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. of Mech. Engineering 29 University of Kentucky ME 510 Vibro-Acoustic Design Exponential Form
Recall that any harmonic function may be expressed as either the real or imaginary part of a similar but complex function, e.g.,
jωt jωt Fo cos(ωt) = Re{Foe } or Fo sin(ωt) = Im{Foe } e jθ = cosθ + j sinθ e j(ωt−kx ) = cos(ωt − kx)+ j sin(ωt − kx) e j(ωt−kx ) = e jωte− jkx (This is a plane harmonic wave traveling in the +x direction.)
e j(ωt+kx ) = e jωte+ jkx = cos(ωt + kx)+ jsin(ωt + kx)
(This is a plane harmonic wave traveling in the -x direction.)
Dept. of Mech. Engineering 30 University of Kentucky ME 510 Vibro-Acoustic Design Harmonic Solution of 1D Wave Equation
d 2 p(x,t) 1 d 2 p(x,t) The 1-D wave equation for plane waves: = dx 2 c2 dt 2 For harmonic waves: p(x,t)= ~p(x)e jωt
2 2 jωt d p x,t d e 2 Substituting: jωt ( ) 1 ( ) 1 jωt e 2 = 2 p (x) 2 = 2 p (x)( jω) e dx c dt c 2 d p (x) 2 − jkx jkx 2 + k p (x) = 0 solution p (x) = C1e +C2e dx − jkx jωt jkx jωt − j(kx−ωt) j(kx+ωt) p(x,t) = C1e e +C2e e = C1e +C2e
Re{p(x,t)} = −C1 cos(kx −ωt) +C2 cos(kx +ωt) p(x,t) = Acos(kx −ωt) + Bcos(kx +ωt) +x wave -x wave
Dept. of Mech. Engineering 31 University of Kentucky ME 510 Vibro-Acoustic Design Harmonic Solution of 1D Wave Equation
In Complex Exponential Form j(ωt−kx) j(ωt+kx) p (x,t) = p+e + p−e
Particle Velocity
p+ j(ωt−kx) p− j(ωt+kx) ux (x,t) = e − e ρ0c ρ0c
Dept. of Mech. Engineering 32 University of Kentucky ME 510 Vibro-Acoustic Design Example
ux = cos ωt ( ) ux = 0 jωt ux =1e
L
p+ j(ωt−kx) p− j(ωt+kx) ux (x,t) = e − e ρ0c ρ0c
at x = 0 p p ρ0c 1 + − p+ = = − 1 e2 jkL ρ0c ρ0c −
at x = L p jkL p jkL 0 = + e − − e− p− = p+ − ρ0c ρ0c ρ0c
Dept. of Mech. Engineering 33 University of Kentucky ME 510 Vibro-Acoustic Design Example - Magnitude f = 1000 Hz, c = 343 m/s, L = 1 m
Dept. of Mech. Engineering 34 University of Kentucky ME 510 Vibro-Acoustic Design Example - Phase
Dept. of Mech. Engineering 35 University of Kentucky ME 510 Vibro-Acoustic Design Specific Impedance
p Z = ux
For the the free plane wave case (no reflected wave)
+ Z0 = ρ0c
Dept. of Mech. Engineering 36 University of Kentucky ME 510 Vibro-Acoustic Design Example
u = cos ωt x ( ) Z (L) = ρc jωt ux =1e
L
j(ωt−kx) j(ωt+kx) p (x,t) = p+e + p−e
p+ j(ωt−kx) p− j(ωt+kx) ux (x,t) = e − e ρ0c ρ0c
Dept. of Mech. Engineering 37 University of Kentucky ME 510 Vibro-Acoustic Design Example at x = 0 p p 1= + − − ρ0c ρ0c p− = 0 at x = L − jkL + jkL p c p e + p e + = ρ0 Z L = ρ c = + − ( ) 0 p p + e− jkL − − e+ jkL ρ0c ρ0c
Solution − jkx jωt p(x,t) = ρ0ce e
Dept. of Mech. Engineering 38 University of Kentucky ME 510 Vibro-Acoustic Design Sound Intensity and Power
The rate at which work is done by a sound wave as it travels is: I dE S = F ⋅u = (pS )⋅u (J / s = W) p,u dt Here S is the area over which the sound wave acts. Since S depends on the situation we define the sound intensity I as: dE I = S = pu (W / m2 ) dt The average intensity (a more useful quantity) is therefore: 1 T p2 p 2 I = ∫ p(x,t)u(x,t)dt = + = + T 0 2ρoc ρoc where P is the amplitude of the sound wave and T is the period.
Dept. of Mech. Engineering 39 University of Kentucky ME 510 Vibro-Acoustic Design Sound Intensity and Power
The sound power transferred by a sound wave of amplitude p+ is:
I p2 W = IS = + S (W) S 2ρoc
Example: A fan on a household furnace produces 0.001 W of sound power that propagates along the supply duct having cross- sectional area of 0.06 m2. What is the sound pressure of the sound produced by the fan in the supply duct?
