M&AE 3050 October 3, 2008

Incompressible Potential Flow

D. A. Caughey

Sibley School of Mechanical & Aerospace Engineering

Cornell University

Ithaca, New York 14853-7501

These notes provide, as a supplement to our textbook [2], a description of two-dimensional, in-compressible, potential flows, and the associated development of several fundamental aerodynamicresults.

1 Introduction

The analysis of high Reynolds number flows past streamlined bodies is greatly simplified by the factthat viscous effects are generally important only within thin regions immediately adjacent to solidboundaries. Outside these viscous boundary layers the flow is well approximated as inviscid . Thefollowing sections will provide some background for analyzing these inviscid flows, at least in theincompressible limit.

2 Incompressible Potential Flow

The vorticity ζ is the curl of the velocity field V

ζ = ∇× V . (1)

It is not too difficult to show that the vorticity at any point in the flow corresponds to twice therotation rate of an infinitesimal fluid particle centered at that point. To demonstrate this for a two-dimensional flow, we let u and v represent the x- and y- components of the velocity field, relativeto the fluid velocity at a given point. Then the instantaneous rates of rotation of two mutually-orthogonal axes (taken, for convenience, to be parallel to the x- and y-axes, respectively), can becomputed as (see Fig. 1)

θ̇x =∂v

∂x(2)

and

θ̇y = −∂u

∂y, (3)

whence the average rotation rate θ̇ is

θ̇ =1

2

(

θ̇x + θ̇y

)

=1

2

(

∂v

∂x−∂u

∂y

)

. (4)

The quantity above within the parentheses is exactly the vorticity of this two-dimensional flow,which can be written1

ζz =∂v

∂x−∂u

∂y. (5)

1For a two-dimensional flow only one component of the vorticity vector is non-zero: the component normal to theplane of the motion.

2 INCOMPRESSIBLE POTENTIAL FLOW 2

x

y∆

∆ x

y

Figure 1: Relative velocity field in the vicinity of a point, illustrating the relationship betweenderivatives of relative velocity and instantaneous rotation rates of a pair of mutually perpendicularaxes.

A flow in which the vorticity is everywhere zero is said to be irrotational , and these irrotationalflows play an important role in aerodynamics.

Now, if viscous stresses are negligible there is no mechanism to introduce rotation of fluid elements.Thus, an initially irrotational flow must remain irrotational for all time.2 This statement, calledKelvin’s Theorem, is a fundamental basis of much of aerodynamic theory.

Any irrotational flow field can be represented as the gradient of a potential function, so we canrepresent the velocity as

V = ∇φ , (6)

where the scalar function φ is called the velocity potential . It is clear from the vector identity

∇× (∇φ) = 0 (7)

that any potential field is irrotational. Since an irrotational flow can be represented by a velocitypotential, the terms irrotational and potential are synonymous.

In Cartesian coordinates (x, y), Eq. (6) gives the components (u, v)T of the velocity vector as

u =∂φ

∂xand v =

∂φ

∂y(8)

for flows in two dimensions.

For constant density flows, recall that the continuity equation requires

∇ · V = 0 , (9)

which, upon introducing the velocity potential, becomes

∇ · (∇φ) = ∇2φ = 0 . (10)

2In addition to having negligibly small viscous stresses, we must also ensure that the pressure forces do not inducerotation. This simply requires that the pressure forces acting on a fluid element must act through the center of mass,which is guaranteed if the density is constant. In the more general case of compressible flows, all that is required is thatsurfaces of constant pressure also be surfaces of constant density; so compressible flows also satisfy this requirementif they are well approximated as isentropic.

2 INCOMPRESSIBLE POTENTIAL FLOW 3

Thus, the velocity potential for an incompressible flow is seen to satisfy Laplace’s Equation. In twodimensions, the Cartesian-coordinate version of this can be written

∂2φ

∂x2+∂2φ

∂y2= 0 . (11)

Two-dimensional, constant-density flows can also be described in terms of the stream function ψ,which is defined to be constant along stream lines. That is, along a streamline we must have

dψ =∂ψ

∂xdx+

∂ψ

∂ydy = 0 , (12)

which requires

dy

dx

)

ψ=const.

=−∂ψ∂x∂ψ∂y

=v

u. (13)

Thus, if we define the stream function such that

u =∂ψ

∂yand v = −

∂ψ

∂x, (14)

the function ψ will be constant along stream lines.

