DM23862: Applied fluid mechanics

3/58 Differential approach In order to know the detailed knowledge of velocities and pressure inside the control volume, we need a relationship that a...

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DM23862: Applied fluid mechanics Ch.6 Differential analysis of fluid flow Spring, 2019

June Kee Min

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Ch.6.1-3. Flow kinematics : Conservation of mass & momentum

(Review) Control volume approach

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• Reynolds transport theorem  Enables the shift from system concepts to control volume concepts B = mb B: the extensive property, b: the intensive property, m: mass

D :materialderivative (substantial derivative) Dt

    bdV    bV  nˆ dA Dt t cv cs

DBsys

 Example: conservation of mass (b = 1, B = m)

Dmsys Dt



  dV    V  nˆ dA t cv cs

  V  nˆ dA   cs

• Limit of control volume approach  By control volume analysis we are not able to get information about what's going on in a or b. ()  It deals with only the global information of a control volume, cannot provide the detailed information of the flow at certain point.

  dV t cv

(continuity equation)

Differential approach  In order to know the detailed knowledge of velocities and pressure inside the control volume, we need a relationship that applies at a point (or very small infinitesimal region).  Control volume approach  integral equations Differential approach  differential equations (non-linear PDEs with boundary conditions)

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Material derivative

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• Material derivative is the infinitesimal version of Reynolds transport theorem.  Reynolds transport theorem : Time rate of change of control mass  Time rate of change of control volume  Material derivative : Time rate of change of particle  Time rate of change of location (Lagrangian approach) (Eulerian approach)

D      u v  w Dt t x y z    (V ) t Local unsteady term

Convection term

Fluid kinematics

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• Kinematics: mathematical description of fluid motion (cf. kinetics)

Eulerian representation

 Velocity field

V ( x, y , z, t )  uˆi  vˆj  wkˆ

 Acceleration field

a

DV V   ( V  ) V Dt t V V V V  u v w t x y z

Translation and linear deformation

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• Translation: no velocity gradient, no deformation

• Linear deformation: due to velocity gradient Taylor series expansion

• Rate of volume change per unit volume





 x  u  x  t   y z   x y z 1 d ( V ) 1 u x   lim    V dt  x y z  t 0 t x

• Similarly in y- and z- directions 1 d ( V ) u v w     V  V dt x y z

Stretching in x-direction

(volumetric dilatation rate) =0 for incompressible flow

Rotation

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• Rotation: angular velocity • Angular velocity of line OA

  t 0  t

OA  lim

v  x   t  v x tan      t

x

x

  x 2x    tan x  x   3 15   v  t x  v (positive in CCW) Therefore OA  lim t 0 t x u  t u y OB  lim  (positive in CW) Likewise  t 0 t y 3

• Rotation about z-axis • Similarly

5

1  v u  z     2  x y  1  w

v 

x    2  y z  1  u

w 

y     2  z x 

• Angular rotation vector 1 1 ω   x ˆi   y ˆj   z kˆ  curlV    V 2 2

• Vorticity vector ς  2ω    V

   V  0 incompressible flow  irrotational flow   V  0

Angular deformation

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• Shear strain (angular deformation)     

• Rate of angular deformation (strain rate)



 v

    lim  lim   t 0  t  t 0  

• If v   u x

y

x

  t  u y   t  t

v u   x  y  

Related with shear stress

: solid block rotation (no angular deformation, only rotation)

 v

u 

cf)  xy       x y 

Mass conservation

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• RTT

Dmsys     dV    V  nˆ dA  0 Dt t cv cs

mcv   mout   min  0 t

• Mass change inside the control volume

   dV   x y z  t cv t

• Net mass outflow along the x-axis

  u   (  u )  x   y z    u   (  u )  x   y z   (  u )  x y z   x 2  x 2  x

• Net mass outflow in all directions

  (  u )  (  v)  (  w)   x  y  z   x y z  

• Rearranging

  (  u )  (  v )  (  w)    0 t x y z

or

    (  V)  0 t

steady (compressible)    V  0     V  0 incompressible ( =const)

