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Introduction to Fluid Dynamics
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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)
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List the fluid systems in …
Typical Home
Car
Aircraft
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Leaking crude oil from the grounded tanker Argo Merchant
(Nantucket Shoals 1976)
ctsy: An album
of fluid motionby Van Dyke
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Smoke plumes
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Turbulent Jet impinging into fresh water
ctsy: An album
of fluid motionby Van Dyke
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Trapped plume in a stratified ambient flow
ctsy: CORMIX
picture gallery
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A salt wedge propagating into fresh water
ctsy: Gravity currents in the environment and
the laboratory, author : John E. Simpson
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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)
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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
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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 ?
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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
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(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
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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”
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– 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).
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– 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
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• 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
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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
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Liquids Gases
Almost incompressible
Forms a free surface
Relatively easy to
compress
Completely fills any vesselin which it is placed
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Hydrostatic Equation
gz
p
z
ghgzp
h
IF = a
constant
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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
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Continuity
AU = constant = m
Mass is conserved
What flows in = what flows out !
1 A1U1=
2 A2U2
.
5
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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
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Reynolds Number
n
m
ULULRe
2
22
L
LU
areaxstressshear
momentumof changeof rate
forceviscous
forceinertia
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Venturi
ttanconsgzUp 221
222111 U AU A
p1 p2
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Discharge Coefficients
QC'Q D
Actual Flowrate = CD x Predicted flowrate
8
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Orifice PlateD D/2
dD
Orifice
plate
Vena
contracta
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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
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A Mathematical Review
Taylor series
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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)
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( ,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
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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
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Kinematics of Fluid Motion
• Kinematics refers to the study of describing
fluid motion
• Traditionally two methods employed todescribe fluid motion
– Lagrangian
– Eularian
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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.
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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
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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
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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
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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
directionthealongvelocity
directionthealongvelocity
where
g
Material derivative
Dt
Df
z
f
w y
f
v x
f
ut
f
dt
df
Therefore
Stream lines
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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
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Path lines
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Conservation of momentsRotational acceleration
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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
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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
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Conservation of momentum equations
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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
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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
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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)