Viscosity: Key Concepts in Fluid Mechanics

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Explore the fundamental principles of viscous fluids, including the Navier-Stokes equation, viscous stress tensor, and effects of viscosity on sound waves. Delve into the equations for motion of non-viscous fluids and viscosity effects in a comprehensive lecture series.

  • Fluid Mechanics
  • Viscous Fluids
  • Navier-Stokes
  • Viscosity Effects
  • Newton-Euler Equation
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  1. PHY 711 Classical Mechanics and Mathematical Methods 10-10:50 AM MWF in Olin 103 Notes on Lecture 35: Chap. 12 in F & W Viscous fluids 1. Viscous stress tensor 2. Navier-Stokes equation 3. Example for incompressible fluid Stokes law 4. Viscous effects on sound waves => next time 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 1

  2. 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 2 2

  3. 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 3

  4. Equations for motion of non-viscous fluid N ew ton -Eul er eq uation f mo on: o t i v ( ) + = v v f p applied t Continuity equation: + ( ) ( ) = + = v v v 0 0 t t Add the two equa + t ion s: v ( ) ( ) + + v ( x ) = v v v v f p applied t t v jv 3 ( ) v = t 1 j j 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 4

  5. Equations for motion of non-viscous fluid -- continued Modified Newton-Euler equation in terms of fluid momentum: = ( ) ( ) v v x v 3 j + = f p applied t 1 j j ( ) ( ) v v x v 3 j + + = f p applied t = 1 j j v Fluid momentum: + Stress tensor: T vv p ij i j ij th component of Newton-Euler equation: v i ( ) T x 3 i j + = i f i t = 1 j j 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 5

  6. Now consider the effects of viscosity In terms of stress tensor: + = ideal ij T = viscous ij p + T T ij = ideal ij T ideal ji v v T i j ij As an example of a viscous effect, consider -- Newton's "law" of viscosity v y A Fx F = x x vx(y) y A x material dependent parameter 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 6

  7. Effects of viscosity Argue that viscosity is due to shear forces in a fluid of the v y A form: drag F = x Formulate viscosity stress tensor with tracel = + ess and diagonal terms: 2 3 v x v x ( ) ( ) viscous kl T v v k l kl k l l k bulk viscosity viscosity Total stres = = + deal kl T viscous kl T i s tensor: + T kl dea kl T i l k l v v p kl 2 3 v x v x ( ) ( ) = + visc kl T ous v v k l kl k l l k 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 7

  8. Effects of viscosity -- continued Incorporating generalized str v T t x = ess tensor into Newton-Eu ler e q u ations ( ) 3 ij + = i f i 1 i j ( ) ( ) v x v 2 v v t 2 3 3 3 1 3 v p x i j j + = + + + i f i i 2 j x x x = = = 1 1 1 j j j j i i j Continuity equ a tion ( ) v 3 j + = 0 t x = 1 j j Vector form (Navier-Stokes equation) + = Continuity equation + = v 1 1 1 3 ( ) ( ) + + + 2 v v f v v p t ( ) v 0 11/15/2024 t PHY 711 Fall 2024 -- Lecture 35 8

  9. Newton-Euler equations for viscous fluids Navier-Stokes equation 1 t v 1 1 3 ( ) ( ) + = f + + + 2 v v v v p Continuity condition + Typical viscosities at 20o C and 1 atm: ( ) = v 0 t (m2/s) 1.00 x 10-6 14.9 x 10-6 1.52 x 10-6 1183 x 10-6 (Pa s) 1 x 10-3 0.018 x 10-3 1.2 x 10-3 1490 x 10-3 Fluid Water Air Ethyl alcohol Glycerine 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 9

