1 Fundamental Solutions to the Wave Equation
Physical insight in the sound generation mechanism can be gained by considering simple analytical solutions to the wave equation. One example is to consider acoustic radiation with spherical symmetry about a point ~y = {yi}, which without loss of generality can be taken as the origin of coordinates. If t stands for time and ~x = {xi} represent the observation point, such solutions of the wave equation, ∂2 ( − c2∇2)φ = 0, (1) ∂t2 o will depend only on the r = |~x − ~y|. It is readily shown that in this case (1) can be cast in the form of a one-dimensional wave equation ∂2 ∂2 ( − c2 )(rφ) = 0. (2) ∂t2 o ∂r2 The general solution to (2) can be written as
f(t − r ) g(t + r ) φ = co + co . (3) r r The functions f and g are arbitrary functions of the single variables τ = t± r , respectively. ± co They determine the pattern or the phase variation of the wave, while the factor 1/r affects only the wave magnitude and represents the spreading of the wave energy over larger surface as it propagates away from the source. The function f(t − r ) represents an outwardly co going wave propagating with the speed c . The function g(t + r ) represents an inwardly o co propagating wave propagating with the speed co.
2 The Pulsating Sphere
Consider a sphere centered at the origin and having a small pulsating motion so that the equation of its surface is r = a(t) = a0 + a1(t), (4) where |a1(t)| << a0. The fluid velocity at the sphere surface is dr v = =a ˙(t). (5) r dt At the surface of the sphere ∂φ ( ) =a ˙(t). (6) ∂r a A Taylor expansion of (6) gives ∂φ ∂φ ∂2φ ( ) = ( ) + (a − a )( ) + ··· (7) ∂r a ∂r a0 0 ∂r2 a0 If ω and λ are representative of the frequency and wave length associated with the acoustic 2 ∂ φ a1ω a1 |a1|ω a1 wave, then (a−a0)( 2 )a ≈ larger{ a,˙ a˙}. Hence, if << 1, or equivalently << 1 ∂r 0 c0 a0 c0 λ
1 ∂2φ , then |(a − a0)( ∂r2 )a0 | << |a˙|. This allows us to linearize the boundary condition along the sphere by transferring it to the mean position at a0, ∂φ ( ) =a ˙(t). (8) ∂r a0 The velocity potential can be expressed as in (3). Moreover since the sphere pulsating motion is the source of acoustic waves, the principle of causality suggests that g ≡ 0. Thus f(t − r ) φ = co . (9) r Applying the condition (8) at the sphere mean location,
f(t − a0 ) f˙(t − a0 ) ∂φ co co = − 2 − =a ˙(t) (10) ∂r a0 a0co Integration of (10) gives
Z t a co 0 0 0 − a (t−t ) 0 f(t) = −a0co a˙(t + )e 0 dt . (11) −∞ c0
Note that if T is a representative period of the sphere pulsation, coT/a0 = λ/a0, where λ is a representative of the sound wave length. If λ/a0 >> 1, then most of the contribution to 0 the integral (11) is when t ≈ t. Neglecting terms of O(a0/λ), we get
2 f(t) = −a0a˙(t), (12) and the acoustic field potential function is given by a2a˙(t − r ) φ = − 0 co . (13) r The expression for the acoustic pressure is a¨(t − r ) p0 = ρ a2 co (14) 0 0 r It is convenient to cast (13,14) in terms of the mass flow rate crossing the sphere of radius 2 m a0, m(t) = 4πρ0a a˙. f(t) = − and 0 4πρ0
m(t − r ) φ = − co , (15) 4πρ0r m˙ (t − r ) p0 = co , (16) 4πr m˙ (t − r ) m(t − r ) 1 c0 c0 vr = ( + 2 ) (17) 4πρ0 rc0 r This suggests that in the farfield, i.e., r >> λ, the acoustic pressure and velocity are in 0 phase and the specific acoustic impedance z = p /vr = ρ0c0.
