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MATH45101/MATH65101

J. S. B. [email protected]

http://www.maths.manchester.ac.uk/∼gajjar

School of MathematicsUniversity of Manchester,Manchester M13 9PL UK.

MATH45101/65101

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Contents

1 Perturbation Methods in Fluid MechanicsCylinder in a uniform flow with small vorticity

Solution of ε0 problemSolution of O(ε) problem

2 Flow at low Reynolds numbersSolution of O(1) problem

3 High Reynolds number fluid mechanics

4 Solution for non-uniformity near leading edge

5 High Reynolds number flows- boundary layer theory.

6 BL eqns for flow over a curved surface

7 Goldstein (1930) near wake solution

2 / 190

Perturbation Methods in Fluid Mechanics

3 / 190


• In the remainder of the lectures we will apply some of thetechniques mentioned in the earlier sections to study someproblems drawn from fluid mechanics.

• A number of these are discussed in the book by Van-Dyke.• The first example concerns inviscid flow past a cylinder in

which the oncoming uniform is slightly perturbed.

4 / 190

Outline

<beamer>1 Perturbation Methods in Fluid Mechanics

Cylinder in a uniform flow with small vorticitySolution of ε0 problemSolution of O(ε) problem

2 Flow at low Reynolds numbersSolution of O(1) problem

3 High Reynolds number fluid mechanics

4 Solution for non-uniformity near leading edge

5 High Reynolds number flows- boundary layer theory.

6 BL eqns for flow over a curved surface

7 Goldstein (1930) near wake solution

5 / 190


We will consider a circular cylinder of radius a in a uniform flowwith the velocity at∞ given by

U∞ = U(1 + εy

a, 0)

where 0 < ε << 1 and U∞ is the velocity in cartesiancoordinates (x, y) and U is a constant.

a

(r, θ)

θ

U∞(y) i

Figure: Circle in uniform flow

6 / 190


Note that the vorticity at∞ is

ω∞ = ∇×U∞ = −Uεak.

We will assume the fluid is incompressible, that the effects ofviscosity can be neglected and that the motion is steady andplanar.

7 / 190

Governing equations

Let u = (u, v) be the velocity components in terms of Cartesiancoordinates (x, y). We can introduce the streamfunction ψ

u = (u, v) = (∂ψ

∂y,−∂ψ

∂x). (1)

This ensures that the continuity equation is identically satisfied.Then since the vorticity ω = ∇×u we obtain

ω = ωk,

where

ω = −(∂u

∂y− ∂v

∂x

)= −∇2ψ.

8 / 190

Governing equations

Note that the Euler equations are

∇(1

2u2) + ω × u = −∇p (2)

and taking the curl of (2) shows that

∇×(ω × u) = 0. (3)

Expanding this gives

∂(uω)

∂x+∂(vω)

∂y= 0,

or∂ψ

∂y

∂ω

∂x− ∂ψ

∂x

∂ω

∂y= 0, (4)

9 / 190

Governing equations

The equation (4) shows that the vorticity ω is conserved alongstreamlines and

ω = ω(ψ)

only. From the flow at∞ we can evaluate ω = −εU/a.Thus the governing equation for this flow is

∇2ψ = −ω(ψ) =εU

a. (5)

10 / 190

Governing equations

We will work in polar coordinates (r, θ) so that(x, y) = r(cos θ, sin θ). The boundary conditions we require arethose of zero normal velocity on the cylinder surface andmatching with the flow at∞ giving

ψ(a, θ) = 0, (6a)

ψ → U(r sin θ +1

4

εr2

a(1− cos 2θ)) as r →∞.(6b)

11 / 190

Solution for ε << 1.

To solve (5) we will seek a solution for ψ with ε small in the form

ψ = ψ0 + εψ1 + . . . , (7)

where ψ = ψ(r, θ). Substituting (7) into (5, 6a,6b) gives

∇2ψ0 = 0, ∇2ψ1 =U

a,

ψ0(a, θ) = 0, ψ0 → Ur sin θ as r →∞,

ψ1(a, θ) = 0, ψ1 →Ur2

4a(1− cos 2θ) as r →∞.

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In the above ∇2 is the Laplacian operator which in polarcoordinates is given by

∇2 =∂2

∂r2+

1

r

∂

∂r+

1

r2

∂2

∂θ2.

13 / 190


The solution of the leading order problem

∇2ψ0 = 0, ψ0(a, θ) = 0, ψ0 → Ur sin θ r →∞, (8)

may be obtained by looking for a solution using separation ofvariables in the form

ψ0 = f(r)g(θ).

Then (8) give

∇2ψ0 = gf ′′ +gf ′

r+fg′′

r2= 0.

14 / 190


Hence

r2

(f ′′

f+f ′

rf

)= −g

′′

g. (9)

The left-hand side of (9) is a function of r only and theright-hand side a function of θ only. Hence both must be equalto a constant k2 say. This gives

−g′′

g= k2 = r2

(f ′′

f+f ′

rf

).

15 / 190


Thusg′′ = −k2g

and sog(θ) = A+Bθ if k = 0

andg(θ) = A sin kθ +B cos kθ if k 6= 0.

The equation for f(r) is

r2f ′′ + rf ′ − k2f = 0.

This is an Euler type differential equation and we can seek asolution of the form f(r) = rα giving

α(α− 1) + α− k2 = 0.

α2 = k2, α = ±k.16 / 190


So a general solution for ∇2ψ0 = 0 can be expressed in theform

ψ0(r, θ) = A0 +B0θ+∑k

(Ckrk +Dkr

−k)(Ek sin kθ+ Fk cos kθ).

(10)The boundary conditions (8) suggest that A0 = B0 = 0 andCk = 0 for k > 1. Thus

17 / 190


ψ0(r, θ) =

(C1r +

D1

r

)(E1 sin θ + F1 cos θ).

For r >> 1 we require ψ0 → Ur sin θ and so F1 = 0, and

C1E1 = U

Hence we obtain

ψ0(r, θ) =

(Ur +

E1D1

r

)sin θ.

Since ψ0 = 0 when r = a we must have(Ua+

E1D1

a

)sin θ = 0

giving E1D1 = −Ua2. Hence

ψ0(r, θ) = U

(r − a2

r

)sin θ. (11)

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Next consider the problem for ψ1 given by

∇2ψ1 =U

a, ψ1(a, θ) = 0, ψ1 →

Ur2

4a(1− cos 2θ) r →∞.

We can seek a particular solution to the equation in the formψ1p = h(r) and then

h′′ +h′

r=U

a,

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(rh′)′ =Ur

a,

rh′ =Ur2

2a+A, h′ =

Ur

2a+A

r

h(r) =Ur2

4a+A log r.

The form of the boundary conditions as r →∞ suggests thatthe solution for ψ1 takes the form

ψ1(r, θ) =Ur2

4a+A log r+A0+B0θ+

(C2r

2 +D2

r2

)(E2 sin 2θ+F2 cos 2θ).

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We require

ψ1(r, θ)→ Ur2

4a(1− cos 2θ) as r →∞

and so

A = 0, A0 = 0, E2 = 0, C2F2 = − U4a.

Thus

ψ1(r, θ) =U

4a(r2 − a2)− U

4a(r2 +

D2

r2) cos 2θ,

where we have redefined the unknown constant D2. To satisfyψ1(a, θ) = 0 we require D2 = −a4.

21 / 190


Hence

ψ1(a, θ) =U

4a(r2 − a2)− U

4a(r2 − a4

r2) cos 2θ.

It can be checked that ψ = ψ0 + εψ1 gives an exact solution tothe full problem.

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Flow at Low Reynolds numbers

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Flow at Low Reynolds numbers- Introduction

• In this section we will discuss flows at low Reynoldsnumbers. The Reynolds number is defined by

Re = U∞L/ν.

• Re gives a measure of the the ratio of inertial to viscousforces.

• For the flows we will be studying in this section theReynolds number will be small so that to leading orderinertial may be neglected.

• There are many flow situations where this type ofapproximation is useful. Typically microrganisms in motionhave Reynolds numbers of O(10−6).

• In microfluidics motion of micro-particles can be describedwith a low Reynolds number approximation.

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Governing equations

The continuity and Navier-Stokes equations for anincompressible fluid may be written in the form:

∇′.u′ = 0,∂u′

∂t′+ u′.∇′u′ = −1

ρ∇′p′ + ν∇′2u′. (12)

Here u′ is the velocity, p′ the pressure, ν the kinematic viscosityand ρ the uniform density.

25 / 190

Governing equations

We will first non-dimensionalise by introducing

u′ = U∞u, p′ = ρU2∞p, x′ = Lx, y′ = Ly, t′ =

L

U∞t.

(13)Here U∞ is a typical velocity scale and L a typical length scale.Substituting (13) into (12) gives the non-dimensional form of theNavier-Stokes equations

∇.u = 0,∂u

∂t+ u.∇u = −∇p+

1

Re∇2u, (14)

whereRe =

U∞L

ν

is a non-dimensional parameter known as the Reynoldsnumber.