• In cylindrical coordinate

 1  ( r  vr ) 1  (  v )  (  v z )    0 t r r r  z

Continuity equation

Examples • Ex 6.2.) For incompressible and steady flow u  x2  y 2  z 2 v  xy  yz  z w? • Continuity

u v w   0 x y z w 2x  ( x  z)  0 z w   2 x  ( x  z )   3 x  z z z2  w  3 xz   f ( x, y ) 2

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Stream function and streamline • Consider incompressible, two-dimensional flow

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* (in cylindrical coordinate)

u v  0 x y

1 rvr 1 v  0 r r r 

• Define a stream function which satisfies

u

vr 

  , v y x

1   , v   r  r

• Then the continuity becomes identically satisfied

     0 x y y x •

(2 unknowns  1 unknown)

  const line  Streamline: a line to which, at each instant, the velocity vectors are tangent

d  0 

  dx  dy   vdx  udy x y

• Flow rate between  and   d   dq  udy  vdx  dy  dx  d y x 2

q   d   2   1 1

dy v  dx u

Examples

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• For steady, incompressible, two-dimensional flow field, determine the corresponding streamline and sketch the flow pattern including the flow direction. (a) (Ex. 6.3)

u  2y v  4x  u  2y    y 2  f ( x) y  v  4 x    2 x 2  g ( y ) x   2x 2  y 2  C

(b)

u  a( x 2  y 2 ) v  2axy

Conservation of linear momentum • Newton’s second law on the infinitesimal control volume

F 

D ( V m ) DV m   ma   Fb   Fs Dt Dt

• Body force: gravity

 Fb   mg • Surface force: normal and shear stress

Using the Taylor‘s series expansion, the net surface force in x-direction

      Fsx   xx  yx  zx   x y z y z   x Likewise in y- and z- directions

      Fsy   xy  yy  zy   x y z y z   x       Fsz   xz  yz  zz   x y z  x

y

z 

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Conservation of linear momentum

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• Equation of motion  Fx   ma x ,  Fy   ma y ,  Fz   ma z  m   x y z • Rearranging

 xx  yx  zx u u u   u     u v w  x y z x y z   t    v v v   v  g y  xy  yy  zy     u  v  w  x y z x y z   t

 gx 

 gz 

• In vector form

g    τ  

 xz  yz  zz w w w   w     u v  w  x y z x y z   t

• Still the equations of motion is too complicated and we need additional modeling of the stress.

DV Dt

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Ch.6.4-7. Potential flow analysis

Inviscid flow

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• Inviscid flow  Flow field in which the shearing stresses are assumed to be negligible. (nonviscous or frictionless)  No shear stress and the normal stress becomes the pressure which is independent of the direction.

 p   xx   yy   zz

(Pascal’s law)

 ij  0 for i  j

• Euler’s equation  Assuming an inviscid flow, the equation of motion becomes the Euler’s equation.

 gx 

p u u u   u   u v  w  x x y z   t

p v v v   v gy      u  v  w  y x y z   t

 gz 

p w w w   w   u v w  z x y z   t

or

V  g  p      V   V   t 

gravity pressure (body force) gradient term term

unsteady term

convection term

Inviscid Flow: Euler’s Equations Leonhard Euler (1707 – 1783) Famous Swiss mathematician who pioneered work on the relationship between pressure and flow.

There is no general method of solving these equations for an analytical solution. The Euler’s equation, for special situations can lead to some useful information about inviscid flow fields.

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Bernoulli’s Equation Daniel Bernoulli (1700-1782)

Swiss mathematician, son of Johann Bernoulli, who showed that as the velocity of a fluid increases, the pressure decreases, a statement known as the Bernoulli principle. He won the annual prize of the French Academy ten times for work on vibrating strings, ocean tides, and the kinetic theory of gases. For one of these victories, he was ejected from his jealous father's house, as his father had also submitted an entry for the prize. His kinetic theory proposed that the properties of a gas could be explained by the motions of its particles.

