Appendix E. STORM DRAINAGE AND STORMWATER MANAGEMENT DESIGN MANUAL

§ X. Design example.

Latest version.

The example problem is the design of a hypothetical storm drain system in a residential area and will be analyzed for a 10-year storm frequency.

The proposed layout of the system was sketched on a contour map of the area to obtain ground elevations, design slopes, and drainage patterns (See Exhibit I.).

All inlets at low points and other inlets in critical areas where debris may be a problem should have extra capacity as a safeguard against flooding. To partially compensate for loss capacity, inlets in such locations should be designed for a flow 20 percent greater than the calculated tributary flow. At critical low points, supplemental inlets may be required.

DRAINAGE INLET SIZING

Beginning Inlet No. 1

Form C is used for sizing inlets, starting at the upstream end of the system.

L = 200 ft

A = 2.0 Ac

S = 0.04 ft/ft (4.0%)

C = 0.45

T _c = 10.0 min. from Figure 1

CA = 0.45 × 2.0 = 0.90

i = 6.1 in./hr from Table 4

Q _r = CiA = 0.9 × 6.1 = 5.49 cfs

Q _g = 5.49 cfs

Gutter Slope, S _g = Not a factor in sump

Depth of Flow at Entrance, H = 0.75

Ref: Figure 9

h = 8 in. (from Details Type A Curb Inlet)

Hh	=	0.75	=	1.125
		0.67

Q	=	1.62 cfs/ft
L

L	=	Q _g × 1.20	=	5.49 × 1.20	=	4.07
		1.62		1.62

Use a standard Type A curb inlet with L = 4′ - 0″.

Inlet Capacity with Blockage Allowance =

4 × 1.62	=	5.40 cfs
1.20

If Q _g was much larger, the next largest inlet would have been considered.

Inlet No. 2

For Inlet No. 2, compute Q _r = 2.28 cfs in the same manner, and find that required L = 2.28 × 120/1.62 = 1.69 ft. Use a Type A Standard Curb Inlet with L = 4′-0″.

Inlet No. 3

For Inlet No. 3, compute Q _r = Q _g = 7.64 cfs. From Figure 7, select an L = 4′-0″ Type A standard curb inlet without deflectors with Q;sub \sub; = 5.80, bypassing 1.84 cfs to Inlet No. 4. Note on Figure 8 that a 4-foot inlet with deflectors on a 4.0 percent gutter grade has only slightly greater capacity and the additional cost would not be justified.

Inlet No. 4

For Inlet No. 4, compute Q _r = 3.40 cfs and Q _g = 3.40 + 1.84 (bypass) = 5.24 cfs. From Figure 7, select an L = 4′-0″ Type A Standard Curb Inlet without deflectors having a capacity of 5.80 cfs, thereby intercepting 100 percent of the flow, a requirement at the arterial street.

Inlet No. 5

For Inlet No. 5, compute Q _r = Q _g = 8.00 cfs. From Figure 8, select an L = 8′-0″ Type A Standard Curb Inlet with deflectors having a capacity of 13.5 cfs at the 5 percent gutter slope and intercepting 100 percent of the gutter flow, a requirement at the arterial street.

Storm Drain Hydraulics and Sizing

Beginning Inlet — Manhole No. 1

Form B is used for sizing the drains, starting with the beginning headwall upstream from Manhole No. 1. The drainage area (Area No. 1) was found to be 3.0 acres with a composite runoff coefficient, C, of 0.45. The percent imperviousness is 1.35 (3.0 × 0.45). From a contour map, L = 600 ft. and the average slope, S, was found to be 6 percent.

T _c = 15 min. from Figure 1

i = 5.3 in. per hour from Table 4

Q = 1.35 × 5.3 = 7.2 cfs

Try a 15 in. RCP

Length, L, of drain (from map) = 100 ft

Hydraulic slope required = 0.015 ft per ft

Drain slope provided = 0.015 ft per ft

The capacity of the 15-inch pipe at a slope of 1.5 percent is 7.2 cfs and the velocity is 5.8 fps from Figure 5.

