Invisid Incompressible Flow

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Fundamentals of Inviscid, Incompressible Flow

Dr.ir. M.I. Gerritsma & Dr.ir. B.W. van Oudheusden Delft University of Technology Department of Aerospace Engineering Section Aerodynamics

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Overview

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Stationary, inviscid, incompressible flow in the direction of a streamline

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The x-component of the momentum equation for a stationary, invisvid, incompressible flow reads

u

∂u ∂u ∂u 1 ∂p +v +w =− ∂x ∂y ∂z ρ ∂x

Multiplying this equation by dx gives

u

∂u ∂u ∂u 1 ∂p dx + v dx + w dx = − dx ∂x ∂y ∂z ρ ∂x

Now bear in mind that along a streamline we have

u dz − w dx = 0 v dx − u dy = 0

Using this in the momentum equation gives

 ∂u ∂u ∂u  1 ∂p u  dx + dy + dz  = u du = − dx ∂y ∂z  ρ ∂x  ∂x TUD

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Stationary, inviscid, incompressible flow in the direction of a streamline u du =

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1 2 1 ∂p du = − dx 2 ρ ∂x

Similar manipulations can also be performed in the y- and z-component of the momentum equation:

1 2 1 ∂p dv = − dy 2 ρ ∂y 1 2 1 ∂p dw = − dz 2 ρ ∂z

Adding these three relations together gives

1 1 1  ∂p ∂p ∂p  1 2 2 2 2 dV = d ( u + v + w ) = −  dx + dy + dz  = − dp 2 2 ρ  ∂x ∂y ∂z  ρ 1 1 dV 2 = − dp 2 ρ TUD

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Stationary, inviscid, incompressible flow in the direction of a streamline

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For an incompressible flow the density is constant so if we integrate these infinitisimal change between two points on the same streamline we get p2

V2

p1

V1

∫ dp = − ρ ∫ V dV ⇔

 V22 V12  p2 − p1 = − ρ  −  2 2   ⇔ 1 1 2 p1 + ρV1 = p2 + ρV22 2 2

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Bernouilli’s Equation p+

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1 ρV 2 = const. 2

So, although the momentum equation consists of 3 partial differential equations, the equation along a streamline reduces to a single algebraic relation. The use of ‘Bernoulli’ is limited. Necessary requirements are • stationary flow • inviscid, no body forces • incompressible • only valid along a streamline! ‘Bernoulli’ is valid for rotational flows also. However, when the flow is irrotational ‘Bernoulli’ is valid everywhere in the flow field and there is no restriction that it is only satisfied along streamlines

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Proof Bernoulli for irrotational flows Home work: Show that Bernoulli holds everywhere in the flow field (I.e. the constant is the same everywhere) if the flow is irrotational. In order to prove this either use • The definition of a irrotational flow • The following vector identity

(

) (

) (

)

(

)

(

∇ a ⋅ b = a ⋅∇ b + b ⋅∇ a + b × ∇ × a + a × ∇ × b

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Application of Bernoulli

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1. Airfoil at se level conditions: Consider an airfoil in a flow at sea level conditions with a freestream velocity of 50 m/s. At a given point on the airfoil, the pressure is 5 2 equal to 0.9 ×10 N/m . Calculate the velocity at this point. Solution: At Standard see level conditions ρ∞ = 1.23 kg/m3 and p∞ = 1.01×10 5 N/m 2 . Hence

p∞ +

1 1 ρV∞2 = p + ρV 2 2 2

5 2 ( p∞ − p ) 2 1.01 − 0.9 × 10 ( ) 2 V= + V∞2 = + ( 50 ) ρ 1.23

⇒ V = 142.8 m/s

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Flow through a converging-diverging nozzle

A1 , p1 , V1

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A2 , p2 , V2

Assumption: We assume that the flow is quasi 1 dimensional, therefore: p = p ( x ) , V = V ( x ) , etc. Continuity

