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

Chapter 11: Drop Structures

Drop structures are commonly used for flow control and energy dissipation. Changing the channel slope from steep to mild, by placing drop structures at intervals along the channel reach, changes a continuous steep slope into a series of gentle slopes and vertical drops. Instead of slowing down and transferring high erosion producing velocities into low non-erosive velocities, drop structures control the slope of the channel in such a way that the high, erosive velocities never develop. The kinetic energy or velocity gained by the water as it drops over the crest of each structure is dissipated by a specially designed apron or stilling basin.

The drop structures discussed here (see Figure 11.1) require an aerated nappe and are, in general, for subcritical flow in the upstream as well as downstream channel. The effect of upstream supercritical flow on drop structure design is discussed in a later section. The stilling basin protects the channel against erosion below the drop and dissipates energy. This is accomplished through the impact of the falling water on the floor, redirection of the flow, and turbulence. The stilling basin used to dissipate the excess energy can vary from a simple concrete apron to an apron with flow obstructions such as baffle blocks, sills, or abrupt rises. The length of the concrete apron required can be shortened by the addition of these appurtenances.

Figure 11.1. Flow Geometry of a Straight Drop Spillway

Profile view summarizing variable definitions provided in the text.

The drop number gives a quantitative measure for drop:

(11.1)

N sub d equals q squared divided by (g times h sub 3 cubed)

where,

Nd = drop number

q = unit discharge, m3/s/m (ft3/s/ft)

g = acceleration due to gravity, 9.81 m/s2 (32.2 ft/s2)

ho = drop height, m (ft)

Another commonly used quantitative measure for drop is given by:

(11.2)

D sub r equals h sub o divided by y sub c

where,

Dr = relative drop

yc = critical depth at the drop, m (ft)

ho = drop height, m (ft)

Drop structures may be categorized based on either Equations 11.1 or 11.2. Drops for which Nd is greater than 1 or Drop is less than 1 are considered "low drop" structures. Two dissipators are discussed in this chapter: the straight drop structure and the box inlet drop structure. Neither of these is considered low drop structures.

11.1 Straight Drop Structure

A straight drop structure is characterized by flow through a rectangular weir followed by a drop into a stilling basin. The stilling basin may be a flat apron or an apron with various baffles and sills depending on the site conditions. First a simple stilling basin is considered followed by discussion of other features available to modify the drop structure performance.

11.1.1 Simple Straight Drop

The basic flow geometry of a straight drop structure is shown in Figure 11.1. The discharge passes through critical depth as it flows over the drop structure crest. The free-falling nappe reverses its curvature and turns smoothly into supercritical flow on the apron at the distance L1 from the drop wall. The mean velocity at the distance L is parallel to the apron; the depth y2 is the smallest depth in the downstream channel, and the pressure is nearly hydrostatic. The depth of supercritical flow in the downstream direction increases due to channel resistance, and at some point will reach a depth sufficient for the formation of a hydraulic jump.

For a given drop height, ho, and discharge, q, the sequent depth, y3, in the downstream channel and the drop length, L1, may be computed. The length of jump Lj, is discussed in Chapter 6. The drop number can be used to estimate the dimensions of a simple straight drop structure.

(11.3a)

L sub 1 equals 4.30 times h sub o times N sub d to the 0.27 power

(11.3b)

y sub 1 equals 1.0 times h sub o times N sub d to the 0.22 power

(11.3c)

y sub 2 equals 0.54 times h sub o times N sub d to the 0.425 power

(11.3d)

y sub 3 equals 1.66 times h sub o times N sub d to the 0.27 power

where,

L1 = drop length (the distance from the drop wall to the position of the depth y2), m (ft)

y1 = pool depth under the nappe, m (ft)

y2 = depth of flow at the toe of the nappe or the beginning of the hydraulic jump, m (ft)

y3 = tailwater depth sequent to y2, m (ft)

By comparing the channel tailwater depth, TW, with the computed, y3, one of the following cases will occur. The case will determine design modifications necessary to the structure.

  1. TW > y3. The hydraulic jump will be submerged and the basin length may need to be increased.
  2. TW = y3. The hydraulic jump begins at depth y2, no supercritical flow exists on the apron and the distance L1 is a minimum. In this case, the basin will function without additional design modifications.
  3. TW < y3. The hydraulic jump will recede downstream and the basin will not function.

For case 3, when the tailwater depth is less than y3, it is necessary to modify the basin to force the hydraulic jump to stay in the basin. Two alternatives to achieve this are to provide an apron:

  1. at the bed level with an end sill or baffles to trigger the jump in the basin, or
  2. depressed below the downstream bed level to effectively increase tailwater with an end sill.

The choice of design type and the design dimensions will depend, for a given unit discharge, q, on the drop height, ho, and on the downstream depth, TW. The apron may be designed to extend to the end of the hydraulic jump. However, including an end sill allows the use of a shorter and more economical stilling basin.

