Introduction to Fluid Dynamics

Applications of Fluid Mechanics

• Fluids are the principle transport media and hence play a central role in nature (winds, rivers, ocean currents, blood etc.)

• Fluids are a source of energy generation (power dams)• They have several engineering applications

– Mechanical engineering (hydraulic brakes, hydraulic press etc.)– Electrical engineering (semi-conductor industries)– Chemical Engineering (centrifuges)– Aerospace engineering (aerodynamics)

List the fluid systems in …

Typical Home

Car

Aircraft

Leaking crude oil from the grounded tanker Argo Merchant (Nantucket Shoals 1976)

ctsy: An album of fluid motion by Van Dyke

Smoke plumes

Turbulent Jet impinging into fresh water

ctsy: An album of fluid motion by Van Dyke

Trapped plume in a stratified ambient flow

ctsy: CORMIX picture gallery

A salt wedge propagating into fresh water

ctsy: Gravity currents in the environment and the laboratory, author : John E. Simpson

What is a fluid ?

• Fluids (liquids and gases) cannot resist shearing forces (tangential stresses) and will continue to deform under applied stress no matter how small.

• Solids can resist tangential stress and will deform only by the amount required to reach static equilibrium.

This class will concentrate on the dynamics of fluid motion Note

– There are several examples where the distinction between solids and fluids blur (e.g. silly putty).

– Individual fluid types have distinct characteristics that will play a critical role in their behavior (e.g. water, oil, air)

What is a Fluid?… a substance which deforms continuously under

the action of shearing forces however small.

… unable to retain any unsupported shape; it takes up the shape of any enclosing container.

... we assume it behaves as a continuum

Continuum Hypothesis

Properties of fluids result from inter-molecular interactions.

Individual interactions are very difficult to quantify. Hence fluid properties are studied by their lumped effects

Continuum hypothesis states that “macroscopic behavior of fluid is perfectly continuous and smooth, and flow properties inside small volumes will be regarded as being uniform”

Is this hypothesis valid ?

Continuum hypothesis breaks down at molecular scales

Example: density is defined as the mass / unit volume

0

vv

m

Measure of density is determined by the volume over which it is calculated.

How small should δv be to get a true local estimate ?

At the molecular scale v

molecule) / (mass*molecules) ofNumber (

Density will depend upon the number of molecules in the sample volume

To study fluctuations in density (e.g. variation in air density with altitude) we need a local estimate

(adapted from Batchelor)

For continuum hypothesis to hold macroscopic properties should not depend on microscopic fluctuations

True for most fluid states (exception : gas flows at very low pressure have mean free path of molecular motion approaching length scales of physical problems)

Scales of fluid processes (1mm ~ 1000 km) >> molecular scales (10-8 cm)

Thus fluid flow and its various properties will be considered continuous

Properties of fluids

Common fluid properties are described below– Mass

• Denoted by the density of the fluid (ρ). It is a scalar quantity• Density varies with temperature, pressure (described below) and

soluble compounds (e.g. sea water as opposed to fresh water)– Velocity

• Is a vector quantity, and together with the density determine the momentum of the flow

– Stresses• Are the forces per unit area acting on the fluid particles. They are of

two types – Normal stress – Shear stress

• Normal stresses in liquids are generally compressive and are referred by a scalar quantity “pressure”

– Viscosity• It is the ability of the fluid to flow freely• Mathematically it is the property of fluid that relates applied shear

stress to rate of deformation (shall be studied in detail later)• Viscosity usually varies with temperature (to a greater extent) and

pressure (to a lesser extent). Note, viscosity in a liquid decreases with increase in temperature but in a gas increases with increase in temperature.

– Thermal conductivity• It is the ability of the fluid to transfer heat through the system• Mathematically it is similar to viscosity (viscosity is the ability of the

fluid to transfer momentum).

– Bulk modulus of elasticity and compressibility• Compressibility is the change in density due to change in normal

pressure• Reciprocal of compressibility is known as the bulk modulus of

elasticity• Liquids have very high values of Ev in comparison to gases. (Thus, for

most practical problems liquids are considered incompressible. This is a major difference between liquids and gases)

– Coefficient of thermal expansion• Thermal expansion is the change in fluid density due to change in

temperature• Liquids in general are less sensitive to thermal expansion than gas• In some cases coefficient of thermal expansion may be negative (e.g.

water inversion near freezing point)

d

dpEv

dT

d

• Fluid properties are inter-related by equations of state.• In gases these equations of state are determined by the collision of

molecules and are given by the kinetic theory of gases (the perfect gas law)

• In liquids these equations of state are very difficult to achieve due to forces of inter-molecular attraction and are thus determined empirically.