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ρ1V1 A1 = ρ 2V2 A2

Bernoulli

p1 ( x ) +

Eliminate

V2

( ρ1 = ρ2 ) ⇒ V1 A1 = V2 A2

1 1 ρV12 ( x ) = p2 ( x ) + ρV22 ( x ) 2 2

⇒ V12 =

2 ( p2 − p1 )   A 2  ρ 1 −  1     A2    

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The pressure coefficient for incompressible flow

C p :=

Bernoulli:

p+

p − p∞ p − p∞ = 1 q∞ ρ ∞V∞2 2

1 1 ρV 2 = p∞ + ρ ∞V∞2 2 2

V  Cp = 1−    V∞ 

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⇒ Cp =

V  p − p∞ = 1−   1  V∞  ρ ∞V∞2 2

2

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Inviscid, incompressible, irrotational flows • Continuity

• Irrotational

∇ ⋅V = 0 ⇔

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∂u ∂v ∂w + + =0 ∂x ∂y ∂z

 ∂w ∂v   ∂y − ∂z     ∂u ∂w  ∇ ×V = 0 ⇔ ξ =  − ∂z ∂x    ∂ v ∂ u  −   ∂x ∂y    V = ∇φ

⇔ u=

∂φ ∂φ ∂φ ,v= ,w= . ∂x ∂y ∂z

• Laplace equation (insert potential equation in continuity equation)

∂ 2φ ∂ 2φ ∂ 2φ + 2 + 2 = 0 ⇔ ∆φ = ∇ ⋅∇φ = 0 2 ∂x ∂y ∂z TUD

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Potential Equation

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∂ 2φ ∂ 2φ ∂ 2φ + 2 + 2 =0 2 ∂x ∂y ∂z • This equation is also applicable to unsteady flows in whichφ = φ ( x, y, z , t ) • By introducing the potential, the irrotationality requirement is identically satisfied • Every inviscid, incompressible, irrotational flow is described by a potential which satisfies the above potential equation. • Conversely, every solution of the Laplace equation generates a valid inviscid, incompressible, irrotational flow. • The Laplace equation is linear, therefore we can use the principle of superposition. So if φ1 and φ2 are solutions of the Laplace equation, so is φ = α1φ1 + α 2φ2 . So complicated flow paterns can be obtained by a suitable combination of elementary flows. (Although it is not known in advance how and which elementary flow patterns to combine) • Once the equation for φ has been solved the velocity components are obtained from ∂φ ∂φ ∂φ u=

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∂x

, v=

∂y

, w=

∂z

.

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Potential Equation • If the potential flow is stationary (why?) the pressure coefficient is given by 2

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 ∂φ   ∂φ   ∂φ  +  +    2 V   ∂x   ∂y   ∂z  Cp = 1−  2  = 1− V∞2  V∞ 

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• The Laplace equation in Cylindrical coordinates

1 ∂  ∂φ  1 ∂ 2φ ∂ 2φ ∆φ = + 2 r + 2 2 r ∂r  ∂r  r ∂θ ∂z • The Laplace equation in Spherical coordinates

∆φ =

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1 r 2 sin θ

∂  2 ∂φ  ∂  ∂φ  ∂  1 ∂φ   r sin θ + sin θ       + ∂ r ∂ r ∂ θ ∂ θ ∂ ϕ sin θ ∂ ϕ        Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Stream function for incompressible flow, 2D • Stream function

ψ ( x, y , t )

⇒ u=

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∂ψ ∂ψ , v=− ∂y ∂x

• The velocity components obtained from a stream function automatically satisfy the incompressibility constraint: ∂  ∂ψ  ∂  ∂ψ   + − =0 ∂x  ∂y  ∂y  ∂x  •For 2D irrotational flow we have

∂v ∂u − =0 ∂x ∂y • Inserting the velocity components obtained from the stream function gives the 2D Laplace equation

∂  ∂ψ − ∂x  ∂x

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 ∂  ∂ψ −   ∂y  ∂y

  = −∆ψ = 0 

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Boundary conditions

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• Note: flows in all kinds of different geometries (a sphere, airfoil, cone) are governed by the same equation ∆φ = 0 • Question: How can one equation generate solutions for so many different flow problems? • Answer: The difference between the various geometries and flows is the domain in which the Laplace equation has to be solved and the the boundary conditions that are imposed at the boundary of the domain. • Boundary conditions at infinity: We assume that any disturbances caused by a object placed in the flow have vanished at “infinity”, so we set:

x = ±∞ u = V∞ , v = 0, y = ±∞ u = V∞ , v = 0.

