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AE 2020Chapter II
L. Sankar
School of Aerospace Engineering
Copyright L. N. Sankar 2008 2
Preliminary Remarks
• In aerodynamics, or fluid mechanics, there are six properties of the flow an engineer is usually interested in:– pressure p (lbf / ft2 or N/ m2)– density (slug/ft3 or kg/m3) – three velocity components u,v,w (ft/sec or m/sec) – temperature T (degrees K or degrees R)
• We need six equations for these six unknowns.• One of them is algebraic.
– This is the equation of state: p = R T• The other 5 equations are partial differential equations
(PDEs)– PDEs are equations that contain partial derivatives of properties
of interest with respect to x, y, z or time.
Copyright L. N. Sankar 2008 3
Lagrangean vs. Eulerian• These equations may be derived using a Lagrangean approach, or
an Eulerian approach. • In the Lagrangean approach, we follow a fixed set of fluid particles
(e.g. a cloud, a tornado, tip vortices from an aircraft) and write down equations governing their motion. – This is somewhat like tracking satellites and missiles in space, using
equations to describe their position in space and the forces acting on them.
• In the Eulerian approach, we look at a (usually) fixed or (sometimes) moving volume in space surrounded by permeable boundaries. – We develop equations describing what happens to the fluid inside the
control volume as new fluid enters and old fluid particles leave. – Eulerian approach is often the preferred approach in most fluid
dynamics applications. – This is what we will follow in our derivations.
Copyright L. N. Sankar 2008 4
Examples of Lagrangean Approach
• Helicopter wake is sometimes tracked in space and time.
– This is done to determine where the wake goes, how strong are the vortices (spinning fluid particles), etc.
• Oceanographers track surface currents, oil spills, algae, etc. and associated velocity and temperatures.
• Weather reporters track migration of clouds, rain, snow, etc over a period of time.
• In football terms, this is like the man-to-man defense.
– The defender follows and tracks where the receiver is, how fast he/she is moving, etc.
Y
Z
X
Y
Z
X
Initially Prescribed Wakes
Wakes after 4 Revolutions
Y
Z
X
Y
Z
X
Initially Prescribed Wakes
Wakes after 4 Revolutions
Copyright L. N. Sankar 2008 5
Examples of Eulerian Approach
• A complex fluid domain inside a duct is computed.
• The flow domain is broken into smaller volumes (called control volume) that remain fixed in space or moves at a constant velocity.
• The properties of the fluid as they enter and leave the domain are tracked.
• In football terms, this is like zonal defense.
Copyright L. N. Sankar 2008 6
Governing Equations• Equation of State: p = RT• We need to come up five additional equations linking the 6
properties. • These five equations are PDEs and turn out to be:
– Conservation of Mass or Continuity– Conservation of u- momentum– Conservation of v-momentum– Conservation of w-momentum– Conservation of energy
• If we are dealing with low speed incompressible flows, we can drop one of these equations (equation of state).
• If we are not interested in temperature distribution, heat transfer etc., we can drop another equation (energy equation).
• We are left with 4 equations (conservation of mass, conservation of u, v, and w momentum) and four unknowns (p, u , v, w)
Copyright L. N. Sankar 2008 7
Conservation of Mass(Also known as Continuity Equation)• The conservation of mass stems
from the principle that mass can not be created or destroyed inside the control volume.
• Obviously, we have situations (e.g. nuclear reactions) involving the conversion of mass into energy.
• In steady flows (i.e. flows where properties do not change with time), what goes in must come out.
• Otherwise there will be an accumulation of mass within the control volume and properties will change with time.
Copyright L. N. Sankar 2008 8
Conservation of Mass (Continued)
• Let V be a control volume, a balloon like shape in space.
– The use symbol V for volume and velocity may be confusing. Sorry!
• We will assume that the control volume V, and its surface S, remain fixed in space.
• The surface is permeable so that fluid can freely enter in and leave.
• The continuity equation says: The time rate of change of Mass within the control volume V = Rate at which mass enters V through the boundary S
• Analogy: If your bank balance is increasing at $1000 per month (you wish), you are earning $1000 more per month than you are spending.
