Fluid dynamics 1

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

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

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

To study fluctuations in density (e.g. variation in m
air density with altitude) we need a local estimate
v v 0

How small should δv be to get a true local estimate ?
( Number of molecules) * (mass / molecule)
At the molecular scale
v
Density will depend upon the number of molecules in the sample volume

(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
dp
Ev • Reciprocal of compressibility is known as the bulk modulus of
d 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
d temperature
dT • 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)

• 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) p RT

• 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

Equation of state for an ideal gas
2
1

p
pV mRT or RT

Sometimes written …
pV = nRuT
n= number of moles of gas (kmol/kg)
Ru= Universal gas constant (kJ/kmol K)

Liquids Gases
Almost incompressible Relatively easy to
compress
Forms a free surface Completely fills any vessel
in which it is placed

Hydrostatic Equation
p
g
z
z

h

IF = a p gz gh
constant

Shear stress in moving fluids
y
Newtonian fluid U
3
U
y

= viscosity (or dynamic viscosity) kg/ms
= kinematic viscosity m2/s 4

Continuity
.
AU = constant = m

Mass is conserved

What flows in = what flows out !
5
1A1U1=
2A2U2

Bernoulli
6

p 2 U2
1 gz …
cons tan t if no losses
7

2
p1 U1 p2 U2
2
z1 z2 1H 2
g 2g g 2g

Where H2
1 is the head loss (m)

Reynolds Number
UL UL
Re

inertia f orce rate of change of momentum U2L2
v iscousf orce shear stress x area L2

Venturi2 A 2U2
1A 1U1

p 2 U2
1 gz cons tan t

p1
p2

  

Discharge Coefficients
Q' CD Q

8

Actual Flowrate = CD x Predicted flowrate

Orifice Plate
D D/2

D d
  

Vena
contracta
Orifice
plate

Discharge Coefficients
Q' CD Q

Venturi meter Orifice plate

CD CD
0.65 Increasing
0.98
values of d/D

0.6
0.94

104 106 Re 10 10 106 Re
D 4 5
D

Taylor series
Taylor stated that in the neighborhood of x a the function f(x) can be given by
Primes denote differentials
2 3
x a x a
f x f a f a x a f a f a 
2 3!
p0 x

p1 x
p2 x

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

Note:
p0 x A constant value at x a
p1 x A straight line fit at x a
p2 x A parabolic fit at x a
For x a 1 each additional term is smaller than the previous term

Example
x
Find the Taylor series expansion for f x e about x 1
1
p0 x f a e
1 1
p1 x f a f a x a e e x 1
2 2
x a 1 1 1 x 1
p2 x f a f a x a f a e e x 1 e
2 2

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

for x 1.2 for x 1.5
1.2 1.5
f (1.2) e 0.3012 f (1.5) e 0.2231
1 1
p0 1.2 e 0.3679 p0 1.5 e 0.3679
1 1 1 1
p1 1.2 e e 0.2 0.2943 p1 1.5 e e 0.5 0.1839
2 2
1 1 1 0.2 1 1 1 0.5
p2 1.2 e e 0.2 e 0.3017 p2 1.5 e e 0.5 e 0.2299
2 2

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
Mathematically a vector field is denoted by the magnitude along each
orthogonal coordinate axis used to describe the system

V ui vˆ wk
ˆ j ˆ (Vector notation for Cartesian coordinates)
ˆ j ˆ
i , ˆ, k unit vecto along the x, y, z directionrespective
rs ly
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
z

w y  
v m agV V u2 v2 w2
u x

Dot Product

V    
V U V U cos

U
 
For θ = 90 (perpendicular vectors) V U 0
   
For θ = 0 (collinear vectors) V U VU
For a Cartesian coordinate system, if

V j ˆ
v xi v y ˆ vz k
ˆ iˆ ˆ
j ˆ
j ˆ ˆ
iˆ iˆ k k iˆ j ˆ ˆ j
ˆ k k ˆ 0

 iˆ iˆ ˆ
j j ˆ ˆ
ˆ k k 1
U ˆ j ˆ
u x i u y ˆ uz k
Then  
V U j ˆ ˆ j ˆ
vxi v y ˆ vz k uxi u y ˆ uz k
ˆ
vxux v yuy vzuz
Note : Dot product of two vectors is a scalar

