Post on 04-Apr-2018
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ME 311: Fluid Mechanics
Differential Analysis
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Course Outline Navier-Stokes equations for Laminar Flow
Characterization of Laminar and Turbulent Flow Reynold Stresses
Boundary layer theory
Flow over flat plate and in pipes
Lift and Drag Forces
Applying energy, momentum and continuityequations of Thermofluids to turbo-machinery,
Performance of Turbo-Machines.
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Recommended Text Books
1 Fundamentals of Fluid
Mechanics
Munson, Young
& Okiishi
2 Mechanics of Fluids B. S. Massey
3. Fluid Mechanics Victor L. Streeter
and E. Benjamin
Wylie
4. Mechanics of Fluids Merle C. Potter;
David C. Wiggert
5 Introduction to Fluid
Mechanics
Robert W. Fox ;
Alan T.
McDonald
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General Classroom Rules
Mutual respect (golden rule)
Punctuality
Minimal disturbance to fellow students and teacher
Turn off your cell phone
No chewing /tobacco
Questions are encouraged
No question is stupid
Your question is valuable to others in learning
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My Preference
Learning happens both inside and
outside the classroom
Inside classroom: interactive,participation
Outside classroom: Any time
Welcome feedback anytime during the
quarter (class format/materials/pace)
you are welcome to come to see me /
call me any time any where.
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Outline
Introduction
Kinematics Review Conservation of Mass
Stream Function
Linear Momentum Inviscid Flow
Viscous Flows
Navier-Stokes Equations
Exact Solutions
Intro. to Computational Fluid Dynamics
Examples
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Differential Analysis: Introduction
Some problems require more detailed analysis.
We apply the analysis to an infinitesimal control
volume or at a point. The governing equations are differential equations
and provide detailed analysis.
Around only 80 exact solutions to the governingdifferential equations.
We look to simplifying assumptions to solve theequations.
Numerical methods provide another avenue forsolution (Computational Fluid Dynamics)
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Kinematic Velocity FieldContinuum Hypothesis: the flow is made of tightly packed fluid particles that
interact with each other. Each particle consists of numerous molecules, and we
can describe velocity, acceleration, pressure, and density of these particles at a
given time.
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Kinematic Acceleration FieldLagrangian Frame:
Eulerian Frame: we describe the acceleration in terms of position and time
without following an individual particle. This is analogous to describing the
velocity field in terms of space and time.
A fluid particle can accelerate due to a change in velocity in time (unsteady)
or in space (moving to a place with a greater velocity).
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Kinematic Acceleration Field: Material (Substantial) Derivative
time dependence spatial dependence
We note:
Then, substituting:
The above is good for any fluid particle, so we drop A:
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Kinematic Acceleration Field: Material (Substantial) Derivative
Writing out these terms in vector components:
x-direction:
y-direction:
z-direction:
Writing these results in short-hand:
where,
kz
jy
ix
()
,
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Kinematics: Deformation of a Fluid Element
General deformation of fluid element is rather complex, however, we can
break the different types of deformation or movement into a
superposition of each type.
Linear Motion Rotational Motion
Linear deformation Angular Deformation
General
Motion
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Kinematics: Linear Motion and DeformationLinear Motion/Translation due to u and vvelocity:
Simplest form of motion the element
moves as a solid body. Unlikely to be the
only affect as we see velocity gradients inthe fluid.
Deformation: Velocity gradients can cause deformation, stretching
resulting in a change in volume of the fluid element.
Rate of Change for one direction:
For all 3 directions: The shape does not change, linear deformation
The linear deformation is zero for incompressible fluids.
= 0
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Kinematics:Angular Motion and DeformationAngular Motion/Rotation:
Angular Motion results from
cross derivatives.
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Kinematics:Angular Motion and Deformation
The rotation of the element about the z-axis is the average of the angular
velocities :
Likewise, about the y-axis, and the x-axis:
Counterclockwise rotation is considered positive.
and
The three components gives the rotation vector:
Using vector identities, we note, the rotation vector is one-half the curl of the
velocity vector:
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Kinematics:Angular Motion and Deformation
The definition, then of the vector operation is the following:
The vorticity is twice the angular rotation:
Vorticity is used to describe the rotational characteristics of a fluid.The fluid only rotates as and undeformed block when ,
otherwise, the rotation also deforms the body.
