Elementary Mechanics of FluidsEuler EquationEx (5.1)EX (5.3)HW (5.7)Ex (5.10)HW (5.11)Bernoulli EquationEx (5.47)ExampleExample – Venturi TubeEx (5.48)Ex (5.49)HW (5.51)Stagnation TubeStagnation Tube in a PipePitot TubePitot Tube Application (p.170)HW (5.69)HW (5.75)HW (5.84)HW (5.93)Elementary Mechanics of FluidsCE 319 FDaene McKinneyBernoulliEquationEuler Equation•Fluid element accelerating in l direction & acted on by pressure and weight forces only (no friction)•Newton’s 2nd LawgazpdldadldzgdldplalpppAalWAppApMaFllllll)(sin)(sin)(Ex (5.1)2.0)5.03.0(30sin)3.0()(dldpggdldzagdldpgazpdldollFlow30ol•Given: Steady flow. Liquid is decelerating at a rate of 0.3g.•Find: Pressure gradient in flow direction in terms of specific weight.EX (5.3)•Given: = 10 kN/m3, pB-pA=12 kPa. •Find: Direction of fluid acceleration.AB1 mverticalup)ision(accelerat0)12.1()1000,10000,12()1()1()(gagappgadzdzdzdpgagazpdzdzzBAzzzHW (5.7)•Ex (5.6) What pressure is needed to accelerate water in a horizontal pipe at a rate of 6 m/s2?323/6000/6*/1000)(mNdldpsmmkgadldpadldzdldpgazpdldlllEx (5.10)•Given: Steady flow. Velocity varies linearly with distance through the nozzle.•Find: Pressure gradient ½-way through the nozzleftftlbfftsftsftftslugsdxdVVadxdpgazpdxdxx//355,5)//50(*)/55(*)/94.1()()(23V1/2=(80+30)/2 ft/s = 55 ft/sdV/dx = (80-30) ft/s /1 ft = 50 ft/s/ftHW (5.11)Bernoulli Equation•Consider steady flow along streamline•s is along streamline, and t is tangent to streamlineConstant202211)(222gVzpgVzpdsdgVdsddsdVVgagzpdsdtheaddynamicVelocitygVheadcPiezometrizp)(22gVzpgVzp2222222111Ex (5.47)•Given: Velocity in outlet pipe from reservoir is 6 m/s and h = 15 m.•Find: Pressure at A.•Solution: Bernoulli equationkPapgVhpgVpghgVzpgVzpAAAAAAAA2.129)81.91815(9810)2(20200222222111Point 1Point AExample•Given: D=30 in, d=1 in, h=4 ft•Find: VA•Solution: Bernoulli equationsftghVgVghgVzpgVzpAAAAA/16220020022222111Point APoint 1Example – Venturi Tube•Given: Water 20oC, V1=2 m/s, p1=50 kPa, D=6 cm, d=3 cm•Find: p2 and p3•Solution: Continuity Eq. •Bernoulli Eq.2121122211dDVAAVVAVAVDDd123  kPapPaVdDpVVppgVzpgVzp1202]3/61[21000000,150]/1[2)(222224214122211222222111Similarly for 2  3, or 1  3Pressure drop is fully recovered, since we assumed no frictional losseskPap 1503Nozzle: velocity increases, pressure decreasesDiffuser: velocity decreases, pressure increases ]/1[)(24212DdppVKnowing the pressure drop 1  2 and d/D, we can calculate the velocity and flow rateEx (5.48)•Given: Velocity in circular duct = 100 ft/s, air density = 0.075 lbm/ft3.•Find: Pressure change between circular and square section.•Solution: Continuity equation•Bernoulli equation)(2222222csscssscccVVppgVzpgVzpsftVDVDAVAVsssscc/54.78)4(100)4(100222223/46.4)10054.78(/2.32*2/075.0ftlbfsluglbmftlbmppscAir conditioning (~ 60 oF)Ex (5.49)•Given:  = 0.0644 lbm/ft3 V1= 100 ft/s, and A2/A1=0.5, m=120 lbf/ft3•Find: h•Solution: Continuity equation•Bernoulli equationsftAAVVAVAV/2002*10021122211)(22221222122222111VVppgVzpgVzpHeating (~ 170 oF)222321/30)100200(/2.32*2/0644.0ftlbfsluglbmftlbmpp•Manometer equationfthsftsluglbmftlbmhhppairm25.0/2.32*/2.32/)0644.0120(30)(2321HW (5.51)Stagnation TubeglVddlppVpgVpgVzpgVzp2))((2)(222211221221122222111Stagnation Tube in a PipepgV22zFlowPipe0zgVzpH2212Pitot Tube)(2)()[(22222211111222221122222111hhgVzpzpgVgVpgVpgVzpgVzpPitot Tube Application (p.170)12Vylz1-z2sftgyVyhhyzzppzzypppylylzzpkHgkHgkkHgkkkHgkHgkk/3.24)1/(2)1/()()()()()(21212121212211HW (5.69)HW (5.75)HW (5.84)HW