The geometry of the undisturbed flow should be taken into consideration in the design of a straight drop structure. If the overfall crest length is less than the width of the approach channel, it is important that a transition be properly designed by shaping the approach channel to reduce the effect of end contractions. Otherwise the contraction at the ends of the spillway notch may be so pronounced that the jet will land beyond the stilling basin and the concentration of high velocities at the center of the outlet may cause additional scour in the downstream channel (see Chapter 4).

11.1.2 Grate Design

A grate or series of rails forming a "grizzly" may be used in conjunction with drop structures as illustrated in Figure 11.2. The incoming flow is divided into a number of jets as it passes through the grate. These fall almost vertically to the downstream channel resulting in good energy dissipator action. This type of design is also utilized as a debris ejector where the debris rides over the grate and falls into a holding area for later removal and the water passes through the grate.

The Bureau of Reclamation has published design recommendations for grates (USBR, 1987) for use where the incoming flow is subcritical. The length of the grate is calculated from:

(11.4)

L sub G equals Q divided by (C times W times N times (2 times g times y sub o) to the one-half power)

where,

C = experimental coefficient equal to 0.245

W = width of the slots, m (ft)

N = number of slots (spaces) between beams

yo = approach flow depth, m (ft)

Figure 11.2. Drop Structure with Grate

Isometric view showing grate length, L sub G, and beam width, W. Slot width equals two-thirds of beam width.

The design process is iterative based on the number of slots and the slot width. The USBR (1987) recommends the following guidelines:

  1. Select an initial slot width. Provide a full slot width at each wall.
  2. Beam width should be approximately 1.5W
  3. Estimate the number of slots, N, for use in Equation 11.4
  4. Calculate the grate length using Equation 11.4. Adjust the slot width until an acceptable beam length, LG, is obtained.
  5. Tilt the grate about 3° downstream to be self-cleaning.

Examination of the beam length equation indicates the relative effect of higher approach velocities on the design of drop structures. Assuming the slot width, W, approaches the channel width making N equal to 1, and considering a constant flow rate, then the relationship in Equation 11.4 reveals that the grate length is inversely proportional to the square root of the approach depth. For constant Q, as the approach velocity increases the approach depth decreases and the length LG increases. Therefore, for high velocity flow, above critical velocity, the length of drop structure required, to contain the jet, may very rapidly exceed practical limits.

11.1.3 Straight Drop Structure Design Features

A general design for a stilling basin at the toe of a drop structure was developed by the Agricultural Research Service, St. Anthony Falls Hydraulic Laboratory, University of Minnesota (Donnelly and Blaisdell, 1954). The basin consists of a horizontal apron with blocks and sills to dissipate energy as shown in Figure 11.3. Tailwater also influences the amount of energy dissipated. The stilling basin length computed for the minimum tailwater level required for good performance may be inadequate at high tailwater levels. Scour of the downstream channel may occur if the nappe is supported sufficiently by high tailwater so that it lands beyond the end of the stilling basin. A method for computing the stilling basin length for all tailwater levels is presented.

Figure 11.3. Straight Drop Structure (Rand, 1955)

Profile and plan view summarizing dimensions described in text.

The recommended design is limited to the following conditions:

  1. Total drop, ho, less than 4.6 m (15 ft) with sufficient tailwater.
  2. Relative drop, ho/yc, between 1.0 and 15.
  3. Crest length, Wo, greater than 1.5yc.

The elements that must be considered in the design of this stilling basin include the length of basin, the position and size of floor blocks, the position and height of end sill, the position of the wingwalls, and the approach channel geometry. Figure 11.3 illustrates a straight drop structure that provides adequate protection from scour in the downstream channel.

Many of the design parameters for the straight drop structure are based on the critical depth. Critical depth in a rectangular channel or culvert is calculated from the unit discharge (discharge divided by culvert/chute width, B).

(11.5)

y sub c equals (q squared divided by g) to the one-third power

where,

yc = critical depth, m (ft)

q = unit discharge (Q/B), m (ft)

Critical flow for an open channel of any shape will occur when:

(11.6)

Q squared times T sub c divided by (g times A sub c cubed) equals 1

where,

Tc = water surface width at critical flow condition, m (ft)

Ac = flow area at critical flow condition, m (ft)

As discussed earlier, the tailwater must neither be too high nor too low. Therefore, the following relationships must be achieved in the design. First, the tailwater depth above the floor of the stilling basin must be calculated from Equation 11.7.

(11.7)

y sub 3 equals 2.15 times y sub c

where,

y3 = tailwater depth above the floor of the stilling basin, m (ft)

The tailwater also needs to be a distance below the crest to maintain the aerated nappe trajectory as given below. Using the crest as the reference point, this distance is a negative number.

