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AERO 301: Spring 2011 II(a): Euler Eqn. & ~ω = 0 Page 1
PART II: 2D Potential Flow
II(a): Euler’s Equation & Irrotational Flow
We have now completed our tour through the fundamental conservation laws that apply to
fluid mechanics in general, and are now about to embark on problems specific to incom-
pressible aerodynamics.
We start by focusing on 2D problems, and work our way toward calculating L′ and M′ for
airfoils.
Recall that incompressible aerodynamics is relevant when M → 0. This allows us to considerρ to be constant. In addition, we are also concerned with the case when Re → ∞. This elim-inates viscosity from the governing equations, which is a dramatic simplification, although
the remaining equations are still extraordinarily complex.
With these assumptions our governing equations are
AERO 301: Spring 2011 II(a): Euler Eqn. & ~ω = 0 Page 2
∇ ·~v =∂u∂x
+∂v∂y
+∂w∂ z
= 0
ρD~vDt
= ρ(
∂~v∂ t
+u∂~v∂x
+ v∂~v∂y
+w∂~v∂ z
)
= −∇p+ρ~g
p+12
ρ V 2 = const along a streamline
Our first task is to develop some means of dealing with the nonlinear nature of the momen-
tum and Bernoulli’s equation. (As written they are near-impossible to solve by hand.)
AERO 301: Spring 2011 II(a): Euler Eqn. & ~ω = 0 Page 3
Vorticity
To do this, we invoke the concept of vorticity, ω, which is the tendency of fluid elements
to rotate. Take the ∂/∂x of the (2D for simplicity) v-momentum equation and subtract∂/∂y of the 2D u-momentum equation and using the continuity (mass) equation to eliminatesome terms
−∂∂y
[
ρ(
∂u∂ t
+u∂u∂x
+ v∂u∂y
)
+∂ p∂x
−ρgx
]
+∂∂x
[
ρ(
∂v∂ t
+u∂v∂x
+ v∂v∂y
)
+∂ p∂y
−ρgy
]
can be manipulated to give
ρ[
∂∂ t
(∂v∂x
−∂u∂y
)
+u∂∂x
(∂v∂x
−∂u∂y
)
+ v∂∂y
(∂v∂x
−∂u∂y
)]
= 0
where the vorticity ω is
AERO 301: Spring 2011 II(a): Euler Eqn. & ~ω = 0 Page 4
ω =∂v∂x
−∂u∂y
From this, we can immediately see that
DωDt
= 0
We could go through an equivalent procedure to the 2D development in 3D and we would
find that ~ω = ∇×~v and that D~ω/Dt = 0 in an inviscid fluid in 3D as well as 2D.
It turns out that ω = constis a very useful restriction. Why?
• At some initial time, we imagine that u = v = 0 everywhere and therefore ω = 0 ev-erywhere at that initial time and at all future times. So in a sense, our momentum
equations reduce to a statement that the vorticity everywhere is zero.
AERO 301: Spring 2011 II(a): Euler Eqn. & ~ω = 0 Page 5
• We have the same condition if we have a uniform upstream velocity, u =U∞, v = 0.
A fluid flow which ω = 0 everywhere is called irrotational.
Two examples (check whether ∇·~v = 0 and whether ∇×~v = 0):
• Solid Body Rotation: vr = 0, vθ =C r → u =−C y, v =C x
• Free (Bathtub) Vortex: vr = 0, vθ =C/r
AERO 301: Spring 2011 II(a): Euler Eqn. & ~ω = 0 Page 6
So, by working with our momentum equations, we are able to develop the idea of vorticity
and, if the initial (or perhaps upstream) vorticity is zero our governing equations become
∇ ·~v = 0 →∂u∂x
+∂v∂y
= 0 in 2D
∇×~v = 0 →∂v∂x
−∂u∂y
= 0 in 2D
p+12
ρ V 2 = const along a streamline
What is useful about this arrangement is that
1. The continuity and (angular) momentum equations are both linear and therefore solv-
able.
