1

# Is Q=CiA Dimensionally Consistent?

Shed runoff flow rate Q has been computed under the Rational Method using the equation:

Q = CiA

for a very long time, but ever so often, a new user would ask if this equation is dimensionally consistent.

In this equation, the units of Q is cfs (cubic feet per second, or ft3/sec). C is the runoff coefficient, which is dimensionless, i is the rainfall intensity in inches per hour (in/hr)), and A is the tributary watershed area, in acres. If the equation were dimensionally consistent, then shouldn’t the units of Q be (in*ac/hr)? If it is, then is in*ac/hr = ft3/sec? This is examined below.

Convert in*ac/hr to ft3/sec as follows:

 in*ac/hr = (1/3600)*in*ac/sec = (1/12)*(1/3600)*ft*ac/sec = (1/12)*(1/3600)*43560*ft*ft2/sec = (43560/(12 * 3600)) ft3/sec = 1.0083*ft3/sec.

Thus, strictly speaking, Q = 1.0083*CiA for English units, which is usually rounded to Q = CiA. Thus, Q = CiA is indeed dimensionally consistent. The Ration Method equation, however, slightly underestimates Q, by 0.83 percent, for English units.

Similarly, it may shown that the metric equation:

Q = (1/360)*CiA,

where Q is the watershed runoff flow rate with units of m3/sec, C is the dimensionless runoff coefficient, i is the rainfall rate in mm/hr, and A is the tributary watershed area in hectares, is dimensionally consistent, without the need for rounding. The Rational Method equation does not underestimate (or overestimate) Q for metric units.

# HEC-22 PAVEMENT DRAINAGE

4. PAVEMENT DRAINAGE

(Based on HEC-22, Third Edition, revised August 2013)

Effective drainage of highway pavements is essential to the maintenance of highway service level and to traffic safety. Water on the pavement can interrupt traffic, reduce skid resistance, increase potential for hydroplaning, limit visibility due to splash and spray, and cause difficulty in steering a vehicle when the front wheels encounter puddles.(18)

Pavement drainage requires consideration of surface drainage, gutter flow, and inlet capacity. The design of these elements is dependent on storm frequency and the allowable spread of storm water on the pavement surface. This chapter presents design guidance for the design of these elements. HEC-12, Drainage of Highway Pavements,(19) and AASHTO’s Model Drainage Manual(18) originally published most of the information presented in this chapter.

Two of the more significant variables considered in the design of highway pavement drainage are the frequency of the design runoff event and the allowable spread of water on the pavement. A related consideration is the use of an event of lesser frequency to check the drainage design.

Spread and design frequency are not independent. The implications of the use of a criteria for spread of one-half of a traffic lane is considerably different for one design frequency than for a lesser frequency. The interdependency of spread and design frequency also have different implications for a low-traffic, low-speed highway than for a higher classification highway. These subjects are central to the issue of highway pavement drainage and important to highway safety.

4.1.1 Selection of Design Frequency and Design Spread

The objective of highway storm drainage design is to provide for safe passage of vehicles during the design storm event. The design of a drainage system for a curbed highway pavement section is to collect runoff in the gutter and convey it to pavement inlets in a manner that provides reasonable safety for traffic and pedestrians at a reasonable cost. As spread from the curb increases, the risks of traffic accidents and delays, and the nuisance and possible hazard to pedestrian traffic increase.

The process of selecting the recurrence interval and spread for design involves decisions regarding acceptable risks of accidents and traffic delays and acceptable costs for the drainage system. Risks associated with water on traffic lanes are greater with high traffic volumes, high speeds, and higher highway classifications than with lower volumes, speeds, and highway classifications.

The primary consideration in selecting the design frequency and design spread is the safety of the travelling public. Beyond that imperative, the following summarizes major considerations:

1. Classification of the highway is a good starting point in the selection process since it defines the public’s expectations regarding water on the pavement surface. Ponding on traffic lanes of high-speed, high-volume highways is contrary to the public’s expectations and thus the risks of accidents and the costs of traffic delays are high.
2. Design speed is important to the selection of drainage design criteria considering that speed has been shown to be a primary factor in the cause of hydroplaning when water is on the pavement.
3. Projected traffic volumes are an indicator of the economic importance of keeping the highway open to traffic. The costs of traffic delays and accidents increase with increasing traffic volumes.
4. Intensity of rainfall events may significantly affect the selection of design frequency and spread. Risks associated with the spread of water on pavements may be less in arid areas subject to high intensity thunderstorm events than in areas accustomed to frequent but less intense events.
5. Capital costs are neither the least nor last consideration. Cost considerations make it necessary to formulate a rational approach to the selection of design criteria. “Tradeoffs” between desirable and practicable criteria are sometimes necessary because of costs. In particular, the costs and feasibility of providing for a given design frequency and spread may vary significantly between projects. In some cases, it may be practicable to significantly upgrade the drainage design and reduce risks at moderate costs. In other instances, such as where extensive outfalls or pumping stations are required, costs may be very sensitive to the criteria selected for use in design.

Other considerations include inconvenience, hazards, and nuisances to pedestrian traffic. These considerations should not be minimized and, in some locations such as in commercial areas, may assume major importance. Local design practice may also be a major consideration since it can affect the feasibility of designing to higher standards, and it influences the public’s perception of acceptable practice.

The relative elevation of the highway and surrounding terrain is an additional consideration where water can be drained only through a storm drainage system, as in underpasses and depressed sections. The potential for ponding to hazardous depths should be considered in selecting the frequency and spread criteria and in checking the design against storm runoff events of lesser frequency than the design event.

Spread on traffic lanes can be tolerated to greater widths where traffic volumes and speeds are low. Spreads of one-half of a traffic lane or more are usually considered a minimum type design for low-volume local roads.

Selection of design criteria for intermediate types of facilities may be the most difficult. For example, some arterials with relatively high traffic volumes and speeds may not have shoulders which will convey the design runoff without encroaching on the traffic lanes. In these instances, an assessment of the relative risks and costs of various design spreads may be helpful in selecting appropriate design criteria. Table 4-1 provides suggested minimum design frequencies and spread based on the type of highway and traffic speed.

Recommended design frequency for depressed sections and underpasses where ponded water can be removed only through the storm drainage system is a 50-year frequency event. The use of a lesser frequency event, such as a 100-year storm, to assess hazards at critical locations where water can pond to appreciable depths is commonly referred to as a check storm or check event.

 Road Classification Design Frequency Design Spread High Volume or Divided or Bi-Directional < 70 km/hr (45 mph) 10-year Shoulder + 1 m (3 ft) > 70 km/hr (45 mph) 10-year Shoulder Sag Point 50-year Shoulder + 1 m (3 ft) Collector < 70 km/hr (45 mph) 10-year 1/2 Driving Lane > 70 km/hr (45 mph) 10-year Shoulder Sag Point 10-year 1/2 Driving Lane Local Streets Low ADT 5-year 1/2 Driving Lane High ADT 10-year 1/2 Driving Lane Sag Point 10-year 1/2 Driving Lane

4.1.2 Selection of Check Storm and Spread

A check storm should be used any time runoff could cause unacceptable flooding during less frequent events. Also, inlets should always be evaluated for a check storm when a series of inlets terminates at a sag vertical curve where ponding to hazardous depths could occur.

The frequency selected for the check storm should be based on the same considerations used to select the design storm, i.e., the consequences of spread exceeding that chosen for design and the potential for ponding. Where no significant ponding can occur, check storms are normally unnecessary.

Criteria for spread during the check event are: (1) one lane open to traffic during the check storm event, and (2) one lane free of water during the check storm event. These criteria differ substantively, but each sets a standard by which the design can be evaluated.

4.2 Surface Drainage

When rain falls on a sloped pavement surface, it forms a thin film of water that increases in thickness as it flows to the edge of the pavement. Factors which influence the depth of water on the pavement are the length of flow path, surface texture, surface slope, and rainfall intensity. As the depth of water on the pavement increases, the potential for vehicular hydroplaning increases. For the purposes of highway drainage, a discussion of hydroplaning is presented and design guidance for the following drainage elements is presented:

• Longitudinal pavement slope
• Cross or transverse pavement slope
• Curb and gutter design
• Bridge decks
• Median barriers
• Impact attenuators

Additional technical information on the mechanics of surface drainage can be found in Reference 20.

4.2.1 Hydroplaning

As the depth of water flowing over a roadway surface increases, the potential for hydroplaning increases. When a rolling tire encounters a film of water on the roadway, the water is channeled through the tire tread pattern and through the surface roughness of the pavement. Hydroplaning occurs when the drainage capacity of the tire tread pattern and the pavement surface is exceeded and the water begins to build up in front of the tire. As the water builds up, a water wedge is created and this wedge produces a hydrodynamic force which can lift the tire off the pavement surface. This is considered as full dynamic hydroplaning and, since water offers little shear resistance, the tire loses its tractive ability and the driver has a loss of control of the vehicle.Hydroplaning can occur at speeds of 89 km/hr (55 mph) with a water depth of 2 mm (0.08 in).(20) However, depending on a variety of criteria, hydroplaning may occur at speeds and depths less than these values.

Hydroplaning is a function of the water depth, roadway geometrics, vehicle speed, tread depth, tire inflation pressure, and conditions of the pavement surface. The 2005 AASHTO Model Drainage Manual(90) provides guidance in calculating when it can occur. It also reports that the driver is responsible for using caution and good judgment when driving in wet conditions similar as when driving in ice and snow.(90) In problem areas, hydroplaning may be reduced by the following:

1. Design the highway geometries to reduce the drainage path lengths of the water flowing over the pavement. This will prevent flow build-up.
2. Increase the pavement surface texture depth by such methods as grooving of portland cement concrete. An increase of pavement surface texture will increase the drainage capacity at the tire pavement interface.
3. Use of open graded asphaltic pavements has been shown to greatly reduce the hydroplaning potential of the roadway surface. This reduction is due to the ability of the water to be forced through the pavement under the tire. This releases any hydrodynamic pressures that are created and reduces the potential for the tire to hydroplane.
4. Use of drainage structures along the roadway to capture the flow of water over the pavement will reduce the thickness of the film of water and reduce the hydroplaning potential of the roadway surface.

4.2.2 Longitudinal Slope

Experience has shown that the recommended minimum values of roadway longitudinal slope given in the AASHTO Policy on Geometric Design(21) will provide safe, acceptable pavement drainage. In addition, the following general guidelines are presented:

1. A minimum longitudinal gradient is more important for a curbed pavement than for an uncurbed pavement since the water is constrained by the curb. However, flat gradients on uncurbed pavements can lead to a spread problem if vegetation is allowed to build up along the pavement edge.
2. Desirable gutter grades should not be less than 0.5% for curbed pavements with an absolute minimum of 0.3%. Minimum grades can be maintained in very flat terrain by use of a rolling profile, or by warping the cross slope to achieve rolling gutter profiles.
3. To provide adequate drainage in sag vertical curves, a minimum slope of 0.3% should be maintained within 15 meters (50 ft) of the low point of the curve. This is accomplished where the length of the curve in meters divided by the algebraic difference in grades in percent (K) is equal to or less than 50 (167 in English units). This is represented as:
 K = L / (G2 – G1) (4-1)

where:

K = Vertical curve constant m/percent (ft/percent)
L = Horizontal length of curve, m (ft)

4.2.3 Cross (Transverse) Slope

Table 4-2 indicates an acceptable range of cross slopes as specified in AASHTO’s policy on geometric design of highways and streets.(21) These cross slopes are a compromise between the need for reasonably steep cross slopes for drainage and relatively flat cross slopes for driver comfort and safety. These cross slopes represent standard practice. Reference 21 should be consulted before deviating from these values.

 Surface Type Range in Rate of Surface Slope m/m (ft/ft) High-Type Surface 2-lanes 0.015 – 0.020 3 or more lanes, each direction 0.015 minimum; increase 0.005 to 0.010 per lane; 0.040 maximum Intermediate Surface 0.015 – 0.030 Low-Type Surface 0.020 – 0.060 Shoulders Bituminous or Concrete 0.020 – 0.060 With Curbs ≥ 0.040

As reported in Pavement and Geometric Design Criteria for Minimizing Hydroplaning,(22) cross slopes of 2% have little effect on driver effort in steering or on friction demand for vehicle stability. Use of a cross slope steeper than 2% on pavements with a central crown line is not desirable. In areas of intense rainfall, a somewhat steeper cross slope (2.5%) may be used to facilitate drainage.

On multi-lane highways where three (3) lanes or more are sloped in the same direction, it is desirable to counter the resulting increase in flow depth by increasing the cross slope of the outermost lanes. The two (2) lanes adjacent to the crown line should be pitched at the normal slope, and successive lane pairs, or portions thereof outward, should be increased by about 0.5 to 1%. The maximum pavement cross slope should be limited to 4% (Table 4-2).

Additional guidelines related to cross slope are:

1. Although not widely encouraged, inside lanes can be sloped toward the median if conditions warrant.
2. Median areas should not be drained across travel lanes.
3. Number and length of flat pavement sections in cross slope transition areas should be minimized. Consideration should be given to increasing cross slopes in sag vertical curves, crest vertical curves, and in sections of flat longitudinal grades.
4. Shoulders should be sloped to drain away from the pavement, except with raised, narrow medians and superelevations.

4.2.4 Curb and Gutter

Curbs are normally used at the outside edge of pavements for low-speed, highway facilities, and in some instances adjacent to shoulders on moderate to high-speed facilities. They serve the following purposes:

• Contain the surface runoff within the roadway and away from adjacent properties
• Prevent erosion on fill slopes
• Provide pavement delineation

Gutters formed in combination with curbs are available in 0.3 through 1.0 meter (12 through 39 inch) widths. Gutter cross slopes may be the same as that of the pavement or may be designed with a steeper cross slope, usually 80 mm per meter (1 inch per foot) steeper than the shoulder or parking lane (if used). AASHTO geometric guidelines state that an 8% slope is a common maximum cross slope.

A curb and gutter combination forms a triangular channel that can convey runoff equal to or less than the design flow without interruption of the traffic. When a design flow occurs, there is a spread or widening of the conveyed water surface. The water spreads to include not only the gutter width, but also parking lanes or shoulders, and portions of the traveled surface.

Spread is what concerns the hydraulic engineer in curb and gutter flow. The distance of the spread, T, is measured perpendicular to the curb face to the extent of the water on the roadway and is shown in Figure 4-1. Limiting this width becomes a very important design criterion and will be discussed in detail in Section 4.3.

Where practical, runoff from cut slopes and other areas draining toward the roadway should be intercepted before it reaches the highway. By doing so, the deposition of sediment and other debris on the roadway as well as the amount of water which must be carried in the gutter section will be minimized. Where curbs are not needed for traffic control, shallow ditch sections at the edge of the roadway pavement or shoulder offer advantages over curbed sections by providing less of a hazard to traffic than a near-vertical curb and by providing hydraulic capacity that is not dependent on spread on the pavement. These ditch sections are particularly appropriate where curbs have historically been used to prevent water from eroding fill slopes.

Roadside channels are commonly used with uncurbed roadway sections to convey runoff from the highway pavement and from areas which drain toward the highway. Due to right-of-way limitations, roadside channels cannot be used on most urban arterials. They can be used in cut sections, depressed sections, and other locations where sufficient right-of-way is available and driveways or intersections are infrequent.

To prevent drainage from the median areas from running across the travel lanes, slope median areas and inside shoulders to a center swale. This design is particularly important for high speed facilities and for facilities with more than two lanes of traffic in each direction.

4.2.6 Bridge Decks

Bridge deck drainage is similar to that of curbed roadway sections. Effective bridge deck drainage is important for the following reasons:

• Deck structural and reinforcing steel is susceptible to corrosion from deicing salts
• Moisture on bridge decks freezes before surface roadways
• Hydroplaning often occurs at shallower depths on bridges due to the reduced surface texture of concrete bridge decks.

Bridge deck drainage is often less efficient than roadway sections because cross slopes are flatter, parapets collect large amounts of debris, drainage inlets or typical bridge scuppers are less hydraulically efficient and more easily clogged by debris, and bridges lack clear zones. Because of the difficulties in providing for and maintaining adequate deck drainage systems, gutter flow from roadways should be intercepted before it reaches a bridge. For similar reasons, zero gradients and sag vertical curves should be avoided on bridges. Additionally, runoff from bridge decks should be collected immediately after it flows onto the subsequent roadway section where larger grates and inlet structures can be used. Reference 23 includes detailed coverage of bridge deck drainage systems.

4.2.7 Median Barriers

Slope the shoulder areas adjacent to median barriers to the center to prevent drainage from running across the traveled pavement. Where median barriers are used, and particularly on horizontal curves with associated superelevations, it is necessary to provide inlets or slotted drains to collect the water accumulated against the barrier. Additionally, some highway department agencies use a piping system to convey water through the barrier.

4.2.8 Impact Attenuators

The location of impact attenuator systems should be reviewed to determine the need for drainage structures in these areas. With some impact attenuator systems it is necessary to have a clear or unobstructed opening as traffic approaches the point of impact to allow a vehicle to impact the system head on. If the impact attenuator is placed in an area where superelevation or other grade separation occurs, grate inlets and/or slotted drains may be needed to prevent water from running through the clear opening and crossing the highway lanes or ramp lanes. Curb, curb-type structures or swales cannot be used to direct water across this clear opening as vehicle vaulting could occur.

4.3 Flow in Gutters

A pavement gutter is defined, for purposes of this circular, as a section of pavement adjacent to the roadway which conveys water during a storm runoff event. It may include a portion or all of a travel lane. Gutter sections can be categorized as conventional or shallow swale type as illustrated in Figure 4-1. Conventional curb and gutter sections usually have a triangular shape with the curb forming the near-vertical leg of the triangle. Conventional gutters may have a straight cross slope (Figure 4-1, a.1), a composite cross slope where the gutter slope varies from the pavement cross slope (Figure 4-1, a.2), or a parabolic section (Figure 4-1, a.3). Shallow swale gutters typically have V-shaped or circular sections as illustrated in Figure 4-1, b.1, b.2, and b.3, respectively, and are often used in paved median areas on roadways with inverted crowns.

4.3.1 Capacity Relationship

Gutter Flow calculations are necessary to establish the spread of water on the shoulder, parking lane, or pavement section. A modification of the Manning’s equation can be used for computing flow in triangular channels. The modification is necessary because the hydraulic radius in the equation does not adequately describe the gutter cross section, particularly where the top width of the water surface may be more than 40 times the depth at the curb. To compute gutter flow, the Manning’s equation is integrated for an increment of width across the section.(24) The resulting equation is:

 Q = (Ku/n)•Sx1.67•SL0.5•T2.67 (4-2)

or in terms of T

 T = [(Q•n) / (Ku•Sx1.67•SL0.5)]0.375 (4-2)

where:

Ku = 0.376 (0.56 in English units)
n = Manning’s coefficient (Table 4-3)
Q = Flow rate, m3/s (ft3/s)
T = Width of flow (spread), m (ft)
Sx = Cross slope, m/m (ft/ft)
SL = Longitudinal slope, m/m (ft/ft)

Equation 4-2 neglects the resistance of the curb face since this resistance is negligible.

Spread on the pavement and flow depth at the curb are often used as criteria for spacing pavement drainage inlets. Design Chart 1 in Appendix A is a nomograph for solving Equation 4-2. The chart can be used for either criterion with the relationship:

 d = T•Sx (4-3)

where:

d = Depth of flow, m (ft)

 Type of Gutter or Pavement Manning’s n Concrete gutter, troweled finish 0.012 Asphalt Pavement: Smooth texture 0.013 Rough texture 0.016 Concrete gutter-asphalt pavement: Smooth 0.013 Rough 0.015 Concrete pavement: Float finish 0.014 Broom finish 0.016 For gutters with small slope, where sediment may accumulate, increase above values of “n” by 0.002 Reference: USDOT, FHWA, HDS-3(36)

Chart 1 can be used for direct solution of gutter flow where the Manning’s n value is 0.016. For other values of n, divide the value of Qn by n. Instructions for use and an example problem solution are provided on the chart.

4.3.2 Conventional Curb and Gutter Sections

Conventional gutters begin at the inside base of the curb and usually extend from the curb face toward the roadway centerline a distance of 0.3 to 1 meter (1.0 to 3.0 ft). As illustrated in Figure 4-1, gutters can have uniform, composite, or curved sections. Uniform gutter sections have a cross-slope which is equal to the cross-slope of the shoulder or travel lane adjacent to the gutter. Gutters having composite sections are depressed in relation to the adjacent pavement slope. That is, the paved gutter has a cross-slope which is steeper than that of the adjacent pavement. This concept is illustrated in Example 4-1. Curved gutter sections are sometimes found along older city streets or highways with curved pavement sections. Procedures for computing the capacity of curb and gutter sections follows.

4.3.2.1 Conventional Gutters of Uniform Cross Slope

The nomograph in Chart 1 solves Equation 4-2 for gutters having triangular cross sections. Example 4-1 illustrates its use for the analysis of conventional gutters with uniform cross slope.

Example 4-1

 Given: Gutter section illustrated in Figure 4-1 a.1. SL = 0.010 m/m (ft/ft) Sx = 0.020 m/m (ft/ft) n = 0.016 Find: (1) Spread at a flow of 0.05 m3/s (1.8 ft3/s) (2) Gutter flow at a spread of 2.5 m (8.2 ft)
 Solution (1): SI Units Step 1. Compute spread, T, using Equation 4-2 or Chart 1A. T = [(Q•n)/(Ku•Sx1.67•SL0.5)]0.375 T = [(0.05•(0.016)/{(0.376)•(0.020)1.67•(0.010)0.5}]0.375 T = 2.7 m English Units Step 1. Compute spread, T, using Equation 4-2 or Chart 1B. T = [(Q•n)/(Ku•Sx1.67•SL0.5)]0.375 T = [(1.8)•(0.016)/{(0.56)•(0.020)1.67•(0.010)0.5}]0.375 T = 9.0 ft
 Solution (2): SI Units Step 1. Using Equation 4-2 or Chart 1 with T = 2.5 m and the information given above, determine Qn. Q•n = Ku•Sx1.67•SL0.5•T2.67 Q•n = (0.376)•(0.020)1.67•(0.010)0.5•(2.5)2.67 Q•n = 0.00063 m3/s Step 2. Compute Q from Q•n determined in Step 1. Q = Q•n / n Q = 0.00063 / 0.016 Q = 0.039 m3/s English Units Step 1. Using Equation 4-2 or Chart 1 with T = 8.2 ft and the information given above, determine Qn. Q•n = Ku•Sx1.67•SL0.5 •T2.67 Q•n = (0.56)•(0.020)1.67•(0.010)0.5•(8.2)2.67 Q•n = 0.022 ft3 /s Step 2. Compute Q from Qn determined in Step 1. Q = Q•n / n Q = 0.022 / .016 Q = 1.4 ft3/s

4.3.2.2 Composite Gutter Sections

The design of composite gutter sections requires consideration of flow in the depressed segment of the gutter, Qw. Equation 4-4, displayed graphically as Chart 2, is provided for use with Equations 4-5 and 4-6 below and Chart 1 to determine the flow in a width of gutter in a composite cross section, W, less than the total spread, T. The procedure for analyzing composite gutter sections is demonstrated in Example 4-2.

 Eo = 1 / {1 + Sw / Sx / ([1 + Sw / Sx / (T / W – 1)]2.67 – 1)} (4-4) Qw = Q – Qs (4-5) Q = Qs / (1 – Eo) (4-6)
 where: Qw = Flow rate in the depressed section of the gutter, m3/s (ft3/s) Q = Gutter flow rate, m3/s (ft3/s) Qs = Flow capacity of the gutter section above the depressed section, m3/s (ft3/s) Eo = Ratio of flow in a chosen width (usually the width of a grate) to total gutter flow (Qw  /Q) Sw = Sx + a / W (figure 4-1 a.2)

Figure 4-2 illustrates a design chart for a composite gutter with a 0.60 m (2 foot) wide gutter section with a 50 mm depression at the curb that begins at the projection of the uniform cross slope at the curb face. A series of charts similar to Figure 4-2 for “typical” gutter configurations could be developed. The procedure for developing charts for depressed gutter sections is included as Appendix B. Example 4-2

 Given: Gutter section illustrated in Figure 4-1, a.2 with W = 0.6 m (2 ft) SL = 0.01 Sx = 0.020 n = 0.016 Gutter depression, a = 50 mm (2 in) Find: (1) Gutter flow at a spread, T, of 2.5 m (8.2 ft) (2) Spread at a flow of 0.12 m3/s (4.2 ft3/s)
 Solution (1): SI Units Step 1. Compute the cross slope of the depressed gutter, Sw, and the width of spread from the junction of the gutter and the road to the limit of the spread, Ts. Sw = a / W + Sx Sw = [(50)/(1000)]/(0.6) + (0.020) = 0.103 m/m Ts = T – W = 2.5 m – 0.6 m Ts = 1.9 m Step 2. From Equation 4-2 or from Chart 1 (using Ts) Qs•n = Ku•Sx1.67•SL0.5•T2.67 Qs•n = (0.376)•(0.02)1.67•(0.01)0.5•(1.9)2.67 Qs•n = 0.00031 m3/s, and Qs = (Qs•n) / n = 0.00031 / 0.016 Qs = 0.019 m3/sec Step 3. Determine the gutter flow, Q, using Equation 4-4 or Chart 2 T / W = 2.5 / 0.6 = 4.17 Sw / Sx = 0.103 / 0.020 = 5.15 Eo = 1/ {1 + [(Sw/Sx)/(1 + (Sw/Sx)/(T/W – 1))2.67 – 1]} Eo = 1/ {1 + [5.15/(1 + (5.15)/(4.17 – 1))2.67 – 1]} Eo = 0.70 Or from Chart 2, for W/T = 0.6/2.5 = 0.24 Eo = Qw / Q = 0.70 Q = Qs / (1 – Eo) Q = 0.019 / (1 – 0.70) Q = 0.06 m3/sec English Units Step 1. Compute the cross slope of the depressed gutter, Sw, and the width of spread from the junction of the gutter and the road to the limit of the spread, Ts. Sw = a / W + Sx Sw = [(2)/(12)]/(2) + (0.020) = 0.103 ft/ft Ts = T – W = 8.2 – 2.0 Ts = 6.2 ft Step 2. From Equation 4-2 or from Chart 1 (using Ts) Qs•n = Ku•Sx1.67•SL0.5•T2.67 Qs•n = (0.56)•(0.02)1.67•(0.01)0.5•(6.2)2.67 Qs•n = 0.011 ft3/s, and Qs = (Qs•n) / n = 0.011 / 0.016 Qs = 0.69 ft3/sec Step 3. Determine the gutter flow, Q, using Equation 4-4 or Chart 2 T / W = 8.2 / 2= 4.10 Sw/Sx = 0.103 / 0.020 = 5.15 Eo = 1/ {1 + [(Sw/Sx)/(1 + (Sw/Sx)/(T/W – 1))2.67 – 1]} Eo = 1/ {1 + [5.15/(1 + (5.15)/(4.10 – 1))2.67 – 1]} Eo = 0.70 Or from Chart 2, for W/T = 2/8.2 =0.24 Eo = Qw / Q = 0.70 Q = Qs / (1 – Eo) Q = 0.69 / (1 – 0.70) Q = 2.3 ft3/sec
 Solution (2): Since the spread cannot be determined by a direct solution, an iterative approach must be used. SI Units Step 1. Try Qs = 0.04 m3/sec Step 2. Compute Qw Qw = Q – Qs = 0.12 – 0.04 Qw = 0.08 m3/sec Step 3. Using Equation 4-4 or from Chart 2, determine W / T ratio Eo = Qw / Q = 0.08 / 0.12 = 0.67 Sw / Sx = 0.103 / 0.020 = 5.15 W / T = 0.23 from Chart 2 Step 4. Compute spread based on the assumed Qs T = W / (W / T) = 0.6 / 0.23 T = 2.6 m Step 5. Compute Ts based on assumed Qs Ts = T – W = 2.6 – 0.6 = 2.0 m Step 6. Use Equation 4-2 or Chart 1 to determine Qs for computed Ts Qs•n = Ku•Sx1.67•SL0.5•T2.67 Qs•n = (0.376)•(0.02)1.67•(0.01)0.5•(2.0)2.67 Qs•n = 0.00035 m3/s Qs = Qs•n / n = 0.00035 / 0.016 Qs = 0.022 m3/s Step 7. Compare computed Qs with assumed Qs. Qs assumed = 0.04 > 0.022 (Qs computed). Not close – try again. Step 8. Try a new assumed Qs and repeat Steps 2 through 7. Assume Qs = 0.058 m3/s Qw = 0.12 – 0.058 = 0.062 m3/s Eo = Qw / Q = 0.062 / 0.12 = 0.52 Sw / Sx = 5.15 W / T = 0.17 T = 0.60 / 0.17 = 3.5 m Ts = 3.5 – 0.6 = 2.9 m Qs•n = 0.00094 m3/s Qs = 0.00094 / 0.016 = 0.059 m3/s Qs assumed = 0.058 m3/s. Close to 0.059 m3/s (Qs computed). English Units Step 1. Try Qs = 1.4 ft3/s Step 2. Compute Qw Qw = Q – Qs = 4.2 -1.4 Qw = 2.8 ft3/s Step 3. Using Equation 4-4 or from Chart 2, determine W / T ratio Eo = Qw / Q = 2.8 / 4.2 = 0.67 Sw / Sx = 0.103 / 0.020 = 5.15 W / T = 0.23 from Chart 2 Step 4. Compute spread based on the assumed Qs T = W / (W / T) = 2.0 / 0.23 T = 8.7 ft Step 5. Compute Ts based on assumed Qs Ts = T – W = 8.7-2.0 = 6.7 ft Step 6. Use Equation 4-2 or Chart 1 to determine Qs for computed Ts Qs•n = Ku•Sx1.67•SL0.5•T2.67 Qs•n = (0.56)•(0.02)1.67•(0.01)0.5•(6.7)2.67 Qs•n = 0.0131 ft3/s Qs = Qs•n / n = 0.0131/ 0.016 Qs = 0.82 ft3/s Step 7. Compare computed Qs with assumed Qs. Qs assumed = 1.4 > 0.82 (Qs computed). Not close – try again. Step 8. Try a new assumed Qs and repeat Steps 2 through 7. Assume Qs = 1.9 ft3/s Qw = 4.2 -1.9 = 2.3 ft3/s Eo = Qw / Q = 2.3/4.2 = 0.55 Sw / Sx = 5.15 W / T = 0.18 T = 2.0/0.18 = 11.1 ft Ts = 11.1 – 2.0 = 9.1 ft Qs•n = 0.30 ft3/s Qs =0.30 / 0.016 = 1.85 ft3/s Qs assumed = 1.9 ft3/s. Close to 1.85 ft3/s (Qs computed).

4.3.2.3 Conventional Gutters with Curved Sections

Where the pavement cross section is curved, gutter capacity varies with the configuration of the pavement. For this reason, discharge-spread or discharge-depth-at-the-curb relationships developed for one pavement configuration are not applicable to another section with a different crown height or half-width.