(11.8)

h sub 2 equals negative (h minus y sub o)

where,

h2 = vertical distance of the tailwater below the crest, m (ft)

h = vertical drop between the approach and tailwater channels, m (ft)

yo = normal depth in the tailwater channel (equals normal depth in approach channel assuming same channel characteristics), m (ft)

To achieve sufficient tailwater and to maintain adequate drop from the crest to the tailwater, it is sometimes necessary to depress the floor below the elevation of the tailwater channel. The total drop from the crest to the stilling basin floor is given by:

(11.9)

h sub o equals h sub 2 minus y sub 3

where,

ho = drop from crest to stilling basin floor, m (ft)

The horizontal dimensions of the basin must also be established. From Figure 11.3 it can be seen that the total basin length is the sum of three components.

(11.10)

L sub B equals L sub 1 plus L sub 2 plus L sub 3

where,

LB = stilling basin length, m (ft)

L1 = distance from the headwall to the point where the surface of the upper nappe strikes the stilling basin floor, m (ft)

L2 = distance from the point where the surface of the upper nappe strikes the stilling basin floor to the upstream face of the floor blocks, m (ft)

L3 = distance from the upstream face of the floor blocks to the end of the stilling basin, m (ft)

L1 is given by:

(11.11)

L sub 1 equals (L sub f plus L sub s) divided by 2

where,

Lf = length given by Equation 11.12, m (ft)

Ls = length given by Equation 11.13, m (ft)

(11.12)

L sub f equals (-0.406 plus square root of (3.195 minus 4.368 times (h sub o divided by y sub c))) times y sub c

(11.13)

L sub s equals (0.691 plus 0.228 times (L sub t divided by y sub c) squared minus (h sub o divided by y sub c)) times y sub c divided by (0.185 plus 0.456 times (L sub t divided by y sub c))

where,

(11.14)

L sub t equals (-0.406 plus square root of (3.195 minus 4.368 times (h sub 2 divided by y sub c))) times y sub c

L2 and L3 are determined by:

(11.15)

L sub 2 equals 0.8 y sub c

(11.16)

L sub 3 is greater than or equal to 1.75 times y sub c

In comparison with the simple straight drop structure discussed in Section 11.1.1, the addition of floor blocks and a sill, allows for a shorter basin as given by Equation 11.10. The floor blocks should be proportioned to have a height of 0.8yc with a width and spacing of 0.4yc. The basin will perform acceptably if the width and spacing varies within plus or minus 0.15yc. The blocks should be square in plan and should occupy between 50 percent and 60 percent of the stilling basin width.

The end sill height should be 0.4yc. Longitudinal sills, as shown in Figure 11.3, are optional from a hydraulic perspective. If needed, they reinforce the basin structurally, but should pass through the blocks, not between them.

Final consideration is given to the configuration of the exit of the basin as well as the transition from the approach channel to the basin. With respect to the exit, the sidewall height at the basin exit should be above the tailwater elevation by 0.85yc. Wingwalls should be located at an angle of 45° with the outlet centerline and have a top slope of 1 to 1.

With respect to the approach channel, the crest of spillway should be at same elevation as the invert of the approach channel. The bottom width of the approach channel should be equal to the spillway notch length, Wo, at the headwall. Because of the acceleration as the flow approaches the crest, riprap or paving should be provided for a distance upstream from the headwall equal to 3yc. (See Section 10.3 for sizing riprap.)

The design procedure for the straight drop structure may be summarized in the following steps.

Step 1. Estimate the elevation difference required between the approach and tailwater channel, h. This may be to address a drop at the outlet of a culvert resulting from erosion or headcutting or it may be to flatten a channel to a series of subcritical slopes and drops.

Step 2. Calculate normal flow conditions approaching the drop to verify subcritical conditions. If not subcritical, repeat step 1.

Step 3. Calculate critical depth over the weir (usually rectangular) into the drop structure. Calculate the vertical dimensions of the stilling basin using Equations 11.7 through 11.9.

Step 4. Estimate the basin length using Equations 11.10 though 11.16.

Step 5. Design the basin floor blocks and end sill.

Step 6. Design the basin exit and entrance transitions.

Design Example: Straight Drop Structure (SI)

Find the dimensions for a straight drop structure with a rectangular weir used to reduce channel slope. Given:

  • Q = 7.1 m3/s
  • h = 1.83 m
  • Wo = 3.10 m
  • Upstream and downstream channel (trapezoidal)
  • B = 3.10 m
  • Z = 1V:3H
  • So = 0.002 m/m (after providing for drop)
  • n = 0.030

Solution

Step 1. Estimate the required approach and tailwater channel elevation difference, h. This is estimated and given above as 1.83 m. This drop forces the slope of the upstream and downstream channel to 0.002 m/m, as given.

