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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 motionby Van Dyke



Smoke plumes



Turbulent Jet impinging into fresh water

ctsy: An album

of fluid motionby 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 underapplied 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 playa 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; ittakes 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)of Number(

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) andsoluble compounds (e.g. sea water as opposed to fresh water)

– Velocity

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

– Stresses

• Are the forces per unit area acting on the fluid particles. They are oftwo 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 E v in comparison to gases. (Thus, for

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

– Coefficient of thermal expansion

• Thermal expansion is the change in fluid density due to change intemperature

• 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

dp E

v

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

RT p



Equation of state for an ideal gas

Sometimes written …

pV = nR uT

n= number of moles of gas (kmol/kg)R u= 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 vesselin which it is placed



Hydrostatic Equation

gz

p

z

ghgzp

h

IF = a

constant



Shear stress in moving fluids

Newtonian fluid

m = viscosity (or dynamic viscosity) kg/ms

n = kinematic viscosity m2/s

y

U

m

yU

m

n

3

4



Continuity

AU = constant = m

Mass is conserved

What flows in = what flows out !

1 A1U1=

2 A2U2

.

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

n

m

ULULRe

2

22

L

LU

areaxstressshear

momentumof changeof rate

forceviscous

forceinertia



Venturi

ttanconsgzUp 221

222111 U AU A

p1 p2



Discharge Coefficients

QC'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

106 Re

D

Venturi meter Orifice plate

QC'Q D



A Mathematical Review

Taylor series



Taylor series

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

!32

32

a xa f a xa f a xa f a f x f

x p0

x p1

x p2

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:

x p0 A constant value at a x x p

1 A straight line fit at a x x p2 A parabolic fit at a x

1 a xFor each additional term is smaller than the previous term

Example



Example

Find the Taylor series expansion for xe x f

about 1 x

2

11

2

1

2

111

2

2

11

1

1

0

xe xee

a xa f a xa f a f x p

xeea xa f a f x p

ea f x p

Each additional term provides a more

accurate fit to the true solution

3017.022.02.02.1

2943.02.02.1

3679.02.1

3012.0)2.1(

2.1for

2

1112

11

1

1

0

2.1

eee p

ee p

e p

e f

x

2299.025.05.05.1

1839.05.05.1

3679.05.1

2231.0)5.1(

5.1for

2

1112

11

1

1

0

5.1

eee p

ee p

e p

e f

x

S l i ld



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

lyrespectivedirection,,thealongrsunit vectoˆ,ˆ,ˆ

ˆˆˆ

z y xk ji

k w jviuV

(Vector notation for Cartesian coordinates)

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

v

w 222 wvuV V mag

D t P d t



Dot Product

cosU V U V

0U V

k u juiuU

k v jvivV

z y x

z y x

ˆˆˆ

ˆˆˆ

Note : Dot product of two vectors is a scalar

For a Cartesian coordinate system, if

V

U

1ˆˆˆˆˆˆ

0ˆˆˆˆˆˆˆˆˆˆˆˆ

k k j jii

jk k jik k ii j ji

U V U V

For θ = 90 (perpendicular vectors)

For θ = 0 (collinear vectors)

k u juiuk v jvivU V z y x z y xˆˆˆˆˆˆ

z z y y x x uvuvuv

Then

D t ti f t i t th l t



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 U U U

Define a vector

V

V v

ˆ

Represents a unit vector along the direction of V

Now coscosˆˆ U vU vU

θ

Component of U

along V

Therefore vvU U ˆˆ1

provides the direction of the vector

and 12 U U U

2

111111112 cos U U U U U U U U U U U U

also

But cos1 U U

Substituting we get 0coscos 2

22

2

12 U U U U

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)



Divergence (Example of dot product)

V

U

V

U

volumeenclosed

yarbitrarilaninfielda vectorConsider

dA

n

ˆ

Unit vector normal to surface

Unit vector tangential to surface

nU ˆ

Therefore

systemtheof into/outflowTotal

of surfacethetonormalflowTotalˆ

V dAnU

A

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



For a Cartesian system

k w jviuU

dx dy

dz dxdydz V 000 ,, z y x

At the front face: in ˆˆ

A

dydz z ydx xudAnU ),,2/(ˆ

000

At the back face: in ˆˆ

A

dydz z ydx xudAnU ),,2/(ˆ000

Thus, contribution from these two faces to the divergence is

x

u

dx

z ydx xu z ydx xu

dx

0lim

000000 ,,2/,,2/

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



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

yield

z

w

y

v

x

uU div

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

z k

y j

xi

ˆˆˆ

known as the “dell” operator

Thus

U U div

Note : A vector operator is different from a vector field

G di t



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

k s j si s s z y xˆˆˆˆ

R

s s R ˆ

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 scalar

z y x s z

T s

y

T s

x

T

sT ˆ

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 T sT sT ds

dT

angle between the gradient and direction of change

ds

dT

Maximum value of occurs when 0or 1cos

Thus T

lies along direction of maximum change



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

0ˆ sT ds

dT

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

0ˆ T ds

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

eV U V U ˆsin

is a unit vector perpendicular to the planeof the two vectors and its direction is

determined by the right hand rule as

shown in the figure

eˆ

k u juiuU

k v jvivV

z y x

z y x

ˆˆˆ

ˆˆˆ

For a Cartesian coordinate system, if

i jk ik j jik

jk ik i jk ji

k k j jii

ˆˆˆ;ˆˆˆ;ˆˆˆ

ˆˆˆ;ˆˆˆ;

