On the sensitivity of jets and shear layers or the...

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On the sensitivity of jets and shearlayers

orthe solution of the Orr-Sommerfeld equation by the compound matrix

method (CMM)

György Paál, Péter Tamás Nagy

Motivation 1.

On the sensitivity of jets and shear layers

Determine the stability properties of self-similar flows

Jets Boundary layers

Motivation 2.


Applications

Modelling self-excited flowsEdge tone Cavity tone

Organ pipes, flueinstrumentsY branches inpipes

car door gapspantograph of trainsweapon bays onaircrafts

Analytical approximation of the jet velocity profile


𝑈 𝑦 = sech2 𝑦𝑛

𝑛=1: - the self-similar profile in the far field- non-dimensional Bickley profile

Plane jet

The perturbed NS equations


𝛻 ∙ 𝑢𝑡 = 0

The non-dimensional Navier-Stokes equations(conservation of mass and momentum):

𝜕𝑢𝑡

𝜕𝑡+ 𝑢𝑡 ∙ 𝛻𝑢𝑡 = −𝛻𝑝𝑡 +

1

𝑅𝑒∆ 𝑢𝑡

Apply perturbation , 𝑢𝑝, 𝑝𝑝

𝑢𝑡 = 𝑈 + 𝑢𝑝 𝑝𝑡 = 𝑃 + 𝑝𝑝

If U, P are known, time-independent variables, describe the base flow, fulfil the governing equations. up = (up, vp, wp)

If U(y) is independent of x (parallel flow, or self-similar flow), theperturbation can be assumed in a complex wave form:

𝑣𝑝 = 𝑣(𝑦) 𝑒𝑖(α𝑥−ω𝑡) α: wavenumber, ω: angular frequency

Temporal and spatial instability

𝜕𝑈

𝜕𝑥= 0, 𝑉 = 0


𝜔 = 5𝛼 = 5

𝑣𝑝 = 𝑣(𝑦) 𝑒𝑖(α𝑥−ω𝑡)

𝜔 = 5 + 0.3𝑖𝛼 = 5𝜇𝑡 = 𝐼𝑚 ω

ω = 5𝛼 = 5 − 0.3𝑖𝜇𝑥 = −𝐼𝑚 𝛼

Here, spatial stability investigation: 𝜇 = 𝜇𝑥 = −𝐼𝑚 𝛼

𝜇 : growth rate

The Orr-Sommerfeld equation


Substituting the perturbation into the NS equation, linearising and simplifying, the equation system can be expressed in terms of v, leading to a fourth order linearordinary differential equation:

𝑈 𝑦 ,𝑈′′(𝑦) describe the flow (known)𝑅𝑒 is a parameter (known)𝑣 𝑦 is the eigenfunction (unknown)α,ω are the eigenvalue pairs (one is a parameter, the otherone is unknown)

𝑣𝑖𝑣 − 2𝛼2𝑣′′ + 𝛼4𝑣 = 𝑖 𝑅𝑒 [ 𝛼𝑈 − 𝜔 𝑣′′ − 𝛼2𝑣 − 𝛼𝑈′′𝑣]

𝑣𝑝 = 𝑣(𝑦) 𝑒𝑖(α𝑥−ω𝑡)

The OS equation in the far field


Fal

The Orr-Sommerfeld equation:

𝑣𝑖𝑣 − 2𝛼2𝑣′′ + 𝛼4𝑣 = i 𝑅𝑒 [ 𝛼𝑈 − 𝜔 𝑣′′ − 𝛼2𝑣 − 𝛼𝑈′′𝑣]

In the far field (𝑦 → ∞) it becomes a constant coefficient linear

differential equation because 𝑈(𝑦) is a constant:

𝑈∞ = lim𝑦→∞

𝑈(𝑦) , lim𝑦→∞

𝑈′′ 𝑦 = 0

𝑣𝑖𝑣 − 2𝛼2 𝑣′′ + 𝛼4 𝑣 = 𝑖 𝑅𝑒 [ 𝛼𝑈∞ − 𝜔 𝑣′′ − 𝛼2 𝑣 ]

Analytical solution in the far-field:

𝑣 y =

𝑖=1

4

෨𝑘𝑖 𝑣𝑖𝑦 =

𝑖=1

4

෨𝑘𝑖 𝑒𝜆𝑖𝑦

𝜆 = − 𝛼,−𝑄, 𝛼, 𝑄 T

𝑄:= 𝛼2 − i𝑅𝑒(𝛼𝑈∞ − 𝜔)

The OS equation in the far field 2.