Deconstruction of a vector into orthogonal components
  
Consider a pair of vectors U and V U
θ 
 V

Aim is to separate U into two components such that one of them is collinear with V
   
U U1 U2  Represents a unit vector along the direction of V
V
Define a vector v ˆ  
V Component of U along V
  
Now U ˆ
v ˆ
U v cos U cos
 
Therefore U1 ˆ ˆ
U vv provides the direction of the vector
  
and U2 U U1
            2
also U 2 U1 U U1 U1 U U1 U1 U1 U U1 cos U1
 
But U1 U cos
  2 2 2 2
Substituting we get U2 U1 U cos U cos 0
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)

 Consider a vectorfield U in an arbitraril
y
U enclosed volumeV

V ˆ
n Unit vector normal to surface

dA ˆUnit vector tangential to surface

ˆ
U n Component of flow normal to surface dA

Therefore ˆ
U ndA T ot alflow normalto thesurface of V
A

T ot alflow into/outof thesyst em
Then Represents the total flow

U ndAˆ across a closed system per

div U A unit volume
V limitV 0

For a Cartesian system
   
U ui vj wk
x0 , y0 , z0 dz V dxdydz
dx dy
At the front face: ˆ ˆ
n i

ˆ
U ndA u( x0 dx / 2, y0 , z0 )dydz
A

At the back face: ˆ
n ˆ
i

ˆ
U ndA u( x0 dx / 2, y0 , z0 )dydz
A

Thus, contribution from these two faces to the divergence is

u x0 dx / 2, y0 , z0 u x0 dx / 2, y0 , z0 u
dx limdx 0 x

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

 u v w
div U
x y z

This can be written in a more concise form by introducing an operator

ˆ
i ˆ
j ˆ
k
x y z
known as the “dell” operator

Thus
  
div U U

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
C
Let the unit vector along this path be
ˆ ˆ sy ˆ ˆ P
s sxi j sz k  
R R ˆ
ss
Then by definition
 
dT T R ˆ
ss T R
ds s lim s 0

Using differentiation by parts we get
dT T dx T dy T dz
ds x ds y ds z ds
T T T
sx sy sz
x y z
Gradient of a scalar

ˆ
T s


Thus ˆ ˆ
T s represents the change in scalar T along the vector s

T Represents a vector of the scalar quantity T. What is its
direction?
angle between the gradient and direction of change
dT   
T ˆ
s ˆ
T s cos T cos
ds

dT
Maximum value of occurs when cos 1 or 0
ds

Thus T lies along direction of maximum change

Now consider that the path s lies along a surface of constant T
dT 
ˆ
T s 0
ds
A vector along a surface is denoted by the tangent to the surface
dT 
T ˆ 0
ds

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
   
U V U V sin e ˆ
ˆ
e 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
   
For θ = 90 (perpendicular vectors) U V UV
 
For θ = 0 (collinear vectors) U V 0
For a Cartesian coordinate system, if

V ˆ
v xi vy ˆ
j ˆ
vz k ˆ ˆ ˆ ˆ ˆ ˆ
i i j j k k 0

U ˆ
uxi uy ˆ
j ˆ
uz k i ˆ
ˆ j ˆ
k; ˆ i
j ˆ ˆ ˆ ˆ
k;i k ˆ
j
ˆ ˆ
k i j j ˆ
ˆ; ˆ k ˆ ˆ
i;k ˆ
j ˆ
i
T hen
 
U V ˆ
i u y vz uz v y ˆ ux vz
j v x uz ˆ
k ux v y vxu y

A convenient rule is

ˆ j ˆ
i ˆ k
 
U V u x u y uz ˆ
i u y vz uz v y ˆ ux vz
j v x uz ˆ
k ux v y vxu y
v x v y vz

Curl of a vector (an example of the cross product)

Consider a velocity field given by V ˆ j ˆ
ui vˆ wk
  
Then curl V V
iˆ j ˆ
ˆ k
  w v w u v u
V iˆ ˆ
j ˆ
k
x y z y z x z x y
u v w
What does this represent?

