RECITATION 10 - SOLUTIONS -PROBLEM N° 1: Since the critical depth depends only on the geometry of the section and the flowrate, it has the same value for all four stretches: hc = (Q²/g)²/³ = 1.36 m The normal depths (hn) are calculated from Manning's equation by iteration: hn = (Q/nS)^(3/5) / (1+2/R) Since the channel is rectangular, the conjugate depth of hn is given by: hn_conj = [-1 + √(1+8Fn²)] / 2hn where Fn = V/√(g * hn) Applying these expressions, we construct the following table: | STRETCH | S_0 | hn [m] | hn_conj [m] | hc [m] | M or S? | |---|---|---|---|---| | 1 & 2 & 4 | 0.001 | 1.411 | 0.93 | 1.36 | M | | 3 | 0.006 | 1.06 | 1.72 | 1.36 | S | | 2 | 0.02 | 0.72 | 2.33 | 1.36 | S | With this information and using the boundary conditions for each stretch, we draw the following sketch: In the outflow from the gate, since hn3, conj = 0.93m > 0.72m = Cv.h6, there is a hydraulic jump from H3 to H2 (Otherwise, we would have a drowned outflow). • In the transition between stretches 3 and 4, since hn3, conj = 0.93m < hn4 = 1.06 m (or equivalently, since hn3, conj = 1.72m < hn4, conj = 1.91m), hn4 has more "MP" than hn3 and it gets into the steep slope. So the hydraulic jump happens in the step above, from S3 to S4. h6 = 1.36m 1.91m 0.93m S2 1.72m S1 M1 0.72m 1.91m h2m Finally, if ho is the alternate depth of hve, i.e., ho + V²/2g = hve + Vve²/2g => hc = 3/2(hve) => hc = 3/2(0.72) = 1.08 m - PROBLEM N°2: Manning: S_0 = n²V²/R^(4/3) S_0 = 0 => n²V²/R^(4/3) = 0 => n²V² = 0 for all R No Z1 A1 Z2 A2 Other than that, the curves Z2, Z3 and A2, A3 look similar to M2, M3, etc. - PROBLEM N°3: n=0015 n=003 STRETCH So hn hn M or S? hc 1 0.005 0.65 m S 1.04 0.79 m 2 0.008 0.56 m S 0.89 m 0.79 m 3 0.015 0.44 m M 0.79 m 0.79 m 4 0.03 0.34 m M 0.79 m 0.79 m a) n=0015 Both stretches are S. Hypothesis: Flow in stretch 1 will proceed with hc=hn1 if it has enough energy to pass the step.