2ρ cW 2 ⋅ 415⋅1×10−3 p = o = = 3.72 Pa I + S 0.06
Dept. of Mech. Engineering 40 University of Kentucky ME 510 Vibro-Acoustic Design Reference Quantities and dB Scale
2 ⎛ prms ⎞ Recall: LP (dB)=10 log10 ⎜ ⎟ pref = 20µPa ⎜ p ⎟ ⎝ ref ⎠
⎛ W ⎞ −12 Sound Power Level: LW (dB)=10 log10 ⎜ ⎟ Wref =1×10 watts ⎜W ⎟ ⎝ ref ⎠ Example: what are the sound pressure and power levels of the fan in the previous problem? ⎛ 3.72 / 2 ⎞ L = 20 log ⎜ ⎟ =102.4 dB re 20 µPa P 10 ⎜ −6 ⎟ ( ) ⎝ 20 ×10 ⎠
⎛ 0.001⎞ −12 LW =10 log10 ⎜ 12 ⎟ = 90 dB (re 10 W) ⎝ 10 − ⎠
Dept. of Mech. Engineering 41 University of Kentucky ME 510 Vibro-Acoustic Design Approximate Relationship between LP and LW
2ρ cW ρ cW P = o → p = o S rms S 2 Approximation! ⎛ p ⎞ W S L =10 log ⎜ rms ⎟ =10 log P 10 ⎜ p ⎟ 10 p2 c ⎝ ref ⎠ ref ρo 2 −6 2 −12 pref ρoc = (20 ×10 Pa) 415 = 0.96 ×10 watts ≈ Wref W S W LP =10 log10 =10 log10 −10 log10 S Wref Wref 2 LP = LW −10 log10 S (S in m )
Comparing to the previous (exact) calculation:
LP = 90 −10 log10 (0.06)=102.2 dB (re 20 µPa)
Dept. of Mech. Engineering 42 University of Kentucky ME 510 Vibro-Acoustic Design Spherical Wave Propagation
z 2 r,θ,φ 2 1 ∂ p ( ) ∇ p − = 0 2 2 θ c ∂t r y
∂p 1 ∂p 1 ∂p φ ∇p = eˆ + eˆ + eˆ x r ∂r θ r ∂θ φ rsinθ ∂φ
1 ∂ 2 1 ∂ 1 ∂uφ ∇⋅u = 2 r ur + (uθ sinθ) + r ∂r ( ) rsinθ ∂θ rsinθ ∂φ 2 2 1 ∂ $ 2 ∂p' 1 ∂ $ ∂p ' 1 ∂ p ∇ p = &r )+ &sinθ )+ r2 ∂r % ∂r ( r2 sinθ ∂θ % ∂θ ( r2 sin2 θ ∂φ 2
Dept. of Mech. Engineering 43 University of Kentucky ME 510 Vibro-Acoustic Design Spherical Wave Propagation
The wave equation for no angular z dependence (i.e. a point source or (r,θ,φ) monopole) θ r ∂2 p 2 ∂p 1 ∂2 p + = y ∂r2 r ∂r c2 ∂t 2 x φ Equation of motion ∂u ∂p ρ r + = 0 0 ∂t ∂r
Dept. of Mech. Engineering 44 University of Kentucky ME 510 Vibro-Acoustic Design Harmonic Solution for Free Spherical Wave
Sound pressure a distance r from the point source:
A j t kr A j t kr p(r,t) = + e (ω − ) + − e (ω + ) r r
r
|p(r,t)|
Free field (no reflections) r This is similar to a plane wave, but for spherical waves the sound pressure amplitude decreases with distance from the source of sound.
Dept. of Mech. Engineering 45 University of Kentucky ME 510 Vibro-Acoustic Design Harmonic Solution for Free Spherical Wave
Assume no reflected wave
A j t kr p(r,t) = + e (ω − ) r r 1 ∂p(r,t) ur (r,t) = − ∫ dt ρ0 ∂r
A $−1 −ik ' j ωt−kr = − + + e ( ) dt Free field (no reflections) ∫ & 2 ) ρ0 % r r (
A+ ! 1 $ j(ωt−kr) ur (r,t) = #1+ &e ρ0cr " ikr %
Dept. of Mech. Engineering 46 University of Kentucky ME 510 Vibro-Acoustic Design Impedance of a Free Spherical Wave
p 1 ikr Z = = ρ c = ρ c u 0 1 0 1 ikr r 1+ + ikr
For kr << 1 (the nearfield) r
Z (r) ≈ jρ0ckr
For kr >> 1 (the farfield) Free field (no reflections)
Z (r) ≈ ρ0c
Dept. of Mech. Engineering 47 University of Kentucky ME 510 Vibro-Acoustic Design Sound Intensity and Sound Power
1 A2 I r Re pu* + r ( ) = ( ) = 2 2 2ρ0cr A2 W = I (r)4πr2 = 2π + ρ0c
S = 4πr2
r
Imaginary sphere enclosing source
Dept. of Mech. Engineering 48 University of Kentucky ME 510 Vibro-Acoustic Design