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Chapter 4 – the Wave Equation and Its Solution in Gases and Liquids

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Chapter 4 – the Wave Equation and Its Solution in Gases and Liquids Chapter 4 – The Wave Equation and its Solution in Gases and Liquids Slides to accompany lectures in Vibro-Acoustic Design in Mechanical Systems © 2012 by D. W. Herrin Department of Mechanical Engineering University of Kentucky Lexington, KY 40506-0503 Tel: 859-218-0609 [email protected] Simplifying Assumptions • The medium is homogenous and isotropic • The medium is linearly elastic • Viscous losses are negligible • Heat transfer in the medium can be ignored • Gravitational effects can be ignored • Acoustic disturbances are small Dept. of Mech. Engineering 2 University of Kentucky ME 510 Vibro-Acoustic Design Sound Pressure and Particle Velocity pt (r,t) = p0 + p(r,t) Total Pressure Undisturbed Pressure Disturbed Pressure Particle Velocity ˆ ˆ ˆ u(r,t) = uxi + uy j + uzk Dept. of Mech. Engineering 3 University of Kentucky ME 510 Vibro-Acoustic Design Density and Temperature ρt (r,t) = ρ0 + ρ (r,t) Total Density Undisturbed Density Disturbed Density Absolute Temperature T (r,t) Dept. of Mech. Engineering 4 University of Kentucky ME 510 Vibro-Acoustic Design Equation of Continuity A fluid particle is a volume element large enough to contain millions of molecules but small enough that the acoustic disturbances can be thought of as varying linearly across an element. Δy # ∂ρ &# ∂u & ρ + t Δx u + x Δx ρ u % t (% x ( x x Δz $ ∂x '$ ∂x ' Δx Dept. of Mech. Engineering 5 University of Kentucky ME 510 Vibro-Acoustic Design Equation of Continuity Δy # ∂ρt &# ∂ux & %ρt + Δx(%ux + Δx( ρ u $ x '$ x ' x x Δz ∂ ∂ Δx ∂ $ ∂ρt '$ ∂ux ' (ρtΔxΔyΔz) = ρtuxΔyΔz −&ρt + Δx)&ux + Δx)ΔyΔz ∂t % ∂x (% ∂x ( Dept. of Mech. Engineering 6 University of Kentucky ME 510 Vibro-Acoustic Design Equation of Continuity ∂ $ ∂ρt '$ ∂ux ' (ρtΔxΔyΔz) = ρtuxΔyΔz −&ρt + Δx)&ux + Δx)ΔyΔz ∂t % ∂x (% ∂x ( ∂ (ρtΔxΔyΔz) = ρtuxΔyΔz − ρtuxΔyΔz ∂t $ ∂ρt ∂ux ∂ρt ∂ux 2 ' −&ux Δx + ρt Δx + (Δx) )ΔyΔz % ∂x ∂x ∂x ∂x ( ∂ρ ∂ρ ∂u t = −u t − ρ x ∂t x ∂x t ∂x ∂ ρ 0 + ρ ∂ ρ 0 + ρ ∂u ( ) = −u ( ) − ρ + ρ x ∂t x ∂x ( 0 ) ∂x ∂ρ ∂u = −ρ x ∂t 0 ∂x Dept. of Mech. Engineering 7 University of Kentucky ME 510 Vibro-Acoustic Design Equation of Continuity ∂ρ ∂u + ρ x = 0 ∂t 0 ∂x 3D Equation of Continuity ∂ρ + ρ ∇⋅u = 0 ∂t 0 Dept. of Mech. Engineering 8 University of Kentucky ME 510 Vibro-Acoustic Design Equation of Motion Δy # & ∂(p0 + p) % p + p + Δx(ΔyΔz (p0 + p)ΔyΔz ( 0 ) Δz $ ∂x ' Δx Dept. of Mech. Engineering 9 University of Kentucky ME 510 Vibro-Acoustic Design Equation of Motion $ ' ∂(p0 + p) Fx = (p0 + p)ΔyΔz −&(p0 + p) + Δx)ΔyΔz % ∂x ( ∂ p + p =− ( 0 ) ΔxΔyΔz ∂x ∂p = − ΔxΔyΔz ∂x Dept. of Mech. Engineering 10 University of Kentucky ME 510 