Since the stream function, as defined by Eqs. (14), can be used to define rotational, as well as irrota-tional flows, the condition of irrotationality supplies an additional constraint. Thus, for irrotationalflows, we must have

∂u

∂y−∂v

∂x=

∂

∂y

(

∂ψ

∂y

)

−∂

∂x

(

−∂ψ

∂x

)

=∂2ψ

∂y2+∂2ψ

∂x2= 0 .

(15)

Thus, for irrotational flows the stream function is seen also to satisfy Laplace’s equation.

Note the duality between the velocity potential and the stream function. The velocity potentialautomatically satisfies irrotationality and continuity requires that it satisfy the Laplace equation.On the other hand, the stream function automatically satisfies continuity, and it is irrotationalitythat requires it to satisfy the Laplace equation.

For some of the elementary solutions to be discussed in the following sections, it is convenient todescribe the flow in the cylindrical coordinates defined by

r =√

x2 + y2 and θ = tan−1

(y

x

)

, (16)

orx = r cos θ and y = r sin θ . (17)

In terms of r and θ the components of the gradient of the velocity potential, corresponding to theradial and circumferential components of velocity, are

vr =∂φ

∂rand vθ =

1

r

∂φ

∂θ, (18)

and Laplace’s equation for the velocity potential φ becomes

∂2φ

∂r2+

1

r

∂φ

∂r+

1

r2∂2φ

∂θ2= 0 . (19)

2 INCOMPRESSIBLE POTENTIAL FLOW 4

V

R

n

Figure 2: Flow field represented in cylindrical coordinates, matching the curvature of the streamlineat a point.

2.1 The Bernoulli Equation

Recall that, for inviscid, incompressible flow the Bernoulli equation requires that

p+1

2ρV 2 = pT , (20)

where pT is constant along a streamline. We show here that for irrotational flows the total pressurepT is the same for all streamlines; i.e., the total pressure is the same constant throughout the flow .

If we differentiate the Bernoulli equation in the direction n normal to a streamline we have

∂p

∂n+ ρV

∂V

∂n=∂pT∂n

. (21)

At the same time, the force balance normal to the streamline requires

∂p

∂n=ρV 2

R, (22)

where R is the radius of curvature of the streamline. Combining these two equations, we see thatthe derivative of the total pressure normal to the streamline must be given by

∂pT∂n

=ρV 2

R+ ρV

∂V

∂n= ρV

(

V

R+∂V

∂n

)

. (23)

But, in cylindrical coordinates the vorticity can be seen to be given by

ζ =V

R+∂V

∂n, (24)

This quantity can be seen to be twice the average rate of rotation of two orthogonal axes (parallelto the r- and θ-directions, respectively) of a polar coordinate system, as suggested by Fig. 2.

Thus, for an irrotational flow we must have

∂pT∂n

= 0 . (25)

Since the total pressure pT is constant along a streamline for inviscid, incompressible flows, and thederivative of pT normal to streamlines must be zero for irrotational flows, Eq. (20) must hold, withthe same constant pT , everywhere in the flow in this case.

3 SOLUTIONS OF LAPLACE’S EQUATION 5

3 Solutions of Laplace’s Equation

Since Laplace’s equation is linear , solutions may be superimposed to produce new solutions. Thatis, if φ1 and φ2 are both solutions, i.e.,

∂2φ1

∂x2+∂2φ1

∂y2= 0 =

∂2φ2

∂x2+∂2φ2

∂y2, (26)

then it is easy to show thataφ1 + bφ2

is also a solution of Laplace’s equation for arbitrary values of the constants a and b. This principlecan be used to construct solutions of arbitrary complexity by superimposing simpler solutions.

3.1 Elementary Solutions

A number of simple solutions to Laplace’s equation are useful for aerodynamic problems. We describesome of these elementary solutions in the following sections.

3.1.1 Uniform stream

The velocity potential for a uniform stream having velocity magnitude U , inclined at the angle αrelative to the x-axis, is given by

φ = U (x cosα+ y sinα) . (27)

It is easily seen that this function satisfies Laplace’s equation, given by Eq. (11), and that thecorresponding Cartesian velocity components are given by

u =∂φ

∂x= U cosα and v =

∂φ

∂y= U sinα. (28)

3.1.2 Source (or Sink)

The velocity potential for a source (or sink) of strengthm located at the origin of the (r, θ) coordinatesystem is given by

φ =m

2πln r . (29)

It is easily verified by direct substitution that this function satisfies Laplace’s equation.