Procedures for developing conveyance curves for parabolic pavement sections are included in Appendix B.

4.3.3 Shallow Swale Sections

Where curbs are not needed for traffic control, a small swale section of circular or V-shape may be used to convey runoff from the pavement. As an example, the control of pavement runoff on fills may be needed to protect the embankment from erosion. Small swale sections may have sufficient capacity to convey the flow to a location suitable for interception.

4.3.3.1 V-Sections

Chart 1 can be used to compute the flow in a shallow V-shaped section. When using Chart 1 for V-shaped channels, the cross slope, Sx is determined by the following equation:

 Sx = (Sx1•Sx2) / (Sx1 + Sx2) (4-7)

Example 4-3 demonstrates the use of Chart 1 to analyze a V-shaped shoulder gutter. Analysis of a V-shaped gutter resulting from a roadway with an inverted crown section is illustrated in Example 4-4.

Example 4-3

 Given: V-shaped roadside gutter (Figure 4-1 b.1.) with SL = 0.01 Sx1 = 0.25 Sx3 = 0.02 n = 0.016 Sx2 = 0.04 BC = 0.6 m (2.0 ft) Find: Spread at a flow of 0.05 m3/s (1.77 ft3/s)
 Solution: SI Units Step 1. Calculate Sx using Equation 4-7 assuming all flow is contained entirely in the V-shaped gutter section defined by Sx1 and Sx2. Sx = Sx1•Sx2 / (Sx1 + Sx2 ) = (0.25)•(0.04) / (0.25 + 0.04) Sx = 0.0345 Step 2. Using Equation 4-2 or Chart 1 find the hypothetical spread, T’, assuming all flow contained entirely in the V-shaped gutter. T’ = [(Q•n) / (Ku•Sx1.67•SL0.5)]0.375 T’ = [(0.05)•(0.016) / {(0.376)•(0.0345)1.67•(0.01)0.5}]0.375 T’ = 1.94 m Step 3. To determine if T’ is within Sx1 and Sx2 , compute the depth at point B in the V-shaped gutter knowing BC and Sx2. Then knowing the depth at B, the distance AB can be computed. dB = BC•Sx2 = (0.6)•(0.04) = 0.024 m AB = dB / Sx1 = (0.024) / (0.25) = 0.096 m AC = AB + BC = 0.096 + 0.60 = 0.7 m 0.7 m < T’ therefore, spread falls outside V-shaped gutter section. An iterative solution technique must be used to solve for the section spread, T, as illustrated in the following steps. Step 4. Solve for the depth at point C, dc, and compute an initial estimate of the spread, TBD along BD, dc = dB – BC•(Sx2) From the geometry of the triangle formed by the gutter, an initial estimate for dB is determined as (dB /0.25) + (dB /0.04) = 1.94 dB = 0.067 m (0.22 ft) dc = 0.067 – (0.60)•(0.04) = 0.043 m (0.14 ft) Ts = dc / Sx3 = 0.043 / 0.02 = 2.15 m TBD = Ts + BC = 2.15 + 0.6 = 2.75 m Step 5. Using a spread along BD equal to 2.75 m and develop a weighted slope for Sx2 and Sx3. 0.6 m at Sx2 (0.04) and 2.15 m at Sx3 (0.02) [(0.6)•(0.04) + (2.15)•(0.02)] / 2.75 = 0.0243 Use this slope along with Sx1, find Sx using Equation (4-7) Sx = (Sx1•Sx2) / (Sx1 + Sx2) [(0.25)•(0.0243)] / [0.25 + 0.0243] = 0.0221 Step 6. Using Equation 4-2 or Chart 1, compute the gutter spread using the composite cross slope, Sx. T = [(Q•n)/(Ku•Sx1.67•SL0.5)]0.375 T= [(0.05)•(0.016)/{(0.376)•(0.0221)1.67•(0.01)0.5}]0.375 T = 2.57 m This (2.57 m) is lower than the assumed value of 2.75 m. Therefore, assume TBD = 2.50 m and repeat Step 5 and Step 6. Step 5. 0.6 m at Sx2 (0.04) and 1.95 m at Sx3 (0.02) [(0.6)•(0.04) + (1.90)•(0.02)] / 2.50 = 0.0248 Use this slope along with Sx1, find Sx using Equation 4-7 [(0.25)•(0.0248)] / (0.25 + 0.0248) = 0.0226 Step 6. Using Equation 4-2 or Chart 1 compute the spread, T. T = [(Q•n)/(Ku•Sx1.67•SL0.5)]0.375 T = [(0.05)•(0.016)/{(0.376)•(0.0226)1.67•(0.01)0.5}]0.375 T = 2.53 m This value of T = 2.53 m is close to the assumed value of 2.50 m, therefore, OK.
 SI Units Step 1. Calculate Sx using Equation 4-7 assuming all flow is contained entirely in the V-shaped gutter section defined by Sx1 and Sx2. Sx = Sx1•Sx2 / (Sx1 + Sx2 ) = (0.25)•(0.04) / (0.25 + 0.04) Sx = 0.0345 Step 2. Using Equation 4-2 or Chart 1 find the hypothetical spread, T’, assuming all flow contained entirely in the V-shaped gutter. T’ = [(Q•n) / (Ku•Sx1.67•SL0.5)]0.375 T’ = [(1.77)•(0.016)/{(0.56)•(0.0345)1.67•(0.01)0.5}]0.375 T’ = 6.4 ft Step 3. To determine if T’ is within Sx1 and Sx2 , compute the depth at point B in the V-shaped gutter knowing BC and Sx2. Then knowing the depth at B, the distance AB can be computed. dB = BC•Sx2 = (2)•(0.04) = 0.08 ft AB = dB / Sx1 = (0.08) / (0.25) = 0.32 ft AC = AB + BC = 0.32 + 2.0 = 2.32 ft 2.32 ft < T’ therefore, spread falls outside V-shaped gutter section. An iterative solution technique must be used to solve for the section spread, T, as illustrated in the following steps. Step 4. Solve for the depth at point C, dc, and compute an initial estimate of the spread, TBD along BD, dc = dB – BC•(Sx2) From the geometry of the triangle formed by the gutter, an initial estimate for dB is determined as (dB /0.25) + (dB /0.04) = 6.4 ft dB = 0.22 ft dc = 0.22 – (2.0)•(0.04) = 0.14 ft Ts = dc / Sx3 = 0.14 / 0.02 = 7 ft TBD = Ts + BC = 7 + 2 = 9 ft Step 5. Using a spread along BD equal to 2.75 m and develop a weighted slope for Sx2 and Sx3. 2.0 ft at Sx2 (0.04) and 7.0 ft at Sx3 (0.02) [(2.0)•(0.04) + (7.0)•(0.02)] / 0.90 = 0.024 Use this slope along with Sx1, find Sx using Equation (4-7) Sx = (Sx1•Sx2) / (Sx1 + Sx2) [(0.25)•(0.024)] / [0.25 + 0.024] = 0.022 Step 6. Using Equation 4-2 or Chart 1, compute the gutter spread using the composite cross slope, Sx. T = [(Q•n)/(Ku•Sx1.67•SL0.5)]0.375 T= [(1.77)•(0.016)/{(0.56)•(0.022)1.67•(0.01)0.5}]0.375 T = 8.5 ft This 8.5 ft is lower than the assumed value of 9.0 ft. Therefore, assume TBD = 8.3 ft and repeat Step 5 and Step 6. Step 5. 2.0 ft at Sx2 (0.04) and 6.3 ft at Sx3 (0.02) [(2.0)•(0.04) + (6.3)•(0.02)] / 8.3 = 0.0248 Use this slope along with Sx1, find Sx using Equation 4-7 [(0.25)•(0.0248)] / (0.25 + 0.0248) = 0.0226 Step 6. Using Equation 4-2 or Chart 1 compute the spread, T. T = [(Q•n)/(Ku•Sx1.67•SL0.5)]0.375 T = [(1.77)•(0.016)/{(0.56)•(0.0226)1.67•(0.01)0.5}]0.375 T = 8.31 m This value of T = 8.31 ft is close to the assumed value of 8.3 ft, therefore, OK.

Example 4-4

 Given: V-shaped gutter as illustrated in Figure 4-1, b-2 with AB = 1 m (3.28ft) BC = 1 m (3.28 ft) SL = 0.01 n = 0.016 Sx1 = Sx2 = 0.25 Sx3 = 0.04 Find: (1) Spread at a flow of 0.7 m3/s (24.7 ft3/s) (2) Flow at a spread of 7 m (23.0 ft)
 Solution (1): SI Units Step 1. Assume spread remains within middle “V” (A to C) and compute Sx Sx = (Sx1•Sx2 ) / (Sx1 + Sx2 ) Sx = (0.25)•(0.25) / (0.25 + 0.25) Sx = 0.125 Step 2. From Equation 4-2 or Chart 1 T = [(Q•n)/(Ku•Sx1.67•SL0.5)]0.375 T = [(0.70)•(0.016)/{(0.376)•(0.125)1.67•(0.01)0.5}]0.375 T = 2.34 m Since “T” is outside Sx1 and Sx2, an iterative approach (as illustrated in Example 4-3) must be used to compute the spread. Step 3. Treat one-half of the median gutter as a composite section and solve for T’ equal to one-half of the total spread. Q’ for T’ = ½•Q = 0.5•(0.7) = 0.35 m3/s Step 4. Try Q’s = 0.05 m3/s Q’w = Q’ – Q’s = 0.35 – 0.05 = 0.30 m3/s Step 5. Using Equation 4-4 or Chart 2 determine the W/T’ ratio E’o = Q’w/Q’ = 0.30/0.35 = 0.86 Sw / Sx = Sx2 / Sx3 = 0.25 / 0.04 = 6.25 W/T’ = 0.33 from Chart 2 Step 6. Compute spread based on assumed Q’s T’ = W / (W/T’) = 1 / 0.33 = 3.03 m Step 7. Compute Ts based on assumed Q’s Ts = T’ – W = 3.03 – 1.0 = 2.03 m Step 8. Use Equation 4-2 or Chart 1 to determine Q’s for Ts Q’s•n = Ku•Sx31.67•SL0.5•Ts2.67 = (0.376)•(0.04)1.67•(0.01)0.5 •(2.03)2.67 Q’s•n = 0.00115 Q’s = 0.00115 / 0.016 = 0.072 m3/s Step 9. Check computed Q’s with assumed Q’s Q’s assumed = 0.05 < 0.072 = Q’s computed. Therefore, try a new assumed Q’s and repeat steps 4 through 9. Assume Q’s = 0.01 Q’w = 0.34 E’o = 0.97 Sw / Sx = 6.25 W / T’ = 0.50 T’ = 2.0 m Ts = 1.0 Qs•n = 0.00017 Qs = 0.01 Qs computed = 0.01 = 0.01 = Qs assumed T = 2•T’ = 2•(2.0) = 4.0 m Solution (2): Analyze in half-section using composite section techniques. Double the computed half-width flow rate to get the total discharge: Step 1. Compute half-section top width T’ = T/2 = 7.0 / 2 = 3.5 m Ts = T’ – 1.0 = 2.5 m Step 2. From Equation 4-2 or Chart 1 determine Qs Qs•n = Ku•Sx1.67•SL0.5•Ts2.67 Qs•n = (0.376)•(0.04)1.67•(0.01)0.5•(2.5)2.67 Qs•n = 0.0020 Qs = 0.0020 / 0.016 = 0.126 Step 3. Determine flow in half-section using Equation 4-4 or Chart 2 T’/W = 3.5 / 1.0 = 3.5 Sw / Sx = 0.25 / 0.04 = 6.25 Eo = 1 / {1 +(Sw/Sx) / [(1 + (Sw/Sx) /(T’/W – 1))2.67 – 1]} Eo = 1 / {1 +(6.25) / [(1 + (6.25) /(3.5 -1))2.67 – 1]} Eo = 0.814 = Q’w / Q = 1 – Q’s / Q’ Q’ = Q’s / (1 – 0.814) = 0.126 / (1 – 0.814) Q’ = 0.68 m3/s Q = 2•Q’ = 2•(0.68) = 1.36 m3/s
 English Units Step 1. Assume spread remains within middle “V” (A to C) and compute Sx Sx = (Sx1•Sx2 ) / (Sx1 + Sx2 ) Sx = (0.25)•(0.25) / (0.25 + 0.25) Sx = 0.125 Step 2. From Equation 4-2 or Chart 1 T = [(Q•n)/(Ku•Sx1.67•SL0.5)]0.375 T = [(24.7)•(0.016)/{(0.56)•(0.125)1.67•(0.01)0.5}]0.375 T = 7.65 ft Since “T” is outside Sx1 and Sx2, an iterative approach (as illustrated in Example 4-3) must be used to compute the spread. Step 3. Treat one-half of the median gutter as a composite section and solve for T’ equal to one-half of the total spread. Q’ for T’ = ½•Q = 0.5•(24.7) = 12.4 ft3/s Step 4. Try Q’s = 1.8 ft3/s Q’w = Q’ – Q’s = 12.4 – 1.8 = 10.6 ft3/s Step 5. Using Equation 4-4 or Chart 2 determine the W/T’ ratio E’o = Q’w/Q’ = 010.6/12.4 = 0.85 Sw / Sx = Sx2 / Sx3 = 0.25 / 0.04 = 6.25 W/T’ = 0.33 from Chart 2 Step 6. Compute spread based on assumed Q’s T’ = W / (W/T’) = 3.28 / 0.33 = 9.94 ft Step 7. Compute Ts based on assumed Q’s Ts = T’ – W = 9.94 – 3.28 = 6.66 ft Step 8. Use Equation 4-2 or Chart 1 to determine Q’s for Ts Q’s•n = Ku•Sx31.67•SL0.5•Ts2.67 = (0.56)•(0.04)1.67•(0.01)0.5 •(6.66)2.67 Q’s•n = 0.041 Q’s = 0.041 / 0.016 = 2.56 ft3/s Step 9. Check computed Q’s with assumed Q’s Q’s assumed = 1.8 < 2.56 = Q’s computed. Therefore, try a new assumed Q’s and repeat steps 4 through 9. Assume Q’s = 0.014 Q’w = 12.0 ft3/s E’o = 0.97 Sw / Sx = 6.25 W / T’ = 0.50 T’ = 6.56 ft Ts = 1.0 Qs•n = 0.0062 Qs = 0.39 ft3/s Qs computed = 0.39, which is close to 0.40 (Qs assumed). T = 2•T’ = 2•(6.56) = 13.12 ft Solution (2): Analyze in half-section using composite section techniques. Double the computed half-width flow rate to get the total discharge: Step 1. Compute half-section top width T’ = T/2 = 23 / 2 = 11.5 ft Ts = T’ – 3.28 = 8.22 ft Step 2. From Equation 4-2 or Chart 1 determine Qs Qs•n = Ku•Sx1.67•SL0.5•Ts2.67 Qs•n = (0.56)•(0.04)1.67•(0.01)0.5•(8.22)2.67 Qs•n = 0.073 Qs = 0.073 / 0.016 = 4.56 ft3/s Step 3. Determine flow in half-section using Equation 4-4 or Chart 2 T’/W = 11.5 / 3.28 = 3.51 Sw / Sx = 0.25 / 0.04 = 6.25 Eo = 1 / {1 +(Sw/Sx) / [(1 + (Sw/Sx) /(T’/W – 1))2.67 – 1]} Eo = 1 / {1 +(6.25) / [(1 + (6.25) /(3.5 -3.28))2.67 – 1]} Eo = 0.814 = Q’w / Q = 1 – Q’s / Q’ Q’ = Q’s / (1 – 0.814) = 4.56 / (1 – 0.814) Q’ = 24.5 ft3/s Q = 2•Q’ = 2•(24.5) = 49 ft3/s

4.3.3.2 Circular Sections

Flow in shallow circular gutter sections can be represented by the relationship:

 (d/D) = Ku•[(Q•n) / (D2.67•SL0.5)]0.488 (4-8)
 where: d = Depth of flow in circular gutter, m (ft) D = Diameter of circular gutter, m (ft) Ku = 1.179 (0.972 in English units)

which is displayed on Chart 3. The width of circular gutter section Tw is represented by the chord of the arc which can be computed using Equation 4-9.

 TW = 2•(r2 – (r – d)2)0.5 (4-9)
 where: Tw = Width of circular gutter section, m (ft) r = Radius of flow in circular gutter, m (ft)

Example 4-5 illustrates the use of Chart 3.

Example 4-5

 Given: A circular gutter swale as illustrated in Figure 4-1 b (3) with a 1.5 meter (4.92 ft) diameter and SL = 0.01 m/m (ft/ft) n = 0.016 Q = 0.5 m3/s (17.6 ft3/s) Find: Flow depth and top width
 Solution: SI UNITS Step 1. Determine the value of Q•n / (D2.67•SL0.5) = (0.5)•(0.016)/[(1.5)2.67•(0.01)0.5] = 0.027 Step 2. Using Equation 4-8 or Chart 3, determine d/D d/D = Ku•[(Q•n)/(D2.67•SL0.5)]0.488 d/D = (1.179)•[0.027]0.488 d/D = 0.20 d = D•(d/D) = 1.5•(0.20) = 0.30 m Step 3. Using Equation 4-9, determine Tw Tw = 2•[r2 – (r – d)2]1/2 = 2•[(0.75)2 – (0.75 – 0.3)2]1/2 = 1.2 m English Units Step 1. Determine the value of Q•n / (D2.67•SL0.5) = (17.6)•(0.016)/[(4.92)2.67•(0.01)0.5] = 0.04 Step 2. Using Equation 4-8 or Chart 3, determine d/D d/D = Ku•[(Q•n)/(D2.67•SL0.5)]0.488 d/D = (0.972)•[0.04]0.488 d/D = 0.20 d = D (d/D) = 4.92•(0.20) = 0.98 ft Step 3. Using Equation 4-9, determine Tw Tw = 2 [r2 – (r – d)2]1/2 = 2•[(2.46)2 – (2.46 – 0.98)2]1/2 = 3.93 ft

4.3.4 Flow in Sag Vertical Curves

As gutter flow approaches the low point in a sag vertical curve the flow can exceed the allowable design spread values as a result of the continually decreasing gutter slope. The spread in these areas should be checked to insure it remains within allowable limits. If the computed spread exceeds design values, additional inlets should be provided to reduce the flow as it approaches the low point. Sag vertical curves and measures for reducing spread are discussed further in Section 4.4.

4.3.5 Relative Flow Capacities

Examples 4-1 and 4-2 illustrate the advantage of a composite gutter section. The capacity of the section with a depressed gutter in the examples is 70% greater than that of the section with a straight cross slope with all other parameters held constant.

Equation 4-2 can be used to examine the relative effects of changing the values of spread, cross slope, and longitudinal slope on the capacity of a section with a straight cross slope.

To examine the effects of cross slope on gutter capacity, Equation 4-2 can be transformed as follows into a relationship between Sx and Q as follows:

Let K1 = n / (Kc•SL0.5•T2.67)

then Sx1.67 = K1•Q

and

 (Sx1 / Sx2)1.67 = [(K1•Q1) / (K1•Q2)] = Q1 / Q2 (4-10)

Similar transformations can be performed to evaluate the effects of changing longitudinal slope and width of spread on gutter capacity resulting in Equations 4-11 and 4-12 respectively.

 (SL1 / SL2)0.5 = Q1 / Q2 (4-11) (T1 / T2)2.67 = Q1 / Q2 (4-12)

Equations 4-10, 4-11, and 4-12 are illustrated in Figure 4-3. As illustrated, the effects of spread on gutter capacity are greater than the effects of cross slope and longitudinal slope, as would be expected due to the larger exponent of the spread term. The magnitude of the effect is demonstrated when gutter capacity with a 3 meter (9.8 ft) spread is 18.8 times greater than with a 1 meter (3.3 ft) spread, and 3 times greater than a spread of 2 meters (6.6 ft).

The effects of cross slope are also relatively great as illustrated by a comparison of gutter capacities with different cross slopes. At a cross slope of 4%, a gutter has 10 times the capacity of a gutter of 1% cross slope. A gutter at 4% cross slope has 3.2 times the capacity of a gutter at 2% cross slope.

Little latitude is generally available to vary longitudinal slope in order to increase gutter capacity, but slope changes which change gutter capacity are frequent. Figure 4-3 shows that a change from S = 0.04 to 0.02 will reduce gutter capacity to 71% of the capacity at S = 0.04.

4.3.6 Gutter Flow Time

The flow time in gutters is an important component of the time of concentration for the contributing drainage area to an inlet. To find the gutter flow component of the time of concentration, a method for estimating the average velocity in a reach of gutter is needed. The velocity in a gutter varies with the flow rate and the flow rate varies with the distance along the gutter, i.e., both the velocity and flow rate in a gutter are spatially varied. The time of flow can be estimated by use of an average velocity obtained by integration of the Manning’s equation for the gutter section with respect to time. The derivation of such a relationship for triangular channels is presented in Appendix B.

Table 4-4 and Chart 4 can be used to determine the average velocity in triangular gutter sections. In Table 4-4, T1 and T2 are the spread at the upstream and downstream ends of the gutter section respectively. Ta is the spread at the average velocity. Chart 4 is a nomograph to solve Equation 4-13 for the velocity in a triangular channel with known cross slope, gutter slope, and spread.

 T1/T2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Ta/T2 0.65 0.66 0.68 0.7 0.74 0.77 0.82 0.86 0.9
 V = (Ku/n)•SL0.5•Sx0.67•T0.67 (4-13)
 where: Ku = 0.752 (1.11 in English units) V = Velocity in the triangular channel, m/s (ft/s)

Example 4-6 illustrates the use of Table 4-4 and Chart 4 to determine the average gutter velocity.

Example 4-6

 Given: A triangular gutter section with the following characteristics: T1 = 1 m (3.28 ft) T2 = 3 m (9.84 ft) SL = 0.03 m/m (ft/ft) Sx = 0.02 m/m (ft/ft) n = 0.016 Inlet spacing anticipated to be 100 meters (330 ft). Find: Time of flow in gutter
 Solution: SI Units Step 1. Compute the upstream to downstream spread ratio. T1 / T2 = 1 / 3 = 0.33 Step 2. Determine the spread at average velocity interpolating between values in Table 4-4. (0.30 – 0.33)/ (0.3 – 0.4) =X / (0.74-0.70) X=0.01•Ta / T2 = 0.70 + 0.01 = 0.71 Ta = (0.71)•(3) = 2.13 m Step 3. Using Equation 4-13 or Chart 4, determine the average velocity Va = (Ku /n)•SL0.5•Sx0.67•T0.67 Va = [0.752/(0.016)]•(0.03)0.5•(0.02)0.67•(2.13)0.67 Va = 0.98 m/s Step 4. Compute the travel time in the gutter. t = L / V = (100) / (0.98) / 60 = 1.7 minutes English Units Step 1. Compute the upstream to downstream spread ratio. T1 / T2 = 3.28 / 9.84 = 0.33 Step 2. Determine the spread at average velocity interpolating between values in Table 4-4. (0.30 – 0.33)/(0.3 – 0.4) = X/(0.74-0.70)•X=0.01•Ta / T2 = 0.70 + 0.01 = 0.71 Ta = (0.71)•(9.84) = 6.99 ft Step 3. Using Equation 4-13 or Chart 4, determine the average velocity Va = (Ku / n)•SL0.5•Sx0.67•T0.67 Va = [1.11/(0.016)]•(0.03)0.5•(0.02)0.67•(6.99)0.67 Va = 3.21 ft Step 4. Compute the travel time in the gutter. t = L / V = (330) / (3.21) / 60 = 1.7 minutes

4.4 Drainage Inlet Design

The hydraulic capacity of a storm drain inlet depends upon its geometry as well as the characteristics of the gutter flow. Inlet capacity governs both the rate of water removal from the gutter and the amount of water that can enter the storm drainage system. Inadequate inlet capacity or poor inlet location may cause flooding on the roadway resulting in a hazard to the traveling public.

4.4.1 Inlet Types

Storm drain inlets are used to collect runoff and discharge it to an underground storm drainage system. Inlets are typically located in gutter sections, paved medians, and roadside and median ditches. Inlets used for the drainage of highway surfaces can be divided into the following four classes:

1. Grate inlets
2. Curb-opening inlets
3. Slotted inlets
4. Combination inlets

Grate inlets consist of an opening in the gutter or ditch covered by a grate. Curb-opening inlets are vertical openings in the curb covered by a top slab. Slotted inlets consist of a pipe cut along the longitudinal axis with bars perpendicular to the opening to maintain the slotted opening. Combination inlets consist of both a curb-opening inlet and a grate inlet placed in a side-by-side configuration, but the curb opening may be located in part upstream of the grate. Figure 4-4 illustrates each class of inlets. Slotted drains may also be used with grates and each type of inlet may be installed with or without a depression of the gutter.

4.4.2 Characteristics and Uses of Inlets

Grate inlets, as a class, perform satisfactorily over a wide range of gutter grades. Grate inlets generally lose capacity with increase in grade, but to a lesser degree than curb opening inlets. The principal advantage of grate inlets is that they are installed along the roadway where the water is flowing. Their principal disadvantage is that they may be clogged by floating trash or debris. For safety reasons, preference should be given to grate inlets where out-of-control vehicles might be involved. Additionally, where bicycle traffic occurs, grates should be bicycle safe.

Curb-opening inlets are most effective on flatter slopes, in sags, and with flows which typically carry significant amounts of floating debris. The interception capacity of curb-opening inlets decreases as the gutter grade steepens. Consequently, the use of curb-opening inlets is recommended in sags and on grades less than 3%. Of course, they are bicycle safe as well.

Combination inlets provide the advantages of both curb opening and grate inlets. This combination results in a high capacity inlet which offers the advantages of both grate and curb-opening inlets. When the curb opening precedes the grate in a “Sweeper” configuration, the curb-opening inlet acts as a trash interceptor during the initial phases of a storm. Used in a sag configuration, the sweeper inlet can have a curb opening on both sides of the grate.

Slotted drain inlets can be used in areas where it is desirable to intercept sheet flow before it crosses onto a section of roadway. Their principal advantage is their ability to intercept flow over a wide section. However, slotted inlets are very susceptible to clogging from sediments and debris, and are not recommended for use in environments where significant sediment or debris loads may be present. Slotted inlets on a longitudinal grade do have the same hydraulic capacity as curb openings when debris is not a factor.

4.4.3 Inlet Capacity

Inlet interception capacity has been investigated by several agencies and manufacturers of grates. Hydraulic tests on grate inlets and slotted inlets included in this document were conducted by the Bureau of Reclamation for the Federal Highway Administration. Four of the grates selected for testing were rated highest in bicycle safety tests, three have designs and bar spacing similar to those proven bicycle-safe, and a parallel bar grate was used as a standard with which to compare the performance of others.

References 25, 26, 27, 28, and 30 are reports resulting from this grate inlet research study. Figures 4-6, through 4-10 show the inlet grates for which design procedures were developed. For ease in identification, the following terms have been adopted:

 P-50 Parallel bar grate with bar spacing 48 mm (1-7/8 in) on center (Figure 4-5). P-50 x 100 Parallel bar grate with bar spacing 48 mm (1-7/8 in) on center and 10 mm (3/8 in) diameter lateral rods spaced at 102 mm (4 in) on center (Figure 4-5). P-30 Parallel bar grate with 29 mm (1-1/8 in) on center bar spacing (Figure 4-6). Curved Vane Curved vane grate with 83 mm (3-1/4 in) longitudinal bar and 108 mm (4-1/4 in) transverse bar spacing on center (Figure 4-7). 45° – 60 Tilt Bar 45° tilt-bar grate with 57 mm (2-1/4 in) longitudinal bar and 102 mm (4 in) transverse bar spacing on center (Figure 4-8). 45° – 85 Tilt Bar 45° tilt-bar grate with 83 mm (3-1/4 in) longitudinal bar and 102 mm (4 in) transverse bar spacing on center (Figure 4-8). 30° – 85 Tilt Bar 30° tilt-bar grate with 83 mm (3-1/4 in) longitudinal bar and 102 mm (4 in) transverse bar spacing on center (Figure 4-9). Reticuline “Honeycomb” pattern of lateral bars and longitudinal bearing bars (Figure 4-10)

The interception capacity of curb-opening inlets have also been investigated by several agencies. Design procedures adopted for this Circular are largely derived from experimental work at Colorado State University for the Federal Highway Administration, as reported in Reference 24 and from Reference 29.

4.4.3.1 Factors Affecting Inlet Interception Capacity and Efficiency on Continuous Grades

Inlet interception capacity, Qi, is the flow intercepted by an inlet under a given set of conditions. The efficiency of an inlet, E, is the percent of total flow that the inlet will intercept for those conditions. The efficiency of an inlet changes with changes in cross slope, longitudinal slope, total gutter flow, and, to a lesser extent, pavement roughness. In mathematical form, efficiency, E, is defined by the following equation:

 E = Q1 / Q (4-14)
 where: E = Inlet efficiency Q = Total gutter flow, m3/s (ft3/s) Qi = Intercepted flow, m3/s (ft3/s)

Flow that is not intercepted by an inlet is termed carryover or bypass and is defined as follows:

 Qb = Q – Qi (4-15)
 where: Qb = Bypass flow, m3/s (ft3/s)

The interception capacity of all inlet configurations increases with increasing flow rates, and inlet efficiency generally decreases with increasing flow rates. Factors affecting gutter flow also affect inlet interception capacity. The depth of water next to the curb is the major factor in the interception capacity of both grate inlets and curb-opening inlets. The interception capacity of a grate inlet depends on the amount of water flowing over the grate, the size and configuration of the grate and the velocity of flow in the gutter. The efficiency of a grate is dependent on the same factors and total flow in the gutter.

Interception capacity of a curb-opening inlet is largely dependent on flow depth at the curb and curb opening length. Flow depth at the curb and consequently, curb-opening inlet interception capacity and efficiency, is increased by the use of a local gutter depression at the curb-opening or a continuously depressed gutter to increase the proportion of the total flow adjacent to the curb. Top slab supports placed flush with the curb line can substantially reduce the interception capacity of curb openings. Tests have shown that such supports reduce the effectiveness of openings downstream of the support by as much as 50% and, if debris is caught at the support, interception by the downstream portion of the opening may be reduced to near zero. If intermediate top slab supports are used, they should be recessed several inches from the curb line and rounded in shape.

Slotted inlets function in essentially the same manner as curb opening inlets, i.e., as weirs with flow entering from the side. Interception capacity is dependent on flow depth and inlet length. Efficiency is dependent on flow depth, inlet length and total gutter flow.

The interception capacity of an equal length combination inlet consisting of a grate placed alongside a curb opening on a grade does not differ materially from that of a grate only. Interception capacity and efficiency are dependent on the same factors which affect grate capacity and efficiency. A combination inlet consisting of a curb-opening inlet placed upstream of a grate inlet has a capacity equal to that of the curb-opening length upstream of the grate plus that of the grate, taking into account the reduced spread and depth of flow over the grate because of the interception by the curb opening. This inlet configuration has the added advantage of intercepting debris that might otherwise clog the grate and deflect water away from the inlet.

