1.060 Engineering Mechanics II Spring 2006Recitation 9 - ProblemsApril 27th and 28thProblem 1Figure 1 shows the cross-section of a circular gate (called Tainter gate) of radius R = 5 m. Thegate is placed in a rectangular channel of width b = 50 m and spans the entire width of thechannel. The depth upstream of the gate is h1= 6 m. The gate is partially opened, leavinga gap of height hg= 1.3 m between the gate and the bottom of the channel. The dischargeunder the gate per unit width of the channel is q = 10 m2/s.a) Determine the depth of flow h2, a short distance downstream of the gate opening.b) Determine the horizontal force from the fluid on the gate, per unit width of the gate.c) Determine the contraction coefficient for the flow under the gate.d) Classify the flow upstream and downstream of the gate as super- or subcritical.Figure 1: Flow under a Tainter gate in Problem 1.Problem 2Figure 2 shows a triangular channel cross-section. For a channel slope S0= 0.01, a dischargeQ = 50 m3/s, and a Manning’s n = 0.02 [SI units]:a) Determine the normal depth, h = hn, corresponding to uniform steady flow, and the associ-ated average velocity, V = Vn.b) Show that normal flow is supercritical.c) Determine the critical depth, h = hc.Due to the presence of a gate downstream, a hydraulic jump takes place in the channel, inwhich the depth transitions from h1= hnto a new value of the depth, h2.(Please turn over.)Recitation 9-1d) Determine the value of the depth h2and the corresponding average velocity, V2.e) Calculate the headloss and the rate of energy dissipation associated with the hydraulic jump.Figure 2: Triangular channel cross-section in Problem 2.Problem 3A rectangular channel of width b = 50 m carries a discharge of Q = 300 m3/s and has aManning’s n = 0.02 [SI units].a) If normal depth of flow in the channel is hn= 3.0 m, determine the slope of the channel.b) Is the normal flow sub- or supercritical?The channel is crossed by a bridge, which is supported by two bridge piers, each b/5 = 10 mwide, and placed in the channel as shown in Figure 3. The obstruction to the flow created by thebridge piers results in a depth of h1=3.2 m, a relatively short distance upstream of the piers (at1-1), whereas the depth downstream of the piers (at 2-2) is h2= 3.0 m (i.e., equal to the normaldepth). The bottom slope may be assumed sufficiently small to neglect differences in bottomelevation over the short distance between 1-1 and 2-2 and, consistent with this assumption, weneglect shear stresses acting on the wetted perimeter of the channel between 1-1 and 2-2.c) Determine the force, FP, exerted by the flow on each of the two bridge piers.d) Determine the headloss, ∆HP, from section 1-1 to 2-2.e) Establish an equation for the depth hMat the point denoted by M in the sketch anddetermine hM.Figure 3: Bridge piers in a rectangular channel in Problem 3.Recitation