2 2.1 Harmonic Motion If we have a harmonic motion a˙ =ve ¯ −iωt, (18) wherev ¯ is the amplitude of the pulsation velocity and ω its frequency. Substituting (18) into (11) and carrying out the integration, we get
−iω(t+ a0 ) e c0 f(t) = −a0c0v¯ . (19) c0 − iω a0 The expressions for the potential function, velocity and the pressure can be readily obtained by substituting (19) into (9),
m¯ φ = − √ ei(k(r−a0)+ϕ−ωt), (20) 2 4πρ0r 1 +ω ˜ m¯ −ikr + 1 v = √ ( ) ei(k(r−a0)+ϕ−ωt), (21) r 2 4πρ0r 1 +ω ˜ r iωm¯ p0 = − √ ei(k(r−a0)+ϕ−ωt). (22) 4πr 1 +ω ˜2
−1 2 where we have introducedω ˜ = ωa0/c0, ϕ = tan ω˜, k = ω/c0, andm ¯ = 4πa0vρ¯ 0. Note that the velocity is the sum of two terms. One terms is in-phase with the pressure and at large distance decays as 1/r. The other term is out of phase with the pressure and decays at large distance as 1/r2.
The specific acoustic impedance p0 kr z = = R + iX = ρ0c0 2 2 (kr − i), (23) vr 1 + k r where R is the resistance and X the reactance. In the farfield where r >> λ or, kr >> 1, the specific acoustic impedance is dominated by R which has the same value as a plane wave, z = ρ0c0. However, in the near field the reactance |X| is comparable to R for kr = O(1) and |X| >> R for kr << 1.
The expression for the instantaneous acoustic intensity is given by
|p0|2 X I = Rsin2(k(r − a ) + ϕ − ωt) + sin[2(k(r − a ) + ϕ − ωt))] . (24) R2 + X2 0 2 0
2 2 2 2 Note R/(R + X ) = 1/(ρ0c0) and X/(R + X ) = −1/(ρ0c0kr), thus (24) simplifies to
0 2 |p | 2 1 I = sin (k(r − a0) + ϕ − ωt) − sin[2(k(r − a0) + ϕ − ωt))] . (25) ρ0c0 2kr The term associated with the reactance X results from coupling the pressure with the out- of-phase term of the velocity. Its time average is zero and hence it does not contribute to
3 the propagation of acoustic energy. Only the term associate with the resistance, which is the result of coupling the in-phase velocity with the pressure produces radiated acoustic energy. The average acoustic intensity are,
2 2 ¯ m¯ ω I = 2 2 2 , (26) 32π ρ0c0(1 +ω ˜ )r m¯ 2ω2 P = 2 . (27) 8πρ0c0(1 +ω ˜ )
For a sphere of small radius compared to the wave length, i.e., a0 << λ, the acoustic impedance |z| << ρ0c0 in the near field. Moreover, since R << X, the impedance is strongly reactive and the surrounding fluid acts mainly as an inertial mass. The velocity is practically out of phase with the pressure. Thus, a source of small size is an inefficient radiator of acoustic energy.
3 The Simple Source
The limit of the pulsating sphere solution as the sphere radius a0 becomes very small rep- resents the simple source or monopole solution. In this case, the source is characterized by the source mass flow rate 2 m(t) = 4πa0a˙(t), and the exact solution is given by (15). If the source is located at the point |~y|, then m(t − r ) φ = − co , (28) 4πρ0r where r = |~x − ~y|. Equation (28) states that, at the observation point ~x and time t, the sound signal received was emitted from the source point ~y at the retarded time τ = t − r . co
The velocity and pressure are given by
"m˙ (t − r ) m(t − r )# ∂φ 1 c0 c0 vr = = + 2 (29) ∂r 4πρ0 rc0 r ∂φ m˙ (t − r ) p0 = −ρ = c0 (30) 0 ∂t 4πr 3.1 Harmonic Sources:
In this case
m =me ¯ −iωt (31)
m¯ m¯ φ = − e−iωτ = − ei(kr−ωt), (32) 4πρ0r 4πρ0r
4 wherem ¯ represent the strength of the source and k = ω/c0. Similarly, the the velocity expression is given by " # ∂φ 1 m˙ (t) m(t) ikr vr = = + 2 e (33) ∂r 4πρ0 rc0 r or
m¯ −iω 1 i(kr−ωt) vr = + 2 e (34) 4πρ0 rc0 r Noting that ω = 2π , c0 λ m¯ −i2π 1 i(kr−t) vr = + 2 e (35) 4πρ0 rλ r
iωm¯ p0 = − ei(kr−ωt). (36) 4πr At large distance, r λ, the acoustic intensity is given by
2 2 0 m¯ ω 2 r I = p vr = 2 2 sin (ω(t − )) (37) 16π ρoc0r c0 and 2 2 ¯ 1 0 m¯ ω I = Re(p v¯r) = 2 2 (38) 2 32π ρoc0r m¯ 2ω2 P¯ = (39) 8ρoπc0
3.2 Simple source distribution:
Suppose we have N sources located at ~yi with strength mi, then the principle of superposition states that:
N N ri mi(t − ) X 1 X c0 φ = φi = − (40) i=1 4π i=1 ri where ri = |~x − ~yi|.