4.4.3.2 Factors Affecting Inlet Interception Capacity in Sag Locations

Grate inlets in sag vertical curves operate as weirs for shallow ponding depths and as orifices at greater depths. Between weir and orifice flow depths, a transition from weir to orifice flow occurs. The perimeter and clear opening area of the grate and the depth of water at the curb affect inlet capacity. The capacity at a given depth can be severely affected if debris collects on the grate and reduces the effective perimeter or clear opening area.

Curb-opening inlets operate as weirs in sag vertical curve locations up to a ponding depth equal to the opening height. At depths above 1.4 times the opening height, the inlet operates as an orifice and between these depths, transition between weir and orifice flow occurs. The curb-opening height and length, and water depth at the curb affect inlet capacity. At a given flow rate, the effective water depth at the curb can be increased by the use of a continuously depressed gutter, by use of a locally depressed curb opening, or by use of an increased cross slope, thus decreasing the width of spread at the inlet.

Slotted inlets operate as weirs for depths below approximately 50 mm (2 in) and orifices in locations where the depth at the upstream edge of the slot is greater than about 120 mm (5 in). Transition flow exists between these depths. For orifice flow, an empirical equation derived from experimental data can be used to compute interception capacity. Interception capacity varies with flow depth, slope, width, and length at a given spread. Slotted drains are not recommended in sag locations because they are susceptible to clogging from debris.

4.4.3.3 Comparison of Interception Capacity of Inlets on Grade

In order to compare the interception capacity and efficiency of various inlets on grade, it is necessary to fix two variables that affect capacity and efficiency and investigate the effects of varying the other factor. Figure 4-11 shows a comparison of curb-opening inlets, grates, and slotted drain inlets with gutter flow fixed at 0.09 m3/s (3.2 ft3/s), cross slope fixed at 3%, and longitudinal slope varied up to 10%. Conclusions drawn from an analysis of this figure are not necessarily transferable to other flow rates or cross slopes, but some inferences can be drawn that are applicable to other sets of conditions. Grate configurations used for interception capacity comparisons in this figure are described in Section 4.4.3.

Figure 4-11 illustrates the effects of flow depth at the curb and curb-opening length on curb-opening inlet interception capacity and efficiency. All of the slotted inlets and curb-opening inlets shown in the figure lose interception capacity and efficiency as the longitudinal slope is increased because spread on the pavement and depth at the curb become smaller as velocity increases. It is accurate to conclude that curb-opening inlet interception capacity and efficiency would increase with steeper cross slopes. It is also accurate to conclude that interception capacity would increase and inlet efficiency would decrease with increased flow rates. Long curb-opening and slotted inlets compare favorably with grates in interception capacity and efficiency for conditions illustrated in Figure 4-11.

The effect of depth at the curb is also illustrated by a comparison of the interception capacity and efficiency of depressed and undepressed curb-opening inlets. A 1.5 m (5 ft) depressed curb-opening inlet has about 67% more interception capacity than an undepressed inlet at 2% longitudinal slope, 3% cross slope, and 0.085 m3/s (3 ft3/s) gutter flow, and about 79% more interception capacity at an 8% longitudinal slope.

At low velocities, all of the water flowing in the section of gutter occupied by the grate, called frontal flow, is intercepted by grate inlets. Only a small portion of the flow outside of the grate, termed side flow, is intercepted. When the longitudinal slope is increased, water begins to skip or splash over the grate at velocities dependent on the grate configuration. Figure 4-11 shows that interception capacity and efficiency are reduced at slopes steeper than the slope at which splash-over begins. Splash-over for the less efficient grates begins at the slope at which the interception capacity curve begins to deviate from the curve of the more efficient grates. All of the 0.6 m by 0.6 m (2 ft by 2 ft) grates have equal interception capacity and efficiency at a flow rate of 0.085 m3/s (3 ft3/s), cross slope of 3%, and longitudinal slope of 2%. At slopes steeper than 2%, splash-over occurs on the reticuline grate and the interception capacity is reduced. At a slope of 6%, velocities are such that splash-over occurs on all except the curved vane and parallel bar grates. From these performance characteristics curves, it can be concluded that parallel-bar grates and the curved vane grate are relatively efficient at higher velocities and the reticuline grate is least efficient. At low velocities, the grates perform equally. However, some of the grates such as the reticuline grate are more susceptible to clogging by debris than the parallel bar grate.

The capacity and efficiency of grates increase with increased slope and velocity if splash-over does not occur. This is because frontal flow increases with increased velocity, and all frontal flow will be intercepted if splash-over does not occur.

Figure 4-11 also illustrates that interception by longer grates would not be substantially greater than interception by 0.6 m by 0.6 m (2 ft by 2 ft) grates. In order to capture more of the flow, wider grates would be needed.

Figure 4-12 can be used for further study and comparisons of inlet interception capacity and efficiency. It shows, for example, that at a 6% slope, splash-over begins at about 0.02 m3/s (0.7 ft3/s) on a reticuline grate. It also illustrates that the interception capacity of all inlets increases and inlet efficiency decreases with increased discharge.

This comparison of inlet interception capacity and efficiency neglects the effects of debris and clogging on the various inlets. All types of inlets, including curb-opening inlets, are subject to clogging, some being more susceptible than others. Attempts to simulate clogging tendencies in the laboratory have not been notably successful, except to demonstrate the importance of parallel bar spacing in debris handling efficiency. Grates with wider spacings of longitudinal bars pass debris more efficiently. Except for reticuline grates, grates with lateral bar spacing of less than 0.1 m (4 in) were not tested so conclusions cannot be drawn from tests concerning debris handling capabilities of many grates currently in use. Problems with clogging are largely local since the amount of debris varies significantly from one locality to another. Some localities must contend with only a small amount of debris while others experience extensive clogging of drainage inlets. Since partial clogging of inlets on grade rarely causes major problems, allowances should not be made for reduction in inlet interception capacity except where local experience indicates an allowance is advisable.

4.4.4 Interception Capacity of Inlets on Grade

The interception capacity of inlets on grade is dependent on factors discussed in Section 4.4.3.1. In this section, design charts for inlets on grade and procedures for using the charts are presented for the various inlet configurations. Remember that for locally depressed inlets, the quantity of flow reaching the inlet would be dependent on the upstream gutter section geometry and not the depressed section geometry.

Charts for grate inlet interception have been made and are applicable to all grate inlets tested for the Federal Highway Administration (references 25 through 28). The chart for frontal flow interception is based on test results which show that grates intercept all of the frontal flow until a velocity is reached at which water begins to splash over the grate. At velocities greater than “Splash-over” velocity, grate efficiency in intercepting frontal flow is diminished. Grates also intercept a portion of the flow along the length of the grate, or the side flow. A chart is provided to determine side-flow interception.

One set of charts is provided for slotted inlets and curb-opening inlets, because these inlets are both side-flow weirs. The equation developed for determining the length of inlet required for total interception fits the test data for both types of inlets.

A procedure for determining the interception capacity of combination inlets is also presented.

4.4.4.1 Grate Inlets

Grates are effective highway pavement drainage inlets where clogging with debris is not a problem. Where clogging may be a problem, see Table 4-5 where grates are ranked for susceptibility to clogging based on laboratory tests using simulated “leaves.” This table should be used for relative comparisons only.

Table 4-5. Average Debris Handling Efficiencies of Grates Tested.
Rank Grate Longitudinal Slope
0.005 0.04
1 Curved Vane 46 61
2 30°- 85 Tilt Bar 44 55
3 45°- 85 Tilt Bar 43 48
4 P – 50 32 32
5 P – 50 x 100 18 28
6 45°- 60 Tilt Bar 16 23
7 Reticuline 12 16
8 P – 30 9 20

When the velocity approaching the grate is less than the “splash-over” velocity, the grate will intercept essentially all of the frontal flow. Conversely, when the gutter flow velocity exceeds the “splash-over” velocity for the grate, only part of the flow will be intercepted. A part of the flow along the side of the grate will be intercepted, dependent on the cross slope of the pavement, the length of the grate, and flow velocity.

The ratio of frontal flow to total gutter flow, Eo, for a uniform cross slope is expressed by Equation 4-16:

 Eo = QW / Q = 1 – (1- W / T)2. 67 (4-16)
 where: Q = Total gutter flow, m3/s (ft3/s) Qw = Flow in width W, m3/s (ft3/s) W = Width of depressed gutter or grate, m (ft) T = Total spread of water, m (ft)

Example 4-2 and Chart 2 provide solutions of Eo for either uniform cross slopes or composite gutter sections.

The ratio of side flow, Qs, to total gutter flow is:

 Qs / Q = 1 – (Qw / Q) = 1 – Eo (4-17)

The ratio of frontal flow intercepted to total frontal flow, Rf, is expressed by Equation 4-18:

 Rf = 1 – Ku•(V – Vo) (4-18)
 where: Ku = 0.295 (0.09 in English units) V = Velocity of flow in the gutter, m/s Vo = Gutter velocity where splash-over first occurs, m/s

(Note: Rf cannot exceed 1.0)

This ratio is equivalent to frontal flow interception efficiency. Chart 5 provides a solution for Equation 4-18 which takes into account grate length, bar configuration, and gutter velocity at which splash-over occurs. The average gutter velocity (total gutter flow divided by the area of flow) is needed to use Chart 5. This velocity can also be obtained from Chart 4.

The ratio of side flow intercepted to total side flow, Rs, or side flow interception efficiency, is expressed by Equation 4-19. Chart 6 provides a solution to Equation 4-19.

 Rs = 1 / (1 + Ku•V1.8) / (Sx•L2.3)) (4-19)

where: Ku = 0.0828 (0.15 in English units)

A deficiency in developing empirical equations and charts from experimental data is evident in Chart 6. The fact that a grate will intercept all or almost all of the side flow where the velocity is low and the spread only slightly exceeds the grate width is not reflected in the chart. Error due to this deficiency is very small. In fact, where velocities are high, side flow interception may be neglected without significant error.

The efficiency, E, of a grate is expressed as provided in Equation 4-20:

 E = Rf•Eo + Rs•(1 – Eo) (4-20)

The first term on the right side of Equation 4-20 is the ratio of intercepted frontal flow to total gutter flow, and the second term is the ratio of intercepted side flow to total side flow. The second term is insignificant with high velocities and short grates.

It is important to recognize that the frontal flow to total gutter flow ratio, Eo, for composite gutter sections assumes by definition a frontal flow width equal to the depressed gutter section width. The use of this ratio when determining a grate’s efficiency requires that the grate width be equal to the width of the depressed gutter section, W. If a grate having a width less than W is specified, the gutter flow ratio, Eo, must be modified to accurately evaluate the grate’s efficiency. Because an average velocity has been assumed for the entire width of gutter flow, the grate’s frontal flow ratio, E’o, can be calculated by multiplying Eo by a flow area ratio. The area ratio is defined as the gutter flow area in a width equal to the grate width divided by the total flow area in the depressed gutter section. This adjustment is represented in the following equations:

 E’o = Eo•(A’w / Aw) (4-20a)
 where: E’o = Adjusted frontal flow area ratio for grates in composite cross sections A’w = Gutter flow area in a width equal to the grate width, m2 (ft2) Aw = Flow area in depressed gutter width, m2 (ft2)

The interception capacity of a grate inlet on grade is equal to the efficiency of the grate multiplied by the total gutter flow as represented in Equation 4-21. Note that E’o should be used in place of Eo in Equation 4-21 when appropriate.

 Qi = E•Q = Q•[Rf•Eo + Rs•(1 – Eo)] (4-21)

The use of Charts 5 and 6 are illustrated in the following examples.

Example 4-7

 Given: Given the gutter section from Example 4-2 (illustrated in Figure 4-1 a.2) with T = 2.5 m (8.2 ft) W = 0.6 m (2.0 ft) n = 0.016 SL = 0.010 Sx = 0.02 Continuous Gutter depression, a = 50 mm (2 in or 0.167 ft) Find: The interception capacity of a curved vane grate 0.6 m by 0.6 m (2 ft by 2 ft)
 Solution: From Example 4-2, Sw = 0.103 m/m (ft/ft) Eo = 0.70 Q = 0.06 m3/sec (2.3 ft3/sec) SI Units Step 1. Compute the average gutter velocity V = Q / A = 0.06 / A A = 0.5•T2•Sx + 0.5•a•W A = 0.5•(2.5)2•(0.02) + 0.5•(0.050)•(0.6) A = 0.08 m2 V = 0.06 / 0.08 = 0.75 m/s Step 2. Determine the frontal flow efficiency using Chart 5. Rf = 1.0 Step 3. Determine the side flow efficiency using Equation 4-19 or Chart 6. Rs = 1 / [1+ (Ku•V1.8) / (Sx•L2.3)] Rs = 1 / [1+ (0.0828)•(0.75)1.8 / [(0.02)•(0.6)2.3] Rs = 0.11 Step 4. Compute the interception capacity using Equation 4-21. Qi = Q•[Rf•Eo + Rs•(1 – Eo)] = (0.06)•[(1.0)•(0.70) + (0.11)•(1 – 0.70)]   Qi = 0.044 m3/s English Units Step 1. Compute the average gutter velocity V = Q / A = 2.3 / A A = 0.5•T2•Sx + 0.5•a•W A = 0.5•(8.2)2•(0.02) + 0.5•(0.167)•(2.0) A = 0.84 ft2 V = 2.3 / 0.84= 2.74 ft/s Step 2. Determine the frontal flow efficiency using Chart 5. Rf = 1.0 Step 3. Determine the side flow efficiency using Equation 4-19 or Chart 6. Rs = 1 / [1+ (Ku•V1.8) / (Sx•L2.3)] Rs = 1 / [1+ (0.15)•(2.74)1.8 / [(0.02)•(2.0)2.3] Rs = 0.10 Step 4. Compute the interception capacity using Equation 4-21. Qi = Q•[Rf•Eo + Rs•(1 – Eo)] = (2.3)•[(1.0)•(0.70)+(0.10)•(1 – 0.70)] Qi = 1.68 ft3/s

Example 4-8

 Given: Given the gutter section illustrated in Figure 4-1 a.1 with T = 3 m (9.84 ft) SL = 0.04 m/m (ft/ft) Sx = 0.025 m/m (ft/ft) n = 0.016 Bicycle traffic not permitted Find: The interception capacity of the following grates: a. P-50; 0.6 m x 0.6 m (2.0 ft x 2.0 ft) b. Reticuline; 0.6 m x 0.6 m (2.0 ft x 2.0 ft) c. Grates in a. and b. with a length of 1.2 m (4.0 ft)
 Solution: SI Units Step 1. Using Equation 4-2 or Chart 1 determine Q. Q = (Ku / n)•Sx1.67•SL0.5•T2.67 Q = {(0.376) / (0.016)}•(0.025)1.67•(0.04)0.5•(3)2.67 Q = 0.19 m3/s Step 2. Determine Eo from Equation 4-4 or Chart 2. W / T = 0.6 / 3 = 0.2 Eo = Qw / Q Eo = 1 – (1 – W / T)2.67 Eo = 1 – (1 – 0.2)2.67 Eo = 0.45 Step 3. Using Equation 4-13 or Chart 4 compute the gutter flow velocity. V = (Ku / n)•SL0.5•Sx0.67•T0.67 V = {0.752 / (0.016)}•(0.04)0.5•(0.025)0.67•(3)0.67 V = 1.66 m/s Step 4. Using Equation 4-18 or Chart 5, determine the frontal flow efficiency for each grate. Using Equation 4-19 or Chart 6, determine the side flow efficiency for each grate. Using Equation 4-21, compute the interception capacity of each grate. English Units Step 1. Using Equation 4-2 or Chart 1 determine Q. Q = (Ku / n)•Sx1.67•SL0.5•T2.67 Q = {(.56) / (0.016)}•(0.025)1.67•(0.04)0.5•(9.84)2.67 Q = 6.62 ft3/s Step 2. Determine Eo from Equation 4-4 or Chart 2. W / T = 2.0 / 9.84 = 0.2 Eo = Qw / Q Eo = 1 – (1 – W / T)2.67 Eo = 1 – ( 1 – 0.2)2.67 Eo = 0.45 Step 3. Using Equation 4-13 or Chart 4 compute the gutter flow velocity. V = (Ku / n)•SL0.5•Sx0.67•T0.67 V = {(1.11) / (0.016)}•(0.04)0.5•(0.025)0.6•9.84)0.67 V = 5.4 ft/s Step 4. Using Equation 4-18 or Chart 5, determine the frontal flow efficiency for each grate. Using Equation 4-19 or Chart 6, determine the side flow efficiency for each grate. Using Equation 4-21, compute the interception capacity of each grate.

The following table summarizes the results.

 Grate Size (width by length) Frontal Flow Efficiency, Rf Side Flow Efficiency, Rs Interception Capacity, Qi P – 50 0.6 m by 0.6 m 1.0 0.036 0.091 m3/s (2.0 ft by 2.0 ft) (3.21 ft3/s) Reticuline 0.6 m by 0.6 m 0.9 0.036 0.082 m3/s (2.0 ft by 2.0 ft) (2.89 ft3/s) P – 50 0.6 m by 1.2 m 1.0 0.155 0.103 m3/s (2.0 ft by 4.0 ft) (3.63 ft3/s) Reticuline 0.6 m by 1.2 m 1.0 0.155 0.103 m3/s (2.0 ft by 4.0 ft) (3.63 ft3/s) The P-50 parallel bar grate will intercept about 14% more flow than the reticuline grate or 48% of the total flow as opposed to 42% for the reticuline grate. Increasing the length of the grates would not be cost-effective, because the increase in side flow interception is small.

With laboratory data, agencies could develop design curves for their standard grates by using the step-by-step procedure provided in Appendix B.

4.4.4.2 Curb-Opening Inlets

Curb-opening inlets are effective in the drainage of highway pavements where flow depth at the curb is sufficient for the inlet to perform efficiently, as discussed in Section 4.4.3.1. Curb openings are less susceptible to clogging and offer little interference to traffic operation. They are a viable alternative to grates on flatter grades where grates would be in traffic lanes or would be hazardous for pedestrians or bicyclists.

Curb opening heights vary in dimension, however, a typical maximum height is approximately 100 to 150 mm (4 to 6 in). The length of the curb-opening inlet required for total interception of gutter flow on a pavement section with a uniform cross slope is expressed by Equation 4-22:

 LT = Ku•Q0.42•SL0.3•[1 / (n•Sx)]0.6 (4-22)
 where: Ku = 0.817 (0.6 in English units) LT = Curb opening length required to intercept 100% of the gutter flow, m (ft) SL = Longitudinal slope Q = Gutter flow, m3/s (ft3/s)

The efficiency of curb-opening inlets shorter than the length required for total interception is expressed by Equation 4-23:

 E = 1 – [ 1 – (L / LT)]1.8 (4-23)

where: L = Curb-opening length, m (ft)

Chart 7 is a nomograph for the solution of Equation 4-22, and Chart 8 provides a solution of Equation 4-23.

The length of inlet required for total interception by depressed curb-opening inlets or curb-openings in depressed gutter sections can be found by the use of an equivalent cross slope, Se, in Equation 4-22 in place of Sx. Se can be computed using Equation 4-24.

 Se = Sx + S’w•Eo (4-24)
 where: S’w = Cross slope of the gutter measured from the cross slope of the pavement, Sx, m/m (ft/ft) S’w = a / [1000•W]•W, for W in m; (a / [12•W], for W in ft) or = Sw – Sx a = Gutter depression, mm (in) Eo = Ratio of flow in the depressed section to total gutter flow determined by the gutter configuration upstream of the inlet

Figure 4-13 shows the depressed curb inlet for Equation 4-24. Eo is the same ratio as used to compute the frontal flow interception of a grate inlet.

As seen from Chart 7, the length of curb opening required for total interception can be significantly reduced by increasing the cross slope or the equivalent cross slope. The equivalent cross slope can be increased by use of a continuously depressed gutter section or a locally depressed gutter section.

Using the equivalent cross slope, Se, Equation 4-22 becomes:

 LT = KT•Q0.42•SL0.3•[1 / (n•Se)]0.6 (4-22a)

where: KT = 0.817 (0.6 in English Units)

Equation 4-23 is applicable with either straight cross slopes or composite cross slopes. Charts 7 and 8 are applicable to depressed curb-opening inlets using Se rather than Sx.

Equation 4-24 uses the ratio, Eo, in the computation of the equivalent cross slope, Se. Example 4-9a demonstrates the procedure to determine spread and then the example uses Chart 2 to determine Eo. Example 4-9b demonstrates the use of these relationships to design length of a curb opening inlet.

Example 4-9a

 Given: A curb-opening inlet with the following characteristics: SL = 0.01 m/m (ft/ft) Sx = 0.02 m/m (ft/ft) Q = 0.05 m3/s (1.77 ft3/s) n = 0.016 Find: (1) Qi for a 3 m (9.84 ft) curb-opening. (2) Qi for a depressed 3 m (9.84 ft) curb opening inlet with a continuously depressed curb section. a = 25 mm (1 in) W = 0.6 m (2 ft)
 Solution (1): SI Units Step 1. Determine the length of curb opening required for total interception of gutter flow using Equation 4-22 or Chart 7. LT = Ku•Q0.42•SL0.3•(1 / (n•Sx))0.6 LT = 0.817•(0.05)0.42•(0.01)0.3•(1 / [(0.016)•(0.02)])0.6 LT = 7.29 m Step 2. Compute the curb-opening efficiency using Equation 4-23 or Chart 8. L / LT = 3 / 7.29 = 0.41 E = 1 – (1 – L / LT)1.8 E = 1 – (1 – 0.41)1.8 E = 0.61 Step 3. Compute the interception capacity. Qi = E•Q = (0.61)•(0.05) Qi = 0.031 m3/s English Units Step 1. Determine the length of curb opening required for total interception of gutter flow using Equation 4-22 or Chart 7. LT = Ku•Q0.42•SL0.3•(1 / (n•Sx))0.6 LT = 0.6•(1.77)0.42•(0.01)0.3•(1 / [(0.016)•(0.02)])0.6 LT = 23.94 ft Step 2. Compute the curb-opening efficiency using Equation 4-23 or Chart 8. L / LT = 9.84 / 23.94 = 0.41 E = 1 – (1 – L / LT)1.8 E = 1 – (1 – 0.41)1.8 E = 0.61 Step 3. Compute the interception capacity. Qi = E•Q = (0.61)•(1.77) Qi = 1.08 ft3/s
 Solution (2): SI Units Step 1. Use Equation 4-4 (Chart 2) and Equation 4-2 (Chart 1) to determine the W/T ratio. Determine spread, T, (Procedure from Example 4­ 2, solution 2) Assume Qs = 0.018 m3/s Qw = Q – Qs = 0.05 – 0.018 = 0.032 m3/s Eo = Qw / Q = 0.032 / 0.05 = 0.64 Sw = Sx + a / W = 0.02 + (25 / 1000) / 0.6 Sw = 0.062 Sw / Sx = 0.062 / 0.02 = 3.1 Use Equation 4-4 or Chart 2 to determine W/T W / T = 0.24 T = W / (W / T) = 0.6 / 0.24 = 2.5 m Ts = T – W = 2.5 – 0.6 = 1.9 m Use Equation 4-2 or Chart 1 to obtain Qs Qs = (Ku / n)•Sx1.67•SL0.5•Ts2.67 Qs = {(0.376) / (0.016)}•(0.02)1.67•(0.01)0.5•(1.9)2.67 Qs = 0.019 m3/s (equals Qs assumed) Step 2. Determine efficiency of curb opening Se = Sx + S’w•Eo = Sx + (a / W)•Eo = 0.02 +[(25 / 1000) / (0.6)]•(0.64) Se = 0.047 Using Equation 4-25 or Chart 7 LT = KT•Q0.42•SL0.3•[1 / (n•Se)]0.6 LT = (0.817)•(0.05)0.42•(0.01)0.3•[1•(0.016)•(0.047)]0.6 LT = 4.37 m Using Equation 4-23 or Chart 8 to obtain curb inlet efficiency L / LT = 3 / 4.37 = 0.69 E = 1 – (1 – L / LT)1.8 E = 1 – (1 – 0.69)1.8 E = 0.88 Step 3. Compute curb opening inflow using Equation 4-14 Qi = Q•E = (0.05)•(0.88) Qi = 0.044 m3/s The depressed curb-opening inlet will intercept 1.5 times the flow intercepted by the undepressed curb opening. English Units Step 1. Use Equation 4-4 (Chart 2) and Equation 4-2 (Chart 1) to determine the W/T ratio Determine spread, T, (Procedure from Example 4­ 2, solution 2) Assume Qs = 0.64 ft3/s Qw = Q – Qs = 1.77 – 0.64 = 1.13 ft3/s Eo = Qw / Q = 1.13 / 1.77 = 0.64 Sw = Sx + a / W = 0.02 + (0.83 / 2.0) Sw = 0.062 Sw / Sx = 0.062 / 0.02 = 3.1 Use Equation 4-4 or Chart 2 to determine W / T W / T = 0.24 T = W / (W / T) = 2.0 / 0.24 = 8.3 ft Ts = T – W = 8.3 – 2.0 = 6.3 ft Use Equation 4-2 or Chart 1 to obtain Qs Qs = (Ku / n)•Sx1.67•SL0.5•Ts2.67 Qs = {(0.56) / (0.016)}•(0.02)1.67•(0.01)0.5•(6.3)2.67 Qs = 0.69 ft3/s (close to Qs assumed) Step 2. Determine efficiency of curb opening. Se = Sx + S’w•Eo = Sx + (a / W)•Eo = 0.02 + [(0.083) / (2.0)]•(0.64) Se = 0.047 Using Equation 4-25 or Chart 7 LT = KT•Q0.42•SL0.3•[1/(n•Se)]0.6 LT = (0.6)•(1.77)0.42•(0.01)0.3•[1 / ((0.016)•(0.047))]0.6 LT = 14.34 ft Using Equation 4-23 or Chart 8 to obtain curb inlet efficiency L / LT = 9.84 / 14.34 = 0.69 E = 1 – (1 – L / LT)1.8 E = 1 – (1 – 0.69)1.8 E = 0.88 Step 3. Compute curb opening inflow using Equation 4-14 Qi = Q•E = (1.77)•(0.88) Qi = 1.55 ft3/s The depressed curb-opening inlet will intercept 1.5 times the flow intercepted by the undepressed curb opening.

Example 4-9b

 Given: From Example 4-7, the following information is given: SL = 0.01 m/m (ft/ft) Sx = 0.02 m/m (ft/ft) T = 2.5 m (8.2 ft) Q = 0.064 m3/s (2.26 ft3/s) n = 0.016 W = 0.6 m (2.0 ft) a = 50 mm (2.0 in) Eo = 0.70 Find: The minimum length of a locally depressed curb opening inlet required to intercept 100% of the gutter flow.
 Solution: SI Units Step 1. Compute the composite cross slope for the gutter section using Equation 4-24. Se = Sx + S’w•Eo Se = 0.02 + 50 / [(1000)•(0.6)]•(0.70) Se = 0.08 Step 2. Compute the length of curb opening inlet required from Equation 4-25. LT = KT•Q0.42•SL0.3•(1 / n•Se)0.6 LT = (0.817)•(0.064)0.42•(0.01)0.3•[1/ (0.016)•(0.08)]0.6 LT = 3.81 m English Units Step 1. Compute the composite cross slope for the gutter section using Equation 4-24. Se = Sx + S’w•Eo Se = 0.02 + 2 / [(12)•(2.0)]•(0.70) Se = 0.08 Step 2. Compute the length of curb opening inlet required from Equation 4-25. LT = KT•Q0.42•SL0.3•(1 / n•Se)0.6 LT = (0.60)•(2.26)0.42•(0.01)0.3•[1/ (0.016)•(0.08)]0.6 LT = 12.5 ft

4.4.4.3. Slotted Inlets

Wide experience with the debris handling capabilities of slotted inlets is not available. Deposition in the pipe is the problem most commonly encountered. The configuration of slotted inlets makes them accessible for cleaning with a high pressure water jet.

Slotted inlets are effective pavement drainage inlets which have a variety of applications. They can be used on curbed or uncurbed sections and offer little interference to traffic operations. An installation is illustrated in Figure 4-14.

Flow interception by slotted inlets and curb-opening inlets is similar in that each is a side weir and the flow is subjected to lateral acceleration due to the cross slope of the pavement. Analysis of data from the Federal Highway Administration tests of slotted inlets with slot widths = 45 mm (1.75 in) indicates that the length of slotted inlet required for total interception can be computed by Equation 4-22. Chart 7, is therefore applicable for both curb-opening inlets and slotted inlets. Similarly, Equation 4-23 is also applicable to slotted inlets and Chart 8 can be used to obtain the inlet efficiency for the selected length of inlet.

When slotted drains are used to capture overland flow, research has indicated that with water depths ranging from 9.7 mm (0.38 in) to 14.2 mm (0.56 in) the 25, 44, and 63 mm (1, 1.75 and 2.5 in) wide slots can accommodate 0.0007 m3/s/m (0.025 ft3/s/ft) with no splash over for slopes from 0.005 to 0.09 m/m (ft/ft).

At a test system capacity of 0.0011 m3/s/m (0.40 ft3/s/ft), a small amount of splash over occurred.

4.4.4.4. Combination Inlets

The interception capacity of a combination inlet consisting of a curb opening and grate placed side-by-side, as shown in Figure 4-15, is no greater than that of the grate alone. Capacity is computed by neglecting the curb opening. A combination inlet is sometimes used with a part of the curb opening placed upstream of the grate as illustrated in Figure 4-16. The curb opening in such an installation intercepts debris which might otherwise clog the grate and is called a “sweeper” inlet. A sweeper combination inlet has an interception capacity equal to the sum of the curb opening upstream of the grate plus the grate capacity, except that the frontal flow and thus the interception capacity of the grate is reduced by interception by the curb opening.

The following example illustrates computation of the interception capacity of a combination curb-opening grate inlet with a portion of the curb opening upstream of the grate.

