CE 4313: CE CAD---Exam 3 Notes

These are some supporting notes and calculations for the exam 3.

These are now in a spreadsheet: OOo xls.

  • Assume each storm sewer branch collects from 400' laterally.
  • The central branch has two sags
    • The further sag at 24+00 collects runoff from 39+00 - 17+00
      • 2200'x400' = 20.2 acres
      • average L ≈ 1500'
      • average slope ≈ 30/1500 = 2% ⇒ v= 1.0 ft/s
      • t_c=L/{60v} = 1500/{60*1}=25
    • The nearer sag at 9+00 from 17+00 - 0+00
      • 15.6 acres
      • average L ≈ 800'
      • average slope ≈ 10/800 = 1.25% ⇒ v=0.7
      • t_c=L/{60v} = 800/{60*0.7}=19
  • The western branch also has two sags
    • the further sag at 21+00 collects 24+00 - 18+00
      • 5.5 acres
      • average L ≈ 300'
      • average slope ≈ 1% ⇒ v=0.6
      • t_c=L/{60v} = 300/{60*0.6}=8.3→ Use 10
    • the near sag at 5+00 collects 18+00 - 0+00
      • 16.5 acres
      • average L ≈ 1600'
      • average slope ≈ 20/1600 = 1.25% ⇒ v=0.7
      • t_c=L/{60v} = 1600/{60*0.7}=38
  • The eastern branch has only one sag at 11+00
    • collects 24+00 - 0+00
      • 22.0 acres
      • average L ≈ 1200'
      • average slope ≈ 20/1200= 1.7% ⇒ v=0.8
      • t_c=L/{60v} = 1200/{60*0.8}=25

c = 0.3

b25 = 95 ; d25 = 9.6 ; e25 = 0.781

b50 = 103 ; d50 = 9.6 ; e50 = 0.781

t_c=L/{60v}

I={b}/{(t_c+d)^e} (in/hr)

I_25={95}/{(t_c+9.6)^0.781} =

I_50={103}/{(t_c+9.6)^0.781} =


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