2. Continuity and momentum are decoupled from the pressure.
AERO 301: Spring 2011 II(a): Euler Eqn. & ~ω = 0 Page 7
So the solution procedure becomes
1. Solve for the velocity field using continuity and momentum.
2. Plug the velocities into Bernoulli’s equation to get the pressure field.
3. Integrate the pressure field on the body (wing, tail, elevator, whatever) to find the net
force and moment on the body.
To do this, we need to introduce just a few more concepts...
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 8
II(b): Scalar Velocity-Field Specifiers: ψ, φ and Γ
How do we actually solve the equations on the previous page?
We define two new scalar quantities:
Streamfunction: ψ
Velocity Potential: φ
Velocity Potential, φ
There is a vector identity in mathematics that says: “the curl of the gradient of a scalar
is zero”
∇× (∇φ) = 0
If the fluid is irrotational then ∇×~v = 0
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 9
Putting these ideas together, if a flow is irrotational then the velocity field can be ex-
pressed as the gradient of a scalar:
~v = ∇φ or
u = ∂φ/∂x
v = ∂φ/∂y
w = ∂φ/∂ z
We call φ the velocity potential
The attractiveness about this approach is that instead of needing to keep track of three
scalar velocity components, u, v, and w, we wrap them all up in a single scalar, φ .
Some examples to illustrate this concept:
Uniform Flow: u =U∞, v = 0
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 10
Stagnation-Point Flow: u = kx, v =−ky
Note that lines of constant φ are perpendicular to streamlines!
When the flowfield is irrotational, thereby allowing φ to be defined, then we need to solve onlyone equation (it would have been three for (u,v,w)). This is why in theoretical aerodynamics,there is such an importance placed on whether or not the flow is irrotational or rotational.
Because irrotational flows can be described by a velocity potential φ , then we call such flowspotential flows.
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 11
But writing ~v in terms of ∇φ is only useful if we can find an equation that we can solve forφ .
We have already used the momentum equation in defining φ — the fact that φ exists meansthat we conserve angular momentum.
However, we have not yet used continuity, ∇ ·~v = 0, so let us put our definition ~v = ∇φ intothat equation. We obtain
∇2φ =∂ 2φ∂x2 +
∂ 2φ∂y2 +
∂ 2φ∂ z2 = 0
This equation is the famous Laplace’s Equation, and it is the governing equation for
incompressible, irrotational flows.
Fortunately, it is only a linear, 2nd-order PDE so its solutions are well known. (In fact, its
probably the most-studied equation in all of mathematics.)
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 12
To find a unique solution to this differential equation we need boundary conditions. We have
physical boundary conditions on the velocities. . . so let us translate these to conditions on φ :
Condition Velocities Velocity Potential
Freestream u =U∞, v = 0
No Penetration
No Slip
The overarching philosophy is that once we have a PDE and boundary conditions we simply
superpose (add together) some simple flows (e.g., uniform flow, vortex flow etc.) that
individually satisfy the equation and together satisfy the boundary conditions to get much
more complicated (and relevant) solutions.
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 13
Streamfunction, ψ
Another approach, besides using φ , is to define a quantity ψ, called the streamfunction,that satisfies ∇ ·~v = 0 automatically in 2D:
u =∂ψ∂y
and v =−∂ψ∂x
Note that the velocities are formed with derivatives of the other direction and that there is
a minus sign associated with the v component.
Check continuity
Note that this approach can only work for 2D flows.
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 14
Examples:
Uniform Flow: u =U∞, v = 0
Stagnation-Point Flow: u = kx, v =−ky
To find the governing equation for ψ, substitute for u and v definitions in the (2D) momen-tum equation ∇×~v = 0:
∇2ψ =∂ 2ψ∂x2 +
∂ 2ψ∂y2 = 0
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 15
Unlike φ that can be defined in 3D, ψ only works in 2D. However, ψ has some other niceproperties:
• ψ can be used in rotational (therefore we can consider viscous) flows while φ is strictlyfor irrotational flows.