The Bernoulli equation (1) • Derivation of Bernoulli equation from Euler’s equation

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The Bernoulli equation (2)

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Bernoulli equation for i) inviscid, ii) steady and iii) incompressible flow iv) along a streamline

Ref) Vector identities

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Irrotational flow (1)

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• Irrotational flow

• The Bernoulli equation for irrotational flow

Bernoulli equation for i) inviscid, ii) incompressible, iii) steady and iv) irrotatinal flow v) throughout the flow field

Irrotational flow (2)  Sometimes, it is advantageous to separate the flow into a region of negligible shear (inviscid and irrotational) and a region having significant shear stress.

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The velocity potential

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Ex. 6.4

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Stream function and velocity potential (1)

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 For two-dimensional flow, the velocity components can be defined by both the velocity potential and the stream function (cf) cylindrical coordinate  1  vr  v  r r  1   vr  v   r  r  From the irrotationality condition

 From the continuity equation

Stream function and velocity potential  The streamline (ψ=const) and the equi-potential line (φ=const) are perpendicular to each other.

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Uniform flow in x-direction

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 The simplest plane flow having uniform velocity magnitude in x-direction

(cf) inclined direction

  U ( x cos   y sin  )   U ( y cos   x sin  )

Line source or sink at the origin

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 Line source: Fluid flowing radially outward from a line through the origin Let m is the volume flow rate (m>0: source, m<0:sink)

4.3. ELEMENTARY COMPLEX POT ENT IALS iy y

y

y

f

y

y

y

f

f

f

y

x

y

Figure 4.3: Velocity vectors and equipotential lines for sourc

Line vortex  Vortex: A flow field in which the streamlines are concentric circles.

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Circulation Γ  Circulation Γ: The line integral of the tangential component of the velocity taken around a closed curve in the flow field.

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Ex. 6.6

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Combined source and sink  Superposition of source plus an equal sink

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Doublet

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 Doublet is formed by letting the source and sink approach one another (a0) while CHAPT ER 4. POT ENT IAL increasing the strength 264 m (m∞). iy

f=a

f=b

4

y=a y=b

2

f=c 0

x

-2

-4

-4

-2

0

2

4

Figure 4.8: Streamlines and equipotential lines for a doublet. In polar coordinates, we then say

FLO

Superposition of potential flows

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• Method of superposition  Since the governing equation of potential flow (the Laplace equation  2  0) is linear PDE, a linear superposition is possible, that is. if φ1(x,y,z) and φ2(x,y,z) are solutions of the governing equation, then φ1=φ1+φ2 is also a solution.  Any combination of basic velocity potentials can be new potentials.  Any streamline in an inviscid flow field can be considered as a solid boundary. 266

 Ex) - Source + Sink  doublet - Source + Uniform flow  Flow around a Rankine half body - Source + Sink + Uniform flow  Rankine oval - Doublet + Uniform flow  Flow around a cylinder

CHAPT ER 4. POT ENT iy 2

y y

1

y y f

y

0

x y y

-1

Considered as a solid boundary f

y f

y

-2 -2

-1

f 0

f

f 1

2

Figure 4.10: Streamlines and equipotential lines for flow over a cylinder withou

Rankine half body (1) • Uniform flow + source

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Rankine half body (2) • Uniform flow + source

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Ex. 6.7

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Rankine oval • Uniform flow + source + sink

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Circular cylinder (1) • Uniform flow + doublet

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Circular cylinder (2)

•d’Alembert paradox - The drag on bodies immersed in inviscid fluid is zero. - In reality, there is a significant drag developed on a cylinder due to the viscous force.