Step 2. Calculate normal flow conditions approaching the drop to verify subcritical conditions. By trial and error,

yo = 1.025 m, Vo = 1.123 m/s, Fro = 0.35; therefore, flow is subcritical. Proceed to step 3.

Step 3. Calculate critical depth over the weir into the drop structure. Calculate the vertical dimensions of the stilling basin. Start by finding the critical depth over the weir using Equation 11.5 based on the unit discharge, q = Q/B = 7.10/3.10 = 2.29 m2/s.

y sub c equals (q squared divided by g) to the one-third power equals (2.29 squared divided by 9.81) to the one-third power equals 0.812 meters

The required tailwater depth above the floor of the stilling basin is calculated from Equation 11.7.

y3 = 2.15yc = 2.15(0.812) = 1.745 m

The distance from the crest down to the tailwater needs to be calculated using Equation 11.8. (The negative indicates the tailwater elevation is below the crest.)

h2 = -(h-yo) = -(1.83 - 1.025) = -0.805 m

The total drop from the crest to the stilling basin floor is given by Equation 11.9:

ho = h2 - y3 = -0.805-1.745 = -2.55 m

Since the nominal drop, h, is 1.83 m, the floor must be depressed by 0.72 m

Step 4. Estimate the basin length. Start by using Equations 11.12, 11.13, and 11.14.

L sub f equals 3.01 meters (solution for Equation 11.12)

L sub t equals 1.90 meters (solution for Equation 11.13)

L sub s equals 3.30 meters (solution for Equation 11.14)

L1 is given by Equation 11.11:

L sub 1 equals (L sub f plus L sub s) divided by 2 equals (3.01 plus 3.30) divided by 2 equals 3.15 meters

L2 and L3 are determined by Equations 11.15 and 11.16:

L2 = 0.8yc = 0.8(0.812) = 0.65 m

L3 ≥ 1.75yc = 1.75(0.812) = 1.43 m

Total basin length required is given by Equation 11.10:

LB = L1 +L2 +L3 = 3.15 +0.65 + 1.43 = 5.23 m

Step 5. Design the basin floor blocks and end sill.

Block height = 0.8yc = 0.8(0.812) = 0.65 m

Block width = block spacing = 0.4yc = 0.4(0.812) = 0.325 m

End sill height = 0.4yc = 0.4(0.812) = 0.325 m

Step 6. Design the basin exit and entrance transitions.

Sidewall height above tailwater elevation = 0.85yc = 0.85(0.812) = 0.69 m

Armour approach channel above headwall to length = 3yc = 3(0.812) = 2.44 m

Design Example: Straight Drop Structure (CU)

Find the dimensions for a straight drop structure with a rectangular weir used to reduce channel slope. Given:

  • Q = 250 ft3/s
  • h = 6.0 ft
  • Wo = 10.0 ft
  • Upstream and downstream channel (trapezoidal)
  • B = 10.0 ft
  • Z = 1V:3H
  • So = 0.002 ft/ft (after providing for drop)
  • n = 0.030

Solution

Step 1. Estimate the required approach and tailwater channel elevation difference, h. This is estimated and given above as 6.0 ft. This drop forces the slope of the upstream and downstream channel to 0.002 ft/ft, as given.

Step 2. Calculate normal flow conditions approaching the drop to verify subcritical conditions. By trial and error,

yo = 3.36 ft, Vo = 3.71 ft/s, Fro = 0.36; therefore, flow is subcritical. Proceed to step 3.

Step 3. Calculate critical depth over the weir into the drop structure. Calculate the vertical dimensions of the stilling basin. Start by finding the critical depth over the weir using Equation 11.5 based on the unit discharge, q = Q/B = 250/10 = 25 ft2/s.

y sub c equals (q squared divided by g) to the one-third power equals (25 squared divided by 32.2) to the one-third power equals 2.69 feet

The required tailwater depth above the floor of the stilling basin is calculated from Equation 11.7.

y3 = 2.15yc = 2.15(2.69) = 5.77 ft

The distance from the crest down to the tailwater needs to be calculated using Equation 11.8. (The negative indicates the tailwater elevation is below the crest.)

h2 = -(h - yo) = -(6.0 - 3.36) = -2.64 ft

The total drop from the crest to the stilling basin floor is given by Equation 11.9:

ho = h2 -y3 = -2.64 - 5.77 = -8.41 ft (round to -8.4)

Since the nominal drop, h, is 6.0 ft, the floor must be depressed by 2.4 ft

Step 4. Estimate the basin length. Start by using Equations 11.12, 11.13, and 11.14.