ˆˆˆ

0ˆˆˆˆˆˆ

V U V U

0V U

For θ = 90 (perpendicular vectors)

For θ = 0 (collinear vectors)

y x y x z x z x y z z y uvvuk uvvu jvuvuiV U ˆˆˆ

Then

A convenient rule is



A convenient rule is

y x y x z x z x y z z y

z y x

z y x uvvuk uvvu jvuvuivvvuuu

k ji

V U ˆˆˆ

ˆˆˆ

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

Consider a velocity field given by k w jviuV ˆˆˆ

Then V V curl

y

u

x

vk

z

u

x

w j

z

v

y

wi

wvu

z y x

k ji

V ˆˆˆ

ˆˆˆ

What does this represent?

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



( ,y)

y

u

x

vk V 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

Δ x

Δ y

Angular velocity of edge OA =

k x

v

x

vv

x

ˆOA

0lim

Angular velocity of edge OB =

k y

u

y

uu

x

ˆBO

0lim

Angular velocity of fluid element = k y

u

x

vˆ

2

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 aTaylor 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 ofthe vector along the direction of the unit vector

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

• The gradient of a scalar represents the direction and magnitude ofmaximum 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 todescribe fluid motion

– Lagrangian

– Eularian



Lagrangian Method

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

t t X F X ,, 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

00

2

00

,,

,,

dt

t t X F d

dt

ud a

dt

t t X F d

dt

X d u

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 Method

Alternatively fluid velocity can be described as functions of space and time

t x f u ,

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

A ele ti i E l i di te te



Acceleration in an Eularian coordinate system

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

U 0 d 0 d( x)U(x)

In a Eularian framework the flow field is given by

i xU u

)(

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

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

xd xU d U 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 ?

C id ti l t i id th h l ith l it



Consider a particle at x0 inside the channel with a velocity

After time Δt the particle moves to x0 + Δ x

0 xU u

where t u x

The new velocity is now given by t t u xU t x xU uu ,, 00

Applying Taylor series expansion we get

t

U t

x

U t u xU uu

0

ort

U

x

U u

t

u

In the limit 0t

x

U u

t

U

t

u

t

Dt

Du onaccelerati

0

Material derivative

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



Consider a property f of a fluid particle t t z t yt x g f ,,,

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

dt

df

According to the chain rule of differentiation

w xdt

dz

v x

dt

dy

u xdt

dx

dt

dz

z

f

dt

dy

y

f

dt

dx

x

f

t

f

dt

df

directionthealongvelocity



where

g

Material derivative

Dt

Df

z

f

w y

f

v x

f

ut

f

dt

df

Therefore

Stream lines



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

0 dxU

Substituting k w jviuU ˆˆˆ

and k dz jdyidxdx ˆˆˆ

We get

w

dz

v

dy

u

dx Differential equation that needs to be solved to

determine stream lines



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 j yi xU ˆˆ

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

yvdt

dy

xudt

dx

;

or dt y

dydt

x

dx;

Which yieldst t eb yea x

00;

Constants of integration

Note that constant00 ba xy

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

w j

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

dxdt xudx

dydt y

vdy

dxdt x

v

udydt

y

dxdxdtu

dx



The stretching (or extensional strain) is given bydx

dxdxdt x

dxdt xx

or x

u xx

Similarly we get y

v yy

z

w zz

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

u xy

2

1

And similarly

y

w

z

v yz

2

1

x

w

z

u xz

2

1 and

Also note that the shear strain is symmetric xy yx

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



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

z

w

y

w

x

w

z v

yv

xv

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

v

x

u

D

stretching shearingrotating

Deformation tensor

stretching + shearing = strain tensor

Acceleration in a rotating reference frame



g

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

frames

In many situations the reference frames themselves are moving. For newtoniandynamics 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 r eˆ