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Governing equations - alternative forms

The continuity and Navier-Stokes equations may be expressedin the alternative form

∇.u = 0, (15)

∂u

∂t+∇(

1

2u2) + ω × u = −1

ρ∇p− 1

Re(∇×ω −∇(∇.u)). (16)

27 / 190

Governing equations

If we eliminate the pressure by taking the curl of (16) and usethe continuity equation we obtain

∂ω

∂t+∇×(ω × u) = − 1

Re(∇×(∇×ω)), (17)

where use has been made of the identity that ∇×(∇p) isidentically zero.

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Stokes’s equations Re << 1

We will consider steady flows and take Re << 1. If we seek asolution of (17) in the form

ω = ω0 +Reω1 + . . . , u = u0 +Reu1 + . . . .

Then we obtain

0 = −(∇×(∇×ω0)), u0 = ∇×ω0 (18)

and∇×(ω0 × u0) = −(∇×(∇×ω1)), u1 = ∇×ω1 (19)

The equations (18) are known as the Stokes’s equations.

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Stokes’s solution for flow past a sphere

Consider three-dimensional axi-symmetric flow past a sphereof non-dimensional radius unity. Velocity at large distancesfrom the sphere is U∞ = k, where (R, θ, λ) denote

1

(R, θ)

θ

U∞ = k

Figure: Sphere in uniform flow.

cylindrical polar coordinates and the unit vector k is alignedwith the direction from which θ is measured, see figure 2.We will see a solution independent of λ.

30 / 190

Stokes’s streamfunction

Introduce the Stokes’s streamfunction ψ(R, θ) where thevelocity may be written as

u =1

R2 sin θ

∂ψ

∂θR− 1

R sin θ

∂ψ

∂Rθ, (20)

where (R, θ, λ) denote unit vectors in the respective coordinatedirections. The form (20) ensures that the continuity equation isidentically satisfied.

31 / 190


∇.u =1

R2

∂

∂R(R2ur) +

1

R sin θ

∂

∂θ(uθ sin θ) +

1

R sin θ

∂uλ∂λ

=1

R2

∂

∂R(

1

sin θ

∂ψ

∂θ) +

1

R sin θ

∂

∂θ(− 1

R

∂ψ

∂R) = 0.

32 / 190


Also the vorticity ω is given by

ω = ∇×u =1

R2 sin θ

∣∣∣∣∣∣R Rθ R sin θλ∂∂R

∂∂θ

∂∂λ

uR Ruθ R sin θuλ

∣∣∣∣∣∣ ,

=1

R2 sin θ

∣∣∣∣∣∣R Rθ R sin θλ∂∂R

∂∂θ

∂∂λ

1R2 sin θ

∂ψ∂θ − 1

sin θ∂ψ∂r 0

∣∣∣∣∣∣ .

33 / 190


Henceω = − 1

R sin θD2ψ λ.

where the D2 operator is defined by

D2 =∂2

∂R2+

sin θ

R2

∂

∂θ

(1

sin θ

∂

∂θ

).

34 / 190


If we write

ω = Ωλ, where Ω = − 1

R sin θD2ψ,

then∇×ω =

1

R sin θ

∂

∂θ(Ω sin θ)R− 1

R

∂

∂R(RΩ)θ.

35 / 190


And

∇×(∇×ω) =1

R2 sin θ

∣∣∣∣∣∣∣R Rθ R sin θλ∂∂R

∂∂θ

∂∂λ

1R sin θ

∂(sin θΩ)∂θ −∂(RΩ)

∂R 0

∣∣∣∣∣∣∣ ,= − 1

R

[∂2

∂R2(RΩ) +

1

R

∂

∂θ

(1

sin θ

(∂(Ω sin θ)

∂θ

))]λ.

36 / 190


Hence substituting for Ω shows that

∇×(∇×ω) =1

R sin θD2(D2ψ)λ.

At lowest order we need to solve Stokes’s equations (18) ie

0 = −(∇×(∇×ω0)), u0 = ∇×ω0 (21)

together with boundary conditions of no slip on the cylindersurface and matching with the flow at large distances from thecylinder.

37 / 190


Expanding ψ asψ = ψ0 +Reψ1 + . . .

the equation (21) reduces to solving

D2(D2ψ0) = 0. (22)

For no slip we require u = 0 on R = 1 and hence from (20) werequire

∂ψ

∂R(1, θ) =

∂ψ

∂θ(1, θ) = 0. (23)

38 / 190


Now since k = cos θR− sin θθ we require

u0 =1

R2 sin θ

∂ψ0

∂θR− 1

R sin θ

∂ψ0

∂Rθ → cos θR−sin θθ as R→∞.

Henceψ0 →

1

2R2 sin2 θ as R→∞. (24)

39 / 190

Solution of O(1) problem

Given the form of the boundary conditions we can try and seeka solution to (22) of the form

ψ0(R, θ) = f(R) sin2 θ.

Then

D2(ψ0) =∂2ψ0

∂R2+

sin θ

R2

∂

∂θ

(1

sin θ

∂ψ0

∂θ

),

= sin2 θf ′′(R) +f

R2

∂

∂θ(2 sin θ cos θ

sin θ) = sin2 θ(f ′′(R)− 2f(R)

R2).

40 / 190


Hence

D2(D2ψ0) = sin2 θ(d2

dR2− 1

R2)(f ′′ − 2f

R2) = 0,

or

(d2

dR2− 1

R2)(f ′′ − 2f

R2) = 0.

41 / 190


One can check that the equation for f(R) is an Euler equationand so if we look for a solution of the form f(R) = Rα we obtain(

d2

dR2− 2

R2

)[α(α− 1)Rα−2 − 2Rα−2],

=

(d2

dR2− 2

R2

)[α(α−1)−2]Rα−2 = (α(α−1)−2)((α−2)(α−2−1)−2)Rα−4 = 0

= (α+ 1)(α− 2)(α− 1)(α− 4)Rα−4 = 0.

Hence possible values of α are 2,-1,1, and 4 .

42 / 190


Thus

ψ0(R, θ) = (AR2 +B

R+ CR+DR4) sin2 θ.

Applying the no slip conditions (23) shows that∂ψ0

∂R= (2AR− B

R2+ C + 4DR3) sin2 θ = 0

when R = 1 giving

2A−B + C + 4D = 0.

Also ∂ψ∂θ (1, θ) = 0 gives

A+B + C +D = 0.

The matching condition (24) implies D = 0 and A = 1/2. Hencesolving the above equations we find that B = 1/4 andC = −3/4 giving

ψ0 =1

4(2R2 +

1

R− 3R) sin2 θ. (25)

43 / 190

Second order solution for Stokes’s flow past a sphere

With the vorticity given by

ω = Ωλ = − 1

R sin θD2(D2ψ)λ.

and the velocity by the Stokes’s stream function we find that forsteady axis-symmetric flows the governing equation

∇×(ω × u) = − 1

Re∇×(∇×ω)

reduces to

1

ReD2(D2ψ) =

1

R2 sin θ

[∂ψ

∂θ

∂

∂R− ∂ψ

∂R

∂

∂θ+ 2 cot θ

∂ψ

∂R− 2

R

∂ψ

∂θ

]D2ψ.

(26)

44 / 190

Stokes’s streamfunction - O(ε) solution

For the second order problem we need to solve

D2(D2ψ1) =1

R2 sin θ

[∂ψ0

∂θ

∂

∂R− ∂ψ0

∂R

∂

∂θ+ 2 cot θ

∂ψ0

∂R− 2

R

∂ψ0

∂θ

]D2ψ0.

(27)

45 / 190


If we substitute for ψ0 = f(R) sin2 θ wheref(R) = 1

4(2R2 + 1R − 3R) then

D2(ψ0) = sin2 θ(f ′′ − 2

R2)

= sin2 θ

(1 +

1

2R3− 2

R2

(R2

2+

1

4R− 3R

4

)),

=3

2Rsin2 θ.

46 / 190


Next

D2(D2ψ1) =1

R2 sin2 θ

[2 sin θ cos θf(R)

∂

∂R

(sin2 θ

3

2R

)− sin2 θf ′(R)

∂

∂θ

(3

2Rsin2 θ

)

+2 cot θ sin2 θf ′(R) sin2 θ3

2R− 4 sin3 θ cos θf

3

2R2

],

=3

R3sin2 θ cos θ

[− 1

R(R2

2+

1

4R− 3R

4)− (R− 1

4R2− 3

4) + (R− 1

4R2− 3

4)− 2

R

R2

2+

1

4R− 3R

4)

],

= −9

4sin2 θ cos θ

(2

R2+− 1

R5− 3

R3

).

47 / 190


Hence we need to solve

D2(D2ψ1) = −9

4sin2 θ cos θ

(2

R2− 1

R5− 3

R3

). (28)

48 / 190


Because of the forcing term we my seek a solution to (28) in theform

ψ1 = g(R) sin3 θ cos θ.