If , then there is no rotation, and the flow is said to be irrotational.
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Kinematics:Angular Motion and Deformation
Angular deformation:
The associated rotation gives rise to angular deformation, which results in the
change in shape of the element
Shearing Strain:
Rate of Shearing Strain:
If , the rate of shearing strain is zero.
The rate of angular deformation is related to the shear stress.
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Conservation of Mass: Cartesian Coordinates
System: Control Volume:
Now apply to an infinitesimal control volume:
For an infinitesimal control volume:
Now, we look at the mass flux in the x-direction:
Out: In:
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Conservation of Mass: Cartesian Coordinates
Net rate of mass in the outflow y-direction:
Net rate of mass in the outflow z-direction:
Net rate of mass in the outflow x-direction:
Net rate of mass flow for all directions:
+
Now, combining the two parts for the infinitesimal control volume:
= 0
Divide out
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Conservation of Mass: Cartesian CoordinatesFinally, the differential form of the equation for Conservation of Mass:
a.k.a. The Continuity Equation
In vector notation, the equation is the following:
If the flow is steady and compressible:
If the flow is steady and incompressible:
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Conservation of Mass: Cylindrical-Polar Coordinates
If the flow is steady and compressible:
If the flow is steady and incompressible:
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Conservation of Mass: Stream Functions
Stream Functions are defined for steady, incompressible, two-dimensional flow.
Continuity:
Then, we define the stream functions as follows:
Now, substitute the stream function into continuity:
It satisfies the continuity condition.
The slope at any point along a streamline:
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Conservation of Mass: Stream Functions
Streamlines are constant, thus d = 0:
Now, calculate the volumetric flow rate between streamlines:
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Conservation of Mass: Stream Functions
In cylindrical coordinates:
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Conservation of Linear Momentum
P is linear momentum,System:
ControlVolume:
We could apply either approach to find the differential form. It turns out the System
approach is better as we dont bound the mass, and allow a differential mass.
By system approach,m is constant.
If we apply the control volume approach to an infinitesimal control volume, wewould end up with the same result.
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Conservation of Linear Momentum: Forces Descriptions
Body forces or surface forces act on the differential element: surface forces act
on the surface of the element while body forces are distributed throughout the
element (weight is the only body force we are concerned with).
Body Forces:
Surface Forces: Normal Stress:
Shear Stress:
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Conservation of Linear Momentum: Forces Descriptions
Looking at the various sides of the differential element, we must usesubscripts to indicate the shear and normal stresses (shown for an x-face).
The first subscript indicates the plane on which the stress acts
and the second subscript the direction
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Conservation of Linear Momentum: Forces
Descriptions
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Now, the surface forces acting on a small cubicle element in each
direction.
Then the total forces:
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Conservation of Linear Momentum: Equations of Motion
Now, we both sides of the equation in the system approach:
In components:
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Conservation of Linear Momentum: Equations of Motion
Writing out the terms for the Generalize Equation of Motion:
The motion is rather complex.
Material derivative foraForce Terms
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Inviscid Flow
An inviscid flow is a flow in which viscosity effects or shearing effects become
negligible.
If this is the case,
And, we define
A compressive force give a positive pressure.
The equations of motion for this type of flow then becomes the following:
Eulers
Equations
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Inviscid Flow: Euler s Equations
Leonhard Euler
(1707 1783)
Famous Swiss mathematician who pioneered work on the
relationship between pressure and flow.
In vector notation Eulers Equation:
The above equation, though simpler than the generalized equations, are still
highly non-linear partial differential equations:
There is no general method of solving these equations for an analytical solution.
The Eulers equation, for special situations can lead to some useful information
about inviscid flow fields.
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Surface Stress Terms for a General Newtonian Fluid
General Stress Elements:
Normal Stresses:
Shear Stresses:
Note, and is known as the second viscosity coefficient
is the viscosity of the fluid and for the general form is allowed to be non-constant.
xxxx p
yyyy p
zzzz p
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Viscous Flows: Surface Stress Terms
Now, we allow viscosity effects for an incompressible Newtonian Fluid:
Normal Stresses:
Shear Stresses:
Cartesian
Coordinates:
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Viscous Flows: Surface Stress Terms
Normal Stresses:
Shear Stresses:
Cylindrical
Coordinates:
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Viscous Flows: Navier-Stokes Equations
Now plugging the stresses into the differential
equations of motion for incompressible flow giveNavier-Stokes Equations:
French Mathematician, L. M. H. Navier (1758-1836) andEnglish Mathematician Sir G. G. Stokes (1819-1903)
formulated the Navier-Stokes Equations by including
viscous effects in the equations of motion.
