Oscillatory Gap Damping

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Άργος Ουζουνίδης
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1 Oscillatory Gap Damping Find the damping due to the linear motion of a viscous gas in in a gap with an oscillating size: ) Find the motion in a gap due to an oscillating external force; ) Recast the solution in terms of an effective lateral bulk drag term; 3) Solve the Fourier-periodic solution driven by an equivalent uniform mass source. ) Here's the solution for oscillatory flow in a gap driven by an externally applied force: g < z < g d dt v d D v + f e iωt D γ f dz ρ external applied force per mass cosh( kz) vzt (, ) v e i cosh k g ωt i k δ γ f v δ ωρ iω cosh ( i) z δ δ :.5 φ : πframe z :.5, vz : i cosh i exp( iφ) δ.4. z..4 Re( v( z) ) For large drag (δ), the velocity is in phase with the force. Small drag has the velocity lagging the force by π/ phase. ) Now get the average velocity:

2 g v av g g cosh( kz) vz dz v z g cosh k g d v tanhk g kg Describe by a bulk drag coefficient b (drag force per mass) dv av dt f bv av + f v av b iω Choose b to force the two solutions agree: f b iω f iω tanhk g kg b( ω) iω i tanh i δ g γ ωρ g b( ) γ G( ) G( ) : ρg i i coth i small... G( ) i b γω ρg ( i) δ g ω( i) The imaginary part reduces the average equivalent flow by removing δ from the undamped flow. large... G( ) 6 b γ large drag ρg Wave Propagating into the Gap Now consider a D compressible wave propagation into in a narrow gap... e i( kxωt) fluid acceleration δv δp δt ρ δx bv

3 conservation of mass δp c δt ρ δv δx ik iωv P ρ bv iω c P ρikv ik iω ( b iω) ρ ik ρc P v k ρc + iω[ ( b iω) ρ] k ω + iωb( ω) c ibω k k + k ω i ω + G( ) k c Q( ) Re( k) Im( k) j j :.. :. j rf i. j ( j ) Re rf j : + G( j ) Q : j Im rf j Q j... j

4 3) Finally, drive the motion by an equivalent mass influx uniformly distributed in the gap. Rectangular Flow-Driven Gap Gap oscillating as gt () g + η( x) e iωt η( x) η in the gap Convert the oscillating gap to a fixed gap with an equivalent oscillatory gas source. fluid acceleration δv δt δp ρ δx b( ω) v conservation of mass δp c δt ρ δv δx + δg g δt The driven solution with the source and the pressure at the gap boundaries will be a Fourier series that is constant in the gap and opposite just outside the gap, periodically reversing for the Fourier series... Square wave function, η for < x < going to zero at the booundaties: η( x) 4η ( n ( ) nπ n cos nπx sum over n odd P P n cos q x e iωt v v n n sin q x e iωt n δg δt iωgη( x) e iωt iωv n sin( q x n ) ρ q P q n n sin( q x n ) b( ω) v n sin( q x n ) iω c P n cos( q x n ) ρ q v n n cos( q x n ) + iωgη cos q g ( x n ( n ) ) iωv ρ qp b( ω) v Temporarily drop the n index. iω c P ρqv + ( iωgη) g

5 q ρ ω i ρc iω b q P v iωη P v ( ω + iωb) ρ c iωqc η c q ω iωb Check out the case of b for getting our bearings... P v ω ρc iωqc η c q ω η( x) 4η ( n ( ) nπ n cos nπx Px 4η ω ρc ( n ( ) n nπ c nπ ω cos nπx Force per length on the plate... F Px dx 8 η ω ρc ( n ( nπ) c nπ ω F( ω) 8 η ρc π ( n n n ω ω πc πc

6 n :, 3.. R( ) : n :, n n 5 R( ) At low frequencies, the gas flows into the gap for positive η. As the first resonance is approached, the amplitude of the gas in-flow increases, resulting in the positive pressure-to-η phase relation. After the resonance, the phase is reversed. This negative phase relation is what would be expected for trapped gas indergoing rarefaction (negative pressure) with increasing. Now include the damping term b... ρ ω + iωb c P η n c q ω n iωb Px 4η ω + iωb Force per length on the plate... ρc ( n n ( ) nπ c nπ ω iωb cos nπx F Px dx 8 η ω + iωb ρc ( n ( nπ) c nπ ω iωb

7 b ω γ G( ) G( ) πcg + i G( ) i tanh i b + i ω tanh i i γ πcρg F( ω) 8 η ρc π tanh i i ( n n n tanh i i Now put in some real numbers... P p p :.5 ATm Fits to NIST Thermodata p cp : γ( p) p : + ρ( p) : 4.45p p.5 :.3 w :. g : 6 ω : π34 γ( p) :.3 πcp ρ( p) g : ω πcp.45 M p : π wη ρc F H( ) π

8 : u H, Σ for n, Σ Σ + uσ tanh i i n n u u :, ( ) ( (, )) Re H, u Im H u 3 4 u This is the force with viscous loss, behaving the same as the b case, but now with the imaginary part. The real part of the scaled force is opposite to the displacement for our situation (around. This is an extra restoring force due to the gas compression. The averagy loss comes from the positive imaginary part of the force: p :.,... Effective Mass f Mω x

9 8 wη ρc π Re H, ( ) Mω gη ( ) 8 3 wρ Re H, M π 4 g μ( p) 8 3 wρ ( p ) γ( p) : π 4 gm p πcp ρ( p) g ω πcp Re H, ( ).5. μ( p) p Damping F 8 wη ρc π H( ) W Re( Fv ) Re F ( iωη g ) ω η g Im( F) E M p ωgη

10 Q ω E M W ω p ( ωgη ) ω η 8 wη ρc gim H( ) π π 4 gm p 4w ρ 3 Im( H( ) ) Qp : π 4 gm p 4w ρ( p) 3 γ( p) πcp ρ( p) g ω πcp ( ) Im H, 5 4 Qp 3.. p

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