Example 4-10

 Given: A combination curb-opening grate inlet with a 3 m (9.8 ft) curb opening, 0.6 m by 0.6 m (2 ft by 2 ft) curved vane grate placed adjacent to the downstream 0.6 m (2 ft) of the curb opening. This inlet is located in a gutter section having the following characteristics: W = 0.6 m (2 ft) Q = 0.05 m3/s (1.77 ft3/s) SL = 0.01 m/m (ft/ft) Sx = 0.02 m/m (ft/ft) SW = 0.062 m/m (ft/ft) n = 0.016 Find: Interception capacity, Qi
 Solution SI Units Step 1. Compute the interception capacity of the curb-opening upstream of the grate, Qic. L = 3m – 0.6 m = 2.4 m From Example 4-9 a, Solution 2, Step 2 LT = 4.37 m L / LT = 2.4 / 4.37 = 0.55 Using Equation 4-23 or Chart 8 E = 1 – (1 – L / LT) E = 1 – (1 – 0.55)1.8 E = 0.76 Qic = E•Q = (0.76)(0.05) = 0.038 m3/s Step 2. Compute the interception capacity of the grate. Flow at grate Qg = Q – Qic = 0.05 – 0.038 Qg = 0.012 m3/s Determine Spread, T (Procedure from Example 4-2, Solution 2) Assume Qs = 0.0003 m3/s Qw = Q – Qs = 0.0120 – 0.0003 = 0.0117 m3/s Eo = Qw / Q = 0.0117 / 0.0120 = 0.97 Sw / Sx = 0.062 / 0.02 = 3.1 From Equation 4-4 or Chart 2 W / T = 1 / {(1 / [(1 / (1 / Eo – 1))•(Sw / Sx) + 1]0.375 – 1)•(Sw / Sx) +1} W / T = 1 / {(1 / [(1 / (1 / 0.97 -1 ))•(3.1) + 1]0.375 – 1)•(3.1) +1} W / T = 0.62 T = W / (W / T) = 0.6 / 0.62 = 0.97 m Ts = T – W = 0.97 – 0.60 = 0.37 m From Chart 1 or Equation 4-2 Qs = 0.0003 m3/s Qs assumed = Qs calculated Determine velocity, V V = Q / A = Q / [0.5•T2•Sx + 0.5•a•W] V = 0.012 / [(0.5)•(0.97)2•(0.02) + (0.5)•(25 / 1000)•(0.6)] V = 0.68 m/s From Chart 5 Rf = 1.0 From Equation 4-19 or Chart 6 Rs = 1 / (1 + (Ku•V1.8) / (Sx•L2.3)) Rs = 1 / (1 + [(0.0828)•(0.68)1.8] / [(0.02)•(0.6)2.3] Rs = 0.13 From Equation 4-21 Qig = Qg•[Rf•Eo + Rs•(1 – Eo)] Qig = 0.012•[(1.0)(0.97) + (0.13)(1 – 0.97)] Qig = 0.011 m3/s Step 3. Compute the total interception capacity. (Note: Interception capacity of curb opening adjacent to grate was neglected.) Qi = Qic + Qig = 0.038 + 0.011 Qi = 0.049 m3/s (approximately 100% of the total initial flow) English Units Step 1. Compute the interception capacity of the curb-opening upstream of the grate, Qic. L = 9.84 – 2.0 = 7.84 ft From Example 4-9 a, Solution 2, Step 2 LT = 14.34 ft L / LT = 7.84 / 14.34 = 0.55 Using Equation 4-23 or Chart 8 E = 1 – (1 – L / LT) E = 1 – (1 – 0.55)1.8 E = 0.76 Qic = E•Q = (0.76)(1.77) = 1.35 ft3/s Step 2. Compute the interception capacity of the grate. Flow at grate Qg = Q – Qic = 1.77 – 1.35 Qg = 0.42 ft3/s Determine Spread, T (Procedure from Example 4-2, Solution 2) Assume Qs = 0.01 ft3/s Qw = Q – Qs = 0.42 – 0.01 = 0.41 ft3/s Eo = Qw / Q = 0.41 / 0.42 = 0.97 Sw / Sx = 0.062 / 0.02 = 3.1 From Equation 4-4 or Chart 2 W / T = 1 / {(1 / [(1 / (1 / Eo – 1))•(Sw / Sx) + 1]0.375 – 1)•(Sw / Sx) +1} W / T = 1 / {(1 / [(1 / (1 / 0.97 -1 ))•(3.1) + 1]0.375 – 1)•(3.1) +1} W / T = 0.62 T = W / (W / T) = 2.0 / 0.62 = 3.2 ft Ts = T – W = 3.2 – 2.0 = 1.2 ft From Chart 1 or Equation 4-2 Qs = 0.01 ft3/s Qs assumed = Qs calculated Determine velocity, V V = Q / A = Q / [0.5•T2•Sx + 0.5•a•W] V = 0.42 / [(0.5)•(3.2)2•(0.02) + (0.5)•(0.083)•(2.0)] V = 2.26 ft/s From Chart 5 Rf = 1.0 From Equation 4-19 or Chart 6 Rs = 1 / (1 + (Ku•V1.8) / (Sx•L2.3)) Rs = 1 / (1 + [(0.15)•(2.26)1.8] / [(0.02)•(2.0)2.3] Rs = 0.13 From Equation 4-21 Qig = Qg•[Rf•Eo + Rs•(1 – Eo)] Qig = 0.42•[(1.0)(0.97) + (0.13)(1 – 0.97)] Qig = 0.41 ft3/s Step 3. Compute the total interception capacity. (Note: Interception capacity of curb opening adjacent to grate was neglected.) Qi = Qic + Qig = 1.35 + 0.41 Qi = 1.76 ft3/s (approximately 100% of the total initial flow)

The use of depressed inlets and combination inlets enhances the interception capacity of the inlet. Example 4-7 determined the interception capacity of a depressed curved vane grate, 0.6 m by 0.6 m (2 ft by 2 ft), Example 4-9 for an undepressed curb opening inlet, length = 3.0 m (9.8 ft) and a depressed curb opening inlet, length = 3.0 m (9.8 ft), and Example 4-10 for a combination of 0.6 m by 0.6 m (2 ft by 2 ft) depressed curve vane grate located at the downstream end of 3.0 m (9.8 ft) long depressed curb opening inlet. The geometries of the inlets and the gutter slopes were consistent in the examples and Table 4-6 summarizes a comparison of the intercepted flow of the various configurations.

 Table 4-6. Comparison of Inlet Interception Capacities. Inlet Type Intercepted Flow, Qi Curved Vane – Depressed 0.033 m3/s (1.2 ft3/s) (Example 4-7) Curb Opening – Undepressed 0.031 m3/s (1.1 ft3/s) (Example 4-9 (1)) Curb Opening – Depressed 0.045 m3/s (1.59 ft3/s) (Example 4-9 (2)) Combination – Depressed 0.049 m3/s (1.76 ft3/s) (Example 4-10)

From Table 4-6, it can be seen that the combination inlet intercepted approximately 100% of the total flow whereas the curved vane grate alone only intercepted 66% of the total flow. The depressed curb opening intercepted 90% of the total flow. However, if the curb opening was undepressed, it would have only intercepted 62% of the total flow.

4.4.5. Interception Capacity of Inlets In Sag Locations

Inlets in sag locations operate as weirs under low head conditions and as orifices at greater depths. Orifice flow begins at depths dependent on the grate size, the curb opening height, or the slot width of the inlet. At depths between those at which weir flow definitely prevails and those at which orifice flow prevails, flow is in a transition stage. At these depths, control is ill-defined and flow may fluctuate between weir and orifice control. Design procedures presented here are based on a conservative approach to estimating the capacity of inlets in sump locations.

The efficiency of inlets in passing debris is critical in sag locations because all runoff which enters the sag must be passed through the inlet. Total or partial clogging of inlets in these locations can result in hazardous ponded conditions. Grate inlets alone are not recommended for use in sag locations because of the tendencies of grates to become clogged. Combination inlets or curb-opening inlets are recommended for use in these locations.

4.4.5.1. Grate Inlets in Sags

A grate inlet in a sag location operates as a weir to depths dependent on the size of the grate and as an orifice at greater depths. Grates of larger dimension will operate as weirs to greater depths than smaller grates.

 Qi = Cw•P•d1.5 (4-26)
 where: P = Perimeter of the grate in m (ft) disregarding the side against the curb Cw = 1.66 (3.0 in English units) d = Average depth across the grate; 0.5 (d1 + d2), m (ft)

The capacity of a grate inlet operating as an orifice is:

 Qi = Co•Ag•(2•g•d)0.5 (4-27)
 where: Co = Orifice coefficient = 0.67 Ag = Clear opening area of the grate, m2 (ft2) g = 9.81 m/s2 (32.16 ft/s2)

Use of Equation 4-27 requires the clear area of opening of the grate. Tests of three grates for the Federal Highway Administration(27) showed that for flat bar grates, such as the P-50×100 and P-30 grates, the clear opening is equal to the total area of the grate less the area occupied by longitudinal and lateral bars. The curved vane grate performed about 10% better than a grate with a net opening equal to the total area less the area of the bars projected on a horizontal plane. That is, the projected area of the bars in a curved vane grate is 68% of the total area of the grate leaving a net opening of 32%, however the grate performed as a grate with a net opening of 35%. Tilt-bar grates were not tested, but exploration of the above results would indicate a net opening area of 34% for the 30-degree tilt-bar and zero for the 45-degree tilt-bar grate. Obviously, the 45-degree tilt-bar grate would have greater than zero capacity. Tilt-bar and curved vane grates are not recommended for sump locations where there is a chance that operation would be as an orifice. Opening ratios for the grates are given on Chart 9.

Chart 9 is a plot of Equations 4-26 and 4-27 for various grate sizes. The effects of grate size on the depth at which a grate operates as an orifice is apparent from the chart. Transition from weir to orifice flow results in interception capacity less than that computed by either the weir or the orifice equation. This capacity can be approximated by drawing in a curve between the lines representing the perimeter and net area of the grate to be used.

Example 4-11 illustrates use of Equations 4-26 and 4-27 and Chart 9.

Example 4-11

 Given: Under design storm conditions a flow to the sag inlet is 0.19 m3/s (6.71 ft3/s). Also, Sx = SW = 0.05 m/m (ft/ft) n = 0.016 Tallowable = 3 m (9.84 ft) Find: Find the grate size required and depth at curb for the sag inlet assuming 50% clogging where the width of the grate, W, is 0.6 m (2.0 ft).
 Solution: SI Units Step 1. Determine the required grate perimeter. Depth at curb, d2 d2 = T•Sx = (3.0)•(0.05) d2 = 0.15 m Average depth over grate d = d2 – (W / 2)•SW d = 0.15 – (0.6 / 2)•(0.05) d = 0.135 m From Equation 4-26 or Chart 9 P = Qi / [Cw•d1.5] P = (0.19) / [(1.66)•(0.135)1.5] P = 2.31 m Some assumptions must be made regarding the nature of the clogging in order to compute the capacity of a partially clogged grate. If the area of a grate is 50% covered by debris so that the debris-covered portion does not contribute to interception, the effective perimeter will be reduced by a lesser amount than 50%. For example, if a 0.6 m by 1.2 m (2 ft by 4 ft) grate is clogged so that the effective width is 0.3 m (1 ft), then the perimeter, P = 0.3 + 1.2 + 0.3 = 1.8 m (6 ft), rather than 2.31 m (7.66 ft), the total perimeter, or 1.2 m (4 ft), half of the total perimeter. The area of the opening would be reduced by 50% and the perimeter by 25%. Therefore, assuming 50% clogging along the length of the grate, a 1.2 m by 1.2 m (4 ft by 4 ft), 0.6 m by 1.8 m (2 ft by 6 ft), or a .9 m by 1.5 m (3 ft by 5 ft) grate would meet requirements of a 2.31 m (7.66 ft) perimeter 50% clogged. Assuming 50% clogging along the grate length, Peffective = 2.4 m = (0.5)•(2)•W + L if W = 0.6 m then L ≥ 1.8 m if W = 0.9 m then L ≥ 1.5 m Select a double 0.6 m by 0.9 m grate. Peffective = (0.5)•(2)•(0.6) + (1.8) Peffective = 2.4 m Step 2. Check depth of flow at curb using Equation 4-26 or Chart 9. d = [Q / (Cw•P)]0.67 d = [0.19 / ((1.66)•(2.4))]0.67 d = 0.130 m Therefore, ok Conclusion: A double 0.6 m by 0.9 m grate 50% clogged is adequate to intercept the design storm flow at a spread which does not exceed design spread. However, the tendency of grate inlets to clog completely warrants consideration of a combination inlet or curb-opening inlet in a sag where ponding can occur, and flanking inlets in long flat vertical curves. English Units Step 1. Determine the required grate perimeter. Depth at curb, d2 d2 = T•Sx = (9.84)•(0.05) d2 = 0.49 ft Average depth over grate d = d2 – (W / 2)•SW d = 0.49 – (2.0 / 2)•(0.05) d = 0.44 ft From Equation 4-26 or Chart 9 P = Qi / [Cw•d1.5] P = (6.71) / [(3.0)•(0.44)1.5] P = 7.66 ft Some assumptions must be made regarding the nature of the clogging in order to compute the capacity of a partially clogged grate. If the area of a grate is 50% covered by debris so that the debris-covered portion does not contribute to interception, the effective perimeter will be reduced by a lesser amount than 50%. For example, if a 0.6 m by 1.2 m (2 ft by 4 ft) grate is clogged so that the effective width is 0.3 m (1 ft), then the perimeter, P = 0.3 + 1.2 + 0.3 = 1.8 m (6 ft), rather than 2.31 m (7.66 ft), the total perimeter, or 1.2 m (4 ft), half of the total perimeter. The area of the opening would be reduced by 50% and the perimeter by 25%. Therefore, assuming 50% clogging along the length of the grate, a 1.2 m by 1.2 m (4 ft by 4 ft), 0.6 m by 1.8 m (2 ft by 6 ft), or a .9 m by 1.5 m (3 ft by 5 ft) grate would meet requirements of a 2.31 m (7.66 ft) perimeter 50% clogged. Assuming 50% clogging along the grate length, Peffective = 8.0 ft = (0.5)•(2)•W + L if W = 2 ft then L ≥ 6 ft if W = 3 ft then L ≥ 5 ft Select a double 2 ft by 3 ft grate. Peffective = (0.5)•(2)•(2.0) + (6.0) Peffective = 8 ft Step 2. Check depth of flow at curb using Equation 4-26 or Chart 9. d = [Q / (Cw•P)]0.67 d = [6.71 / ((3.0)•(8.0))]0.67 d = 0.43 ft Therefore, ok Conclusion: A double 2 ft by 3 ft grate 50% clogged is adequate to intercept the design storm flow at a spread which does not exceed design spread. However, the tendency of grate inlets to clog completely warrants consideration of a combination inlet or curb-opening inlet in a sag where ponding can occur, and flanking inlets in long flat vertical curves.

4.4.5.2. Curb-Opening Inlets

The capacity of a curb-opening inlet in a sag depends on water depth at the curb, the curb opening length, and the height of the curb opening. The inlet operates as a weir to depths equal to the curb opening height and as an orifice at depths greater than 1.4 times the opening height. At depths between 1.0 and 1.4 times the opening height, flow is in a transition stage.

Spread on the pavement is the usual criterion for judging the adequacy of a pavement drainage inlet design. It is also convenient and practical in the laboratory to measure depth at the curb upstream of the inlet at the point of maximum spread on the pavement. Therefore, depth at the curb measurements from experiments coincide with the depth at curb of interest to designers. The weir coefficient for a curb-opening inlet is less than the usual weir coefficient for several reasons, the most obvious of which is that depth measurements from experimental tests were not taken at the weir, and drawdown occurs between the point where measurement were made and the weir.

The weir location for a depressed curb-opening inlet is at the edge of the gutter, and the effective weir length is dependent on the width of the depressed gutter and the length of the curb opening. The weir location for a curb-opening inlet that is not depressed is at the lip of the curb opening, and its length is equal to that of the inlet, as shown in Chart 10.

The equation for the interception capacity of a depressed curb-opening inlet operating as a weir is:

 Qi = Cw•(L + 1.8•W)•d1.5 (4-28)
 where: Cw = 1.25 (2.3 in English Units) L = Length of curb opening, m (ft) W = Lateral width of depression, m (ft) d = Depth at curb measured from the normal cross slope, m (ft), i.e., d = T•Sx

The weir equation is applicable to depths at the curb approximately equal to the height of the opening plus the depth of the depression. Thus, the limitation on the use of Equation 4-28 for a depressed curb-opening inlet is:

 d ≤ h + a / (1000) (d ≤ h + a / 12, in English units) (4-29)
 where: h = Height of curb-opening inlet, m (ft) a = Depth of depression, mm (in)

Experiments have not been conducted for curb-opening inlets with a continuously depressed gutter, but it is reasonable to expect that the effective weir length would be as great as that for an inlet in a local depression. Use of Equation 4-28 will yield conservative estimates of the interception capacity.

The weir equation for curb-opening inlets without depression becomes:

 Qi = Cw•L•d1.5 (4-30)

Without depression of the gutter section, the weir coefficient, Cw, becomes 1.60 (3.0, English system). The depth limitation for operation as a weir becomes d ≤ h.

At curb-opening lengths greater than 3.6m (12 ft), Equation 4-30 for non-depressed inlet produces intercepted flows which exceed the values for depressed inlets computed using Equation 4-28. Since depressed inlets will perform at least as well as non-depressed inlets of the same length, Equation 4-30 should be used for all curb opening inlets having lengths greater than 3.6 m (12 ft).

Curb-opening inlets operate as orifices at depths greater than approximately 1.4 times the opening height. The interception capacity can be computed by Equation 4-31a and Equation 4-31b. These equations are applicable to depressed and undepressed curb-opening inlets. The depth at the inlet includes any gutter depression.

 Qi = Co•h•L•(2•g•do)0.5 (4-31a)

or

 Qi = Co•Ag•{2•g•[di – (h / 2)]}0.5 (4-31b)
 where: Co = Orifice coefficient (0.67) do = Effective head on the center of the orifice throat, m (ft) L = Length of orifice opening, m (ft) Ag = Clear area of opening, m2 (ft2) di = Depth at lip of curb opening, m (ft) h = Height of curb-opening orifice, m (ft)

The height of the orifice in Equations 4-31a and 4-31b assumes a vertical orifice opening. As illustrated in Figure 4-18, other orifice throat locations can change the effective depth on the orifice and the dimension (di – h / 2). A limited throat width could reduce the capacity of the curb-opening inlet by causing the inlet to go into orifice flow at depths less than the height of the opening.

For curb-opening inlets with other than vertical faces (see Figure 4-18), Equation 4-31a can be used with:

 h = orifice throat width, m (ft) do = effective head on the center of the orifice throat, m (ft)

Chart 10 provides solutions for Equations 4-28 and 4-31 for depressed curb-opening inlets, and Chart 11 provides solutions for Equations 4-30 and 4-31 for curb-opening inlets without depression. Chart 12 is provided for use for curb openings with other than vertical orifice openings.

Example 4-12 illustrates the use of Charts 11 and 12.

Example 4-12

 Given: Curb opening inlet in a sump location with L = 2.5 m (8.2 ft) h = 0.13 m (0.43 ft) (1) Undepressed curb opening Sx = 0.02 T = 2.5 m (8.2 ft) (2) Depressed curb opening Sx = 0.02 a = 25 mm (1 in) local W = 0.6 m (2 ft) T = 2.5 m (8.2 ft) Find: Qi
 Solution (1): Undepressed SI Units Step 1. Determine depth at curb. d = T•Sx = (2.5)•(0.02) d = 0.05 m d = 0.05 m ≤ h = 0.13 m, therefore weir flow controls. Step 2. Use Equation 4-30 or Chart 11 to find Qi. Qi = Cw•L•d1.5 Qi = (1.60)•(2.5)•(0.05)1.5 = 0.045 m3/s English Units Step 1. Determine depth at curb. d = T•Sx = (8.2)•(0.02) d = 0.16 ft d = 0.16 ft ≤ h = 0.43 ft, therefore weir flow controls. Step 2. Use Equation 4-30 or Chart 11 to find Qi. Qi = Cw•L•d1.5 Qi = (3.0)•(8.2)•(0.16)1.5 = 1.6 ft3/s
 Solution (2): Depressed SI Units Step 1. Determine depth at curb, di di = d + a di = Sx•T + a di = (0.02)•(2.5) + 25 / 1000 di = 0.075 m di = 0.075 m < h =0.13 m, therefore weir flow controls. Step 2. Use Equation 4-28 or Chart 10 to find Qi. P = L + 1.8•W P = 2.5 m + (1.8)•(0.6) P = 3.58 m Qi = Cw•(L + 1.8•W)•d1.5 Qi = (1.25)•(3.58)•(0.05)1.5 Qi = 0.048 m3/s The depressed curb-opening inlet has 10% more capacity than an inlet without depression. English Units Step 1. Determine depth at curb, di di = d + a di = Sx•T + a di = (0.02)•(8.2) + 1 / 12 di = 0.25 ft di = 0.25 ft < h =0.43 ft, therefore weir flow controls. Step 2. Use Equation 4-28 or Chart 10 to find Qi. P = L + 1.8•W P = 8.2 + (1.8)•(2.0) P = 11.8 ft Qi = Cw•(L + 1.8•W)•d1.5 Qi = (2.3)•(11.8)•(0.16)1.5 Qi = 1.7 ft3/s The depressed curb-opening inlet has 10% more capacity than an inlet without depression.

4.4.5.3 Slotted Inlets

Slotted inlets in sag locations perform as weirs to depths of about 0.06 m (0.2 ft), dependent on slot width. At depths greater than about 0.12 m, (0.4 ft), they perform as orifices. Between these depths, flow is in a transition stage. The interception capacity of a slotted inlet operating as a weir can be computed by an equation of the form:

 Qi = Cw•L•d1.5 (4-32)
 where: Cw = Weir coefficient; various with flow depth and slot length; typical value is approximately 1.4 (2.48 for English units) L = Length of slot, m (ft) d = Depth at curb measured from the normal cross slope, m (ft)

The interception capacity of a slotted inlet operating as an orifice can be computed by Equation 4-33:

 Qi = 0.8•L•W•(2•g•d)0.5 (4-33)
 where: W = Width of slot, m (ft) L = Length of slot, m (ft) d = Depth of water at slot for d > 0.12 m (0.4 ft), m (ft) g = 9.81 m/s2 (32.16 ft/s2 in English units)

For a slot width of 45 mm (1.75 in), Equation 4-33 becomes:

 Qi = CD•L•d0.5 (4-34)

where: CD = 0.16 (0.94 for English units)

Chart 13 provides solutions for weir and orifice flow conditions as represented by Equations 4­ 32 and 4-33. As indicated in Chart 13, the transition between weir and orifice flow occurs at different depths. To conservatively compute the interception capacity of slotted inlets in sump conditions in the transition area, orifice conditions should be assumed. Due to clogging characteristics, slotted drains are not recommended in sag locations.

Example 4-13

 Given: A slotted inlet located along a curb having a slot width of 45 mm (1.75 in). The gutter flow at the upstream end of the inlet is 0.14 m3/s (4.9 ft3/s). Find: The length of slotted inlet required to limit maximum depth at the curb to 0.09 m (3.6 in) assuming no clogging.
 Solution: SI Units From Chart 13A with Q = 0.14 m3/s and d = 0.09, L = 3.66 m say 4.0 m English Units From Chart 13B with Q = 4.9 ft3/s and d = 3.6 in, L = 10 ft Note: Since the point defined by Q and d on Chart 13 falls in the weir flow range, Equation 4-32 defines the flow condition. However, Equation 4-32 cannot be directly applied since Cw varies with both flow depth and slot length.

4.4.5.4 Combination Inlets

Combination inlets consisting of a grate and a curb opening are considered advisable for use in sags where hazardous ponding can occur. Equal length inlets refer to a grate inlet placed along side a curb opening inlet, both of which have the same length. A sweeper inlet refers to a grate inlet placed at the downstream end of a curb opening inlet. The curb opening inlet is longer than the grate inlet and intercepts the flow before the flow reaches the grate. The sweeper inlet is more efficient than the equal length combination inlet and the curb opening has the ability to intercept any debris which may clog the grate inlet. The interception capacity of the equal length combination inlet is essentially equal to that of a grate alone in weir flow. In orifice flow, the capacity of the equal length combination inlet is equal to the capacity of the grate plus the capacity of the curb opening.

Equation 4-26 and Chart 9 can be used for grates in weir flow or combination inlets in sag locations. Assuming complete clogging of the grate, Equations 4-28, 4-30, and 4-31 and Charts 10, 11 and 12 for curb-opening inlets are applicable.

Where depth at the curb is such that orifice flow occurs, the interception capacity of the inlet is computed by adding Equations 4-27 and 4-31a as follows:

 Qi = 0.67•Ag•(2•g d)0.5 + 0.67•h•L•(2•g•do)0.5 (4-35)
 where: Ag = Clear area of the grate, m2 (ft2) g = 9.81 m/s2 (32.16 ft/s2 in English units) d = Average depth over the grate, m (ft) h = Height of curb opening orifice, m (ft) L = Length of curb opening, m (ft) do = Effective depth at the center of the curb opening orifice, m (ft)

Trial and error solutions are necessary for determining the depth at the curb for a given flow rate using Charts 9, 10, and 11 for orifice flow. Different assumptions for clogging of the grate can also be examined using these charts as illustrated by the following example.

Example 4-14

 Given: A combination inlet in a sag location with the following characteristics: Grate – 0.6 m by 1.2 m (2 ft by 4 ft) P-50 Curb opening – L = 1.2 m (4 ft) h = 0.1 m (3.9 in) Q = 0.15 m3/s (5.3 ft3/s) Sx = 0.03 m/m (ft/ft) Find: Depth at curb and spread for: (1) Grate clear of clogging (2) Grate 100% clogged
 Solution (1): SI Units Step 1. Compute depth at curb. Assuming grate controls interception: P = 2•W + L = 2•(0.6) + 1.2 P = 2.4 m From Equation 4-26 or Chart 9 davg = [Qi / (Cw•P)]0.67 davg = [(0.15) / {(1.66)•(2.4)}]0.67 = 0.11 m Step 2. Compute associated spread. d = davg + Sx•W / 2 d = 0.11 + .03•(0.6) / 2 =0.119 m T = d / Sx = (0.119) / (0.03) T = 3.97 m English Units Step 1. Compute depth at curb. Assuming grate controls interception: P = 2•W + L = 2•(2) + 4 P = 8 ft From Equation 4-26 or Chart 9 davg = [Qi / (Cw•P)]0.67 davg = [(5.3) / {(3.0)•(8.0)}]0.67 = 0.36 ft Step 2. Compute associated spread. d = davg + Sx•W / 2 d = 0.36 + .03•(2) / 2 =0.39 T = d / Sx = (0.39) / (0.03) T = 13 ft
 Solution (2): SI Units Step 1. Compute depth at curb. Assuming grate clogged. Using Chart 11 or Equation 4-31b with Q = 0.15 m3/s d = {Qi / (Co•h•L)}2 / (2•g) + h / 2 d = {(0.15) / [(0.67)•(0.10)•(1.2)]}2 / [(2)•(9.81)] + (0.1 / 2) d = 0.24 m Step 2. Compute associated spread. T = d / Sx T = (0.24) / (0.03) T = 8.0 m English Units Step 1. Compute depth at curb. Assuming grate clogged. Using Chart 11 or Equation 4-31b with Q = 5.3 ft3/s d = {Qi / (Co•h•L)}2 / (2•g) + h / 2 d = {(5.3) / [(0.67)•(0.325)•(4)]}2 / [(2)•(32.2)] + (0.325 / 2) d = 0.74 ft Step 2. Compute associated spread. T = d / Sx T = (0.74) / (0.03) T = 24.7 ft

Interception by the curb-opening only will be in a transition stage between weir and orifice flow with a depth at the curb of about 0.24 m (0.8 ft). Depth at the curb and spread on the pavement would be almost twice as great if the grate should become completely clogged.

4.4.6. Inlet Locations

The location of inlets is determined by geometric controls which require inlets at specific locations, the use and location of flanking inlets in sag vertical curves, and the criterion of spread on the pavement. In order to adequately design the location of the inlets for a given project, the following information is needed:

• Layout or plan sheet suitable for outlining drainage areas
• Typical cross sections
• Superelevation diagrams
• Contour maps

4.4.6.1 Geometric Controls

There are a number of locations where inlets may be necessary with little regard to contributing drainage area. These locations should be marked on the plans prior to any computations regarding discharge, water spread, inlet capacity, or flow bypass. Examples of such locations follow.

• At all low points in the gutter grade
• Immediately upstream of median breaks, entrance/exit ramp gores, cross walks, and street intersections, i.e., at any location where water could flow onto the travelway
• Immediately upgrade of bridges (to prevent pavement drainage from flowing onto bridge decks)
• Immediately downstream of bridges (to intercept bridge deck drainage)
• Immediately up grade of cross slope reversals
• Immediately up grade from pedestrian cross walks
• At the end of channels in cut sections
• On side streets immediately up grade from intersections
• Behind curbs, shoulders or sidewalks to drain low area

In addition to the areas identified above, runoff from areas draining towards the highway pavement should be intercepted by roadside channels or inlets before it reaches the roadway. This applies to drainage from cut slopes, side streets, and other areas alongside the pavement. Curbed pavement sections and pavement drainage inlets are inefficient means for handling extraneous drainage.

4.4.6.2 Inlet Spacing on Continuous Grades

Design spread is the criterion used for locating storm drain inlets between those required by geometric or other controls. The interception capacity of the upstream inlet will define the initial spread. As flow is contributed to the gutter section in the downstream direction, spread increases. The next downstream inlet is located at the point where the spread in the gutter reaches the design spread. Therefore, the spacing of inlets on a continuous grade is a function of the amount of upstream bypass flow, the tributary drainage area, and the gutter geometry.

For a continuous slope, the designer may establish the uniform design spacing between inlets of a given design if the drainage area consists of pavement only or has reasonably uniform runoff characteristics and is rectangular in shape. In this case, the time of concentration is assumed to be the same for all inlets. The following procedure and example illustrates the effects of inlet efficiency on inlet spacing.

In order to design the location of inlets on a continuous grade, the computation sheet shown in Figure 4-19 may be used to document the analysis. A step by step procedure for the use of Figure 4-19 follows.

 Step 1. Complete the blanks at the top of the sheet to identify the job by state project number, route, date, and your initials. Step 2. Mark on a plan the location of inlets which are necessary even without considering any specific drainage area, such as the locations described in Section 4.4.6.1. Step 3. Start at a high point, at one end of the job if possible, and work towards the low point. Then begin at the next high point and work backwards toward the same low point. Step 4. To begin the process, select a trial drainage area approximately 90 to 150 m (300 to 500 ft) long below the high point and outline the area on the plan. Include any area that may drain over the curb, onto the roadway. However, where practical, drainage from large areas behind the curb should be intercepted before it reaches the roadway or gutter. Step 5. Col. 1, Col. 2, Col. 19 Describe the location of the proposed inlet by number and station and record this information in columns 1 and 2. Identify the curb and gutter type in column 19 remarks. A sketch of the cross section should be prepared. Step 6. Col. 3 Compute the drainage area (hectares) (acres) outlined in step 4 and record in column 3. Step 7. Col. 4 Determine the runoff coefficient, C, for the drainage area. Select a C value provided in Table 3-1 or determine a weighted C value using Equation 3-2 and record the value in column 4. Step 8. Col. 5 Compute the time of concentration, tc, in minutes, for the first inlet and record in column 5. The time of concentration is the time for the water to flow from the most hydraulically remote point of the drainage area to the inlet, as discussed in Section 3.2.2.3. The minimum time of concentration is 5 minutes. Step 9. Col. 6 Using the time of concentration, determine the rainfall intensity from the Intensity-Duration-Frequency (IDF) curve for the design frequency. Enter the value in column 6. Step 10. Col. 7 Calculate the flow in the gutter using Equation 3-1, Q = C•I•A / Ku. The flow is calculated by multiplying column 3 times column 4 times column 6 divided by Ku. Using the SI system of units, Ku = 360 (= 1 for English units). Enter the flow value in column 7. Step 11. Col. 8 From the roadway profile, enter in column 8 the gutter longitudinal slope, SL, at the inlet, taking into account any superelevation. Step 12. Col. 9, Col. 13 From the cross section, enter the cross slope, Sx, in column 9 and the grate gutter width, W, in column 13.

