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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 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)