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Boundary conditions at a solid wall

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• At a solid wall we assume that the flow cannot enter the object, nor will fluid emerge from the object. Since we assume that the flow is inviscid, the fluid is allowed to slide along the solid. If viscous effects are taken into account the friction will prevent the fluid from sliding along the surface of the object (the so-called no-slip condition). In the latter case a so-called boundary layer will develop. However, these viscous phenomena cannot be described by potential equations (why?). • So at a solid interface the velocity component perpendicular to the surface must be set to zero, i.e.

( V ⋅ n) = 0

•In terms of the potential function this condition can be written as

(

)

∇φ ⋅ n =

∂φ =0 ∂n

•In terms of the stream function this can be written as (the wall is a streamline!) ∂ψ = 0 ⇔ ψ = const. along streamline ∂s

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Inviscid, incompressible, irrotational flow

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F visc = 0, ρ = const., ∇ × V = 0, ⇒ ∂ 2φ ∂ 2φ ∂ 2φ + 2 + 2 =0 2 ∂x ∂y ∂z Solution Strategy: • Solve the Laplace equation for φ or conditions.

ψ which satisfy the appropriate boundary

• Determine the velocity components using

V = ∇φ

or u =

∂ψ ∂ψ ,v=− ∂y ∂x

• Determine the pressure distribution using Bernoulli

p+

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1 1 ρV 2 = p∞ + ρ ∞V∞2 2 2

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φ = const.

y

Uniform parallel flow V∞

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Consider a uniform parallel flow given by

u = V∞ , v = 0 x

ψ = const.

V∞ = const.

• Satisfies the continuity condition • Irrotational, therefore a potential flow

∂φ  = V∞  ∂x   ⇒ φ ( x, y ) = V∞ x + const. ∂φ v= =0  ∂y 

u=

• Stream function

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u = V∞ =

∂ψ ∂ψ , v=0=− ⇒ ψ ( x, y ) = V∞ y ∂y ∂x

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a

C

D

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Uniform parallel flow Circulation:

V∞

Γ=Ñ ∫ V ⋅ds B

C

D

A

A

B

C

D

Γ=∫ +∫ +∫ +∫ = V∞ a + 0 − V∞ a + 0 = 0

B

A

⇒ Γ=0

Uniform parallel flow, under an angle α y

∂φ ∂ψ = ∂x ∂y ∂φ ∂ψ v = V∞ sin α = =− ∂y ∂x

u = V∞ cos α =

V∞

α x

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φ = V∞ ( x cos α + y sin α ) , ψ = V∞ ( y cos α − x sin α ) Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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uθ φ = const.

y

ur

Source flow

r

I am looking for a potential flow solution which only depends on r (in polar coordinates), so

θ x

φ = φ ( r)

ψ = const.

inserting this in the two dimensional Laplace equation gives

1 ∂  ∂φ  1 ∂ 2φ ∆φ = =0 r + 2 2 r ∂r  ∂r  r ∂θ So the velocity components are given by:

∂φ c = , ∂r r 1 ∂φ uθ = = 0. r ∂θ ur =

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⇒ φ ( r ) = c1 ln r + c2 • The volume flow through a circle with radius R is equal to Q = 2π R ⋅ ur ( R ) = 2π c Q is called the source strength and c =

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Q 2π

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Source flow (cont.) Q Q ⇒ φ ( r) = ln r 2π 2π Q ur = , uθ = 0 2π r c=

If Q>0 the flow is called a source flow (fluid is emanating from the origin) and when Q<0 the flow is called a sink flow (fluid disappears at the origin)

uθ

The stream function belonging to this flow can be found by solving

φ = const.

y

1 ∂ψ Q Q = ⇒ ψ ( r ,θ ) = θ + f ( r) r ∂θ 2π r 2π ∂ψ Q uθ = − = 0 ⇒ ψ ( r ,θ ) = ψ ( θ ) = θ ∂r 2π ur =

Note that: ψ = const. ⇒

ur

r θ

x

ψ = const.

rays ( θ = const.)