L. M. H. Navier(1758-1836)
Sir G. G. Stokes(1819-1903)
(x direction)
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Viscous Flows: Navier Stokes Equations
Local Acceleration Advective Acceleration
(non-linear terms)
Pressure term Weight term
Viscous terms
Terms in the x-direction:
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Viscous Flows: Navier-Stokes Equations
The governing equations can be written in cylindrical coordinates as well:
(r-direction)
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Viscous Flows: Navier-Stokes Equations
There are very few exact solutions to Navier-Stokes Equations, maybe a
total of 80 that fall into 8 categories. The Navier-Stokes equations are
highly non-linear and are difficult to solve.
Some simple exact solutions presented in the text are the following:
1. Steady, Laminar Flow Between Fixed Parallel Plates
2. Couette Flow3. Steady, Laminar Flow in Circular Tubes
4. Steady, Axial Laminar Flow in an Annulus
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Assumptions:
1. Plates are infinite and parallel/horizontal
2. The flow is steady and laminar3. Fluid flows 2D, in the x-direction only u=u(y)
only, v and w = 0
4. Fully develop
5. 5. Incompressible
32
2
21
3
3
333
33 3
33 4
3
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Viscous Flows: Exact Solutions/Parallel Plate Flow
Navier-Stokes Equations Simplify Considerably:
Applying Boundary conditions (no-slip conditions at y = h) and solve:
The pressure gradient must be specified and is
typically constant in this flow! The sign is
negative.
(Integrate Twice)
3 3
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Viscous Flows: Exact Solutions/Parallel Plate Flow
Solution is Parabolic:
Can determine Volumetric Flow Rate:
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Viscous Flows: Exact Solutions/Parallel Plate Flow
Navier-Stokes Equations Simplify Considerably:
Applying Boundary conditions (no-slip conditions at y = h) and solve:
The pressure gradient must be specified and is
typically constant in this flow! The sign is
negative.
(Integrate Twice)
3 3
Solution of flow between two flat plates (Couette flow)
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Solution of flow between two flat plates (Couette flow)
The differential equation may be solved by integration
2
2
1d u dpdy dydy dx
Hence 1du dp y Ady dx
And a further integration wrt y yields21
2
dp yu Ay B
dx
Boundary conditions
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Boundary conditions
Due to molecular bonding between the fluid and the wall
it may be assumed that the fluid velocity on the wall is zero
u=0 at y=0u=0 at y=c
This is known as the no-slip condition.
To satisfy the first boundary condition, B=0Then the second b.c. gives
210
2
dpcAc
dx
1
2
dp cA
dx
Quadratic velocity profile for flow in a channel
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Quadratic velocity profile for f low in a channel
Substituting the values for A and B into the previous
equation gives the quadratic equation:
212
dpu y ycdx
For a long, straight channel, of length l, p decreases
with length at a constant rate, so
dp p
dx l
21
2
pu y yc
l
Graph of velocity profile
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G ap o e oc y p o e
0
0.2
0.4
0.6
0.8
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
y
u
Volume flow rate
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Volume flow rate
To calculate the volume flow rate, integrate from y=0 to y=c
y=0
y=c
dy
dq udy 202
cp
q yc y dy
l
3
12
c pq
l
per unit width (z direction)
Maximum and mean velocity
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y
Max velocity occurs at y=c/2, the centre of the channel
2
max8
c pu l
Mean velocity is gained by dividing the flow rate by the
channel width
/u q c
2
max
2
12 3
c pu u
l
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Viscous Flows: Exact Solutions/Couette FlowAgain we simplify Navier-Stokes Equations:
Same assumptions as before except the no-slip
condition at the upper boundary is u(b) = U.
