Lecture 5

Equation of motions for a liquid. Inviscid and Viscous flowNavier-Stockes equation

Differential analysis of Fluid Flow

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

Description of forces

Forces

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

b 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

Stress tensor

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

=

= =

=

Inviscid flow

• no shearing stress in inviscid flow, so xx yy zzp σ σ σ− = = =

• equation of motion is reduced to Euler equations

( )

( )

( )

xxx

yyy

zzz

u u u ug u v wx t x y z

v v v vg u v wy t x y z

w w w wg u v wx t x y z

σρ ρ

σρ ρ

σρ ρ

∂ ∂ ∂ ∂ ∂− = + + +

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂

− = + + +∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂

− = + + +∂ ∂ ∂ ∂ ∂

(pt

ρ ρ ∂⎡ ⎤− = + ⋅⎢ ⎥∂⎣ ⎦Vg V V∇ ∇)

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

(pρ ρ− = ⋅g V V∇ ∇)

g z= − ∇g 12( ) ( ) ( )⋅ = ⋅ − × ×V V V V V V∇ ∇ ∇

( ) ( )2

z p ρρ 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∇ ∇ ∇ ∇

221 ( ) 0

2 2dp p Vd V gdz or gz constρ ρ+ + = + + =

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

assume if the flow is irrotational:1 02

; ;

zv ux y

v u w v u wx y y z z x

ω⎛ ⎞∂ ∂

= − =⎜ ⎟∂ ∂⎝ ⎠∂ ∂ ∂ ∂ ∂ ∂

= = =∂ ∂ ∂ ∂ ∂ ∂

• Example: uniform flow in x-direction:

Bernoulli equation for irrotational flow

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

( ) ( )2

z p ρρ ρ− − = ⋅ − × ×g V V V V∇ ∇ ∇ ∇

always =0, not only along a stream line

2

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:

u v wx y zφ φ φ∂ ∂ ∂

= = =∂ ∂ ∂

φ=V ∇

• As for incompressible flow conservation of mass leads to:20, 0φ⋅ = =∇ V ∇

Laplace equation2 2 2

2 2 2 0x y zφ φ φ∂ ∂ ∂+ + =

∂ ∂ ∂

• This type of flow is called potential flow

• 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 each other producing another solution;

• stream lines (y=const) and equipotential lines (f=const) are mutually perpendicular

along streanline

along const

dy vdx u

dy ud dx dy udx vdyx y dx vφ

φ φφ=

=

∂ ∂= + = + ⇒ = −∂ ∂

Both f and y satisfy Laplace’s equation

u vy x y y x x

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

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

parallel0

0

Ux y

Ux

Uy x

Uy

φ φ

φψ ψ

ψ

∂ ∂= =

∂ ∂=

∂ ∂= =

∂ ∂=

( cos sin )( cos sin )

U x yU y x

φ α αψ α α= += −

Source and Sink

• Let’s consider fluid flowing radially outward from a line through the origin perpendicular to x-y planefrom mass conservation:

(2 ) rr v mπ =

1 02m

r r rφ φ

π θ∂ ∂

= =∂ ∂

1 02m

r r rψ ψθ π

∂ ∂= =

∂ ∂

ln2m rφπ

=

2mψ θπ

=

singularity point

Vortex• now we consider situation when the

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

lnK

K rφ θψ== −

01

rvKv

r r rθφ ψθ

=∂ ∂

= = − =∂ ∂

• circulation

0C C C

ds ds dφ φΓ = ⋅ = ∇ ⋅ = =∫ ∫ ∫V

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

0

( ) 2

ln2 2

K rd Kr

r

π

θ π

φ θ ψπ π

Γ = =

Γ Γ= = −

∫

for potential flow

Shape of a free vortex

2φ θ

πΓ

=

2

2p V gz constρ+ + =

2 21 2

2 2V V z

g g= +

at the free surface p=0:

2

2 28z

r gπΓ

= −

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

1 2( )2mψ θ θπ

= − −

12 2

2 sintan ( )2m ar

r aθψ

π−= −

−

if the source and sink are close to each other:

sin

cos

Kr

Kr

θψ

θφ

= −

=

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 a small hole. It is observed that for this combination the streamline passing through the stagnation point 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 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 strength

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

= − − = − −

Flow around circular cylinder

• when the distance between source and sink approaches 0, shape of Rankine oval approaches a circular shape

sinsin

coscos

KUrr

KUrr

θψ θ

θφ 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, viscous flow between closely spaced parallel plates. For such a system, dye injected upstream reveals an approximate potential 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 follow the body shape. Generally, however, the flow would separate at the rear of the body, an important phenomenon not accounted for with potential theory.

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

• 6.8 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

=

• 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