Post on 26-Dec-2019
Sound propagation in a lined duct with flow
Martien Oppeneer
supervisors: Sjoerd Rienstra and Bob Mattheij
CASA dayEindhoven, April 7, 2010
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Outline
1 Introduction & Background
2 Modeling the problem
3 Numerical implementation
4 (Numerical) results
5 Future plans
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Introduction & Background
Outline
1 Introduction & BackgroundProject motivationAcoustic linersGeneral project goalBrush-up: modes
2 Modeling the problemPridmore-Brown equation (ODE)Boundary conditions
3 Numerical implementationMethod 1: bvp4c / BVP SOLVERMethod 2: COLSYSContinuation in Z
4 (Numerical) resultsImpedance wall, no flowImpedance wall, uniform mean flowSome numerical problems
5 Future plans3 / 47
Introduction & Background Project motivation
Project motivation
APU: Auxiliary Power Unit
produces power when main enginesare switched offto start main engines, AC, ...major source of ramp noise
Goal: APU noise reduction Figure: APU onan Airbus A380.
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Introduction & Background Acoustic liners
Acoustic liners
Figure: Locally reacting liner(impedance wall).
Figure: Metallic foam (bulk absorber).
Figure: Spiralling (non-locally reacting)liner.
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Introduction & Background General project goal
General project goal
hard wall resistive sheet
liner cavity
cool air inlet
exhaust
mean flow profile u(r)
Figure: APU geometry.
Model sound propagation /attenuation
sheared flownon-locally reacting linerssegmented / non-uniformlinersstrong temperature gradients(swirling flow)(annular hub)(varying duct radius)
Sufficiently fast for liner designcalculations⇒ semi-analytical model, basedon modes
6 / 47
Introduction & Background Brush-up: modes
Modes
Motivation: Direct Navier-Stokes(DNS) not practical / feasible (esp.for design)
Eigensolution of a BVP
Characterized by:
eigenfunction Pmµ(r)‘eigenvalue’ kmµ ∈ C
Traveling waves of the form:
pmµ(r) =Pmµ(r) exp(−iωt+ikmµx+imθ)
Total field is superposition (orintegral) of modes
Figure: ω = 20, m = 1, Z = 3 + 3i,µ = 3.
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Modeling the problem
Outline
1 Introduction & BackgroundProject motivationAcoustic linersGeneral project goalBrush-up: modes
2 Modeling the problemPridmore-Brown equation (ODE)Boundary conditions
3 Numerical implementationMethod 1: bvp4c / BVP SOLVERMethod 2: COLSYSContinuation in Z
4 (Numerical) resultsImpedance wall, no flowImpedance wall, uniform mean flowSome numerical problems
5 Future plans8 / 47
Modeling the problem
Duct geometry
d
h x
r
θ
Figure: Duct geometry, velocity components: v = ux+ vr + wθ.