Solving,
If there is no Pressure Gradient:
The termDetermines effects of
pressure gradient
Dimensionless,
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Viscous Flows: Exact Solutions/Pipe Flow
Assumptions:
Steady Flow and Laminar FlowFlow is only in the z-direction
vz = f(r)
Navier-Stokes
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Note that exactly the same result for the velocity
distribution could be derived by solving the Navier-
Stokes equations in radial coordinates.
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Cylindrical Coordinates
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Cylindrical Coordinates
The Navier-Stokes equation in the r-direction is:
rzrr
r
z
rr
r
r
g
z
vv
r
v
rr
rv
rrr
p
z
vv
r
vv
r
v
r
vv
t
v
2
2
22
2
2
2
21)(1
Cylindrical Coordinates
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Cylindrical Coordinates
The Navier-Stokes equation in the -direction is:
gz
vv
r
v
rr
rv
rr
p
r
z
v
vr
vvv
r
v
r
v
vt
v
r
z
r
r
2
2
22
2
2
21)(11
Cylindrical Coordinates
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y
The Navier-Stokes equation in the z-direction is:
zzzz
z
z
zz
r
z
gz
vv
rr
vr
rrz
p
z
v
v
v
r
v
r
v
vt
v
2
2
2
2
2
11
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We will return to the pipe flow problem from the start ofthe lecture and solve it using the Navier-Stokes
equations.
Continuity:
0
0
v
vr
01)(1
z
vv
rr
rv
rzr
0 0
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r-direction Navier-Stokes:
0
21)(1
2
2
2
2
22
2
2
2
z
v
r
p
gzvv
rv
rrrv
rrrp
zvv
rvv
rv
rvv
tv
z
rzrr
rz
rrr
r
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-direction Navier-Stokes:
0
21)(112
2
22
2
2
p
gzvv
rv
rrrv
rrp
r
z
vv
r
vvv
r
v
r
vv
t
v
r
zr
r
z direction Navier Stokes
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z-direction Navier-Stokes
zz
zzzz
zz
zzr
z
gr
vr
rrz
p
gzvv
rrvr
rrzp
zvvv
rv
rvv
tv
2
2
2
2
211
Integrate:
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Integrate:
2
2
1
1
1
2
4
00
2
2
cr
gL
ppv
ratfinite
r
vkeeptoc
r
v
r
crgL
pp
rvrcrg
Lpp
gr
vr
rrz
p
zoL
z
z
zzoL
zz
oL
zz
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Viscous Flows: Exact Solutions/Pipe Flow
Solving the equations with the no slip
conditions applied at r = R (the walls of the
pipe).
Parabolic Velocity Profile
Viscous Flows: Exact Solutions/Pipe Flow
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Viscous Flows: Exact Solutions/Pipe Flow
The volumetric flow rate:
The mean velocity:
Pressure drop per length of pipe:
The maximum velocity:
Non-Dimensional velocity profile:
For Laminar Flow:
Substituing Q,
Conservation of Energy
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gy
The energy equation is developed similar to the momentum equation for aninfinitesimal control volume.
Conservation of Energy
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The energy equation is developed similar to the momentum equation for an
infinitesimal control volume.
(Heat and Work)
Internal Kinetic Potential
(Time rate of change
following the particle)
Differentiate:To get the L.H.S:
Now for the R.H.S., define the fluid properties of Heat and Work:
Heat Conduction into the element, Fouriers Law
Heat per Unit Area
Heat:
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Conservation of Energy
Now, we do a control volume analysis on our control element:
Heat Flow into the left x-face of the element
Heat Flow out of the right x-face of the element
The above can be written for all six faces of the cube with the net result
between the in and out:
The net heat flow is transferred to the element, neglecting production terms
Heat: Heat Conduction into the element, Fouriers Law
Heat per Unit Area
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Conservation of EnergyWork: Work is done on the element per unit area.
on the left x-face
on the right x-face
We can do the same for the other faces, and the net rate of work done is:
In condensed form:
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Conservation of EnergyWe can rewrite the equation using and identity:
We note, then, that from the momentum equation:
Now, the rate of change of work is the following:
Kinetic Potential
=
Now, when we substitute work and heat back into the governing equation:
We note potential and kinetic energy portions cancelled on each side!
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Conservation of EnergyNow, we can split the stress tensor into pressure and viscous terms:
Using continuity, we can rewrite the pressure term:
Now, rewriting the Conservation of Energy:
Noting, the definition of fluid enthalpy:
And, defining the dissipation function:
This term always takes energy from the flow!
