Lecture 4

Equation of motions for a liquidEquation of motions for a liquid. Inviscid and Viscous flowN i St k tiNavier-Stockes equation

Differential analysis of Fluid Flow

• The aim: to produce differential equation e a o p oduce d e e a equa odescribing the motion of fluid in detail

Fluid Element Kinematics

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

Velocity and acceleration field

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

• Acceleration

DVVVVV ∂∂∂∂

• Acceleration

DtD

zw

yv

xu

ttr VVVVV

=∂∂

+∂∂

+∂∂

+∂∂

=),(a

• Material derivative

∂∂∂∂∂D )V ∇•+∂∂

=∂∂

+∂∂

+∂∂

+∂∂

= (tz

wy

vx

utDt

D

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

gradient in one direction

u⎡ ∂ ⎤⎛ ⎞Volumetric dilatation rate:

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→

⎡ ∂ ⎤⎛ ⎞⎜ ⎟⎢ ⎥ ∂∂ ⎛ ⎞⎝ ⎠⎢ ⎥= = ⎜ ⎟∂⎝ ⎠⎢ ⎥0tV dt x y z t xδδ δ δ δ δ→ ∂⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦

1 ( )d Vδ ⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞1 ( )d V u v wV dt x y z

δδ

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

Angular motion and deformation

Fluid elements located in a moving fluid move with the fluid and generally undergo a change in shape (angular deformation)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 changeundergoes 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

( / ) 1lim v x x t vδα δ δω ∂ ∂ ∂= ≈ =

0limOA t

OB

t x t xu

δω

δ δ δ

ω

→= ≈ =

∂∂

=∂

• Rotation is defined as the average of those velocities:OB y∂

12z

v ux y

ω⎛ ⎞∂ ∂

= −⎜ ⎟∂ ∂⎝ ⎠

Angular motion and deformation1 1 12 2 2z x y

v u w v u wx y y z z x

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

ˆ ˆ ˆi j k1 12 2 x y z

∂ ∂ ∂= ∇× =

∂ ∂ ∂ω V =

1 1 1ˆ ˆ ˆ

u v w

w v u w v u⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞1 1 12 2 2

w v u w v uy z z x x y

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= − + − −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠i j + k

• Vorticity is defined as twice the rotation vector

2ζ ω= = ∇×V2ζ 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 CV

d ρ δ δ δ∂ ∂∫ V ( )uρ δ δ δ∂

CV

d x y zt t

ρρ δ δ δ=∂ ∂∫ V Flow of mass in x-direction:

( ) x y zxρ 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

∂∂ ∂∂+ + + =

∂ ∂ ∂ ∂t r r r zθ∂ ∂ ∂ ∂

• Incompressible flow in cylindrical coordinates

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

θ

θ∂∂ ∂

+ + =∂ ∂ ∂r r r zθ∂ ∂ ∂

Stream function• 2D incompressible flow

∂ ∂ 0u vx y∂ ∂

+ =∂ ∂

• We can define a scalar function such thatu vψ ψ∂ ∂= =∂ ∂ Stream functiony x∂ ∂

• Lines along which stream function is const are

Stream function

stream lines:dy vdx u

=

0d dx dy vdx udyψ ψψ ∂ ∂= + = − + =∂ ∂Indeed: y yx y

ψ∂ ∂Indeed:

Stream function

• Flow between streamlines

d d ddq udy vdx= −

d d d dψ ψ∂ ∂dq dy dx dy xψ ψ ψ∂ ∂

= + =∂ ∂

2 1q ψ ψ= −

Description of forces

Body forces – distributed through the element, e.g. b mδ δ=F gthrough the element, e.g. Gravity

b

Normal stress

S face fo ces es lt

Normal stress

Forces

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

Sh i t

Linear forces: Surface tension

Shearing stresses

Stress acting on a fluidic element

• normal stress0

lim nn A

FAδ

δσδ→

=0A Aδ δ→

• shearing stresses• shearing stresses

1 21 2lim limF F

A Aδ δτ τδ δ

= =1 20 0A AA Aδ δδ 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

Stress tensor• To define stress at a point we need to define “stress vector”

for all 3 perpendicular planes passing through the point

11 12 13

21 22 23

σ τ ττ τ σ τ

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟

21 22 23

31 32 33τ τ σ⎜ ⎟⎜ ⎟⎝ ⎠

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( g) g

F zxyxxx δδδττσδ )( ∂+

∂+

∂ zyxzyx

F zxyxxxx δδδδ )(

∂+

∂+

∂=

Differential equation of motion

aF mδδ = ,x x

y y

F maF ma m x y z

δ δδ δ δ δ δ δ

== =y y

z zF maδ δ=

)()( uuuuyx ∂∂∂∂∂∂∂ ττσ )()(zuw

yuv

xuu

tu

zyxg zxyxxx

x ∂∂

+∂∂

+∂∂

+∂∂

=∂∂

+∂

+∂∂

+ ρτσρ

Acceleration (“material derivative”)