Vibro-Acoustic Design Equation of Motion ux (x + Δx,t + Δt) − ux (x,t) ax = lim Δt→0 Δt ∂ux ∂ux ux + (x + Δx − x) + (t + Δt − t) +− ux ∂x ∂t ax = lim Δt→0 Δt Δx = uxΔt ∂u ∂u u + x u Δt + x Δt +− u x ∂x x ∂t x ax = lim Δt→0 Δt ∂u ∂u a = u x + x x x ∂x ∂t Dept. of Mech. Engineering 11 University of Kentucky ME 510 Vibro-Acoustic Design Equation of Motion ∂u ∂u a = u x + x x x ∂x ∂t Assuming small disturbances the 1st term is small. ∂u a = x x ∂t Dept. of Mech. Engineering 12 University of Kentucky ME 510 Vibro-Acoustic Design Equation of Motion ∂u a = x x ∂t ∂p F = − ΔxΔyΔz x ∂x ∂p ∂ux − ΔxΔyΔz = (ρ0 + ρ)ΔxΔyΔz ∂x ∂t ∂u ∂p ρ x + = 0 0 ∂t ∂x 3D Equation of Motion ∂u ρ + ∇p = 0 0 ∂t Dept. of Mech. Engineering 13 University of Kentucky ME 510 Vibro-Acoustic Design Thermodynamic Equation of State The pressure-density relation for a fluid is usually non-linear: pt = f (ρt ) pt p = pt − p0 p ∂p o p = ρ t ∂ρt ρt =ρ0 pt , ρt ρ = ρt − ρ0 ρo ρt For small (acoustic) changes about the ambient state: ∂p p = ρ t ∂ρt ρt =ρ0 Dept. of Mech. Engineering 14 University of Kentucky ME 510 Vibro-Acoustic Design Thermodynamic Equation of State Consider a fluid of fixed mass M occupying volume Vo M = ρ0V0 M = (ρ0 + ρ)(V0 +δV) ≈ ρ0V0 +V0ρ + ρ0δV M , Vo V0ρ + ρ0δV = 0 δV ρ = −ρ0 V0 Dept. of Mech. Engineering 15 University of Kentucky ME 510 Vibro-Acoustic Design Thermodynamic Equation of State Bulk Modulus Volumetric Strain ∂pt ∂pt δV p = ρ = −ρ0 ∂ρt ∂ρt V0 ρt =ρ0 ρt =ρ0 (compare with σ = Eε for solids, where E is Young’s modulus) The fluid compressibility is the reciprocal of the bulk modulus. ∂pt β = ρ0 ∂ρt ρt =ρ0 Dept. of Mech. Engineering 16 University of Kentucky ME 510 Vibro-Acoustic Design For an Ideal Gas p Equation of State: t = RT ρt For an adiabatic process: γ γ p + p ! ρ + ρ $ p ! ρ $ 0 = # 0 & ''→ t = # t & p0 " ρ0 % p0 " ρ0 % c γ = p cv ! $γ−1 ∂pt ρt 1 γ p0 = p0γ # & ≈ = γRT0 ∂ρt ρ0 ρ0 ρ0 ρt =ρ0 " % Dept. of Mech. Engineering 17 University of Kentucky ME 510 Vibro-Acoustic Design The Speed of Sound ∂p -2 kg m s-2 m-2 2 Units of t Pa N ⋅ m ( ⋅ ⋅ )⋅ " m % 3 = -3 = -3 =$ ' ∂ρt ρt =ρ0 kg m kg⋅ m kg⋅ m # s & ∂p β c = t = ∂ρt ρ0 ρt =ρ0 For an ideal gas undergoing adiabatic compression and expansion: R 8.315 J ∂p γ p γRT = mol K c = t = 0 = 0 ∂ρt ρ0 M 0 ρt =ρ0 Mass of a mole of gas Dept. of Mech. Engineering 18 University of Kentucky ME 510 Vibro-Acoustic Design The Speed of Sound γ p γRT c = 0 = ρ0 M 0 5 For: p0 = 1.013x10 Pa (one atmosphere) 3 o ρ0 = 1.21 kg/m (air at 20 C) γ = 1.402 (air) 1.402 ⋅1.013×105 m m c = = 343.0 1.21 s s Dept. of Mech. Engineering 19 University of Kentucky ME 510 Vibro-Acoustic Design The Speed of Sound The speed of sound is a function of temperature γRT c = M 0 The speed of sound at one temperature is easily related to another T c = c0 T0 Dept. of Mech. Engineering 20 University of Kentucky ME 510 