  10. Example steady flow of an incompressible fluid in a long pipe with a circular cross section of radius R Navier-Stokes equation 1 p t v 1 1 3 ( ) ( ) + = f + + + 2 v v v v Continuity condition + Incompressible flu = 2 v v v Note that ( ) ( ) ( ) = v 0 t = v id 0 v = Steady flow 0 t = v Irrotational flow No applied force =0 0 f ( ) v = 2 Neglect non-linear terms 0 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 10

  11. Example steady flow of an incompressible fluid in a long pipe with a circular cross section of radius R -- continued Navier-Stokes equation becomes: 1 0 p ( ) Assu me that , p v r z p z p v r L = + v 2 R ( ) = v r z t v r L z v(r) = 2 ( ) (independent of ) z z p = Suppo se that L (uniform pressure gradient) = 2 ( ) z 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 11

  12. Example steady flow of an incompressible fluid in a long pipe with a circular cross section of radius R -- continued p v r L d dv r p r r dr dr L pr v r C r L = = = = 2 ( ) z 1 ( ) R = z 2 L = + + v(r) ( ) ln( ) C 1 2 z 4 2 pR + 0 ( ) 0 C C v R 1 2 z 4 L p L ( ) = 2 2 ( ) v r R r z 4 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 12

  13. Comment on boundary condition ( ) R zv R = 0 L Fluid approximately stationary at boundary v(r) R 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 13

  14. Example steady flow of an incompressible fluid in a long pipe with a circular cross section of radius R -- continued p L ( ) = 2 2 ( ) v r R r z 4 R Mass flow rate through the pipe: 4 dM dt p R R = = L 2 ( ) rdrv r v(r) z 8 L 0 Poiseuille formula; Method for measuring 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 14

  15. Example steady flow of an incompressible fluid in a long tube with a circular cross section of outer radius R and inner radius R p L = 2 ( ) v r z R 1 r dr ( ) dr d dv r p L R = z r 2 pr = + + L ( ) ln( ) C v r C r 1 2 z 4 L 2 pR = = + + ( ) 0 ln( ) C v R C R 1 2 z 4 L p 2 2 R L = = + + ( ) 0 ln( ) C v R C R 1 2 z 4 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 15

  16. Example steady flow of an incompressible fluid in a long tube with a circular cross section of outer radius R and inner radius R -- continued Solving for an d : C C 1 2 2 2 2 1 pR r R r R R = ( ) 1 ln v r R z 4 l n L L Mass flow rate through th e pipe: ( ) 2 2 1 4 dM dt p R R = = + 4 2 ( ) 1 rdrv r z 8 ln L R 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 16

  17. More discussion of viscous effects in incompressible fluids Stokes' analysis of viscous drag on a sphere of radius moving at speed in medium with viscosity : 6 D F Ru = R u ( ) u FD Plan: 1. Consider the general effects of viscosity on fluid equations 2. Consider the solution to the linearized equations for the case of steady-state flow of a sphere of radius R 3. Infer the drag force needed to maintain the steady-state flow 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 17

  18. Have you ever encountered Stokes law in previous contexts? a. Milliken oil drop experiment b. A sphere falling due to gravity in a viscous fluid, reaching a terminal velocity c. Other? 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 18

  19. Newton-Euler equation for incompressible fluid, modified by viscous contribution (Navier-Stokes equation): p t v ( ) + = + 2 v v f v applied Kinematic vis cosity Typical kinematic viscosities at 20o C and 1 atm: (m2/s) 1.00 x 10-6 14.9 x 10-6 1.52 x 10-6 1183 x 10-6 Fluid Water Air Ethyl alcohol Glycerine 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 19

  20. Stokes' analysis of viscous sphere a on drag of radius R moving = speed at u medium in with visc osity : ( ) 6 F Ru u D F FD Effects of motion on force drag of particle of mass m with constant force : F du 6 = = with ) 0 ( 0 F R u m u dt 6 R F t = ( ) 1 u t e m 6 R 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 20

  21. Effects of motion on force drag of particle of mass m with constant force : F du 6 = = with ) 0 ( 0 F R u m u dt u 6 R F t = F ( ) 1 u t e m FD 6 R u t 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 21

