Finite Control Volume AnalysisMoving from a System to a Control VolumeConservation of MassSlide 4Continuity Equation for Constant Density and Uniform VelocityExample: Conservation of Mass?Linear Momentum EquationSlide 8Steady Control Volume Form of Newton’s Second LawResultant Force on the Solid SurfacesSlide 11Example: Reducing ElbowExample: What is p2?Example: Reducing Elbow Horizontal ForcesExample: Reducing Elbow Vertical ForcesExample: Fire nozzleFire nozzle: SolutionFind the VelocitiesSlide 19Example: Momentum with Complex Geometry5 Unknowns: Need 5 EquationsSolve for Q2 and Q3Slide 23Solve for FssxVector solutionVector AdditionMoment of Momentum EquationApplication to TurbomachineryExample: SprinklerSlide 30Energy EquationdE/dt for our System?General Energy EquationSimplify the Energy EquationEnergy Equation: Kinetic EnergyEnergy Equation: steady, one-dimensional, constant densitySlide 37Thermal Components of the Energy EquationExample: Energy Equation (energy loss)Example: Energy Equation (pressure at pump outlet)Example: Energy Equation (pressure at pump outlet)Kinetic Energy Correction Term: aSlide 43Example: Energy Equation (Hydraulic Grade Line - HGL)Example: Energy Equation (Energy Grade Line - EGL)EGL (or TEL) and HGLSlide 47Example HGL and EGLBernoulli vs. Control Volume Conservation of EnergySlide 50Power and EfficienciesExample: HydroplantEnergy Equation ReviewConservation of Energy, Momentum, and MassHead Loss: Minor LossesHead Loss due to Sudden Expansion: Conservation of EnergyHead Loss due to Sudden Expansion: Conservation of MomentumHead Loss due to Sudden ExpansionExample: Losses due to Sudden Expansion in a Pipe (Teams!)ScoopSlide 61Scoop Problem: ‘The Real Scoop’SummarySlide 64Scoop ProblemScoop Problem: Change your PerspectiveScoop Problem: Be an Extremist!Example: Conservation of Mass (Team Work)Example Conservation of Mass Constant VolumeExample Conservation of Mass Changing VolumeExample Conservation of MassPump HeadExample: VenturiSlide 74Example VenturiReflectionsTemperature Rise over Taughanock FallsHydropowerSolution: Losses due to Sudden Expansion in a PipeMonroe L. Weber-Shirk School of Civil and Environmental EngineeringFinite Control Volume AnalysisFinite Control Volume AnalysisCEE 331January 13, 2019Application of Reynolds Transport TheoremMoving from a System to a Control VolumeMassLinear MomentumMoment of MomentumEnergyPutting it all together!Conservation of MassB = Total amount of ____ in the systemb = ____ per unit mass = __ ˆsyscv csDMdV dADt tr r�= + ��� �V nˆcs cvdA dVtr r�� =-�� �V nmass1massBut DMsys/Dt = 0!cv equationmass leaving - mass entering = - rate of increase of mass in cvˆsyscv csDBbdV b dADt tr r�= + ��� �V nContinuity EquationConservation of Mass1 21 1 1 2 2 2ˆ ˆ 0cs csdA dAr r� + � =� �V n V n12V1A1If mass in cv is constantUnit vector is ______ to surface and pointed ____ of cvˆcs cvdA dVtr r�� =-�� �V nˆnnormaloutˆcsdAr � =�V nm�&ˆcsdAVA�=�V nVAr� =ˆnWe assumed uniform ___ on the control surface is the spatially averaged velocity normal to the csV[M/T]Continuity Equation for Constant Density and Uniform Velocity1 21 1 2 20V A V Ar r- + =1 21 2V A V A Q= =1 21 1 1 2 2 2ˆ ˆ0cs csdA dAr r� + � =� �V n V nDensity is constant across csDensity is the same at cs1 and cs2[L3/T]Simple version of the continuity equation for conditions of constant density. It is understood that the velocities are either ________ or _______ ________.1 1 2 2V A V A Q= =uniform spatially averagedExample: Conservation of Mass?The flow out of a reservoir is 2 L/s. The reservoir surface is 5 m x 5 m. How fast is the reservoir surface dropping?resAQdtdhhdtdhAQresoutExampleConstant densityˆcs cvdA dVtr r�� =-�� �V nˆcsVdAt�� =-��V nout indVQ Qdt- =-Velocity of the reservoir surfaceLinear Momentum Equationm=B Vmm=Vbcv equationmomentummomentum/unit massSteady stateˆsyscv csDBbdV b dADt tr r�= + ��� �V nˆcv csDmdV dADt tr r�= + ��� �VV V V nˆcsDmdADtr= ��VV V n0F ��This is the “ma” side of the F = ma equation!Vectors!Linear Momentum Equation( ) ( )1 1 1 1 2 2 2 2DmV A V ADtr r=- +VV V   111111VVM QAV   222222VVM QAVAssumptionsVectors!!!Uniform densityUniform velocityV  ASteadyˆcsDmdADtr= ��VV V n1 21 1 1 1 2 2 2 2ˆ ˆcs csDmdA dADtr r= � + �� �VV V n V V nV fluid velocity relative to cv( )1 2D mDt= = +�VF M MSteady Control Volume Form of Newton’s Second LawWhat are the forces acting on the fluid in the control volume?21MMF 1 2 wall wallp p p t= + + + +�F F F F FWGravityShear at the solid surfacesPressure at the solid surfacesPressure on the flow surfacesWhy no shear on control surfaces? _______________________________No velocity tangent to control surfaceResultant Force on the Solid SurfacesThe shear forces on the walls and the pressure forces on the walls are generally the unknownsOften the problem is to calculate the total force exerted by the fluid on the solid surfacesThe magnitude and direction of the force determinessize of _____________needed to keeppipe in placeforce on the vane of a pump or turbine...1 2p p ss= + + +�F F F FWwallwallFFpssF=force applied by solid surfacesthrust blocksLinear Momentum Equation1 2p p ss= + + +�F F F FW1 2m = +a M M1 21 2 p p ss+ = + + +M M F F FWThe momentum vectors have the same direction as the velocity vectorsFp1Fp2WM1M2FssyFssx( )1 1Qr=-M V( )2 2Qr=M VForces by solid surfaces on fluidReducing elbow in vertical plane with water flow of 300 L/s. The volume of water in the elbow is 200 L.Energy loss is negligible.Calculate the force of the elbow on the fluid.W = ________________________section 1 section 2D 50 cm 30 cmA ____________ ____________V ____________ ____________p 150 kPa ____________M ____________ ____________Fp____________ ____________Example: Reducing Elbow1 21 2 p p ss+ = + + +M M F F FW121 mzxDirection of V vectors0.196 m20.071 m21.53 m/s ↑ 4.23 m/s →-459 N ↑ 1270 N →29,400 N ↑g*volume=-1961 N ↑??←Example: What is p2?2 21 1 2 21 21 22 2p V p Vz zg gg g+ + = + +2 21 22 1 1 22 2V Vp p z zg gg� �= + - + -� �� �( ) ( )( ) ( )( )2 23 3221.53 m/s 4.23 m/s150 x 10 Pa 9810 N/m 0 1 m2 9.8 m/sp� �-� �= + - +� �� �P2 = 132 kPaFp2 = 9400 NExample: