SW6201202_Calcs (Ditch)_20210222

1691- Faithwill Farms Subdivision
Ditch Calculations Q = 2.53 cfs
Ditch #1 Ditch #5
Faithwill Ct Sta. 4+00 to 8+00 right Faithwill Ct Sta. 2+50 to 4+00 left
Tc=31.6 min; use30 min Tc=5 min
Kerby: L = 430 If; s = 0.005; n = 0.4 DA = 0.12 ac
DA=1.22ac C=0.8
C = 0.45 1 = 7.90 i n/h r
I = 3.86 in/hr
Q = 0.76 cfs
Q = 2.12 cfs
Ditch #6
Ditch #2 Faithwill Ct Sta. 0+00 to 2+50 left
Faithwill Ct Sta. 2+50 to 4+00 right
Tc=5 min
Tc = 21.7 min; use 15 min DA = 0.15 ac
Kerby: L = 191 If; s = 0.005; n = 0.4 C = 0.8
DA = 0.51 ac I = 7.90 in/hr
C = 0.45
1 = 5.33 i n/h r Q = 0.95 cfs
Q = 1.22 cfs Ditch #7
Ditch #3
Tc
= 30 min
Faithwill Ct Sta. 0+00 to 2+50 right
DA
= 3.60 ac
C =
0.45
Tc = 45 min
I =
3.86 i n/h r
DA = 3.09 ac
C=0.45
Q=6.25cfs
1 = 3.19 i n/h r
Ditch #8
Q = 4.44 cfs
Tc=45 min
Ditch #4
DA = 4.25 ac
Faithwill Ct Sta. 4+00 to 8+00 left
C = 0.45
1 = 3.19 i n/h r
Tc=5 min
DA=0.40
Q=6.10cfs
C = 0.8
1 = 7.90 in/hr

I = 3.19 i n/h r
Tc=5 min
DA=0.40 Q=6.10cfs
C = 0.8
1 = 7.90 in/hr
raiinwiii ui zjia. 4+uu io 6+uu ieu u = u.4.7
1 = 3.19 i n/h r
Tc=5 min
DA=0.40 Q=6.10cfs
C = 0.8
1 = 7.90 in/hr
ralinwiii ui Jia. 4+uu io 6+uu leii u = U.4.7
1 = 3.19 i n/h r
Tc=5 min
DA=0.40 Q=6.10cfs
C = 0.8
1 = 7.90 in/hr
ralinwiii ui Jia. 4+uu io 6+uu leii u = U.4.7
1 = 3.19 i n/h r
Tc=5 min
DA=0.40 Q=6.10cfs
C = 0.8
1 = 7.90 in/hr
ralinwiii ui Jia. 4+uu io 6+uu leii u = U.4.7
1 = 3.19 i n/h r
Tc=5 min
DA=0.40 Q=6.10cfs
C = 0.8
1 = 7.90 in/hr