– If your bank balance is steady, then income equals expenditure.`
V
Surface S
Copyright L. N. Sankar 2008 9
Conservation of Mass (Continued)
• We can assume that the control volume V is made of several infinitely small (infinitesimal) volume elements dV.
• The mass of the fluid inside each of these elements is dV, where is the fluid density.
• The density is free to change from point to point, from one sub-element dV to another within V. Thus,
Total mass within V = dVV
dV
Time rate of change of mass within the control volume V = d
dtdV
V
Copyright L. N. Sankar 2008 10
Conservation of Mass (Continued)• We next look at the rate at which mass is
entering and leaving the control volume V through the surface S.
• For this purpose, we assume that the surface S is made of many quilt-like infinitesimal patches dS.
– Imagine a patch on your Jeans or an elbow patch on some fashionable (?)jackets.
• At the center of each patch is a unit normal vector (i.e. a vector of length unity, normal to the surface) pointing away from the surface dS.
– Imagine needles on a cactus.• The normal component of fluid velocity
pointing towards the control volume (entering the control volume) is the negative of the dot product of the velocity vector and the normal vector.
• Notice the negative sign. We are interested in the component of velocity pointing towards the control volume, not away from it.
V n
V
n
dSnV- =
dS through voumecontrol theenters massat which Rate
S
dSnV- =S surface entire ethrough th
volumecontrol theenters massat which Rate
Copyright L. N. Sankar 2008 11
Conservation of Mass (Concluded)
The time rate of change of Mass within the control volume V = Rate at which mass enters V through the boundary S
t
dV V ndSSV
t
dV V ndSSV
0
Volume integral Surface integral
Copyright L. N. Sankar 2008 12
Simple Application of Continuity
n
n
222111 UAUA
Copyright L. N. Sankar 2008 13
Conservation of u- Momentum
• The u- momentum equation is an extension of Newton’s law:– the rate of change of momentum of a particle (i.e. a system
with fixed mass) is equal to the force acting on the particle.• Newton was thinking of fixed mass particles (apples..) • In our work, we are dealing with open systems, i.e.
control volumes. • The mass may change with time as a result of fluid
entering or leaving the control volume. • We therefore generalize Newton’s law.
– The rate of change of u-momentum of particles within a control volume = Forces along the x-Direction acting on the control volume + Net rate at which u-momentum enters the control volume through the boundaries.
Copyright L. N. Sankar 2008 14
Conservation of u- Momentum(Continued)
dV
V
V
V
dVt
u
udV
udV
= V within momentum -u of change of Rate
volume,control fixed afor operationsation differenti andn integratio theingInterchang
dt
d = V within momentum -u of Change of Rate
= V volumecontrol entire e within thMomentum-u
udV = dV within Momentum -u
dV = dV within Mass
Copyright L. N. Sankar 2008 15
Body Forces acting on the fluid within our control volume V
• The forces may be divided into two broad categories – body forces and surface forces.
• Body Forces forces acting on every fluid particle within V. – Examples of body forces include gravity, electrostatic forces,
and magnetic forces. – Let us assume the symbol ax represents the x- component of all
the acceleration due to these effects acting on the fluid particles within V.
– The quantity ax may vary with x, y, z and t.
V
=V wthin paticles allon actingdirection - x thealong forceBody
dV = dV wthin paticles allon actingdirection - x thealong forceBody
dVa
a
x
x
Copyright L. N. Sankar 2008 16
Surface Forces acting on the fluid within our control volume V
• At the surface of the control volume there are surface forces acting on the fluid within V.
• These forces may be divided into two categories: pressure forces and viscous forces.
• We will consider each of these separately.
Copyright L. N. Sankar 2008 17
Surface Pressure Forces• The control volume V is surrounded by the surface S. • Pressure from surrounding fluid (or solid) acts on this surface S.
Pressure forces are always directed towards the fluid within, and is always normal to the surface.
• To compute this quantity, we divide the surface S into patches dS• On each patch, the pressure force is pdS• It acts towards the fluid within V.
The pressure forces acting on S may therefore be written as pndSS
. Notice the
negative sign. It is there because the normal vector n is pointing outwards, whereas the
pressure forces are acting inwards. The x- component of these pressure forces is found by
performing a dot product of this expression with i .