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

 
Then ˆ v
V k
u
x y y
A fluid particle in motion has both translation B
and rotation Δy
A rigid body rotates without change of shape, O Δx A
but a deformable body can get sheared x
Thus for a fluid element, angular velocity is defined as the average angular velocity
of two initially perpendicular elements
v A vO v ˆ
Angular velocity of edge OA = k
x lim x 0 x
uO u B u ˆ
Angular velocity of edge OB = k
y lim x 0 y
1 v u ˆ
Angular velocity of fluid element = k
2 x y

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) 
• ˆ
The dot product of a vector and a unit vector V s represents the magnitude of
the vector along the direction of the unit vector
 
• The divergence V represents the net flow out / in to a closed system per
unit volume

• The gradient of a scalar T 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 T s represents the change of
ˆ
the scalar along the direction of the unit vector
 
• The curl of a velocity vector V represents twice the angular velocity of the
fluid

Kinematics of Fluid Motion
• Kinematics refers to the study of describing
fluid motion
• Traditionally two methods employed to
describe 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
  
X F X 0 , t0 , t
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   
 dX dF X 0 , t0 , t
u
dt dt
  
 du d 2 F X 0 , t0 , t
a
dt dt2
Lagrangian framework is useful because fundamental laws of mechanics are
formulated for particles of fixed identity.

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

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

  in domain
u f x, t

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

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

U0 d0 U(x) d(x)

In a Eularian framework the flow field is given by
 
u U ( x)i
Where, according to the conservation of mass (to be discussed later)
U0d0 U x d x

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

u U x0
After time Δt the particle moves to x0 + Δx
where x u t
The new velocity is now given by u u U x0 x, t U x0 u t, t

Applying Taylor series expansion we get
U U
u u U x0 u t t
x t
u U U
or u
t x t
In the limit t 0
u Du U U
accelerati
on u
t t 0 Dt t x

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 f g x t , y t , z t ,t

Rate of change of f with respect to time is given by
df
dt
According to the chain rule of differentiation

df f f dx f dy f dz
dt t x dt y dt z dt
where
dx
velocit yalong t he x direct ion u
dt
dy
velocit yalong t he x direct ion v
dt
dz
velocit yalong t he x direct ion w
dt
Therefore
Material derivative
df f f f f Df
u v w
dt t x y z Dt

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

U dx 0

Substituting U uiˆ vˆ
j ˆ
wk and dx ˆ
dxi dyˆ
j ˆ
dzk
We get

dx dy dz Differential equation that needs to be solved to
u v w determine stream lines

Example:

Consider a flow field given by U ˆ
xi yˆ
j
Substituting in our equation for stream lines we get

dx dy
x y
Integrating we get ln xy constant
or xy constant

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 U ˆ
xi yˆ
j
Path lines can be obtained by integrating the flow field in time

dx dy
u x; v y
dt dt
dx dy
or dt; dt
x y Constants of integration
Which yields x a0et ; y b0e t

Note that xy a0b0 constant

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

ˆ
1   i w v ˆ
j w u ˆ
k v u
V
2 2 y z 2 x z 2 x y
  
V

Referred to as vorticity

Apart from rotation the fluid particle is also stretched and distorted
u
dydt
y

v
dy dydt
dy y
v
dxdt
x

dx u
dx dxdt
x

u
dx dxdt dx
The stretching (or extensional strain) is given by dt x
xx
dx
u
or xx
x
v w
Similarly we get yy and zz
y z

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
1 u v
From the figure this yields xy
2 y x
1 v w 1 u w
And similarly yz and xz
2 z y 2 z x

Also note that the shear strain is symmetric yx xy

The total rate of deformation of a moving fluid particle can be written in matrix form
u u u
x y z
v v v
D
x y z
Deformation tensor
w w w
x y z