)()( vvvvzyxyyy ∂+∂+∂+∂∂+

∂+

∂+

ττσ)()(

zw

yv

xu

tzxyg yyyy

y ∂+

∂+

∂+

∂=

∂+

∂+

∂+ ρρ

wwww ∂∂∂∂∂∂∂ ττσ )()(zww

ywv

xwu

tw

xyzg xzyzzz

z ∂∂+

∂∂+

∂∂+

∂∂=

∂∂

+∂∂

+∂∂+ ρττσρ

Inviscid flow

• no shearing stress in inviscid flow, so

xx yy zzp σ σ σ− = = =

• equation of motion is reduced to Euler equations

( )xxx

u u u ug u v wt

σρ ρ∂ ∂ ∂ ∂ ∂− = + + +

∂ ∂ ∂ ∂ ∂

( )

x

yyy

x t x y zv v v vg u v w

y t x y zσ

ρ ρ

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂

− = + + +∂ ∂ ∂ ∂ ∂

( )zzz

y t x y zw w w wg u v w

x t x y zσρ ρ

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂

− = + + +∂ ∂ ∂ ∂ ∂x t x y z∂ ∂ ∂ ∂ ∂

(∂⎡ ⎤V V V∇ ∇)(pt

ρ ρ ⎡ ⎤− = + ⋅⎢ ⎥∂⎣ ⎦g V V∇ ∇)

Bernoulli equation• let’s write Euler equation for a steady flow along a streamline

(V V∇ ∇)(pρ ρ− = ⋅g V V∇ ∇)1( ) ( ) ( )V V V V V V∇ ∇ ∇g z= − ∇g 12( ) ( ) ( )⋅ = ⋅ − × ×V V V V V V∇ ∇ ∇

( ) ( )z p ρρ ρ− − = ⋅ − × ×g V V V V∇ ∇ ∇ ∇( ) ( )2

z pρ ρ= × ×g V V V V∇ ∇ ∇ ∇

• now we multiply it by ds along the streamline

( ) ( )2

z d p d d dρρ ρ− ⋅ − ⋅ = ⋅ ⋅ − × × ⋅g s s V V s V V s∇ ∇ ∇ ∇2

221 ( ) 0dp p Vd V gdz or gz const+ + = + + =( ) 0

2 2d V gdz or gz const

ρ ρ+ + + +

Irrotational Flow• Analysis of inviscide flow can be further simplified if we

assume if the flow is irrotational:1 ⎛ ⎞∂ ∂1 02z

v ux y

v u w v u w

ω⎛ ⎞∂ ∂

= − =⎜ ⎟∂ ∂⎝ ⎠∂ ∂ ∂ ∂ ∂ ∂; ;v u w v u wx y y z z x∂ ∂ ∂ ∂ ∂ ∂

= = =∂ ∂ ∂ ∂ ∂ ∂

• Example: uniform flow in x-direction:• Example: uniform flow in x-direction:

Bernoulli equation for irrotational flow

( ) ( )z p ρρ ρ− − = ⋅ − × ×g V V V V∇ ∇ ∇ ∇( ) ( )2

pρ ρg

always =0, not only along a stream line

• Thus, Bernoulli equation can be applied between any two points in the flow fieldpoints in the flow field

22

2p V gz constρ+ + =

Velocity potential• equations for irrotational flow will be satisfied automatically if

we introduce a scalar function called velocity potential such that:that:

u v wx y zφ φ φ∂ ∂ ∂

= = =∂ ∂ ∂

φ=V ∇• This type of flow is called potential flow

• As for incompressible flow conservation of mass leads to:20 0φ∇

• This type of flow is called potential flow

20, 0φ⋅ = =∇ V ∇Laplace equation2 2 2

0φ φ φ∂ ∂ ∂+ + =2 2 2 0

x y z+ + =

∂ ∂ ∂

I li d i l 2 21 1 1φ φ φ φ φ φ∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞• In cylindrical coordinates:

2 2

2 2 2

1 1 1 0r z rr r z r r r r zφ φ φ φ φ φφ θ

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

∇

Some basic potential flows

• As Laplace equation is a linear one, the solutions can be added to eachthe solutions can be added to each other producing another solution;