L sub f equals 9.94 feet (solution for Equation 11.12)

L sub t equals 6.26 feet (solution for Equation 11.13)

L sub s equals 10.89 feet (solution for Equation 11.14)

L1 is given by Equation 11.11:

L sub 1 equals (L sub f plus L sub s) divided by 2 equals (9.94 plus 10.89) divided by 2 equals 10.4 feet

L2 and L3 are determined by Equations 11.15 and 11.16:

L2 = 0.8yc = 0.8(2.69) = 2.2 ft

L3 ≥ 1.75yc = 1.75(2.69) = 4.7 ft

Total basin length required is given by Equation 11.10:

LB = L1 +L2 +L3 = 10.4 +2.2 + 4.7 = 17.3 ft

Step 5. Design the basin floor blocks and end sill.

Block height = 0.8yc = 0.8(2.69) = 2.1 ft

Block width = block spacing = 0.4yc = 0.4(2.69) = 1.1 ft

End sill height = 0.4yc = 0.4(2.69) = 1.1 ft

Step 6. Design the basin exit and entrance transitions.

Sidewall height above tailwater elevation = 0.85yc = 0.85(2.69) = 2.3 ft

Armour approach channel above headwall to length = 3yc = 3(2.69) = 8.1 ft

11.2 Box Inlet Drop Structure

The box inlet drop structure may be described as a rectangular box open at the top and downstream end (Figure 11.4). Water is directed to the crest of the box inlet by earth dikes and a headwalls. Flow enters over the upstream end and two sides. The long crest of the box inlet permits large flows to pass at relatively low heads.

The outlet structure can be adjusted to fit a wide variety of field conditions. It is possible to lengthen the straight section and cover it to form a highway culvert. The sidewalls of the stilling basin section can be flared if desired, thus permitting use with narrow channels or wide flood plains. Flaring the sidewalls also makes it possible to adjust the outlet depth to that in the natural channel.

The design information is based on an extensive experimental program performed by the Soil Conservation Service, St. Anthony Falls Hydraulic Laboratory, Minneapolis (Blaisdell and Donnelly, 1956). The recommended design is limited to the following conditions:

  1. Total drop, ho, less than 3.7 m (12 ft) and greater than 0.6 m (2 ft). The total drop is that amount required to reduce the channel slope to a desired stable slope.
  2. Downstream width of structure should be less than or equal to the width of the tailwater channel.
  3. Approach channel is level with the crest of the box inlet

Figure 11.4. Box Inlet Drop Structure

Profile and plan view summarizing dimensions described in text.

One of two different sections will control the flow: the crest of the box inlet or the opening in the headwall. The flow at which the control changes from one point to the other is dependent upon a number of factors, the principal factors being the box inlet depth and its length. The design of the box inlet drop structure involves determining which section (crest or headwall opening) controls at the design flow.

First, assume crest control and calculate the head, yo, at the crest of the box inlet drop structure required to pass the design discharge. The general equation relating discharge to head for a rectangular weir is:

(11.17)

Q equals C sub w times the square root of (2 times g) times L times h to the two-thirds power

where,

Cw = dimensionless weir coefficient = 0.43

L = weir length, m (ft)

h = head on the weir crest, m (ft)

Calling the weir length, Lc, and the head, yo, and solving for head yields the following relationship.

(11.18)

y sub o equals (Q divided by (C sub w times square root of (2 times g) times L sub c)) to the two-thirds power

where,

yo = required head on the weir crest to pass the design flow, m (ft)

Lc = length of box inlet crest, m (ft)

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

By inspection of Figure 11.4, it is apparent that the weir crest length is:

(11.19)

L sub c equals W sub 2 plus 2 times L sub 1

where,

Lc = length of box inlet crest, m (ft)

W2 = width of box inlet, m (ft)

L1 = length of box inlet, m (ft)

Various lengths of crest, Lc, are evaluated in Equation 11.18 to obtain a head consistent with the hydraulic conditions in the approach channel.

Several corrections to the weir coefficient used in Equation 11.18 are appropriate if the crest does control the structure hydraulics. However, for determining crest versus headwall control it is not necessary to make these corrections. Four multiplicative corrections are considered:

  1. Correction for low relative head given in Figure 11.5.
  2. Correction for box inlet shape given in Figure 11.6. This correction is only applicable for W1/Lc≥ 3.
  3. Correction for approach channel width given in Figure 11.7. This correction is only applicable for W1/Lc < 3.
  4. Correction for dike proximity to the box inlet crest given in Table 11.1. These values have a low precision.

The precision of the correction curves is within 7 percent when there is no dike present and within 15 percent when dikes are used.

Second, assume headwall control and calculate the head, yo, to determine if this head is greater than that obtained for the box inlet crest control. The general equation relating discharge and head for a rectangular weir was described in Equation 11.17. For this case, the weir length is W2, the head is yo + CH, and the weir coefficient is C2. Solving for head yields the following relationship.

