CIV E 444 - APPLIED HYDRAULICS
SPRING 2016
LAB No. 11 SOLUTION

For a culvert type square edge with headwalls, use K_e = 0.5. Use K_E = 1 for free outlet. Use ONLINE CULVERT for the given data.
Results: Culvert diameter: D = 5 ft; critical depth y_c = 4.735 ft; normal depth y_n = 4.653 ft.
By trial and error, at the discharge Q = 375 cfs, the required culvert diameter would be D = 5.5 ft.
H_D = 1.0 m.
P = 170 - 155 = 15 m.
P / H_D = 15
From Fig. 7-10 (a) below: C_D = 0.492

Q = C (2g)^1/2 L H^3/2 = 53.15 C H^3/2
Spillway rating is shown in the following table. Head H (Col. 2) is elevation above the crest.
The ratio C/C_D (Col. 4) is obtained from Fig. 7-10 (b) (below) for each H / H_D.

Elevation (m)
(1)
H (m)
(2)
H / H_D
(3)
C / C_D
(4)
C
(5)
H^3/2
(6)
Q (m³/s)
(7)
170.0 0.0 0.0 0.00 0.000 0.00 0.000
170.2 0.2 0.2 0.842 0.414 0.0894 1.967
170.4 0.4 0.4 0.895 0.440 0.253 5.917
170.6 0.6 0.6 0.935 0.460 0.465 11.369
170.8 0.8 0.8 0.968 0.476 0.715 18.089
171.0 1.0 1.0 1.000 0.492 1.000 26.150
171.2 1.2 1.2 1.025 0.504 1.314 30.377
171.4 1.4 1.4 1.048 0.516 1.656 45.416
171.6 1.6 1.6 1.070 0.526 2.024 56.585

The same results are obtained when running ONLINE OGEE RATING.
ΔE = 0.674 ft, verified with ONLINE CHANNEL 12.
D_o/D₁ = 24 / 8 = 3
The pressure coefficient is defined as: V₁ = V_o (1 - C_p)^1/2
C_p = 1 - (V₁/V_o)² = 1 - (D_o/D₁)⁴ = - 80.
The free-vortex incipient cavitation index is: σ_{i, A} = 1 - 2 C_p = 1 - [2 × (- 80)] = 161
p_v = 37 psf
σ_i = (p_o - p_v) / (ρV_o² /2)
161 = [(25 × 144 in²/ft²) - 37)] / [ 1.94 V_o²/2]
V_o = 4.78 fps
Q = V_o A_o = V_o (π/4) D_o² = 4.78 (0.7854) (2)² = 15 cfs.