Q = C L Hw 3/2 (1)
| ||
Where:
| ||
Q
|
=
|
Discharge, in cubic feet per second
|
C
|
=
|
Weir coefficient
|
L
|
=
|
Length of the weir, in feet; for circular riser pipes, L is
the pipe circumference
|
Hw
|
The depth of flow over the weir crest, in feet
|
Q = C A (2gHo)0.5 (2)
| ||
Where:
| ||
Q
|
=
|
Discharge, in cubic feet per second
|
C
|
=
|
Orifice coefficient
|
A
|
=
|
Cross-sectional area of the pipe, in square feet
|
g
|
=
|
Acceleration of gravity, 32.2 feet per second squared
|
Ho
|
=
|
Head above the centerline of the pipe, in feet
|
Where:
| ||
Q
|
=
|
Discharge, in cubic feet per second
|
A
|
=
|
Cross-sectional area of the pipe, in square feet
|
g
|
=
|
Acceleration of gravity, 32.2 feet per second squared
|
H
|
=
|
The difference between headwater and tailwater elevations, in
feet
|
kb
|
=
|
Bend loss coefficient, use 0.6
|
ke
|
=
|
Entrance loss coefficient, use 0.5
|
kf
|
=
|
Friction loss coefficient = 185n2/D4/3
|
n
|
=
|
Manning’s roughness coefficient
|
D
|
=
|
Diameter of pipe, in feet
|
L
|
=
|
Length of pipe, in feet
|
Table 1. Acceptable Grasses for Vegetated Spillways
|
---|
Western Wheatgrass
|
Buffalograss
|
Bermudagrass
|
Tall Fescue
|
Blue Grama
|
|
Example 1. Modified Rational Method
|
Given:
|
Area: A = 2.0 acres
|
|
Type of development: commercial parking lot, fully paved, C
= 0.88
|
|
Time of concentration: tc = 8 minutes
|
|
Design frequency = 25 years, Cf = 1.10
|
|
Use intensity-duration-frequency curves, figure 2-1.
|
Find:
|
Develop family of curves representing Modified Rational Method
hydrographs for the 8-, 10-, 15-, 20-, 30- and 40-minute rainfall
durations.
|
Solution:
|
Qp = C Cf i A
|
|
Example (.88)(1.1)(7.97)(2.0) = 15.4 cfs
|
Rainfall Duration
(min)
|
Rainfall Intensity
(in/hr)
|
Peak Runoff Rate
(cfs)
|
---|---|---|
8
|
7.97
|
15.4
|
10
|
7.44
|
14.4
|
15
|
6.36
|
12.3
|
20
|
5.88
|
11.4
|
30
|
4.92
|
9.5
|
40
|
4.40
|
8.5
|
Vr = 60 D Qp (4)
| ||
Where:
| ||
Vr
|
=
|
Storm runoff volume, in cubic feet
|
D
|
=
|
Storm duration, in minutes
|
Qp
|
=
|
Peak runoff rate of the inflow hydrograph, in cubic feet per
second
|
The equation for the required storage volume, Vs, can also be
simplified as:
|
Vs = 60 D (Qp -
Qo) (5)
| ||
Where:
| ||
Vs
|
=
|
Required storage volume, in cubic feet
|
D
|
=
|
Storm duration, in minutes
|
Qp
|
=
|
Peak runoff rate of the inflow hydrograph, in cubic feet per
second
|
Qo
|
=
|
Maximum release rate, in cubic feet per second
|
Example 2. Critical Storage Volume
| |
Given:
|
Drainage basin and other hydrologic information presented in
Example 1.
|
|
Allowable release rate: Qo = 6.0 cfs
|
Find:
|
Determine the critical storage volume.
|
Solution:
|
V = 60 D Qp
Vs = 60 D (Qp -
Qo)
|
Storm Duration
(min)
|
Storm Runoff Volume
(ft3)
|
Storage Volume
(ft3)
|
---|---|---|
8
|
7,392
|
4,512
|
10
|
8,640
|
5,040
|
15
|
11,070
|
5,670
|
20
|
13,680
|
6,480 Maximum
|
30
|
17,100
|
6,300
|
40
|
20,400
|
6,000
|
Ex: (60)(8)(15.4 - 6.0) = 4,512
|
I - O = ?S/?T (6)
| ||
Where:
| ||
I
|
=
|
Average inflow rate
|
O
|
=
|
Average outflow rate
|
DS
|
=
|
Change in storage volume
|
?t
|
=
|
Routing period
|
Where:
| ||
I1
|
=
|
Inflow rate at time interval 1
|
I2
|
=
|
Inflow rate at time interval 2
|
O1
|
=
|
Outflow rate at time interval 1
|
O2
|
=
|
Outflow rate at time interval 2
|
S1
|
=
|
Storage volume at time interval 1
|
S2
|
=
|
Storage volume at time interval 2
|
?t
|
=
|
Routing period
|
Table 2. Development of a (2S/?t) + O versus O Relationship
| |||
---|---|---|---|
Depth
|
Storage, S
|
Outflow, O
|
(2S/?t) + O
|
(ft) (1)
|
(acre-feet) (2)
|
(cfs) (3)
|
(cfs) (4)
|
0
|
0.0
|
0
|
0
|
2
|
0.1
|
40
|
69
|
4
|
0.6
|
138
|
313
|
6
|
3.0
|
274
|
1,147
|
8
|
11.0
|
426
|
3,627
|
10
|
32.0
|
560
|
9,872
|
12
|
72.0
|
671
|
21,623
|
14
|
131.0
|
765
|
38,886
|