南开大学：《力学》课程教学资源（PPT课件，英文讲稿）Chapter 5 Fluid Mechanics

Chapter.5Fluid Mechanics

Chapter 5 Fluid Mechanics

$1.Ideal fluidl-1 Definition Fluid Statics(density) and pWe consider the relation between P (pressure).We should establish an simple model①p=constantincompressible（△visnegligible)samein solidNoviscous force“dryfluidNo shearing stress like the pressure in static fluid"P is independent of the“orientation of area "", butdepend on the position and velocity (when it flows)

§1. Ideal fluid  1-1 Definition Fluid Statics We consider the relation between (density) and P (pressure). We should establish an simple model:  v  =constant incompressible ( is negligible ) same in solid No viscous force “dry fluid”. No shearing stress like the pressure in “static fluid”. P is independent of the “ orientation of area ” , but depend on the position and velocity (when it flows). P

1-2 Streamline and stream tubuleFlow field V = v(x, y,z)Curve: tangential line indicate the direction of like “electric line of force""For any point , y is uniqueStream tubeNo crossNoflow through

1-2 Streamline and stream tubule v = v(x, y,z)  v  Flow field Curve: tangential line like “electric line of force”. indicate the direction of For any point , is unique. Stream tube: v  No cross. No flow through. 

1-3 Steady flow and unsteady flowSteady flow = v(x, y,z) path lineUnsteady flowstream lineds1-4 Fluid fluxdv=vcosdsdtdm = pvcos dsdt

v = v(x, y,z)  1-3 Steady flow and unsteady flow Unsteady flow stream line path line Steady flow 1-4 Fluid flux dv = v cos dsdt dm = v cos dsdt n v ds 

Volume flux: dQ, = vcosOds = ·dsMass flux:dQm = pvcosOds = pv·dsI = pe,vdcurrentFinite surfaceOm=JdOm=JpvdsSimilar:of Ed(s)(S)Φβ=[BdQ,= [dQ, = [vds(s)(s)

dQ v ds v ds v   = cos =  dQ v ds v ds m   =  cos =   I en vd  =  =  =  (s) (s) v v Q dQ vds   =  =  (s) (s) m m Q dQ vds    Volume flux: Mass flux: current Finite surface Similar: E =  Ed  B =  Bd 

1-5Principleof continuitySteady flow:dQmi= dQm2 PP,ds, =P,P,ds,一pvds,=Cvdsr = CFluid is incompressible. P = P2EVdsVNo.crossNo flow throughds

1-5 Principle of continuity 1 1 1 2 2 2 v ds v ds      =  vds⊥ =C    1 = 2 vds⊥ = C  Steady flow: dQm1 = dQm2 Fluid is incompressible. V V 1 2 s2 d No cross. No flow through. 

Fvds=0important. Stream tube:orSdsdLV Low of conservationdsofEnergy,charge, mass,..number of particleelectromagnetic fieldquantum mechanics

vds = 0 s   or important. Stream tube: V V 1 2 s2 d d dL L 1 2 Low of conservation Energy,charge, mass, .number of particle, electromagnetic field, quantum mechanics. of: 

Remember:① p=C IncompressibleIndeed fluid2No viscous “dry"Streamline: No crossSteady flow V = V(r)Stream tube: No flow throughO, = Iv.dsVolume flux(s)Om= Jvp.dsMass flux:(S)

 = C  =  (s) v Q v ds    =  (S ) m Q v ds    (r) v v   = Remember: Incompressible. Indeed fluid No viscous “dry”. Streamline: No cross. Steady flow Stream tube: Volume flux: Mass flux:   No flow through

Examplehigh speed1-6 Reaction of fluiddm=Qmdt=pv,s,dtIn dt intervaldP=OmV,dtIn:dP,=Qm2V,dtOut:Oml = Qm2Steady flow:V

Example : high speed 1-6 Reaction of fluid In dt interval dm Q dt v s dt m1 1 1   = =  In: dP Q v dt 1 m1 1   = Out: dP Q v dt 2 m2 2   = Steady flow: Qm1 = Qm2 F' V1 V2

-Force exert on fluiddp,-dpF-Om(V2 -)dt一 F'=-F =Om(2 -)XmRocket:F=Qm(,-)=Qm(0-u)=-Qmudm

( ) 2 1 ' ' F F Q v v m     = − = − Force exert on fluid: ( ) 2 1 2 1 Q v v dt dP dP F m      = − − = Rocket: ( ) 2 1 → → → F = Q v − v m Qm u Qm u   = (0 − ) = − F' V1 V2 m m dm x