Storm water flows at a depth of 4 ft in the natural channel as shown in the figure. The channel has a slope of 0.0l ft/ft and n = 0.025.
1. Determine the flow area:
Area 1: Tan σ^{1} = opp/adj adj = opp/Tan σ^{1} adj = 4/1.19 = 3.36 ft
(1/2)(b)(h) = (1/2)(3.36)(4) = 6.7 ft^{2}
Area 2 = (b)(h) = (9)(4) = 36 ft^{2}
Area 3 = Tan σ^{2} = opp/adj adj = opp/Tan σ^{2} adj = 4/0.84 = 4.76 ft
(1/2)(b)(h) = (1/2)( 4.76)( 4) = 9.5 ft^{2}
Total Area= Area 1 +Area 2 +Area 3 = 6.7 ft^{2} + 36 ft^{2} + 9.5 ft^{2} = 52 ft^{2}2.
2. Determine the flow velocity:
v = (1.49/n) (R)^{0.67} (S)^{0.5} R = A/P P_{total} = P_{Area 1} + P_{Area 2} + P_{Area 3}
P_{Area 1} = hyp_{Area 1} sin σ_{1} = opp_{Area 1} /hyp_{Area 1} hypArea 1 = 4/sin σ_{1} = 5.2 ft
P_{Area 3}= hyp_{Area 3} sin σ_{2} = opp_{Area 3} /hyp_{Area 3} hyp_{Area 3} = 4/ sin σ_{1} = 6.2 ft
P = P_{Area 1} + P_{Area 2} P_{Area 3} = 5.2 ft + 9 ft + 6.2 ft = 20.4 ft
R = A/P = 52 ft^{2} / 20.4 ft = 2.5 ft
v = (1.49/0.025)(2.5)^{0.67} (0.01)^{0.5} = (59.6)(1.88)(0.1) = 11.2 ft/sec
3. Determine the flow rate: Q = vA = (11.2 ft/sec)(52 ft^{2}) = 582 ft^{3}/sec
4. Determine the hydraulic grade line (HGL) at this channel section:
The hydraulic grade line is the depth of flow HGL = 4 ft
5. Determine the energy grade line (EGL) at this channel section:
P/γ = 4ft Z = 0 ft (bottom) v^{2}/2g = (11.2)^{2}/64.4 = 1.95 ft
EGL = 4ft + 0 ft + 1.95 ft = 5.95 ft