PERMFILE66486

• Permit No: C1981020A
• IBM Index Class Name: Permit File
• Doc Date: 5/13/2002
• Section_Exhibit Name: Volume 6 APPENDIX As TO Cs
• Media Type: D
• Archive: Yes

A - 3
sg
5. Area 2 = 0.54 ac Slope = 50% Slope adjustment factor = 1.17
• Hyd. length = 500' Equivalent Area = 4.27 ac CN = 85
Runoff = 1.10" Peak Discharge = 4.6 CFS/inch of runoff
Equivalent Discharge = 1.10 X 4.6 X 1.17 = 5.92 CFS
Actual Discharge = 5.92 X 0.54 / 4.27 = 0.75 CFS
4100 = 0.75 cfs
6. Area 3 = 0.37 ac Slope = 507. Slope adjustment factor = 1.17
Hyd. length = 135' Equivalent Area = 0.48 ac CN = 85
Runoff = 1.10" Peak Discharge = l.6 CFS/inch of runoff
Equivalent Discharge = 1.10 X 1.6 X 1.17 = 2.06 CFS
Actual Discharge = 2.06 X 0.37 / 0.48 = 1.58 CFS
4100 - 1.58 cfs
7. Area 4 = 0.97 ac Slope = 189. Slope adjustment factor = 1.03
Hyd. length = 580' Equivalent Area = 5.48 ac CN = 80
Runoff = 0.82" Peak Discharge = 10.5 CFS/inch of runoff
Equivalent Discharge = 0.82 X 10.5 X 1.03 = 8.86 CFS
Actual Discharge = 10.07 X 0.97 / 5.48 = 1.78CFS
4100 1.78 cfs
8. Area 5 = 1.42 ac Slope = 35
• Hyd. length = 700' Equivalent
Runoff = 0.82" Peak Discharge
Equivalent Discharge = 0.82 X
Actual Discharge = 11.83 X 1.
Slope adjustment factor = l.ll
Area = 7.50 ac CN = 80
= 13.0 CFS/inch of runoff
13.0 X l.ll = 11.83 CFS
42 / 7.50 = 2.24CF5
4100 - 2.24 cfs
9. Design Segment 1 to contain the combined flows of Area 1,
Area 2. and Area 4.
4100 Segment l = 1.35 + 0.75 + 1.78 = 3.88 CFS
l0. Design Segment 2 to contain all flows.
4100 Segment 2 = 3.88 + 1.58 +2.24 = 7.70 CFS

2/16/89