  22. Effects of drag force on motion of particle of mass with an initial velocity with (0) du R u mdt = m = and no external force u U 0 = 6 u 6 R t ( ) u t U e FD m 0 u t 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 22

  23. Recall: PHY 711 -- Assignment #21 Oct. 30, 2024 Determine the form of the velocity potential for an incompressible fluid representing uniform velocity in the z direction at large distances from a spherical obstruction of radius a. Find the form of the velocity potential and the velocity field for all r > a. Assume that for r = a, the velocity in the radial direction is 0 but the velocity in the azimuthal direction is not necessarily 0. 0 = = + In the present viscous case, we will assume that v(a)=0. 2 3 a ( ) , cos r v r 0 2 2 r 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 23

  24. This treatment follows Landau & Lifshitz, Fluid Mechanics Newton-Euler equation for incompressible fluid, modified by viscous contribution (Navier-Stokes equation): p t = v v ( ) + = + 2 v v f v applied Continuity equation: 0 v = Assume steady state: 0 t Assume non-linear effects small Initially set applied p = v = f 0; 2 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 24

  25. = 2 v p Take curl of both sides of equation: 0 p = = ( ) ( ) 2 v Assume (with a little insight from Landau): = + v u ( ) ( ) f r u where ( ) 0 f r r Note that: ( ) ( ) = 2 A A A 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 25

  26. Digression ( ) ( ) f r = + v u u Some comment on assumption: ( ) f r ( ) = 2 A A A ( ) ( = = = = A u Here ( ) ) ( ) = = 2 v A A 2 v Also note: p ( ) = 2 2 v v 0 or 0 p ( ) ( ) = 2 2 4 A A = 0 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 26

  27. ( ( ) r ) = u + v u u f = u z ( v ( ) 0 ) ( ( ) ) ) = 2 z z z ) r ( f r f r f ( = = 2 v = 0 ( ) ( ) C = = 4 4 4 z z ) r ( 0 ( ) 0 ( ) 0 f f r f r = + + + 2 ( ) f r C r C r C 4 1 2 3 r 2 r dr 2 2 df C r C r = = 1 4 + cos 1 cos v u u C 2 4 1 r 3 2 1 r dr d f dr df C r C r = = 1 4 sin 1 sin v u u C 2 4 3 1 2 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 27

  28. Some details: 2 2 2 r dr d dr d = + = 4 ( ) f r 0 ( ) f r 0 2 C r z = + + + ) 2 ( ) f r 4 C r C r C 1 2 3 ( ) ( ( ) f r = + v z u ( ) ( ) ( ) ( ) f r + 2 z z z = ( ) f r u = r z Note that: = cos sin )( ) df dr ( ( ) 2 r v cos ( ) f r 1 cos si n u 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 28

  29. 2 r dr 2 2 df C r C r = = 1 4 + cos 1 cos v u u C 2 4 1 r 3 2 1 r dr u d f dr df C r C r = = 1 4 sin 1 sin v u u C 2 4 3 1 2 = = = v v To satisfy ( To satisfy ( ) ) : 0 r R C 1 0 solve for , C C 2 4 3 3 2 R r R = + cos 1 v u r 3 2 r 3 3 4 R r R = sin 1 v u 3 4 r 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 29

  30. 3 3 2 R r R = + cos 1 v u r 3 2 r 3 3 4 R r R = sin 1 v u 3 4 r Determining pressure: 3 2 R r = = 2 v cos p u 2 3 2 R r = ( ) p r cos p u 0 2 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 30

  31. 3 2 R r = ( ) p r cos p u 0 2 Corresponds to: cos D F ( ) = = cos (6 2 ( ) p R 4 ) p R u R 0 ( ) F = D 6 u R u FD 11/15/2024 PHY 711 Fall 2024 -- Lecture 35 31

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