Question

Does a 60” RCP culvert work for the 25 year flow rate design controls and prevent flooding of the critical structure given the following conditions: 200’ long culvert @0.30%. Invert elevation at tail end of pipe = 10.00. ODOT ½ headwall. Pavement elevation = 17.95. Critical Structure FFE = 18.65.

Q25 = 140 cfs, Q50 = 172 cfs, TW25 = 2.5’ TW50 = 2.9’

Answer 1

ANSWER
Reinforced concrete pipe Current Besign! A) 1) Eischarge : 295 = 140CFS 050 = 1490FS. Area of culvert = I xor A = 4602 A = 2897.43 in 2 A=1963 91.ft. Velocity 1- V- 140 19.63 V=7-13 fus (ii) using inlet ContooNomograph Chadt:- for 60" pipe and 225 =140643. HW = 11 Depth & head watey= HW HWEH X 60 12 HW-1-15 HW=5.5) Hw depth for CN pipe cutest with inlet Refey control chart.

@ L200 Om 60 pu! ! Discharge CFS 100 'Hw depth. (ilo) Length Culvest = 2009 slope 0.9 100 -0.003 chart head for concrete pipe cuherts floring n=0.018 Jan 1963 H value =) Head water for outlet Contool. HWCH tho-13

H-value=5 hos Tw or ((critical depth 0)) 200 50K-6-0-25 60 100+ V100 DETO) Head (4) ho=2.5 ho = (3.5+4) from cortical depth -67 circular pipe chart. ho=25 (01) $ (3-575) to =9.5 (01)4-85-→ Heae 4.25 is selected. Hw+5+4.25-(200X0.003) HW= 8.65 ft. (iv) maximum Head water Expected chigher of S-5' and 8.65' HW = 8.63) < 10.00 hence 60" pipe is acceptable. 6) () 650 = 142 Cks. =X602 A = 19.63 99.ft.

4 (a) using inlet Control homograph. Holo 1.3 Hw=1-35% bom HwE 6.75 Ft (11) H-value = 4.5 Hwe 4.5+4.95- (900X0-003) HW-3.15 (acceptable). ".