φ = const. ⇒ circles ( r = const.) TUD

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Uniform flow + source flow

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Since the Laplace equation is linear, we are able to add elemenatary flows together. So one may ask what the resulting flow would be if we define a source at the origin in a uniform parallel flow along the x-axis.

⇒ ψ ( r ,θ ) =

Q θ + V∞ y 2π

Converting y to polar coordinates gives

Q ⇒ ψ ( r ,θ ) = θ + V∞ r sin θ 2π Source

What does this flow look like?? – Calculate the flow field – Determine stagnation points – Special streamlines

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Uniform flow

1 ∂ψ Q ur = = + V∞ cosθ r ∂θ 2π r ∂ψ uθ = − = −V∞ sin θ ∂r

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Uniform flow + source flow * Stagnation points:

ur = uθ = 0

Q  θ = 0, r = −Q ( Q > 0 ) ur = + V∞ cos θ = 0  2π V∞ 2π r  ⇒ uθ = −V∞ sin θ = 0  Q  θ =π , r = 2π V∞  * Streamlines

A

θ =π

B

Q ψ= 2 TUD

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θ =0

Q 2V∞

C

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Uniform flow + source flow

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The streamline ABC gives the contour of a semi-infinite body. The streamline passes through the stagnation point B, so

Q B :θ = π , r = 2π V∞

Q ⇒ ψ= 2

So the streamline passing through ABC is given by

Q Q θ + V∞ r sin θ = 2π 2

 θ 1−   Q  π ⇒ r= . 2V∞ sin θ

Q

The half-width of the semi-infinite body tends to (prove this). 2V∞ for x → ∞

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Uniform parallel flow + source (+Q) + sink (-Q)

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• Consider a uniform parallel flow in the x-direction • and a source and a sink placed at a distance 2b from eachother in the x-direction.

Q Q ψ = V∞r sin θ + θ1 − θ2 2π 2π

P ( r ,θ )

Rankine oval

θ1

b

θ b

θ2 r sin θ tan θ 2 = r cos θ − b

r sin θ tan θ1 = b + r cos θ

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Uniform parallel flow + source (+Q) + sink (-Q) Q Q ψ = V∞ r sin θ + θ1 − θ 2 2π 2π 1 ∂ψ Q  ∂θ1 ∂θ 2  = V∞ cos θ + −   r ∂θ 2π  ∂θ ∂θ  ∂ψ Q  ∂θ1 ∂θ 2  uθ = − = −V∞ sin θ − −   ∂r 2π  ∂r ∂r 

* Velocity components:

ur =

* Stagnation points:

Qb A: θ =π , r = b + π V∞ 2

Qb B : θ = 0, r = b + π V∞ 2

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Uniform parallel flow + source (+Q) + sink (-Q) Remarks: Because the total strength of the source and the sink (+Q-Q) is equal to zero, a closed streamline through the stagnation points A and B will appear. All the mass created by the source is consumed by the sink Since the flow is assumed to be inviscid, the closed streamline can be cosidered as the shape of the Rankine oval placed in a uniform flow. Materializing the inner domain does not effect the outer flow. Note that the Rankine oval is not an ellips!

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Doublet flow Consider the flow of a source and a sink placed at a distance l at either side of the origin.

