Question

   number 7 is what I'm asking about

here's Macon Ga table data

dit View History Bookmarks People Window Help 19% E) Sat Mar 23 4:5 Invalid xESCPXWater Filel/Users/Steven/Downloads/ENVE%204435%20Fall%202016%20Exam%20%232%20Practice%20Problems%20[Post].pdf a You have been asked to design a culvert to handle the discharge for a 50-year storm on a 19-acre catchment in Macon, Georgia Neighborhoods constitute 60% of the land use with the remainder being open space lawns, average condition, with clay soils and average slope. The catchment is characterized hydraulically by a 1235 ft. trapezoidal-shaped channel covered with dense grass along a 2.1% slope (Bottom Width-3 ft.; water Depth -0.75 ft.; Side Slopes 3:1) [See below cross-section for trapezoidal-shaped channel]) and 120 feet of overland flow in the upper sections where the cover is open space lawn/grass, sandy soil, and 2.5% slope. Calculate the design discharge. 7. Flow Depth Side Slopes Bottom Width 8. Stormwater runoff from an urbanized area has been measured in the field to be 162.9 ft3/sec and discharges into a nearby stream. The stream has a bottom width of 15 feet and a slope of 1.8%. It has also been determined that the flow cannot exceed 2.1 feet. Determine an appropriate stream channel material that can be used in order to manage the discharge so that it flows at the maximum depth. 9. A gravel bottom trapezoidal open channel with formed concrete sides is to convey 188.44 cfs. The channel has a bottom width of 10 ft., flow depth of 2 ft., and side slopes of 2:1. At what slope must the channel be prin https:/hdsc.nws.noa. govhdsc pfds/pfds Pre Precipitation Frequency Data Server NOAA Atlas 14, Volume 9, Version 2 Location name: Macon, Georgia, USA Latitude: 32.8386, Longitude: -83.6203 Elevation: 282.04 ft" sourte ESRI source USG POINT PRECIPITATION FREQUENCY ESTIMATES Sana Perca Deborsh Masin Sandra Paviovie, ishani Roy, Mchsel St. Laurent, Carl Trypeuk, Dsle Unnuh Mchael Yes, Geoftery Bonnin NOMA Narional Weather Service Siver Spring Maryland PF tabular I PF graphical | Maps & aenials PF tabular PDS-based point precipitati on frequency estimates with 90 % confidence intervals in inches/hour)! Average recurrence interval (years) 200 1000 25 50 3 80-6.11) 4 43-7.13 15 47-884) 6.32-10.3) 734-12(8 11-143 (878-16.2) 19 36-18 2) 10 10-min3.53 8.91 (2 78-d 47) | (З 25-522; | 14,00.6.47) | |4 63-754) | I5 38-9 201 | (5 94-10 5) | (6 43-11 8| | (6 85-13. 31, | 〔748-153). I 17.95-16 8) 2 26-363) (2.64-424) 326-526) 1376-6.13) 4.37-748) (483-8 50) (5.22-9 62) 5 57-10 8) (6.08-12.4) (646-136) 30-min 1.70-273) (1.98-3 19) 12.43-3.96 284-4.62 (3.30-54) (365642) (35-726) (4.21-8.17) (4 59-9.38) (4.88-10.3) 60-min 1.2-1.80) (1.31-210) 1.62-2.62) 188-3 06) (2 20-3.77) (244-4.30) (266-4.90 (285-5.54) (3.14-6.41) 3 35-7.06) 0.883 |(0 706-11 2-hr | 10 822-1 2911 (112-161| |(1191 BBI | (139-233) 11.55-2.57) 111.69-3.05) | (182,346) 1 (201-402) 1 (2164 44) 0.656 0.946 3-hr 0 528-0 813) ko 813-0 946) (0759-1 18) (0 863-1.38) 1.04-1.72 (1.17-197) (120-2.26) 1.38-2 58) (1.53-302) (1.65-3 35) 0 321-0482) (ro 370-0 555) 10 453 0 685) (0 526-0 801) 10.623-1.00) (0697-1 15) (0766-1.33 (0831-152) 0 928-1.79) 1.00-1.99) o 193-0 282)0219-0.321) 0114-0 162) ko 130.0 1841 TO 157 0.224) ka 180-0 260X0 212:0 Ζ3) Ito neo 370) 10 255.0 4251 ko 280.0 4861031,-0 sm D D66-0.091)|(0 075-0 104) 110 091-0 127)10 105-0 148)10 123-0. 183)10 137-0.209) io 150.0 240) k0.162-0274) 10 179.0 321) 0 049-0 055) 0 055-0 075)o 086-0091)0076-0.105) 0.089-0.129) 10.c98-0.148) (0 107-0 1701 ( 0 039-0 053) 00 044-0 060)0053-0.02) 100510 083) k0071-0102)8-0116085-0 133) 0 027-0 035,0 030-0 039) 035-0 047) 10.040-0 054) (0.047-0 085) o 052-0.075)0 056-0 088) 0.061-0 008 (0.067 0.021-0 028) 0 024-0.031) (0.028 0 037) (0 032-0042 10 037-0 051) ko.040-0.058) 10 044-0 056) 0047-0075) 0052-0.088 (0056-0.097) 0.015-0.018) (0 016-0 020) 0 019-0.024 (0 021-0 027) o.024-0.032) o 026-0.036) 10 028-00400029-0.045)(0032-0.052)(0.034-0 057) 0 012 0 015) Ho 013-0.016) 0 015-0019) k0.017-0 021) 10.019-0.025)(0 020-0.027) 0 021-0.031 (0 022-0.034 0024-0 039 (0.025-0.042) 0 010-0012) o 011-0 013) ko 013-0.015) (0 014-0017) (0015-0020 (0 016-0 021 0017-0 024 (0017-0 026) 10018-0029) 0.019-0 031) 0.009-0.0 0.903 0.323 0.266-0 39% |[O 307.0454) ko 362.0 566) |(0 404-о eso) Iro 443.0 7481 ko 481-0 857) |

Answer 1

Step-1 : Calculate the culvert weight

culvert weight = D+T1

Step-2: Find the horizontal distance of W1 perpendicular to the road alignment using the batter slope :

horizontal distance perpendicular to the road,(W1) = (D+T1)×S - T2

Step-3 : Find the horizontal distance of W2 perpendicular to the road alignment using the batter slope :

horizontal distance perpendicular to the road,(W2) = (D+T1)×S – T2 – (D2 – T1) sin⁡∅

Step-4 : For square culverts :

W1 angle = Skew angle + α from the table of angles

W2 angle = skew angle - β from the table of angles

Step-5 : Calculate W1 :

W1 = [(D+T1)×S-T2]÷Cos (∅+∝)

Step-6 : For skew culverts, adjust the initial perpendicular calculation of W2 for the offset created by the part of the headwall that is square to the culvert cell.

Check that is at least 1.5 x the culvert height. If 1.5 x (D+T1) is greater than the above horizontal distance, then choose 1.5 x (D+T1) as the horizontal distance.

Step-7 : Calculate W2 :

W2 = Horizontal distance from above / cos( ∅−