• stream lines (ψ=const) and i t ti l li ( t)equipotential lines (φ=const) are

mutually perpendiculard

along streanline

dy vdx u

dy uφ φ

=

∂ ∂

along const

dy ud dx dy udx vdyx y dx vφ

φ φφ=

∂ ∂= + = + ⇒ = −∂ ∂

Both φ and ψ satisfy Laplace’s equation

u v ψ ψ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞u vy x y y x x

ψ ψ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= ⇒ = −⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠

Uniform flow• constant velocity, all stream lines are straight and

parallelφ φ∂ ∂ 0Ux y

Ux

φ φ

φ

∂ ∂= =

∂ ∂=

( cos sin )( cos sin )

U x yU y x

φ α αψ α α= += −

0Uy x

Uy

ψ ψ

ψ

∂ ∂= =

∂ ∂=

( )yψ

yψ

Source and Sink

• Let’s consider fluid flowing radially outward from a line through th i i di l t lthe origin perpendicular to x-y planefrom mass conservation:

(2 )r v mπ =(2 ) rr v mπ =

1 02mφ φ

θ∂ ∂

= =∂ ∂2r r rπ θ∂ ∂

lnm rφ =

1 ∂ ∂

ln2

rφπ

1 02m

r r rψ ψθ π

∂ ∂= =

∂ ∂ singularity point

2mψ θπ

=

Vortex• now we consider situation when the

stream lines are concentric circles i.e. we interchange potential and streamwe interchange potential and stream functions:

lK

Kφ θ= 0

1rv

Kv φ ψ=

∂ ∂lnK rψ = − v

r r rθ θ= = − =

∂ ∂

• circulation for potential flow

0C C C

ds ds dφ φΓ = ⋅ = ∇ ⋅ = =∫ ∫ ∫VC C C

• in case of vortex the circulation is zero along any contour except ones enclosing origin 2πenclosing origin 2

0

( ) 2K rd Kr

π

θ πΓ = =∫

ln2 2

rφ θ ψπ πΓ Γ

= = −

Shape of a free vortex

2φ θΓ=

2π2p V g const+ +

2p gz constρ+ + =

t th f f 0

2 21 2V V z= +

at the free surface p=0:

2 2z

g g+

2

2 28z

r gπΓ

= −8 r gπ

Doublet• let’s consider the equal strength, source-sink pair:

1 2( )2mψ θ θπ

= − −

2 sinθ12 2

2 sintan ( )2m ar

r aθψ

π−= −

−

if the source and sink are close to each other:

sinKrθψ = −

K – strength of a doubletcosKrθφ =

K strength of a doublet

Summary

Superposition of basic flows• basic potential flows can be combined to form new

potentials and stream functions. This technique is called the method of superposition

• superposition of source and uniform flow

sin cos ln2 2m mUr Ur rψ θ θ φ θπ π

= + = +

Superposition of basic flows

• Streamlines created by injecting dye in steadily flowing water show a uniform flow. Source flow is created by injecting water through created by injecting water through a small hole. It is observed that for this combination the streamline passing through the stagnation point could be replaced by a solidpoint could be replaced by a solid boundary which resembles a streamlined body in a uniform flow. The body is open at the downstream end and is thus calleddownstream end and is thus called a halfbody.

Rankine Ovals

• a closed body can be modeled as a combination of a uniform flow and source and a sink of equal strengthq g

1 2 1 2sin ( ) cos (ln ln )2 2m mUr Ur r rψ θ θ θ φ θ= − − = − −2 2π π

Flow around circular cylinder

• when the distance between source and sink approaches 0, h f R ki l h i l hshape of Rankine oval approaches a circular shape

iK θsinsin KUrr

K

θψ θ

θ

= −

coscos KUrrθφ θ= +

Potential flows• Flow fields for which an incompressible

fluid is assumed to be frictionless and the motion to be irrotational are commonly referred to as potential flows.

• Paradoxically, potential flows can be simulated by a slowly moving viscoussimulated by a slowly moving, viscous flow between closely spaced parallel plates. For such a system, dye injected upstream reveals an approximate potential flow pattern around apotential flow pattern around a streamlined airfoil shape. Similarly, the potential flow pattern around a bluff body is shown. Even at the rear of the bluff body the streamlines closely followbluff body the streamlines closely follow the body shape. Generally, however, the flow would separate at the rear of the body, an important phenomenon not

t d f ith t ti l thaccounted for with potential theory.

