Detention Basin Design Example
6
APPENDIX H Detention Basin Design Utilizing the Rational Method of Runoff Determination The following procedure utilizing the Rational Method of Design has been adopted from the Erosion Control Manual of Oakland County, Michigan. This method has been revised to include the particular requirements of the Washtenaw County Drain Commissioner. A. Determine the acreage tributary to the detention basin and the composite runoff coefficient. The runoff coefficient must be based on the area and surface types tributary to the basin. Example Extended Detention Pond Volume Calculations and Outlet Design The following is an example for one method of extended detention pond design that meets the minimum standards of the Washtenaw County Drain Commissioner. More complex systems are preferred and shall be provided where feasible. The example is based on a 7.3 acre site with a runoff coefficient of 0.75. The release rate specified in the example is based on a pond which outlets to an existing watercourse that is adequate to effectively handle a concentrated flow of water from the proposed development. C= 0.75 A= 7.3 Ac. The allowable release rate is 0.15 cfs/ac: ( ( ( ( cfs ac a ac cfs a ac cfs a Q Q A Q 10 . 1 3 . 7 15 . 0 15 . 0 = = = 100-Year Flood Volume Required ( ( imp ac cfs o o a o Q Q AC Q Q ac cfs - = = = 20 . 0 75 . 0 3 . 7 10 . 1 min 1 . 202 2 . 0 5 . 10312 25 5 . 10312 25 = - = - = - T T Q T imp ac cfs o
description
Calculatıon for basins
Transcript of Detention Basin Design Example
APPENDIX H
Detention Basin Design Utilizing the Rational Method of Runoff Determination The following procedure utilizing the Rational Method of Design has been adopted from the Erosion Control Manual of Oakland County, Michigan. This method has been revised to include the particular requirements of the Washtenaw County Drain Commissioner.
A. Determine the acreage tributary to the detention basin and the composite runoff coefficient. The runoff coefficient must be based on the area and surface types tributary to the basin.
Example Extended Detention Pond Volume Calculations and Outlet Design The following is an example for one method of extended detention pond design that meets the minimum standards of the Washtenaw County Drain Commissioner. More complex systems are preferred and shall be provided where feasible. The example is based on a 7.3 acre site with a runoff coefficient of 0.75. The release rate specified in the example is based on a pond which outlets to an existing watercourse that is adequate to effectively handle a concentrated flow of water from the proposed development. C= 0.75 A= 7.3 Ac. The allowable release rate is 0.15 cfs/ac:
( )( )( )( )
cfs
ac
a
accfs
a
accfs
a
Q
Q
AQ
10.1
3.715.0
15.0
=
=
=
100-Year Flood Volume Required
( )( )impac
cfso
o
ao
Q
Q
ACQ
Q
ac
cfs
−=
=
=
20.0
75.03.710.1
min1.202
2.05.10312
25
5.1031225
=
+−=
+−=
−
T
T
QT
impaccfs
o
( )( ) ( )( )( )
impaccf
s
impaccfs
s
os
V
V
TQT
TV
−
−
=
−+
=
−+
=
8.13066
1.2022.040251.2021.20216500
4025
16500
minmin
min
( )( )( )cf
ac
t
impaccf
t
st
V
V
ACVV
71541
75.03.78.13066
=
=
=
−
Bankfull Flood Volume The bankfull storm is defined as the 24 hour, 1.5-year storm event:
( )( )( )( )( )( )[ ]( )( ) ( )( )( )
cf
ac
bf
acft
bf
acft
bf
V
V
ACV
44731
75.03.7817075.03.7"25.2
"25.2
143560
"12'1
143560
"12'1
2
2
=
==
=
First Flush Volume The first flush storm is defined as the first 0.5” of rain over the entire watershed:
( )( )( )( )( )( )[ ]( )( ) ( )( )( )
cfV
V
ACV
ff
acft
ff
acft
ff
ac
9937
75.03.7181575.03.7"5.0
"5.0
143560
"12'1
143560
"12'1
2
2
=
==
=
Storage Provided Elevation Area (sf) Depth (ft) Volume (cf) Total Volume (cf) 94.0 28,960 1 26,640 80,080 93.0 24,320 1 22,240 53,440 92.0 20,160 1 18,000 31,200 91.0 15,840 1 13,200 13,200 90.0 10,560 1 0 0 Storage Elevations First flush: Bankfull: 100 year:
( )6.92.61.923120044731
0.92
31200534400.920.93
approxbf
bf
x
x
=−
−=
−−
( )8.90.75.9009937
0.90
0132000.900.91
approxff
ff
x
x
=−
−=
−−
( )7.93.68.935344071541
0.935344080080
0.930.94
100
100
approxx
x
=−−
=−−
Outlet Control Structure
First flush of runoff: The average allowable release rate for runoff resulting from 0.5” of rain over watershed area in 24 hours:
Place openings in standpipe at bottom of basin (90.0): To determine the appropriate size orifice to release the first flush volume, an average head value can be used in the orifice equation. If the basin is designed to be trapezoidal in shape, 2/3 of the total head is an acceptable approximation for the average head.
