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.

4.1 Design Frequency and Spread

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.

Table 4-1. Suggested Minimum Design Frequency and Spread.
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
  • Roadside and median ditches
  • 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)
Gi = Grade of roadway, percent

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.

Table 4-2. Normal Pavement Cross Slopes.
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
  • Enable the orderly development of property adjacent to the roadway

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.

Figure 4-1. Typical gutter sections.

4.2.5 Roadside and Median Channels

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)

Table 4-3. Manning’s n for Street and Pavement Gutters.
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.

HEC-22 Figure 4-2

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
S/ 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 = dBBC•(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 = dBBC•(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.

HEC-22 Fig 4-3.
Figure 4-3. Relative effects of spread, cross slope, and longitudinal slope, on gutter capacity.

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.

Table 4-4. Spread at Average Velocity in a Reach of Triangular Gutter.
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 0.70 0.74 0.77 0.82 0.86 0.90
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 = (K/ 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.

HEC-22 Figure 4-4.
Figure 4-4. Classes of storm drain inlets.

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)

HEC-22 Figure 4-5.
Figure 4-5. P-50 and P-50 x 100 grate (P-50 is this grate without 10 mm (3/8″) transverse rods).
HEC-22 Figure 4-6.
Figure 4-6. P-30 grate.
HEC-22 Figure 4-7.
Figure 4-7. Curved vane grate.
HEC-22 Figure 4-8.
Figure 4-8. 45-degree – 60 (2.25″) and 45-degree – 85 (3.25″) tilt-bar grates.
HEC-22 Figure 4-9.
Figure 4-9. 30-degree – 85 (3.25″) tilt-bar grates.
HEC-22 Figure 4-10.
Figure 4-10. Reticuline grate.

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.

HEC-22 Figure 4-11.
Figure 4-11. Comparison of inlet interception capacity, slope variable.

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.

HEC-22 Figure 4-12.
Figure 4-12. Comparison of inlet interception capacity, flow rate variable.

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)] 
&nbsp; 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.

HEC-22 Figure 4-13.
Figure 4-13. Depressed curb opening 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.

HEC-22 Figure 4-14.
Figure 4-14. Slotted drain inlet at an intersection.

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.

HEC-22 Figure 4-15.
Figure 4-15. Combination curb-opening, 45 degree tilt-bar grate inlet.
HEC-22 Figure 4-16.
Figure 4-16. Sweeper combination inlet.

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)
HEC-33 Figure 4-17
Figure 4-17. Definition of depth.

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.

HEC-22 Figure 4-18.
Figure 4-18. Curb-opening inlets.

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)•(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)•(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
  • Road profiles
  • Typical cross sections
  • Grading 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.
HEC-22 Figure 4-19.
Figure 4-19. Inlet spacing computation sheet.
Step 13. Col. 11, Col. 10 For the first inlet in a series, enter the value from column 7 into column 11 since there was no previous bypass flow. Additionally, if the inlet is the first in a series, enter 0 into column 10.
Step 14. Col. 14, Col. 12 Determine the spread, T, by using Equations 4-2 and 4-4 or Charts 1 and 2 and enter the value in column 14. Also, determine the depth at the curb, d, by multiplying the spread by the appropriate cross slope, and enter the value in column 12. Compare the calculated spread with the allowable spread as determined by the design criteria outlined in Section 4.1. Additionally, compare the depth at the curb with the actual curb height in column 19. If the calculated spread, column 14, is near the allowable spread and the depth at the curb is less than the actual curb height, continue on to step 15. Else, expand or decrease the drainage area up to the first inlet to increase or decrease the spread, respectively. The drainage area can be expanded by increasing the length to the inlet and it can be decreased by decreasing the distance to the inlet. Then, repeat steps 6 through 14 until appropriate values are obtained.
Step 15. Col. 15 Calculate W / T and enter the value in column 15.
Step 16. Col. 16 Select the inlet type and dimensions and enter the values in column 16.
Step 17. Col. 17 Calculate the flow intercepted by the grate, Qi, and enter the value in column 17. Use Equations 4-16 and 4-13 or Charts 2 and 4 to define the gutter flow. Use Chart 5 and Equation 4-19 or Chart 6 to define the flow intercepted by the grate. Use Equations 4-22 and 4-23 or Charts 7 and 8 for curb opening inlets. Finally, use Equation 4-21 to determine the intercepted flow.
Step 18. Col. 18 Determine the bypass flow, Qb, and enter into column 18. The bypass flow is column 11 minus column 17.
Step 19. Col. 1-4 Proceed to the next inlet down the grade. To begin the procedure, select a drainage area approximately 90 m to 120 m (300 to 400 ft) below the previous inlet for a first trial. Repeat steps 5 through 7 considering only the area between the inlets.
Step 20. Col. 5 Compute the time of concentration for the next inlet based upon the area between the consecutive inlets and record this value in column 5.
Step 21. Col. 6 Determine the rainfall intensity from the IDF curve based upon the time of concentration determined in step 19 and record the value in column 6.
Step 22. Col. 7 Determine the flow in the gutter by using Equation 3-1 and record the value in column 7.
Step 23. Col. 11 Record the value from column 18 of the previous line into column 10 of the current line. Determine the total gutter flow by adding column 7 and column 10 and record in column 11.
Step 24. Col. 12, Col. 14 Determine the spread and the depth at the curb as outlined in step 14. Repeat steps 18 through 24 until the spread and the depth at the curb are within the design criteria.
Step 25. Col. 16 Select the inlet type and record in column 16.
Step 26. Col. 17 Determine the intercepted flow in accordance with step 17.
Step 27. Col. 18 Calculate the bypass flow by subtracting column 17 from column 11. This completes the spacing design for the inlet.
Step 28.   Repeat steps 19 through 27 for each subsequent inlet down to the low point.

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

HEC-22 Figure 4-20.
Figure 4-20. Storm drainage system for Example 4-15.
HEC-22 Figure 4-21a.
Figure 4-21a. Inlet spacing computation sheet for Example 4-15 SI Units.

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 / K(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 
HEC-22 Figure 2-21b
Figure 4-21b. Inlet spacing computation sheet for Example 4-15 English Units.
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 ft/ 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 / K(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.

HEC-22 Figure 4-22.
Figure 4-22. Example of flanking inlets.

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.

HEC-22 Figure 4-23.
Figure 4-23. Median drop inlet.(85)

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.

HEC-22 Figure 4-24.
Figure 4-24. Embankment 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

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