Viscous Flow

• Moving fluid develops additional components of stress due to viscosity. For incompressible fluids:

ddupxx μσ 2+−= )(

dxdv

dydu

yxxy +== μττ

ddvp

dxp

yy

xx

μσ

μ

2+−= )(ddw

ddv

dxdy

zyyz +== μττ

dwp

dypyy

μσ

μ

2+−= )(

)(

dwdudydzzyyz

+== μττ

μ

dzpzz μσ 2+= )(

dxdzxzzx +μττ

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

Navier-Stokes Equations

)()(222 ∂

+∂

+∂

++∂∂

+∂

+∂

+∂ uuupuuuu

)()(

)()(

222

222

∂+

∂+

∂++

∂−=

∂+

∂+

∂+

∂

∂+

∂+

∂++

∂−=

∂+

∂+

∂+

∂

vvvgpvwvvvuvzyx

gxp

zw

yv

xu

t x

μρρ

μρρ

)()(

)()(

222

222

∂+

∂+

∂++

∂−=

∂+

∂+

∂+

∂

∂+

∂+

∂++

∂−=

∂+

∂+

∂+

∂

wwwgpwwwvwuwzyx

gyz

wy

vx

ut y

μρρ

μρρ

0

)()( 222

=∂∂+

∂∂+

∂∂

∂+

∂+

∂++

∂∂+

∂+

∂+

∂wvu

zyxg

zzw

yv

xu

t z μρρ

∂∂∂ zyx

• 4 equations for 4 unknowns (u,v,w,p)A l ti l l ti k f l f• Analytical solution are known for only few cases

General form of the Navier-Stokes Equation

⎛ ⎞

2

2 23ij ij ijp V eτ μ δ μ⎛ ⎞= − + ∇ ⋅ +⎜ ⎟

⎝ ⎠⎛ ⎞2

2 13

ii

i i

Du p g V VDt x x

ρ ρ μ⎛ ⎞∂ ∂

= − + + ∇ + ∇⋅⎜ ⎟∂ ∂⎝ ⎠

• incompressible fluidp

DV 2DV p g VDt

ρ ρ μ= −∇ + + ∇

Steady Laminar Flow between parallel plates

xuwv =∂∂

⇒== 00;0)()( 2

2

2

2

2

2

∂∂

+∂∂

+∂∂

++∂∂

−=∂∂

+∂∂

+∂∂

+∂∂ uuugpuwuvuu

tu

x μρρ

yu

xp

x

∂∂

+∂∂

−=

∂

)(0 2

2

μ)()(

)()(

2

2

2

2

2

2

222

∂∂

+∂∂

+∂∂

++∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

∂∂∂∂∂∂∂∂

zv

yv

xvg

yp

zvw

yvv

xvu

tv

zyxg

xzyxt

y

x

μρρ

μρρ

p

gyp

∂

−∂∂

−=0 ρ)()( 2

2

2

2

2

2

∂∂∂∂∂

+∂∂

+∂∂

++∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

wvuzw

yw

xwg

zp

zww

ywv

xwu

tw

z μρρ

zp∂∂

−=00=∂∂

+∂∂

+∂∂

zw

yv

xu

xuwv =∂∂

⇒== 00;0

yu

xp

x

∂∂

+∂∂

−=

∂

)(0 2

2

μ11

00;0

2 CCpCpduxuwv

∂∂

=∂∂

⇒==

p

gyp

∂

−∂∂

−=

0

0 ρ)(

2 212

1

xfgyp

CyCyxpuCy

xp

dy+−=

++∂

=⇒+∂

=

ρμμ

zp∂

−=0

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 ysolve the differential equations governing fluid motion. One condition is that any viscous fluid sticks to any solid surface that it touches.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 itcan 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 itpast 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

00;0xuwv =∂∂

⇒==

)(211

212

1

f

CyCyxpuCy

xp

dydu

x

++∂∂

=⇒+∂∂

=

∂

μμ)(xfgyp +−= ρ

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

;8R pQ

lπ

μΔ

=Poiseuille law:

2

8 lR p

μΔ

8R pv

lμΔ

=

Laminar flow• The velocity distribution is

parabolic for steady laminar flowparabolic for steady, laminar flow in circular tubes. A filament of dye is placed across a circular tube containing a very viscous liquid

hi h i i iti ll t t With thwhich is initially at rest. With the opening of a valve at the bottom of the tube the liquid starts to flow, and the parabolic velocityand the parabolic velocity distribution is revealed. Although the flow is actually unsteady, it is quasi-steady since it is only slowly h i Th t i t t ichanging. Thus, at any instant in

time the velocity distribution corresponds to the characteristic steady-flow parabolic distribution.steady flow parabolic distribution.

Problems6 74 Oil SAE30 15 6C dil fl b fi d h i l• 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 isand 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;

• 6 8 An incompressible viscous

and direction of the shearing stress on the bottom plate; velocity along the centerline of the channel

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

vorticity;yu U=– vorticity;

– rate of angular deformation. u U

b=