• Lines of constant ψ are streamlines
• The ∆ψ between two streamlines is equal to the volume flux per unit depth betweenthe streamlines (Huh?)
Again, boundary conditions for ψ:Condition Velocities Velocity Potential
Freestream u =U∞, v = 0
No Penetration
No Slip
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 16
Polar Coordinates
We have introduced ourselves to φ and ψ in Cartesian coordinates, but it is often easier touse polar coordinates.
In polar coordinates:
vr
vθ
vz
=
∂φ∂ r
1r
∂φ∂θ
∂φ∂ z
And 2D continuity is∂∂ r
(r vr) +∂vθ∂θ
= 0
so
(
vr
vθ
)
=
(1r
∂ψ∂θ
−∂ψ∂ r
)
These transformations will be useful when we consider vortices etc. later on.
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 17
Circulation, Γ
Everything we have used to describe velocity fields (~v, ω, φ , ψ) are local quantities— theydescribe conditions at a point.
Yet local definitions fall short and sometimes we need a global quantification.
For instance, in the case of the (bathtub) vortex we know that the vorticity is zero everywhere
except at the origin where it is infinite. Therefore, we only know the strength of the vortex
in terms of the constant, C, that is used in the vortex’s definition: vθ =C/r.
Because we will soon be dealing with more complicated flows, we need a more general
definition of overall vortex strength. (Vortex strength will soon be related to lift.)
Let us define circulation:
Γ =−∮
C~v· n̂ds
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 18
where ~v is the velocity and ds is a differential length on the boundary of a given contourregion C. The integral must be taken counterclockwise around a closed path.
The concept of circulation can be misleading. In aerodynamics, it simply means that the line
integral about C is finite—it does not necessarily mean that the fluid elements are movingin circles in the flowfield.
Examples:
Uniform Flow u =U∞, v = 0
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 19
Vortex Flow (where C includes the origin) vr = 0, vθ =C/r
Therefore, we usually redefine a vortex flow to be vθ =−Γ /2πr and call Γ the vortexstrength.
Vortex Flow (where C doesn’t include the origin)
Circulation also has an intimate connection with vorticity. From Stokes’ theorem we have
∮
C~v· n̂ds =
∫ ∫
S(∇×~v) · n̂dS
AERO 301: Spring 2011 II(b): φ , ψ and Γ Page 20
Hence, the (negative) circulation about a curve C is equal to the vorticity integrated overany open surface bounded by C.
Thus, if the flow is irrotational everywhere within the contour integral (∇×~v = 0) thenit follows that Γ = 0 everywhere.
For a vortex in particular, because the vorticity is zero everywhere except the origin, then
any region that does not include it has zero circulation.
Any region that does include the origin includes an infinitesimally small point with infinite
vorticity and 0×∞ = Γ .
AERO 301: Spring 2011 II(c): Elementary Flows Page 21
II(c): Elementary 2-D Flows and Superposition
Now that we know about φ , ψ, and Γ , how do we use them?
Our approach will be to find several simple flows, each of which satisfy ∇2φ = 0 and ∇2ψ = 0,and superpose them so that the satisfy boundary conditions. We can do this because the
governing equation (Laplace’s Eqn.) is linear.
The simple flows are building blocks for any possible inviscid, incompressible flow, and if we
are clever, then we will never need anything but the simple pieces (and we are clever).
Again, the boundary conditions we are concerned with are Uniform Freestream and No
Penetration of Solid Surfaces
What are the simple building blocks?...
AERO 301: Spring 2011 II(c): Elementary Flows Page 22
1. Uniform flow
Our first elementary flow represents a uniform flow, i.e. a flow with no object.
u =U∞ v = 0 → φ =U∞ x ψ =U∞ y
• Incompressible?
• Irrotational?