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Modeling the problem
Modeling outline
BVP
No viscosityNo heat conduction
Cylindrical coordinates
Navier-Stokeseqns
LinearizedEuler eqns
Time-harmonicsolutions
Boundaryconditions
Eulereqns
Smallperturbations
ODE forP (r)
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Modeling the problem Pridmore-Brown equation (ODE)
No viscosityNo heat conduction
Cylindrical coordinatesNavier-Stokeseqns
LinearizedEuler eqns
Time-harmonicsolutions
Eulereqns
Smallperturbations
ODE forP (r)
Navier-Stokes (conservation laws)
∂
∂tρ+∇ · (ρv) = 0
∂
∂t(ρv) +∇ · (ρvv) = −∇p+∇ · τ
∂
∂t(ρE) +∇ · (ρEv) = −∇ · q −∇ · (pv) +∇ · τv
inviscid: ∇ · τ = 0 non-heat-conducting: ∇ · q = 0
11 / 47
Modeling the problem Pridmore-Brown equation (ODE)
No viscosityNo heat conduction
Cylindrical coordinatesNavier-Stokeseqns
LinearizedEuler eqns
Time-harmonicsolutions
Eulereqns
Smallperturbations
ODE forP (r)
Euler Equations (in primary variables)
∂
∂tρ+∇ · (ρv) = 0
ρ
(∂v
∂t+ v · ∇v
)= −∇p
∂p
∂t+ v · ∇p+ γp∇ · v = 0
Ideal gas: p = ρRT
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Modeling the problem Pridmore-Brown equation (ODE)
Small perturbations
Sound is due to small pressure perturbationsAssumptions:
main sound source: turbine engine (rotor / stator interaction)
negligible sound source: turbulence
⇒ Total field = mean flow + perturbations:
(u, v, w, ρ, p) = (u, v, w, ρ, p) + (u, v, w, ρ, p)
Linearize: neglect quadratic terms (since perturbations are small)
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Modeling the problem Pridmore-Brown equation (ODE)
Mean flow
Mean flow assumptions:
independent of x: ∂u∂x = 0, ∂v∂x = 0, ∂w∂x = 0
radial velocity v = 0circumferential velocity independent of θ: ∂w
∂θ = 0
14 / 47
Modeling the problem Pridmore-Brown equation (ODE)
No viscosityNo heat conduction
Cylindrical coordinatesNavier-Stokeseqns
LinearizedEuler eqns
Time-harmonicsolutions
Eulereqns
Smallperturbations
ODE forP (r)
Linearized Euler Equations
∂ρ
∂t+ u
∂ρ
∂x+
1r
∂(rρv)∂r
+w
r
∂ρ
∂θ+ ρ
(1r
∂w
∂θ+∂u
∂x
)= 0
ρ
(∂v
∂t+ u
∂v
∂x+w
r
∂v
∂θ− 2w
rw
)− w2
rρ = −∂p
∂r
ρ
(∂w
∂t+ u
∂w
∂x+w
r
∂w
∂θ+dw
drv +
w
rv
)= −1
r
∂p
∂θ
ρ
(∂u
∂t+ u
∂u
∂x+w
r
∂u
∂θ+ v
∂u
∂r+w
r
∂u
∂θ
)= −∂p
∂x
∂p
∂t+ u
∂p
∂x+w
r
∂p
∂θ+ρw2
rv + γp
(1r
∂(rv)∂r
+1r
∂w
∂θ+∂u
∂x
)= 0
15 / 47
Modeling the problem Pridmore-Brown equation (ODE)
No viscosityNo heat conduction
Cylindrical coordinatesNavier-Stokeseqns
LinearizedEuler eqns
Time-harmonicsolutions
Eulereqns
Smallperturbations
ODE forP (r)
We seek time-harmonic solutions:
(u, v, w, ρ, p) = (U, V,W,R, P ) exp(−iωt+ ikx+ imθ)
⇒ ODE in P (r):
P ′′ + β(r, k)P ′ + γ(r, k)P = 0, on h ≤ r ≤ d
where β(r, k) and γ(r, k) are functions of:
mean flow parameters: u(r), w(r), ρ(r), p(r)m,ω (given)
r, k
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Modeling the problem Pridmore-Brown equation (ODE)
No viscosityNo heat conduction
Cylindrical coordinatesNavier-Stokeseqns
LinearizedEuler eqns
Time-harmonicsolutions
Eulereqns
Smallperturbations
ODE forP (r)
Simplifications:
No swirl: w(r) = 0, p(r) constantρ(r) constant
Pridmore-Brown equation
P ′′ + β(r, k)P ′ + γ(r, k)P = 0, on h ≤ r ≤ dwhere
β(r, k) =1r
+2ku′
ω − ku
γ(r, k) =(ω − ku)2
c2− k2 − m2
r2
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Modeling the problem Boundary conditions
Three types of conditions
We need 3 types of conditions
1 Impedance wall BC at r = d (and r = h when h 6= 02 Regularization condition at r = 0 (when h = 0)
3 Normalization condition
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Modeling the problem Boundary conditions
1. Impedance wall BC
Assume: locally reacting liner with impedance Z
Due to vanishing mean-flow boundary layer:
−iωvn =(−iω + u
∂
∂x+w
r
∂
∂θ
) (p
Z
)
(Ingard-Myers condition)
Resulting boundary condition for locally reacting liner:
P ′ + κh(r, k)P = 0 at r = h
P ′ + κd(r, k)P = 0 at r = d
where
κh(k) =iρ (ω − ku)2
ωZh, κd(k) = − iρ (ω − ku)2
ωZd.