The following example illustrates the use of this procedure and Figure 4-19.

Example 4-15

 Given: The storm drainage system is illustrated in Figure 4-20 with the following roadway characteristics: n = 0.016 Sx = 0.02 m/m (ft/ft) SL = 0.03 m/m (ft/ft) Allowable spread = 2.0 m (6.6 ft) Gutter and shoulder cross slope = 0.04 m/m (ft/ft) W = 0.6 m (2.0 ft) For maintenance reasons, inlet spacing is limited to 110 m (360 ft) Find: The maximum design inlet spacing for a 0.6 m wide by 0.9 m long (2 ft by 3 ft) P 50 x 100 grate, during a 10 – year storm event.

Solution: Use the inlet computation sheet shown in Figure 4-19. The entries are shown in Figure 4-21a (SI) and Figure 4-21b (English).

 SI Units Steps 1-4 The computations begin at inlet located at station 20+00. The initial drainage area consists of a 13 m wide roadway section with a length of 200 m. The top of the drainage basin is located at station 22+00. Step 5 Col. 1 Inlet # 40 Col. 2 Station 20+00 Col. 19 composite gutter with a curb height = 0.15 m Step 6 Col. 3 Distance from top of drainage area to first inlet = 22+00 – 20+00 = 200 m. Width = 13 m. Drainage area = (200)•(13) = 2600 m2 = 0.26 ha

 Step 7 Col. 4 Runoff coefficient, C = 0.73 (Table 3-1) Step 8 Col. 5 First calculate velocity of gutter flow using Equation 3-4 and Table 3-3. V = K•Sp0.5 = (0.619)•(3.0)0.5 = 1.1 m/s Calculate the time of concentration, tc, using Equation 3-6. tc = L / [60•V] = (200) / [(60)•(1.1)] = 3.0 min (use 5 min minimum) Step 9 Col. 6 Determine rainfall intensity, I, from IDF curve. I = 180 mm/hr (Figure 3-1) Step 10 Col. 7 Determine gutter flow rate, Q, using Equation 3-1. Q = C•I•A / Ku = (0.73)•(180)•(0.26) / (360) = 0.095 m3/s Step 11 Col. 8 SL = 0.03 m/m Step 12 Col. 9 Sw = 0.04 m/m Step 13 Col. 13 W = 0.6 m Step 14 Col. 14 Determine spread, T, using Equation 4-2 or Chart 1. T = [{Q•n} / {K •Sx1.67•SL0.5}]0.375 T = [{(0.095)•(0.016)} / {(0.376)•(0.04)1.67•(0.03)0.5}]0.375 T = 1.83 m (6.0 ft) (less than allowable so therefore proceed to next step) Col. 12 Determine depth at curb, d, using Equation 4-3. d = T•Sx = (1.83)•(0.04) = 0.073 m (less than actual curb height so proceed to next step) Step 15 Col. 15 W / T = 0.6 / 1.83 = 0.33 Step 16 Col. 16 Select a P 50 x 100 grate measuring 0.6 m wide by 0.9 m long Step 17 Col. 17 Calculate intercepted flow, Qi. Eo = 1 – (1 – W / T)2.67 (Equation 4-16 or Chart 2) Eo = 1 – (1 – 0.33)2.67 Eo = 0.66 V = 0.752 / n•SL0.5•Sx0.67•T0.67 (Equation 4-13 or Chart 4) V = 0.752 / (0.016)•(0.03)0.5•(0.04)0.67•(1.83)0.67 V = 1.41 m/s Rf = 1.0 (Chart 5) Rs = 1 / [1 + (0.0828•V1.8) / (Sx•L2.3)] (Equation 4-19 or Chart 6) Rs = 1 / [1 + {(0.0828)•(1.41)1.8} / {(0.04)•(0.9)2.3}] Rs = 0.17 Qi = Q•[Rf•Eo + Rs•(1 – Eo)] (Equation 4-21) Qi = (0.095)•[(1.0)•(0.66) + (0.17)•(1 – 0.66)] Qi = 0.068 m3/s Step 18 Col. 18 Qb = Q – Qi = 0.095 – 0.068 = 0.027 m3/s Step 19 Col. 1 Inlet # 41 Col. 2 Station 18+90 Col. 3 Drainage area = (110 m)•(13 m) = 1430 m2 = 0.14 ha Col. 4 Runoff coefficient, C = 0.73 (Table 3-1) Step 20 Col. 5 V = 1.1 m/s (step 8) tc = L / [60•V] = 110 / [(60)•(1.1)] tc = 2 min (use 5 min minimum) (Equation 3-6) Step 21 Col. 6 I = 180 mm/hr (Figure 3-1) Step 22 Col. 7 Q = C•I•A / Ku (Equation 3-1) Q = (0.73)•(180)•(0.14) / (360) = 0.051 m3/s Step 23 Col. 11 Col. 11 = Col. 10 + Col. 7 = 0.027 + 0.051 = 0.078 m3/s Step 24 Col. 14 T = 1.50 m (Equation 4-2 or Chart 1) T < Tallowable Col. 12 d = 0.06 m d < curb height Since the actual spread is less than the allowable spread, a larger invert spacing could be used here. However, in this case, maintenance considerations limit the spacing to 110 m. Step 25 Col. 16 Select P 50 x 100 grate 0.6 m wide by 0.9 m long Step 26 Col. 17 Qi = 0.057 m3/s (step 17) Step 27 Col. 18 Qb = Q – Qi Col. 18 = Col. 11 – Col. 17 Col. 18 = 0.078 – 0.057 = 0.021 m3/s Step 28 Repeat steps 19 through 27 for each additional inlet.

English Units

 Steps 1-4 The computations begin at the inlet located at station 20+00. The initial drainage area consists of a 42.7 ft wide roadway section with a length of 656 ft. The top of the drainage basin is located at station 26+56. Step 5 Col. 1 Inlet # 40 Col. 2 Station 20+00 Col. 19 Composite gutter with a curb height = 0.50 ft

 Step 6 Col. 3 Distance from top of drainage area to first inlet = 26+56 – 20+00 = 656 ft. Width = 42.7 ft. Drainage area = (656)(42.7)/43560 = 0.64 ac Step 7 Col. 4 Runoff coefficient, C = 0.73 (Table 3-1) Step 8 Col. 5 First calculate velocity of gutter flow using Equation 3-4 and Table 3-3. V = K•Sp0.5 = (3.28)•(3.0)0.5 = 3.5 ft/s Calculate the time of concentration, tc, using Equation 3-6. tc = L / [60•V] = (656) / [(60)•(3.5)] = 3.1 min (use 5 min minimum) Step 9 Col. 6 Determine rainfall intensity, I, from IDF curve. I = 7.1 in/hr (Figure 3-1) Step 10 Col. 7 Determine gutter flow rate, Q, using Equation 3-1. Q = C•I•A / Ku = (0.73)•(7.1)•(0.64) / (1) = 3.32 ft3 / s Step 11 Col. 8 SL = 0.03 ft / ft Step 12 Col. 9 Sw = 0.04 ft / ft Step 13 Col. 13 W = 2.0 ft Step 14 Col. 14 Determine spread, T, using Equation 4-2 or Chart 1. T = [{Q•n} / {K •Sx1.67•SL0.5}]0.375 T = [{(3.32)•(0.016)} / {(0.56)•(0.04)1.67•(0.03)0.5}]0.375 T = 5.99 ft (6.0 ft) (less than allowable so therefore proceed to next step) Col. 12 Determine depth at curb, d, using Equation 4-3. d = T•Sx = (5.99)•(0.04) = 0.24 ft (less than actual curb height so proceed to next step) Step 15 Col. 15 W / T = 2.0 / 5.99 = 0.33 Step 16 Col. 16 Select a P 50 x 100 grate measuring 2 ft wide by 3 ft long Step 17 Col. 17 Calculate intercepted flow, Qi. Eo = 1 – (1 – W / T)2.67 (Equation 4-16 or Chart 2) Eo = 1 – (1 – 0.33)2.67 Eo = 0.66 V = Ku / n•SL0.5•Sx0.67•T0.67 (Equation 4-13 or Chart 4) V = 1.11 / (0.016)•(0.03)0.5•(0.04)0.67•(1.83)0.67 V = 4.61 ft/s Rf = 1.0 (Chart 5) Rs = 1 / [1 + (Ku•V1.8) / (Sx•L2.3)] (Equation 4-19 or Chart 6) Rs = 1 / [1 + {(0.15)•(4.6)1.8} / {(0.04)•(0.9)2.3}] Rs = 0.18 Qi = Q•[Rf•Eo + Rs•(1 – Eo)] (Equation 4-21) Qi = (3.32)•[(1.0)•(0.66) + (0.18)•(1 – 0.66)] Qi = 2.39 ft3/s Step 18 Col. 18 Qb = Q – Qi = 3.32 – 2.39 = 0.93 ft3/s Step 19 Col. 1 Inlet # 41 Col. 2 Station 18+90 Col. 3 Drainage area = (360)•(42.6) / 43560 = 0.35 ac Col. 4 Runoff coefficient, C = 0.73 (Table 3-1) Step 20 Col. 5 V = 3.5 ft/s (step 8) tc = L / [60•V] = 360 / [(60)•(3.5)] tc = 1.7 min (use 5 min minimum) (Equation 3-6) Step 21 Col. 6 I = 7.1 in / hr (Figure 3-1) Step 22 Col. 7 Q = C•I•A / Ku (Equation 3-1) Q = (0.73)•(7.1)•(0.35) / (1) = 1.81 ft3/s Step 23 Col. 11 Col. 11 = Col. 10 + Col. 7 = 0.9 + 1.81 = 2.74 ft3/s Step 24 Col. 14 T = 5.6 ft (Equation 4-2 or Chart 1) T < Tallowable Col. 12 d = (5.6)•(0.4) = 0.22 ft d < curb height Since the actual spread is less than the allowable spread, a larger invert spacing could be used here. However, in this case, maintenance considerations limit the spacing to 360 ft. Step 25 Col. 16 Select P 50 x 100 grate 2 ft wide by 3 ft long Step 26 Col. 17 Qi = 2.05 ft3/s (step 17) Step 27 Col. 18 Qb = Q – Qi Col. 18 = Col. 11 – Col. 17 Col. 18 = 2.74 – 2.05 = 0.69 ft3/s Step 28 Repeat steps 19 through 27 for each additional inlet.

For inlet spacing in areas with changing grades, the spacing will vary as the grade changes. If the grade becomes flatter, inlets may be spaced at closer intervals because the spread will exceed the allowable. Conversely, for an increase in slope, the inlet spacing will become longer because of increased capacity in the gutter sections. Additionally, individual transportation agencies may have limitations for spacing due to maintenance constraints.

4.4.6.3 Flanking Inlets

As discussed in the previous section, inlets should always be located at the low or sag points in the gutter profile. In addition, it is good engineering practice to place flanking inlets on each side of the low point inlet when in a depressed area that has no outlet except through the system. This is illustrated in Figure 4-22. The purpose of the flanking inlets is to act in relief of the inlet at the low point if it should become clogged or if the design spread is exceeded.

Flanking inlets can be located so they will function before water spread exceeds the allowable spread at the sump location. The flanking inlets should be located so that they will receive all of the flow when the primary inlet at the bottom of the sag is clogged. They should do this without exceeding the allowable spread at the bottom of the sag. If the flanking inlets are the same dimension as the primary inlet, they will each intercept one-half the design flow when they are located so that the depth of ponding at the flanking inlets is 63% of the depth of ponding at the low point. If the flanker inlets are not the same size as the primary inlet, it will be necessary to either develop a new factor or do a trial and error solution using assumed depths with the weir equation to determine the capacity of the flanker inlet at the given depths.(18)

The spacing required for various depths at curb criteria and vertical curve lengths is defined as follows:

 K = L / (G2 – G1) (4-36)
 where: L = Length of the vertical curve in feet G1, G2 = Approach grades in percent

The AASHTO policy on geometrics specifies maximum K values for various design speeds and a maximum K of 167 considering drainage.

 X = (74•d•K)0.5 (4-37)
 where: X = Maximum distance from bottom of sag to flanking inlet d = Depth of water over inlet in bottom of sag as shown in Figure 4-22 K = As defined above
 Step 1. Determine the K value for the sag curve. Step 2. Determine the depth at design spread, d = Sx•T (Sx = cross slope, T = gutter spread) Step 3. Establish X from Equation 4-37. Note this distance is the maximum distance one can use.

Table 4-7 provides a check on the distance established by the above procedure.

Example 4-16

 Given: A 150 m (500 ft)(L) sag vertical curve at an underpass on a 4-lane divided highway with begin and end slopes of -2.5% and +2.5% respectively. The spread at design Q is not to exceed the shoulder width of 3.0 m (9.8 ft). Sx = 0.02 Find: The location of the flanking inlets if located to function in relief of the inlet at the low point when the inlet at the low point is clogged.

Solution:

 SI Units Step 1. Find the rate of vertical curvature, K. K = L / (Send – Sbeginning) K = 150 m / (2.5% – (-2.5%)) K = 30 m / % Step 2. Determine depth at curb for design spread. d = Sx•T = (0.02)•(3.0) d = 0.06 m Step 3. Determine the flanker locations. X = X = (74•d•K)0.5 = (74•(0.06)•(30))0.5 = 11.5 m Inlet spacing = 11.5 m from the sag point;
 English Units Step 1. Find the rate of vertical curvature, K. K = L / (Send – Sbeginning) K = 500 ft / (2.5% – (-2.5%)) K = 100 ft / % Step 2. Determine depth at curb for design spread. d = Sx•T = (0.02)•(9.84) d = 0.2 ft Step 3. Determine the flanker locations. d = X = (74•d•K)0.5 = (74•(0.2)•(100))0.5 = 34.5 ft

Inlet spacing = 34.5 ft from the sag point

Example problem solutions in Section 4.4.5 illustrate the total interception capacity of inlets in sag locations. Except where inlets become clogged, spread on low gradient approaches to the low point is a more stringent criterion for design than the interception capacity of the sag inlet. AASHTO(21) recommends that a gradient of 0.3% be maintained within 15 m (50 ft) of the level point in order to provide for adequate drainage. It is considered advisable to use spread on the pavement at a gradient comparable to that recommended by the AASHTO Committee on Design to evaluate the location and excessive spread in the sag curve. Standard inlet locations may need to be adjusted to avoid excessive spread in the sag curve.

Inlets may be needed between the flankers and the ends of the curves also. For major sag points, the flanking inlets are added as a safety factor, and are not considered as intercepting flow to reduce the bypass flow to the sag point. They are installed to assist the sag point inlet in the event of clogging.

Table 4-7. Distance to Flanking Inlets in Sag Vertical Curve.(19)

SI Units

K (m/%) 4 8 11 15 20 25 30 37 43< 50
d (m)
0.01   1.7 2.4 2.8 3.3 3.8 4.3 4.7 5.2 5.6 6.0
0.02   2.4 3.4 4.0 4.7 5.4 6.0 6.6 7.4 7.9 8.6
0.03   2.9 4.2 4..9 5.7 6.6 7.4 8.1 9.0 9.7 10.5
0.05   3.8 5.4 6.3 7.4 8.6 9.6 10.5 11.7 12.6 13.6
0.08   4.8 6.8 8.0 9.4 10.8 12.1 13.3 14.8 15.9 17.2
0.10   5.4 7.6 9.0 10.5 12.1 13.6 14.8 16.5 17.8 19.2
0.15   6.6 9.4 11.0 12.9 14.8 16.6 18.2 20.2 21.8 23.5
0.18   7.2 10.3 12.1 14.1 16.3 18.2 19.9 22.2 23.9 25.8
0.21   7.8 11.1 13.0 15.2 17.6 19.7 21.5 23.9 25.8 27.8
 NOTES: 1. x = (74•d•K)0.5, where x = distance from sag point. 2. d = depth of ponding at curb as shown in Figure 4.22 3. Drainage maximum K = 50

English Units

K (ft/%) 20 30 40 50 70 90 110 130 160 167
d (ft)
0.1   12 14 17 19 22 25 28 31 34 35
0.2   17 21 24 27 32 36 40 43 48 49
0.3   21 25 29 33 39 44 49 53 59 60
0.4   24 29 34 38 45 51 57 62 68 70
0.5   27 33 38 43 50 57 63 69 76 78
0.6   29 36 42 47 55 63 69 75 84 86
0.7   32 39 45 50 60 68 75 82 91 93
 NOTES: 1. x = (74•d•K)0.5, where x = distance from sag point. 2. d = depth of ponding at curb as shown in Figure 4.22 3. Drainage maximum K =167

4.4.7 Median, Embankment, and Bridge Inlets

Flow in median and roadside ditches is discussed briefly in Chapter 5 and in Hydraulic Engineering Circular 15(34) and Hydraulic Design Series 4.(7) It is sometimes necessary to place inlets in medians at intervals to remove water that could cause erosion. Inlets are sometimes used in roadside ditches at the intersection of cut and fill slopes to prevent erosion downstream of cut sections.

Where adequate vegetative cover can be established on embankment slopes to prevent erosion, it is preferable to allow storm water to discharge down the slope with as little concentration of flow as practicable. Where storm water must be collected with curbs or swales, inlets are used to receive the water and discharge it through chutes, sod or riprap swales, or pipe downdrains.

Bridge deck drainage is similar to roadway drainage and deck drainage inlets are similar in purpose to roadway inlets. Bridge decks lack a clear zone, adding to the need to consider proper inlet design. Reference 23 discusses bridge deck drainage.

4.4.7.1 Median and Roadside Ditch Inlets

Median and roadside ditches may be drained by drop inlets similar to those used for pavement drainage, by pipe culverts under one roadway, or by cross drainage culverts which are not continuous across the median. Figure 4-23 illustrates a traffic-safe median inlet. Inlets, pipes, and discontinuous cross drainage culverts should be designed so as not to detract from a safe roadside. Drop inlets should be flush with the ditch bottom and traffic-safe bar grates should be placed on the ends of pipes used to drain medians that would be a hazard to errant vehicles, although this may cause a plugging potential. Cross drainage structures should be continuous across the median unless the median width makes this impractical. Ditches tend to erode at drop inlets; paving around the inlets helps to prevent erosion and may increase the interception capacity of the inlet marginally by acceleration of the flow.

Pipe drains for medians operate as culverts and generally require more water depth to intercept median flow than drop inlets. No test results are available on which to base design procedures for estimating the effects of placing grates on culvert inlets. However, little effect is expected.

The interception capacity of drop inlets in median ditches on continuous grades can be estimated by use of Charts 14 and 15 to estimate flow depth and the ratio of frontal flow to total flow in the ditch.

Chart 14 is the solution to the Manning’s equation for channels of various side slopes. The Manning’s equation for open channels is:

 Q = (Ku / n)•A•R0.67•SL0.5 (4-38)
 where: Q = Discharge rate, m3/s (ft3/s) Ku = 1.0 (1.486) n = Hydraulic resistance variable A = Cross sectional area of flow, m2 (ft2) R = Hydraulic radius = area / wetted perimeter, m (ft) SL = Bed slope, m/m (ft/ft)

For the trapezoidal channel cross section shown on Chart 14, the Manning’s equation becomes:

 Q = (Ku / n)•(B•d + z•d2) {(B•d + Z•d2) / [B + 2•d•(Z2 + 1)0.5]}0.67•SL0.5 (4-39)
 where: B = Bottom width, m (ft) z = Horizontal distance of side slope to a rise of 1 m (ft) vertical, m (ft)

Equation 4-39 is a trial and error solution to Chart 14.

Chart 15 is the ratio of frontal flow to total flow in a trapezoidal channel. This is expressed as:

 Eo = W / (B + d•z) (4-40)

Charts 5 and 6 are used to estimate the ratios of frontal and side flow intercepted by the grate to total flow.

Small dikes downstream of drop inlets (Figure 4-23) can be provided to impede bypass flow in an attempt to cause complete interception of the approach flow. The dikes usually need not be more than a few inches high and should have traffic safe slopes. The height of dike required for complete interception on continuous grades or the depth of ponding in sag vertical curves can be computed by use of Chart 9. The effective perimeter of a grate in an open channel with a dike should be taken as 2•(L + W) since one side of the grate is not adjacent to a curb. Use of Chart 9 is illustrated in Section 4.4.4.1.

The following examples illustrate the use of Charts 14 and 15 for drop inlets in ditches on continuous grade.

Example 4-17

 Given: A median ditch with the following characteristics: B = 1.2 m (3.9 ft) n = 0.03 z = 6 S = 0.02 The flow in the median ditch is to be intercepted by a drop inlet with a 0.6 m by 0.6 m (2 ft by 2 ft) P-50 parallel bar grate; there is no dike downstream of the inlet. Q = 0.28 m3/s (9.9 ft3/s) Find: The intercepted and bypassed flows (Qi and Qb)

Solution:

 SI Units Step 1. Compute the ratio of frontal to total flow in trapezoidal channel. Q•n = (0.28)•(0.03) Q•n = 0.0084 m3/s From Chart 14 d / B = 0.12 d = (B)•(d / B) = (0.12)•(1.20) = 0.14 m Using Equation 4-38 or Chart 15 Eo = W / (B + d•z) = (0.6) / [1.2 + (0.14)•(6)] = 0.30 Step 2. Compute frontal flow efficiency V = Q / A A = (0.14)•[(6)•(0.14) + 1.2) A = 0.29 m2 V = (0.28) / (0.29) = 0.97 m/s From Chart 5 Rf = 1.0 Step 3. Compute side flow efficiency Since the ditch bottom is wider than the grate and has no cross slope, use the least cross slope available on Chart 6 or use Equation 4-19 to solve for Rs. Using Equation 4-19 or Chart 6 Rs = 1 / [1 + (Ku•V1.8)/(Sx•L2.3)] Rs = 1 / [1 + (0.0828)•(0.97)1.8/{(0.01)•(0.6)2.3}] = 0.04 Step 4. Compute total efficiency. E = Eo•Rf + Rs•(1 – Eo) E = (0.30)•(1.0) + (0.04)•(1 – 0.30) = 0.33 Step 5. Compute interception and bypass flow. Qi = E•Q Qi = (0.33)•(0.28) Qi = 0.1 m3/s Qb = Q – Qi = (0.28) – (0.1) Qb = 0.18 m3/s In the above example, a P-50 inlet would intercept about 33% of the flow in a 1.2 m (3.9 ft) bottom ditch on continuous grade. For grate widths equal to the bottom width of the ditch, use Chart 6 by substituting ditch side slopes for values of Sx, as illustrated in Example 4-18.
 English Units Step 1. Compute the ratio of frontal to total flow in trapezoidal channel. Q•n = (9.9)•(0.03) Q•n = 0.30 ft3/s From Chart 14 d / B = 0.12 d = (B)•(d / B) = (0.12)•(3.9) = 0.467 ft Using Equation 4-38 or Chart 15 Eo = W / (B + d•z) = (2.0) / [3.9 + (0.47)•(6)] = 0.30 Step 2. Compute frontal flow efficiency V = Q / A A = (0.47)•[(6)•(0.47) + 3.9) A = 3.18 ft2 V = 9.9 / 3.18 = 3.11 ft/s From Chart 5 Rf = 1.0 Step 3. Compute side flow efficiency Since the ditch bottom is wider than the grate and has no cross slope, use the least cross slope available on Chart 6 or use Equation 4-19 to solve for Rs. Using Equation 4-19 or Chart 6 Rs = 1 / [1 + (Ku•V1.8)/(Sx•L2.3)] Rs = 1 / [1 + (0.15)•(3.11)1.8/{(0.01)•(2.0)2.3}] = 0.04 Step 4. Compute total efficiency. E = Eo•Rf + Rs•(1 – Eo) E = (0.30)•(1.0) + (0.04)•(1 – 0.30) = 0.33 Step 5. Compute interception and bypass flow. Qi = E•Q Qi = (0.33)•(9.9) Qi = 3.27 ft3/s Qb = Q – Qi = (9.9) – (3.27) Qb = 6.63 ft3/s In the above example, a P-50 inlet would intercept about 33% of the flow in a 1.2 m (3.9 ft) bottom ditch on continuous grade. For grate widths equal to the bottom width of the ditch, use Chart 6 by substituting ditch side slopes for values of Sx, as illustrated in Example 4-18.

Example 4-18

 Given: A median ditch with the following characteristics: Q = 0.28 m3/s (9.9 ft3/s) W = 0.6 m (2 ft) z = 6 S = 0.03 m/m (ft/ft) B = 0.6 m (2 ft) n = 0.03 Sx = 1/6 = 0.17 m/m (ft/ft) The flow in the median ditch is to be intercepted by a drop inlet with a 0.6 m by 0.6 m (2 ft by 2 ft) P-50 parallel bar grate; there is not dike downstream of the inlet. Find: The intercepted and bypassed flows (Qi and Qb).

Solution:

 SI Units Step 1. Compute ratio of frontal to total flow in trapezoidal channel. Q•n = (0.28)•(0.03) Q•n = 0.0084 m3/s From Chart 14 d / B = 0.25 d = (0.25)•(0.6) = 0.15 m Using Equation 4-38 or Chart 15 Eo = W / (B + d•z) = (0.6) / [0.6 + (0.15)•(6)] = 0.40 Step 2. Compute frontal flow efficiency V = Q / A A = (0.15)•[(6)•(0.15) + 0.6)] A = 0.23 m2 (2.42 ft2) V = (0.28) / (0.23) = 1.22 m/s From Chart 5 Rf = 1.0 Step 3. Compute side flow efficiency Using Equation 4-19 or Chart 6 Rs = 1 / [1 + (Ku•V1.8) / (Sx•L2.3)] Rs = 1 / [1 + (0.0828)•(1.22)1.8 / {(0.17)•(0.6)2.3}] = 0.30 Step 4. Compute total efficiency. E = Eo•Rf + Rs•(1 – Eo) E = (0.40)•(1.0) + (0.30)•(1-0.40) E = 0.58 Step 5. Compute interception and bypass flow. Qi = E•Q Qi = (0.58)•(0.28) Qi = 0.16 m3/s Qb = Q – Qi = 0.28 – 0.16 Qb = 0.12 m3/s The height of dike downstream of a drop inlet required for total interception is illustrated by Example 4-19.
 English Units Step 1. Compute ratio of frontal to total flow in trapezoidal channel. Q•n = (9.9)•(0.03) Q•n = 0.30 m3/s From Chart 14 d / B = 0.25 d = (0.25)•(2.0) = 0.50 ft Using Equation 4-38 or Chart 15 Eo = W / (B + d•z) = (2.0) / [2.0 + (0.5)•(6)] = 0.40 Step 2. Compute frontal flow efficiency V = Q / A A = (0.5)•[(6)•(0.5) + 2.0)] A = 2.5 ft2 V = 9.9 / 2.5 = 4.0 ft/s From Chart 5 Rf = 1.0 Step 3. Compute side flow efficiency Using Equation 4-19 or Chart 6 Rs = 1 / [1 + (Ku•V1.8) / (Sx•L2.3)] Rs = 1 / [1 + (0.15)•(4.0)1.8 / {(0.17)•(2.0)2.3}] = 0.32 Step 4. Compute total efficiency. E = Eo•Rf + Rs•(1 – Eo) E = (0.40)•(1.0) + (0.32)•(1-0.40) E = 0.59 Step 5. Compute interception and bypass flow. Qi = E•Q Qi = (0.59)•(9.9) Qi = 5.83 ft3/s Qb = Q – Qi = 9.9 – 5.83 Qb = 4.07 ft3/s The height of dike downstream of a drop inlet required for total interception is illustrated by Example 4-19.

Example 4-19

Given: Data from Example 4-18.

Find: The required height of a berm to be located downstream of the grate inlet to cause total interception of the ditch flow.

Solution:

 SI Units P = 2•(L + W) P = 2•(0.6 + 0.6) = 2.4 m Using Equation 4-26 or Chart 9 d = [Qi / (Cw•P)]0.67 d = [(0.28) / {(1.66)•(2.4)}]0.67 d = 0.17 m A dike will need to have a minimum height of 0.17 m (0.55 ft) for total interception. Due to the initial velocity of the water which may provide adequate momentum to carry the flow over the dike, an additional 0.15 m (0.5 ft) may be added to the height of the dike to insure complete interception of the flow.
 English Units P = 2•(L + W) P = 2•(2.0 + 2.0) = 8.0 ft Using Equation 4-26 or Chart 9 d = [Qi / (Cw•P)]0.67 d = [(9.9) / {(3.0)•(8.0)}]0.67 d = 0.55 ft A dike will need to have a minimum height of 0.17 m (0.55 ft) for total interception. Due to the initial velocity of the water which may provide adequate momentum to carry the flow over the dike, an additional 0.15 m (0.5 ft) may be added to the height of the dike to insure complete interception of the flow.

4.4.7.2 Embankment Inlets

Drainage inlets are often needed to collect runoff from pavements in order to prevent erosion of fill slopes or to intercept water upgrade or downgrade of bridges. Inlets used at these locations differ from other pavement drainage inlets in three respects. First, the economies which can be achieved by system design are often not possible because a series of inlets is not used; second, total or near total interception is sometimes necessary in order to limit the bypass flow from running onto a bridge deck; and third, a closed storm drainage system is often not available to dispose of the intercepted flow, and the means for disposal must be provided at each inlet. Intercepted flow is usually discharged into open chutes or pipe downdrains which terminate at the toe of the fill slope.

Example problem solutions in other sections of this circular illustrate by inference the difficulty in providing for near total interception on grade. Grate inlets intercept little more than the flow conveyed by the gutter width occupied by the grate. Combination curb-opening and grate inlets can be designed to intercept total flow if the length of curb opening upstream of the grate is sufficient to reduce spread in the gutter to the width of the grate used. Depressing the curb opening would significantly reduce the length of inlet required. Perhaps the most practical inlets or procedure for use where near total interception is necessary are sweeper inlets, increase in grate width, and slotted inlets of sufficient length to intercept 85-of inlets on embankments. Figure 4-24 illustrates a combination inlet and downdrain.

Downdrains or chutes used to convey intercepted flow from inlets to the toe of the fill slope may be open or closed chutes. Pipe downdrains are preferable because the flow is confined and cannot cause erosion along the sides. Pipes can be covered to reduce or eliminate interference with maintenance operations on the fill slopes. Open chutes are often damaged by erosion from water splashing over the sides of the chute due to oscillation in the flow and from spill over the sides at bends in the chute. Erosion at the ends of downdrains or chutes can be a problem if not anticipated. The end of the device may be placed low enough to prevent damage by undercutting due to erosion. Well-graded gravel or rock can be used to control the potential for erosion at the outlet of the structure. However, some transportation agencies install an elbow or a “tee” at the end of the downdrains to re-direct the flow and prevent erosion. See HEC-14(35) for additional information on energy dissipator designs. 100% of the gutter flow. Design charts and procedures in Section 4.4.4 are applicable to the design.