P

∆θ

r θ1

Q

Q Q ψ ( r ,θ ) = ( θ1 − θ 2 ) = − ∆θ 2π 2π

1 44 2 4 43 l

θ2 −Q

 −Q   −κ ∆θ  ψ doublet = lim ∆ θ = lim  l →0   l → 0  2π 2π l     Λ fixed Λ fixed

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Now let the distance l shrink to zero and let the source strength Q grow to infinity, such that the productQl = κ = const. This will result in a doublet. ∆θ sin θ Since lim it follows that = l →0 l r

ψ doublet = −

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κ sin θ 2π r Ad

∂φ 1 ∂ψ = ∂r r ∂θ 1 ∂φ ∂ψ uθ = =− r ∂θ ∂r

Using: ur =

• Streamlines:

Doublet flow we find that φdoublet

=

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κ cos θ 2π r

sin θ ψ = const ⇒ =c ⇒ r 2

1   1   2 ⇒ x + y −  =   2c   2c  

y = c ( x2 + y 2 )

2

So the streamlines consist of circles with midpoint (0,1/2c) and radius 2c • The doublet is oriented in the direction of the x-axis. The tilted doublet may be obtained by rotating the frame of reference. This yields:

ψ doublet TUD

κ sin ( θ − α ) =− 2π r

φdoublet

κ cos ( θ − α ) = 2π r

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Uniform flow over a circular cylinder

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Now we add together a uniform parallel flow in the x-direction and a doublet at the origin oriented in the x-direction.

φ = V∞ r cos θ +

 κ cos θ κ  = V∞ r cos θ  1 + 2  2π r 2 π V r  ∞ 

 κ sin θ κ  ψ = V∞ r sin θ − = V∞ r sin θ 1 − 2  2π r 2 π V r  ∞ 

2 Set R =

κ gives 2π V∞

 R2  φ = V∞ r cos θ 1 + 2  r    R2  ψ = V∞ r sin θ 1 − 2  r   TUD

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Uniform flow over a circular cylinder

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 R2  ψ = V∞ r sin θ 1 − 2  r   Note that for r=R the stream function vanishes identically, therefore the circle with radius R is a streamline. If we ‘materialize’ the region inside the cylinder, we obtain the potential flow solution over a circular cylinder.

 R2  1 ∂ψ • The velocity field ur = = V∞ cos θ 1 − 2  r ∂θ r    R2  ∂ψ uθ = − = −V∞ sin θ 1 + 2  ∂r r   • Stagnation points

ur = uθ = 0 ⇒ A : θ = 0, r = R B: θ =π , r = R ψ A =ψ B = 0

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Flow over a circular cylinder Streamline ψ = 0 , then either

θ =0 r=R θ =π

: the positive x-axis : the circular cylinder : the negative x-axis

The velocity at the cylinder (r=R) are given by

 R  1 ∂ψ ur = = V∞ cosθ 1 − 2  = 0 r ∂θ r    R2  ∂ψ uθ = − = −V∞ sin θ 1 + 2  = −2V∞ sin θ ∂r r   2

Was this to be expected??

The pressure coefficient now provides the pressure over the cylinder 2

V  C p = 1 −   = 1 − 4sin 2 θ  V∞  TUD

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Source in 3D

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In order to find the source in 3D we insert a potential function, which only depends on the radius into the Laplace equation in sperical coordinates

∆φ =

1 r 2 sin θ

⇒ ⇒ ∂φ ∂θ

∂  2 ∂φ  ∂  ∂φ  ∂  1 ∂φ   r sin θ + sin θ     = 0   + ∂r  ∂θ  ∂θ  ∂ϕ  sin θ ∂ϕ    ∂r 

c1 ( θ ) ∂φ ∂φ r sin θ = c1 ( θ ) ⇒ = 2 ∂r ∂r r sin θ c1 ( θ ) φ ( r ,θ ) = − + c2 ( θ ) r sin θ c Compare in 2D: φ ( r ) = c ln r = 0 ⇒ φ ( r) = − r 2

Velocity components:

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∂φ c 1 ∂φ 1 ∂φ ur = = , uθ = = 0 , uϕ = = 0. ∂r r 2 r ∂θ r sin θ ∂ϕ Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Source in 3D

The volume flow:

λ = 4π R 2 ⋅ ur ( R ) = 4π c ⇒ c = ⇒

λ ur = 4π r 2

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λ 4π

λ and φ = − 4π r

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Doublet (dipole) in 3D z

P

r1

λ l

z

r r

θ

θ −λ

3D-source: φ = −

x

x

λ 4π r

⇒ φ =−

λ 4π

ϕ

 1 1 λ r − r1  − =− 4π rr1  r1 r 

Now let l → 0 with λl = µ = const.

r − r1 ≈ l cos θ  µ cos θ ⇒φ = − 2 r ≈ r1 4 π r  TUD

y

 ∂φ µ cos θ u = =  r ∂r 2π r 3  1 ∂φ µ sin θ  u = =  θ 3 r ∂ θ 4 π r  1 ∂φ  u =  ϕ r sin θ ∂ϕ = 0 

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Flow over a Sphere Uniform parallel flow

V = −V∞ e z ⇒

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ur = −V∞ cos θ uθ = V∞ sin θ uϕ = 0

Addition of the doublet

µ cos θ µ   = − V − cos θ  ∞ 3 3  Stagnation points: 2π r 2π r     µ sin θ  µ  uθ = 0 ⇒ θ = 0, θ = π uθ = V∞ sin θ + = V + sin θ  ∞ 4π r 3  4π r 3 µ 3 3  14 2 43  u = 0 ⇒ r = = R r > 0   2π V∞ uϕ = 0

ur = −V∞ cos θ +

µ R=3 2π V∞ TUD

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37

Flow over a Sphere Stagnation points: (R,0) and (R,π) Note: For r=R :

ur = 0

Materialize region

r ≤ R (sphere)

Parallel flow + doublet = incompressible, inviscid, irrotational, steady flow over a sphere For r=R :

µ   uθ =  V∞ + sin θ 3  4π R  

= ↑

1 R

= 3

2π V∞ µ 2

3 V∞ sin θ 2

V  3  ⇒ C p = 1 −   = 1 −  sin θ  2   V∞  9 C p = 1 − sin 2 θ 4

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2

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Comparison between the flow over a cylinder and a sphere Cylinder

38

Sphere

κ R= 2π V∞

R=

 R  ur = V∞ 1 − 2  cos θ r    R2  uθ = −V∞ 1 + 2  sin θ r   2

C p = 1 − 4sin 2 θ

3

µ 2π V∞

 R3  ur = −V∞ 1 − 3  cos θ r    R3  uθ = V∞ 1 + 3  sin θ r   9 C p = 1 − sin 2 θ 4

Note the different definitions of θ

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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39

Vortex flow Consider a 2 dimensional potential function in order to satisfy the Laplace equation

1 ∂  ∂φ  1 ∂ 2φ ∆φ = =0 ⇒ r + 2 2 r ∂r  ∂r  r ∂θ Velocity components:

φ = c1θ + c2 = cθ uθ

∂φ ur = =0 ∂r 1 ∂φ c uθ = = r ∂θ r

ur

Circulation

Γ = −Ñ ∫ V ⋅ds = 2π

Note this flow irrotational! (Why?)

TUD

φ = φ (θ )

??????

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

c − ∫ ⋅ rdθ = −2π c r 0

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Vortex flow Γ = −2π c

40

Note that unless c=0 the circulation will be non-zero, that means the flow field cannot be irrotational according to Stokes. However, if c=0 then there will be no flow at all! How do we resolve this problem?? If we calculate the vorticity, we will find that the vorticity is zero everywhere in the flow field, except at the origin where the vorticity is infinitely large. So all contours which enclose will the origin will have a non-zero circulation. In complex function theory where similar phenomena occur the plane is usually cut to prevent contours around the origin.

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Vortex flow Γ = −2π c

41

φ = cθ Instead of the integration constant c we usually express the strength of the vortex in terms of its circulation.

Γ φ =− θ 2π Note that the velocity is constant along the streamlines and therefore the pressure is constant along the streamlines.

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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42

Vortex flow Calculation of the stream function

1 ∂ψ  =0  Γ  r ∂θ ⇒ ψ = ln r  ∂ψ −Γ  2π uθ = − = ∂r 2π r  ur =

φ = const. ⇒ rays ( θ = const.)