X component of Surface Forces acting on V= pn i dS
Copyright L. N. Sankar 2008 18
Surface Viscous Forces• The surrounding fluid can exert an
additional type of force on the surface S, called a “viscous” force.
• This force will have a component normal to the surface S, and a component tangential to the surface, both.
– Put some glue on your finger tip. – Touch a piece of paper.– If you move it tangentially on the
paper, you are exerting a viscous shear force.
– If you try to lift it up vertically to the plane of the paper, you are exerting a normal viscous force.
• We will study the effects of viscosity in more detail later.
• For now, we will call this contribution exerted by surroundings on the fluid within our control volume as Fx-Viscous.
Viscous forceMay be tangential
It may be normalTo the surface
Copyright L. N. Sankar 2008 19
Rate at which u-Momentum enters the Control Volume V
• We can finally turn our attention to the rate at which the u- momentum is brought into V though the surface S.
• Let dS be an infinitesimal element (or patch) on S. • Then rate at which u-momentum enters the control
volume through dS is simply the rate at which mass enters the control volume through dS times the u-component of velocity.
• Summing up contributions over the entire surface S, we get:
Rate at which u - momentum enters V through S = -S
uV ndS
Copyright L. N. Sankar 2008 20
Assemble it all together..
V S V
ViscousxBodyx FdVadSnVuipdVt
u
Rate of changeof u-momentumwithin the controlVolume V
This term representsPressure force along theX-direction exerted byThe surrounding on the fluid within V
This term represents the rate at which u- momentum enters through the boundaries
Body forcesAlong x
ViscousForces along X-direction
Copyright L. N. Sankar 2008 21
Conservation of v- and w- Momentum
• We can write down v- and w- momentum equations similarly.
V S V
ViscousyBodyy FdVadSnVjpdVt
vv
V S V
ViscouszBodyz FdVadSnVwkpdVt
w
Copyright L. N. Sankar 2008 22
Summary: We got the four equations we asked for.
V S V
ViscousxBodyx FdVadSnVuipdVt
u
V S V
ViscousyBodyy FdVadSnVjpdVt
vv
t
dV V ndSSV
0
V S V
ViscouszBodyz FdVadSnVwkpdVt
w
Moral of the story: Be careful what you ask for..
Copyright L. N. Sankar 2008 23
Some Vector CalculusDel Operator in Cartesian coordinates:
=x y z
i j k
It can operate on scalar functions. The result is called the gradient of that function.
e.g. Let F(x,y,z) = 27x2 + 48xy + 92 xyz
kxyjxzx
kj
929248
i92yz48y54x
z
F
y
F
x
FiF F ofGradient
Copyright L. N. Sankar 2008 24
Vector Calculus (Continued)Since “del” operator is a vector, it may be applied on a vector function.
One can perform either a dot product operation or a cross product operation.
ky
F
x
Fj
x
F
z
F
z
F
y
F
FFFzyx
kj
kFjFiFF
1231
23
321
321
321
i
i
=F=F of Curl
z
F
y
F
x
F=F=F of Divergence
Then,
Let
Copyright L. N. Sankar 2008 25
Examples
447
5472
5472
100106
20103
20103x z)y,(x,FLet
zyxxzx
zyxz
xyzy
xx
F
kzyxjxyzi
jyzixz
xyzxyyzyx
kji
G
kxyzjxyiy
66
622
622
2x
2xz)y,(x,GLet
Copyright L. N. Sankar 2008 26
Polar Coordinate System
x
y
r
P
ere
The coordinates of a point Pare described by the radialdistance from the origin “r”and the angle with respectto the x- axis.
x = r cos y = r sin
e isr unit vecto the, Along
e isr unit vecto ther, Along r
Copyright L. N. Sankar 2008 27
Del Operator in Polar System
r
er
er
1
k
A
r
rB
rrBA
r
kere
rF
B
rrA
rrF
ee
r
r
1
0
01
11
BA FLet
Copyright L. N. Sankar 2008 28
Cylindrical Co-Ordinate System
• In a cylindrical coordinate system, a point P is defined by the z ordinate (the the vertical distance between the point and the x-y plane), the radius r (the distance between the origin and the projection of P onto the x-y plane), and the angle that the projection makes with x-axis.