The head, HW, or depth of water above the inlet flow line required for the drain to operate effectively can be obtained from Figure 11. The purpose of this head is to overcome entrance losses and to generate the required velocity.

HW	=	1.47 from Figure 11
D

HW = 1.25 × 1.75 = 2.18 ft

The 15-inch pipe will be used in this example; however, if the HW would have been too high for local inlet conditions, a larger pipe would have been selected.

The time of flow from the beginning inlet to Manhole No. 1 is calculated as follows:

t	=	L	=	100	=	0.3 minute
		V _f × 60		5.8 × 60

Manhole No. 1 - Inlet No. 1

From Manhole No. 1 to Inlet No. 1, there is no additional runoff introduced, so Q remains 7.2 cfs.

Try a 15 in. RCP

L = 150 ft

Hydraulic slope required = 0.015 per ft

Drain slope provided = 0.017 ft per ft

Capacity 7.4 cfs

Velocity 6.0 cfs

t	=	150	=	0.4 minute
		6.0 × 60

The manhole losses through Manhole No. 1 can now be calculated with the aid of Figure 10.

Curve A	Entrance and exit losses (V = 6.0 fps)	=	0.20 ft

Curve B	Loss due to velocity change

	(V ₂ = 6.0 and V ₁ = 5.8) = 0.55 - 0.52	=	0.03 ft

Curve C	Loss due to change in direction
	(45º ± bend = 0.10 × 1.0)	=	0.10 ft

Curve D	Loss due to entrance of secondary flow	=	None
	Total Manhole Losses (Manhole No. 1)	=	0.33 ft

These losses may be compensated for in the manhole by lowering the outlet flow line by lowering the outlet flow line by at least the amount of the losses below the incoming flow line.

Inlet No. 1 - Inlet No. 2

Accumulative CA = 0.90 + 1.35 = 2.25

T _c = 15.3 + 0.4 = 15.7 minutes

i = 5.2 in. per hour from Table 4

Q = 2.25 × 5.2 = 11.7 cfs

Try an 18 in. RCP

Hydraulic slope required = 0.016 ft per ft from Figure 5

Drain slope provided = 0.016 ft per ft

Capacity = 11.7 cfs from Figure 5

Velocity = 6.7 ft per sec. From Figure 5

Time of flow = 36/6.7 × 60 = 0.1 minute

The manhole losses at Inlet No. 1 can now be computed.

Curve A	(V = 6.7 fps)	=	0.24 ft
Curve B	(V ₂ = 6.7 and V ₁ = 6.0) = 0.70 - 0.57	=	0.13 ft
Curve C	(V = 6.7 fps (45º ± bend))	=	0.13 ft
Curve D			None
	Total Manhole Losses at Inlet No. 1	=	0.50 ft

At Inlet No. 1 the flow into the inlet from the street gutter is:

Q _I = 0.90 × 5.2 = 4.7 cfs

This inflow represents 40 percent of the total flow leaving the structure. Where velocities are high and in critical locations where structures are shallow, the position of the hydraulic gradient in the structure should be checked by means found in the references. The position of the hydraulic gradient in the inlet would then confirm that the amount of freeboard above the standing water surface elevation was adequate for this flow to free fall into the inlet.

As a minimum where the inflow from the street is a significant part of the total flow out of a combination inlet-manhole, the elevation of the top of the outgoing pipe should be below the gutter entrance weir by the following amount:

d	=	1.78	V	+	0.50, where g = 32.2
			2g

In this instance, d + D is equal to the HW that can be obtained from Figure 11, plus a 0.50-foot freeboard. If surcharge of the storm drain should be avoided, the outgoing pipe should be placed at a depth below the incoming pipe equal to the manhole losses, Column 25, plus 1.78 V /29.

