FALL 2013 - MIDTERM 2 - SOLUTION
So = f F12 F12 = So / f = 0.0025 / 0.0025 = 1 F1 = 1 ANSWER. F12 = v2/(gy1) = q2/(gy13) = 1 y1 = (q2/g)1/3 = (0.42/9.81)1/3 = 0.254 m. ANSWER. v1 = q / y1 = 0.4 / 0.254 = 1.575 m/s ANSWER. Because the u/s flow is critical, the water-surface flow profile type is C1. ANSWER. So = Δy/Δx Δx = Δy / So = (y2 - y1) / So = (1.5 - 0.254) / 0.0025 Δx = 498.4 m. ANSWER. Problem 2
So = f F12 F12 = So / f = 0.0025 / 0.0025 = 1 F1 = 1 ANSWER. F12 = v2/(gy1) = q2/(gy13) = 1 y1 = (q2/g)1/3 = (0.42/9.81)1/3 = 0.254 m. ANSWER. v1 = q / y1 = 0.4 / 0.254 = 1.575 m/s ANSWER. Because the u/s flow is critical, the water-surface flow profile type is C1. ANSWER. So = Δy/Δx Δx = Δy / So = (y2 - y1) / So = (1.5 - 0.254) / 0.0025 Δx = 498.4 m. ANSWER.
F12 = So / f = 0.0025 / 0.0025 = 1
F1 = 1 ANSWER.
y1 = (q2/g)1/3 = (0.42/9.81)1/3 = 0.254 m. ANSWER.
Δx = Δy / So = (y2 - y1) / So = (1.5 - 0.254) / 0.0025
Δx = 498.4 m. ANSWER.