• Response of a fluid to external forcings is to a large extent determined by its properties.

External Forcings + Fluid properties + laws of physics = Fluid motion

RTp

Equation of state for an ideal gas

Sometimes written …

pV = nRuT

n= number of moles of gas (kmol/kg)

Ru= Universal gas constant (kJ/kmol K)

RTp

mRTpV or

12

Liquids Gases

Almost incompressible

Forms a free surface

Relatively easy to compress

Completely fills any vessel in which it is placed

Hydrostatic Equation

gzp

z

ghgzp

h

IF r = a constant

Shear stress in moving fluids

Newtonian fluid

m = viscosity (or dynamic viscosity) kg/ms

n = kinematic viscosity m2/s

y

U

y U

t

3

4

Continuity

rAU = constant = m

Mass is conserved

What flows in = what flows out !

r1A1U1= r2A2U2

.

5

Bernoulli

… if no losses

Where is the head loss (m)

ttanconsgzUp 221

212

222

1

211 Hz

g2

U

g

pz

g2

U

g

p

21H

6

7

Reynolds Number

ULUL

Re

2

22

L

LU

area x stress shear

momentum of change of rate

force viscous

force inertia

Venturi

ttanconsgzUp 221

222111 UAUA

p1 p2

Discharge CoefficientsQC'Q D

Actual Flowrate = CD x Predicted flowrate

8

Orifice PlateD D/2

dD

Orifice plate

Vena contracta

Discharge Coefficients

CD

104 106

0.98

0.94

CD

Re

D

104

105

0.65

0.6

Increasing values of d/D

106ReD

Venturi meter Orifice plate

QC'Q D

A Mathematical Review

Taylor series

Taylor stated that in the neighborhood of ax the function f(x) can be given by

!32

32 axaf

axafaxafafxf

xp0

xp1

xp2

A Taylor series expansion replaces a complex function with a series of simple polynomials. This works as long as the function is smooth (continuous)

Primes denote differentials

Note: xp0 A constant value at ax xp1 A straight line fit at ax xp2 A parabolic fit at ax

1 axFor each additional term is smaller than the previous term

Example

Find the Taylor series expansion for xexf about 1x

2

11

2

12

1112

2

111

10

xexee

axafaxafafxp

xeeaxafafxp

eafxp

Each additional term provides a more accurate fit to the true solution

3017.0

2

2.02.02.1

2943.02.02.1

3679.02.1

3012.0)2.1(

2.1for

2111

2

111

10

2.1

eeep

eep

ep

ef

x

2299.0

2

5.05.05.1

1839.05.05.1

3679.05.1

2231.0)5.1(

5.1for

2111

2

111

10

5.1

eeep

eep

ep

ef

x

Scalar Field

Variables that have magnitude, but no direction (e.g. Temperature)

Vector Field

Variables that have both magnitude and direction (e.g. velocity). A vector field is denoted by the coordinate system used. There are two ways to denote a vector field

lyrespective direction ,, thealong rsunit vectoˆ,ˆ,ˆ

ˆˆˆ

zyxkji

kwjviuV

(Vector notation for Cartesian coordinates)

Geometrically a vector field is denoted by an arrow along the direction of the

field, with the magnitude given by the length of the arrow

Mathematically a vector field is denoted by the magnitude along each orthogonal coordinate axis used to describe the system

x

y

z

u

vw 222 wvuVVmag

Dot Product

cosUVUV

0UV

kujuiuU

kvjvivV

zyx

zyx

ˆˆˆ

ˆˆˆ

Note : Dot product of two vectors is a scalar

For a Cartesian coordinate system, if

V

U

1ˆˆˆˆˆˆ

0ˆˆˆˆˆˆˆˆˆˆˆˆ

kkjjii

jkkjikkiijji

UVUV

For θ = 90 (perpendicular vectors)

