Question

Problem #2 Design a culvert that is to be placed under a new roadway for a 25-year peak discharge of 180 cfs. Use a circular pipe culvert with beveled ring edges. Assume 45° bevels and a Manning's roughness coefficient of 0.025 for the culvert). The site conditions include: natural stream bed elevation at culvert inlet at 100 ft: culvert and culvert barrel slope of 1 %; tailwater for 25-vear flood equal to 3.5 ft; approximate culvert length of 200 ft; shoulder elevation of 110 ft. Base the design headwater on the shoulder elevation with a 2-ft freeboard (i.e. elevation of 108 ft). Set the inlet invert at the natural stream bed elevation Assume a culvert diameter of 5 ft. Determine if your design is adequate for the given conditions. (Hints: Check for both inlet and outlet control conditions. Use a graphical approach to determine yo.)

Answer 1

Solution : Given thal: A New dood way for a 25-year. peak des charge - 180CFS. Assume 45° bevels and a manning's ooughney co-efficient 20.025 for the culvent → The site conditions include: Nalural stream bed elevalion of Culverd intel 2100 Ft Calvert an d culvert barrel slope = 1) talwaler for 25 year flood equal to 3.567 Approrúmale Cellvert length = 20016. shoulder eleveition = 110 Ft. Bose the design head valer on the Shoulder elevo. tion with free board = $- Ft Cie elevation - 108 tel. Assume a calvest diameter - 5ft. - Take the inlet Control Submerged condition calculate the head maler. Scanned with CamScanner

Here, the table shows the constant value constants for inlet Control Design equatiory. un submerged. Submerged charta no Shape and Nomo greeph material scale Inlet edge description equalion 5 M Form 1. 0.0098 2-0 0-0398 067 Circular concrete 418 cellay Square edge whead wall Groove and whead Wall Groove and projecting 0.0078 2-0 0.0292 0.74 0.0045 2.0 0-317 0.69 circular CMP m - dm Head hall mitered to Slope projecting 0.0078 20 0879 0.69 0.0210 1.33 0.0463 0.75 0.8340 150 0-0553 0-54 3. circular A B CScanned with CamScanner Beveled ping, 45° bevels Beveled ring, 33-1 bevely 0.0018 2.500.0300 0.74 0:0018 2-50 0.0243 0-83 !

Therefore, НА tytz ADO 5 L (H) 2000 euro 09-04-6 HW = 6.2Ft. → check the normal depth. = tega & 43 5'z Ā - p7352 213 =149 LMP) 180 1-49 0.025 Vz - 180 3.1.4 Cyn?) 3. 14 CV) Yn = 5.56 ft a Yr – eft Revise the diameter of the culvert og 6ft. Scanned with CamScanner

calculate the head water. +y+z 18 0 SHW T = 0.0 Pancas tou Color 0.03 1 (3.14 1 +0.74 - → HW = 5.6377 Find Critical depth in the circular using the condition for minimum Specific energy @?. A g To Substitute, Ac = $ loc-Sin@c] do? Te = do sin (0) do =6Ft g = 32.214/82 0 — 18о се ☆ x(@ - Sinoc )x627 6 xsin 2 180 32. 2 Scanned with CamScanner

1803x3483 (oc - sinoc 13 30.2 x (6²)3 sin (oc). Oc = 150° 59'13.6" → Circular Cellvert is partially fello Draw the water level above the centre Calvert (Oc 7180°), at a coitical flow conditione as in tigure. do=6ff a 154°59113.6" < Denote the vertical distance bělween the centre of calvent and water surface as a → use the right angled triangle to computer og in figure 6FL Seons '49.64 Scanned with CamScanner

→ Find the value of a using the toomula. cos (80°58'49.62') = /6 → x= 0.94 Ft Find the critical depth using the delation Ye = dosta Yc +20.94) Y = 3.94ft → The critical depth of flow in the circular Culvert is 3.guft which is less than the 6ft, hence ok. the welvert is inlet controls. Note: Hope this was helpful please do leave positive rāling if you liked answer thanks and have a good day Scanned with CamScanner