Analytical solution in the far field:

If 𝑅𝑒 ≫ 1 then 𝑄 ≫ 𝛼 the differential equation becomes stiff.

Special method: the Compound Matrix Method

𝑣 y =

𝑖=1

4

෨𝑘𝑖 𝑣𝑖𝑦 =

𝑖=1

4


𝜆 = − 𝛼,−𝑄, 𝛼, 𝑄 T

𝑄:= 𝛼2 − i𝑅𝑒(𝛼𝑈∞ − 𝜔)

Transformation of variables


A compound matrix method alkalmazása az Orr-Sommerfeld egyenletre

In the far field the perturbation vanishes 𝑣 𝑦 → ∞ = 𝑣′ 𝑦 → ∞ = 0

The „true” solution reduces to

In the far field: 𝑣𝑖𝑦 → ∞ = 𝑣

𝑖𝑦

Let us transcribe the 4th order diff. eq. into a first order diff. eq.

system:

𝜙 = [𝑣, 𝑣′, 𝑣′′, 𝑣′′′]

𝜆 = − 𝛼,−𝑄, 𝛼, 𝑄 T ෨𝑘3 = ෨𝑘4 = 0𝑣 y =

𝑖=1

4


𝑣 𝑦 =

𝑖=1

2

𝑘𝑖 𝑣𝑖𝑦

Transformation of variables 2.


A compound matrix method alkalmazása az Orr-Sommerfeld egyenletre

Use the so-called compound matrix variables:

𝜂1 = det𝜙1(1)

𝜙2(1)

𝜙1(2)

𝜙2(2)

= 𝜙1(1)𝜙2(2)

− 𝜙1(2)

𝜙2(1)

= 𝑣1𝑣′

2− 𝑣

2𝑣′

1…

𝜂2 = det𝜙1(1)

𝜙3(1)

𝜙1(2)

𝜙3(2)

, 𝜂3 = det𝜙1(1)

𝜙4(1)

𝜙1(2)

𝜙4(2)

, 𝜂4 = det𝜙2(1)

𝜙3(1)

𝜙2(2)

𝜙3(2)

, …, 𝜂6

In the far field:

The problem of stiffness is solved! Furthermore, by introducing the 𝜂 =

𝜂

− 𝑄+𝛼 𝑒− 𝛼+𝑄 𝑦 variable, non-homogenous bc.s can be prescribed at „infinity”

𝑣1

𝑦 ~𝑒−𝛼𝑦

𝑣2

𝑦 ~𝑒−𝑄𝑦

𝜂1 = 𝑣1𝑣′

2− 𝑣

2𝑣′

1~ −(𝑄 + 𝛼) 𝑒− 𝛼+𝑄 𝑦

𝜂2 ~ − (𝑄2 − 𝛼2)𝑒− 𝛼+𝑄 𝑦, …, 𝜂6 ~ 𝑒− 𝛼+𝑄 𝑦

𝜙 = 𝑣, 𝑣′, 𝑣′′, 𝑣′′′ = [𝜙1, 𝜙2, 𝜙3, 𝜙4]

Boundary conditions in the near field


The far field boundary conditions (𝑣 𝑦 → ∞ = 𝑣′ 𝑦 → ∞ = 0) are

fulfilled after transformation. Two other boundary conditions are

necessary. Near field boundary conditions

Symmetric profile 𝑣′(0) = 𝑣′′′(0) = 0 → 𝜂5 0 = 0

Antisymmetric profile 𝑣(0) = 𝑣′′(0) = 0 → 𝜂2 0 = 0

Wall 𝑣(0) = 𝑣′(0) = 0 → 𝜂1 0 = 0

Jet

Shooting method


𝑦 = 0𝑦 = 𝐿

Integration (�ු�, 𝜔) Error

Desired BC

(𝛼, 𝜔)

The boundary value problem should be solved as an initial value

problem. The solution can be initialised at a point (𝑦0 = L), where the

velocity is approximately constant 𝑈 𝑦 → ∞ ≈ 𝑈(𝐿)

The integration should start from there towards 𝑦 = 0.