For θ = 0 (collinear vectors)

kujuiukvjvivUV zyxzyxˆˆˆˆˆˆ

zzyyxx uvuvuv

Then

Deconstruction of a vector into orthogonal components

Consider a pair of vectors U

U

V

and V

Aim is to separate U

into two components such that one of them is collinear with V

21 UUU

Define a vector

V

Vv

ˆ

Represents a unit vector along the direction of V

Now coscosˆˆ UvUvU

θ

Component of U

along V

Therefore vvUU ˆˆ1

provides the direction of the vector

and 12 UUU

2

111111112 cos UUUUUUUUUUUU

also

But cos1 UU

Substituting we get 0coscos 2222

12 UUUU

Thus, a vector can be split into two orthogonal components, one of which is at an arbitrary angle to the vector.

Divergence (Example of dot product)

V

U

V

U

volumeenclosed

yarbitraril an in fielda vector Consider

dA

n

Unit vector normal to surface

Unit vector tangential to surface

nU ˆ

Therefore

system theofinto/out flow Total

of surface the tonormalflow Totalˆ

VdAnUA

Then

0limit

ˆ

div

V

A

V

dAnU

U

Represents the total flow across a closed system per unit volume

Component of flow normal to surface dA

For a Cartesian system

kwjviuU

dx dy

dz dxdydzV 000 ,, zyx

At the front face: in ˆˆ

A

dydzzydxxudAnU ),,2/(ˆ 000

At the back face: in ˆˆ

A

dydzzydxxudAnU ),,2/(ˆ 000

Thus, contribution from these two faces to the divergence is

x

u

dx

zydxxuzydxxu

dx

0lim

000000 ,,2/,,2/

Similar exercise can be carried out for the remaining faces as well to yield

z

w

y

v

x

uUdiv

This can be written in a more concise form by introducing an operator

zk

yj

xi

ˆˆˆ

known as the “dell” operator

Thus

UUdiv

Note : A vector operator is different from a vector field

Gradient

Consider the differential of a scalar T along the path PC, which we shall denote as s

P

CLet the unit vector along this path be

ksjsiss zyxˆˆˆˆ

R ssR ˆ

Δ s

lim 0

ˆ

s

T R ss T RdT

ds s

Then by definition

Using differentiation by parts we get

ds

dz

z

T

ds

dy

y

T

ds

dx

x

T

ds

dT

Gradient of a scalarzyx s

z

Ts

y

Ts

x

T

sT ˆ

Thus represents the change in scalar T along the vector sT ˆ

s

T

Represents a vector of the scalar quantity T. What is its direction?

coscosˆˆ TsTsTds

dTangle between the gradient and direction of change

ds

dTMaximum value of occurs when 0or1cos

Thus T

lies along direction of maximum change

Now consider that the path s lies along a surface of constant T

0ˆ sTds

dT

A vector along a surface is denoted by the tangent to the surface

0ˆ Tds

dT

From the definition of dot product, this means that

T

is perpendicular to surface of constant T

In summary, the gradient of a scalar represents the direction of maximum change of the scalar, and is perpendicular to surfaces of constant values of the scalar

Cross Product

A cross product of two vectors is defined as

eVUVU ˆsin

is a unit vector perpendicular to the plane of the two vectors and its direction is determined by the right hand rule as shown in the figure

e

kujuiuU

kvjvivV

zyx

zyx

ˆˆˆ

ˆˆˆ

For a Cartesian coordinate system, if

ijkikjjik

jkikijkji

kkjjii

ˆˆˆ;ˆˆˆ;ˆˆˆ

ˆˆˆ;ˆˆˆ;ˆˆˆ

0ˆˆˆˆˆˆ

VUVU

0VU

For θ = 90 (perpendicular vectors)

For θ = 0 (collinear vectors)

yxyxzxzxyzzy uvvukuvvujvuvuiVU ˆˆˆ

Then

A convenient rule is

yxyxzxzxyzzy

zyx

zyx uvvukuvvujvuvui

vvv

uuu

kji

VU ˆˆˆ

ˆˆˆ

Curl of a vector (an example of the cross product)

Consider a velocity field given by kwjviuV ˆˆˆ

Then VVcurl

y

u

x

vk

z

u

x

wj

z

v

y

wi

wvu

zyx

kji

V ˆˆˆ

ˆˆˆ

What does this represent?