(11.20)

y sub o equals (Q divided by (C sub 2 times square root of (2 times g) times W sub 2)) to the two-thirds power minus C sub H

where,

yo = required head on the headwall crest to pass the design flow, m (ft)

W2 = width of the box inlet, m (ft)

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

C2 = dimensionless weir coefficient (discharge coefficient)

CH = head correction, m (ft)

The discharge coefficient, C2, is obtained from Figure 11.8. The head correction, CH, is given in Figure 11.9. If ho/W2 is between 1/4 and 1, CH may be more readily determined from Figure 11.10. The precision of the design curves for headwall control is estimated to be within 10 percent.

When the box inlet drop structure operates under submerged conditions, reference should be made to Blaisdell and Donnelly (1956) to determine the submerged design. However, this is not a desirable design condition.

The outlet for a box inlet drop structure should be designed as follows. Critical depth in the straight section is:

(11.21)

y sub c equals ((Q divided by W sub 2) squared divided by g) to the one-third power

Similarly, critical depth at the exit of the stilling basin is:

(11.22)

y sub c3 equals ((Q divided by W sub 3) squared divided by g) to the one-third power

The minimum length of the straight section for values of L1/W2 equal to or greater than 0.25 is:

(11.23)

L sub 2 equals y sub c times (0.2 times W sub 2 divided by L sub 1 plus 1)

As shown in Figure 11.4, the sidewalls of the stilling basin may flare from z=0 (parallel extensions of the section walls) to z=0.5.

Figure 11.5. Discharge Coefficients/Correction for Head with Control at Box Inlet Crest

Curve for determining the discharge coefficient, C sub 1, as a function of Y sub o divided by W sub 2

Figure 11.6. Correction for Box Inlet Shape with Control at Box Inlet Crest

Curve for determining the shape correction, C sub s, as a function of L sub 1 divided by W sub 2

Figure 11.7. Correction for Approach Channel Width with Control at Box Inlet Crest

Curve for determining the approach channel width correction, C sub A, as a function of W sub 1 divided by L sub C

Table 11.1. Correction for Dike Effect, CE, with control at Box Inlet Crest
L1/W1W4/W2
0.00.10.20.30.40.50.6
0.50.900.961.001.021.041.051.05
1.00.800.880.930.960.981.001.01
1.50.760.830.880.920.940.960.97
2.00.760.830.880.920.940.960.97

Figure 11.8. Coefficient of Discharge with Control at Headwall Opening

Curve for determining the discharge coefficient, C sub 2, as a function of h sub o divided by W sub 2

Figure 11.9. Relative Head Correction with Control at Headwall Opening

Curves for determining the relative head correction, C sub H. The abscissa is L sub 1 divided by h sub o and the ordinate is C sub H divided by W sub 2. A family of curves is presented which are also a function of L sub 1 divided by W sub 2 and h sub o divided by W sub 2.

Figure 11.10. Relative Head Correction for ho/W2 ≥ 1/4 with Control at Headwall Opening

Curve for determining the relative head correction, C sub H divided by h sub o, as a function of L sub 1 divided by h sub o

The minimum length of the final, potentially flared portion of the stilling basin is taken as the larger of the following two equations. However, Equation 11.24b is only valid for L1/W2 values equal to or greater than 0.25.

(11.24a)

L sub 3 equals L sub c times W sub 2 divided by (2 times L sub 1)

or

(11.24b)

L sub 3 equals (W sub 3 minus W sub 2) divided by (2 times z)

Frequently, it is desirable to design the stilling basin outlet width to equal the width of the tailwater channel. When the stilling basin width at the exit is less than 11.5yc3, the minimum tailwater depth over the basin floor is:

(11.25)

y sub 3 equals 1.6 times y sub c3

When the stilling basin width at the exit is greater than 11.5 yc3, the minimum tailwater depth over the basin floor is calculated from Equation 11.26. However, such a stilling basin may make inefficient use of the outlet.

(11.26)

y sub 3 equals y sub c3 plus (0.52 times W sub 3)

The height of the end sill is:

(11.27)

h sub 4 equals y sub 3 divided by 6

Longitudinal sills will improve the flow distribution in the outlet. Considerations for their use are:

  • When the stilling basin sidewalls are parallel (z=0), the longitudinal sills may be omitted.
  • The center pair of longitudinal sills should start at the exit of the box inlet and extend through the straight section and stilling basin to the end sill.
  • When W3 is less than 2.5W2, only two sills are needed. These sills should be located at a distance W5, each side of the centerline.
  • When W3 exceeds 2.5W2 two additional sills are required. These sills should be located parallel to the outlet centerline and midway between the center sills and the sidewalls at the exit of the stilling basin.
  • The height of the longitudinal sills should be the same as the height of the end sill.

The minimum height of the sidewalls above the water surface at the exit of the stilling basin is calculated from Equation 11.28. The sidewalls should extend above the tailwater surface under all conditions.