• Do φ and ψ satisfy Laplace’s equation?
• Satisfy uniform freestream boundary condition?
2. Source flow
Our second elementary flow, is called a source flow. Consider injecting mass into a 2D
system at the origin, and this injected fluid flows radially outward from the origin. If the
volume flux per unit depth of the injected mass is Λ, the velocities are:
AERO 301: Spring 2011 II(c): Elementary Flows Page 23
vr =Λ
2π rvθ = 0
• Incompressible? (Hard way = Cartesian. Easy way = polar.)
∇ ·~v =1r
∂∂ r
(r vr)+1r
∂vθ∂θ
• Irrotationality? (Hard way = Cartesian. Easy way = polar.)
ω =1r
∂∂ r
(r vθ )−1r
∂vr
∂θ
• Would this screw up the uniform freestream boundary condition were it added to a
uniform flow?
• What is the circulation around any closed contour (that does not pass through the
origin)?
• What are φ and ψ?
φ =Λ2π
lnr and ψ =Λ2π
θ
AERO 301: Spring 2011 II(c): Elementary Flows Page 24
3. Vortex Flow
The vortex flow we already considered earlier is another elementary flow. We established
that it is incompressible and irrotational (except at the origin where it is infinite). Because
of its connection to circulation, we define its strength using Γ and write
φ =−Γ2π
θ and ψ =Γ2π
ln r
• Does this affect the freestream boundary condition?
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 25
II(d): Flows About Bodies without Circulation
Superposition example #1: Uniform + source flow
Having established some elementary flows, we superpose them to make more interesting
and relevant flowfields. Imagine we first superpose a source at the origin with a uniform
freestream flow. What does this produce?
Find: velocity fields, stagnation point(s), and the stagnation streamline (i.e., the body
shape).
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 26
Stagnation point: (r,θ)=(
Λ2πU∞
,π)
The value of streamfunction for the stagnation streamline ψstag= Λ/2. This is importantbecause it gives us an equation for the body surface
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 27
The body shape that results from this is shown
on the right. (All positions have been
nondimensionalized by the distance from the
source at the origin to the stagnation point.)
-3 -2 1
-2
-1
1
2
Sta
gn
atio
n S
tream
line
Stag
nation S treamline
Stagnation StreamlineAll the fluid outside the body/stagnation streamline
originates upstream, and the fluid inside the body
streamline originates at the source.
This means that the body streamline really does act
as a body surface and we have modelled something
that looks very much like the leading edge of a wing.
However, the body streamlines never close as they travel toward x → +∞ and we have notdone a good job yet of modelling the trailing edge.
To fix this, we add another basic flow to the problem: a sink (i.e., a source with a negative
source strength) with a Λ that’s equal and opposite to the strength of the existing source.The combination will add no net mass to the system.
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 28
Superposition example #2: Uniform + source + sink
x
y
l l
x
y
l l
θ2
θ1
(x,y)
Notice that neither the source nor the sink are at the origin.
The combined streamfunction field is
ψ =U∞y+Λ2π
θ1−Λ2π
θ2
where
θ1 = tan−1(
yx+ ℓ
)
and θ2 = tan−1(
yx− ℓ
)
Because of the way the system is setup it’s easier to use Cartesian coordinates this time.
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 29
Again, find the stagnation point(s), and the stagnation (body) streamline.
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 30
1. ψ =U∞ y+Λ2π
[
tan−1(
yx+ ℓ
)
− tan−1(
yx− ℓ
)]
2. Stagnation points: (x,y)=
(
±ℓ
√
1+Λ
π U∞ ℓ,0
)
3. Streamfunction of the stagnation streamline: ψstag= 0
4. It’s tricky to plot streamlines because of the tan−1 functions. A quick-and-dirty Math-
ematica plot is below. This assumes that the lengths have been nondimensionalized by
ℓ and that Λ = 2π U∞ ℓ.
