FALL 2008 - MIDTERM 2 - SOLUTION
So = Sc F12 F12 = So / Sc = 0.0035 / 0.0035 = 1 F1 = 1 ANSWER. F12 = v2/(gy1) = q2/(gy13) = 1 y1= (q2/g)1/3 = (2.52/9.81)1/3 = 0.86 m. ANSWER. v1 = q/y1 = 2.5 / 0.86 = 2.91 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 = (4 - 0.86) / 0.0035 = 897.14 m. ANSWER. Problem 2
So = Sc F12 F12 = So / Sc = 0.0035 / 0.0035 = 1 F1 = 1 ANSWER. F12 = v2/(gy1) = q2/(gy13) = 1 y1= (q2/g)1/3 = (2.52/9.81)1/3 = 0.86 m. ANSWER. v1 = q/y1 = 2.5 / 0.86 = 2.91 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 = (4 - 0.86) / 0.0035 = 897.14 m. ANSWER.
F12 = So / Sc = 0.0035 / 0.0035 = 1
F1 = 1 ANSWER.
y1= (q2/g)1/3 = (2.52/9.81)1/3 = 0.86 m. ANSWER.
Δx = Δy / So = (y2 - y1) / So = (4 - 0.86) / 0.0035 = 897.14 m. ANSWER.