(11.28)

h sub 3 equals y sub 3 divided by 3

The wingwalls should be triangular in elevation and have top slope of 45° with the horizontal. Top slopes as flat as 30° are permissible. The wingwalls should flare in plan at an angle of 60° with the outlet centerline. Flare angles as small as 45° are permissible; however, wingwalls parallel to the outlet centerline are not recommended.

The design procedure for the box inlet drop structure may be summarized in the following steps.

Step 1. Select the initial box inlet trial dimensions, ho, L1, and W2.

Step 2. Assume crest control and estimate the crest control head using Equation 11.18.

Step 3. Assume headwall control and estimate the headwall control head using Equation 11.20.

Step 4. Select the largest head from steps 2 and 3. If the largest head is crest control, adjust the crest control head with the correction factors from Figures 11.5, 11.6, and 11.7 as well as Table 11.1.

Step 5. Calculate critical depths in both the straight and flared basin sections using Equations 11.21 and 11.22, respectively.

Step 6. Determine the basin length from Equations 11.23 and 11.24.

Step 7. Calculate the outlet depth from Equations 11.25 and 11.26 and compare this depth with the tailwater depth.

Step 8. Calculate the sill height using Equation 11.27 and determine the need for longitudinal sills.

Step 9. Determine the height of the sidewalls using Equation 11.28. Lay out the wingwalls.

Design Example: Box Inlet Drop Structure (SI)

Find the dimensions for a box inlet drop structure used to reduce channel slope. Given:

  • Q = 7.1 m3/s
  • ho = 1.20 m
  • W2 = 1.20 m
  • L1 = 1.20 m
  • Upstream and downstream channel (trapezoidal)
  • B = 6.0 m
  • Z = 1V:3H
  • So = 0.002 m/m
  • n = 0.030

Solution

Step 1. The initial box inlet trial dimensions, ho, L1, and W2 were given.

Step 2. Assume crest control and estimate the crest control head using Equation 11.18. Equation 11.19 gives us the crest length for Equation 11.18.

Lc = W2 + 2L1 = 1.2 + 2(1.2) = 3.6 m

y sub o equals (Q divided by (C sub w times square root of (2 times g) times L sub c)) to the two-thirds power equals (7.1 divided by (0.43 times square root of (2 times 9.81) times 3.6)) to the two-thirds power equals 1.024 meters

Step 3. Assume headwall control and estimate the headwall control head using Equation 11.20. First we need to determine C2 and CH.

From Figure 11.8 for a value of ho/W2=1.2/1.2=1.0 we determine C2=0.43.

From Figure 11.10 for a value of L1/ho=1.2/1.2=1.0 we determine CH/ho=0.49. Therefore, CH=0.588 m

y sub o equals (Q divided by (C sub 2 times square root of (2 times g) times W sub 2)) to the two-thirds power minus C sub H equals (7.1 divided by (0.43 times square root of (2 times 9.81) times 1.2)) to the two-thirds power minus 0.588 equals 1.541 meters

Step 4. Select the largest head from steps 2 and 3. In this case, the headwall controls and yo=1.541 m

Step 5. Calculate critical depths in the straight section using Equation 11.21.

y sub c equals ((Q divided by W sub 2) squared divided by g) to the one-third power equals ((7.1 divided by 1.2) squared divided by 9.81) to the one-third power equals 1.528 meters

Calculate critical depth at the exit using Equation 11.22 and taking the structure width equal to the channel width.

y sub c3 equals ((Q divided by W sub 3) squared divided by g) to the one-third power equals ((7.1 divided by 6.0) squared divided by 9.81) to the one-third power equals 0.523 meters

Step 6. Determine the length of the straight section beyond the inlet from Equation 11.23.

L sub 2 equals y sub c times (0.2 times W sub 2 divided by L sub 1 plus 1) equals 1.528 times (0.2 times 1.2 divided by 1.2 plus 1) equals 1.834 meters(round to 1.8 m)

The length of the flared section is determined by the maximum of Equations 11.24a and 11.24b.

L sub 3 equals L sub c times W sub 2 divided by (2 times L sub 1) equals 3.6 times 1.2 divided by (2 times 1.2) equals 1.8 meters

L sub 3 equals (W sub 3 minus W sub 2) divided by (2 times z) equals (6.0 minus 1.2) divided by (2 times 0.5) equals 4.8 meters

Therefore, L3 = 4.8 m

Step 7. Calculate the outlet depth from Equations 11.25 or 11.26 depending on whether or not W3 is less than 11.5yc3. 11.5yc3 = 11.5(0.523) = 6.01 m. Therefore, use Equation 11.25.

y3 = 1.6yc3 = 1.6(0.523) = 0.84 m

Normal depth in the tailwater channel is 0.80 m. Since y3 is slightly greater than the tailwater channel depth, some form of channel protection at the exit may be advisable to protect against erosion from the accelerating flow.

