Repeat Problem 16-19 using a 6.5 ft diameter circular pipe culvert with ke = 0.5.

Question

What will be an ideal response?

Answer 1

Determine the required conditions for inlet control.
Using Figure 16.17, draw a line connecting 78 in to 200 ft3/sec to obtain the
headwater depth at culvert face (HW / D).
HW / D = 0.88 ft/ft
Calculate the required headwater, HW.
HW = (HW / D) × depth of culvert
HW = (0.88)(6.5 ft)
HW = 5.72 ft
Neglect the approach velocity head in this problem. Therefore;
HWi = 5.72 ft
Calculate the required depression. Use Equation 16.13 to determine the design
headwater depth, HWd.
HWd = ELhd – ELsf
HWd = 105 – 99.55
HWd = 5.45 ft
Determine the fall using Equation 16.14.
Fall = HWi – HWd
Fall = 5.72 – 5.45
Fall = 0.27 ft
Calculate the culvert invert elevation
invert elevation = 99.55 – 0.27 = 99.28 ft
Determine the required conditions for outlet control.
Determine the critical depth with Q = 200 ft3/sec, from Figure 16.24.
dc = 3.75 ft
Calculate depth from outlet invert to hydraulic grade line and determine if TW is
greater.
(dc + D) / 2 = (3.75 + 6.5) / 2 = 5.13 ft
Since TW is not greater, use ho = 5.13 ft
Determine the total head loss (H) from Figure 16.22
= 1.05 ft
Calculate the required outlet headwater elevation using Equation 16.23.
ELho = Elo + H + ho
ELho = (99.28 – (0.015)(200)) + 1.05 + 5.13
ELho = 102.46 ft
The required outlet headwater elevation (102.46 ft) is less than the design
headwater elevation (105 ft); therefore the 6.5 ft diameter circular culvert is
acceptable and inlet control governs.