Lecture 5

Differential Analysis of Fluid FlowNavier-Stockes equation

Differential analysis of Fluid Flow

• The aim: to produce differential equation describing the motion of fluid in detail

Fluid Element Kinematics

• Any fluid element motion can be represented as consisting of translation, linear deformation, rotation and angular deformation

Velocity and acceleration field

• Material derivative

DtD

zw

yv

xu

ttr VVVVV

=∂∂

+∂∂

+∂∂

+∂∂

=),(a

)V ∇•+∂∂

=∂∂

+∂∂

+∂∂

+∂∂

= ( tz

wy

vx

utDt

D

• Acceleration

• Velocity fieldˆ ˆ ˆu v w= + +V i j k

Linear motion and deformation• Let’s consider stretching of a fluid element under velocity

gradient in one direction

0

1 ( ) limt

u x t y zd V ux

V dt x y z t xδ

δ δ δ δδ

δ δ δ δ δ→

⎡ ∂ ⎤⎛ ⎞⎜ ⎟⎢ ⎥ ∂∂ ⎛ ⎞⎝ ⎠⎢ ⎥= = ⎜ ⎟∂⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦

1 ( )d V u v wV dt x y z

δδ

⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞= + + = ∇⋅⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠V

Volumetric dilatation rate:

Angular motion and deformation

Fluid elements located in a moving fluid move with the fluid and generally undergo a change in shape (angular deformation).A small rectangular fluid element is located in the space between concentric cylinders. The inner wall is fixed. As the outer wall moves, the fluid element undergoes an angular deformation. The rate at which the corner angles change (rate of angular deformation) is related to the shear stress causing the deformation

Angular motion and deformation

• Rotation is defined as the average of those velocities:

0

( / ) 1limOA t

OB

v x x t vt x t x

uy

δ

δα δ δωδ δ δ

ω

→

∂ ∂ ∂= ≈ =

∂∂

=∂

12z

v ux y

ω⎛ ⎞∂ ∂

= −⎜ ⎟∂ ∂⎝ ⎠

Angular motion and deformation

• Vorticity is defined as twice the rotation vector

1 1 12 2 2z x y

v u w v u wx y y z z x

ω ω ω⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

ˆ ˆ ˆ

1 12 2

1 1 1ˆ ˆ ˆ2 2 2

x y zu v w

w v u w v uy z z x x y

∂ ∂ ∂= ∇× =

∂ ∂ ∂

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= − + − −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

ω

i j k

V =

i j + k

2ζ ω= = ∇×V• If rotation (and vorticity) is zero flow is called irrotational

Angular motion and deformation

• Rate of shearing strain (or rate of angular deformation) can be defined as sum of fluid element rotations:

v ux y

γ ∂ ∂= +∂ ∂

Conservation of mass• As we found before: 0Sys

CV CV

DMd dA

Dt tρ ρ∂

= + ⋅ =∂ ∫ ∫ V nV

CV

d x y zt t

ρρ δ δ δ∂ ∂=

∂ ∂∫ V Flow of mass in x-direction:( )u x y z

xρ δ δ δ∂∂

( ) ( ) ( ) 0 0u v wt x y z tρ ρ ρ ρ ρ ρ∂ ∂ ∂ ∂ ∂+ + + = +∇⋅ =

∂ ∂ ∂ ∂ ∂V

Conservation of mass• Incompressible flow

0 0u v wx y z∂ ∂ ∂

∇ ⋅ = + + =∂ ∂ ∂

V

• Flow in cylindrical coordinates

( )( ) ( )1 1 0r zvr v vt r r r z

θρρ ρρθ

∂∂ ∂∂+ + + =

∂ ∂ ∂ ∂

( )( ) ( )1 1 0r zvrv vr r r z

θ

θ∂∂ ∂

+ + =∂ ∂ ∂

• Incompressible flow in cylindrical coordinates

Stream function• 2D incompressible flow

0u vx y∂ ∂

+ =∂ ∂

• We can define a scalar function such thatu v

y xψ ψ∂ ∂

= =∂ ∂

• Lines along which stream is const are stream lines:

Stream function

dy vdx u

=

0d dx dy vdx udyx yψ ψψ ∂ ∂

= + = − + =∂ ∂Indeed:

Stream function

• Flow between streamlines

dq udy vdx= −

dq dy dx dy xψ ψ ψ∂ ∂

= + =∂ ∂

2 1q ψ ψ= −

Description of forces

Forces

Body forces – distributed through the element, e.g. Gravity

mδ δ=F g

Surface forces – result of interaction with the surrounding elements: e.g. Stress

Linear forces: Surface tension

Normal stress

Shearing stresses

Stress acting on a fluidic element

• normal stress0

lim nn A

FAδ

δσδ→

=

• shearing stresses

1 21 20 0

lim limA A

F FA Aδ δ

δ δτ τδ δ→ →

= =

Stresses: double subscript notation• normal stress: xxσ

• shearing stress: xy xzτ τ

normal to the plane

direction of stress

sign convention: positive stress is directed in positive axis directions if surface normal is pointing in the positive direction

Vectors and Tensors

11 12 13

21 22 23

31 32 33

σ τ ττ τ σ τ

τ τ σ

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

• To define stress at a point we need to define “stress vector”for all 3 perpendicular planes passing through the point

zyxzyx

F zxyxxxx δδδττσδ )(

∂∂

+∂∂

+∂∂

=

Force on a fluid element• To find force in each direction we need to sum all forces