Inlet No. 2 - Inlet No. 3

Calculations for this drain segment follow the above procedures.

The manhole losses at Inlet No. 2 are as follows:

Curve A	(V = 7.9 fps)	=	0.32 ft

Curve B	(V ₂ = 7.9 and V ₁ = 6.7) = 1.0 - 0.7	=	0.30 ft

Curve C	(V = 7.9 fps) (90º bend) 0.19 × 2	=	0.38 ft

Curve D			None
	Total Manhole Losses at Inlet No. 2	=	1.00 ft

Inlet No. 3 - Manhole No. 2

Calculate following the above procedures.

Manhole No. 2 - Drain Outlet

Calculate pipe hydraulics and size following above procedures. Note inflow from Inlets No. 4 and 5.

Manhole losses at Manhole No. 2 are as follows:

Curve A	(V = 8.7 fps)	=	0.40 ft

Curve B	(V ₂ = 6.5 and V ₁ = 8.7) = 0.65 - 1.13	=	-0.48 ft

Curve C	(V = 8.7 fps (90º±)) = 0.22 × 2	=	0.44 ft

Curve D	Q ₁ = 20.6 cfs Q ₃ = 10.3 = 0.50

	Q ₂ = 30.9 cfs Q ₁ 20.6
	Q ₃ = 10.3 cfs

Use Curve for	Q ₃	=	50%
	Q ₁

	Curve D loss	=	0.38 ft
	Total Manhole Losses @ Manhole No. 2	=	0.74 ft

Open Channel

By using the total accumulated CA and the intensity for the total T _c , find Q at the upstream end of the open channel to be 30.2 cfs. Try a channel with a 4′-0″ bottom and 2:1 side slopes.

T _c = 17.6 minutes and CA = 6.31

Assume vegetal lining, n - 0.035 (Table 2)

Assume erosion-resistant soils

Slope S = 0.02 ft per ft

Assume depth of flow = 1.0 ft

A = 4 × 1.0 + 2 × 1.0 = 6.0 sq ft

W = 4 + 2 × 2.236 = 8.47

R	=	A	=	6	=	0.708
		W		8.47

V = 4.8 fps

Q = AV = 6.0 × 4.8 = 28.8 cfs

Try S = 0.025 ft per ft

V = 5.4 fps (less than 7 fps, Table 8)

Q = 6.0 × 5.4 = 32.4 cfs

Length of Channel = 250 ft

For this example, assume no inflow along the channel.

t (in channel)	=	250	=	0.8 minute
		5.4 × 60

T _c = 17.6 + 0.8 = 18.4 minutes

i = 4.8 in. per hour

Q = 6.31 × 4.8 = 30.3 cfs (for Culvert design)

Culvert (See Form E)

The culvert under the arterial street will be sized so that the headwater is limited to 3.5 ft. above the upstream invert.

Try a 30 in. RCP, square edge with headwall

Slope, S _o , = 0.02 ft per ft

Length = 100 ft

Q = 30.3 cfs

From Figure 11, HW/D = 1.3

HW = 1.3 × 2.5 = 3.25 ft (This is the headwater for inlet control.)

From Table 6, select K _e = 0.5

H - 1.48 from Figure 15

d _c , Critical Depth, = 1.8 (Figure 18)

d _c + D	-	1.8 + 2.5	=	2.15
2		2

For this example, use TW = 0 ft (This will generally be the case in the City, except for culverts in flood plains).

LS _o = 100 × 0.02 = 2.0 ft

HW = H + h _o - LS _o = 1.48 + 2.15 - 2.0 = 1.63 ft

(This is the HW for outlet control.)

The culvert will operate under inlet control passing Q = 30.3 cfs at a head (HW) of 3.25 ft on the entrance.

(Ord. No. 4119, § 1(Exh. A), 6-16-08)

                    
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