Vibro-Acoustic Design Deriving the Wave Equation The Continuity Equation ∂ρ ∂u t + ρ x = 0 ∂t 0 ∂x Differentiate with time ∂2ρ ∂2u t + ρ x = 0 ∂t 2 0 ∂x∂t Dept. of Mech. Engineering 21 University of Kentucky ME 510 Vibro-Acoustic Design Deriving the Wave Equation The Equation of Motion ∂u ∂p ρ x + = 0 0 ∂t ∂x Differentiate with position ∂2u ∂2 p ρ x + = 0 0 ∂t∂x ∂x2 Dept. of Mech. Engineering 22 University of Kentucky ME 510 Vibro-Acoustic Design Deriving the Wave Equation ∂2ρ ∂2u t + ρ x = 0 ∂t 2 0 ∂x∂t ∂2u ∂2 p ρ x + = 0 0 ∂x∂t ∂x2 Subtract first from second equation ∂2 p ∂2ρ − t = 0 ∂x2 ∂t 2 Dept. of Mech. Engineering 23 University of Kentucky ME 510 Vibro-Acoustic Design Deriving the Wave Equation ∂2 p ∂2ρ − t = 0 ∂x2 ∂t 2 Use equation of state to eliminate ρt ∂pt ρ β = ρ0 or p = β ∂ρt ρ ρt =ρ0 0 Then ∂2 p ρ ∂2 p − 0 = 0 ∂x2 β ∂t 2 ∂2 p 1 ∂2 p − = 0 ∂x2 c2 ∂t 2 Dept. of Mech. Engineering 24 University of Kentucky ME 510 Vibro-Acoustic Design The Wave Equation In 1D ∂2 p 1 ∂2 p − = 0 ∂x2 c2 ∂t 2 In 3D 2 2 1 ∂ p ∇ p − = 0 c2 ∂t 2 2 2 2 2 $ ∂ ∂ ∂ ' ∇ p = & + + ) p %∂x2 ∂y2 ∂z2 ( Dept. of Mech. Engineering 25 University of Kentucky ME 510 Vibro-Acoustic Design Solutions to the Wave Equation In One Dimension t = t 1 λ Section of duct Piston oscillates with frequency f Higher pressure Lower pressure (compression) (rarefaction) Waves move with speed of sound c t = t2 Dept. of Mech. Engineering 26 University of Kentucky ME 510 Vibro-Acoustic Design Functional Description Δx = c Δt 1.5 1 t = 0 t = Δt 0.5 0 (a traveling wave) -0.5 Sound Pressure Sound(Pa) Pressure -1 λ -1.5 Distance - x Wavelength ! 2π $ !ω $ c 2πc For the wave at t = 0 : p(x) = Po sin# x& = Po sin# x& λ = cT = = " λ % " c % f ω !ω $ For the wave at t = Δt : p(x, Δt) = Po sin# (x − cΔt)& " c % Wavenumber !ω $ ω For the wave at any t : p(x,t) = Po sin# (x − ct)& = Po sin (kx −ωt) k = " c % ( ) c Dept. of Mech. Engineering 27 University of Kentucky ME 510 Vibro-Acoustic Design Functional Description A stationary observer at any point sees the sound pressure oscillate between –Po and +Po exactly f times per second Δt = Δx/c 1.5 1 x= x 1 x = x2 0.5 0 Phase shift of ωΔt -0.5 Time - t Sound Pressure (Pa) Sound -1 T = 1/f -1.5 p(x1,t) = Po sin(kx1 −ωt) # ωΔx & p(x2,t) = Po sin k(x1 + Δx) −ωt = Po sin%kx1 + −ωt( ( ) $ c ' p(x2,t) = Po sin(kx1 +ωΔt −ωt) = Po sin(kx1 −ω (t − Δt)) Dept. of Mech. Engineering 28 University of Kentucky ME 510 Vibro-Acoustic Design Acoustic Particle Velocity Equation of Motion dp x,t NO FLOW!! Fluid particles only oscillate! ρ u = − ( ) o dx (series of springs and masses) d P sin kx t 1 ( o ( −ω )) Pok u(x,t) = − ∫ dt = − ∫ cos(kx −ωt)dt ρo dx ρo P k u(x,t) = − o sin(kx −ωt) ρ0 (−ω) P u(x,t) = o sin(kx −ωt) (i.e., a momentum wave in phase with ρoc the sound pressure wave) Dept.
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