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Conservation of EnergyWriting out the terms of Viscous Dissipation for a Newtonian Fluid:
Now, with the substitutions, the energy equation take the following form:
We note,
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Conservation of Energy
Now, lets assume the flow is incompressible:
Enthalpy:
Then,
If the flow velocity is low relative to Heat Transfer then terms of order
U disappear.
is the thermal expansion coefficient, for aper fect gas the second term goes to zero!
If, we assume constant thermal conductivity:
Heat Convection Equation
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Summary of Governing Equations
Mass:
Momentum:
Energy:
Most General forms of the Equations:
Only Assumptions:
(1) The fluid is a continuum
(2) the particles are essentially in thermodynamics equilibrium
(3) Only body forces are gravity
(4) The Heat conduction follows Fouriers Law
(5) There are no internal heat sources.
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Summary of Governing Equations
Some general comments on the general form of the governing equations:
1. They are a coupled system of non-linear partial differential equations
you must solve energy, continuity, and linear momentumsimultaneously. No closed form solution exists!
2. For Newtonian flow, the shear and normal stresses can be written in
terms of the velocity gradients introducing no new unknowns.
3. There appear to be five equations and nine unknowns in the system of
equations: , k, p, u, v, w, h, and T.4. However, we note the following:
5. Now, we have five unknowns and five equations
),(),,(
),(),,(
TpkkTphh
TpTp
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Summary of Governing EquationsIn, general in Fluid Mechanics/CFD we often work with a simplified form of the
equations known as the Navier-Stokes Equations:
Additional Assumptions:
(1) The fluid is Newtonian(2) Incompressible
(3) Constant properties (k, )
where,
uncoupled equations: The fluid flowcan be solved independent of the Heat
Transfer
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Summary of Governing EquationsSome general comments on the Navier-Stokes governing equations:
1. They are non-linear partial differential equations which are uncoupled
in energy, and linear momentum. We can solve linear momentum andcontinuity equations separately for the flow field without knowledge of
the Temperature field (4 Equations, 4 unknowns, u, v, w, p).
2. For Newtonian flow, the shear and normal stresses can be written in
terms of the velocity gradients introducing no new unknowns.
3. There appear to be five equations and 5 unknowns in the system of
equations: p, u, v, w, and T.
4. If the convective term disappears we have a linear solution.
5. If the convective term remains we have a non-linear solution.
6. The Energy equations relies on the solution of the flow field for itssolution.
Viscous Flow Equations
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Summary of Governing EquationsSummary of the Euler form of the governing equations: Inviscid Flow Equations
Linear Momentum:
Continuity and Energy are the same as for Navier-Stokes Equations
Some general remarks:
(1) The system of equations have five unknowns and five equations (same as
Navier-Stokes)
(2) Flow is Inviscid (frictionless), Pressure is the only normal stress, and
there are no shear stresses.
(3) A specialized case of inviscid flow is irrotational flow.(4) The energy and momentum equations are also uncoupled in this set of
equations.
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Physical Boundary ConditionsTypes of Boundary Conditions: Fluid/Gas-Solid Interface
Fluid-Fluid Interface
Gas-Fluid Interface
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Physical Boundary Conditions
No Slip Condition:
At the fluid-boundary interface the velocities must be equal. If the boundary
is stationary, then u, v, w = 0.
The temperature of the fluid has to equal the temperature
of boundary at the interface.
Heat Flux in the fluid must equal the heat flux of the solid at the interface
At a solid boundary:
No Temperature Jump:
Equality of Heat Flux:
Examples:
Stationary Solid BoundaryMoving Boundary:
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Computational Fluid Dynamics: Differential AnalysisGoverning Equations:
Navier-Stokes:
Continuity:The above equations can not be solved for most practical problems with analytical
methods so Computational Fluid Dynamics or experimental methods are
employed.
The numerical methods employed are the following:1. Finite difference method
2. Finite element (finite volume) method
3. Boundary element method.
These methods provide a way of writing the governing equations in discreteform that can be analyzed with a digital computer.