ψ = const. ⇒ circles ( r = const.)

TUD

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

uθ

ur

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Elementary Potential Flows

Uniform flow in the x-direction Source flow Doublet flow Vortex flow

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Potential

Stream function

φ = V∞ x

ψ = V∞ y

Q ln r 2π κ cos θ φ= 2π r −Γ φ= θ 2π

Q θ 2π −κ sin θ ψ= 2π r Γ ψ= ln r 2π

φ=

43

ψ=

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Flow over a cylinder with circulation

44

Consider again the flow over a cylinder, but now we add a vortex at the origin. The stream function for this flow is given by

κ sin θ Γ Γ ψ = V∞ r sin θ + + ln r − ln R 2π r 2π 2π Uniform flow

Doublet

Vortex

Just a constant

This can be succinctly written as

 R2  Γ r ψ = V∞ r sin θ 1 − 2  + ln r  2π R 

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Flow over a cylinder with circulation

45

 R2  Γ r ψ = V∞ r sin θ 1 − 2  + ln r  2π R  * Velocity field:

 R2  1 ∂ψ ur = = V∞ cos θ 1 − 2  r ∂θ r    R2  Γ ∂ψ uθ = − = −V∞ sin θ 1 + 2  − ∂r r  2π r 

* Velocity at the cylinder:

* Stagnation points:

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1 ∂ψ =0 r ∂θ ∂ψ Γ uθ = − = −2V∞ sin θ − ∂r 2π R ur =

r = R , sin θ = −

Γ 4π RV∞

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Flow over a cylinder with circulation * Stagnation points (cont.) • Γ=0

46

 −Γ  r = R , θ = arcsin   4 π RV  ∞ 

: Two stagnation points, θ = 0 , θ = π

• 0 < Γ < 4π RV∞ Two stagnation points underneath the cylinder

3 Γ = 4 π RV π ∞ One stagnation point at r = R , θ = • 2 • Γ > 4π RV∞ Two stagnation points, one outside the cylinder, one inside the cylinder, 2

 Γ  3 Γ 2 θ= π, r= ±   −R 2 4π V∞  4π V∞  Note that for every value of Γ, the resulting flow will be the flow around a cylinder, so the potential solution allows for infinitely many cylinder flows.

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Flow over a cylinder with circulation

47

1 ∂ψ =0 r ∂θ ∂ψ Γ uθ = − = −2V∞ sin θ − ∂r 2π R

The velocity field at the cylinder was found to be ur =

So the pressure distribution over the cylinder is given by 2    ur2 + uθ2    2Γ sin θ Γ 2 Cp = 1−  +  = 1 − 4sin θ +   2 V π RV 2 π RV  ∞  ∞  ∞    

2  1 2Γ sin θ  Γ   2 p = p∞ + ρV∞ 1 − 4sin θ − −   2 π RV∞  2π RV∞    

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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48

Calculation of the Drag Drag: Only the pressure contributes to the total drag, since body and viscous forces have been neglected.

D=

2π

∫ − pRdθ cosθ 0

D 1 CD = =− q∞ ⋅ 2 R ⋅1 2

2π

∫ 0

p 1 cos θ dθ = − q∞ 2

2π

∫C

p

cos θ dθ

0

⇒ CD = 0 Independent of Γ.

Use:

TUD

2π

2π

2π

0

0

0

∫ cosθ dθ = 0,

∫ sin θ cosθ dθ = 0 ,

2 sin ∫ θ cosθ dθ = 0

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Calculation of the Lift over the Cylinder CL = L=

L q∞ 2 R ⋅1

p

y

θ

2π

∫ − pRdθ sin θ

⇒ CL = −

0

1 2Γ CL = 2 Rπ V∞

49

2π

2 sin ∫ θ dθ = 0

1 2

2π

∫C

p

sin θ dθ

R

Rdθ

x

0

Γ RV∞

Using the definition of the lift coefficient, we get

L = ρV∞ Γ The Kutta-Joukowski Theorem Use:

TUD

2π

2π

0

0

2π

∫ sin θ dθ = 0, ∫ sin θ dθ = 0, ∫ sin 3

2

θ dθ = π

0

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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V∞ , ρ ∞

50

The Kutta-Joukowski Theorem L

Contour B

Contour A

L = ρ ∞V∞ j∫ V ⋅ d s A

This expression is generally valid for 2D shapes in an icompressible, inviscid and irrotational flow. (Proof by means of complex potentials) In order to describe the flow around an airfoil, one usually employes not one vortex, but a vortex distribution. The sum of the circulation induced by all these individual vortices appears in the KJ-Theorem.