Consider a 2-dimensional flow field (x,y)

y

u

x

vkV ˆ

Then

O A

BA fluid particle in motion has both translation and rotation

A rigid body rotates without change of shape, but a deformable body can get sheared

Thus for a fluid element, angular velocity is defined as the average angular velocity of two initially perpendicular elements

x

y

Δx

Δy

Angular velocity of edge OA =

kx

v

x

vv

x

ˆOA

0lim

Angular velocity of edge OB =

ky

u

y

uu

x

ˆBO

0lim

Angular velocity of fluid element = ky

u

x

v ˆ2

1

Thus, the curl at a point represents twice the angular velocity of the fluid at that point

Recap

• For continuous functions, properties in a neighborhood can be represented by a Taylor series expansion

• Fluid flow properties are represented as either scalars (magnitude) or vectors (magnitude and direction)

• The dot product of a vector and a unit vector represents the magnitude of the vector along the direction of the unit vector

• The divergence represents the net flow out / in to a closed system per unit volume

• The gradient of a scalar represents the direction and magnitude of maximum change of the scalar. It is oriented perpendicular to surfaces of constant values

• The dot product of a gradient and a unit vector represents the change of the scalar along the direction of the unit vector

• The curl of a velocity vector represents twice the angular velocity of the fluid

sV ˆ

V

T

sT ˆ

V

Kinematics of Fluid Motion

• Kinematics refers to the study of describing fluid motion

• Traditionally two methods employed to describe fluid motion– Lagrangian– Eularian

Lagrangian MethodIn the Lagrangian method, fluid flow is described by following fluid particles. A “fluid particle” is defined by its initial position and time

ttXFX ,, 00

For example : Releasing drifters in the ocean to study currents

Velocity and acceleration of each particle can be obtained by taking time derivatives of the particle trajectory

2

002

00

,,

,,

dt

ttXFd

dt

uda

dt

ttXFd

dt

Xdu

However, due to the number of particles that would be required to accurately describe fluid flow, this method has limited use and we have to rely on an alternative method

Lagrangian framework is useful because fundamental laws of mechanics are formulated for particles of fixed identity.

Eularian MethodAlternatively fluid velocity can be described as functions of space and time

txfu ,

Describing the velocity as a function of space and time is convenient as it precludes the need to follow hundreds of thousands of fluid particles. Hence the Eularian method is the preferred method to describe fluid motion.

Note : Particle trajectories can be obtained from the Eularian velocity field

in domain

The trick is to apply Lagrangian principles of conservation in an Eularian framework

Acceleration in an Eularian coordinate system

Consider steady (not varying with time) 1-D flow in a narrowing channel

U0d0 d(x)U(x)

In a Eularian framework the flow field is given by

ixUu

)(

Standing at one point and taking the time derivative we find that the acceleration is zero

Where, according to the conservation of mass (to be discussed later)

xdxUdU 00

However a fluid particle released into the channel would move through the narrow channel and its velocity would increase. In other words its acceleration would be non zero

What are we missing in the Eularian description ?

Consider a particle at x0 inside the channel with a velocity

After time Δt the particle moves to x0 + Δx

0xUu

where tux

The new velocity is now given by ttuxUtxxUuu ,, 00

Applying Taylor series expansion we get

t

Ut

x

UtuxUuu

0

ort

U

x

Uu

t

u

In the limit 0t

x

Uu

t

U

t

u

t

Dt

Du onaccelerati

0

Material derivative

Consider a property f of a fluid particle ttztytxgf ,,,

Rate of change of f with respect to time is given by dt

df

According to the chain rule of differentiation

wxdt

dz

vxdt

dy

uxdt

dx

dt

dz

z

f

dt

dy

y

f

dt

dx

x

f

t

f

dt

df

direction thealongvelocity

direction thealongvelocity

direction thealongvelocity

where

We can derive a 3D form of the material derivative using the chain rule of differentiation

Material derivative

Dt

Df

z

fw

y

fv

x

fu

t

f

dt

df

Therefore

Stream lines

Streamlines are imaginary lines in the flow field which, at any instant in time, are

tangential to the velocity vectors.