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Computational Fluid Dynamics: Finite ElementThese methods discretize the domain of the flow of interest (Finite Element
Method Shown):
The discrete governing equations are solved in every element. This
method often leads to 1000 to 10,000 elements with 50,000 equationsor more that are solved.
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Computational Fluid Dynamics: Finite DifferenceThese methods discretize the domain of the flow of interest as well (FiniteDifference Method Shown):
Finite Difference Mesh:
Comparison between
Experiment and CFD
Analysis:
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Computational Fluid Dynamics: Pitfalls
Numerical Solutions can diverge or exhibit unstable wiggles.
Finer grids may cause instability in the solution rather than better
results.
Large flow domains can be computationally intensive.
Turbulent flows have yet to be well described with CFD.
Inviscid Flow: Bernoull i Equation
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Daniel Bernoulli
(1700-1782)
Earlier, we derived the Bernoulli Equation from a direct
application of Newtons Second Law applied to a fluid particle
along a streamline.
Now, we derive the equation from the Euler Equation
First assume steady state:
Select, the vertical direction as up, opposite gravity:
Use the vector identity:
Now, rewriting the Euler Equation:
Rearrange:
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Inviscid Flow: Bernoulli EquationNow, take the dot product with the differential length ds along a streamline:
ds and Vare parrallel, , is perpendicular to V, and thus to ds.
We note,
Now, combining the terms:
Integrate:
Then,1) Inviscid flow
2) Steady f low
3) Incompressible flow
4) Along a streamline
I i id Fl I t ti l Fl
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Inviscid Flow: Irrotational FlowIrrotational Flow: the vorticity of an irrotational flow is zero.
= 0
For a flow to be irrotational, each of the vorticity vector components must be
equal to zero.
The z-component:
The x-component lead to a similar result:
The y-component lead to a similar result:
Uniform flow will satisfy these conditions:
There are no shear forces in irrotational flow.
Inviscid Flow: Irrotational Flow
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Example flows, where inviscid flow theory can be used:
Viscous RegionInviscid Region
Inviscid Flow: Bernoull i Irrotational Flow
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Recall, in the Bernoulli derivation,
However, for irrotational flow, .
Thus, for irrotational flow, we do not have to follow a streamline.
Then,
1) Inviscid flow
2) Steady flow
3) Incompressible flow4) Irrotational Flow
Potential Flow: Velocity Potential
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For irrotational flow there exists a velocity potential:
Take one component of vorticity to show that the velocity potential is irrotational:
Substitute u and v components:
02
1 22
xyyx
we could do this to show all vorticity components are zero.
Then, rewriting the u,v, and w components as a vector:
For an incompressible flow:
Then for incompressible irrotational flow:
And, the above equation is known as Laplaces Equation.
Potential Flow: Velocity Potential
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Laplacian Operator in Cartesian Coordinates:
Laplacian Operator in Cylindrical Coordinates:
Where the gradient in cylindrical coordinates, the gradient operator,
Then,
May choose cylindrical
coordinates based on the
geometry of the flow problem,
i.e. pipe flow.
If a Potential Flow exists, with
appropriate boundary
conditions, the entire velocity
and pressure field can bespecified.
Potential Flow: Plane Potential Flows
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Laplaces Equation is a Linear Partial Differential Equation, thus there are
know theories for solving these equations.
Furthermore, linear superposition of solutions is allowed:
where andare solutions to Laplaces equation
For simplicity, we consider 2D (planar) flows:
Cartesian:
Cylindrical:
We note that the stream functions also exist for 2D planar flows
Cartesian:
Cylindrical:
Potential Flow: Plane Potential Flows
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For irrotational, planar flow:
Now substitute the stream function:
Then, Laplaces Equation
For plane, irrotational flow, we use either the potential or the stream function,
which both must satisfy Laplaces equations in two dimensions.
Lines of constant are streamlines:
Now, the change of from one point (x, y) to a nearby point (x + dx, y + dy):
Along lines of constant we have d = 0,
0
Potential Flow: Plane Potential Flows
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Lines of constant are called equipotential lines.
The equipotential lines are orthogonal to lines of constant , streamlineswhere they intersect.
The flow net consists of a family of streamlines and equipotential lines.
The combination of streamlines and equipotential lines are used to visualize a
graphical flow situation.
The velocity is inversely proportional
to the spacing between streamlines.
Velocity increases
along this streamline.