ThenD2ψ1 = sin2 θ cos θ(g′′ − 6

R2g),

and

D2(D2ψ1) = (d2

dR2− 6

R2)(g′′ − 6

R2) sin2 θ cos θ.

Hence g(R) satisfies

(d2

dR2− 6

R2)(g′′ − 6

R2) = −9

4

(2

R2− 1

R5− 3

R3

). (29)

49 / 190


A solution to the homogeneous form of (29) can be found in theform g(R) = rα and this gives

(d2

dR2− 6

R2)(α(α− 1)− 6)Rα−2,

= (d2

dR2− 6

R2)(α− 3)(α+ 2)Rα−2,

= (α− 5)(α)(α− 3)(α+ 2)Rα−4 = 0.

Hence α = −2, 0, 3, or 5. A particular solution to (29) is

g(R) =3

32R+

9R

32− 3R2

16.

50 / 190


Hence the solution for ψ1 takes the form

ψ1 =

(3

32R+

9R

32− 3R2

16+AR2 +BR5 +

C

R2+D

)sin2 θ cos θ.

The boundary conditions are

∂ψ1

∂R(1, θ) = 0,

∂ψ1

∂θ(1, θ) = 0,

and ψ1(R, θ) = o(R2) as R→∞. This means that we can takeA = B = 0 and we find that C = −3/32, D = −3/32 giving

51 / 190


ψ1(R, θ) =3

32(

1

R+ 3R− 2R2 − 1

R2− 1) sin2 θ cos θ. (30)

Once can also add a term of the form

k(2R2 − 3R+1

R) sin2 θ

to (30) but it is clear that there is no choice of the free constantwhich ensures that ψ1(R, θ) = o(R2) as R→∞.

52 / 190

Stokes-Whitehead paradox

• This difficulty in being able to satisfy the boundaryconditions at infinity is known as the Stokes-Whiteheadparadox.

• The Stoke’s (1851) paradox refers to the non-existence ofthe solution of the Stokes’s equations in planar flow past abody.

• Whitehead (1889) showed with the second order solution(as above) that the same applies to three-dimensionalflows as well.

• Whereas for the flow past a sphere the problems appear athigher order for the planar case, ie flow past a circularcylinder, the problems appear at leading order.

53 / 190

Stokes-Whitehead paradox

• The difficulty stems from neglecting some of the nonlinearterms used in the Stokes approximation.

• One can show that the ratio of the neglected convectiveterms to the retained viscos terms is

O(Convective/Viscous) = O(ReR) as R→∞.

• Hence whilst the terms can be neglected for small R forlarge R = O(1/Re) the neglected terms are no longersmaller than the ones retained. So one needs toincorporate an outer solution where R = O(1/Re).

54 / 190

High Reynolds number fluid mechanics

55 / 190

Re » 1 flows

• The Reynolds number Re, represents the ratio of theinertia terms to the viscous terms in the Navier-Stokesequations.

• For an aircraft wing, for example, estimates of theReynolds number for different types of aircraft and flightregimes are found to vary from about Re ∼ 106 for a gliderto Re ∼ 108 for modern passenger aircraft operating in acruise condition.

• The Reynolds number may be even larger in typical marineapplications involving ships or submarines. For example,using the length of the submarine hull for L yields typicalReynolds numbers in the range Re ∼ 108 to 109.

• Although 1/Re is small, Prandtl in 1904 argued that theinfluence of viscosity (no matter how small) must becomesimportant near solid surfaces. This led to the developmentof boundary layer theory.

56 / 190

Thin aerofoil theory

We will assume that the fluid is incompressible and that the flowis steady and two-dimensional and there are no body forces.Then the governing equations are the continuity andNavier-Stokes equations given by

∇.u = 0, (31)

∂u

∂t+∇(

1

2u2) + ω × u = −∇p− 1

Re(∇×ω −∇(∇.u)). (32)

57 / 190


Consider the uniform slow with speed 1 (non-dimensional) pasta thin aerofoil at a small angle of attack.

α

x

y

1

LETE

Figure: Uniform flow past an aerofoil at angle of attack α. LE and TEdenote the leading and trailing edges respectovely. The dashed lineshows the camber.

58 / 190


Suppose the upper/lower surface of the aerofoil is given by

y = y+(x) 0 ≤ x ≤ 1,

y = y−(x) 0 ≤ x ≤ 1,

59 / 190


The thickness t(x) and camber c(x) functions are defined by

t(x) =1

2(y+(x) + y−(x)), c(x) =

1

2(y+(x)− y−(x)),

and then the upper/lower surfaces may be expressed as

y±(x) = c(x)± t(x). (33)

t(x) Thickness function

c(x) camber position or midline.

Figure: Thickness and camber functions for aerofoil.

60 / 190


We will assume that the Reynolds number Re >> 1 and we willuse Prandtl’s hypothesis that viscous effects are confined to athin region close to the body and in regions of rapid change.We will also assume that the flow is attached, ie no separation.Then the governing equations are just the continuity and Eulerequations given by

∇.u = 0, (34)

∂u

∂t+∇(

1

2u2) + ω × u = −∇p. (35)

61 / 190


Next we will assume that the flow is irrotational so that ω = 0.Hence there exists a velocity potential φ so that

u = (u, v) = ∇φ =

(∂φ

∂x,∂φ

∂y

). (36)

Then if we substitute for u from (??) into (34) we obtain

∇2φ = 0. (37)

62 / 190

Thin aerofoil theory- boundary conditions

We require thatv

u=dy

dx

on the aerofoil surface where y = y(x) is the shape of theaerofoil. Since the fluid velocity is given by (36) the boundarycondition we require is that

∂φ

∂y(x, y) =

∂φ

∂x(x, y)

dy±dx

(x) on y = y±(x). (38)

Other boundary conditions we need are that

u→ cosα i + sinα j as x2 + y2 →∞. (39)

to match with the oncoming uniform stream at an angle ofattack α.

63 / 190


In summary we need to solve

∇2φ = 0, (40a)∂φ

∂y(x, y) =

∂φ

∂x(x, y)

dy±dx

(x) on y = y±(x), (40b)(∂φ

∂x,∂φ

∂y

)→ (cosα, sinα) as x2 + y2 →∞. (40c)

64 / 190


For general α and aerofoil shapes y±(x) the problem (40)needs to be solved numerically. We can make progresshowever if |α| << 1 and |y±(x)| << 1.We will assume that we are dealing with small angles of attackand the aerofoil is thin and set

α = εα, y±(x) = εY±(x), (41)

with 0 < ε << 1. We will also introduce scaled camber andthickness functions as

C(x) = ε−1c(x) =1

2(Y+(x)− Y−(x)),

T (x) = ε−1t(x) =1

2(Y+(x) + Y−(x)).

65 / 190


Far away from the aerofoil we have a uniform stream at angle αand so we can write

φ = x cosα+ y sinα+ . . .

i.e.,φ = x+ εyα+ . . . .

66 / 190


This suggests an expansion

φ = x+ ε(αy + φ1) + ε2φ2 + . . . , (42)

where φ1, φ2 = o(1) as x2 + y2 →∞.

67 / 190


Next we will consider the boundary conditions (40b) on theaerofoil surface. We have

∂φ

∂y(x, y) =

∂φ

∂x(x, y)ε

dY±dx

(x) on y = εY±(x).

Now from (42)

∂φ

∂y(x, y) = ε

(α+

∂φ1

∂y

)+ ε2

∂φ2

∂y+ . . . .

Thus

∂φ

∂y(x, εY±(x)) = ε

(α+

∂φ1

∂y(x, εY±(x))

)+ε2

∂φ2

∂y(x, εY±(x))+. . . .

(43)

68 / 190


Next if we use a Taylor expansion and note that

∂φ

∂y(x, εY±(x)) =

∂φ

∂y(x,±0) + εY±(x)

∂2φ

∂y2(x,±0) + . . . ,

then (43) becomes

φy(x, εY±(x)) = ε(α+φ1y(x,±0)+ε2(φ2y(x,±0)+Y±(x)φ1yy(x,±0))+. . . .(44)

Similarly from (42)

φx(x, y) = 1 + εφ1x(x, y) + ε2φ2x(x, y) + . . . ,

69 / 190


φx(x, εY±(x)) = 1+εφ1x(x,±0)+ε2(φ2x(x,±0)+Y±(x)φ1xy(x,±0))+. . . .(45)

Hence combining (44,45) the boundary on the aerofoil surfaceis

ε(α+ φ1y(0,±0) + . . . ) = εdY±dx

(x)(1 + . . . ).

Equating terms at O(ε) we find

φ1y(x,±0) = Y ′±(x)− α. (46)

70 / 190


Finally substituting for Y±(x) = C(x)± T (x) gives

φ1y(x,±0) = (C ′(x)− α)± T ′(x).

If we substitute (42) into (40a) then we find

∇2φ1 = 0.