The number and size of orifices to meet the area requirements is variable. In general larger holes are preferable, although multiple outlets should be used if possible. For this example, we will choose a 1.25” diameter orifice (which has an area of 0.0085 sf).
76.3#0085.0032.0
#
=
=sf
sf
Therefore, use 3 – 1.25” diameter holes @ elev. 90.0 The detention time for 3 – 1.25” diameter holes is:
( )( )cfs
hr
cf
ff
hrff
ff
Q
Q
TV
Q
115.0
249937
1sec3600
24
=
=
=
( )( )
ft
elevelev
ave
ave
botffave
h
h
h
53.0
0.908.90
..
32
32
=
−=
−=
( )( )sf
cfs
A
ftA
gh
QA
ft
ave
ff
032.0
53.02.32262.0
115.0
262.0
2sec
=
=
=
( )( )( )( ) ( )( )
cfs
sf
newff
ftnewff
avenewff
newff
Q
ftQ
ghAQ
092.0
53.02.32262.00085.03
262.0
2sec
=
=
=
Bankfull flood: The bankfull flood must be detained 36-48 hours; check the discharge through the first flush orifice to see if additional holes are necessary:
( )( )73.1
0.906.92
..
32
32
=
−=
−=
ave
ave
botbfave
h
h
h elevelev
Because the holding time exceeds 48 hours, additional orifices in the standpipe are required. The release rate may be considered occurring in two phases; the release rate when both the ff and bf orifices are contributing, and the release rate when the water elevation is below the bf orifice (ff elevation). The time for the ff volume to release was calculated above, so the remaining volume (bf volume – ff volume) must be released so the total detention time does not exceed 36 – 48 hours. A target detention time of 44 hours was chosen for this example.
( )( )( ) ( )( )( )
( )( )hr
cfs
cf
cfs
sf
T
T
Q
VT
Q
ftQ
ghAQ
hr
bf
ft
aveff
1.73
17.044731
17.0
73.12.3220085.0362.0
262.0
1sec3600
sec2
=
=
=
=
=
=
( )( )hr
cfs
cf
newff
hr
newff
newff
ffnewff
T
T
Q
VT
0.30
092.09937
1sec3600
=
=
=
newfftotrem
rem
rem
ffbfrem
TTT
VV
VVV
cf
cfcf
−=
=−=
−=
34794993744731
hr
hrhr
rem
rem
TT
0.140.3044
=−=
Volume through 3 – 1.25” diameter holes in 14.0 hours:
Q1will be defined as the discharge through the ff orifices when both the ff and bf holes are contributing.
( )( )( ) ( )( )
( )( )( )cf
cfshr
cfs
fsf
V
V
QTV
Q
tQ
ghAQ
hr
rem
ft
ffaveff
9072
18.00.14
18.0
22.3220085.0362.0
262.0
1
1sec3600
1
11
1
sec1
1
2
=
=
==
=
=
The leftover volume will be released by the bankfull orifice. V2 will be defined as the amount of water to be discharged by the bf orifices in 14.0 hours.
( )( )
( )( )
( )( )sf
f
cfs
ft
ftft
elevelev
cfs
hr
cf
cf
cfcf
A
tA
gh
QA
h
h
h
Q
Q
TV
Q
VV
VVV
ft
bfave
bfave
bfave
ffbfbfave
hr
rem
rem
094.0
2.12.32262.0
51.0
262.0
2.1
8.906.92
..
51.0
0.1425722
25722907234794
2
sec
2
22
32
32
2
1sec36002
22
2
2
12
2
=
=
=
=
−=
−=
=
=
=
=−=
−=
( ) ( )( ) ( )fth
h
h
ffave
ffave
botffffbfff
ave elevelevelevelev
2
0.908.908.906.92
....
32
32
=
−+−=
−+−=
A 2” diameter orifice has an area of 0.022 sf
27.4#022.0094.0
#
=
=sfsf
Therefore, use 4 – 2” diameter holes @ elev. 90.8 100-year flood: Qa = 1.10 cfs Qa is a peak or maximum flow. Calculate the maximum flow passing through first flush and bankfull orifices, using the total head, and subtract from Qa to determine the orifice size to release the 100-year storm volume:
( )( )( ) ( )( ) ( )( )( ) ( )( )
( )
( )( )sf
cfs
cfs
cfscfs
cfs
sfsf
A
A
gh
QA
QQQQ
ghAghAQQ
ft
bfffa
bfff
ftftbfff
bftotbftotffbfff
021.0
6.927.932.32262.0
11.0
262.0
11.099.010.1
99.0
8.907.932.322022.0462.00.907.932.3220085.0362.0
262.0262.0
100
sec
100
100
100100
100
100
100
secsec
2
22
=
−=
=
=−=
−−=
=+
−+−=+
+=+
A 1” diameter orifice has an area of 0.0055 sf
82.3#0055.0021.0
#
=
=sf
sf
Therefore, use 3 - 1” diameter holes @ elev. 92.6