Velocity decreases
along this streamline.
Potential Flow: Uniform Flow
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The simplest plane potential flow is a uniform flow in which the streamlines
are all parallel to each other.
Consider a uniform flow in the x-direction:Integrate the two equations:
= Ux + f(y) + C
= f(x) + C
Matching the solution
C is an arbitrary constant, can be set to zero:
Now for the stream function solution:
Integrating the two equations similarto above.
Potential Flow: Uniform Flow
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For Uniform Flow in an Arbitrary direction,
Potential Flow: Source and Sink Flow
S Fl
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Source/Sink Flow is a purely radial flow.
Fluid is flowing radially from a line through
the origin perpendicular to the x-y plane.
Let m be the volume rate emanating from the line (perunit length.
Then, to satisfy mass conservation:
Since the flow is purely radial:
Now, the velocity potential can be obtained:
Integrate
0If m is positive, the flow is radially outward, source flow.
If m is negative, the flow is radially inward, sink flow.
m is the strength of the source or sink!
This potential flow does not exist at r = 0, the origin, because it is not a real flow, but can
approximate flows.
Source Flow:
Potential Flow: Source and Sink Flow
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0
Now, obtain the stream function for the flow:
Then, integrate to obtain the solution:
The streamlines are radial lines and the equipotentiallines are concentric circles centered about the origin:
lines
lines
Potential Flow: Vortex Flow
In vortex flow the streamlines are concentric circles and the equipotential
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In vortex flow the streamlines are concentric circles, and the equipotential
lines are radial lines.
where K is a constant.
Solution:
The sign of K determines whether the flow rotates
clockwise or counterclockwise.
In this case, ,
The tangential velocity varies inversely with the distance from the origin. At the
origin it encounters a singularity becoming infinite.
lines
lines
Potential Flow: Vortex FlowHow can a vortex flow be irrotational?
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How can a vortex flow be irrotational?
Rotation refers to the orientation of a fluid element and not the path
followed by the element.
Irrotational Flow: Free VortexRotational Flow: Forced Vortex
Traveling from A to B, consider two sticks
Initially, sticks aligned, one in the flow direction, and the
other perpendicular to the flow.
As they move from A to B the perpendicular-aligned
stick rotates clockwise, while the flow-aligned stick
rotates counter clockwise.
The average angular velocities cancel each other, thus, the
flow is irrotational.
Irrotational Flow:
Velocity
increases
inward.
Velocity
increases
outward.
Rotational Flow: Rigid Body RotationInitially, sticks aligned, one in the flow
direction, and the other perpendicular to theflow.
As they move from A to B they sticks move
in a rigid body motion, and thus the flow is
rotational.
i.e., water
draining from
a bathtub
i.e., a rotatingtank filled with
fluid
Potential Flow: Vortex Flow
A bi d t fl i i hi h th i f d t t th d
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A combined vortex flow is one in which there is a forced vortex at the core, and
a free vortex outside the core.
A Hurricane is
approximately a
combined vortex
Circulation is a quantity associated with vortex flow. It is defined as the lineintegral of the tangential component of the velocity taken around a closed
curve in the flow field.
For irrotational flow the
circulation is generally
zero.
Potential Flow: Vortex Flow
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However, if there are singularities in the flow, the circulation is not zero if the
closed curve includes the singularity.
For the free vortex:
The circulation is non-zero and constant for the free vortex:
The velocity potential and the stream function can be rewritten in terms of the
circulation:
An example in which the closed surface circulation will be zero:
Beaker Vortex:
Potential Flow: Doublet FlowCombination of a Equal Source and Sink Pair:
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Rearrange and take tangent,
Note, the following:
Substituting the above expressions,
and
Then,
If a is small, then tangent of angle is approximated by the angle:
Potential Flow: Doublet Flow
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Now, we obtain the doublet flow by letting the source and sink approach one
another, and letting the strength increase.
K is the strength of the doublet, and is
equal to ma/is then constant.
The corresponding velocity potential then is the following:
Streamlines of a Doublet:
Potential Flow: Summary of Basic Flows
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Potential Flow: Superposition of Basic Flows
Beca se Potential Flo s are go erned b linear partial differential eq ations
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Because Potential Flows are governed by linear partial differential equations,
the solutions can be combined in superposition.