71 / 190


So the leading order problem to solve is

∇2φ1 = 0, (47a)

φ1y(x,±0) = (C ′(x)− α)± T ′(x), (47b)

andφ1 → 0 as x2 + y2 →∞. (47c)

72 / 190


The above problem is equivalent to the following if we write

φ1 = φ1T + φ1C

with∇2φ1T = 0, ∇2φ1C = 0,

φ1T , φ1C → 0 as x2 + y2 →∞,φ1Ty(x,±0) = ±T ′(x), symmetric aerofoil at zero incidence,

φ1Cy(x,±0) = C ′(x)−α aerofoil with zero thickness at incidence.

We will consider the φ1T problem. The φ1C problem can bedealt with similarly.

73 / 190

Solution for symmetric aerofoil, zero angle of attack

We will consider the solution of the problem for φ1T which isgoverned by the equations

∇2φ1 = 0, (48a)

φ1 → 0 as x2 + y2 →∞, (48b)

φ1(x,±0) = ±T ′(x), (48c)

74 / 190


y = ǫT (x)

y = −ǫT (x)

x

y

1

Figure: Uniform flow past a symmetric aerofoil at zero angle of attack

75 / 190


This problem can be solved using complex variable theoryusing a distribution of sources. The solution is given by

u1(x, y)− iv1(x, y) =∂φ1

∂x(x, y)− i∂φ1

∂y=

1

π

∫ 1

0

T ′(s)

x+ iy − s ds,(49)

76 / 190


Suppose we write the complex velocity in the form

U1(z) = u1(x, y)− iv1(x, y) =

∫ ∞−∞

m(ξ)

z − ξ dξ,

where z = x+ iy and we will find the unknown sourcedistribution 2πm(z). We will study the limit as y → 0± so thatz = x+ iy approaches the real axis. Consider first the limity → 0+ and we deform the contour as shown in figure 6 belowthe point ξ = x taking off a segment Lδ of length 2δ either sideof ξ = x.

y

x

xC

L

Figure: Contour along real axis deformed with semi-circular path ofradius δ below ξ = x

77 / 190


Then

U1(z) = limδ→0

∫L−Lδ

m(ξ)

z − ξ dξ + limδ→0

∫C

m(ξ)

z − ξ dξ.

The first integral reduces to∫−∞

−∞

m(ξ)

x− ξ dξ.

For the second integral if we put ξ − x = δeiθ then∫C

m(ξ)

z − ξ dξ =

∫ 0

−π

iδeiθm(x+ δeiθ)

−δeiθ dθ → −πim(x) as δ → 0.

78 / 190


HenceU1(x, 0+)→

∫−∞

−∞

m(ξ)

x− ξ dξ − πim(x).

But the left hand-side is

U1(x, 0+) = u1(x, 0+)− iv1(x, 0+) = u1(x, 0+)− if(x).

Here we have set f(x) = 0 if x is not on the aerofoil surface andf(x) = T ′(x) if x is on the aerofoil surface. Equating real andimaginary parts shows that

m(x) =f(x)

π, u(x, 0+) =

∫−∞

−∞

m(ξ)

x− ξ dξ.

Hence substituting for m(x) shows that the given U(z) satisfiesthe boundary conditions on the aerofoil surface by construction.

79 / 190


Also

u1(x, 0+) = φ1x(x, 0+) =1

π

∫−

1

0

T ′(ξ)

x− ξ dξ.

A similar argument can be performed for y → 0− by deformingthe contour above the point ξ = x. Hence φ1(x, y) given by (49)satisfies (48).Now the surface speed us(x)is given by

us(x) = u(x, 0±) = 1 + εφ1x(x, 0±) +O(ε2).

80 / 190


Thus

us(x) = 1 +ε

π

∫−

1

0

T ′(s)

(x− s) ds. (50)

Here the notation∫−b

ameans the Cauchy Principal value integral

∫−b

a

f(s)

x− s ds = limδ→0

∫ x−δ

a

f(s)

x− s ds+

∫ b

x+δ

f(s)

(x− s) ds

when a < x < b and f(x) 6= 0. The integral does not convergeas a normal integral.

81 / 190


We will now use the above results to look at a few examples.

82 / 190

Example- Parabolic circular arc aerofoil

Consider a parabolic circular arc aerofoil with

T (x) = x− x2.

Let us calculate

I =1

π

∫−

1

0

T ′(s)

x− s ds =1

π

∫−

1

0

1− 2s

x− s ds.

83 / 190


Now

I =1

π

[∫−

1

0

1− 2x

x− s ds+

∫−

1

0

2x− 2s

x− s ds

],

=1

π

[−(1− 2x) lim

δ→0

[log |x− s|]x−δ0 + [log(|x− s|)]1x+δ

]+

2

π,

= − 1

π(1− 2x) log |x− 1

x|+ 2

π.

Hence

us(x) = 1 + ε

[2

π+

(1− 2x)

πlog | x

1− x |]

+O(ε2).

84 / 190


Clearly there are problems near x = 0 and x = 1. We can seethat the expansion is not uniformly valid close to these points.In fact for x→ 0 the O(ε) term in the expansion for us iscomparable to the first term when

−1 ∼ ε

πlog |x|

ie when x ∼ e−π/ε.

85 / 190

Example Elliptic aerofoil

Consider an elliptic aerofoil

T (x) = (x− x2)1/2.

ThenT ′(x) =

1

2

(1− 2x)

(x− x2)1/2.

For the surface speed we need to evaluate

I =1

π

∫−

1

0

T ′(s)

x− s ds =1

2π

∫−

1

0

(1− 2s)

(x− s)(s− s2)1/2ds.

The trick in evaluating this integral is to make the followingsubstitution. Put s = 1

2(1 + cos θ). Then ds = −12 sin θ, and when

s = 0, θ = π and when s = 1, θ = 0. Also

s(1−s) =1

2(1+cos θ)(1−1

2−1

2cos θ) =

1

4(1+cos θ)(1−cos θ) =

1

4sin2 θ.

86 / 190


Hence

I =1

2π

∫−

0

π

(− cos θ)(−12 sin θ)

12 sin θ(x− 1

2(1 + cos θ))dθ,

= − 1

π

∫−π

0

cos θ

(2x− 1− cos θ)dθ,

=1

π

∫−π

0

2x− 1− cos θ − (2x− 1)

(2x− 1− cos θ)dθ,

=1

π

[π − (2x− 1)

∫−π

0

dθ

(2x− 1)− cos θ.

].

87 / 190


Now ∫−π

0

dθ

a+ b cos θ=

0 if a2 < b2π√a2−b2 if a2 > b2

.

In our case (2x− 1)2 < 1 implies 4x(x− 1) < 0 and hence0 < x < 1 then ∫

−π

0

dθ

(2x− 1)− cos θ= 0.

88 / 190


So for 0 < x < 1

I =1

2π

∫−

1

0

(1− 2s)

(x− s)(s− s2)1/2ds =

1

π[π − 0] = 1.

If x < 0 or x > 1 then (2x− 1)2 > 1 and∫−π

0

dθ

(2x− 1)− cos θ=

π√(2x− 1)2 − 1

=π

2√x(x− 1)

.

89 / 190


Hence

I =1

2π

∫−

1

0

(1− 2s)

(x− s)(s− s2)1/2ds =

1

π

[π − π(2x− 1)

2√x(x− 1)

].

Thus from (50) the surface speed for an elliptic aerofoil is givenby

us(x) =

1 + ε

[1− 2x−1

2√x(x−1)

]x < 0 or x > 1,

1 + ε 0 < x < 1. (51)

90 / 190


The surface speed is constant on the aerofoil and this issomewhat unusual.Notice from (51) that as x→ 0− just before the leading-edge orx→ 1+ just after the trailing edge, the approximation becomesnon-uniform. In fact the O(ε) term in (51) become comparablewhen either x = O(ε2) or when x− 1 = O(ε2). This suggeststhat we need to examine the flow near the leading (and trailing)edge.

91 / 190

Solution for non-uniformity near leading edge

• The non-uniformities in the solution near the leading edgeare typical of other round nosed aerofoils.

• The discussion above suggests examining the regionwhere x = O(ε2) near nose of the aerofil.

92 / 190


Hence introduce a scaled O(1) variable X with x = ε2X.Now for the elliptic wing we have

y = ±ε(x− x2)1/2 = ±ε2(X − ε2X2)1/2.

Thus close to the leading edge the local shape is given by

y = ±ε2X1/2.

93 / 190


This suggests that we put y = ε2Y where Y = O(1). Then (45)suggests that

φ = ε2Φ0(X,Y ) + . . . .