psi = y + lambda(ArcTan[x + 1, y]
- ArcTan[x - 1, y])
For[points = {}; xx = -3; incX = 0.05,xx ¡ 3 + incX/2, xx += incX,
For[yy = -1.5; incY = 0.1,
yy ¡ 1.5 + incY/2, yy += incY,
points = Append[points, {xx,
y/.FindRoot[(psi /.{lambda -¿ 1,
x -¿ xx}) == 0, {y, yy}]}]]]
ListPlot[points, PlotRange-¿{{-3, 3},{-1.5, 1.5}}, AspectRatio -¿ 0.5]
-3 -2 -1 1 2 3
-1.5
-1
-0.5
0.5
1
1.5
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 31
Now consider a special limiting case of a Rankine oval in
which ℓ→ 0. If we simply let ℓ= 0, the source and the sinkcancel out and there’s nothing but freestream flow. But, if
we simultaneously let Λ → ∞ as ℓ→ 0 then we retain someeffect.
To imagine this situation, consider θ1 and θ2 as
ℓ→ 0. Both become approximately equal to θ but θ2
is a little bigger than θ and θ1 is a little bit smaller.
θ2
θ2−θ1
θθ1
(r,θ)
2l
h
As before, we have ψ =U∞y+Λ2π
(θ1−θ2)
Considering the diagram, h ≈ r(θ2−θ1) as ℓ→ 0and h = 2ℓsinθ1 ≈ 2ℓsinθ as ℓ→ 0. Thus
ψ =U∞y−κ sinθ
2πr︸ ︷︷ ︸
doublet
whereκ = 2ℓΛ = constant asℓ→ 0 andΛ → ∞
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 32
κ (kappa) is the doublet strength
The special combination of a source and a sink is called a doublet, and is the equivalent of
a dipole in electric field theory.
The velocities that result from a doublet are:
Notice that these depend on θ and decrease with r . Doublet flow does not have circular
symmetry (as do vortices and sources) and its effect decreases more rapidly as the distance
from the origin increases.
The velocity potential of a doublet is φdoublet=κ cosθ
2πr.
(Get this by integrating the velocities.)
Streamlines of doublet flow trace out
figure-of-eights centered at the origin.
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 33
Superposition example #3: Uniform + doublet
Let us find velocities, stagnation point(s), the stagnation streamfunction value and the body
shape for uniform freestream flow superposed with a doublet of strength κ.
Stagnation points: r =
√κ
2π U∞and θ = 0 orπ
Stagnation streamline value and location:
θ = 0 or π or r =
√κ
2π U∞
A doublet superposed on a uniform flow gives the flow over a cylinder.
Taking the cylinder’s radius to be R we can eliminate κ and write the streamfunction as
ψ =U∞ r
[
1−R2
r2
]
sinθ
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 34
The streamlines about the cylinder are
given to the right. (The coordinates are
nondimensionalized by R.)
-2 -1.5 -1 -0.5 0.5 1 1.5 2
-1.5
-1
-0.5
0.5
1
1.5
This is a closed body that is actually
something of interest. The goal is to know
the lift and drag so we need the pressure
distribution on the body. Use Bernoulli’s
equation:
p(r,θ)+12
ρ{
[vr(r,θ)]2+[vθ (r,θ)]2}
= p∞+12
ρ U2∞
Cp = 1−4sin2θ On the cylinder’s surface.
-π -π/2 π/2 π
θ
-3
-2
-1
1Cp
Trailing EdgeLeading
Edge
AERO 301: Spring 2011 II(d): Γ = 0 Flows Page 35
The pressure distribution is interesting but we really care about net forces. The net lift and
drag (per unit span) are given by
L′ = −
∫ 2π
0p sinθ Rdθ
D′ = −
∫ 2π
0p cosθ Rdθ
Returning to the Cp distribution, p = 12ρ U2
∞Cp+ p∞.
• We do not care about p∞ because it will always cancel as we integrate about the closed
body.
•12ρ U2
∞ is constant and can be moved outside of the integral.
