RECITATION #3 FLOW NET CONSTRUCTION Some Simple Rules Solid boundaries are (special) streamlines Interior streamlines are guided by boundaries but smoother (no kinks or sharp corners) Streamlines can never cross Streamlines can end only at flow areas Region(s) where flow is uniform - straight parallel streamlines - is a good place to start (end). Streamlines divide flow into streamtubes Discharge in a streamtube is constant, DQ. Velocity in a streamtube is V=DQ/Ah Ah = distance between adjacent streamlines Streamlines & Equipotential lines (h-lines) form a square pattern. Example No:1 Flow Contraction 1 (2) Q 1. For to the left of contraction, flow is uniform 9 /3 V * V 0Q streamlines straight parallel to boundaries 1/3 Q h, V 3 streamtubes formed by 4stream= hines Q-V, Sh, Vih/3Q13 2: Far to the right of contraction, flow is uniform ② 40 streamlines shaights parallel to boundaries Q ② Q=ha Va; 3 sheamtubes are 3 stream fubes we connected to the had of 50 each canies a discharge Q his = AQ= h Ve 1/3 she h₂ 2 When Ah → is small Vis longe 104 1/3 * 13. 2 RECITATION #3 Problem No:1 1) Draw the flow net connecting the uniform inflows at 1-1 and the uniform outflow at 2-2, using 4 streamtubes, on attached page 2) Determine the ratio of outflow velocity V₂ to inflow velocity V₁, if the depth of flow is h₁ = 3.5m and h₂ = 0.75m. 3) If it is assumed that P = 0 at the top boundary of the inflow section 1-1 and also at the top of the outflow section 2-2, determine the discharge Q (m³/s per m into paper), V₁, and V₂. 4) Estimate the pressure at the corner C 5) Estimate the pressure on the horizontal bottom immediately below the vertical wall, B. Problem No:2 Sketch the flow net (in <10min) for the ground water flow under the sheet pile wall shown on the attached page. 1) Identify equipotential lines and streamlines before starting. 2) Draw your flow net - stop after <10min. 3) Determine the number of streamtubes Ns obtained from your flow net. 4) Determine the number of equipotential head drops Nh from your flow net. Q = K (h₁ - h₂) * Ns / Nh 5) What is your 10min answer for the important quantity Ns / Nh? RECITATION #3 Problem #1 a) For a perfect flow net (see attached), note that your flow net was for a conduit, whereas the attached is for a free surface flow under a gate. The main difference is that the upper boundary is a free surface, which is a streamline along which p=0 and varies and not a horizontal wall along which p=const. and pressure may vary. b) Well behaved flow at 1-1 and 2-2 means that volume conservation yields Qin = V₁h₁ = Qout = V₂h₂. V₂/V₁ = h₁/h₂ = 3.5/0.75 = 4.67 c) If p=0 at 1-1 and 2-2 on upper boundary (it's a free surface in reality), so p=0 is correct. And flow is well behaved with straight parallel streamlines. The pressure varies hydrostatically along lines L (streamlines) - so the case this direction is 2. So, we have p₁ + ρgz₁ = ρgh₁ anywhere along 1-1 and p₂ + ρgz₂ = ρgh₂ anywhere along 2-2 Bernoulli along any streamline from 1-1 to 2-2 therefore gives

Flownet and Streamline Construction

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