The streamlines for this flow are radial lines from the origin, so the tangential component of velocityvθ is everywhere zero, and the radial component of velocity is given by

vr =∂φ

∂r=

m

2πr. (30)

For this velocity field, the volume rate of flow through any circular contour centered on the originis the same, and is given by

∫

2π

0

vr r dθ =

∫

2π

0

m

2πrr dθ = m. (31)

Thus, the continuity equation is satisfied at any finite value of the radius r, and m can be interpretedas the rate at which fluid volume is introduced by the singularity at r = 0.

3 SOLUTIONS OF LAPLACE’S EQUATION 6

3.1.3 Vortex

The streamlines for a source are seen to be radial lines from the origin, and lines of constant potentialare concentric circles. The flow having constant potential along radial lines and concentric circularstreamlines is called a vortex . The velocity potential for a vortex is

φ = −Γ

2πθ . (32)

For this flow the radial component of velocity vr is everywhere zero, and the tangential componentof velocity is given by

vθ =1

r

∂φ

∂θ= −

Γ

2πr. (33)

The parameter Γ appearing in the definition of the velocity potential for a vortex is identified withthe circulation associated with the vortex. The circulation is defined as (minus) the line integral ofthe velocity around a closed contour3

−

∮

V · ds . (34)

Just as the volume rate of flow for a source is constant when evaluated on a circular contour of anyradius, the circulation associated with a vortex is the same when evaluated on a circle of any radiusr centered on the origin

−

∫

2π

0

vθ r dθ = −

∫

2π

0

−Γ

2πrr dθ = Γ . (35)

This is true so long as circles of successively larger radius do not enclose any new vortices. In fact,it can be shown using Stokes’ Theorem that the circulation is the same, independent of the shapeof the contour, and is equal to the net strength of the vortices enclosed by the contour [1]. Thus,just as a source can be identified as a source of fluid volume, a vortex can be identified as a sourceof circulation.

3.2 Superposition – the Doublet

As described in the introduction to this section, more complicated flows can be constructed bysuperposition of elementary flows. A doublet can be constructed by superposing a source of strengthm at x = −a and a source (sink) of strength −m at x = a, and taking the limit as a → 0 (whileincreasing the source and sink strengths so that the product am remains constant). A sketch of thesituation is provided for finite separation a in Fig. 3(a).

The velocity potential for this flow is then given by

φh = lima→0

[ m

2πrln

√

(x+ a)2 + y2 −m

2πrln

√

(x− a)2 + y2

]

. (36)

Using the fact that

limǫ→0

[ln(1 + ǫ)] = ǫ−ǫ2

2+ǫ3

3+ · · ·

the velocity potential for the doublet can be expressed as

φh =am

π

x

r2= µh

x

r2= µh

cos θ

r, (37)

3The minus sign is introduced here simply for convenience, so that positive lift will correspond to positive circulationwhen the free stream is in the positive x direction.

4 FLOW PAST A CIRCLE 7

x

y

aa

0.25

0.5

0.75

−0.25

−0.5−0.75

−1

−3 −2 −1 0 1 2 3

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

(a) (b)

Figure 3: Superposition of source and sink to form a doublet. (a) Schematic of source at x = −aand sink at x = +a, separated by a finite distance 2a. (b) Selected streamlines for a (horizontal)doublet of unit strength µh = 1.0; value of stream function is indicated for selected streamlines.Flow along x-axis is from right to left.

where we have defined the doublet strength µh = am/π.

Following a similar procedure, it can be shown that the velocity potential for a doublet in which asource located at y = −a and a sink of equal, but opposite, strength located at y = +a, in the limitas a→ 0 while keeping the product am constant is given by

φv =am

π

y

r2= µv

y

r2= µv

sin θ

r, (38)

where the doublet strength is again µv = am/π. The use of the subscripts ()h in Eq. (37) and()v in Eq. (38) is meant to suggest that the limiting process is occurring horizontally or vertically,respectively. The streamline pattern for a (horizontal) doublet of unit strength is indicated inFig. 3(b).

4 Flow past a Circle

The potential flow past a circle can be represented by superposition of a uniform stream parallel tothe x-axis and a (horizontal) doublet. Thus, we let

φ = Ux+ µx

x2 + y2,

or, in polar coordinates,

φ = Ur cos θ + µcos θ

r= cos θ

(

Ur +µ

r

)

. (39)

The radial and tangential components of the velocity are then given by

vr =∂φ

∂r= cos θ

(

U −µ

r2

)

,

vθ =1

r

∂φ

∂θ= − sin θ

(

Ur +µ

r

)

.