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Modeling the problem Boundary conditions
2. Regularization BC
No mean flow (u(r) = 0):Pridmore-Brown → Bessel’s equation
P ′′+1rP ′+
(α2 − m2
r2
)P = 0, α2 =
ω2
c2−k2
General solution:
P = AJm(αr) +BYm(αr)
Note: Ym(αr) is singular at r = 0.⇒ Make sure P (r) <∞ at r = 0
P ′(0) = 0, for m 6= 1P (0) = 0, for m = 1
0 5 10 15 20−0.5
0
0.5
1
012
(a) Jm(x)
0 5 10 15 20−1
−0.5
0
0.5
1
012
(b) J ′m(x)
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Modeling the problem Boundary conditions
3. Normalization
General solution:
P = AJm(αr) +BYm(αr)
Every solution P (r) can be scaled
P (r) can become 0 at r = 0
⇒ Choose P (r) = 1 at r = d
0 5 10 15 20−0.5
0
0.5
1
012
(c) Jm(x)
0 5 10 15 20−1
−0.5
0
0.5
1
012
(d) J ′m(x)
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Numerical implementation
Outline
1 Introduction & BackgroundProject motivationAcoustic linersGeneral project goalBrush-up: modes
2 Modeling the problemPridmore-Brown equation (ODE)Boundary conditions
3 Numerical implementationMethod 1: bvp4c / BVP SOLVERMethod 2: COLSYSContinuation in Z
4 (Numerical) resultsImpedance wall, no flowImpedance wall, uniform mean flowSome numerical problems
5 Future plans22 / 47
Numerical implementation
Numerical solution of BVP
Why numerics? Sheared flow / temperature gradients
Important: good initial guess for k and P (r)Handle singularity at r = 0Handle unknown parameter k
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Numerical implementation Method 1: bvp4c / BVP SOLVER
bvp4c / BVP SOLVER
Based on
Runge-Kutta (MIRKDC)(damped) Newton root-finderMesh adaptation based on error estimation(⇒ more refinement for boundary layers)
Can handle parameters
Can handle 1/r type singularities
bvp4c: Matlab, BVP SOLVER: Fortran
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Numerical implementation Method 1: bvp4c / BVP SOLVER
Transformation to remove 1/r2 singularity
By introducing:P (r) = rmφ(r),
Pridmore-Brown transforms into:
rm[φ′′ + φ′
(2m+ 1
r+ β(r, k)
)+ φ
(mrβ(r, k) + γ(r, k)
)]= 0,
where
β(r, k) =2ku′
ω − ku
γ(r, k) =(ω − ku)2
c2− k2
Convert to first order system, φ(r) = φ1(r) and φ′(r) = φ2(r):[φ1
φ2
]′=
1r
[0 0−βm −(2m+ 1)
] [φ1
φ2
]+
[0 1−γ −β
] [φ1
φ2
]
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Numerical implementation Method 1: bvp4c / BVP SOLVER
Handeling the 1/r singularity
First order system:
φ′(r) =1rSφ(r) +Aφ(r)
Use:
limr→0
Sφ(r)− φ(0)
r − 0= Sφ′(0)
Make sure that Sφ(0) = 0, then:
φ′(0) = Sφ′(0) +Aφ(0)
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Numerical implementation Method 2: COLSYS
COLSYS
Robust BVP solver: COLNEW / COLSYS (Fortran) [1] [2]
Based on
collocation at Gaussian points (⇒ no evaluation in singular pointr = 0)B-splines (piecewise polynomial functions)(damped) Newton root-finderMesh adaptation based on error estimation(⇒ more refinement for boundary layers)
rr = 0 ︸ ︷︷ ︸
subinterval