4.5 Grate Type Selection Considerations

Grate type selection should consider such factors as hydraulic efficiency, debris handling characteristics, pedestrian and bicycle safety, and loading conditions. Relative costs will also influence grate type selection.

Charts 5, 6, and 9 illustrate the relative hydraulic efficiencies of the various grate types discussed here. The parallel bar grate (P-50) is hydraulically superior to all others but is not considered bicycle safe. The curved vane and the P-30 grates have good hydraulic characteristics with high velocity flows. The other grates tested are hydraulically effective at lower velocities.

Debris-handling capabilities of various grates are reflected in Table 4-5. The table shows a clear difference in efficiency between the grates with the 83 mm (3-1/4 inch) longitudinal bar spacing and those with smaller spacings. The efficiencies shown in the table are suitable for comparisons between the grate designs tested, but should not be taken as an indication of field performance since the testing procedure used did not simulate actual field conditions. Some local transportation agencies have developed factors for use of debris handling characteristics with specific inlet configurations.

Table 4-8 ranks the grates according to relative bicycle and pedestrian safety. The bicycle safety ratings were based on a subjective test program as described in Reference 30. However, all the grates are considered bicycle and pedestrian safe except the P-50. In recent years with the introduction of very narrow racing bicycle tires, some concern has been expressed about the P-30 grate. Caution is advised in its use in bicycle area.

Grate loading conditions must also be considered when determining an appropriate grate type. Grates in traffic areas must be able to withstand traffic loads; conversely, grates draining yard areas do not generally need to be as rigid.

 Table 4-8. Ranking with Respect to Bicycle and Pedestrian Safety. Rank Grate Style 1 P-50 x 100 2 Reticuline 3 P-30 4 45° – 85 Tilt Bar 5 45º – 60 Tilt Bar 6 Curved Vane 7 30º – 85 Tilt Bar

# Manning’s Equation

Water flows in a sloping drainage channel because of the force of gravity. Flow is resisted by the friction between the water and wetted surface of the channel. The quantity of water flowing (Q), the depth of flow (y), and the velocity of flow (V) depend upon the channel shape, roughness (n), and slope (S). Various equations have been devised to determine the velocity and discharge in open channels. A useful equation is the one that is named for Robert Manning, an Irish engineer.

Manning’s equation (HDS-4 Section 4.3.1) is valid for steady, uniform and turbulent flow, has the following form:

Q = (k/n)⋅A⋅R2/3⋅S½,

where:

• k = 1.49 (US Customary units) or 1 (metric units),
• n is Manning’s roughness coefficient,
• A is the flow cross-section area (square feet or square meters),
• R is the hydraulic radius (feet or meters), and
• S is the longitudinal slope (feet/feet or meters/meters).

The hydraulic radius R is found by dividing the cross sectional area A by the wetted perimeter P. These cross-section properties are shown below for an open channel.

Over many decades, typical Manning’s n values have been compiled allowing an engineer to estimate the appropriate value by knowing the general nature of the channel boundaries. Most hydraulics textbooks and drainage design manuals provide tables of typical Manning’s n values. An abbreviated list of such Manning’s roughness coefficients is given in Appendix B, Table B.2 of HDS-4. Several pictorial guides are also available showing the Manning’s n value for different types of channels and floodplains (Barnes 1967 and Acrement and Schneider 1984). Special considerations exist for very steep channels (Jarrett 1985).

A numerical approach for n value estimates consists of the selection of a base roughness value for a straight, uniform, and smooth channel in the materials involved, and then adding values for the channel under consideration:

 n = (n0 + n1 + n2 + n3 + n4)•m5 (4.2)

where:

• no = Base value for straight uniform channels
• n1 = Additive value due to cross-section irregularity
• n2 = Additive value due to variations of the channel
• n3 = Additive value due to obstructions
• n4 = Additive value due to vegetation
• m5 = Multiplication factor due to sinuosity

A discussion of this method and coefficients can be found in Cowan (1956) and Chow (1959). This method may be useful for natural channels, but has limited application for most roadway drainage design work.

For rock riprap channels the Manning’s n is often described as some function of the rock size. Several equations are provided in HEC-15.

Roughness characteristics on the floodplain are complicated by the presence of vegetation, natural and artificial irregularities, buildings, undefined direction of flow, varying slopes, and other complexities. Resistance factors reflecting these effects must be selected largely on the basis of past experience with similar conditions. In general, resistance to flow is large on the floodplains. In some instances, conditions are further complicated by deposition of sediment and development of dunes and bars which affect resistance to flow and direction of flow.

The presence of ice affects channel roughness and resistance to flow in various ways. When an ice cover occurs, the open channel is more nearly comparable to a closed conduit. There is an added shear stress developed between the flowing water and ice cover. This surface shear is much larger than the normal shear stresses developed at the air-water interface. The ice-water interface is not always smooth. In many instances, the underside of the ice is deformed so that it resembles ripples or dunes observed on the bed of sand-bed channels. This may cause overall resistance to flow in the channel to be further increased. With total or partial ice cover, the drag of ice retards flow, decreasing the average velocity and increasing the depth.

When a channel cross section is irregular in shape such as one with a relatively narrow deep main channel and wide shallow overbank area, the cross section must be subdivided and the flow computed separately for the main channel and overbank area. The same procedure is used when different parts of the cross section have different roughness coefficients. In computing the hydraulic radius of the subsections, the water depth common to the two adjacent subsections is not counted as wetted perimeter (see Example Problem 4.3).

Conveyance can be computed and a curve drawn for any channel cross section. The area and hydraulic radius are computed for various assumed depths and the corresponding value of K is computed from the equation. Values of conveyance are plotted against the depths of flow and a smooth curve connecting the plotted points is the conveyance curve. If the section was subdivided, the conveyance of each subsection (Ka, Kb,…Kn) is computed and the total conveyance of the channel is the sum of the conveyances of the subsections. Discharge can then be computed using Equation 4.7

Example Problem 4.3 illustrates a conveyance curve for a compound cross section. The concept of channel conveyance is useful when computing the distribution of overbank flood flows in the stream cross section and the distribution through the openings in a proposed stream crossing. The discharge through each opening can be assumed to have the same ratio to the total discharge as the ratio of conveyance of the opening bears to the total conveyance of the channel.

4.3.2 Aids in the Solution of Manning’s Equation

Equations for the computation of Area, A, wetted perimeter, P, and hydraulic radius, R, in rectangular and trapezoidal channels (Figure 4.2) are:

 A = B•y + Z•y2 (4.8) P = B + 2•y•*√(1 + Z2) (4.9) R = (B•y + Z•y2)/(B + 2•y•*√(1 + Z2)) (4.10)

Variables are defined in Figure 4.2.

Figure 4.2. Trapezoidal channel.

# Caltrans Riprap Sizes

This has been superseded by Caltrans New Riprap Classes.

D50 Equivalent Spherical Diameter
RSP Class D50Size1 D50Weight
inches pounds
8 Ton 71 17600
4 Ton 56 8800
2 Ton 45 4400
1 Ton 36 2200
1/2 Ton 28 1100
1/4 Ton 23 550
Light 16 200
Facing 12 75
Backing No 1 12 75
Backing No 2 8 25
Backing No 3 4 2/3 5
Small RSP (7-inch) 3 1 1/3
Small RSP (5-inch) 2 2/5
Small RSP (4-inch) 1 1/20
1Assumes rock density = 165 lb/ft3

Caltrans Standard Specifications

72-2 ROCK SLOPE PROTECTION

72-2.02A Rock

For method A placement and the class of RSP described, comply with the rock grading shown in the following table:

Rock Grading for Method A Placement
Rock size Percentage larger thana
Class
8T 4T 2T 1T 1/2 T
16 Ton 0–5
8 Ton 50–100 0–5
4 Ton 95–100 50–100 0–5
2 Ton 95–100 50–100 0–5
1 Ton 95–100 50–100 0–5
1/2 Ton 95–100 50–100
1/4 Ton 95–100
aFor any class, the percentage of rock smaller than the smallest rock size must be determined on the basis of weight. For all other rock sizes within a class, the percentage must be determined on the basis of the ratio of the number of individual rocks larger than the smallest size shown for that class compared to the total number of rocks.

For method B placement and the class of RSP described, comply with the rock grading shown in the following table:

Rock Grading for Method B Placement
Rock size Percentage larger thana
Class
1 T 1/2 T 1/4 T Light Facing No. 1 No. 2 No. 3
2 Ton 0–5
1 Ton 50–100 0–5
1/2 Ton 50–100 0–5
1/4 Ton 95–100 50–100 0–5
200 lb 95–100 50–100 0–5 0–5
75 lb 95–100 50–100 50–100 0–5
25 lb 95–100 90–100 90–100 25–75 0–5
5 lb 90–100 25–75
1 lb 90–100
aFor any class, the percentage of rock smaller than the smallest rock size must be determined on the basis of weight. For all other rock sizes within a class, the percentage must be determined on the basis of the ratio of the number of individual rocks larger than the smallest size shown for that class compared to the total number of rocks.

Rock must have the values for the material properties shown in the following table:

Rock Material Properties
Property California Test Value
Apparent specific gravity 206 2.5 minimum
Absorption 206 4.2% maximum
Durability index 229 52 minimum

Select rock so that shapes provide a stable structure for the required section. If the slope is steeper than 2:1, do not use rounded boulders and cobbles. Angular shaped rock may be used on any planned slope. Flat or needle shaped rock must not be used unless the individual rock thickness is greater than 0.33 times the length.

72-2.02B Fabric

Fabric must be RSP fabric that complies with the class shown in the following table:

Fabric Class
Class Largest rock grading class used in slope protection
8 1 ton or smaller
10 Larger than 1 ton

72-4 SMALL-ROCK SLOPE PROTECTION

72-4.02 MATERIALS

Rock must be cobble, gravel, crushed gravel, crushed rock, or any combination of these.

If the rock layer is shown as 7 inches thick, comply with grading shown in the following table:

Sieve sizes Percentage passing
5 inch 100
4 inch 90–100
3 inch 25–40
2 inch 0–10

If the rock layer is shown as 5 inches thick, comply with the grading shown in the following table:

Sieve sizes Percentage passing
4 inch 100
3 inch 90–100
2 inch 25–40
1 inch 0–10

If the rock layer is shown as 4-inches thick, comply with grading shown in the following table:

Rock Grading for 4-inch Thick Layer
Sieve sizes Percentage passing
3 inch 100
2 inch 90–100
1 inch 25–40
3/4 inch 0–10

Granular material must contain at least 90 percent crushed particles when tested under California Test 205.

# Circular Pipe Alternate Depth

Flow conditions in a closed conduit can occur as open-channel flow, gravity full flow or pressure flow. In open-channel flow the water surface is exposed to the atmosphere. Gravity full flow occurs at that condition where the conduit is flowing full, but not yet under any pressure. Pressure flow occurs when the conduit is flowing full and under pressure.

Due to the additional wetted perimeter and increased friction that occurs in a gravity full pipe, a partially full pipe will actually carry greater flow. For a circular conduit the peak flow occurs at 93 percent of the height of the pipe, and the average velocity flowing one-half full is the same as gravity full flow (See figure below). Part-full flow relationships for circular pipes.

# FHWA Riprap Size

 Source: FP-14 Standard Specifications Author: FHWA
Table 705-1 Gradation Requirements for Riprap(1)
Class % of Rock Equal or Smaller by Count Range of Intermediate Dimensions,(2) Range of Rock Mass,(3)
DX inches (millimeters) pounds (kilograms)
1 100 9 – 15 (230 – 380) 59 – 270 (27 – 120)
85 7 – 11 (180 – 280) 28 – 110 (13 – 50)
50 5 – 8 (130 – 200) 10 – 42 (5 – 19)
15 3 – 6 (80 – 150) 2 – 18 (1 – 8)
2 100 15 – 21 (380 – 530) 270 – 750 (120 – 340)
85 11 – 15 (280 – 380) 110 – 270 (50 – 120)
50 8 – 11 (200 – 280) 42 – 110 (19 – 50)
15 6 – 8 (130 – 200) 10 – 42 (6 – 19)
3 100 21 – 27 (530 – 690) 750 – 1600 (340 – 730)
85 15 – 19 (380 – 480) 270 – 560 (120 – 250)
50 11 – 14 (280 – 360) 110 – 220 (50 – 100)
15 8 – 10 (200 – 250) 42 – 81 (19 – 37)
4 100 27 – 33 (690 – 840) 1600 – 2900 (730 – 1300)
85 19 – 23 (480 – 580) 560 – 990 (250 – 450)
50 14 – 17 (360 – 430) 220 – 400 (100 – 180)
15 9 – 12 (230 – 300) 59 – 140 (27 – 64)
5 100 33 – 39 (840 – 990) 2900 – 4850 (1300 – 2200)
85 23 – 28 (580 – 710) 990 – 1800 (450 – 820)
50 17 – 20 (430 – 510) 400 – 650 (180 – 290)
15 11 – 15 (280 – 380) 110 – 270 (50 – 120)
6 100 39 – 45 (990 – 1140) 4850 – 7400 (2200 – 3350)
85 28 – 32 (710 – 810) 1800 – 2650 (820 – 1200)
50 20 – 23 (510 – 580) 650 – 990 (290 – 450)
15 13 – 17 (330 – 430) 180 – 400 (82 – 180)
7 100 45 – 54 (1140 – 1370) 7400 – 12,800 (3350 – 5800)
85 32 – 38 (810 – 970) 2650 – 4450 (1200 – 2000)
50 23 – 28 (580 – 710) 990 – 1800 (450 – 820)
15 15 – 20 (380 – 510) 270 – 650 (120 – 290)
8 100 54 – 66 (1370 – 1680) 12,800 – 23,400 (5800 – 10,600)
85 38 – 47 (970 – 1190) 4450 – 8450 (2000 – 3850)
50 28 – 35 (710 – 890) 1800 – 3500 (820 – 1600)
15 19 – 25 (480 – 640) 560 – 250 (250 – 570)
9 100 66 – 78 (1680 – 1980) 23,400 – 38,600 (10,600 – 17,500)
85 47 – 55 (1190 – 1400) 8450 – 13,500 (3850 – 6100)
50 35 – 41 (890 – 1040) 3500 – 5600 (1600 – 2550)
15 22 – 30 (560 – 760) 870 – 2200 (390 – 1000)
10 100 78 – 90 (1980 – 2290) 38,600 – 59,300 (17,500 – 26,900)
85 55 – 64 (1400 – 1630) 13,500 – 21,300 (6100 – 9650)
50 41 – 48 (1040 – 1220) 5600 – 9000 (2550 – 4100)
15 26 – 36 (660 – 910) 1450 – 3800 (660 – 1700)
(1) Gradation includes spalls and rock fragments to provide a stable, dense mass.
(2) The intermediate dimension is the longest straight-line distance across the rock that is perpendicular to the rock’s longest axis on the rock face with the largest projection plane.
(3) Rock mass is based on a specific gravity of 2.65 and 85 percent of the cubic volume as calculated using the intermediate dimension.

# HDS-5 Inlet Control Equations

INLET CONTROL EQUATIONS

A.1 INTRODUCTION

This appendix contains the inlet control equations used to develop the design charts of this publication (HDS-5). Section A.2 contains the equations for the unsubmerged and submerged inlet control equations. Section A.3 demonstrates how the Section A.2 equations are used to create dimensionless design curves for culvert shapes with coefficients (Section A.3.1) and without (Section A.3.2). Section A.4 discusses how the dimensionless design curves are used to develop the nomographs in Appendix C. Section A.5 discusses how the dimensionless design curves are used to develop the polynomial equations used in FHWA software.

A.2 INLET CONTROL EQUATIONS

The equations used to develop the inlet control nomographs in Appendix C are based on the research conducted by the National Bureau of Standards (NBS) under the sponsorship of the Bureau of Public Roads (now the Federal Highway Administration). John L. French of the NBS produced seven progress reports as a result of this research. Of these, the first (NBS 1955) and fourth (NBS 1961) through seventh reports (NBS 1966a, 1966b, and 1967) dealt with the hydraulics of pipe and box culvert entrances, with and without tapered inlets. Herbert G. Bossy of the FHWA provides an excellent synthesis of the research in his paper, “Hydraulic of conventional Highway Culverts” (Bossy 1961). Additional background on the development of the equations is found in HEC-13 (FHWA 1972a) and unpublished notebooks and notes (Bossy 1963 and Normann 1974).

The two basic conditions of inlet control depend upon whether the inlet end of the culvert is or is not submerged by the upstream headwater. If the inlet is not submerged by the headwater, the inlet performs as a weir and the unsubmerged equations are used (Section A.2.1). If the inlet is submerged by the headwater, the inlet performs as an orifice and the submerged equations are used (Section A.2.2).

Between the unsubmerged and the submerged conditions, there is a transition zone for which the NBS research provided only limited information. The transition zone is defined empirically by drawing a curve between and tangent to the curves defined by the unsubmerged and submerged equations. In most cases, the transition zone is short and the curve is easily constructed.

A.2.1 Unsubmerged Inlet Control Equations

The unsubmerged equation has two forms. Form (1) is based on the specific head at critical depth, adjusted with two correction factors. Form (2) is an exponential equation similar to a weir equation. Form (1) is preferable from a theoretical standpoint, but Form (2) is easier to apply and is the only documented form of equation for some of the inlet control equations. Equations (A.1) and (A.2) apply up to about Q/(A*D0.5) = 3.5 (1.93 SI).

 Form (1) HWi/D = Hc/D + K*[(Ku*Q)/(A*D0.5)]M + Ks*S (A.1) Form (2) HWi/D = K*[(Ku*Q)/(A*D0.5)]M (A.2)
 Where: HWi Headwater depth above inlet control section invert, m (ft) D Interior height of culvert barrel, ft (m) Hc Specific head at critical depth (dc + Vc2/(2*g)), ft (m) Q Discharge, ft3/s (m3/s) A Full cross sectional area of culvert barrel, ft2 (m2) S Culvert barrel slope, ft/ft (m/m) K, M, c, Y Constants from Tables A.1, A.2, A.3 Ku Unit conversion 1.0 (1.811 SI) Ks Slope correction, -0.5 (mitered inlets +0.7)

A.2.2 Submerged Inlet Control Equations

The submerged equation (A.3) applies above about Q/(A*D0.5) = 4.0 (2.21 SI). The terms are defined in Section A.2.1.

 HWi/D = c*[(Ku*Q)/(A*D0.5)]2 + Y + Ks*S (A.3)

A.3 INLET CONTROL DIMENSIONLESS DESIGN CURVES

The equations in Section A.2 may be used to develop design curves for any conduit shape or size. Careful examination of the equation constants for a given form of equation reveals that there is very little difference between the constants for a given inlet configuration. Therefore, given the necessary conduit geometry for a new shape from the manufacturer, a similar shape is chosen and the constants are used to develop new design curves. The curves may be quasi- dimensionless, in terms of Q/(A*D0.5) and HWi/D, or dimensional, in terms of Q and HWi for a particular conduit size. To make the curves truly dimensionless, Q/(A*D0.5) must be divided by g0.5, but this produces small decimal numbers. Note that coefficients for rectangular (box) shapes should not be used for nonrectangular (circular, arch, pipe-arch, etc.) shapes and vice- versa. A constant slope value of 2 percent (0.02) is usually selected for the development of design curves. This is because the slope effect is small and the resultant headwater is conservatively high for sites with slopes exceeding 2 percent (except for mitered inlets). The procedure is illustrated in Section A.3.1.

A.3.1 Elliptical Structural Plate Example

Develop a dimensionless design curve for elliptical structural plate corrugated metal culverts, with the long axis horizontal. Assume a thin wall projecting inlet. Use the coefficients and exponents for a corrugated metal pipe-arch, a shape similar to an ellipse.

From Table A.1, Chart 34, Scale 3:

• Unsubmerged: equation Form (1) with K = 0.0340 M = 1.5
• Submerged: c = 0.0496 and Y = 0.53

Unsubmerged, equation Form (1) (Equation A.1):

HWi/D = Hc/D + 0.0340*(Q/(A*D0.5))1.5 – 0.5*0.02

Submerged (Equation A.3):

HWi/D = 0.0496*(Q/(A*D0.5))2 + 0.53 – 0.5*0.02

A direct relationship between HWi/D and Q/(A*D0.5) may be obtained for the submerged condition. For the unsubmerged condition, it is necessary to obtain the flow rate and equivalent specific head at critical depth. At critical depth, the critical velocity head is equal to one-half the hydraulic depth.

Vc2/(2*g) = yh/2 = Ap/2*Tp

Therefore, specific head at critical depth divided by D is:

 Hc/D = dc/D + yh/(2*D) (A.4)

Since the Froude number equals 1.0 at critical depth, Vc can be determuned from the Froude number equation and set equal to Vc in the continuity equation to solve for Qc.

Fr = Vc/(g*yh)0.5 = 1, and Vc = Qc/Ap = (g*yh)0.5, from which Qc = Ap*(g*yh)0.5, or

 Qc/(A*D0.5) = Ap/A*(g•(yh/D))0.5 (A.5)

From geometric data supplied by the manufacturer for a horizontal ellipse (Kaiser 1984), the necessary geometry is obtained to calculate Hc/D and Qc/(A*D0.5).

dc/D yh/D (Equation A.4)
Hc/D
Ap/A (Equation A.5)
Qc/(A*D0.5)
0.1 0.04 0.12 0.04 0.05
0.2 0.14 0.27 0.14 0.30
0.4 0.30 0.55 0.38 1.18
0.6 0.49 0.84 0.64 2.54
0.8 0.85 1.22 0.88 4.60
0.9 1.27 1.53 0.97 6.20
1.0 1.00

From unsubmerged equation and the above table:

Qc/(A*D0.5) 0.0340*
(Qc/(A*D0.5)1.5
+Hc/D -0.5*S = HWi/D
0.05 0.0004 0.12 0.01 0.11
0.30 0.0054 0.27 0.01 0.27
1.18 0.044 0.55 0.01 0.58
2.54 0.138 0.84 0.01 0.97
4.60 0.336 1.22 0.01 1.54
6.20 0.525 1.53 0.01 2.05

For the submerged equation, any value of Q/(A*D0.5) may be selected, since critical depth is not involved:

Qc/(A*D0.5) 0.0496*
(Qc/(A*D0.5))2
+Y -0.5*S = HWi/D
1 0.05 0.53 0.01 *0.57
2 0.20 0.53 0.01 *0.72
4 0.79 0.53 0.01 1.31
6 1.79 0.53 0.01 2.31
8 3.17 0.53 0.01 3.69
*Obviously unsubmerged

Note that overlapping values of HWi/D were calculated in order to define the transition zone between the unsubmerged and the submerged states of flow. The results of the above calculations are plotted in Figure A.1. A transition line is drawn between the unsubmerged and the submerged curves. The scales are dimensionless in Figure A.1, but the figures could be used to develop dimensional curves for any selected size of elliptical conduit by multiplying: Q/(A*D0.5) by A*D0.5 and HWi/D by D. x
Figure A.1. Dimensionless performance curve for structural plate elliptical conduit, long axis horizontal, thin wall projecting entrance.

A.3.2 Dimensionless Design Charts for Culverts without Coefficientc

The dimensionless inlet control design charts provided for long span arches (Chart 52) and for circular and elliptical pipes (Chart 51) were derived using the inlet control equations in Section A.2, selected constants from Table A.1, conduit geometry obtained from various tables and manufacturer’s information (FHWA 1972b, Kaiser 1984, AISI 1983).

Some inlet configurations have no hydraulic tests. In lieu of such tests, the selected edge conditions should approximate the untested configurations and lead to a good estimate of culvert performance. In some cases, it will be necessary to evaluate the inlet edge configuration at a specific flow depth. For example, some inlets may behave as mitered inlets at low headwaters and as thin wall projecting inlets at high headwaters. The designer must apply engineering judgment in selection of the proper relationships for these major structures.

Unsubmerged Conditions. Equation (A.1) was used to calculate HWi/D for selected inlet edge configurations. The following constants were taken from Table A.1, Chart 34 for pipe- arches, except for the 45 degree beveled edge inlet. These constants were taken from Chart 3, Scale A, for circular pipe. No constants were available from tests on pipe-arch models with beveled edges.

Inlet Edge K M Ku*S
Thin Wall Projecting 0.0340 1.5 -0.01
Mitered to Embankment 0.0300 1.0 +0.01
Square Edge in Headwall 0.0083 2.0 -0.01
Beveled Edge (45° Bevels) 0.0018 2.5 -0.01

Geometric relationships for the circular and elliptical (long axis horizontal) conduits were obtained from Tables 4 and 7 (FHWA 1972b), respectively. Geometric relationships for the high and low profile long span arches were obtained from DP-131 (Kaiser 1984) and the results were checked against tables in AISI handbook (AISI 1983).

Submerged Conditions. Equation (A.3) was used to calculate HWi/D for the same inlet configurations using the following constants:

Inlet Edge c Y Ku*S
Thin Wall Projecting 0.0496 0.53 -0.01
Mitered to Embankment 0.0463 0.75 +0.01
Square Edge in Headwall 0.0496 0.57 – 0.01
Beveled Edge (45° Bevels) 0.0300 0.74 – 0.01

In terms of Q/(A*D0.5), all non-rectangular shapes have practically the same dimensionless curves for submerged, inlet control flow. This is not true if Q/(B*D1.5) is used as the dimensionless flow parameter.

To convert Q/(B*D1.5) to Q/(A*D0.5), divide by A/(B*D) for the particular shape of interest as shown in Equation (A.6). This assumes that the shape is geometrically similar, so that A/(B*D) is nearly constant for a range of sizes.

 (Q/(B*D1.5))/(A/(B*D)) = (Q/(B*D1.5))*(B*D/A) = Q/(A*D0.5) (A.6)

Dimensionless Curves. By plotting the results of the unsubmerged and submerged calculations and connecting the resultant curves with transition lines, the dimensionless design curves shown in Charts 51 and 52 of Appendix C were developed. All high and low profile arches can be represented by a single curve for each inlet edge configuration. A similar set of curves was developed for circular and elliptical shapes. It is recommended that the high and low profile arch curves in Chart 52 be used for all true arch shapes (those with a flat bottom) and that the curves in Chart 51 be used for curved shapes including circles, ellipses, pipe-arches, and pear shapes.

A.3.3 Dimensionless Critical Depth Charts

Some of the long span culverts and special culvert shapes have no critical depth charts. These special shapes are available in numerous sizes, making it impractical to produce individual critical depth curves for each culvert size and shape. Therefore, dimensionless critical depth curves were developed for the shapes which have adequate geometric relationships in the manufacturer’s literature. It should be noted that these special shapes are not truly geometrically similar, and any generalized set of geometric relationships will involve some degree of error. The amount of error is unknown since the geometric relationships were developed by the manufacturers.

The manufacturers’ literature contains geometric relationships which include the hydraulic depth divided by the rise (inside height) of the conduit (yh/D) and area of the flow prism divided by the barrel area (Ap/A) for various partial depth ratios, y/D. From Equation (A.5):

 Q/(A*D0.5) = Ap/A*(g•yh/D)0.5 (A.7)

Setting y/D equal to dc/D, it is possible to determine Ap/A and yh/D at a given relative depth and then to calculate Qc/(A*D0.5). Dimensionless plots of dc/D versus Qc/(A*D0.5) have been developed for the following culvert materials and shapes:

• Chart 20, corrugated metal box culverts (see Second edition HDS 5)
• Chart 44, corrugated metal arches (see Second edition HDS 5)
• Chart 53, Structural plate corrugated metal ellipses, long axis horizontal
• Chart 54, Structural plate corrugated metal arches, low and high profile

A.4 INLET CONTROL NOMOGRAPHS

The nomographs in Appendix C were developed using the equations in Section A.2 and the constants shown in Table A.1. The unsubmerged and submerged equations for a given shape, material and edge configuration were plotted using the dimensionless design curve procedures shown in Section A.3.1. A constant slope value of 2 percent (0.02) was used for the development of these design curves. This is because the slope effect is small and the resultant headwater is conservatively high for sites with slopes exceeding 2 percent (except for mitered inlets). A smooth transition was drawn by hand. This curve was the data used for constructing a nomograph. Dr. F. T. Mavis describes the process of making nomographs in “The Construction of Nomographic Charts” (Mavis 1939). Nomographs were used extensively in engineering prior to the introduction of microcomputers in the early 1980s.

In formulating inlet and outlet control design nomographs, a certain degree of error is introduced into the design process. This error is due to the fact that the nomograph construction involves graphical fitting techniques resulting in scales which do not exactly match the equations. Checks by the authors of the first edition and others indicate that all of the nomographs from HEC-5 have precisions of + 10 percent of the equation values in terms of headwater (inlet control) or head loss (outlet control). The nomographs for tapered inlets have errors of less than 5%, again in terms of headwater.

A.5 INLET CONTROL POLYNOMIAL EQUATIONS

The polynomial equations were developed to be used in software. The equations in Section A.2 with the constants shown in the tables of constants for a given shape, material and edge configuration were plotted using the dimensionless design curve procedures shown in Section A.3.1. The coordinates of selected points can be read from the curve and a best fit statistical analysis performed. A polynomial equation of the following form has been found to provide an adequate fit.

HWi/D = A + B*[Q/(B*D1.5)] + C*[Q/(B*D1.5)]2 + … + X*[Q/(B*D1.5)]n + Ks*S

For fitting the polynomial equations, Ks = 0 was used for most equations so that the slope correction could be applied by the software. The flow factor can be based on A*D0.5 rather than B*D1.5. The constants for the best fit equations are found in the HY-8 User Manual provided with HY-8. For equations that included Ks*S, the A term is adjusted so that Ku*S = 0. HY-8 uses the polynomial equations for all shapes that have constants determined in the laboratory or by FHWA. These include:

• Table A.1 – circles, boxes and tapered inlets (NBS, Bossy 1961)
• Table A.2 – pipe-arches, ellipses, metal boxes and arches (Bossy 1961)
• Table A.3 – South Dakota DOT RCB (FHWA 2006c)
• Table A.4 – open bottom concrete boxes (Chase 1999)
• Table A.5 – embedded circular shapes (NCHRP 2011)
• Table A.6 – embedded elliptical shapes (NCHRP 2011)

For shapes without constants, HY-8 uses Chart 52 developed using the procedures of Section A.3.2.

Note From HDS-5 Section 3.1.3 Inlet Control

The original equations for computer software were generally 5th order polynomial curve-fitted equations that were developed to be as accurate as the nomograph solution (plus or minus 10%) within the headwater range of 0.5*D to 3.0*D. These equations are still being used in HY-8, but have been supplemented with a weir equation from 0.0*D to 0.5*D and an orifice equation above 3.0*D.

