Elementary Mechanics of FluidsFlow Past a Flat plateShear Stress CoefficientsEx (9.77)Slide 5Lift of an AirfoilDrag of a Thin PlateDrag CoefficientsHW (11.4, 11.7, 11.56)Ex (11.3)Ex (11.8)Ex (11.57)Elementary Mechanics of FluidsCE 319 FDaene McKinneyResistanceFlow Past a Flat plate•Boundary layer: Region next to an object where fluid has its velocity changed due to shear resistance of the boundary.•Velocity gradient exists between free stream and object, thus shear stress exists at surface which retards the flow.•Boundary layer grows in downstream direction, until the onset of turbulence•Shear stress is high at leading edge and decreases until transition and then increases again000,500ReTransition,RePlates000,2ReTransition,RePipesVxVD   LLoFLoxoUAFCBLUFURe1520Re*06.0ln523.0:tCoefficienRe072.05.0:ResistanceRe058.05.0:StressregionTurbulent225/125/12LosFLoxoUBLFCUBFxURe33.12/:tCoefficienRe664.0:ResistanceRe332.0:StressregionLaminar2Shear Stress Coefficients•Shear stress coefficient = ratio of shear stress at wall to dynamic pressure of free stream•Total shearing force2/20ofUc2/2220osfAfoAsUFCdAcUdAFEx (9.77)Given: Ship prototype 500 ft long, wetted area of 25,000 ft2, and velocity of 30 ft/s in sea water at 10oC. Model is in fresh water at 60oF, model:prototype scale = 1/100, and Froude numbers are matched. Drag is calculated as flat plate with model wetted area and length. A drag of 0.1 lbf is measured in the model tests.Find: Drag on the ship.sftVLLVVgLVgLVFrFrmpmpmppmmpm/330*)10/1(1011001lbfFFFlbfFAVCFCxxLVLsheartotalwavemfmLLfL0361.00639.01.00639.05.2*3*94.1*5.0*)00293.0(200293.0Re1520)Re06.0(ln523.01023.11022.15*3Re22265Froude No. similarityShear resistance on modelEx (9.77)   22222/2/ppmmpmpmppmpVVppVpVpCC lbfxFFFLLAAVVAAppFFpmppmpmpmppmmpmpmpm4633322107.310*94.199.1*0361.010094.199.1100199.194.1Scale up of wave dragForce on PrototypeLift of an Airfoil•Forces acting on airfoil: Velocity over the top of foil is greater than free stream velocity, pressure there is less than freestream. Similarly, the pressure on the bottom of the foil is greater than freestream pressure. This difference in pressure contributes to the lift of the foil.•Shear stress along the foil acts to drag on the foil.Drag of a Thin Plate•For a plate parallel to the flow, shear forces are the only ones acting•For a plate normal to the flow, shear and pressure forces act•For a more general object222ofDVBLCF2)2.18.0(2oDVBLF22opDDVACFDrag Coefficients•Coefficient of Drag22opDDVAFCHW (11.4, 11.7, 11.56)•11.4Ex (11.3)•Given: Pressure distribution is shown, flow is left to right.•Find: Find CD•Solution: CD is based on the projected area of the block from the direction of flow. Force on downstream face is:The total force on each side face is:The drag force on one face is:The total drag force is:Coefficient of Drag is: CD=1 2/5.02/22VAVACFpppDragD2/5.02/22VAVACFpppS 5.0*2/5.0sin2VAFFpSDragS2/2/5.0)5.0*2/5.0(*2)()(2222VACVAVAFFFpDppDragDDragSDragEx (11.8)•Given: Flag pole, 35 m high, 10 cm diameter, in 25-m/s wind, Patm = 100 kPa, T=20oC•Find: Moment at bottom of flag pole•Solution:325/20.1,/1051.1 mkgsmx 551066.11051.11.0*25Re xxVD5)11(figure95.0 DCkNHVACHFMopDD8.21235*225*2.1*35*10.0*95.022222Ex (11.57)•Given: Spherical balloon 2-m diameter, filled with helium @std conditions. Empty weight = 3 N.•Find: Velocity of ascent.•Solution: WBWHeFBFDDDpDDoopDDCCACFVVACF739.0225.1*2*)4/(422.1*22222NDFDFDWWFFFairHeairDHeDairHeBDBy422.126)077,22871(36)(363603333Iteration: Guess CD=0.4smVo/36.14.0739.0Check Re551086.11046.12*36.1Re xxVDIteration: