Determine whether a 5 ft × 5 ft reinforced concrete box culvert with 45o flared wingwalls and beveled edge at top of inlet carrying a 50-year flow rate of 200 ft3/sec will operate under inlet or outlet control for the following conditions. Assume ke = 0.5.

Question

Design headwater elevation (ELhd) = 105 ft
Elevation of stream bed at face of invert = 99.55 ft
Tailwater depth = 4.75 ft
Approximate length of culvert = 200 ft
Slope of stream = 1.5%
n = 0.012

What will be an ideal response?

Answer 1

Determine the required conditions for inlet control.
Calculate the flow rate per ft of width.
Q / NB = 200/5 = 40.0 ft3/sec/ft
Using Figure 16.16, draw a line connecting 5 ft to 40 ft to obtain the headwater
depth at culvert face (HW / D).
HW / D = 1.19 ft/ft
Calculate the required headwater, HW.
HW = (HW / D) × depth of culvert
HW = 1.19 × 5
HW = 5.95 ft
Neglect the approach velocity head in this problem. Therefore;
HWi = 5.95 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 = 5.95 – 5.45
Fall = 0.50 ft
Calculate the culvert invert elevation
invert elevation = 99.55 – 0.50 = 99.05 ft
Determine the required conditions for outlet control.
Determine the critical depth with Q / B = 200 / 5 = 40.0, from Figure 16.23.
dc = 3.7 ft
Calculate depth from outlet invert to hydraulic grade line and determine if TW is
greater.
(dc + D) / 2 = (3.7 + 5) / 2 = 4.35 ft
Since TW is not greater, use ho = 4.75 ft
Determine the total head loss (H) from Figure 16.21
H = 2.2 ft
Calculate the required outlet headwater elevation using Equation 16.23.
ELho = Elo + H + ho
ELho = (99.05 – (0.015)(200)) + 2.2 + 4.75
ELho = 103.00 ft
The required outlet headwater elevation (103 ft) is less than the design headwater
elevation (105 ft); therefore the 5 ft × 5 ft culvert is acceptable and inlet control
governs.