The 𝛼,𝜔 parameters should be tuned to fulfil the boundary conditions

at y=0 and find the correct missing initial condition at y = L.

Thumb curve


𝜇 = 𝐼𝑚(𝛼)

The curves show constant αi contours. The solid curveseparates the stable and the unstable regions. The criticalReynolds number can be read from the curve.

Recrit

Jet: thumb curves


𝑈 𝑦 = sech2 𝑦𝑛

Bickley profile: 𝑛 = 1

Symmetric: Antisymmetric:

Results: sim.–antisym.


Thomas, F. O., Goldschmidt, V. W., 1985. The

possibility of a resonance mechanism in the

developing two-dimensional jet. Phys. Fluids 28

SYMMETRIC

ANTISYMETRIC

SYMMETRIC

Mode

SYMMETRIC

Mode

ANTISYM.

Mode

Numerical velocity profiles I.


The analytical profiles do not describe

• The planar jet with parabolic velocity profile at the orifice

• The effect of the nozzle wall thickness

• The transition of the velocity profiles. The parameter 𝑛 must be integer.

Calculate the base flow with CFD software. (Computational

Fluid Dynamics = numerical flow simulation)

Numerical velocity profiles II.


Prescribed

velocity profile

Long, parallel nozzle:

parabolic

Short, convergent

nozzle:

„top hat” profile

Wall/opening

Opening

𝑈 𝑦 Matlab

Numerical velocity profiles III.

On the sensitivity of planar jets

One result


|𝑢𝑡| = |𝑈 + 𝑢𝑝|

𝑈

|𝑢𝑝|

Growth rates for various velocity profiles


Parabolic „Top hat”

Wall BC, and symmetric perturbation

Symmetric-antisymmetric


Opening Wall

𝑥/𝛿 = 0.1

Boundary condition around the orifice

”Strange” mode


𝜇

ω = 0.1


Why is the jet more sensitive at the exit?

There are three reasons for the increased sensitivity of laminar,

plane jets close to the orifice.

1., the steeper velocity profile amplifies the disturbances at a

higher growth rate.

2., the local length scale is smaller close to the orifice, meaning

that the dimensional growth rate of disturbances is larger than

that in downstream region.

3., the disturbances excited upstream have more space to grow

exponentially.

Application for boundary layers: motivation


Aim: drag reduction

Examples in nature:

Laminar-turbulent transition I.


μ < 0 μ > 0

Laminar-turbulent transition II.


Forrás: YouTube

|𝑢𝑡|

𝑈

|𝑢𝑝|

Simple controller I.


𝑢(𝑡) = ǁ𝑐𝑝τ(𝑡 − Δ𝑡) 𝑢𝑝

|𝑢𝑡|

𝑈

|𝑢𝑝|

𝑐𝑝 = 0.5, Δ𝑡 = 0

Simple controller II.


𝑢(𝑡) = ǁ𝑐𝑝τ(𝑡 − Δ𝑡)

Orr-Sommerfeld equation

Reynolds-Orr equation

Stable Unstable

Stable

Unstable

Unstable

Passive coating


Rigid wall

Coated wall

Material: silicone rubber

𝑙1 = 25.5 𝜇𝑚

𝑙2 = 8653 𝜇𝑚

ℎ1 = 168 𝜇𝑚

ℎ2 = 14.5 𝜇𝑚

𝑤1 = 10 𝜇𝑚

𝑤2 = 33 𝜇𝑚

Thank you for your attention!


On the sensitivity of jets and shear layers or the...

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