(40)

The radial component of velocity is seen to be identically zero for all values of θ when r =√

µ/U .Thus, if we choose µ = U , the circle of radius r = R = 1 will be a streamline – i.e., this combination

4 FLOW PAST A CIRCLE 8

−3 −2 −1 0 1 2 3

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

0.8

0

−3

−3 −2 −1 0 1 2 3

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

(a) (b)

Figure 4: Potential flow past a circle. (a) Selected streamlines and (b) Lines of constant pressurecoefficient; selected contours are labeled with the pressure coefficient value. Note that pressures aresymmetric about both x- and y- axes, so the net aerodynamic force is zero.

of the velocity potentials for a doublet and a uniform stream will represent the flow past a circle ofunit radius. The streamlines and lines of constant pressure for such a flow are illustrated in Fig. 4.

On the surface of the circle, the tangential component of velocity is given by

vθ = −2U sin θ . (41)

The velocity is zero on the circle at the stagnation points θ = 0 and θ = π, and takes its maximumvalue V = 2U at the shoulders θ = ±π. The minimum pressure also occurs at the shoulders θ = ±π.Contours of constant pressure coefficient are displayed in Fig. 4(b). The pressure coefficient is definedas

Cp =p− p∞1

2ρU2

, (42)

where p∞ is the pressure in the free stream, where the velocity is equal to U . For constant-densitypotential flows, the Bernoulli equation can be used to express the pressure coefficient as

Cp = 1 −v2

r + v2

θ

U2. (43)

For this flow the net lift ℓ and drag d forces acting on the circle are zero, by symmetry. Thesymmetry about the x-axis can be broken by adding a vortex to the flow. Since the vortex hascircular streamlines, adding the vortex will not alter the fact that the circle of radius r = 1 remainsa streamline. The velocity potential for the flow past the unit circle with circulation is thereforegiven by

φ = U cos θ

(

r +1

r

)

−Γ

2πθ , (44)

and the velocity components are given by

vr = U cos θ

(

1 −1

r2

)

vθ = −U sin θ

(

1 +1

r2

)

−Γ

2πr.

(45)

The streamlines and contours of constant pressure for this flow, with Γ = πU , are illustrated inFig. 5.

4 FLOW PAST A CIRCLE 9

−3 −2 −1 0 1 2 3

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

−1

−50

0

−3 −2 −1 0 1 2 3

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

(a) (b)

Figure 5: Potential flow past a unit circle with circulation. In this illustration, the doublet andvortex strengths are chosen such that µ = U and Γ = πU . (a) Selected streamlines and (b) Lines ofconstant pressure coefficient; selected contours are labeled with the pressure coefficient value. Notethat pressures remain symmetric about the y- axis, so the drag is still zero, but the symmetry isbroken about the x-axis, resulting in a net lift force.

On the circle r = 1 the pressure is given by

p = pT −1

2ρv2

θ

= pT −ρ

2

[

4U2 sin2 θ + 2UΓ

πsin θ +

Γ2

4π2

]

.(46)

The net lift force ℓ (i.e., the force in the y-direction) on the circle can then be calculated by integratingthe pressure force around the surface of the circle; this is found to be4

ℓ = −

∫

2π

0

p sin θ dθ =ρUΓ

π

∫

2π

0

sin2 θ dθ = ρUΓ , (47)

since∫

2π

0

sin2 θ dθ = π . (48)

Similarly, the drag force d acting on the circle is

d = −

∫

2π

0

p cos θ dθ = 0 . (49)

The result shown in Eqs. (47) and (49) is, in fact, a general result for potential flow past two-dimensional bodies: namely that there is no drag and the lift is proportional to the net circulationgenerated by the body (with the constant of proportionality equal to the product of the free streamvelocity and the fluid density). A proof of this result, which is known as the Kutta-Joukowsky

Theorem, is sketched in the following section.

4Only the middle term in square brackets of Eq. (46) contributes to the integral, since the integral of any oddpower of the sine function over the range from 0 to 2π is identically zero.

5 THE KUTTA-JOUKOWSKY THEOREM 10

5 The Kutta-Joukowsky Theorem

The flow past any closed, two-dimensional body can be represented by the superposition of a uniformstream

φ = Ux ,

a sum of sources and sinks

φ =∑

i

mi

2πln

√

(x− xi)2 + (y − yi)2 ,

where∑

imi = 0 since the body is closed; and a sum of point vortices

φ =∑

i

−Γi2π

tan−1

(

y − yix− xi

)

.