Streamlines are a very useful way of denoting the flow field in a Eularian description

In steady flows (not changing with time) stream lines remain unchanged

Consider an element of the stream line curve given by dx

And the flow field at this element by U

Then the required condition for the element to be tangential (parallel) to the flow is

0dxU

Substituting kwjviuU ˆˆˆ

and kdzjdyidxdx ˆˆˆ

We get

w

dz

v

dy

u

dx Differential equation that needs to be solved to

determine stream lines

Example:

Consider a flow field given by jyixU ˆˆ

Substituting in our equation for stream lines we get

y

dy

x

dx

Integrating we get constantln xy

or constantxy

Graphically this represents flow around a corner

Path lines

Path lines are the lagrangian trajectories of fluid particles in the flow. They represent the locus of coordinates over time for an identified particle

In steady flows, path lines = stream lines

Returning to our previous example jyixU ˆˆ

Path lines can be obtained by integrating the flow field in time

yvdt

dyxu

dt

dx ;

or dty

dydt

x

dx;

Which yields tt ebyeax 00 ;

Constants of integration

Note that constant00 baxy

Which is the same as the stream line curves. The initial position of the particle determines the constants of integration, and thus which stream line the particle is going to be on

Relative motion of a fluid particle

When introducing the concept of a cross product we showed that for a deforming fluid particle the angular velocity of the rotating fluid particle is given

y

u

x

vk

z

u

x

wj

z

v

y

wiV

2

ˆ

2

ˆ

2

ˆ

2

1

V

Referred to as vorticity

Apart from rotation the fluid particle is also stretched and distorted

dx

dy

dxdtx

udx

dydty

vdy

dxdtx

v

udydt

y

The stretching (or extensional strain) is given bydx

dxdxdtxu

dxdtxx

orx

uxx

Similarly we get y

vyy

z

wzz

and

The shear strain for a fluid particle is defined by the average rate at which the two initially perpendicular edges are deviating away from right angles

From the figure this yields

x

v

y

uxy 2

1

And similarly

y

w

z

vyz 2

1

x

w

z

uxz 2

1and

Also note that the shear strain is symmetric xyyx

The total rate of deformation of a moving fluid particle can be written in matrix form

z

w

y

w

x

w

z

v

y

v

x

v

z

u

y

u

x

u

D

Which can be split up to yield

02

1

2

1

2

10

2

1

2

1

2

10

02

1

2

1

2

10

2

1

2

1

2

10

00

00

00

y

w

z

v

x

w

z

u

y

w

z

v

x

v

y

u

x

w

z

u

x

v

y

u

y

w

z

v

x

w

z

u

y

w

z

v

x

v

y

u

x

w

z

u

x

v

y

u

z

w

y

vx

u

D

stretching shearingrotating

Deformation tensor

stretching + shearing = strain tensor

Acceleration in a rotating reference frame

Reference frames that are stationary in space are referred to as inertial reference frames

In many situations the reference frames themselves are moving. For newtonian dynamics it is important to take the motion due to the moving reference frame into account

A very common example is the acceleration due to rotation of the earth

Effect of rotation on a vector reeB

Consider a point B on the surface of a cylinder rotating about its vertical axis with angular velocity Ω

The position vector is given by

O

R

rz erezR ˆˆ

eee rz ˆ,ˆ,ˆUnit vectors along orthogonal coordinate axes

Velocity vector is then given by

eredt

dr

dt

edr

dt

RdV r ˆˆ

ˆ

At the same time ererezeR rzz ˆˆˆˆ

Thus Rdt

Rd

Now consider a fluid particle in an arbitrarily moving coordinate system

R

r

is the position vector of the origin of the moving coordinate system with respect to the inertial coordinate system

r

is the position vector of the fluid particle in the moving coordinate system

Any arbitrary motion can be reduced to a translation and a rotation. Thus the velocity of the fluid particle in the inertial coordinate system can be reduced to three components

R

a) Motion relative to moving framedt

rduR

b) Angular velocity r

c) Motion of moving reference framedt

Rd

Thus the velocity is given by

rudt

Rdu RI

Differentiating once again yields acceleration

rurudt

d

dt

Rda RRI

2

2

Simplifying we get

rurdt

d

dt

ud

dt

Rda R

RI

2

2

2

(1) (2) (3) (4) (5)

(1) Acceleration of moving coordinate system

(2) Acceleration of fluid particle in moving coordinate system

(3) Tangential acceleration (acceleration due to change in rotation rate)