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Numerical implementation Method 2: COLSYS
Problem formulation for COLSYS
Add dif. eq. for parameter k: k′ = 0
k′ = 0P ′′ = −β(r, k)P ′ − γ(r, k)P
Split into real and imaginary parts
k′R = 0, k′I = 0,P ′′R = −βR(r, kR, kI)P ′R + βI(r, kR, kI)P ′I
− γR(r, kR, kI)PR + γI(r, kR, kI)PI ,P ′′I = −βI(r, kR, kI)P ′R − βR(r, kR, kI)P ′I
− γI(r, kR, kI)PR − γR(r, kR, kI)PI
COLSYS solves for {kR, kI , PR, P ′R, PI , P ′I}Calculate JacobiansSimilarly for BCs
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Numerical implementation Continuation in Z
Hard wall, no flow
No flow: P (r) = AJm(αr), α2 = ω2 − k2
Hard walls: P ′(1) = J ′m(α) = 0
−20 −15 −10 −5 0 5 10 15 20−50
−40
−30
−20
−10
0
10
20
30
40
50
Re(k)
Im(k
)
right runningleft running
Figure: h = 0, d = 1,m = 3, ω = 20.
Here: using p = P (r) exp(+iωt− ikx− imθ) convention29 / 47
Numerical implementation Continuation in Z
Continuation in Z
Good initial guess is important ⇒ continuation
Re
Im i∞
−i∞
Z
Z = R+ iX
Keep R constant
Vary X from −∞ to ∞(from hard wall to hard wall)
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(Numerical) results
Outline
1 Introduction & BackgroundProject motivationAcoustic linersGeneral project goalBrush-up: modes
2 Modeling the problemPridmore-Brown equation (ODE)Boundary conditions
3 Numerical implementationMethod 1: bvp4c / BVP SOLVERMethod 2: COLSYSContinuation in Z
4 (Numerical) resultsImpedance wall, no flowImpedance wall, uniform mean flowSome numerical problems
5 Future plans31 / 47
(Numerical) results Impedance wall, no flow
R is large: close to hard wall
−30 −20 −10 0 10 20 30
−25
−20
−15
−10
−5
0
5
10
15
20
25
Figure: Trajectories of k for R = 2, X runs from −∞ to ∞,h = 0, d = 1,m = 3, ω = 20, no mean flow.
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(Numerical) results Impedance wall, no flow
R becomes smaller: trajectories join
−30 −20 −10 0 10 20 30
−25
−20
−15
−10
−5
0
5
10
15
20
25
Figure: Trajectories of k for R = 1.5, X runs from −∞ to ∞,h = 0, d = 1,m = 3, ω = 20, no mean flow.
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(Numerical) results Impedance wall, no flow
R becomes smaller: acoustic surface waves arise
−30 −20 −10 0 10 20 30
−25
−20
−15
−10
−5
0
5
10
15
20
25
Figure: Trajectories of k for R = 1, X runs from −∞ to ∞,h = 0, d = 1,m = 3, ω = 20, no mean flow.
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(Numerical) results Impedance wall, no flow
Surface wave
k far away from hard wall value⇒ magnitude of Im(α) < 0 is largeThen:
|P (r)| =∣∣∣∣Jm(αr)Jm(α)
∣∣∣∣ 'eIm(α)(1−r)√r
⇒ P (r) decays away from wall: surface wave
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(Numerical) results Impedance wall, uniform mean flow
Mean flow: trajectories shift
−40 −30 −20 −10 0 10 20 30
−30
−20
−10
0
10
20
30
Figure: Trajectories of k for R = 2, X runs from ∞ to −∞,h = 0, d = 1,m = 3, ω = 5, u = 0.5.