Table A.1 Constants for Inlet Control Equations for Charts in Appendix G.
Chart   Nomograph   Equation Unsubmerged Submerged
No. Shape and Material Scale Inlet Configuration Form K M c Y References
1 Circular Concrete 1 Square edge w/ headwall 1 0.0098 2.0 0.0398 0.67 1, 2
1 Circular Concrete 2 Groove end w/ headwall 1 0.0018 2.0 0.0292 0.74 1, 2
1 Circular Concrete 3 Groove end projecting 1 0.0045 2.0 0.0317 0.69 1, 2
2 Circular CMP 1 Headwall 1 0.0078 2.0 0.0379 0.69 1, 2
2 Circular CMP 2 Mitered to slope 1 0.0210 1.33 0.0463 0.75 1, 2
2 Circular CMP 3 Projecting 1 0.0340 1.50 0.0553 0.54 1, 2
3 Circular A Beveled ring, 45° bevels 1 0.0018 2.50 0.0300 0.74 2
3 Circular B Beveled ring, 33.7° bevels* 1 0.0018 2.50 0.0243 0.83 2
8 Rect. Box Concrete 1 30° to 75° wingwall flares 1 0.026 1.0 0.0347 0.81 1, 3
8 Rect. Box Concrete 2 90° and 15° wingwall flares 1 0.061 0.75 0.0400 0.80 1, 3
8 Rect. Box Concrete 3 0° wingwall flares 1 0.061 0.75 0.0423 0.82 1, 3
9 Rect. Box Concrete 1 45° wingwall flare d = .043D 2 0.510 0.667 0.0309 0.80 3
9 Rect. Box Concrete 2 18° to 33.7° wingwall flare d = .083D 2 0.486 0.667 0.0249 0.83 3
10 Rect. Box Concrete 1 90° headwall w/3/4″ chamfers 2 0.515 0.667 0.0375 0.79 3
10 Rect. Box Concrete 2 90° headwall w/45° bevels 2 0.495 0.667 0.0314 0.82 3
10 Rect. Box Concrete 3 90° headwall w/33.7° bevels 2 0.486 0.667 0.0252 0.865 3
11 Rect. Box Concrete 1 3/4″ chamfers; 45° skewed headwall 2 0.545 0.667 .04505 0.73 3
11 Rect. Box Concrete 2 3/4″ chamfers; 30° skewed headwall 2 0.533 0.667 0.0425 0.705 3
11 Rect. Box Concrete 3 3/4″ chamfers; 15° skewed headwall 2 0.522 0.667 0.0402 0.68 3
11 Rect. Box Concrete 4 45° bevels; 10°-45° skewed headwall 2 0.498 0.667 0.0327 0.75 3
12 Rect. Box 3/4″ chamf. Conc. 1 45° non-offset wingwall flares 2 0.497 0.667 0.0339 0.803 3
12 Rect. Box 3/4″ chamf. Conc. 2 18.4° non-offset wingwall flares 2 0.493 0.667 0.0361 0.806 3
12 Rect. Box 3/4″ chamf. Conc. 3 18.4° non-offset wingwall flares 30° skewed barrel 2 0.495 0.667 0.0386 0.71 3
13 Rec. Box Top Bev. Conc. 1 45° wingwall flares – offset 2 0.497 0.667 0.0302 0.835 3
13 Rec. Box Top Bev. Conc. 2 33.7° wingwall flares – offset 2 0.495 0.667 0.0252 0.881 3
13 Rec. Box Top Bev. Conc. 3 18.4° wingwall flares – offset 2 0.493 0.667 0.0227 0.887 3
55 Circular 1 Smooth tapered inlet throat 2 0.534 0.555 0.0196 0.90 4
55 Circular 2 Rough tapered inlet throat 2 0.519 0.64 0.0210 0.90 4
56 Elliptical Face 1 Tapered inlet – beveled edges 2 0.536 0.622 0.0368 0.83 4
56 Elliptical Face 2 Tapered inlet – square edges 2 0.5035 0.719 0.0478 0.80 4
56 Elliptical Face 3 Tapered inlet – thin edge projecting 2 0.547 0.80 0.0598 0.75 4
57 Rectangular Concrete 1 Tapered inlet throat 2 0.475 0.667 0.0179 0.97 4
58 Rectangular Concrete 1 Side tapered – less favorable edges 2 0.56 0.667 0.0446 0.85 4
58 Rectangular Concrete 2 Side tapered – more favorable edges 2 0.56 0.667 0.0378 0.87 4
59 Rectangular Concrete 1 Slope tapered – less favorable edges 2 0.50 0.667 0.0446 0.65 4
59 Rectangular Concrete 2 Slope tapered – more favorable edges 2 0.50 0.667 0.0378 0.71 4
1Bossy 1963, 2FHWA 1974, 3NBS 5th, 4HEC-13

Table A.2 Constants for Inlet Control Equations for Discontinued Charts (see HDS-5).
Chart   Nomograph   Equation Unsubmerged Submerged
No. Shape and Material Scale Inlet Configuration Form K M c Y References
16-19 Boxes CM 2 90° headwall 1 0.0083 2.0 0.0379 0.69 1
16-19 Boxes CM 3 Thick wall projecting 1 0.0145 1.75 0.0419 0.64 1
16-19 Boxes CM 5 Thin wall projecting 1 0.0340 1.5 0.0496 0.57 1
29 Horizontal Ellipse Concrete 1 Square edge w/ headwall 1 0.0100 2.0 0.0398 0.67 1
29 Horizontal Ellipse Concrete 2 Groove end w/ headwall 1 0.0018 2.5 0.0292 0.74 1
29 Horizontal Ellipse Concrete 3 Groove end projecting 1 0.0045 2.0 0.0317 0.69 1
30 Vertical Ellipse Concrete 1 Square edge w/ headwall 1 0.0100 2.0 0.0398 0.67 1
30 Vertical Ellipse Concrete 2 Groove end w/ headwall 1 0.0018 2.5 0.0292 0.74 1
30 Vertical Ellipse Concrete 3 Groove end projecting 1 0.0095 2.0 0.0317 0.69 1
34 Pipe Arch 18″ Corner radius CM 1 90° headwall 1 0.0083 2.0 0.0379 0.69 1
34 Pipe Arch 18″ Corner radius CM 2 Mitered to slope 1 0.0300 1.0 0.0463 0.75 1
34 Pipe Arch 18″ Corner radius CM 3 Projecting 1 0.0340 1.5 0.0496 0.57 1
35 Pipe Arch 18″ Corner radius CM 1 Projecting 1 0.0300 1.5 0.0496 0.57 2
35 Pipe Arch 18″ Corner radius CM 2 No Bevels 1 0.0088 2.0 0.0368 0.68 2
35 Pipe Arch 18″ Corner radius CM 3 33.7° Bevels 1 0.0030 2.0 0.0269 0.77 2
36 Pipe Arch 31″ Corner radius CM 1 Projecting 1 0.0300 1.5 0.0496 0.57 2
36 Pipe Arch 31″ Corner radius CM 2 No Bevels 1 0.0088 2.0 0.0368 0.68 2
36 Pipe Arch 31″ Corner radius CM 3 33.7° Bevels 1 0.0030 2.0 0.0269 0.77 2
41-43 Arch CM 1 90° headwall 1 0.0083 2.0 0.0379 0.69 1
41-43 Arch CM 2 Mitered to slope 1 0.0300 1.0 0.0463 0.75 1
41-43 Arch CM 3 Thin wall projecting 1 0.0340 1.5 0.0496 0.57 1
1FHWA 1974, 2Bossy 1963

Table A.3. Constants for Inlet Control Equations for South Dakota Concrete Box (HY-8 User Manual and Table 11 of FHWA 2006).
Sketch Wingwall Flare Top Bevel Top Radius Corner Fillet RCB Inlet Configuration Equation Form Unsub- merged K Unsub- merged M Sub- merged c Sub- merged Y
1 30° 45° Single barrel 2 0.44 0.74 0.040 0.48
2 30° 45° 6″ Multiple barrel (2, 3, and 4 cells) 2 0.47 0.68 0.04 0.62
3 30° 45° Single barrel (2:1 to 4:1 span-to-rise ratio) 2 0.48 0.65 0.041 0.57
4 30° 45° Multiple barrels (15°skewed headwall) 2 0.69 0.49 0.029 0.95
5 30° 45° Multiple barrels (30° to 45° skewed headwall) 2 0.69 0.49 0.027 1.02
6 none Single barrel, top edge 90° 2 0.55 0.64 0.047 0.55
7 45° 6″ Single barrel, (0 and 6-inch corner fillets) 2 0.56 0.62 0.045 0.55
8 45° 6″ Multiple barrels (2, 3, and 4 cells) 2 0.55 0.59 0.038 0.69
9 45° Single barrels 2:1 to 4:1 span-to-rise ratio) 2 0.61 0.57 0.041 0.67
10 8″ 6″ Single barrel (0 and 6-inch fillets) 2 0.56 0.62 0.038 0.67
11 8″ 12″ Single barrel (12-inch corner fillets) 2 0.56 0.62 0.038 0.67
12 8″ 12″ Multiple barrels (2, 3, and 4 cells) 2 0.55 0.6 0.023 0.96
13 8″ 12″ Single barrel (2:1 to 4:1 span-to-rise ratio) 2 0.61 0.57 0.033 0.79

Sketches are shown in the HY-8 documentation and research report. Since sketches 2 and 8 show fillets, a 6-inch fillet is assumed. Sketches 1 through 5 have the first configuration. Sketches 7 through 13 have the second.

Table A.4. Constants for Inlet Control Equations for Concrete Open-Bottom Arch (Chase 1999).
Span to Rise1 Wingwall Flare Top Edge Inlet Configuration Equation Form Unsub- merged K Unsub- merged M Sub- merged c Sub- merged Y
2:1 90° Mitered to conform to slope 2 0.44 0.74 0.040 0.48
2:1 45° 90° Headwall with wingwalls 2 0.47 0.68 0.04 0.62
2:1 90° 90° Headwall 2 0.48 0.65 0.041 0.57
4:1 90° Mitered to conform to slope 2 0.69 0.49 0.029 0.95
4:1 45° 90° Headwall with wingwalls 2 0.69 0.49 0.027 1.02
4:1 90° 90° Headwall 2 0.56 0.62 0.045 0.55
1The 2:1 constants are used for ratios less than or equal to 3:1 and the 4:1 constants for ratios greater than 3:1.

Table A.5. Constants for Inlet Control Equations for Embedded Circular Shapes (NCHRP 15-24).
Embedded Top Edge Inlet Configuration Unsub- merged K Form 1 Unsub- merged M Form 1 Unsub- merged K Form 2 Unsub- merged M Form 2 Sub- merged c Sub- merged Y
0.2D thin Projecting End, Ponded 0.0860 0.58 0.4293 0.64 0.0303 0.58
0.2D thin Projecting End, Channelized 0.0737 0.45 0.4175 0.62 0.0250 0.63
0.2D Mitered End 1.5H:1V 0.0431 0.58 0.4002 0.63 0.0235 0.61
0.2D 90° Square Headwall 0.0566 0.44 0.4001 0.63 0.0198 0.69
0.2D 45° Beveled End 0.0292 0.57 0.3869 0.63 0.0161 0.73
0.4D thin Projecting End, Ponded 0.0840 0.76 0.4706 0.69 0.0453 0.69
0.4D thin Projecting End, Channelized 0.0927 0.59 0.4789 0.66 0.0441 0.52
0.4D Mitered End 1.5H:1V 0.0317 0.77 0.4185 0.68 0.0363 0.65
0.4D 90° Square Headwall 0.0490 0.71 0.4354 0.68 0.0332 0.67
0.4D 45° Beveled End 0.0358 0.62 0.4223 0.67 0.0245 0.75
0.5D thin Projecting End, Ponded 0.1057 0.69 0.4955 0.71 0.0606 0.54
0.5D thin Projecting End, Channelized 0.1055 0.59 0.4955 0.69 0.0570 0.48
0.5D Mitered End 1.5H:1V 0.0351 0.59 0.4419 0.68 0.0504 0.44
0.5D 90° Square Headwall 0.0595 0.59 0.0595 0.59 0.0402 0.65
0.5D 45° Beveled End 0.0464 0.46 0.4364 0.69 0.0324 0.67

Table A.6. Constants for Inlet Control Equations for Embedded Elliptical Shape (NCHRP 15-24).
Embedded Top Edge Inlet Configuration Unsubmerged K Form1 Unsubmerged M Form 1 Unsubmerged K Form 2 Unsubmerged M Form 2 Submerged c Sub- merged Y
0.5D thin Projecting End, Ponded 0.1231 0.51 0.5261 0.65 0.0643 0.50
0.5D thin Projecting End, Channelized 0.0928 0.54 0.4937 0.67 0.0649 0.12
0.5D Mitered End 1.5H:1V 0.0599 0.60 0.4820 0.67 0.0541 0.50
0.5D 90° Square Headwall 0.0819 0.45 0.4867 0.66 0.0431 0.61
0.5D 45° Beveled End 0.0551 0.52 0.4663 0.63 0.0318 0.68

# HEC 14: Riprap Apron

10.2 RIPRAP APRON

The most commonly used device for outlet protection, primarily for culverts 60 in (1500 mm) or smaller, is a riprap apron. An example schematic of an apron taken from the Central Federal Lands Division of the Federal Highway Administration is shown in Figure 10.4.  Figure 10.4. Placed Riprap at Culverts (per Central Federal Lands Highway Division Detail C251-50). Click images to enlarge.

They are constructed of riprap or grouted riprap at a zero grade for a distance that is often related to the outlet pipe diameter. These aprons do not dissipate significant energy except through increased roughness for a short distance. However, they do serve to spread the flow helping to transition to the natural drainage way or to sheet flow where no natural drainage way exists. However, if they are too short, or otherwise ineffective, they simply move the location of potential erosion downstream. The key design elements of the riprap apron are the riprap size as well as the length, width, and depth of the apron.

Several relationships have been proposed for riprap sizing for culvert aprons and several of these are discussed in greater detail in Appendix D of HEC-14. The independent variables in these relationships include one or more of the following variables: outlet velocity, rock specific gravity, pipe dimension (e.g. diameter), outlet Froude number, and tailwater. The following equation (Fletcher and Grace, 1972) is recommended for circular culverts:

 D50 = 0.2⋅D(Q/(√g⋅D2.5))4/3(D/TW) (10.4)

where,

• D50 = riprap size, m (ft)
• Q = design discharge, m3/s (ft3/s)
• D = culvert diameter (circular), m (ft)
• TW = tailwater depth, m (ft)
• g = acceleration due to gravity, 9.81 m/s2 (32.2 ft/s2)

Tailwater depth for Equation 10.4 should be limited to between 0.4D and 1.0D. If tailwater is unknown, use 0.4D.

Whenever the flow is supercritical in the culvert, the culvert diameter is adjusted as follows:

 D’ = (D + yn)/2 (10.5)

where,

• D’ = adjusted culvert rise, m (ft)
• yn = normal (supercritical) depth in the culvert, m (ft)

Equation 10.4 assumes that the rock specific gravity is 2.65. If the actual specific gravity differs significantly from this value, the D50 should be adjusted inversely to specific gravity.

The designer should calculate D50 using Equation 10.4 and compare with available riprap classes. A project or design standard can be developed such as the example from the Federal Highway Administration Federal Lands Highway Division (FHWA, 2003) shown in Table 10.1 (first two columns). The class of riprap to be specified is that which has a D50 greater than or equal to the required size. For projects with several riprap aprons, it is often cost effective to use fewer riprap classes to simplify acquiring and installing the riprap at multiple locations. In such a case, the designer must evaluate the tradeoffs between over sizing riprap at some locations in order to reduce the number of classes required on a project.

Class D50 (mm) D50 (in) Apron Length1 Apron Depth
Table 10.1 Example Riprap Classes and Apron Dimensions
1 125 4 4⋅D 3.5⋅D50
2 150 6 4⋅D 3.3⋅D50
3 250 10 5⋅D 2.4⋅D50
4 350 14 6⋅D 2.2⋅D50
5 500 20 7⋅D 2.0⋅D50
6 550 22 8⋅D 2.0⋅D50
1D is the culvert rise.

The apron dimensions must also be specified. Table 10.1 provides guidance on the apron length and depth. Apron length is given as a function of the culvert rise and the riprap size. Apron depth ranges from 3.5⋅D50 for the smallest riprap to a limit of 2.0⋅D50 for the larger riprap sizes. The final dimension, width, may be determined using the 1:3 flare shown in Figure 10.4 and should conform to the dimensions of the downstream channel. A filter blanket should also be provided as described in HEC 11 (Brown and Clyde, 1989).

For tailwater conditions above the acceptable range for Equation 10.4 (TW > 1.0⋅D), Figure 10.3 should be used to determine the velocity downstream of the culvert. The guidance in Section 10.3 may be used for sizing the riprap. The apron length is determined based on the allowable velocity and the location at which it occurs based on Figure 10.3.

Over their service life, riprap aprons experience a wide variety of flow and tailwater conditions. In addition, the relations summarized in Table 10.1 do not fully account for the many variables in culvert design. To ensure continued satisfactory operation, maintenance personnel should inspect them after major flood events. If repeated severe damage occurs, the location may be a candidate for extending the apron or another type of energy dissipator.

Design Example: Riprap Apron (CU)

Design a riprap apron for the following CMP installation. Available riprap classes are provided in Table 10.1. Given:

• Q = 85 ft3/s
• D = 5.0 ft
• TW = 1.6 ft

Solution

Step 1. Calculate D50 from Equation 10.4. First verify that tailwater is within range.

TW/D = 1.6/5.0 = 0.32. This is less than 0.4⋅D, therefore, use TW = 0.4⋅D = 0.4⋅5 = 2.0 ft.

D50 = 0.2⋅D(Q/(√g⋅D2.5))4/3(D/TW) = 0.2⋅5.0(85/(√32.2⋅5.02.5))4/3(5.0/2.0) = 0.43 ft = 5.2 in.

Step 2. Determine riprap class. From Table 10.1, riprap class 2 (D50 = 6 in) is required.

Step 3. Estimate apron dimensions.

From Table 10.1 for riprap class 2,

• Length, L = 4⋅D = 4⋅5 = 20 ft
• Depth = 3.3⋅D50 = 3.3⋅6 = 19.8 in = 1.65 ft
• Width (at apron end) = 3⋅D + (2/3)⋅L = 3⋅5 + (2/3)⋅20 = 28.3 ft

Design Example: Riprap Apron (SI)

Design a riprap apron for the following CMP installation. Available riprap classes are provided in Table 10.1. Given:

• Q = 2.33 m3/s
• D = 1.5 m
• TW = 0.5 m

Solution

Step 1. Calculate D50 from Equation 10.4. First verify that tailwater is within range.

TW/D = 0.5/1.5 = 0.33. This is less than 0.4⋅D, therefore, use TW = 0.4⋅D = 0.4⋅1.5 = 0.6 m.

D50 = 0.2⋅D(Q/(√g⋅D2.5))4/3(D/TW) = 0.2⋅1.5(2.33/(√9.81⋅1.52.5))4/3(1.5/0.6) = 0.13 m.

Step 2. Determine riprap class. From Table 10.1, riprap class 2 (D50 = 0.15 m) is required.

Step 3. Estimate apron dimensions.

From Table 10.1 for riprap class 2,

• Length, L = 4⋅D = 4⋅1.5 = 6 m
• Depth = 3.3⋅D50 = 3.3⋅0.15 = 0.50 m
• Width (at apron end) = 3⋅D + (2/3)⋅L = 3⋅1.5 + (2/3)⋅6 = 8.5 m

# HEC-15: Vegetative Lining Design

CHAPTER 4: VEGETATIVE LINING AND BARE SOIL DESIGN

Grass-lined channels have been widely used in roadway drainage systems for many years. They are easily constructed and maintained and work well in a variety of climates and soil conditions. Grass linings provide good erosion protection and can trap sediment and related contaminants in the channel section. Routine maintenance of grass-lined channels consists of mowing, control of weedy plants and woody vegetation, repair of damaged areas and removal of sediment deposits.

The behavior of grass in an open channel lining is complicated by the fact that grass stems bend as flow depth and shear stress increase. This reduces the roughness height and increases velocity and flow rate. For some lining materials (bare earth and rigid linings), the roughness height remains constant regardless of the velocity or depth of flow in the channel. As a result, a grass-lined channel cannot be described by a single roughness coefficient.

The Soil Conservation Service (SCS) (1954) developed a widely used classification of grass channel lining that depends on the degree of retardance. In this classification, retardance is a function of the height and density of the grass cover (USDA, 1987). Grasses are classified into five broad categories, as shown in Table 4.1. Retardance Class A presents the highest resistance to flow and Class E presents the lowest resistance to flow. In general, taller and denser grass species have a higher resistance to flow, while short flexible grasses have a low flow resistance.

Kouwen and Unny (1969) and Kouwen and Li (1981) developed a useful model of the biomechanics of vegetation in open-channel flow. This model provides a general approach for determining the roughness of vegetated channels compared to the retardance classification. The resulting resistance equation (see HEC-15 Appendix C.2) uses the same vegetation properties as the SCS retardance approach, but is more adaptable to the requirements of highway drainage channels. The design approach for grass-lined channels was developed from the Kouwen resistance equation.

Grass linings provide erosion control in two ways. First, the grass stems dissipate shear force within the canopy before it reaches the soil surface. Second, the grass plant (both the root and stem) stabilizes the soil surface against turbulent fluctuations. Temple (SCS, 1954) developed a relationship between the total shear on the lining and the shear at the soil surface based on both processes.

A simple field method is provided to directly measure the density-stiffness parameter of a grass cover. Grass linings for roadside ditches use a wide variety of seed mixes that meet the regional requirements of soil and climate. These seed mix designs are constantly being adapted to improve grass-lined channel performance. Maintenance practices can significantly influence density and uniformity of the grass cover. The sampling of established grasses in roadside ditch application can eliminate much of the uncertainty in lining performance and maintenance practices.

Expertise in vegetation ecology, soil classification, hydrology, and roadway maintenance is required in the design of grass-lined channels. Engineering judgment is essential in determining design parameters based on this expert input. This includes factoring in variations that are unique to a particular roadway design and its operation.

Table 4.1. Retardance Classification of Vegetal Covers
Retardance Class Cover1 Condition
A Weeping Love Grass Excellent stand, tall, average 760 mm (30 in)
Yellow Bluestem Ischaemum Excellent stand, tall, average 910 mm (36 in)
B Kudzu Very dense growth, uncut
Bermuda Grass Good stand, tall, average 300 mm (12 in)
Native Grass Mixture (little bluestem, bluestem, blue gamma, and other long and short midwest grasses) Good stand, unmowed
Weeping lovegrass Good stand, tall, average 610 mm (24 in)
Lespedeza sericea Good stand, not woody, tall, average 480 mm (19 in)
Alfalfa Good stand, uncut, average 280 mm (11 in)
Weeping lovegrass Good stand, unmowed, average 330 mm (13 in)
Kudzu Dense growth, uncut
Blue Gamma Good stand, uncut, average 280 mm (11 in)
C Crabgrass Fair stand, uncut 250 to 1200 mm (10 to 48 in)
Bermuda grass Good stand, mowed, average 150 mm (6 in)
Common Lespedeza Good stand, uncut, average 280 mm (11 in)
Grass-Legume mixture–summer (orchard grass, redtop, Italian ryegrass, and common lespedeza) Good stand, uncut, 150 to 200 mm (6 to 8 in)
Centipede grass Very dense cover, average 150 mm (6 in)
Kentucky Bluegrass Good stand, headed, 150 to 300 mm (6 to 12 in)
D Bermuda Grass Good stand, cut to 60 mm (2.5 in) height
Common Lespedeza Excellent stand, uncut, average 110 mm (4.5 in)
Buffalo Grass Good stand, uncut, 80 to 150 mm (3 to 6 in)
Grass-Legume mixture—fall, spring (orchard grass, redtop, Italian ryegrass, and common lespedeza) Good stand, uncut, 100 to 130 mm (4 to 5 in)
Lespedeza sericea After cutting to 50 mm (2 in) height. Very good stand before cutting.
E Bermuda Grass Good stand, cut to height, 40 mm (1.5 in)
Bermuda Grass Burned stubble
1Covers classified have been tested in experimental channels. Covers were green and generally uniform.

4.1 GRASS LINING PROPERTIES

The density, stiffness, and height of grass stems are the main biomechanical properties of grass that relate to flow resistance and erosion control. The stiffness property (product of elasticity and moment of inertia) of grass is similar for a wide range of species (Kouwen, 1988) and is a basic property of grass linings.

Density is the number of grass stems in a given area, i.e., stems per m2 (ft2). A good grass lining will have about 2,000 to 4,000 stems/m2 (200 to 400 stems/ft2). A poor cover will have about one-third of that density and an excellent cover about five-thirds (USDA, 1987, Table 3.1). While grass density can be determined by physically counting stems, an easier direct method of estimating the density-stiffness property is provided in Appendix E of HEC-15.

For agricultural ditches, grass heights can reach 0.3 m (1.0 ft) to over 1.0 m (3.3 ft). However, near a roadway grass heights are kept much lower for safety reasons and are typically in the range of 0.075 m (0.25 ft) to 0.225 m (0.75 ft).

The density-stiffness property of grass is defined by the Cs coefficient. Cs can be directly measured using the Fall-Board test (Appendix E) or estimated based on the conditions of the grass cover using Table 4.2. Good cover would be the typical reference condition.

Table 4.2. Density-stiffness Coefficient, Cs
Conditions Excellent Very Good Good Fair Poor
Cs (SI) 580 290 106 24 8.6
Cs (CU) 49 25 9.0 2.0 0.73

The combined effect of grass stem height and density-stiffness is defined by the grass roughness coefficient.

 Cn=α•Cs0.10h0.528 (4.1)

where:

 Cn = grass roughness coefficient Cs = density-stiffness coefficient h = stem height, m (ft) α = unit conversion constant, 0.35 (SI), 0.237 (CU)

Table 4.3 provides Cn values for a range of cover and stem height conditions based on Equation 4.1. Denser cover and increased stem height result in increased channel roughness.

Table 4.3. Grass Roughness Coefficient, Cn
Stem Height m (ft) Excellent Very Good Good Fair Poor
0.075 (0.25) 0.168 0.157 0.142 0.122 0.111
0.150 (0.50) 0.243 0.227 0.205 0.177 0.159
0.225 (0.75) 0.301 0.281 0.254 0.219 0.197

SCS retardance values relate to a combination of grass stem-height and density. Cn values for standard retardance classes are provided in Table 4.4. Comparing Table 4.3 and 4.4 shows that retardance classes A and B are not commonly found in roadway applications. These retardance classes represent conditions where grass can be allowed to grow much higher than would be permissible for a roadside channel, e.g., wetlands and agricultural ditches. Class E would not be typical of most roadside channel conditions unless they were in a very poor state.

The range of Cn for roadside channels is between 0.10 and 0.30 with a value of 0.20 being common to most conditions and stem heights. In an iterative design process, a good first estimate of the grass roughness coefficient would be Cn = 0.20.

Table 4.4 (SI). Grass Roughness Coefficient, Cn, for SCS Retardance Classes
Retardance Class A B C D E
Stem Height, mm 910 610 200 100 40
Cs 390 81 47 33 44
Cn 0.605 0.418 0.220 0.147 0.093
Table 4.4 (CU). Grass Roughness Coefficient, Cn, for SCS Retardance Classes
Retardance Class A B C D E
Stem Height, in 36 24 8.0 4.0 1.6
Cs 33 7.1 3.9 2.7 3.8
Cn 0.605 0.418 0.220 0.147 0.093

4.2 MANNING’S ROUGHNESS

Manning’s roughness coefficient for grass linings varies depending on grass properties as reflected in the Cn parameter and the shear force exerted by the flow. This is because the applied shear on the grass stem causes the stem to bend, which reduces the stem height relative to the depth of flow and reducing the roughness.

 n = α•Cn•τ-0.4 (4.2)

where,

 τo = mean boundary shear stress, N/m2 (lb/ft2) α = unit conversion constant, 1.0 (SI), 0.213 (CU)

See Appendix C.2 for the derivation of Equation 4.2.

4.3 PERMISSIBLE SHEAR STRESS

The permissible shear stress of a vegetative lining is determined both by the underlying soil properties as well as those of the vegetation. Determination of permissible shear stress for the lining is based on the permissible shear stress of the soil combined with the protection afforded by the vegetation, if any.

4.3.1 Effective Shear Stress

Grass lining moves shear stress away from the soil surface. The remaining shear at the soil surface is termed the effective shear stress. When the effective shear stress is less than the allowable shear for the soil surface, then erosion of the soil surface will be controlled. Grass linings provide shear reduction in two ways. First, the grass stems dissipate shear force within the canopy before it reaches the soil surface. Second, the grass plant (both the root and stem) stabilizes the soil surface against turbulent fluctuations. This process model (USDA, 1987) for the effective shear at the soil surface is given by the following equation.

 τe = τd•(1-Cf)•(ns/n)2 (4.3)

where,

 τe = effective shear stress on the soil surface, N/m2 (lb/ft2) τd = design shear stress, N/m2 (lb/ft2) Cf = grass cover factor ns = soil grain roughness n = overall lining roughness

Soil grain roughness is taken as 0.016 when D75 < 1.3 mm (0.05 in). For larger grain soils, the soil grain roughness is given by:

 ns = α•(D75)1/6 (4.4)

where,

 ns = soil grain roughness (D75 < 1.3 mm (0.05 in)) D75 = soil size where 75% of the material is finer, mm (in) α = unit conversion constant, 0.015 (SI), 0.026 (CU)

Note that soil grain roughness value, ns, is less than the typical value reported in Table 2.1 for a bare soil channel. The total roughness value for bare soil channel includes form roughness (surface texture of the soil) in addition to the soil grain roughness. However, Equation 4.3 is based on soil grain roughness.

The grass cover factor, Cf, varies with cover density and grass growth form (sod or bunch). The selection of the cover factor is a matter of engineering judgment since limited data are available. Table 4.5 provides a reasonable approach to estimating a cover factor based on (USDA, 1987, Table 3.1). Cover factors are better for sod-forming grasses than bunch grasses. In all cases a uniform stand of grass is assumed. Non-uniform conditions include wheel ruts, animal trails and other disturbances that run parallel to the direction of the channel. Estimates of cover factor are best for good uniform stands of grass and there is more uncertainty in the estimates of fair and poor conditions.

Table 4.5. Cover Factor Values for Uniform Stands of Grass
Growth Form Cover Factor, Cf
Excellent Very Good Good Fair Poor
Sod 0.98 0.95 0.90 0.84 0.75
Bunch 0.55 0.53 0.50 0.47 0.41
Mixed 0.82 0.79 0.75 0.70 0.62

4.3.2 Permissible Soil Shear Stress

Erosion of the soil boundary occurs when the effective shear stress exceeds the permissible soil shear stress. Permissible soil shear stress is a function of particle size, cohesive strength, and soil density. The erodibility of coarse non-cohesive soils (defined as soils with a plasticity index of less than 10) is due mainly to particle size, while fine-grained cohesive soils are controlled mainly by cohesive strength and soil density.

New ditch construction includes the placement of topsoil on the perimeter of the channel. Topsoil is typically gathered from locations on the project and stockpiled for revegetation work. Therefore, the important physical properties of the soil can be determined during the design by sampling surface soils from the project area. Since these soils are likely to be mixed together, average physical properties are acceptable for design.