e

B

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

r z er e z R ˆˆ

eee r z ˆ,ˆ,ˆ

Unit vectors along

orthogonal coordinate

axes

Velocity vector is then given by

er e

dt

d r

dt

ed r

dt

Rd V r

ˆˆ

ˆ

At the same time er er e z e R r z z ˆˆˆˆ

Thus R

dt

Rd

Now consider a fluid particle in an arbitrarily moving coordinate system



p y g y

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

r d u R

b) Angular velocity r

c) Motion of moving reference framedt

Rd

Thus the velocity is given by

r udt

Rd u

R I

Differentiating once again yields acceleration



g g y

r ur udt

d

dt

Rd a R R I

2

2

Simplifying we get

r ur dt

d

dt

ud

dt

Rd a R

R I

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

z y

x into the paper

Angular velocity of the earth is-1-5 sec10x7.29

Using cartesian coordinates we have

E

k z j yi xr ˆˆˆ

k R R ˆ

k w jviuu Rˆˆˆ

k j E ˆsinˆcos

Since change in R

Is occurring purely due to rotation, we have

Rdt

Rd E

Rdt

Rd E E

2

2

We thus get



k u juivw

wvu

k ji

u z y x Rˆcos2ˆsin2ˆsin2cos2

ˆˆˆ

22

k z y j y z i x

r r r

ˆcossincosˆsincossinˆ 222

Where we have used the vector identity

C B A AC BC B A

Similarly we get

k R j R R E E ˆ

cosˆ

cossin 222

Assuming a constant rotation rate

0

dt

d

Individual components of acceleration can thus be given by



p g y

sincoscoscoscos2

sincoscossinsin2

sin2cos2

2

2

2

y z Rudt

dwa

y z Rudt 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 sin2 f

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 Stokesequations)

– In each direction

F am

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 = t z y x ,,,

Let velocity = k t z y xw jt z y xvit z y xuV ˆ,,,

ˆ,,,

ˆ,,,

dxdydz

Mass flux into the control volume = wdxdyvdxdz udydz

Mass flux out of the control volume =

dxdydz w z

w

dxdz dyv y

vdydz dxu x

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 w z v yu x

Or, using differentiation by parts, we get



0

z w

z

w

yv

y

v

xu

x

u

t

Rearranging, we get

0

z

w

y

v

x

u

z w

yv

xu

t

Using vector notation

0

1

V Dt

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

Conservation of momentum



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

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

F am

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

θ xx

zz

nn Since there is no net motion forces in

the x and z direction must balance each

other out

Force balance in the x direction

dz dl dz nnnn xx sin

ornn xx

Force balance in the z direction

gdxdz dl dx nn zz 2

1cos ; or gdz nn zz

2

1

Reducing particle to zero size we get nn zz

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

n pnn ˆ Hydrostatic pressure (scalar)

Negative sign indicating compressive forces (convention)

What is the expression for pressure ? 2/dzzp



Consider a finite sized particle cube in

stationary fluid dz

dx 2/dx x p

2/dx x p 2/dz z p

2/dz z p

Force balance in the x direction

02/2/ dx x pdx x p

022

dx

x

p p

dx

x

p p

0 x

p

Force balance in the z direction

gdxdydz dxdydz z pdz z p 2/2/

Similarly 0 y

p

dy

g z

p

Integrating yields C gz p



Conservation of momentsRotational acceleration



z I moments

Moment of inertia

Rotational acceleration

For a rectangular element

y x y x I z 22

12

Therefore

22

12 y x yx xy

if yx xy

then as ;0, y x

Therefore yx xy

Similarly zy yz zx xz ;

Thus the stress tensor

xz xy xx

i id b i



zz zy zx

yz yy yxij

is said to be symmetric

An important property of symmetric tensors is that the sum of diagonals isindependent of the coordinate system used

For a hydrostatic fluid

p

p

p

ij

00

00

00

Sum of diagonals = -3 p Which is independent of coordinate

system

Using the analogy of hydrostatic fluid we define

3

zz yy xx

zz zz

yy yy

xx xx

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



Conservation of momentum equations



For x direction: xx dx x

xx xx

Net normal surface force =

dxdydz x

dxdydz x

p

dxdydz x

dydz dydz dx

x

xx

xx

xx xx

xx

Similarly, net shear surface flow = dxdydz z y

xz xy

Let the body force in x direction = Xdxdydz

Conservation of momentum states that

F ma

dxdydz z y x x

p X

Dt

Dudxdydz xz xy xx

or

puuuu 11



z y x x

p X

z

uw

y

uv

x

uu

t

u xz xy xx

11

Similarly in y and z direction

z y x y

pY z

vw y

vv x

vut

v yz yy yx

11

z y x z

p Z

z

ww

y

wv

x

wu

t

w zz zy zx

11

What about body forces ?

For gravity : g Z Y X ;0Also, for Newtonian fluid

x

w

z

u

x

v

y

u

x

u xz xy xx m m m ;;2

Assuming that μ does not vary we get

2

2

2 2 2

2 2 2

12

xy xx 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

m

m m

Navier Stokes equations

=0

Thus

2221 uuupuuuu m



222

1

z

u

y

u

x

u

x

p

z

uw

y

uv

x

uu

t

u

m

2

2

2

2

2

21

z

v

y

v

x

v

y

p

z

vw

y

vv

x

vu

t

v

m

2

2

2

2

2

21

z

w

y

w

x

w

z

p g

z

ww

y

wv

x

wu

t

w

m

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

2

0

1ˆ

U

U U U gk p U

t n

where

222

m n

Kinematic coefficient of viscosity

Laplace’s operator (or Laplacian)

fluiddynamics1-100425112946-phpapp02

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