of the baffles, h3=y1=0.348 m. (round to 0.35 m)

The basin has no flare so the width in the basin is constant and equal to WB.

The number of baffles blocks is from Equation 8.23:

N sub B equals W sub B2 divided by (1.5 times y sub 1) equals 3.0 divided by (1.5 times 0.348) equals 5.7 which approximates 6

Baffle width and spacing are determined from Equation 8.24. In this case WB2=WB.

W sub 3 equals W sub 4 equals W sub B2 divided by (2 times N sub B) equals 3.0 divided by (2 times 6) equals 0.25 meters

For this design, we have 6 baffles at 0.25 m and 5 spaces between them at 0.25 m. The remaining 0.25 m is divided in half and provided as a space between the sidewall and the first baffle.

The total percentage blocked by baffles is 6(0.25)/3.0=50 percent which falls within the acceptable range of between 40 and 55 percent.

The distance from the downstream face of the chute blocks to the upstream face of the baffle block equals LB/3=3.2/3=1.1 m.

For the sill:

The height of the end sill, is given in Equation 8.25:

h sub 4 equals 0.07 times y sub 2 divided by C equals 0.07 times 2.41 divided by 0.85 equals 0.20 meters

Total basin length = 5.4 + 3.2 + 3.8 = 12.4 m. The basin is shown in the following sketch.

Sketch for SAF Stilling Basin Design Example (SI)

Schematic showing the dimensions calculated for the SI design example

Design Example: SAF Stilling Basin (CU)

Design a SAF stilling basin with no flare for a reinforced concrete box culvert. Given:

  • Q = 417 ft3/s
  • Culvert
  • B = 10.0 ft
  • D = 6.0 ft
  • n = 0.015
  • So = 0.065 ft/ft
  • zo = 100.0 ft
  • Downstream channel (trapezoidal)
  • B = 10.2 ft
  • Z = 1V:2H
  • n = 0.030

Solution

The culvert, design discharge, and tailwater channel are the same as considered for the free hydraulic jump stilling basin addressed in the design example in Section 8.1. Steps 1 through 3 of the general design process are identical for this example so they are not repeated here. The tailwater depth from the previous design example is TW=1.88 ft.

Step 4. Try z1 = 91.40 ft. WB=10.0 ft (no flare), ST=0.5 ft/ft, and SS=0.5 ft/ft. From Equation 8.5:

L sub T equals (z sub o minus z sub 1) divided by S sub T equals (100.0 minus 91.4) divided by 0.5 equals 17.2 feet

By using Equation 8.2 or other appropriate method by trial and error, the velocity and depth conditions entering the basin are:

V1 = 36.8 ft/s, y1 = 1.13 ft

Fr sub 1 equals V sub 1 divided by square root of (g times y sub 1) equals 36.8 divided by square root of (32.2 times 1.13) equals 6.1

Step 5. Calculate the conjugate depth in the basin using Equation 8.4. First estimate C using Equation 8.18. For the calculated Froude number, C=0.85.

y sub 2 equals C times y sub 1 divided by 2 times (square root of (1 plus 8 times Fr sub 1 squared) minus 1) equals 0.85 times 1.13 divided by 2 times (square root of (1 plus 8 times 6.1 squared) minus 1) equals 7.85 feet

From Equation 8.19 basin length is calculated:

L sub B equals 4.5 times y sub 2 divided by C divided by Fr sub 1 to the 0.76 power equals 4.5 times 7.85 divided by 0.85 divided by 6.1 to the 0.76 power equals 10.5 feet

The length of the basin from the floor to the sill is calculated from Equation 8.6:

L sub s equals (L sub T times (S sub T minus S sub o) minus L sub B times S sub o) divided by (S sub s plus S sub o) equals (17.2 times (0.5 minus 0.065) minus 10.5 times 0.065) divided by (0.5 plus 0.065) equals 12.0 feet

The elevation at the entrance to the tailwater channel is from Equation 8.7:

Z3 = LSSS + Z1 = 12.0(0.5) + 91.40 =97.40 ft

Since y2 +z2 (7.85+91.40) < z3 + TW (97.40+1.88), tailwater is sufficient to force a jump in the basin. If tailwater had not been sufficient, repeat step 4 with a lower assumption for z1.

Step 6. Determine the needed radius of curvature for the slope changes entering the basin. See the design example Section 8.1 for this step. It is unchanged.

Step 7. Size the basin elements. For the SAF basin, the elements include the chute blocks, baffle blocks, and an end sill.

For the chute blocks:

The height of the chute blocks, h1=y1=1.13 (round to 1.1 ft).

The number of chute blocks is determined by Equation 8.20:

N sub c equals W sub B divided by (1.5 times y sub 1) equals 10.0 divided by (1.5 times 1.13) equals 5.9 which approximates 6

Block width and block spacing are determined by Equation 8.21:

W sub 1 equals W sub 2 equals W sub B divided by (2 times N sub c) equals 10.0 divided by (2 times 6) equals 0.8 feet

A half block is placed at each basin wall so there is no space at the wall.

For the baffle blocks:

The height of the baffles, h3=y1=1.13 ft. (round to 1.1 ft)

The basin has no flare so the width in the basin is constant and equal to WB.

The number of baffles blocks is from Equation 8.23:

N sub B equals W sub B2 divided by (1.5 times y sub 1) equals 10.0 divided by (1.5 times 1.13) equals 5.9 which approximates 6

Baffle width and spacing are determined from Equation 8.24. In this case WB2=WB.

W sub 3 equals W sub 4 equals W sub B2 divided by (2 times N sub B) equals 10.0 divided by (2 times 6) equals 0.8 feet

For this design, we have 6 baffles at 0.8 ft and 5 spaces between them at 0.8 ft. The remaining 1.2 ft is divided in half and provided as a space between the sidewall and the first baffle.

The total percentage blocked by baffles is 6(0.8)/10.0=48 percent which falls within the acceptable range of between 40 and 55 percent.

The distance from the downstream face of the chute blocks to the upstream face of the baffle block equals LB/3=10.5/3=3.5 ft.

For the sill:

The height of the end sill, is given in Equation 8.25:

h sub 4 equals 0.07 times y sub 2 divided by C equals 0.07 times 7.85 divided by 0.85 equals 0.6 feet

Total basin length = 17.2 + 10.5 + 12.0 = 39.7 ft. The basin is shown in the following sketch.

Sketch for SAF Stilling Basin Design Example (CU)

Schematic showing the dimensions calculated for the CU design example

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Contact:

Cynthia Nurmi
Resource Center (Atlanta)
404-562-3908
cynthia.nurmi@dot.gov

Updated: 04/07/2011
 

FHWA
United States Department of Transportation - Federal Highway Administration