In terms of the rescaled variables the govering equation

∂2φ

∂x2+∂2φ

∂y2= 0,

becomes∂2Φ

∂X2+∂2Φ

∂Y 2= 0. (52)

94 / 190


The boundary condition on the aerofoil surface (48b) becomes

ε−2ε2∂Φ

∂Y(X,Y ) = ±ε−2ε2

∂Φ

∂X(X,Y )F ′(X)

on Y = ±F (X) = ±X1/2,

i.e.,∂Φ

∂Y(X,Y ) = ± ∂Φ

∂XF ′(X) on Y = ±F (X). (53)

95 / 190


Far upstream since as x→ 0−

φx ∼ 1 + ε−1

2√−x +O(ε2),

this implies that

ΦX ∼ 1− 1

2√−X

, as X2 + Y 2 →∞. (54)

96 / 190


Here the boundary conditions (53) cannot be simplied as in thethin aerofoil case.We need to solve for the full potential flow past a parabola. Thiscan be done using complex variable theory and conformalmapping.

97 / 190


LetW (z) = Φ + iΨ

be a complex function of Z = X + iY . We need to find a W (Z)such that =(W (z)) = Ψ(X,Y ) is constant onY = ±F (X) = ±X1/2 and which tends to a uniform flow farupstream.The complex velocity is given by

dW

dZ=∂Φ

∂X− i ∂Φ

∂Y.

98 / 190


We will use conformal mapping and introduce the mappingfunction Z = ζ2 +A with ζ = ξ + iη. Then

Z = X + iY = ξ2 − η2 + 2iξη +A.

Equating real and imaginary parts gives

X = ξ2 − η2 +A, Y = 2ξη.

99 / 190


We will choose A so that when X = Y 2 then η = η0 so that theparabola is mapped to a straight line in the ξ − η plane,So this gives

X = ξ2 − η20 +A = Y 2 = 4ξ2η2

0.

X

YZ−plane

ξ

ηζ−plane

Y = X1/2η = η0

Figure: Sketch of the Z− and ζ− planes. The parabola Y = ±X1/2 ismapped to η = η0 = 1/2

100 / 190


We can choose η20 = 1/4 and η2

0 = A = 1/4 giving η0 = 1/2.Hence

Z2 = ζ2 +1

4, ζ = (Z − 1

4)1/2.

We choose a branch cut such that when Z = x < 0 then(Z − 1

4)1/2 = iη, η > 12 .

Note that for large Z we require

W (Z) ∼ Z, W ∼ ζ2.

101 / 190


Also we require =(W ) to be constant on η = 1/2. Hence try

W = ζ2 + (C + iD)ζ = (ξ + iη)2 + (C + iD)(ξ + iη).

When η = η0 = 1/2 then

=(W ) = 2ξη0 + Cη0 +Dξ = (2η0 +D)ξ + Cη0.

We can choose D = −2η0 = −1 and C = 0. Thus

W = ζ2 − iζ

is the required function.

102 / 190


The complex velocity is

dW

dZ=dW

dζ

dζ

dZ=

(2ζ − i)2ζ

,

= 1− i

2ζ= 1− i

2(Z − 14)1/2

.

Hence on Z = x(< 0) we have

dΦ

dX= 1− 1

2(14 −X)1/2

.

103 / 190


There is now no singularity at X = 0. In terms of x = ε2X wehave

dφ

dx= 1− 1

2(14 − ε−2x)1/2

= 1− ε

2( ε2

4 − x)1/2.

For X >> 1 (ie ε−2x >> 1, x >> ε2)

dφ

dx∼ 1− ε

2(−x)1/2

as before. This matches with the earlier solution as x→ 0−.Note that at the stagnation point dφ/dx = 0 as expected.

104 / 190

High Reynolds number flows- boundary layer theory.

• Intuitively one might think that since 1/Re multiplying theviscous terms in the Navier-Stokes equations is small, theeffects of viscosity may be neglected.

• Prandtl (1904) in his seminal paper was the first to suggestthat viscous effects no matter how small become importantnear solid surfaces and also when the flow is changingrapidly. This led to the development of boundary layertheory.

• We start first with the classical problem of flow past a flatplate aligned with the oncoming uniform stream.

105 / 190

Prandtl (1904) boundary layer theory

Consider two-dimensional steady flow past a flat plate alignedwith a oncoming uniform stream. We will assume that the fluidis incompressible and take non-dimensional cartesiancoordinates (x, y) as shown in figure 8. In nondimensionalterms the plate is of length unity and the oncoming stream isU∞ = (1, 0).

x

y

O

U∞

1

Figure: Flat plate aligned with oncoming uniform stream106 / 190


The continuity and Navier-Stokes equations innon-dimensiomal form are given by

∂u

∂x+∂v

∂y= 0, (55)

u∂u

∂x+ v

∂u

∂y= −∂p

∂x+

1

Re

(∂2u

∂x2+∂2u

∂y2

), (56)

u∂v

∂x+ v

∂v

∂y= −∂p

∂y+

1

Re

(∂2v

∂x2+∂2v

∂y2

). (57)

107 / 190


HereRe =

LU∞ν

is the Reynolds number and L is the length of the plate and U∞the speed far upstream. Also (u, v) are the velocity componentsin cartesian coordinates (x, y) and p is the pressure.

108 / 190


We also need suitable boundary conditions and these are theno-slip conditions

u = v = 0 on y = 0±, x ∈ [0, 1], (58)

and conditions of matching with the oncoming uniform streamexpressed by

u→ 1v → 0p→ 0

as x2 + y2 →∞. (59)

109 / 190


We will seek a solution of the Navier-Stokes equations (103)with

x = O(1), y = O(1), Re→∞,in the form

u(x, y, ;Re) = u0(x, y) + . . . , v(x, y, ;Re) = v0(x, y) + . . . ,

p(x, y;Re) = p0(x, y) + . . . .(60)

x

y

O1

Region 1

Figure: Outer inviscid region 1.

110 / 190


Substitution of (60) into the (103) leads to the Euler equations

∂u0

∂x+∂v0

∂y= 0, (61a)

u0∂u0

∂x+ v0

∂u0

∂y= −∂p0

∂x, (61b)

u0∂v0

∂x+ v0

∂v0

∂y= −∂p0

∂y. (61c)

111 / 190


• Notice that in going from the Navier-Stokes equations tothe Euler equations we have lost the highest derivativeterms in the equations.

• The Euler equations (61) cannot be solved with all theboundary conditions given in (58,59) and we need to dropone of the conditions.

• Inviscid flow theory would suggest that we can relax thecondition u = 0 on the flat plate.

112 / 190


Boundary conditions for Euler equations are thus

u→ 1v → 0p→ 0

as x2 + y2 →∞.

together with

v0 = 0 on y = 0±, x ∈ [0, 1]. (62)

113 / 190


It is easy to see that

u0 = 1, v0 = 0, p0 = 0, (63)

is a solution to the problem (61)-(62).The solution (63) does not satisfy the no-slip condition (58).Hence we need to reconsider the region in the vicinity of theplate. This is called the boundary layer region after Prandtl(1904),

114 / 190


In this thin region we will assume that

y = δ(Re)Y, (64)

with δ(Re) << 1 is to be found.

x

y

O1

Region 1

Region 2

Figure: Outer inviscid region 1 and boundary layer region 2.

115 / 190


We will seek a solution of the Navier-Stokes equations (103)with the limiting procedure

x = O(1), Y = δ−1y = O(1), Re→∞.

Since the outer flow velocity is (1, 0) it suggests the expansion

u(x, y, ;Re) = U0(x, Y ) + . . . , v(x, y;Re) = σ(Re)V0(x, Y ) + . . . ,

p(x, y, ;Re) = λ(Re)P0(x, Y ) + . . . . (65)

Here the coefficients in the expansion σ(Re) and λ(Re) need tobe determined.

116 / 190


Firstly the continuity equation shows that

∂U0

∂x+σ(Re)

δ(Re)

∂V0

∂Y= 0.

This suggests that we need to take

σ(Re) = δ(Re), (66)

and the continuity equation reduces to

∂U0

∂x+∂V0

∂Y= 0. (67)

117 / 190


Next the x− momentum equation gives

U0∂U0

∂x+ V0

∂U0

∂Y= −λ(Re)

∂P0

∂x+

1

Re

∂2U0

∂x2+

1

δ2Re

∂2U0

∂Y 2. (68)

Clearly the second term on the right-hand side of (68) becomessmall as Re→∞. The third term will be comparable with theinertial terms provided

δ(Re) = Re−1/2. (69)

118 / 190


Finally the y− momentum equation (??) gives

U0∂V0

∂x+ V0

∂V0

∂Y= −λ(Re)

Re−1

∂P0

∂x+

1

Re

∂2V0

∂x2+∂2V0

∂Y 2, (70)

We can see that provided λ >> Re−1 the dominant balancegives

∂P0

∂Y= 0. (71)

119 / 190


SoP0(x, Y ) = P0(x)

iet the pressure not change in the boundary layer.if we match the pressure as we leave the boundary layer withY →∞ in region 2 and y → 0± in region 1 then we obtain

P0(Y →∞) = 0.

120 / 190


Hence P0 = 0 and (68) reduces to

U0∂U0

∂x+ V0

∂U0

∂Y=∂2U0

∂Y 2. (72)

121 / 190


We need suitable boundary conditions. Notice that since theequation is parabolic we need

U0 = 1 at x = 0, Y ∈ [0,∞).