So, we have
cl =L′
12ρ U2
∞ 2R= −
12
∫ 2π
0Cp sinθ dθ
cd =D′
12ρ U2
∞ 2R= −
12
∫ 2π
0Cp cosθ dθ
Evaluating these using orthogonality . . . cl = 0 and cd = 0
D’Alembert’s Paradox.
AERO 301: Spring 2011 II(e): Γ 6= 0 Flows Page 36
II(e): Flows About Bodies with Circulation
When we have inviscid flow over a doublet we have top/bottom and fore/aft symmetry and,
therefore, no lift and no drag. To get either we need to break the symmetry. We can do
this by adding a vortex at the origin.
This is simply ψ =U∞ r
(
1−R2
r2
)
sinθ +Γ2π
lnr
The velocities are. . .
. . . and the stagnation points are. . .
AERO 301: Spring 2011 II(e): Γ 6= 0 Flows Page 37
r = R, θ = sin−1(
−Γ
4π U∞ R
)
if
∣∣∣∣
Γ4π U∞ R
∣∣∣∣≤ 1
or
r = R
Γ
4π U∞ R+
√(
Γ4π U∞ R
)2
−1
, θ =−π2
if
∣∣∣∣
Γ4π U∞ R
∣∣∣∣> 1
Adding the vortex at the origin does not change the body shape!
AERO 301: Spring 2011 II(e): Γ 6= 0 Flows Page 38
Γ = 0 Γ = 2π U∞ R
-2 -1 0 1 2
-2
-1
0
1
-2 -1 0 1 2
-2
-1
0
1
Γ = 4π U∞ R Γ = 4.4π U∞ R
-2 -1 0 1 2
-2
-1
0
1
-2 -1 0 1 2
-2
-1
0
1
AERO 301: Spring 2011 II(e): Γ 6= 0 Flows Page 39
When Γ > 4π U∞ R and the stagnation streamline moves off the body there is fluid outsidethe r = R cylinder that is trapped in the neighborhood of the cylinder inside the stagnationstreamline that simply circles the cylinder.
Restricting ourselves to cases when Γ ≤ 4π U∞ R, what is the pressure distribution about thebody?
Cp(R,θ) = 1−4sin2θ︸ ︷︷ ︸
A
−2Γ sinθπ U∞ R
︸ ︷︷ ︸
B
−Γ 2
4π2U2∞ R2
︸ ︷︷ ︸
C
A is the previous result
B is a non-uniform pressure contribution due to the vortex
C is a uniform pressure contribution due to the vortex that does not affect the lift or drag
AERO 301: Spring 2011 II(e): Γ 6= 0 Flows Page 40
So, we have
cl =L′
12ρ U2
∞ 2R= −
12
∫ 2π
0Cp sinθ dθ
cd =D′
12ρ U2
∞ 2R= −
12
∫ 2π
0Cp cosθ dθ
and
cl =Γ
U∞ R→ L′ = ρ U∞ Γ
cd = 0
This is an important result. It states that the lift per unit span on a 2D body is directly
proportional to the circulation around the body.
AERO 301: Spring 2011 II(e): Γ 6= 0 Flows Page 41
The Kutta–Joukowski Theorem
So, despite the fact that circulation is a concept we have invented, we can use it as a means
of calculating lift per unit span in any situation, for any object without a detailed investi-
gation of the pressure field on the object. This is called the Kutta–Joukowski Theorem.
Said another way: In a 2D inviscid flow, the lift per unit span on an object is given by
L′ = ρ U∞ Γ
Just keep in mind that circulation theory is an alternative way of looking at the generation
of lift on an aerodynamic body. In reality, the true physical sources of lift are due to
the pressure and shear stress distributions exerted on the surface. The Kutta-Joukowski
Theorem is merely an alternative way of expressing the consequences of these stresses.
The power of the approach is that it is much easier to calculate the circulation around an
object rather than the individual surface stresses.