At large distances from the body, the contributions to the velocity potential due to the sources andsinks must approach that of effective doublets, since the net source strength is zero; and all thepoint vortices can be represented as a single vortex located at the origin (plus additional doubletcontributions5). Thus, at large distances from the body, the velocity potential can be representedas

φ = Ur cos θ + µhcos θ

r+ µv

sin θ

r−

Γ

2πθ , (50)

where Γ =∑

i Γi is the net circulation produced by the flow past the body and µh and µv are thenet doublet strengths required to represent the displacements of the sources and vortices from theorigin. Thus, far from the body the radial and tangential components of the velocity are given by

vr =∂φ

∂r= U cos θ − µh

cos θ

r2− µv

sin θ

r2,

vθ =1

r

∂φ

∂θ= −U sin θ − µh

sin θ

r2+ µv

cos θ

r2−

Γ

2πr.

(51)

Note that the doublet contributions to the velocities decay as 1/r2, while the contribution from thevortex decays only as 1/r. Thus, if the vortex strength is non-zero, the doublet contributions tovelocities far from the body will always be small compared to that due to the vortex.

Now, we can use this far-field behavior of the velocities to determine the net aerodynamic forceacting on a body in two-dimensional potential flow using the Momentum Theorem applied to acontrol volume consisting of the fluid between the body and a circular contour of radius R in thefar field, taking the limit as R → ∞. The y-component of the sum of the forces acting on the fluidin this control volume can be written

∑

Fy = −ℓ−

∫

2π

0

p sin θ R dθ , (52)

where ℓ is the lift force on the body.6 The Bernoulli equation can be used to express the pressureon the far-field contour as

p = pT −ρ

2

[

U2 +UΓ

π

sin θ

R+

Γ2

4π2

1

R2

]

. (53)

Thus, the net force acting on the fluid in the control volume is7

∑

Fy = −ℓ+ρUΓ

2π

∫

2π

0

sin2 θ

RR dθ = −ℓ+

ρUΓ

2. (54)

5A doublet can also be formed by the limiting process in which two vortices of equal, but opposite, strength aremoved together, keeping the product of their strength and spacing constant; see Exercise 2

6Note that the force exerted by the body on the fluid is minus the force exerted by the fluid on the body.7Again, only the middle term in brackets in Eq. (53) contributes to the integral

REFERENCES 11

According to the Momentum Theorem, this force must be equal to the net rate at which y-momentumis leaving the control volume, which is

∫

2π

0

ρvy(V · n̂)R dθ , (55)

where (far from the body)

vy = vr sin θ + vθ cos θ = −Γ

2π

cos θ

r, (56)

andV · n̂ = vr = U cos θ . (57)

Thus, the net rate at which y-momentum is leaving the control volume is

∫

2π

0

ρvy(V · n̂)R dθ = −

∫

2π

0

ρUΓ

2πcos2 θ dθ = −

ρUΓ

2. (58)

Thus, equating Eqs. (54) and (58) givesℓ = ρUΓ (59)

for the net force acting on the body in the direction normal to the free stream. A similar anal-ysis shows that the x-component of force – i.e., the drag d – is identically zero. Thus, we havedemonstrated the Kutta-Joukowsky Theorem for the potential flow past a general, two-dimensionalbody.

References

[1] John J. Bertin, Aerodynamics for Engineers, Fourth Edition, Prentice-Hall, Upper SaddleRiver, New Jersey, 2002.

[2] Richard S. Shevell, Fundamentals of Flight, Second Edition, Prentice-Hall, Upper SaddleRiver, New Jersey, 1989.

6 Exercises

1. Show that, to within a constant, the stream function for a vortex is the same as the velocitypotential for a source, and vice versa.

2. Show that the doublet formed superposition of a vortex of strength Γ located at y = +a and avortex of strength −Γ located at y = −a, in the limit as a → 0 while keeping the product aΓconstant, is the same as the “horizontal” doublet of Eq. (37). What arrangement of vorticeswould produce, in the proper limit, a flow equivalent to the “vertical” doublet of Eq. (38)?

3. For a critical value of the circulation Γ, the two stagnation points in the circulatory flow pasta circle will merge into a single point on the circle (at θ = −π/2). What is the value ofcirculation for which this happens? Sketch the streamlines for values of circulation greaterthan this critical value.