On time domain methods for Computational Aeroacoustics Fang Q. Hu - PowerPoint PPT Presentation

On time domain methods for Computational Aeroacoustics Fang Q. Hu and Ibrahim Kocaogul Old Dominion University, Norfolk, Virginia Xiaodong Li, Xiaoyan Li and Min Jiang Beihang University, Beijing 100191, China

Noise prediction by linear acoustic wave propagation Noise source modelling + Noise propagation FW−H equation Kirchhoff integral Linearized Euler Equations Green’s function ......

Time Domain Wave Packet (TDWP) method t ) = ∆ t sin ( ω t ) 2 , | 0 e ( ln 0 . 01 )( t / M ∆ t ) t | ≤ M ∆ t t sinusoidal wave wave packet (single frequency) (broadband and mulch -frequency) Proposed broadband acoustic test pulse function for source: Ψ( π

Advantages of Time Domain Wave Packet (TDWP) method ◮ One computation for all frequencies (within numerical resolution) for linear problems ◮ Ability to synthesize broadband noise sources ◮ Acoustic source has a short time duration, so computation is more efficient than driving a time domain calculation to a time periodic state ◮ Separation of acoustic and hydrodynamic instability waves becomes possible ◮ Long numerical transient state is avoided

Application Examples 1. Sound propagation through shear flows 2. Vortical gust-blade interaction 3. Duct sound radiation problem

1. Sound source in a jet flow (a CAA Benchmark problem) y=50 y=15 2 + B 2 ) t ) e − ( ln 2 )( B x y x y Jet S ( x , y , t ) = sin (Ω Flow x=100 Single frequency source function:

Instability wave in a shear flow 60 40 20 y 0 -20 -40 -60 -50 0 50 100 150 x

Instability wave in duct radiation computation

Time Domain Wave Packet (TDWP) method Separation of acoustic and instability waves: t t t 1 < 2 < 3 . t 2 t 3 Shear layer instability wave t 1 Acoustic and instability waves travel at different speeds. An acoustic wave packet has a short time duration, it will be separated from the instability wave in time domain calculation, here

Time Domain Wave Packet approach y=50 2 + B 2 ) t ) e − ( ln 2 )( B x y x y S ( x , y , t ) = sin (Ω y=15 Jet Flow x=100 2 + B 2 ) t ) = Ψ( t ) e − ( ln 2 )( B x y x y S ( x , y , Single frequency source function: TDWP source function:

without suppression with suppression

y = 15 y = 50 Frequency domain solution recovered by FFT (Symbol: analytical; Line: computation) x = 100

2. Vortical gust-blade interaction V β u os ( α x + β y − ω t ) g = − (U ,V ) v V os ( α x + β y − ω t ) 0 0 g = Incident Vortical Gust : α

Vortical gust imposition in TDWP u g ( x , y , t ) = − B Ψ( t − Ax − By ) v g ( x , y , t ) = A Ψ( t − Ax − By ) os ( θ ) sin ( θ ) u v g g A = B = 0 θ u os ( θ )+ ¯ v sin ( θ ) , u os ( θ )+ ¯ v sin ( θ ) = x + ∂ y = 0 0 0 0 vortical wave packet ⇒ ∂ ∂ ∂ ¯ ¯

u 0 . 45 ) 0 = Example of time domain solution (v-velocity and pressure, ¯

Frequency domain solution by FFT (u-velocity and pressure)

Frequency domain solution (pressure magnitudes)

3. Duct mode imposition in TDWP P M L p p u u u 2 x u p mn ( y , z )Ψ( t ) e − σ 0 0 t + ¯ x + γ ¯ x + ∂ y + ∂ z Source plane Pressure equation at in-flow region: ∂ ∂ � ∂ � = φ ∂ ∂ ∂ ∂ ∂

Example of time domain solution

p ( x , y , z , t ) = φ y , z )ˆ P ( x , t ) mn ( 2 X T ′ Duct mode wave packet—exact solution D 2 � 2 − ¯ x ′ 2 P ( x , t ) = α J ( t − ¯ t ′ ) x t ′ ) e − σ dx ′ dt ′ 0 mn 2 Dt − X − T 2 2 2 x √ 2 � X T ′ mn − ω e − ¯ i ω ¯ t ′ D x ′ e t ′ ) e − σ dx ′ dt ′ 0 < ω < ¯ mn (cut-off) 2 − X − T Dt 2 2 mn − ω P ( x , ω ) = 2 2 − ¯ i ¯ x √ 2 mn 2 � X T ′ ie i ω ¯ t ′ D x ′ e t ′ ) e − σ dx ′ dt ′ 0 < ¯ mn < ω (cut-on) 2 − X − T 2 − ¯ 2 Dt mn M ˆ ˆ � � � 2 2 , β = � ˆ ¯ λ 1 − M t ′ = t ′ − β ( x − x ′ ) , ¯ x = | x − x ′ | / α Ψ( 2 , ¯ 1 − M  ¯ λ α � λ √ ¯ ´ ´ Ψ(   λ      ˆ    ω λ  α � λ  ´ ´ √ Ψ(   ω λ � � α = �

Time history (line: numerical, symbol:theoretical)

1 . 6 : 1 = 1 . 9 : 2 = Frequency domain solution 2 . 3 : 3 = ω 2 . 6 : 4 = ω ω ω

NASA/GE Fan Noise Source Diagnostic Test

Aft fan exhaust radiation problem p p u u u 2 z )Ψ( t ) e − σ ( x − x 0 ) u p mn ( y , 0 0 t + ¯ x + γ ¯ x + ∂ y + ∂ z Pressure equation at in-flow region: ∂ ∂ � ∂ � = φ ∂ ∂ ∂ ∂ ∂

Aft fan exhaust radiation problem

Aft fan exhaust radiation problem

Aft fan exhaust radiation problem

Aft fan exhaust radiation problem

Pressure field obtained by FFT at 2BPF, 61.7% design speed Mode (-10,2) introduced at source plane inside duct

Pressure field obtained by FFT at 3BPF, 61.7% design speed Mode (-10,2) introduced at source plane inside duct

Pressure field obtained by FFT at 4BPF, 61.7% design speed Mode (-10,2) introduced at source plane inside duct

Far-field modal transfer function, forward solution Forward Radiation, mode (-10,0) 49 50 45 51 35 40

Far-field modal transfer function, forward solution Forward Radiation, mode (-10,1) 49 45 50 51 40

Far-field modal transfer function, forward solution Forward Radiation, mode (-10,2) 47 45 49 35 50 51

Far-field modal transfer function, forward solution Forward Radiation, mode (-10,3) 40 35 45 49 50 51

Far-field modal transfer function, forward solution Forward Radiation, mode (-10,4)

P mn P mn = A mn Far-field modal transfer function (2BPF) ˆ Forward Radiation 49 (−10,0) 47 (−10,1) (−10,2) 45 (−10,3) 40 50 (−10,4) 35 51

Reciprocity Condition (forward and adjoint problems) ~ p (r’, ) ω p(r’,t) nm P mn ( r ′ , ω ) A ∗ mn mn A mn P ( r ′ , ω ) A ~ nm A mn R 2 A φ dS = 4 π r ) rdr mn = mn mn mn ( k D r mn H ˜ = α ˜ � ∂ω � � � ˆ ˆ ∗ α φ φ ˜ ∂

Adjoint solution, time domain

Adjoint solution, time domain

Adjoint solution, time domain

Adjoint solution, time domain

Adjoint solution, frequency domain at 2BPF

Adjoint solution, frequency domain at BPF

Adjoint solution, frequency domain at 3BPF

A ∗ P mn mn mn A mn = α P ˜ Reciprocity condition (2BPF) ˜ Forward Radiation 49 47 45 (−10,0) 40 (−10,1) 50 (−10,2) 35 (−10,3) 51 (−10,4) Adjoint Solution ~ 45

A ∗ P mn mn mn A mn = α P ˜ Reciprocity condition (2BPF) ˜ Forward Radiation 49 47 45 (−10,0) 40 (−10,1) 50 (−10,2) 35 (−10,3) 51 (−10,4) Adjoint Solution 48 ~

A ∗ P mn mn mn A mn = α P ˜ Reciprocity condition (2BPF) ˜ Forward Radiation 49 47 45 (−10,0) 40 (−10,1) 50 (−10,2) 35 (−10,3) 51 (−10,4) Adjoint Solution 49 ~

A ∗ P mn mn mn A mn = α P ˜ Comparison of modal transfer function ˜ Forward Radiation 49 47 45 (−10,0) 40 (−10,1) 50 (−10,2) 35 (−10,3) 51 (−10,4) Adjoint Solution ~

p ( r , ω ) = ∑ A P r , ω ) mn ( ω )ˆ mn ( m , n A ( r ) iA ( i ) A mn = mn + mn = Amplitude P mn = Modal transfer function Modal detection 1 • Far-field pressure 2 2 = ∑ 2 � | p ( r , ω ) | | A P r , ω ) mn ( ω ) | mn ( m , n 2 51 2 2 � 2 | A P r P MIN mn | mn ( i , ω ) i ( ω ) i = 38 m , n ˆ • Assume duct modes have no interference: � � ˆ (1) � � � • Minimization: � � ∑ ∑ � � ˆ − = (2) � � �

Far-field SPL measurements computation

Summary ◮ Time Domain Wave Packet formulation can be used to eliminate the initial long transient state that is often required in single frequency formulation ◮ Computational time is reduced due to shortened time duration of the wave packet; the wave packet method is preferred for linear propagation problems even if only solutions at a few frequencies are of interest ◮ Solution of the adjoint problem provides a useful tool for verifying the numerical results of Euler equations