The following sections offer detailed methods for determination of soil permissible shear. However, the normal variation of permissible shear stress for different soils is moderate, particularly for fine-grained cohesive soils. An approximate method is also provided for cohesive soils.

4.3.2.1 Non-cohesive Soils

The permissible soil shear stress for fine-grained, non-cohesive soils (D75 < 1.3 mm (0.05 in)) is relatively constant and is conservatively estimated at 1.0 N/m2 (0.02 lb/ft2). For coarse grained, non-cohesive soils (1.3 mm (0.05 in) < D75 < 50 mm (2 in)) the following equation applies.

 τp,soil = α•D75 (4.5)

where,

 τp,soil = permissible soil shear stress, N/m2 (lb/ft2) D75 = soil size where 75% of the material is finer, mm (in) α = unit conversion constant, 0.75 (SI), 0.4 (CU)

4.3.2.2 Cohesive Soils

Cohesive soils are largely fine grained and their permissible shear stress depends on cohesive strength and soil density. Cohesive strength is associated with the plasticity index (PI), which is the difference between the liquid and plastic limits of the soil. The soil density is a function of the void ratio (e). The basic formula for permissible shear on cohesive soils is the following.

 τp,soil = (c1•PI2 + c2•PI + c3)•(c4 + c5•e)2•c6 (4.6)

where,

 τp,soil = soil permissible shear stress, N/m2 (lb/ft2) PI = plasticity index e = void ratio c1, c2, c3, c4, c5, c6 = coefficients (Table 4.6)

A simplified approach for estimating permissible soil shear stress based on Equation 4.6 is illustrated in Figure 4.1. Fine grained soils are grouped together (GM, CL, SC, ML, SM, and MH) and coarse grained soil (GC). Clays (CH) fall between the two groups.

Higher soil unit weight increases the permissible shear stress and lower soil unit weight decreases permissible shear stress. Figure 4.1 is applicable for soils that are within 5 percent of a typical unit weight for a soil class. For sands and gravels (SM, SC, GM, GC) typical soil unit weight is approximately 1.6 ton/m3 (100 lb/ft3), for silts and lean clays (ML, CL) 1.4 ton/m3 (90 lb/ft3) and fat clays (CH, MH) 1.3 ton/m3 (80 lb/ft3).

Table 4.6. Coefficients for Permissible Soil Shear Stress (USDA, 1987)
ASTM Soil Classification(1) Applicable Range c1 c2 c3 c4 c5 c6 (SI) c6 (CU)
GM 10 < PI < 20 1.07 14.3 47.7 1.42 -0.61 4.8×10-3 10-4
20 < PI     0.076 1.42 -0.61 48. 1.0
GC 10 < PI < 20 0.0477 2.86 42.9 1.42 -0.61 4.8×10-2 10-3
20 < PI     0.119 1.42 -0.61 48. 1.0
SM 10 < PI < 20 1.07 7.15 11.9 1.42 -0.61 4.8×10-3 10-4
20 < PI     0.058 1.42 -0.61 48. 1.0
SC 10 < PI < 20 1.07 14.3 47.7 1.42 -0.61 4.8×10-3 10-4
20 < PI     0.076 1.42 -0.61 48. 1.0
ML 10 < PI < 20 1.07 7.15 11.9 1.48 -0.57 4.8×10-3 10-4
20 < PI     0.058 1.48 -0.57 48. 1.0
CL 10 < PI < 20 1.07 14.3 47.7 1.48 -0.57 4.8×10-3 10-4
20 < PI     0.076 1.48 -0.57 48. 1.0
MH 10 < PI < 20 0.0477 1.43 10.7 1.38 -0.373 4.8×10-2 10-3
20 < PI     0.058 1.38 -0.373 48. 1.0
CH 20 < PI     0.097 1.38 -0.373 48. 1.0
 (1) Note: Typical names GM Silty gravels, gravel-sand silt mixtures GC Clayey gravels, gravel-sand-clay mixtures SM Silty sands, sand-silt mixtures SC Clayey sands, sand-clay mixtures ML Inorganic silts, very fine sands, rock flour, silty or clayey fine sands CL Inorganic clays of low to medium plasticity, gravelly clays, sandy clays, silty clays, lean clays MH Inorganic silts, micaceous or diatomaceous fine sands or silts, elastic silts CH Inorganic clays of high plasticity, fat clays Figure 4.1. Cohesive Soil Permissible Shear Stress

4.3.3 Permissible Vegetation/Soil Shear Stress

The combined effects of the soil permissible shear stress and the effective shear stress transferred through the vegetative lining results in a permissible shear stress for the vegetative lining. Taking Equation 4.3 and substituting the permissible shear stress for the soil for the effective shear stress on the soil, τe, gives the following equation for permissible shear stress for the vegetative lining:

 τp=τp,soil/(1-Cf)•(n/ns)2 (4.7)

where,

 τp = permissible shear stress on the vegetative lining, N/m2 (lb/ft2) τp,soil = permissible soil shear stress, N/m2 (lb/ft2) Cf = grass cover factor ns = soil grain roughness n = overall lining roughness

Design Example: Grass Lining Design (SI)

Evaluate a grass lining for a roadside channel given the following channel shape, soil conditions, grade, and design flow. It is expected that the grass lining will be maintained in good conditions in the spring and summer months, which are the main storm seasons.

Given:

 Shape: Trapezoidal, B = 0.9 m, Z = 3 Soil: Clayey sand (SC classification), PI = 16, e = 0.5 Grass: Sod, height = 0.075 m Grade: 3.0 percent Flow: 0.5 m3/s

Solution

The solution is accomplished using procedure given in Section 3.1 for a straight channel.

 Step 1. Channel slope, shape, and discharge have been given. Step 2. A vegetative lining on a clayey sand soil will be evaluated. Step 3. Initial depth is estimated at 0.30 m From the geometric relationship of a trapezoid (see Appendix B): A = B•d + Z•d2 = 0.9•(0.3) + 3•(0.3)2 = 0.540 m2 P = B + 2•d•√(Z2 + 1) = 0.9 + 2•(0.3)•√(32 + 1) = 2.80 m R = A/P = (0.54)/(2.8) = 0.193 m Step 4. To estimate n, the applied shear stress on the grass lining is given by Equation 2.3 τo = γ•R•So = 9810•(0.193)•(0.03) = 56.8 N/m2 Determine a Manning’s n value from Equation 4.2. From Table 4.3, Cn = 0.142 n = α•Cnτ-0.4 = 1.0•(0.142)•(56.8)-0.4 = 0.028 The discharge is calculated using Manning’s equation (Equation 2.1): Q = α/n•A•R2/3•Sf1/2 = 1/0.028•(0.540)•(0.193)2/3•(0.03)1/2 = 1.12 m3/s Step 5. Since this value is more than 5 percent different from the design flow, we need to go back to step 3 to estimate a new flow depth. Step 3 (2nd iteration). Estimate a new depth solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth: d = 0.21 m Revise the hydraulic radius. A = B•d + Z•d2 = 0.9•(0.21) + 3•(0.21)2 = 0.321 m2 P =B + 2•d•√(Z2 + 1) = 0.9 + 2•(0.21)•√(32 + 1) = 2.23 m R = A/P = (0.321)/(2.23) = 0.144 m Step 4 (2nd iteration). To estimate n, the applied shear stress on the grass lining is given by Equation 2.3 τo = γ•R•So = 9810•(0.144)•(0.03) = 42.4 N/m2 Determine a Manning’s n value from Equation 4.2. From Table 4.3, Cn = 0.142 n = α•Cn•τ-0.4 = 1.0•(0.142)•(42.4)-0.4 = 0.032 The discharge is calculated using Manning’s equation (Equation 2.1): Q = α/n•A•R2/3•Sf1/2 = 1/0.032•0.321)•(0.144)2/3•(0.03)1/2 = 0.48 m3/s Step 5 (2nd iteration). Since this value is within 5 percent of the design flow, we can proceed to step 6. Step 6. The maximum shear on the channel bottom is: τd = γ•d•So = 9810•(0.21)•(0.03) = 61.8 N/m2 Determine the permissible soil shear stress from Equation 4.6. τp,soil = (c1•PI2 + c2•PI + c3)•(c4+c5•e)2•c6 = (1.07•(16)2 + 14.3•(16) + 47.7)(1.42 – 0.61•(0.5))2•(0.0048) = 3.28 N/m2 Equation 4.7 gives the permissible shear stress on the vegetation. The value of Cf is found in Table 4.5. τp = τp,soil/(1-Cf)•(n/ns)2 = 3.28/(1 – 0.9)•(0.032/0.016)2 = 131 N/m2 The safety factor for this channel is taken as 1.0. Step 7. The grass lining is acceptable since the maximum shear on the vegetation is less than the permissible shear of 131 N/m2.

Design Example: Grass Lining Design (CU)

Evaluate a grass lining for a roadside channel given the following channel shape, soil conditions, grade, and design flow. It is expected that the grass lining will be maintained in good conditions in the spring and summer months, which are the main storm seasons.

 Shape: Trapezoidal, B = 3.0 ft, Z = 3 Soil: Clayey sand (SC classification), PI = 16, e = 0.5 Grass: Sod, height = 0.25 ft Grade: 3.0 percent Flow: 17.5 ft3/s

Solution

The solution is accomplished using procedure given in Section 3.1 for a straight channel.

 Step 1. Channel slope, shape, and discharge have been given. Step 2. A vegetative lining on a clayey sand soil will be evaluated. Step 3. Initial depth is estimated at 1.0 ft From the geometric relationship of a trapezoid (see Appendix B): A = B•d + Z•d2 = 3.0•(1.0) + 3•(1.0)2 = 6.00 ft2 P = B + 2•d•√(Z2 + 1) = 3.0 + 2•(1.0)•√(32 + 1) = 9.32 ft R = A/P = (6.00)/(9.32) = 0.643 ft Step 4. To estimate n, the applied shear stress on the grass lining is given by Equation 2.3 τo = γ•R•So = 62.4•(0.643)•(0.03) = 1.20 lb/ft2 Determine a Manning’s n value from Equation 4.2. From Table 4.3, Cn = 0.142 n = α•Cnτ-0.4 = 0.213•(0.142)•(1.20)-0.4 = 0.028 The discharge is calculated using Manning’s equation (Equation 2.1): Q = α/n•A•R2/3•Sf1/2 = 1.49/0.028•(6.00)•(0.643)2/3•(0.03)1/2 = 41.2 ft3/s Step 5. Since this value is more than 5 percent different from the design flow, we need to go back to step 3 to estimate a new flow depth. Step 3 (2nd iteration). Estimate a new depth solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth: d = 0.70 ft Revise the hydraulic radius. A = B•d + Z•d2 = 3.0•(0.70) + 3•(0.70)2 = 3.57 ft2 P =B + 2•d•√(Z2 + 1) = 3.0 + 2•(0.70)•√(32 + 1) = 7.43 ft R = A/P = (3.57)/(7.43) = 0.481 ft Step 4 (2nd iteration). To estimate n, the applied shear stress on the grass lining is given by Equation 2.3 τo = γ•R•So = 62.4•(0.481)•(0.03) = 0.90 lb/ft2 Determine a Manning’s n value from Equation 4.2. From Table 4.3, Cn = 0.142 n = α•Cn•τ-0.4 = 0.213•(0.142)•(0.90)-0.4 = 0.032 The discharge is calculated using Manning’s equation (Equation 2.1): Q = α/n•A•R2/3•Sf1/2 = 1.49/0.032•3.57)•(0.481)2/3•(0.03)1/2 = 17.7 ft3/s Step 5 (2nd iteration). Since this value is within 5 percent of the design flow, we can proceed to Step 6. Step 6. The maximum shear on the channel bottom is: τd = γ•d•So = 62.4•(0.70)•(0.03) = 1.31 lb/ft2 Determine the permissible soil shear stress from Equation 4.6. τp,soil = (c1•PI2 + c2•PI + c3)•(c4+c5•e)2•c6 = (1.07•(16)2 + 14.3•(16) + 47.7)(1.42 – 0.61•(0.5))2•(0.0001) = 0.068 lb/ft2 Equation 4.7 gives the permissible shear stress on the vegetation. The value of Cf is found in Table 4.5. τp = τp,soil/(1-Cf)•(n/ns)2 = 0.068/(1 – 0.9)•(0.032/0.016)2 = 2.7 lb/ft2 The safety factor for this channel is taken as 1.0. Step 7. The grass lining is acceptable since the maximum shear on the vegetation is less than the permissible shear of 2.7lb/ft2.

4.4 MAXIMUM DISCHARGE APPROACH

The maximum discharge for a vegetative lining is estimated following the basic steps outlined in Section 3.6. To accomplish this, it is necessary to develop a means of estimating the applied bottom shear stress that will yield the permissible effective shear stress on the soil. Substituting Equation 4.2 into Equation 4.3 and assuming the τo = 0.75•τd and solving for τd yields:

 τd=[α•τe/(1-Cf)•(Cn/ns)2]5/9 (4.8)

where,

 α = unit conversion constant, 1.26 (SI), 0.057 (CU)

The assumed relationship between τo and τd is not constant. Therefore, once the depth associated with maximum discharge has been found, a check should be conducted to verify the assumption.

Design Example: Maximum Discharge for a Grass Lining (SI)

Determine the maximum discharge for a grass-lined channel given the following shape, soil conditions, and grade.

Given:

 Shape: Trapezoidal, B = 0.9 m, z = 3 Soil: Silty sand (SC classification), PI = 5, D75 = 2 mm Grade: 5.0 percent

Solution

The solution is accomplished using procedure given in Section 3.6 for a maximum discharge approach.

 Step 1. The candidate lining is a sod forming grass in good condition with a stem height of 0.150 m. Step 2. Determine the maximum depth. For a grass lining this requires several steps. First, determine the permissible soil shear stress. From Equation 4.5: τp = α•D75=0.75•(2)=1.5 N/m2 To estimate the shear, we will first need to use Equation 4.1 to estimate Cn with Cs taken from Table 4.2 Cn = α•Cs0.10•h0.528 = 0.35•(106)0.10•(0.150)0.528 = 0.205 Next, estimate the maximum applied shear using Equation 4.8. τd = [α•τe/(1 – Cf)•(C

Design Example: Maximum Discharge for a Grass Lining (CU)

Determine the maximum discharge for a grass-lined channel given the following shape, soil conditions, and grade.

Given:

 Shape: Trapezoidal, B = 3.0 ft, z = 3 Soil: Silty sand (SC classification), PI = 5, D75 = 0.08 in Grade: 5.0 percent

Solution

The solution is accomplished using procedure given in Section 3.6 for a maximum discharge approach.

 Step 1. The candidate lining is a sod forming grass in good condition with a stem height of 0.5 ft. Step 2. Determine the maximum depth. For a grass lining this requires several steps. First, determine the permissible soil shear stress. From Equation 4.5: τp = α•D75 = 0.4•(0.08) = 0.032 lb/ft2 To estimate the shear, we will first need to use Equation 4.1 to estimate Cn with Cs taken from Table 4.2 Cn = α•Cs0.10•h0.528 = 0.237•(9.0)0.10•(0.5)0.528 = 0.205 Next, estimate the maximum applied shear using Equation 4.8. τd = [α•τe/(1 – Cf)•(C

4.5 TURF REINFORCEMENT WITH GRAVEL/SOIL MIXTURE

The rock products industry provides a variety of uniformly graded gravels for use as mulch and soil stabilization. A gravel/soil mixture provides a non-degradable lining that is created as part of the soil preparation and is followed by seeding. The integration of gravel and soil is accomplished by mixing (by raking or disking the gravel into the soil). The gravel provides a matrix of sufficient thickness and void space to permit establishment of vegetation roots within the matrix. It provides enhanced erosion resistance during the vegetative establishment period and it provides a more resistant underlying layer than soil once vegetation is established.

The density, size and gradation of the gravel are the main properties that relate to flow resistance and erosion control performance. Stone specific gravity should be approximately 2.6 (typical of most stone). The stone should be hard and durable to ensure transport without breakage. Placed density of uniformly graded gravel is 1.76 metric ton/m3 (1.5 ton/yd3). A uniform gradation is necessary to permit germination and growth of grass plants through the gravel layer. Table 4.7 provides two typical gravel gradations for use in erosion control.

Table 4.7. Gravel Gradation Table, Percentages Passing Nominal Size Designations
Size Very Coarse (D75 = 45 mm (1.75 in)) Coarse (D75 = 30 mm (1.2 in))
50.0 mm (2 in) 90 – 100
37.5 mm (1.5 in) 35 – 70 90 – 100
25.0 mm (1 in) 0 – 15 35 – 70
19.0 mm (0.75 in)   0 – 15

The application rate of gravel mixed into the soil should result in 25 percent of the mixture in the gravel size. Generally, soil preparation for a channel lining will be to a depth of 75 to 100 mm (3 to 4 inches). The application rate of gravel to the prepared soil layer that results in a 25 percent gravel mix is calculated as follows.

 Igravel = α•((1 – igravel)/3)•Ts•γgravel (4.9)

where,

 Igravel = gravel application rate, metric ton/m2 (ton/yd2) igravel = fraction of gravel (equal to or larger than gravel layer size) already in the soil Ts = thickness of the soil surface, m (ft) γgravel = unit weight of gravel, metric ton/m3 (ton/yd3) α = unit conversion constant, 1.0 (SI), 0.333 (CU)

The gravel application rates for fine-grained soils (igravel = 0) are summarized in Table 4.8. If the soil already contains some coarse gravel, then the application rate can be reduced by 1- igravel.

Table 4.8. Gravel Application Rates for Fine Grain Soils
Soil Preparation Depth Application Rate, Igravel
75 mm (3 inches) 0.044 ton/m2 (0.041 ton/yd2)
100 mm (4 inches) 0.058 ton/m2 (0.056 ton/yd2)

The effect of roadside maintenance activities, particularly mowing, on longevity of gravel/soil mixtures needs to be considered. Gravel/soil linings are unlikely to be displaced by mowing since they are heavy. They are also a particle-type lining, so loss of a few stones will not affect overall lining integrity. Therefore, a gravel/soil mix is a good turf reinforcement alternative.

Design Example: Turf Reinforcement with a Gravel/Soil Mixture (SI)

Evaluate the following proposed lining design for a vegetated channel reinforced with a coarse gravel soil amendment. The gravel will be mixed into the soil to result in 25 percent gravel. Since there is no existing gravel in the soil, an application rate of 0.058 ton/m2 is recommended (100 mm soil preparation depth). See Table 4.8.

Given:

 Shape: Trapezoidal, B = 0.9 m, Z = 3 Soil: Silty sand (SC classification), PI = 5, D75 = 2 mm Grass: Sod, good condition, h = 0.150 m Gravel: D75 = 25 mm Grade: 5.0 percent Flow: 1.7 m3/s

Solution

The solution is accomplished using procedure given in Section 3.1 of HEC-15 for a straight channel.

 Step 1. Channel slope, shape, and discharge have been given. Step 2. Proposed lining is a vegetated channel with a gravel soil amendment. Step 3. Initial depth is estimated at 0.30 m From the geometric relationship of a trapezoid (see Appendix B): A = B•d + Z•d2 = 0.9•(0.3) + 3•(0.32 = 0.540 m2 P = B + 2•d•√(Z2 +1) = 0.9 + 2•(0.3)•√(32 +1) = 2.80 m R = A/P = (0.540 m2)/(2.80 m) = 0.193 m Step 4. To estimate n, the applied shear stress on the grass lining is given by Equation 2.3 τo = γ•R•So = 9810•(0.193)•(0.05) = 94.7 N/m2 Determine a Manning’s n value from Equation 4.2. From Table 4.3, Cn = 0.205 n = α•Cn•τ-0.4 = 1.0•(0.205)•(94.7)-0.4 = 0.033 The discharge is calculated using Manning’s equation (Equation 2.1): Q = α/n•A•R2/3•Sf1/2 = 1/(0.033)•(0.540)•(0.193)2/3•(0.05)1/2 = 1.22 m3/s Step 5. Since this value is more than 5 percent different from the design flow, we need to go back to Step 3 to estimate a new flow depth. Step 3 (2nd iteration). Estimate a new depth solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth: d = 0.35 m Revise hydraulic radius. A = B•d + Z•d2 = 0.9•(0.35) + 3•(0.35)2 = 0.682 m2 P = B + 2•d•√(Z2 +1) = 0.9 + 2•(0.35)•√(32 +1) = 3.11 m R = A/P = (0.682 m2)/(3.11 m) = 0.219 m Step 4 (2nd iteration). To estimate n, the applied shear stress on the grass lining is given by Equation 2.3 τo = γ•R•So = 9810•(0.219)•(0.05) = 107 N/m2 Determine a Manning’s n value for the vegetation from Equation 4.2. From Table 4.3, Cn = 0.205 n = α•Cn•τ-0.4 = 1.0•(0.205)•(107)-0.4 = 0.032 The discharge is calculated using Manning’s equation (Equation 2.1): Q = α/n•A•R2/3•Sf1/2 = 1/(0.032)•(0.682)•(0.219)2/3•(0.05)1/2 = 1.73 m3/s Step 5 (2nd iteration). Since this value is within 5 percent of the design flow, we can proceed to Step 6. Step 6. The maximum shear on the channel bottom is: τd = γ•d•So = 9810•(0.35)•(0.05) = 172 N/m2 Determine the permissible shear stress from Equation 4.4. For turf reinforcement with gravel/soil the D75 for the gravel is used instead of the D75 for the soil. τp,soil = α•D75 = 0.75•(25) = 19 N/m2 A Manning’s n for the soil/gravel mixture is derived from Equation 4.4: ns = α•D751/6 = 0.015•(25)1/6 = 0.026 Equation 4.7 gives the permissible shear stress on the vegetation. The value of Cf is found in Table 4.5. τp = τp,soil/(1 – Cf)•(n/ns)2 = 19/(1 – 0.9)•(0.032/0.026)2 = 288 N/m2 The safety factor for this channel is taken as 1.0. Step 7. The grass lining reinforced with the gravel/soil mixture is acceptable since the permissible shear is greater than the maximum shear.

Design Example: Turf Reinforcement with a Gravel/Soil Mixture (CU)

Evaluate the following proposed lining design for a vegetated channel reinforced with a coarse gravel soil amendment. The gravel will be mixed into the soil to result in 25 percent gravel. Since there is no gravel in the soil, an application rate of 0.056 ton/yd2 is recommended (4 inch soil preparation depth). See Table 4.8.

Given:

 Shape: Trapezoidal, B = 3 ft, Z = 3 Soil: Silty sand (SC classification), PI = 5, D75 = 0.08 in Grass: Sod, good condition, h = 0.5 in Gravel: D75 = 1.0 in Grade: 5.0 percent Flow: 60 ft3/s

Solution

The solution is accomplished using procedure given in Section 3.1 of HEC-15 for a straight channel.

 Step 1. Channel slope, shape, and discharge have been given. Step 2. Proposed lining is a vegetated channel with a gravel soil amendment. Step 3. Initial depth is estimated at 1.0 ft From the geometric relationship of a trapezoid (see Appendix B): A = B•d + Z•d2 = 3.0•(1.0) + 3•(1.02 = 6.0 ft2 P = B + 2•d•√(Z2 +1) = 3.0 + 2•(1.0)•√(32 +1) = 9.32 ft R = A/P = (6.0 ft2)/(9.32 ft) = 0.644 ft Step 4. To estimate n, the applied shear stress on the grass lining is given by Equation 2.3 τo = γ•R•So = 62.4•(0.644)•(0.05) = 2.01 lb/ft2 Determine a Manning’s n value from Equation 4.2. From Table 4.3, Cn = 0.205 n = α•Cn•τ-0.4 = 0.213•(0.205)•(2.01)-0.4 = 0.033 The discharge is calculated using Manning’s equation (Equation 2.1): Q = α/n•A•R2/3•Sf1/2 = 1.49/(0.033)•(6.0)•(0.644)2/3•(0.05)1/2 = 45.2 ft3/s Step 5. Since this value is more than 5 percent different from the design flow, we need to go back to Step 3 to estimate a new flow depth. Step 3 (2nd iteration). Estimate a new depth solving Equation 2.2 or other appropriate method iteratively to find the next estimate for depth: d = 1.13 ft Revise hydraulic radius. A = B•d + Z•d2 = 3.0•(1.13) + 3•(1.13)2 = 7.22 ft2 P = B + 2•d•√(Z2 +1) = 3.0 + 2•(1.13)•√(32 +1) = 10.1 ft R = A/P = (7.22 ft2)/(10.1 ft) = 0.715 ft Step 4 (2nd iteration). To estimate n, the applied shear stress on the grass lining is given by Equation 2.3 τo = γ•R•So = 62.4•(0.715)•(0.05) = 2.23 lb/ft2 Determine a Manning’s n value for the vegetation from Equation 4.2. From Table 4.3, Cn = 0.205 n = α•Cn•τ-0.4 = 1.0•(0.205)•(107)-0.4 = 0.032 The discharge is calculated using Manning’s equation (Equation 2.1): Q = α/n•A•R2/3•Sf1/2 = 1.49/(0.032)•(7.22)•(0.715)2/3•(0.05)1/2 = 60.1 m3/s Step 5 (2nd iteration). Since this value is within 5 percent of the design flow, we can proceed to Step 6. Step 6. The maximum shear on the channel bottom is: τd = γ•d•So = 62.4•(1.13)•(0.05) = 3.53 lb/ft2 Determine the permissible shear stress from Equation 4.4. For turf reinforcement with gravel/soil the D75 for the gravel is used instead of the D75 for the soil. τp,soil = α•D75 = 0.4•(1.0) = 0.4 lb/ft2 A Manning’s n for the soil/gravel mixture is derived from Equation 4.4: ns = α•D751/6 = 0.026•(1.0)1/6 = 0.026 Equation 4.7 gives the permissible shear stress on the vegetation. The value of Cf is found in Table 4.5. τp = τp,soil/(1 – Cf)•(n/ns)2 = 0.4/(1 – 0.9)•(0.032/0.026)2 = 6.06 lb/ft2 The safety factor for this channel is taken as 1.0. Step 7. The grass lining reinforced with the gravel/soil mixture is acceptable since the permissible shear is greater than the maximum shear.

# HEC-15: Permissible Shear Stress

HEC-15 Section 6.2 – PERMISSIBLE SHEAR STRESS

Values for permissible shear stress for riprap and gravel linings are based on research conducted at laboratory facilities and in the field. The values presented here are judged to be conservative and appropriate for design use. Permissible shear stress is given by the following equation:

 τp = F*.(γs – γ).D50 (6.7)

where,

• τp = permissible shear stress, N/m2 (lb/ft2)
• F* = Shield’s parameter, dimensionless
• γs = specific weight of the stone, N/m3 (lb/ft3)
• γ = specific weight of the water, 9810 N/m3 (62.4 lb/ft3)
• D50 = mean riprap size, m (ft)

Typically, a specific weight of stone of 25,900 N/m3 (165 lb/ft3) is used, but if the available stone is different from this value, the site-specific value should be used.

Recalling Equation 3.2,

τp ≥ SF.τd

and Equation 3.1,

τd = γ.d.So

Equation 6.7 can be written in the form of a sizing equation for D50 as shown below:

 D50 ≥ (SF.d.So)/(F*.(SG – 1)) (6.8)

where,

• d = maximum channel depth, m (ft)
• SG = specific gravity of rock (γs/γ), dimensionless

Changing the inequality sign to an equality gives the minimum stable riprap size for the channel bottom. Additional evaluation for the channel side slope is given in Section 6.3.2.

Equation 6.8 is based on assumptions related to the relative importance of skin friction, form drag, and channel slope. However, skin friction and form drag have been documented to vary resulting in reports of variations in Shield’s parameter by different investigators, for example Gessler (1965), Wang and Shen (1985), and Kilgore and Young (1993). This variation is usually linked to particle Reynolds number as defined below:

 Re = V*.D50/ν (6.9)

where,

• Re = particle Reynolds number, dimensionless
• V* = shear velocity, m/s (ft/s)
• ν = kinematic viscosity, 1.131×10-6 m2/s at 15.5 deg C (1.217×10-5 ft2/s at 60 deg F)

Shear velocity is defined as:

 V* = √(g.d.S) (6.10)

where,

• g = gravitational acceleration, 9.81 m/s2 (32.2 ft/s2)
• d = maximum channel depth, m (ft)
• S = channel slope, m/m (ft/ft)

Higher Reynolds number correlates with a higher Shields parameter as is shown in Table 6.1. For many roadside channel applications, Reynolds number is less than 4×104 and a Shields parameter of 0.047 should be used in Equations 6.7 and 6.8. In cases for a Reynolds number greater than 2×105, for example, with channels on steeper slopes, a Shields parameter of 0.15 should be used. Intermediate values of Shields parameter should be interpolated based on the Reynolds number.

Table 6.1. Selection of Shields’ Parameter and Safety Factor
Reynolds number F* SF
≤ 4×104 0.047 1.0
4×104<Re<2×105 Linear interpolation Linear interpolation
≥ 2×105 0.15 1.5

Higher Reynolds numbers are associated with more turbulent flow and a greater likelihood of lining failure with variations of installation quality. Because of these conditions, it is recommended that the Safety Factor be also increased with Reynolds number as shown in Table 6.1. Depending on site-specific conditions, safety factor may be further increased by the designer, but should not be decreased to values less than those in Table 6.1.

As channel slope increases, the balance of resisting, sliding, and overturning forces is altered slightly. Simons and Senturk (1977) derived a relationship that may be expressed as follows:

 D50 ≥ SF•d•S•Δ/(F*•(SG – 1)) (6.11)

where,

• Δ = function of channel geometry and riprap size.

The parameter Δ can be defined as follows (see HEC-15 Appendix D for the derivation):

 Δ = (K1•(1 + sin(α + β)•tan Φ)/(2•(cosθ•tanΦ  – SF•sinθ•cosβ)) (6.12)

where,

• α = angle of the channel bottom slope
• β = angle between the weight vector and the weight/drag resultant vector in the plane of the side slope
• θ = angle of the channel side slope
• Φ = angle of repose of the riprap.

Finally, β is defined by:

 β = tan-1(cosα/(2•sinθ/(η•tanΦ) + sinα)) (6.13)

where,

• η = stability number.

The stability number is calculated using:

 η = τs/(F*•(Υs – Υ)•D50) (6.14)

Riprap stability on a steep slope depends on forces acting on an individual stone making up the riprap. The primary forces include the average weight of the stones and the lift and drag forces induced by the flow on the stones. On a steep slope, the weight of a stone has a significant component in the direction of flow. Because of this force, a stone within the riprap will tend to move in the flow direction more easily than the same size stone on a milder gradient. As a result, for a given discharge, steep slope channels require larger stones to compensate for larger forces in the flow direction and higher shear stress.

The size of riprap linings increases quickly as discharge and channel gradient increase. Equation 6.11 (not shown) is recommended when channel slope is greater than 10 percent and provides the riprap size for the channel bottom and sides. Equation 6.8 is recommended for slopes less than 5 percent. For slopes between 5 percent and 10 percent, it is recommended that both methods be applied and the larger size used for design. Values for safety factor and Shields parameter are taken from Table 6.1 for both equations.