Step 8. Calculate the sill height using Equation 11.27.

h sub 4 equals y sub 3 divided by 6 equals 0.84 divided by 6 equals 0.14 meters

We determine if longitudinal sills are necessary by comparing W3 to 2.5W2. Since 2.5W2=2.5(1.2)=3.0 m and this is less than W3, 4 sills are needed.

Step 9. Determine the height of the sidewalls above the water surface elevation using Equation 11.28.

h sub 3 equals y sub 3 divided by 3 equals 0.84 divided by 3 equals 0.28 meters

Design Example: Box Inlet Drop Structure (CU)

Find the dimensions for a box inlet drop structure used to reduce channel slope. Given:

  • Q = 250 ft3/s
  • ho = 4.0 ft
  • W2 = 4.0 ft
  • L1 = 4.0 ft
  • Upstream and downstream channel (trapezoidal)
  • B = 20.0 ft
  • Z = 1V:3H
  • So = 0.002 ft/ft
  • n = 0.030

Solution

Step 1. The initial box inlet trial dimensions, ho, L1, and W2 were given.

Step 2. Assume crest control and estimate the crest control head using Equation 11.18. Equation 11.19 gives us the crest length for Equation 11.18.

Lc = W2 + 2L1 = 4.0 + 2(4.0) = 12.0 ft

y sub o equals (Q divided by (C sub w times square root of (2 times g) times L sub c)) to the two-thirds power equals (250 divided by (0.43 times square root of (2 times 32.2) times 12.0)) to the two-thirds power equals 3.32 ft

Step 3. Assume headwall control and estimate the headwall control head using Equation 11.20. First we need to determine C2 and CH.

From Figure 11.8 for a value of ho/W2=4.0/4.0=1.0 we determine C2=0.43.

From Figure 11.10 for a value of L1/ho=4.0/4.0=1.0 we determine CH/ho=0.49. Therefore, CH=1.96 ft

y sub o equals (Q divided by (C sub 2 times square root of (2 times g) times W sub 2)) to the two-thirds power minus C sub H equals (250 divided by (0.43 times square root of (2 times 32.2) times 12.0)) to the two-thirds power minus 1.96 equals 4.94 feet

Step 4. Select the largest head from steps 2 and 3. In this case, the headwall controls and yo=4.94 ft

Step 5. Calculate critical depths in the straight section using Equation 11.21.

y sub c equals ((Q divided by W sub 2) squared divided by g) to the one-third power equals ((250 divided by 4.0) squared divided by 32.2) to the one-third power equals 4.95 feet

Calculate critical depth at the exit using Equation 11.22 and taking the structure width equal to the channel width.

y sub c3 equals ((Q divided by W sub 3) squared divided by g) to the one-third power equals ((250 divided by 20.0) squared divided by 32.2) to the one-third power equals 1.69 feet

Step 6. Determine the length of the straight section beyond the inlet from Equation 11.23.

L sub 2 equals y sub c times (0.2 times W sub 2 divided by L sub 1 plus 1) equals 4.95 times (0.2 times 4.0 divided by 4.0 plus 1) equals 5.94 feet(round to 5.9 ft)

The length of the flared section is determined by the maximum of Equations 11.24a and 11.24b.

L sub 3 equals L sub c times W sub 2 divided by (2 times L sub 1) equals 12 times 4.0 divided by (2 times 4.0) equals 6.0 feet

L sub 3 equals (W sub 3 minus W sub 2) divided by (2 times z) equals (20.0 minus 4.0) divided by (2 times 0.5) equals 16.0 feet

Therefore, L3 = 16.0 ft

Step 7. Calculate the outlet depth from Equations 11.25 or 11.26 depending on whether or not W3 is less than 11.5yc3. 11.5yc3 = 11.5(1.69) = 19.4 ft. Therefore, use Equation 11.25.

y3 = 1.6yc3 = 1.6(1.69) = 2.7 ft

Normal depth in the tailwater channel is 2.6 ft. Since y3 is slightly greater than the tailwater channel depth, some form of channel protection at the exit may be advisable to protect against erosion from the accelerating flow.

Step 8. Calculate the sill height using Equation 11.27.

h sub 4 equals y sub 3 divided by 6 equals 2.7 divided by 6 equals 0.45 feet

We determine if longitudinal sills are necessary by comparing W3 to 2.5W2. Since 2.5W2=2.5(4.0)=10 ft and this is less than W3, 4 sills are needed.

Step 9. Determine the height of the sidewalls above the water surface elevation using Equation 11.28.

h sub 3 equals y sub 3 divided by 3 equals 2.7 divided by 3 equals 0.90 feet

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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