I want a solution for the same question but instead of Rectangular culvert shape, I want circler culvert pipe

Subject related: - Pavement Drainage System

Department: - Highway and Transportation

What Exactly data you need? just replace Rectangular culvert shape by circler culvert pipe

Example: Determine the headwater elevation for the last Example, if the culvert is flowing full under outlet control, the tailwater depth above the outlet invert at the design flow rate is 6.5 ft, the length of culvert is 200 ft, and the natural stream slopes at 2%. Assume n=0.012, and k=0.5 Solution: . From the chart of Critical depth of rectangular section find the critical depth de from (Q/B) Q/B= 250/5 (from the previous example) =50 dc = 4.275 ft or: de = 0.315 instead the chart) dc+D How is the depth from the outlet to the hydraulic grade line 2 Or tail water depth TW whichever is greater. TW=6.5 ft. > Hiw=4.275+5 = 4.64 ft . Locate the size, length and ke of the culvert at A and B, then draw a straight line from A to B and locate the intersection C. Locate D on the discharge scale and draw a straight line C to D, extend this line to head loss MILD SLOPE scale at E. The total head loss (H) is obtained as 3.3 ft. The required outlet head water elevation ELis EL. = EL + H+Htw ROADWAY ELEVATION EL inv.: invert ele. At outlet. ELo=223.5 -0.02 * 200 =219.5 ft. W EL h=219.5 + 3.4 +6.5 =229.3 ft. ORIGINAL STREAM BD. EL SSTIL S. 229.3 Hiw=4.275+5 = 4.64 ft . Locate the size, length and ke of the culvert at A and B, then draw a straight line from A to B and locate the intersection C. Locate D on the discharge scale and draw a straight line C to D, extend this line to head loss MILD SLOPE scale at E. The total head loss (H) is obtained as 3.3 ft. The required outlet head water elevation ELis EL. = EL + H+Htw ROADWAY ELEVATION EL inv.: invert ele. At outlet. ELo=223.5 -0.02 * 200 =219.5 ft. W EL h=219.5 + 3.4 +6.5 =229.3 ft. ORIGINAL STREAM BD. EL SSTIL S. 229.3