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Potential Flow around Bodies (overview)

51

• Combination of elementary flows • Basic idea: Replace streamlines by solid wall • Example 1.: Uniform flow + source • Example 2.: Uniform flow + source + sink (Rankine oval) • Example 3.: Uniform flow + doublet (flow around cylinder) • Example 4.: Uniform flow + doublet + vortex (flow over cylinder with lift) Extension: The approximation of arbitrary shapes by distributed sources on the body of the contour. The panel method.

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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52

Source panel method • Numerical method for an approximate determination of the flow around bodies of arbitrary shape • Idea: Distribute sources (and sinks) with a yet undetermined strength along the boundary of the object. Use the boundary condition at the wall of the object to determine the strength of the sources and sinks. Finally, determine the flow of the source distribution in a uniform parallel flow. •In order to do this we have to introduce the concept of a source sheet, which is a continuous distribution of sources along a contour.

ds

a

TUD

λ ( s)

s

b s is a parameter along the contour and λ is the source strength per unit length which can be positive (source) or negative (sink)

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Source panel method (cont.)

r ds

a

53

P ( x, y )

b

λ ( s)

s

An infinitisimal elemenent ds has a source strength of λdsand this induces a potential at the point P equal to λ ds dφ = ln r 2π So the total potential induced by the coutour at the point P is given by

λ ds φ ( P) = ∫ ln r 2π a b

Problem: How do we determine the strength λ(s) such that the desired profile is approximated?

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Source panel method (cont.)

54

Approximate the contour by a finite number of straight line segments and assume that λ is constant within each segment, so we have a finite number, N, of source strengths λi to determine. Calculate the contribution at an arbitrary point P to the total potential due to one segment. Add all N contribution to obtain the total potential at a point P. Place the point P at the midpoint of an arbitrary panel and set the derivative of the potential in the direction of the normal of the panel equal to zero (boundary condition), i.e.

∂φ =0 ∂n

This gives one equation for the N unknowns λi . Imposing the boundary conditions at all segments, gives N equations for N unknowns, which in general to a unique solution.

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Source panel method (cont.)

55

The equation to be solved have the following form:

λi N λ j ∂ V∞ cos β i + + ∑ ln rij ) ds j = 0 ( ∫ 2 j =1 2π j ∂ni 1 44 2 4 43 j ≠i Iij

Remarks: The influence coefficients Iij do not depend on the flow, but only of the geometry of the profile. Of course, increasing the number of panels, will improve the approximation (higher accuracy). Modern panel techniques employ curved panels and a non-constant source distribution. Once the source strengths have been obtained we can calculate the velocity along the panels, using

( VP) panel i

N λ ∂φ = = V∞ sin β i + ∑ i ∂s j =1 2π j ≠i

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∂ ∫j ∂s ( ln rij ) ds j

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Source panel method (cont.)

(V )

P panel i

N λ ∂φ = = V∞ sin β i + ∑ i ∂s j =1 2π j ≠i

56

∂ ∫j ∂s ( ln rij ) ds j

Once we have the velocity along the panel I, we can use Bernoulli to obtain the pressure acting on panel I.

C p, i

 ( VP ) panel i = 1−   V∞  N

For closed contour we should have

TUD

∑ s ⋅λ i =1

i

i

   

2

=0

Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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Comparison with real flows

57

This finalizes chapter 3 (and 6) on potential flows. I wish you good luck with the two remaining chapters.

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Aerodynamics-B, AE2-115 I, Chapter III Gerritsma & Van Oudheusden

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