As far as boundary conditions in Y are concerned we requirethe no-slip conditions on the plate and so

U0 = 0, V0 = 0 at Y = 0, x ∈ [0, 1]. (73)

In addition we require matching with the outer flow in region 1and hence

U0 → 1 as Y →∞.

122 / 190


In summary then the analysis of the boundary layer regionrequires the solution of the problem

∂U0

∂x+∂V0

∂Y= 0, (74a)

U0∂U0

∂x+ V0

∂U0

∂Y=∂2U0

∂Y 2, (74b)

and boundary and initial conditions

U0 = 1 at x = 0, Y ∈ [0,∞), (74c)

U0 = 0, V0 = 0 at Y = 0, x ∈ [0, 1], (74d)

U0 → 1 as Y →∞. (74e)

123 / 190

Blasius solution

A self-similar solution to the boundary layer equations ispossible.It is convenient to introduce the streamfunction Ψ0 by

U0(x, Y ) =∂Ψ0

∂Y, V0(x, Y ) = −∂Ψ0

∂x,

so that the continuity equation is indentically satisfied.Suppose we make the ‘ansatz’ that

U0(x, Y ) = f ′(η), η = Y/g(x), (75)

where f(η) and g(x) are functions to be found and primesdenote differentiation with respect to the argument.

124 / 190

Blasius solution

We regard the above as a change of variables from (x, Y ) to(x, η) and using the chain rule

∂

∂x=

∂

∂x+∂η

∂x

∂

∂η,

∂

∂Y=∂η

∂Y

∂

∂η.

Hence we have

Ψ0(x, Y ) = g(x)f(η) + ψ0(x),

and

V0 = −∂Ψ0

∂x= −g′(x)f(η) + g′(x)ηf ′(η)− ψ′0(x). (76)

125 / 190

Blasius solution

Given that U0 = V0 = 0 when Y = 0 and hence when η = 0 werequire that

f(0) = f ′(0) = 0, ψ0(x) = 0.

Substituting for U0, V0 from (75) into (74b) leads to

f ′(−ηg

)f ′′ − (g′f − ηg′f ′)1

gf ′′ =

1

g2f ′′′,

126 / 190

Blasius solution

f ′′′ + gg′ff ′′ = 0. (77)

In order for (77) to be an ordinary differential equation we needto set

gg′ = α,

where α is a constanst. Hence

g(x) = (2αx)1/2. (78)

127 / 190

Blasius solution

The boundary conditions in (74) imply that

f(0) = f ′(0) = 0, f ′(∞) = 1.

Hence a self-similar solution to (74) in the form (75, 76) ispossible provided g(x) is given by (78). The choice α = 1/2leads to

Ψ0(x, Y ) =√xf(η), η =

Y√x

U0(x, Y ) = f ′(η), V0(x, Y ) = − 1

2√x

(f(η)− ηf ′(η)), (79)

128 / 190

Blasius solution

The function f(η) satisfies the Blasius equation

f ′′′ +1

2ff ′′ = 0. (80a)

with boundary conditions

f(0) = f ′(0) = 0, f ′(∞) = 1. (80b)

129 / 190

Blasius solution - properties

First note that for η → 0 we have

f(η) ∼ 1

2λη2,

where λ = f ′′(0) = 0.3321 from the numerical solution. Since

f ′′′ = −1

2ff ′′

we see that for η → 0

f ′′′ ∼ −1

4λ2η2,

and hence

f(η) ∼ 1

2λη2 − λ2

2.5!λ5 + . . . , η → 0.

130 / 190


Next for large η the boundary conditions suggest that

f(η) ∼ η + . . . .

We can seek additional terms by putting

f(η) = η +Aηβ + . . . as η →∞. (81)

Here A is a constant and we require β < 1. If we substitute (81)into (80a) we find that

Aβ(β − 1)(β − 2)ηβ−3 = −1

2Aβ(β − 1)ηβ−1.

131 / 190


For η >> 1 the right-hand side of this equation is much largerthan then left-hand side and so to leading order

Aβ(β − 1) = 0.

If we insist that A 6= 0 then the only possibilities are β = 0, orβ = 1. Since we required β < 1 we have to choose β = 0 andthen

f(η) ∼ η − η0 + . . . as η →∞.Numerical solution of Blasius’s equation shows thatη0 = 1.21678.

132 / 190


0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 60

1

2

3

4

5

φ′(η) φ(η)

η

Figure: Numerical solution of the Blasius equation with the solid lineshowing φ′(η) and the dashed line φ(η).

133 / 190

BL eqns for flow over a curved surface

• In the previous section we saw that the boundary layerequations for flow past a flat plate reduce to a form inwhich the pressure gradient terms was absent.

• In more general flow past bluff bodies the pressuregradient term is no longer zero and plays an important rolein the development of the boundary layer.

134 / 190


Consider the steady uniform flow of an incompressible fluidpast a fixed bluff body.

U∞i

Figure: Uniform flow past a bluff body.

135 / 190


As far as the outer solution is concerned this is still governed bythe Euler equations

∂u0

∂x+∂v0

∂y= 0, (82a)

u0∂u0

∂x+ v0

∂u0

∂y= −∂p0

∂x, (82b)

u0∂v0

∂x+ v0

∂v0

∂y= −∂p0

∂y. (82c)

136 / 190


The boundary conditions are those of zero normal velocity onthe body expressed by

(u0, v0).n = 0 on the body surface, (83)

where n is the normal to the surface of the body. In addition wealso require that

(u0, v0)→ i as x2 + y2 →∞.

137 / 190


• The Euler equations will give rise to a slip velocity sayUe(s) on the surface of the body, where s say measuresdistance along the body.

• To satisfy the full boundary conditions we need to introducea boundary layer. We will take orthogonal coordinatesx1, y1 with corresponding velocity components (u, v).

138 / 190


Now an element of length ds is given by

ds2 = h21dx

21 + h2

2dy21,

where h1, h2 are the scale factors. Suppose we take x1 to bedistance along a curved wall with radius of curvature R(x1) andy1 distance normal to the wall.Consider a point P at location x1, y1 and suppose we take aneighbouring point Q displaced by dx1 and dy1, see figure 13).Now MN represents a line with y1 fixed and x1 changing andso the increment dx1 is given by PS. Similarly if we keepx+ dx1 fixed and vary y then SQ is the increment dy1.

139 / 190


Nowds2 = PS2 + SQ2.

From similar trianglesPS

MN=

PO

MO=R(x1) + y1

R(x1).

P dy1

Q

y1

M Nx1

R(x1)

S

O

Figure:

HencePS = (1 + κ(x1)y1)dx1,

where κ(x1) = 1/R(x1) is the curvature, see figure 14.

140 / 190


Henceds2 = (1 + κ(x1)y1)2dx2

1 + dy21.

This shows that the scale factors are given by

h1 = 1 + κ(x1)y1, h2 = 1.

κ > 0 κ < 0

Figure: Sketch showing negative and positive curvature κ(x).

141 / 190


In order to write down the boundary layer equations we firstneed the Navier-Stokes equations in orthogonal curvilinearcoordinates. Starting from the equations in the form

∇(1

2u2) + ω × u = −∇p+

1

Re[∇(∇.u)−∇× ω], (84)

whereω = ∇× u = (0, 0,Ω),

is the vorticity and

Ω =1

h1h2

[∂

∂x1(h2v)− ∂

∂y1(h1u)

].

142 / 190


The equations may be written as

∂

∂x1(h2u) +

∂

∂y1(h1v) = 0, (85a)

−vΩ = − 1

h1

∂

∂x1

(p+

1

2(u2 + v2)

)− 1

h2Re

∂Ω

∂y, (85b)

uΩ = − 1

h2

∂

∂y1

(p+

1

2(u2 + v2)

)+

1

h1Re

∂Ω

∂x1. (85c)

143 / 190


Next we introduce the boundary layer scalings so that

y1 = Re−1/2Y, x1 = x, h1 = 1 + κ(x)Re−1/2Y,

u = U0 + . . . , v = Re−1/2V0 + . . . , P = P0 + . . . .

Substituting into (85) leads to

144 / 190


∂U0

∂x+

∂

∂Y(h1V0) + · · · = 0, (86a)

1

h1

[U0∂U0

∂x+ V0

∂(h1U0)

∂Y

]=

− 1

h1

∂P0

∂x1+

∂

∂Y

[1

h1

∂

∂Y(h1U0)− 1

h1Re

∂V0∂x

]+ . . . , (86b)

1

Re1/2

[U0

h1

∂V0∂x

+V0

Re1/2∂V

∂Y

]− U2

0

h1

κ

Re1/2=

−Re1/2 ∂P0

∂Y+

1

h1Re

∂

∂x

[1

Re1/2h1

∂V0∂x− 1

h1

∂

∂Y(h1U)

]. (86c)

145 / 190


Taking the limit Re→∞ and assuming that

h1 → 1,∂h

∂x→ 0,

∂h

∂Y→ 0,

then (86) reduce to

∂U0

∂x+∂V0

∂Y= 0, (87)

U0∂U0

∂x+ V0

∂U0

∂Y= −∂P0

∂x+∂2U0

∂Y 2. (88)

0 = −∂P0

∂Y. (89)

146 / 190


In addition to the no slip conditions expressed by

U0 = V0 = 0 on Y = 0, (90)

we have the matching condition with the outer flow given by

U0 → Ue(x) as Y →∞. (91)

147 / 190


• The equations (86) are known as the Prandtl boundarylayer equations.

