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.