Which can be split up to yield

u 1 u v 1 u w 1 u v 1 u w
0 0 0 0
x 2 y x 2 z x 2 y x 2 z x
v 1 u v 1 v w 1 u v 1 v w
D 0 0 0 0
y 2 y x 2 z y 2 y x 2 z y
w 1 u w 1 v w 1 u w 1 v w
0 0 0 0
z 2 z x 2 z y 2 z x 2 z y

stretching shearing
rotating
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
ˆ
e
Effect of rotation on a vector B ˆ
er
Unit vectors along
Consider a point B on the  ˆ ˆ ˆ
ez , er , e
R orthogonal coordinate
surface of a cylinder rotating
axes
about its vertical axis with O
angular velocity Ω

ˆ ˆ
The position vector is given by R zez rer

 dR ˆ
der d
Velocity vector is then given by V r r ˆ
e ˆ
re
dt dt dt
 
At the same time R ˆ
ez ˆ
zez ˆ
rer ˆ
re

dR  
Thus R
dt

Now consider a fluid particle in an arbitrarily moving coordinate system
 
R is the position vector of the origin of the r
moving coordinate system with respect to 
the inertial coordinate system R

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

 dr
a) Motion relative to moving frame uR
dt
 
b) Angular velocity r

dR
c) Motion of moving reference frame
dt
Thus the velocity is given by

 dR   
uI uR r
dt

Differentiating once again yields acceleration

 2
d R d       
aI uR r uR r
dt2 dt
Simplifying we get
  
 2
d R duR d      
aI r 2 uR r
dt2 dt dt
(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 
E 
Angular velocity of the earth is 7.29x 10-5 sec-1 y
z
Using cartesian coordinates we have x into the paper
 ˆ ˆ
r xi yˆ zk
j

R Rkˆ
 ˆ
uR ui vˆ wk
ˆ j
 
cos ˆ
j ˆ
sin k
E

Since change in R Is occurring purely due to rotation, we have

dR  
E R
dt

2
d R   
E E R
dt2

We thus get

ˆ
i j ˆ
ˆ k
 
2 uR 2 j ˆ
2 cos w 2 sin v i 2 sin uˆ 2 cos uk
ˆ
x y z

u v w

        
r r r
x 2iˆ 2
sin z cos y sin ˆ
j 2
cos y sin ˆ
z cos k

Where we have used the vector identity
        
A B C B CA A BC

Similarly we get
  
R 2
R sin cos ˆ
j 2 ˆ
R cos2 k
E E

Assuming a constant rotation rate

d
0
dt

Individual components of acceleration can thus be given by

du 2
ax 2 cos w 2 sin v x
dt
dv 2
ay 2 sin u sin R cos z cos y sin
dt
dw 2
az 2 cos u cos R cos z cos y sin
dt

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
du
ax fv Note : f has opposite signs in the northern and
dt southern hemispheres. This is why hurricanes are
dv counter clockwise in the northern hemisphere and
ay fu
dt clockwise in the southern
where f 2 sin

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
 
ma F
mass
Sum of all forces acting on fluid
acceleration

Conservation of mass

Consider a control volume of infinitesimal size
Let density = x, y, z, t dz

ˆ j ˆ
Let velocity = V u x, y, z, t i v x, y, z, t ˆ w x, y, z, t k dy
dx
Mass inside volume = dxdydz
Mass flux into the control volume = udydz vdxdz wdxdy
Mass flux out of the control volume = u u dx dydz v v dy dxdz
x y

w w dz dxdy
z
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

u v w
x y z t

Or, using differentiation by parts, we get
u v w
u v w 0
t x x y y z z

Rearranging, we get
u v w
u v w 0
t x y z x y z

D
 
V
Dt
Using vector notation

1 D  
V 0 Conservation of mass equation,
Dt 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

Conservation of momentum

Law of conservation of momentum states that for a fluid particle accelerating in water
 
ma F
Let us look at the forces acting on a fluid particle

Forces
Two 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
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
Since there is no net motion forces in
the x and z direction must balance each nn
other out dz
Force balance in the x direction
xx
dx θ
xx dz nn dl sin nn dz
or xx nn zz
Force balance in the z direction
1 1
zz dx nn dl cos gdxdz ; or zz nn gdz
2 2
Reducing particle to zero size we get zz nn

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

nn ˆ
pn Hydrostatic pressure (scalar)

Negative sign indicating compressive forces (convention)