• We note that the normal pressure gradient is unchanged inthe boundary layer and if we let Y →∞ in (88) we obtain

UedUedx

= −dP0

dx.

148 / 190

Displacement thickness and skin friction

The displacement thickness δ∗1(x′) is defined by

δ∗1(x′) =

∫ ∞0

(1− u′(x′, y′)

U ′e(x′)

)dy′ = LRe−1/2δ1(x)

where in non-dimensional terms

δ1(x) =

∫ ∞0

(1− U0(x, Y )

Ue(x)

)dY. (92)

149 / 190


Here Ue(x) is the velocity at the edge of the boundary layer.The quantity δ1(x) gives a measure of the ‘thickness’ of theboundary layer. We can rewrite (92) by replacing U0 in terms ofthe streamfunction by U0 = ∂Ψ0/∂Y to get

δ1(x) = limY→∞

[∫ Y

0

(1− Ψ0Y (x, Y )

Ue(x)

)dY

],

and soδ1(x) ∼ Y − Ψ(x, Y )

Ue(x)for Y >> 1.

ThusY ∼ Ψ0

Ue(x)+ δ1(x).

This shows that the streamlines given by Ψ0 = constant aredisplaced upwards by an amount δ1(x), and hence the termdisplacement thickness.

150 / 190


Now the normal and tangential components of shear stress aregiven by

σ′x′x′ = −p′ + 2µ∂u′

∂x′, σ′y′y′ = −p′ + 2µ

∂v′

∂y′.

Recall that σij is the i component of the force per unit areaexerted across a plane surface normal to the j direction. If wenon-dimensionalise we get

σ′x′x′

ρU2∞

= −p+2

Re

∂U0

∂x∼ −p for Re >> 1,

andσ′y′y′

ρU2∞

= −p+2

Re

∂V0

∂Y∼ −p for Re >> 1.

151 / 190


Also the shear stress exerted by the fluid on surfaces parallel tothe wall is defined by

σ′x′y′ = µ

[∂u′

∂y′+∂v′

∂x′

].

Non-dimensionalising gives

σ′x′y′ =Re1/2µU∞

L

[∂U0

∂Y+

1

Re

∂V0

∂x

].

For Re >> 1 the second term is negligble. We define the skinfriction τw(x) by

τw(x) =Lσ′x′y′

Re1/2µU∞=

(∂U0

∂Y

)Y=0

. (93)

152 / 190

Displacement thickness and skin friction- BlasiusSolution

For the Blasius solution with Ue = 1 and

U0(x, Y ) = f ′(η), η = Y/√x, (94)

where f(η) satisfies Blasius’s equation (80a) we have

δ1(x) =

∫ ∞0

(1− U0(x, y)) dY =√x

∫ ∞0

(1− f ′(η)) dη.

Henceδ1(x) = η0

√x = 1.7208

√x.

153 / 190


For the Blasius flow the scaled skin friction is given by

τw =

(∂U0

∂Y

)Y=0

= x−1/2f ′′(0).

Hence the dimensional skin friction is

τ ′w(x′) =Re1/2µU∞

Lx−1/2f ′′(0),

= ρU∞ν

L

(U∞L

ν

)1/2 1√xf ′′(0),

= ρU2∞

(ν

U∞x′

)−1/2

f ′′(0).

154 / 190


The drag on one side of a plate of length L and unit width isdefined by

D =

∫ L

0τ ′w(x′) dx′ = Re1/2µU∞

∫ 1

0τw(x) dx,

= Re1/2µU∞f′′(0)

∫ 1

0

1√xdx,

= Re1/2µU∞f′′(0)2

√x.

The drag coefficient is defined by

CD =D

12ρU

2∞L

,

=4Re1/2µU∞f

′′(0)√x

ρU2∞L

= 4f ′′(0)Re−1/2 = 1.3282Re−1/2.

155 / 190

Goldstein (1930) near wake solution

• Consider again the flow past a finite flat plate.• In the outer region where

x = O(1), y = O(1), Re→∞,

the governing equations are the Euler equations.• The solution to the leading order problem gave

u0 = 1.

• This solution does not satisfy the no-slip conditions on theplate

• In the inner region, the boundary layer

x = O(1), Y = yRe1/2 = O(1), Re→∞.

the governing equations are the Prandtl boundary layerequations.

156 / 190


For the flat plate flow there is a self-similar solution given byBlasius’s solution in which

U0 = f ′B(η), η =Y√x,

and fB(η) satisfes the Blasius equation

f ′′′B +1

2fBf

′′B = 0, fB(0) = f ′B(0) = 0, f ′B(∞) = 1.

157 / 190

Goldstein (1930) wake solution

• The boundary layer equations are parabolic and so theboundary layer does not see the trailing-edge of the finiteplate.

• The trailing edge the solution is

U0 = f ′B(Y ).

• The question is what happens beyond the trailing-edge?Just behind the trailing-edge we have a change inboundary conditions and therefore the viscous terms willbe important in the vicinity of the trailing-edge.

158 / 190


The boundary layer equations

∂U0

∂x+∂V0

∂Y= 0, (95a)

U0∂U0

∂x+ V0

∂U0

∂Y=∂2U0

∂Y 2, (95b)

now need to be solved subject to the conditions

V0 =∂U0

∂Y= 0 at Y = 0, x > 1, (95c)

U0 → 1 as Y →∞, (95d)

and initial condition

U0 = f ′B(Y ) at x = 1. (95e)

159 / 190


Goldstein (1930) showed that the boundary layer near thetrailing-edge divides into two distinct regions, region 2a andregion 2b as shown in figure 15.

O s

y

1

1

2 2b2a

Figure: Flow structure near trailing-edge of the flat plate where theboundary splits into two regions.

160 / 190


If we set s = x− 1 then for s = 0− we note that

U0 = f ′B(Y ) ∼ λY + . . . , η << 1,

where λ = f ′′(0) = 0.332. For small Y a balance of theconvective and viscous terms shows that

U0∂U0

∂x∼ ∂U0

∂Y 2, =⇒ U2

0

s∼ U0

Y 2, =⇒ Y 2

s∼ Y

Y 2

givingY ∼ O(s

13 ).

161 / 190


This suggets that in region 2a we can look for a solution in theform

Ψ0 ∼ s2/3g(ξ) + . . . , U0 = s13 g′(ξ) + . . . , (96)

whereξ =

Y

(x− 1)13

=Y

s13

= O(1).

162 / 190


Substitution of (96) into (95) and using

∂

∂x=∂s

∂x

∂

∂s+∂ξ

∂x

∂

∂ξ=

∂

∂s− ξ

3s

∂

∂ξ,

∂

∂Y=

∂s

∂Y

∂

∂s+∂ξ

∂Y

∂

∂ξ=

1

s13

∂

∂Y,

163 / 190


shows that

s1/3g′(ξ)

[s−2/3

3g′(ξ)− ξ

3s−2/3g′′(ξ)

]−[

2

3s−1/3g(ξ)− ξ

3s−1/3g′

]g′′

= s−1/3g′′′(ξ) + . . . .

At leading order we find

g′2

3− 2

3gg′′ = g′′′. (97)

164 / 190


The boundary conditions we need are those of symmetry onthe wake centreline

g(0) = g′′(0) = 0.

In addition we must match with the flow outside region 2a asξ >> 1. Since for Y small we have

U0 ∼ λY + . . . ,

U0 ∼ λs1/3ξ + . . . .

Hence from (96) we see that for ξ >> 1 we require

g(ξ) ∼ λ

2ξ2 + . . . for ξ >> 1.

165 / 190


The behaviour of the function g(ξ) for large ξ can be deducedfrom the equation (97). If we write

g(ξ) =λ

2ξ2 +Aξα + . . . ξ >> 1,

Then (97) shows that

Aξα−3α(α− 1)(α− 2) +1

3λ2ξ2 +

2

3Aλξα+

1

3Aλα(α− 1)ξα − 1

3λ2ξ2 − 2

3Aλαξα + · · · = 0.

166 / 190


Since ξ >> 1 the dominant balance gives

Aλ(α2 − 3α+ 2)ξα = 0.

Hence(α− 2)(α− 1) = 0.

Since we require α < 2 this means that α = 1.

g′2

3− 2

3gg′′ = g′′′, (98a)

g(0) = g′′(0) = 0, g(ξ) ∼ λ

2ξ2 + . . . ξ >> 1, (98b)

167 / 190


A numerical solution of the problem (98) shows that

g(ξ) ∼ λ

2ξ2 + cξ ξ >> 1 (99)

wherec = 1.288λ.