P
r
z
x
z
X = r cosy = r sin
Copyright L. N. Sankar 2008 29
Unit Vectors in Cylindrical Coordinates
ordinates-coCartesian in as k isr unit vecto thez, Along
e isr unit vecto the, Along
e isr unit vecto ther, Along
.directions three
thealong rsunit vecto definefirst We
r
r
er
e
x
y
In the picture on the right side, the z-axis is Perpendicular to the plane of the slide.
Copyright L. N. Sankar 2008 30
Del, Curl, and Gradient inCylindrical Co-Ordinates
zk
re
rer
1
CrBAzr
kere
rF
z
CB
rrA
rrF
kCee
r
r
1
11
BA FLet
Copyright L. N. Sankar 2008 31
Divergence Theorem• This is a theorem from vector calculus.• We take it and use it as given, without proof. • Let F be any three-dimensional vector, and is a
general function of (x, y, z, t). • Then, divergence theorem applied to a control
volume V surrounded by a surface S states:
F ndS FdV
VS
Volume integral as beforeSurface integral
As before
Copyright L. N. Sankar 2008 32
Application of Divergence Theoremto Conservation of Mass
t
dV V ndSSV
0
F ndS FdV
VS
Start withConservation of Mass:
Apply Divergence theorem:
t
dV V dVVV
0We get
Or
t
V dVV
0
Copyright L. N. Sankar 2008 33
Application of Divergence Theoremto Conservation of Mass
(Continued)
From the previous slide:
t
V dVV
0
Consider the above volume integral. It must hold for any arbitrarily shaped control volume V, at any instance in time for all flows. The only way this can be true is if the integrand is zero. Therefore,
t
V
0
The above equation is called the PDE form of the continuity equation.
Copyright L. N. Sankar 2008 34
Application of Divergence Theoremto Conservation of Mass (Continued)
• For steady flows, the time derivative vanishes. • For incompressible flows, is a constant. Thus, continuity
equation for incompressible flows becomes:
0
,
0
v
zw
yxu
or
V
Copyright L. N. Sankar 2008 35
Application of Divergence Theorem to u-Momentum Equation
(Neglect viscous forces)
V S S V
xdVadSnipdSnVudVt
u Start with
Use S V
dVAdSnA
S V
S V
dVipdSnip
dVVudSnVu
We get
Throw this into u- momentum equation:
V
x dVaipVut
u0
Copyright L. N. Sankar 2008 36
Application of Divergence Theorem to u-Momentum Equation (Continued)
x
pip
zk
yj
xiip
uwz
uy
ux
kuwjuiuz
ky
jx
iVu
kuwjuiukwjiuuVu
vv
vv
22
2
Finally.. xax
puw
zu
yu
xu
t
v2
Set integrand to zero. This is the only way to ensure that the integral fromat the bottom of the previous slide will hold for all arbitrary control volumes.
Copyright L. N. Sankar 2008 37
SummaryInviscid form
xax
puw
zu
yu
xu
t
v2
We can similarly convert the v- and w- momentum equations into PDEs.
v- Momentum equation in PDE form: yay
pw
zyu
xt
vvvv 2
w- Momentum Equation in PDE form: zaz
pw
zw
yuw
xw
t
2v
u- Momentum equation in PDE form:
Continuity equation in PDE form:
0v
z
w
yx
u
t
Copyright L. N. Sankar 2008 38
Simplification of u- Momentum Equation
xax
puw
zuv
yu
xu
t
2
Differentiate by parts:
z
wu
z
uw
z
uw
y
vu
y
uv
y
uvx
uu
x
uu
x
uu
x
u
tu
t
uu
t
2
Copyright L. N. Sankar 2008 39
Simplification of u- Momentum Equation (Continued)
Using these expansions in the previous slide in the u-momentum equation:
xa
z
w
y
v
x
u
tu
x
p
z
uw
y
uv
x
uu
t
u
Continuity equation says this is zero
xax
p
z
uw
y
uv
x
uu
t
u
Result:
Copyright L. N. Sankar 2008 40
Simplified Forms
yay
p
zw
yxu
t
vv
vvv
xax
p
z
uw
y
u
x
uu
t
u
vU- Momentum:
V- Momentum:
w- Momentum: `za
z
p
z
ww
y
w
x
wu
t
w
v
Caution: These equations are for inviscid flows. We will need to add viscousEffects later.