(4) Coriolis acceleration

(5) Centripetal acceleration

Example: Acceleration in a cartesian coordinate system fixed on the surface of the rotating earth

zy

x into the paper

Angular velocity of the earth is-1-5 sec 10x 7.29

Using cartesian coordinates we have

E

kzjyixr ˆˆˆ

kRR ˆ

kwjviuuRˆˆˆ

kjEˆsinˆcos

Since change in R

Is occurring purely due to rotation, we have

Rdt

RdE

Rdt

RdEE

2

2

We thus get

kujuivw

wvu

kji

u zyxRˆcos2ˆsin2ˆsin2cos2

ˆˆˆ

22

kzyjyzix

rrr

ˆcossincosˆsincossinˆ 222

Where we have used the vector identity

CBAACBCBA

Similarly we get

kRjRREEˆcosˆcossin 222

Assuming a constant rotation rate

0

dt

d

Individual components of acceleration can thus be given by

sincoscoscoscos2

sincoscossinsin2

sin2cos2

2

2

2

yzRudt

dwa

yzRudt

dva

xvwdt

dua

z

y

x

Note that for problems relating to earth’s rotation 2

Thus, centripetal accelerations can usually be ignored.

Also for most problems involving ocean circulation vertical velocities and accelerations are much smaller than horizontal motion (to be covered later)

In their simplest form the accelerations are thus given by

fudt

dva

fvdt

dua

y

x

where sin2f

Note : f has opposite signs in the northern and southern hemispheres. This is why hurricanes are counter clockwise in the northern hemisphere and clockwise in the southern

Equations of motion for fluids

• Conservation of mass (continuity equation)– Net mass loss or gain is zero

• Conservation of momentum (Navier Stokes equations)– In each direction

Fam

mass

acceleration

Sum of all forces acting on fluid

Conservation of mass

Consider a control volume of infinitesimal size

dxdy

dz

Mass inside volume =

Let density = tzyx ,,,Let velocity = ktzyxwjtzyxvitzyxuV ˆ,,,ˆ,,,ˆ,,,

dxdydz

Mass flux into the control volume = wdxdyvdxdzudydz

Mass flux out of the control volume =

dxdydzwz

w

dxdzdyvy

vdydzdxux

u

Conservation of mass states that

Mass flux out of the system – Mass flux into the system = Rate of change of mass inside the system

Thus

t

wz

vy

ux

Or, using differentiation by parts, we get

0

zw

z

w

yv

y

v

xu

x

u

t

Rearranging, we get

0

z

w

y

v

x

u

zw

yv

xu

t

Using vector notation

01

VDt

D

Conservation of mass equation, valid for both water and air

Physically it means that the divergence at any point (net flow out/in) is balanced by the rate of change in density through that point

Dt

D V

ForcesTwo types of forces acting on fluid particles

• Body forces – occur through the volume of the fluid particle (e.g. gravity)• Surface forces – occur at the surfaces of the fluid particle

• Two types of surface forces• Normal stress – acting either in tension or compression• Shear stress – acting to deform the particle

Conservation of momentum

Law of conservation of momentum states that for a fluid particle accelerating in water

Fam

Let us look at the forces acting on a fluid particle

Shear stress

Normal stress

Simple case : static fluid

In static fluid, there is no shear force, since the particles are stationary and not deforming. Hence all the forces occur due to normal compressive forces

Consider a fluid particle as shown

dxdz dl

θxx

zz

nnSince there is no net motion forces in the x and z direction must balance each other out

Force balance in the x directiondzdldz nnnnxx sin

or nnxx Force balance in the z direction

gdxdzdldx nnzz 2

1cos ; or gdznnzz

2

1

Reducing particle to zero size we get nnzz

Thus, in a static fluid, the normal stresses are isotropic and are given by

npnn ˆ Hydrostatic pressure (scalar)

Negative sign indicating compressive forces (convention)

What is the expression for pressure ?