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Rotating circular cylinder (1) • Uniform flow + doublet + vortex  a2   Ur 1  2 r 

  sin   ln r  2  

 a2   Ur 1  2 r 

    cos   2 

•The tangential velocity on the cylinder (r=a) v s  

 r

 2U sin   r a

 2 a

•At stagnation point, q = qstag  where vq = 0

sin  stag 

 4 Ua

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Rotating circular cylinder (2) •Static pressure distribution

1 1    p0  U 2  ps    2U sin    2 2  2 a 

2

 1 2 sin  2 2  2 ps  p0  U 1  4sin    2 2 2 2  aU 4 a U   •Drag and lift Fx   

2

Fy   

2

0

0

ps cos  ad  0 ps sin  ad   U 

 Kutta-Joukowski’s law

•Magnus effect : The development of the lift force on rotating bodies

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Ch.6.8-11. Viscous flow and simple solutions of the Navier-Stokes equations

(Recall) Equation of motion

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• Continuity equation   (  u )  (  v) (  w)    0 t x y z

• Equation of motion  xx  yx  zx u u u   u     u v  w  x y z x y z   t    v v v   v  g y  xy  yy  zy     u  v  w  x y z x y z   t

    (  V)  0 t

 gx 

 gz 

 xz  yz  zz w w w   w     u v w  x y z x y z   t

 g    τ ij  

DV Dt

Stress-deformation relationship

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• For incompressible Newtonian fluid, it is known that the stresses are linearly proportional to the rates of deformation.  Normal stresses  xx   p  2 p

u x

 yy   p  2

v y

 zz   p  2

w z

1  xx   yy   zz  3

 Shear stresses  u v 

 xy   yx       y x 

 v w 

 yz   zy       z y 

w u     x z 

 zx   xz   

The Navier-Stokes equations (1)

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• For incompressible fluid flow  Continuity equation

u v w   0 x y z

or

V  0

 Equation of motion (The Navier-Stokes equations)   2u  2u  2u  u u u  p  u    u  v  w      gx    2  2  2  x y z  x  t  x y z    2v  2v  2v  v v v  p  v    u  v  w     gy    2  2  2  x y z  y  t  x y z   2w 2w 2w  w w w  p  w    u  v  w      gz    2  2  2  x y z  z y z   t  x

or

DV   p   g   2 V Dt

or

V   (V )V   p   g   2 V  t 

 

unsteady term

pressure gravity convection gradient term term term

viscous term

The Navier-Stokes equations (2) • The Navier-Stokes equations • The basic equation of motion for viscous fluid flow.  A “complete” mathematical description of the flow. : Four equations for four unknowns (u, v, w, p)

• Nonlinear second order PDE  No general exact analysis solution except a few instances. French Mathematician, L. M. H. Navier (1758-1836) and English Mathematician Sir G. G. Stokes (1819-1903) formulated the Navier-Stokes Equations by including viscous effects in the equations of motion.

L. M. H. Navier (1758-1836)

Sir G. G. Stokes (1819-1903)

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Flow between parallel plates (1) • Two-dimensional, fully-developed, steady, laminar • 2D

w0

• Parallel fullydeveloped

v0

• Continuity • Momentum

 0 z

u 0 x

• Steady

u  u( y)

 d 2u  p 0   2  x  dy  0

p  g y

 0 t

(1)

p    gy  f1 ( x) (2)

 d 2u  p    2  and from (2), we can know that • From x-momentum equation, x  dy   d 2u  p    2   const x  dy  • Integrating Eq. (1) twice,

u

1 p 2 y  C1 y  C2 2 x

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Flow between parallel plates (2)

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• Using boundary conditions,

u  0 at y  h C1  0, C2  

: no-slip conditions at the wall

1  p  2  h 2  x 

• Finally the velocity profile becomes

u

1 p 2 ( y  h2 ) 2 x

: parabolic profile

Poiseuille flow

• Volume flow rate

1 p 2 2h3  p  2h3p 2 q   udy   ( y  h )dy     h  h 2  x 3  x  3 h

h

where

p

 p      x 

• Average velocity

q h 2 p V  2h 3 • Maximum velocity occurs at y=0

umax

h 2  p  3    V 2  x  2

This analysis is valid only for laminar region

Re 

V (2h)  1400 

Couette flow (1)

