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Particle AccelerationPhysical InterpretationExampleExample (cont.)Slide 5Momentum ConservationMomentum Balance (cont.)Euler’s EquationsNavier and Stokes EquationsParticle AccelerationParticleP@time tTracking the particle as we follow it path:V ( , , , )Note: V(x,y,z,t) is the velocity field of the entire flow, not the velocity of a particle.As the particle moves, its velocity changes tV x y z tr rP@time t dtoV ( , , , )V x dx y dy z dz t dt    r rtt+dtPThe acceleration of a particle (substantial acceleration) is given bya. where u , ,P P P PP P PdV dx dy dzV V V Vdt t x dt y dt z dtdx dy dzV V V Vu v w v wt x y z dt dt dt                      rr r r rrr r r rPhysical InterpretationPDV V V V Va u v wDt t x y z          r r r r rrTotal accelerationof a particleLocalaccelerationConvective accelerationtimevelocityUnsteady flowSteady flowxvelocityaccelerationExampleAn incompressible, inviscid flow past a circular cylinder of diameter d is shown below. The flow variation along the approaching stagnation streamline (A-B) can be expressed as: 2O2 21( , 0) ( ) , where u(x) U (1 ) 1RV x y u x ix x     rrxyABUO=1 m/sR=1 mu x( )x5 4 3 2 1 000.51Along A-B streamline, the velocity drops very fast as the particleapproaches the cylinder. At the surface of the cylinder, the velocity is zero (stagnation point) and the surface pressure is a maximum.Example (cont.)Determine the acceleration experienced by a particle as it flows along the stagnation streamline.x2 30 0, since v w 0 along the stagnation streamline.1 2Therefore, a , 0, (1 )( ) for steady state flowy z xDV V Va uDt t xu uu a a at x x x               r r rr00.4a x( )15 x5 4 3 2 1 00.40.20• The particle slows down due to the strong deceleration as it approaches the cylinder.• The maximum deceleration occurs at x=-1.29R=-1.29 m with a magnitude of a(max)=-0.372(m/s2)Example (cont.)Determine the pressure distribution along the streamline using Bernoulli’s equation. Also determine the stagnation pressure at the stagnation point.222 22 22P(x) u ( )Bernoulli's equation: 2 21 1( ) ( ( )) 1 12 2 2( )1( )2Oatm OatmUPxP x P U u xx xP x PP xx                         P x( )x5 4 3 2 100.20.40.6• The pressure increases as the particle approaches the stagnation point.• It reaches the maximum value of 0.5, that is Pstag-P=(1/2)UO2 as u(x)0 near the stagnation point.Momentum Conservationbelow.shown as zyxelement small aConsider leration)mass)(acce(Force:law second sNewton' FromxyzThe element experiences an accelerationDVm ( )Dtas it is under the action of various forces:normal stresses, shear stresses, and gravitational force.V V V Vx y z u v wt x y z               r r r r rxxxxx y zx      xxy z  yxyxy x zy      yxx z  Momentum Balance (cont.)yxxx zxNet force acting along the x-direction:x x xxx y z x y z x y z g x y z                  Normal stressShear stresses (note: zx: shear stress acting on surfaces perpendicular to the z-axis, not shown in previous slide)Body forceyxxx zxThe differential momentum equation along the x-direction isx x xsimilar equations can be derived along the y & z directionsxu u u ug u v wt x y z                     Euler’s Equationsxx yy zzFor an inviscid flow, the shear stresses are zero and the normal stressesare simply the pressure: 0 for all shear stresses, xSimilar equations forxPP u u u ug u v wt x y z                        y & z directions can be derivedyzyzP v v v vg u v wt x y zP w w w wg u v wt x y z                                  Note: Integration of the Euler’s equations along a streamline will give rise to the Bernoulli’s equation.Navier and Stokes EquationsFor a viscous flow, the relationships between the normal/shear stresses and the rate of deformation (velocity field variation) can be determined by making a simple assumption. That is, the stresses are linearly related to the rate of deformation (Newtonian fluid). (see chapter 5-4.3) The proportional constant for the relation is the dynamic viscosity of the fluid (). Based on this, Navier and Stokes derived the famous Navier-Stokes equations:2 2 22 2 22 2 22 2 22 2 22 2 2xyzxyzu u u u P u u uu v w gt x y z x y zv v v v P v v vu v w gt x y z x y zw w w w P w w wu v w gt x y z x y z                                                                                     


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FSU EML 3016 - Navier-stokes equatons

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