Copyright L. N. Sankar 2008 41
Simplified Forms
zw
yv
xu
tDt
D
where
az
p
Dt
Dw
ay
p
Dt
Dv
ax
p
Dt
Du
z
y
x
,
Copyright L. N. Sankar 2008 42
Physical Significance of D/Dt
The operator D/Dt is called the "substantial derivative" or "material derivative".
A(x,y,z,t)
B(x+x,y+y,z+z,t+t)
Consider a fluid particle at a point A in space, given by the coordinates (x,y,z).
A short time later this particle has moved to a new location B given by the coordinates (x+x, y+y, z+z)
Du/Dt must be thought of as (u at point B – u at point A)/t
Copyright L. N. Sankar 2008 43
Physical Significance of D/Dt(Continued)
wz
uv
y
uu
x
u
t
u
t
uu
Dt
Du
Thus
twz
uv
y
uu
x
u
t
u
tt
z
z
u
t
y
y
u
t
x
x
u
t
u
zz
uy
y
ux
x
ut
t
u
tzyxuttzzyyxxuuu
AB
t
AB
Limit0
,
),,,(),),,(
Copyright L. N. Sankar 2008 44
Streamlines
• Streamlines are defined as curves in space, that are drawn so that at every point on that curve the instantaneous velocity vector is tangential to the curve.
Copyright L. N. Sankar 2008 45
Equation for a Streamline
w
dz
v
dy
u
dx
0 vdx -udy 0 udz - wdx 0vdx-udy
zero. bemust vector thisofcomponent Each
0kvdx)-(udy judz-wdxi vdx)-(udy
0
dzdydx
wvu
kji
0 sdV
direction. same in the
pointingboth ,sd tol tangentiabemust vector
velocity the,streamline theof definitionby Then,
vector. velocity thebe k w j v iu VLet
k dz jdy idx sd
points. end two theseconnectsat vector tha definecan We
dz)z dy,y dx,(x and z)y,(x, are points end twoThe
.streamline a alongsegment line small a be dsLet
(x,y,z)
(x+dx,y+dy,z+dz)
w
dz
v
dy
u
dx
Copyright L. N. Sankar 2008 46
Pathlines
• Pathlines are defined as the path along which fluid partilces travel.
• In unsteady flow, different particles that start at the same spatial location may travel along different paths.
• This is like cars entering a highway.– They may enter at the
same entrance ramp.– Their subsequent trajectory
(or path) differs from one vehicle to the next.
Copyright L. N. Sankar 2008 47
Equation for a PathlineWe integrate the velocity to find the path
t)z,y,w(x,dt
dz
t)z,y,v(x,dt
dy
t)z,y,u(x,dt
dx
This is what astronomers and scientists do to track stars, space shuttle, and satellites
Copyright L. N. Sankar 2008 48
Streamlines are instantaneous images Pathlines are time-elapsed images
Copyright L. N. Sankar 2008 49
Steady Flows
• In steady flows, properties will not change from one time instance to the next.
• Particles that start from the same starting locations will follow the same path, as new groups of particles are released from the same starting locations.– This is like cars traveling on their assigned lanes, without
crisscrossing or changing lanes.– The lanes become pathlines.– The velocity of the cars will be along the lanes , i.e. tangential to
the dotted lines that describe the boundaries of the lanes.
• In steady flows, streamlines and pathlines are one and the same.
Copyright L. N. Sankar 2008 50
Streamtubes
• Stream tubes are streamlines that start from a closed contour.
• Think of it as a bundle of streamlines that form a tube-like shape.
Copyright L. N. Sankar 2008 51
Angular Velocity of the Fluid
• Like solid particles that can spin or rotate, fluid elements (a collection of particles that are closely packed) may spin also.
• The angular velocity is a vector.– This is because the fluid element may spin about the
x-axis, y-axis, and z-axis simultaneously.– Kind of like Tom Glavin’s curve ball on a good day.
• Vorticity is twice the angular velocity.• Vortcity is a vector, since angular velocity is a
vector.