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(Numerical) results Impedance wall, uniform mean flow
Mean flow: poles go to ∞
−40 −30 −20 −10 0 10 20 30 40
−30
−20
−10
0
10
20
30
Figure: Trajectories of k for R = 0.5, X runs from ∞ to −∞,h = 0, d = 1,m = 3, ω = 5, u = 0.5.
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(Numerical) results Impedance wall, uniform mean flow
Mean flow: hydrodynamic surface waves arise
−40 −30 −20 −10 0 10 20 30 40
−30
−20
−10
0
10
20
30
Figure: Trajectories of k for R = 0.2, X runs from ∞ to −∞,h = 0, d = 1,m = 3, ω = 5, u = 0.5.
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(Numerical) results Impedance wall, uniform mean flow
Mean flow: hydrodynamic surface waves arise
−40 −30 −20 −10 0 10 20 30 40
−30
−20
−10
0
10
20
30
Figure: Trajectories of k for R = 0.1, X runs from ∞ to −∞,h = 0, d = 1,m = 3, ω = 5, u = 0.5.
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(Numerical) results Some numerical problems
Some numerical problems
bvp4c: no problems, but slow
COLSYS: some convergence problems
BVP SOLVER: currently working on it
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(Numerical) results Some numerical problems
Everything ok
−15 −10 −5 0 5 10 15
−10
−5
0
5
10
intermediatehard−wallsoft−wall
Figure: Paths of wave number k for several modes, where ω = 5, m = 1,Ma = 0.08, and Z = 1 + iZi where Zi runs from -100 to 100.
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(Numerical) results Some numerical problems
Convergence problems
−15 −10 −5 0 5 10 15
−10
−5
0
5
10
intermediatehard−wallsoft−wall
Figure: Paths of wave number k for several modes, where ω = 5, m = 1,Ma = 0.09, and Z = 1 + iZi where Zi runs from -100 to 100.
42 / 47
(Numerical) results Some numerical problems
More convergence problems
−20 −15 −10 −5 0 5 10 15
−10
−5
0
5
10
15
intermediatehard−wallsoft−wall
Figure: Paths of wave number k for several modes, where ω = 5, m = 1,Ma = 0.3, and Z = 1 + iZi where Zi runs from -100 to 100.
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(Numerical) results Some numerical problems
More convergence problems
−25 −20 −15 −10 −5 0 5 10 15 20 25
−20
−15
−10
−5
0
5
10
15
20
intermediatehard−wallsoft−wall
Figure: Paths of wave number k for several modes, where ω = 5, m = 5,Ma = 0.3, and Z = 1 + iZi where Zi runs from -100 to 100.
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Future plans
Outline
1 Introduction & BackgroundProject motivationAcoustic linersGeneral project goalBrush-up: modes
2 Modeling the problemPridmore-Brown equation (ODE)Boundary conditions
3 Numerical implementationMethod 1: bvp4c / BVP SOLVERMethod 2: COLSYSContinuation in Z
4 (Numerical) resultsImpedance wall, no flowImpedance wall, uniform mean flowSome numerical problems
5 Future plans45 / 47
Future plans
Future plans
1 Create fast and robust solver
2 Add non-uniform flow
3 Add non-locally reacting liners
4 Add segmented liners
5 Add temperature gradients
46 / 47
Future plans
Thank you for your attention
47 / 47
Appendix
Bibliography I
U. Ascher, J. Christiansen, and R.D. Russel.Collocation software for boundary-value odes.ACM Transaction on Mathematical Software, 7(2):209–222, June1981.
Uri M. Ascher, Robert M.M. Mattheij, and Robert D. Russel.Numerical solution of Boundary Value Problems for OrdinaryDifferential Equations.Computational Mathematics. Prentice Hall, 1988.
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