455 10.16 Water-Surface Profiles in Gradually Varied Flow dy + M1 y >yo ye dx + Horiz. M1 Уo M2 М, dy dx yo Yo y> y Yo M1 M3 Ус Yo ye> y M3 J dx Yo Mild slope So< Se So< Sc S1 yo Horiz. Si dy dx y y yo + = + Ус S1 S2 yc> y> yo dy yo + dx Уo St ye yo> y S3 dy Уo dx Steep slope So> S So Se Horiz y >(yo yc) Yo yc C1 = + dy dx Critical slope So= yo yc So Se Нz dy H2 dx + Н y > yc На Hз dy y < Yc dx Нз ye So = 0 Horizontal bed So= 0 dy A2 A2 dx A3 y >y dy dx So
A trapezoidal channel with a 5-m bottom width and side slope m = 1.0 discharges 35 m3/s. The slope is 0.004 and is paved with smooth concrete (n = 0.012).
a. What is the critical depth?
b. What is the normal depth?
c. Is it a mild slope, or steep?
Hint: Compare critical depth with normal depth OR find the critical slope (??) and compare it with bed slope(?0)
d. Is the depth increasing or decreasing upstream? Why would that be (e.g. a spillway, dam, etc.)?
Hint: For the first part, use ????=?0−?1−??2 to show if the depth is decreasing/increasing (either by finding Fr and S, OR by simply using the values of ?, ?? and ?0 to figure out if the fraction is +/-).
e. How would you conceptualize the flow – what regime (e.g., S1, S2, S3, M1, M2, M3 in Figure 10.20)?
f. Determine the depth 3.3 m upstream from a section that has a measured depth of 1.69 m.
Hint: If you are using MATLAB, solve the GVF differential equation (????=?0−?1−??2) and find flow depth y at location x = -3.3 m. If you are using MS Excel, GVF equation #2 (Δ?=?1−?2?−?0) assuming only one reach. Use a solver (e.g. Goal Seek) to change flow depth (y) at location 2 until Δx is equal -3.3 m (see table below – at location 1, y=1.69 m, and at location 2, y=?).
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