Consider a finite sized particle cube in stationary fluid dz

dx 2/dxxp 2/dxxp

2/dzzp

2/dzzp

Force balance in the x direction

02/2/ dxxpdxxp

022

dx

x

pp

dx

x

pp

0

x

p

Force balance in the z direction

gdxdydzdxdydzzpdzzp 2/2/

Similarly 0

y

p

dy

gz

p

Integrating yields Cgzp

At any given plane of a fluid particle there are three forces acting on the fluid particle (one normal and two shear

stresses)

xx

zx

yx(Notation : First subscript refers to direction of force and second subscript to direction of perpendicular to plane)

xz

zz

yz

xy

zy

yy

zzzyzx

yzyyyx

xzxyxx

ij

Referred to as stress tensor

For each axis there are three contributions from the stress tensor

Surface forces in a moving fluid

Viscous forces between the different fluid particles will lead to1. Non – isotropic normal stresses2. Development of shear stresses

How are the shear stresses related ?

2

y

yyx

yx

2

y

yyx

yx

2

x

xxy

xy

2

x

xxy

xy

Consider the moments acting on a fluid particles

Counter clockwise moments

2 2 2 2

xy xyxy xy

xy

x x x xy y

x x

y x

2 2 2 2

yx yxyx yx

yx

y y y yx x

y y

y x

clockwise moments

Conservation of moments

zImoments

Moment of inertia

Rotational acceleration

For a rectangular element

yxyxI z 22

12

Therefore

22

12yxyxxy

if yxxy

then as ;0, yx

Therefore yxxy

Similarly zyyzzxxz ;

Thus the stress tensor

zzzyzx

yzyyyx

xzxyxx

ij

is said to be symmetric

An important property of symmetric tensors is that the sum of diagonals is

independent of the coordinate system used

For a hydrostatic fluid

p

p

p

ij

00

00

00

Sum of diagonals = -3p

Which is independent of coordinate system

Using the analogy of hydrostatic fluid we define

3

zzyyxx

zzzz

yyyy

xxxx

p

p

p

p

Separates normal stress into a compressive part + deviations

Note: This is a mechanical definition of pressure for moving fluids and is not equal to that of hydrostatic fluids

The stress tensor can thus be written as

zzzyzx

yzyyyx

xzxyxx

ij

p

p

p

00

00

00

What is the expression for deviatoric stress tensor ?

For a newtonian fluid, Stokes (1845) hypothesized that• The stress tensor is at most a linear function of strain rates (rate of

deformation)in a fluid• The stresses are isotropic (independent of direction of coordinate system)• When the strain rates are zero (no motion), the stresses should reduce to

hydrostatic conditions

This leads to

(deviatoric stress tensor)

ij

z

w

z

v

y

w

z

u

x

w

z

v

y

w

y

v

x

v

y

u

z

u

x

w

x

v

y

u

x

u

ij

2

1

2

1

2

1

2

1

2

1

2

1

2

Coefficient of viscosity

Conservation of momentum equations

For x direction: xx dxxxx

xx

Net normal surface force =

dxdydzx

dxdydzx

p

dxdydzx

dydzdydzdxx

xx

xx

xxxx

xx

Similarly, net shear surface flow = dxdydzzyxzxy

Let the body force in x direction = Xdxdydz

Conservation of momentum states that

Fma

dxdydzzyxx

pX

Dt

Dudxdydz xzxyxx

or

zyxx

pX

z

uw

y

uv

x

uu

t

u xzxyxx 11

Similarly in y and z direction

zyxy

pY

z

vw

y

vv

x

vu

t

v yzyyyx 11

zyxz

pZ

z

ww

y

wv

x

wu

t

w zzzyzx 11

What about body forces ? For gravity : gZYX ;0

Also, for Newtonian fluid

x

w

z

u

x

v

y

u

x

uxzxyxx ;;2

Assuming that μ does not vary we get

2

2

2 2 2

2 2 2

12xyxx xz u u v u w

x y z x y y x z z x

u u u u v w

x y z x x y z

Navier Stokes equations

=0

Thus

2

2

2

2

2

21

z

u

y

u

x

u

x

p

z

uw

y

uv

x

uu

t

u

2

2

2

2

2

21

z

v

y

v

x

v

y

p

z

vw

y

vv

x

vu

t

v

2

2

2

2

2

21

z

w

y

w

x

w

z

pg

z

ww

y

wv

x

wu

t

w

The equations of motion, can thus be written in vector form as

2

0

1ˆ

U

UU U gk p U

t

where

2

2

2

2

2

22

zyx

Kinematic coefficient of viscosity

Laplace’s operator (or Laplacian)