What is the expression for pressure ? p z dz / 2

Consider a finite sized particle cube in dy
stationary fluid
dz p x dx / 2
p x dx / 2
Force balance in the x direction dx
p z dz / 2
p x dx / 2 p x dx / 2 0
p dx p dx
p p 0
x 2 x 2
p
0
x
p
Similarly 0
y
Force balance in the z direction

p z dz / 2 p z dz / 2 dxdy gdxdydz
p
g
z
Integrating yields p gz C

Surface forces in a moving fluid
Viscous forces between the different fluid particles will lead to
1. Non – isotropic normal stresses
2. Development of shear stresses
zz
At any given plane of a fluid particle
zx there are three forces acting on the fluid
xz particle (one normal and two shear
yz stresses)
zy xx
yx
(Notation : First subscript refers to direction
xy of force and second subscript to direction of
yy perpendicular to plane)

xx xy xz
For each axis there are three
ij yx yy yz contributions from the stress tensor
Referred to as stress tensor zx zy zz

How are the shear stresses related ? yxy
yx
y 2
Consider the moments acting on
a fluid particles
xy x xy x
xy xy
Counter clockwise moments x 2 x 2

xxy x xy x x y
xy y xy y yx
x 2 2 x 2 2 yx
y 2
xy y x
clockwise moments

yx y y yx y y
yx x yx x
y 2 2 y 2 2
yx y x

Conservation of moments
Rotational acceleration

moments I z 

For a rectangular element Moment of inertia

Iz x2 y2 x y
12
Therefore

xy yx x2 y2 
12
if xy yx

then as x, y 0; 

Therefore xy yx

Similarly xz zx ; yz zy

Thus the stress tensor xx xy xz

is said to be symmetric ij yx yy yz

zx zy zz

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

p0 0
For a hydrostatic fluid
ij 0 p 0
Sum of diagonals = -3p
0 0 p
Which is independent of coordinate
system

Using the analogy of hydrostatic fluid we define
xx p xx
Separates normal stress into a compressive part +
yy p yy deviations
zz p zz Note: This is a mechanical definition of pressure for
xx yy zz
moving fluids and is not equal to that of hydrostatic
p fluids
3

The stress tensor can thus be written as

p0 0 xx xy xz

ij 0 p 0 yx yy yz ij
0 0 p zx zy zz (deviatoric stress tensor)

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
u 1 u v 1 w u
x 2 y x 2 x z
1 u v v 1 w v
ij 2
2 y x y 2 y z

Coefficient of viscosity 1 w u 1 w v w
2 x z 2 y z z

Conservation of momentum equations
xx
xx dx
For x direction: xx
x

Net normal surface force = xx
dx dydz dydz
xx xx
x
xx
dxdydz
x
p xx
dxdydz dxdydz
x x
xy xz
Similarly, net shear surface flow = dxdydz
y z
Let the body force in x direction = Xdxdydz
Conservation of momentum states that
ma F
Du p xx xy xz
dxdydz X dxdydz
Dt x x y z

or
u u u u 1 p 1 xx xy xz
u v w X
t x y z x x y z
Similarly in y and z direction
v v v v 1 p 1 yx yy yz
u v w Y
t x y z y x y z
w w w w 1 p 1 zx zy zz
u v w Z
t x y z z x y z
What about body forces ? Navier Stokes equations
For gravity : X Y 0; Z g
Also, for Newtonian fluid
u u v u w
xx 2 ; xy ; xz
x y x z x
Assuming that μ does not vary we get
2
1 xx xy xz u u v u w
2
x y z x2 y y x z z x =0
2 2 2
u u u u v w
x2 y2 z2 x x y z

Thus
2 2 2
u u u u 1 p u u u
u v w
t x y z x x2 y2 z2
2 2 2
v v v v 1 p v v v
u v w
t x y z y x2 y2 z2
2 2 2
w w w w 1 p w w w
u v w g
t x y z z x2 y2 z2

The equations of motion, can thus be written in vector form as
 
U 0

U    1  
U U ˆ
gk p 2
U
t
where Kinematic coefficient of viscosity

Laplace’s operator (or Laplacian)
2
  2 2 2

x2 y2 z2

Fluid dynamics 1

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Fluid dynamics 1