# MAAE 2300A Fluid Mechanics lab 3 .docx - MAAE 2300A Fluid...

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MAAE 2300A FluidMechanicsEXPERIMENT 3: Flow through a Sluice Gate andHydraulic JumpLABREPORTName: Yasser Ammouri Student ID: 100850436Group Members:1.Layth Naji 2.Mohamed Ahmed Lab: L8
Summary: The purpose of this experiment is to apply what was taught in the course to analyze the flow of water through a sluice gate and hydraulic jump, and so test the predicted values that are calculated from equations. To do that two water flows were examined and the data was recorded. Bernoulli’s equation and other derived equations in the lab were used to do the calculations. Nomenclature:SymbolDefinition UnitsPStatic PressurePsfvFluid velocityft/szWater depthftATank cross-sectional areaft2´mMass flow rate of waterlbm/sHHead ftwTank widthftQVolumetric flow rateft3/sHVnotchWater level at zero flowftTable 1: NomenclatureAssumptions:There were some assumptions taken while doing this lab like: 1.The water flow is 1D flow.2.The fluid is incompressible.3.No friction with the walls during wate4r flow.4.Steady state flow: no or little change in conditions over time.5.The pitot tube is directly aligned in the direction of the flow and the bottom of the tube.6.No energy losses
Flow analysis:Figure 1: The flow through a sluice gate and a hydraulic jump showing the control volume .Defining a stream line between points 1 and 2, we start with Bernoulli’s equation across the sluice gate and we obtain:P1+12ρv12+ρg z1=P2+12ρv22+ρg z2Since P1=P2=Patmand we are assuming incompressible flow then we can eliminate the pressure and density terms from equation 1 to get:12v12+g z1=12v22+g z2Now we must use the continuity equation between points 1 and 2 and find v1in terms ofv2: ´m1= ´m2→ ρ1v1A1=ρ2v2A2Assuming incompressible flow and rearranging for v1simplifies equation 3 to:
v1=v2A2A1Substituting A=wzin equation 4 gives: v1=v2z2z1Now we will sub equation 5 into equation 2 to get: 12v22z22z12+g z1=12v22+g z2Rearranging equation 6 for v2we get:z2z1¿¿¿¿¿2g(z2z1)¿v2=¿To calculate the actual v2we need to first calculate the flow rate in the channel using the V-notch weir at the downstream end of the flume:Q=1.38H2.5Now using the continuity equation and rearranging equation 8 for v2:v2=1.38H2.5A2Next we need to derive an expression for the depth downstream of the hydraulic jump. From the controlvolume in figure 1 we see there are forces acting on it. First we will find an expression for forces 2 and 3; since neither pressure is constant, we integrate over the area:
F2=P2dA=(ρg z2)w dz=12ρg z22wF3=P3dA=(ρg z3)w dz=12ρg z32wNow we will find an expression for v3
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