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• Two-dimensional, steady, laminar • Continuity

• Momentum

u 0 x

u  u( y)

 d 2u  p 0   2  x  dy 

(1)

p p    gy  f1 ( x) (2)  g y • Governing equations are same as those of Poiseuille flow. So the velocity profile becomes 0

u

1 p 2 y  C1 y  C2 2 x

• Using the boundary conditions

u  0 at y  0 u  U at y  b • The velocity profile becomes

u U

y 1  p  2    ( y  by ) b 2  x 

C2  0, C1 

U 1 p  b b 2 x

Couette flow (2) or

u y b 2  p  y  y      1   U b 2U  x  b  b 

• If

p 0 x

b 2  p  P   2U  x 

y b linear profile u U

• Example : Flow in narrow gap of journal bearing • If the gap is very small i.e. r0  ri r then the flow in the gap can be approximated by the Couette flow

U  ri, b  r0  ri y b  ri du U     dy b (ro  ri ) u U

Principle of rotational viscometer

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Couette + Poiseuille Usually, this is called the Couette flow

Ex. 6.9

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Steady laminar flow in circular tube (1) • Hagen-Poiseuille flow • Steady, incompressible, laminar flow through a straight circular tube of constant cross-section • Parallel flow

v  0, vr  0

• Continuity

vz 0 z

• Momentum

0    g sin  

vz  vz ( r )

p r 1 p 0    g cos   r  p  1 d  dvz   0  r  z  r dr  dr  

• From z-momentum Eq. (3)

(1) p    gr sin   f1 ( z )    gy  f1 ( z )

(2) (3)

p  1 d  dvz   we can know p  const.  r  z z  r dr  dr  

• Integration yields

vz 

1  p  2   r  C1 ln r  C2 4  z 

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Steady laminar flow in circular tube (2) vz 

1  p  2   r  C1 ln r  C2 4  z 

• vz should be finite at r=0: C1  0

C2  

• vz =0 at r=R:

1  p  2  R 4  z 

vz 

1  p  2 2   (r  R ) 4  z 

: parabolic profile Hagen-Poiseuille flow

• Volume flow rate

Q

R

0

 R 4  p   R 4 p vz 2 rdr     8  z  8

• Mean velocity

V

Poiseuille law

This analysis is valid only for laminar region

Q R p   R 2 8 2

Re 

• Maximum velocity occurs at r=0

vmax  

R  p  R p   4  z  4 2

2

vmax

v r  2V and z  1    vmax R

2

V (2 R)  2100 

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Flow in annulus

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• Steady, axial, Laminar • Using the same approach as in the previous section, but different BCs

vz  0 at r  r0 vz  0 at r  ri • The velocity profile

1  p   2 2 ri 2  r02 r vz  ln     r  r0  4  z   ln(r0 / ri ) r0  • The volume flow rate

Q

ro

ri

  p   4 4 (ro 2  ri 2 ) 2  vz 2 rdr      ro  ri  8  z   ln(r0 / ri )  p  4 4 (ro 2  ri 2 )2   ro  ri  8  ln(r0 / ri ) 

• The maximum velocity occurs at the radius r=rm where vz / r  0 1/2

 ro2  ri 2  rm     2 ln(r0 / ri ) 

The maximum velocity does not occur at the midpoint of annular space, But rather it occurs nearer the inner cylinder.

Hydraulic diameter

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• Effective (equivalent) diameter for non-circular duct • For non-circular duct, how to define the Reynolds number?  Hydraulic diameter!

Dh 

4A P

A : Cross-sectional area P : Wetted perimeter

• For the case of annulus

4 (ro 2  ri 2 ) Dh   2(ro  ri ) 2 (ro  ri ) • For rectangular duct (fully filled)

Dh 

4 LW 2 LW  2( L  W ) L  W

The analysis of annulus in the previous slide is valid only for laminar region VDh Re   2100



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Thank you!