Copyright L. N. Sankar 2008 52
Vortex is a just collection of spinning fluid elements
Copyright L. N. Sankar 2008 53
In the case of solids, we can define the angular velocity by drawing a line on the solid and
watching how that line moves as the solid
rotates.
Not so with fluid elementsthat not only rotate, but alsoUndergo deformation with time.
Think of a jello (a very viscousAnd dense fluid) as you throw itAcross the room.
The different faces of the jellomay rotate at different velocities
Or think of a smoke ring, which startsdeforming quickly when it encountersTurbulent air.
Fluid-dynamicists thereforemeasure the angular velocity of twoPerpendicular lines and average them
Copyright L. N. Sankar 2008 54
Smoke Ring
Copyright L. N. Sankar 2008 55
Angular Velocity of a Fluid ElementShown in 2-D for simplicity
A B
C D
A’B’
C’D’
We measure or compute the angular velocityof the face AB and that of face AC and take average.
(x,y) (x+dx,y)
(x,y+dy)
Copyright L. N. Sankar 2008 56
Angular Rotation of Face ABover a small instance in time dt
A B
A’
B’
vA times dt vB times dt
(vA – vB) times dt
The rotation of the line AB was caused because point B moved faster in the y-direction compared to point A, over this interval dt.
Copyright L. N. Sankar 2008 57
Angular Rotation of Face ABover a small instance in time dt
A B
A’
B’
To find out how much AB rotated, we compare the initial and finalorientation of the line AB. To do this, we bring A’B’ to AB and see how muchWas the rotation.
(vA – vB) times dt
dx
d
The angle by which the face AB rotated is d
Copyright L. N. Sankar 2008 58
Angular velocity of Face AB
x
v
dx
vv
dx
dtvv
dx
dtvvtan
AB
ABAB1
dt
d
d
A B
A’
B’
(vA – vB) times dt
dx
d
Copyright L. N. Sankar 2008 59
We next look at the face AC
A B
C D
A’B’
C’D’
(x,y) (x+dx,y)
(x,y+dy)
Copyright L. N. Sankar 2008 60
Angular Velocity of Face AC
A
C
A’
C’
u at A times dt
u at C times dt (uC-uA)dt
dy Angle is approximately(uC-uA) dt dived by dy
Angluar velocity is (uC-uA) /dyAs dy goes to zero, this is ∂u/ ∂y
The face AC rotates clockwise at an angular velocity of ∂u/ ∂y
Copyright L. N. Sankar 2008 61
Take average of angular velocity of faces AB and AC
A B
C D
A’B’
C’D’
(x,y) (x+dx,y)
(x,y+dy)
Angular velocity of this element about the Z-axis (perpendicular to the plane of the paper) is ½ (∂v/ ∂x - ∂u/ ∂y) if we take the sign of rotation of the two faces into consideration.
Copyright L. N. Sankar 2008 62
In 3-D, Angular velocity by a similar logic is, then..
V2
1
jx
w
z
ui
z
v
y
wk
y
u
x
v
2
1
Vorticity is twice the angular velocity.
Copyright L. N. Sankar 2008 63
Regions of High Vorticity
• In aerodynamics, we find high levels of voriticty if the fluid is moving at different velocities relative to each other.– One example is a jet. The particles inside a jet
move faster than those outside. The fluid elements spin.
– Another example is boundary layer, a thin viscous region close to the solid surface. The aprticles close to the surface move slowly, while particles above move more rapidly.
Copyright L. N. Sankar 2008 64
Jets have a lot of vorticity, especially near the edges of the jets
These particles are spinning counter-clockwise
These particles are in the clockwise direction
Copyright L. N. Sankar 2008 65
Boundary Layers have Vorticity as well
Which way will the particles spin? Clockwise or counter-clockwise?
Copyright L. N. Sankar 2008 66
Boundary layer over an Airfoil
Copyright L. N. Sankar 2008 67
Wake Behind a Bluff Body (Truck)
http://www.eng.fsu.edu/~shih/succeed/cylinder/vorvec.gif
Copyright L. N. Sankar 2008 68
Rotational Flow
• A flow in which there is a lot of vorticity is called a rotational flow.
• In rotational flows, the fluid elements will rotate as they move from upstream to downstream. – This is like a bowling ball rolling along a
bowling lane.