Any streamline in an inviscid flow acts as solid boundary, such that there is no
flow through the boundary or streamline.
Thus, some of the basic velocity potentials or stream functions can be
combined to yield a streamline that represents a particular body shape.
The superposition representing a body can lead to describing the flow aroundthe body in detail.
Superposition of Potential Flows: Rankine Half-Body
The Rankine Half-Body is a combination of a source and a uniform flow.
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The Rankine Half Body is a combination of a source and a uniform flow.
Stream Function (cylindrical coordinates):
Potential Function (cylindrical coordinates):
There will be a stagnation point, somewhere along the negative x-axis wherethe source and uniform flow cancel (
For the source: For the uniform flow:
Evaluate the radial velocity:
cosUvr
For Uvr
Then for a stagnation point, at some r = -b, = :
2
mvr and
Superposition of Potential Flows: Rankine Half-Body
Now, the stagnation streamline can be defined by evaluating at r = b, and
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Now, the stagnation streamline can be defined by evaluating at r b, and = .
Now, we note that m/2 = bU, so following this constant streamline givesthe outline of the body:
Then, describes the half-body outline.
So, the source and uniform can be used to describe an aerodynamic body.
The other streamlines can be obtained by setting constant and plotting:
Half-Body:
Superposition of Potential Flows: Rankine Half-Body
The width of the half-body:
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The width of the half-body:
Total width then,
The magnitude of the velocity at any point in the flow:
Noting,
and
Knowing, the velocity we can now determine the pressure field using the BernoulliEquation:
Po and U are at a point far away from the body and are known.
Superposition of Potential Flows: Rankine Half-Body
Notes on this type of flow:
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Notes on this type of flow:
Provides useful information about the flow in the front part of streamlined body.
A practical example is a bridge pier or a strut placed in a uniform stream
In a potential flow the tangent velocity is not zero at a boundary, it slips The flow slips due to a lack of viscosity (an approximation result).
At the boundary, the flow is not properly represented for a real flow.
Outside the boundary layer, the flow is a reasonable representation.
The pressure at the boundary is reasonably approximated with potential flow.
The boundary layer is to thin to cause much pressure variation.
Superposition of Potential Flows: Rankine Oval
Rankine Ovals are the combination a source, a sink and a uniform flow,
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producing a closed body.
Some equations describing the flow: The body half-length
The body half-width
Iterative
Potential and Stream Function
Superposition of Potential Flows: Rankine Oval
Notes on this type of flow:
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yp
Provides useful information about the flow about a streamlined body.
At the boundary, the flow is not properly represented for a real flow.
Outside the boundary layer, the flow is a reasonable representation. The pressure at the boundary is reasonably approximated with potential flow.
Only the pressure on the front of the body is accurate though.
Pressure outside the boundary is reasonably approximated.
Superposit ion of Potential Flows: Flow Around a Circular Cylinder
Combines a uniform flow and a doublet flow:
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and
Then require that the stream function is constant for r = a, where a is the
radius of the circular cylinder:
K = Ua2
Then, and
Then the velocity components:
Superposit ion of Potential Flows: Flow Around a Circular Cylinder
At the surface of the cylinder (r = a):
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y ( )
The maximum velocity occurs at the top and bottom of the cylinder,
magnitude of 2U.
Superposit ion of Potential Flows: Flow Around a Circular Cylinder
Pressure distribution on a circular cylinder found with the Bernoulli equation
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Then substituting for the surface velocity:
Theoretical and experimental agree
well on the front of the cylinder.
Flow separation on the back-half in the
real flow due to viscous effects causes
differences between the theory and
experiment.
Superposit ion of Potential Flows: Flow Around a Circular Cylinder
The resultant force per unit force acting on the cylinder can be determined
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by integrating the pressure over the surface (equate to lift and drag).
(Drag)
(Lift)
Substituting,
Evaluating the integrals:
Both drag and lift are predicted to be zero on fixed cylinder in a uniform flow?
Mathematically, this makes sense since the pressure distribution is
symmetric about cylinder, ahowever, in practice/experiment we see
substantial drag on a circular cylinder (dAlemberts Paradox, 1717-
1783).Viscosity in real flows is the Culprit Again!
Jean le Rond
dAlembert
(1717-1783)