(normal and shearing) acting in the same direction

)()(zuw

yuv

xuu

tu

zyxg zxyxxx

x ∂∂+

∂∂+

∂∂+

∂∂=

∂∂

+∂∂

+∂∂

+ ρττσρ

aF mδδ =

Acceleration (“material derivative”)

)()(zvw

yvv

xvu

tv

zxyg zyxyyy

y ∂∂+

∂∂+

∂∂+

∂∂=

∂∂

+∂∂

+∂∂

+ ρττσ

ρ

)()(zww

ywv

xwu

tw

xyzg xzyzzz

z ∂∂+

∂∂+

∂∂+

∂∂=

∂∂

+∂∂

+∂∂+ ρττσρ

Differential equation of motion

,x x

y y

z z

F maF ma m x y z

F ma

δ δδ δ δ δ δ δ

δ δ

=

= =

=

Viscous Flow

dzdwp

dydvp

dxdup

zz

yy

xx

µσ

µσ

µσ

2

2

2

+−=

+−=

+−=

)(

)(

)(

dxdw

dzdu

dydw

dzdv

dxdv

dydu

xzzx

zyyz

yxxy

+==

+==

+==

µττ

µττ

µττ

for viscous flow normal stresses are not necessary the same in all directions

Navier-Stokes Equations

0

)()(

)()(

)()(

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

=∂∂+

∂∂+

∂∂

∂∂

+∂∂

+∂∂

++∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

++∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

++∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

zw

yv

xu

zw

yw

xwg

zp

zww

ywv

xwu

tw

zv

yv

xvg

yp

zvw

yvv

xvu

tv

zu

yu

xug

xp

zuw

yuv

xuu

tu

z

y

x

µρρ

µρρ

µρρ

• 4 equations for 4 unknowns (u,v,w,p)• Analytical solution are known for only few cases

Steady Laminar Flow between parallel plates

zp

gyp

yu

xp

xuwv

∂∂

−=

−∂∂

−=

∂∂

+∂∂

−=

=∂∂

⇒==

0

0

)(0

00;0

2

2

ρ

µ

0

)()(

)()(

)()(

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

=∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

++∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

++∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

++∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

zw

yv

xu

zw

yw

xwg

zp

zww

ywv

xwu

tw

zv

yv

xvg

yp

zvw

yvv

xvu

tv

zu

yu

xug

xp

zuw

yuv

xuu

tu

z

y

x

µρρ

µρρ

µρρ

zp

gyp

yu

xp

xuwv

∂∂

−=

−∂∂

−=

∂∂

+∂∂

−=

=∂∂

⇒==

0

0

)(0

00;0

2

2

ρ

µ

)(211

00;0

212

1

xfgyp

CyCyxpuCy

xp

dydu

xuwv

+−=

++∂∂

=⇒+∂∂

=

=∂∂

⇒==

ρµµ

Boundary condition (no slip) 0)()( =−= huhu

)(21 22 hy

xpu −∂∂

=µ

Velocity profile

Flow rate )(32)(

21 3

22

xphdyhy

xpudyq

h

h

h

h ∂∂−=−

∂∂== ∫∫

−

−− µµ

What is maximum velocity (umax) and average velocity?

No-slip boundary condition

• Boundary conditions are needed to solve the differential equations governing fluid motion. One condition is that any viscous fluid sticks to any solid surface that it touches.

• Clearly a very viscous fluid sticks to a solid surface as illustrated by pulling a knife out of a jar of honey. The honey can be removed from the jar because it sticks to the knife. This no-slip boundary condition is equally valid for small viscosity fluids. Water flowing past the same knife also sticks to it. This is shown by the fact that the dye on the knife surface remains there as the water flows past the knife.

Couette flow

)(211

00;0

212

1

xfgyp

CyCyxpuCy

xp

dydu

xuwv

+−=

++∂∂

=⇒+∂∂

=

=∂∂

⇒==

ρµµ

Boundary condition (no slip) Ubuu == )(;0)0(

Please find velocity profile and flow rate

Boundary condition (no slip) Ubuu == )(;0)0(

)(21 2 byy

xp

byUu −

∂∂

+=µ

Velocity profile )1)((2

2

by

by

xp

Ub

by

Uu

−∂∂

−=µ

Dimensionless parameter P

Hagen-Poiseuille flow

4

2

;8

8

R pQl

R pvl

πµ

µ

∆=

∆=

Poiseuille law:

Laminar flow• The velocity distribution is

parabolic for steady, laminar flow in circular tubes. A filament of dye is placed across a circular tube containing a very viscous liquid which is initially at rest. With the opening of a valve at the bottom of the tube the liquid starts to flow, and the parabolic velocity distribution is revealed. Although the flow is actually unsteady, it is quasi-steady since it is only slowly changing. Thus, at any instant in time the velocity distribution corresponds to the characteristic steady-flow parabolic distribution.

Problems

• 6.2 A certain flow field is given by equation:

Determine expression for local and convective components of the acceleration in x and y directions

2(3 1) 6V x i xyj= + −

• 6.8 An incompressible viscous fluid is placed between two large parallel plates. The bottom plate is fixed and the top moves with the velocity U. Determine: – volumetric dilation rate; – rotation vector; – vorticity; – rate of angular deformation.

yu Ub

=

Problems• 6.22 The stream function for an

incompressible flow

sketch the streamline passing through the origin; determine of flow across the strait path AB

• 6.74 Oil SAE30 at 15.6C steadily flows between fixed horizontal parallel plates. The pressure drop per unit length is 20kPa/m and the distance between the plates is 4mm, the flow is laminar.Determine the volume rate of flow per unit width; magnitude and direction of the shearing stress on the bottom plate; velocity along the centerline of the channel