Copyright L. N. Sankar 2008 69
Irrotational Flow
• An irrotational flow is a flow in which the voriticity is zero.
• Many practical flows (e.g. regions outside the boundary layer over an airfoil) do not have significant angular velocity or vorticity (which is twice the angular velocity).– These regions outside the thin viscous region
may be approximated as irrotational flows.
Copyright L. N. Sankar 2008 70
Irrotational Flow
A B
C D
A B
C D
A B
C D A B
C D
In irrotational flow, the fluid elements do not spin about, but maintain their upright orientation.
Copyright L. N. Sankar 2008 71
Potential Flow
• : A irrotational flow (that is, a flow in which vorticity is zero) is also called a potential flow.
• This is because we can define a function called the velocity potential such that
zw
yv
xu
Or
V
,
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Potential Flow
0222222
xyyx
kzxxz
jyzzy
i
zyx
zyx
kji
V
0
,
0
2
2
2
2
2
2
zyxzk
xj
xi
zk
xj
xi
Or
V
Continuity becomes:
Copyright L. N. Sankar 2008 73
Potential Flow
02
The operator 2 is called the Laplacian operator. It is simply three second derivatives added together as shown below:
2
2
2
2
2
22
zyx
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Why did we define ?
• Why did we introduce a new variable called ?
• It is because we would rather solve a single linear PDE for than solve 4 nonlinear PDEs– -conservation of mass, – Conservation of u- v-, w-
momentum
• If we can somehow solve Laplace’s equation, we can find u, v and w.
• Finally, we can find p from the Bernoulli equation.
zw
yv
xu
Or
V
,
02
Copyright L. N. Sankar 2008 75
Derivation of Bernoulli’s Equation
dxadxx
pdx
z
uwdx
y
udx
x
uu x
v
xax
p
z
uw
y
u
x
uu
t
u
v
We start with inviscid u-momentum equation:
We assume steady flow. Multiply the above equation by dx
Along a streamline (see slide 45):
udzwdx
udyvdxw
dz
v
dy
u
dx
Copyright L. N. Sankar 2008 76
Derivation of Bernoulli’s Equation
dxadxx
pdz
z
uudy
y
udx
x
uu x
u
dxadxx
pdz
u
zdy
u
ydx
u
x x
222
222
We get:
Or
From calculus, for any function F(x,y,z):
dzz
Fdy
y
Fdx
x
FzyxFdzzdyydxxFdF
),,(),,(
Copyright L. N. Sankar 2008 77
Derivation of Bernoulli’s Equation
dxadxx
pud x
2
2
dyadyy
pd y
2
v2
Thus u-momentum equation in the previous slide becomes:
If we multiply the v-momentum equation by dy, and use stream function, we get
If we multiply the w-momentum equation by dz, and use stream function, we get
dzadzz
pd z
2
w 2
Copyright L. N. Sankar 2008 78
Derivation of Bernoulli’s Equation
dzadyadxadyx
pdy
y
pdx
x
pd zzx
2
wvu 222
dzadyadxadpd zzx
2
wvu 222Add:
Or:
If we only have gravity as the body force, then ax =0 , ay = 0, az = -g
Integrate:
Cgzpwvu
2
222
Copyright L. N. Sankar 2008 79
Bernoulli’s Equation
Cgzpwvu
2
222
Kinetic energy
Pressure energy
Potential energy
Since we used equation for a streamline in deriving this, this equation holds Along a streamline. We also neglected viscosity, and assumed steady flow.If all the streamlines start with the same upstream pressure and velocity, thenThey will differ only in their starting z- position as far as the total energy goes.
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Stream Function ψ
• Stream function ψ is also a useful variable.• Unlike velocity potential which applies for 3-D
flows, stream function is defined only for 2-D flows and axi-symmetric flows.
• In 2-D flows, the stream function is defined as:
xv
yu
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Stream Function satisfies continuity
• Recall the continuity equation for incompressible flows:
0
y
v
x
u
If we plug-in xy
u
v
We notice that we satisfy continuity automatically.
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Stream Function ψ, when combined with irrotationality, gives a Laplace’s
equation
0uv
yxIn irrotational flows, the vorticity is zero. In 2-D,
If we plug in xy
u
v
We get: 022
2
2
2
yx