Hence using (96) we see that

Ψ0 = s2/3g(ξ) + . . . ∼ s2/3(λ

2ξ2 + cξ) + . . . , as ξ →∞,

giving

Ψ0 ∼λ

2Y 2 + s1/3cY + . . . as Y → 0 + . (100)

168 / 190


This suggests that in region 2b where Y = O(1) and s→ 0+ weset

Ψ0 = fB(Y ) + s1/3f1(Y ) + . . . , (101a)

U0 = f ′B(Y ) + s1/3f ′1(Y ) + . . . , (101b)

where fB(Y ) is the Blasius function. Thus (101) represents aperturbation to the oncoming Blasius solution.

169 / 190


If we substitute (101) into (95) then we obtain[f ′B + s1/3f ′1 + . . . ][

1

3s−2/3f ′1 + . . .

]− 1

3s−2/3f1f

′′B + · · · = f ′′′B .

The leading order problem is

1

3f ′Bf1 −

1

3f1f′′B = 0.

Hence

f ′2B

(f1

f ′B

)′= 0,

givingf1 = Df ′B(Y ).

170 / 190


Hence

Ψ0 ∼λ

2Y 2 + s1/3DλY + . . . as Y → 0 + .

Matching with region 2a as Y → 0 and comparing with (100)shows that

λD = c =⇒ D = 1.288.

SinceΨ0 = fB(Y ) + s1/3Df ′B(Y ) + . . . ,

andfB(Y )→ Y − η0 + . . . ,

where η0 = 1.21678 we see that the displacement thicknessbehaves like

171 / 190


β(x) ∼ β(1)−Ds1/3 + . . . , as s→ 0+,

i.e.β(x) ∼ β(1)−D(x− 1)1/3 + . . . as x→ 1+,

where β(1) = 1.7208. Also

β(x) ∼ 1.7208√x+ . . . as x→ 1− .

172 / 190


Clearly there is a discontinuity in the slope of the displacementthickness as x→ 1. The acceleration of fluid particles neartrailoing-edge is so sharp that this is the reason why the slopeof the displacement thickness is infinite at the trailing edge,

Figure: Displacement thickness near the trailing-edge of the flat plate.

173 / 190


Next note that in the boundary layer region just before thetrailing-edge the solution takes the form

u(x, y, ;Re) = U0(x, Y )+. . . , v(x, y;Re) = Re−1/2V0(x, Y )+. . . ,

where

V0 = − 1

2√x

(fB(η)− ηf ′B(η)), η =Y√x. (102a)

In the region 2b just after the trailing edge edge we have

V0(x, Y ) =1

3(x− 1)−2/3Df ′B(Y ) + . . . . (102b)

174 / 190


We observe that the normal velocity in the vicinity of thetrailing-edge as we leave the boundary layer behaves like

v(x, y : Re) ∼ Re−1/2 1

2η0 as x→ 1−,

and

v(x, y : Re) ∼ Re−1/2 1

3(x− 1)−2/3D as x→ 1 + .

175 / 190


• The normal velocity component becomes very large in thevicinity of the trailing-edge which contradicts theassumptions made in boundary layer theory.

• That there is a drastic change in flow properties in theimmediate vicinity of the trailing-edge may also bededuced from the second order solution in the inviscidregion outside the boundary layer.

176 / 190


The solution in region 1 may be expanded in the form

u(x, y;Re) = 1 +Re−1/2u1(x, y) + . . . , (103a)

v(x, y;Re) = Re−1/2v1(x, y) + . . . , (103b)

p(x, y;Re) = Re−1/2p1(x, y) + . . . . (103c)

177 / 190


Substitution of (103) into the Navier-Stokes equations gives thefollowing set of equations

∂u1

∂x+∂v1

∂y= 0,

∂u1

∂x= −∂p1

∂x,

∂v1

∂x= −∂p1

∂y. (104)

The equations are just the same as those for thin aerofoiltheory and lead to

∇2u1 = ∇2v1 = ∇2p1 = 0.

The free-stream boundary conditions are that

u1 → 0, v1 → 0, p1 → 0 as x2 + y2 →∞. (105)

178 / 190


• We need one more boundary condition.• Whereas in thin aerofoil theory we used the condition of

impermeability on the body surface, here instead of a bodysurface we have a modified body shape because ofdisplacement thickness effects.

• Matching the normal component of the velocity in region 1as y → 0± with the boundary layer solution in region 2 asY →∞. implies

v1(x, y → 0±) = ±G(x) x ∈ [0,∞), (106)

whereG(x) =

1

2√xη0 x ∈ [0, 1], (107)

and

G(x) ∼∼ 1

3(x− 1)−2/3D as x→ 1+, (108)

in the Goldstein wake region.179 / 190


The solution to the problem (??) can be obtained usingcomplex variable theory. Since p1, v1 satisfy theCauchy-Riemann equations, let

W (z) = p1(x, y) + iv1(x, y)

be a function of the complex variable z = x+ iy. We require ananlytic function W (z) such that

=(W (z)) = ±G(x) on y = 0±,

andW (z)→ 0 as z →∞.

180 / 190


We can deduce the local behaviour of p(x, y) in the vicinity ofthe trailing-edge as follows.The inviscid flow in the vicinity of the trailing-edge may beconsidered independently of the rest of the flow in the inviscidregion. In fact we have

=(W (z))|y=0 −

A−2 + . . . as x→ 1−,

−A+

3 (x− 1)−2/3 + . . . as x→ 1+,(109)

where A− = η0 and A+ = D.

181 / 190


The imaginary part of the function W (z) remains finite on theplate but is singular just downstream. This suggests that

W (z) = C(z − 1)−2/3 + . . . , as z → 1. (110)

Here C = Cr + iCi is a complex constant. If we put z = 1 + reiθ

then we can write (110) as

W (z) = (C1 + iCi)r−2/3e−i

23θ. (111)

182 / 190


Downstream of the trailing-edge we have θ = 0, andr = x− 1 = s. Thus

W (z) = p1(x, 0) + iv1(x, 0) = (Cr + iCi)s−2/3. (112)

The imaginary parts give

Ci = −A+

3.

Next on the plate we have θ = π, r = 1− x and then

W (z) = (Cr + iCi)(1− x)−2/3e−23iπ

= (Cr + iCi)

(cos

2

3π − i sin

2

3π

)(1− x)−2/3 + . . . .

183 / 190


The real and imaginary parts give

p1(x, 0) =

(Cr cos

2π

3+ Ci sin

2π

3

)(1− x)−2/3 + . . . , (113)

v1(x, 0) =

(Ci cos

2π

3− Cr sin

2π

3

)(1− x)−2/3 + . . . . (114)

184 / 190


Since from (109) the imaginary part of W (z) is finite as x→ 1−we must have

Ci cos2π

3− Cr sin

2π

3= 0.

This gives

Cr =Ci

tan 2π3

=A+

3√

3.

185 / 190


Thus from (113) we find that on the flat plate just upstream ofthe trailing-edge we have

p1(x, 0) = −2A+

3√

3(1− x)−2/3 + . . . as x→ 1− . (115a)

Immediately downstream of the trailing-edge we have

p1(x, 0) =A+

3√

3(x− 1)−2/3 + . . . as x→ 1 + . (115b)

186 / 190

Triple-deck scales

• The expressions (115) show that the induced pressureterm p1(x, y) becomes large in the vicinity of thetrailing-edge.

• The whole Prandtl (1904) hierarchical strategy in which wefirst calculate the inviscid flow and then the boundary layerflow followed by the flow due to displacement thickness,breaks down in the vicinity of the trailing-edge.

187 / 190


In the derivation of the boundary layer equations in region weretained terms of the form

U0∂U0

∂x∼ ∂2U0

∂Y 2∼ (x− 1)−1/3.

However that neglected next order pressure gradient terms areof order

Re−1/2∂p1

∂x∼ Re−1/2(x− 1)−5/3.

188 / 190


Clearly the neglected induced pressure gradient terms becomecomparable with the other terms retained in the equation when

Re−1/2(x− 1)−5/3 ∼ (x− 1)−1/3

x− 1 = O(Re−3/8).

When x = 1 +O(Re−3/8) the assumptions that we have madebreak down and we need to reconsider the solution in thevicinity of the trailing-edge when this happens.

189 / 190


Other scalings also follow readily. Note that in the region 2a ofthe Goldstein wake, the thickness is given by

Re−1/2Y ∼ Re−1/2(x− 1)1/3 ∼ Re−5/8

when x− 1 = O(Re−3/8). The streamwise velocity componentof velocity U0 is then of order

U0 = O((x− 1)1/3 ∼ O(Re−1/8).

The induced pressure term is now or order

Re−1/2p1 ∼ O(Re−1/2(x− 1)−2/3) ∼ O(Re−1/4),

and similarly the induced normal velocity is of order

Re−1/2v1 ∼ O(Re−1/4).

190 / 190

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