5 4 The Dipole:
Consider two sources of equal and opposite strength ±m located at ~y ± ~`/2. The potential of the two sources is
1 m(t − r+ ) m(t − r− ) φ = − [ c0 − c0 ], (41) 4πρ0 r+ r− ~ ~ where r = |~x − ~y| and r± = |~x − (~y ± `/2)|. We further assume |`| r, then "m(t − r )# 1 ∂ c0 φ = `i + ... (42) 4πρ0 ∂xi r or ~ r 1 `m(t − c ) φ = ∇ · 0 + .... (43) 4πρ0 r The pressure is given by
~ r 0 1 f(t − c ) p = − ∇ · 0 + ..., (44) 4π r or r 1 ∂ fi(t − ) p0 = − ( c0 ), (45) 4π ∂xi r ~ ~ ~ where f = {fi} =m ˙ `. Note that f has the dimension of a force. The pressure can be more explicitly expressed as
r r ∂fi(t − ) fi(t − ) 0 1 1 c0 c0 (xi − yi) p = [ + 2 ] · , (46) 4π rc0 ∂t r r We can always assume that dipole is located at a finite distance, i.e., |~y| is finite. The far field acoustic velocity and pressure are the leading terms of those fields as the observer ~x → ∞,
1 df~(t − r ) ~x p0 = c0 · . (47) 4πc0|~x| dt |~x| or 1 df(t − r ) p0 = c0 cosθ, (48) 4πc0|~x| dt where θ is the angle between f~ and ~x and f = |f~|. Note that for |~y| << |~x|, r has the following expansion
~y · ~x 1 r = |~x| − + O( ). (49) |~x| |~x|
6 Note that the general expression for the velocity can be obtained by from (43). However, in the farfield, the acoustic velocity reduces to its radial component, vr, whose expression is simply given in terms of the acoustic pressure
p0 vr = . (50) ρ0c0 The acoustic intensity has the following expression
I = p0v. (51) The average acoustic intensity is then given by
1 p0p0 I = . (52) 2 ρoco. Using (48), we get
f˙(t − r )f˙(t − r )cos2θ c0 c0 I = 2 3 2 . (53) 32π ρ0c0r The total power radiated is obtained by integration of a sphere of radius r, this gives
f˙(t − r )f˙(t − r ) 1 c0 c0 P = 3 . (54) 24π ρ0c0 In many applications, the force representing the dipole is created by the interaction of flow nonuniformities and turbulence interaction with structural components such as wings, fan and compressor blades. The fluctuating pressure along the surface of these structure rep- resent a dipole distribution. The farfield acoustics due to this unsteady pressure can be calculated by summing the contribution of all the dipoles. As an example, we consider a flat plate airfoil in a nonuniform flow composed of a uniform flow V in the x1 direction and a transverse gust disturbance
v =ve ¯ i(k1x1−ωt), (55) ω Where k1 = V . The surface pressure jump is given by v u c 0 2 v¯ u 2 − y1 ∗ −iωt ∆ps = −2ρ0V t c S(ω )e , (56) V 2 + y1 where ω∗ = ωc/(2V ) is the reduced frequency and S(ω∗) is the Sears function. The elemen- 0 tary force applied on the elementary surface dA = dy1dy2 is df = −∆psdA. The acoustic pressure is given by
v c Z +b/2 Z +c/2 u |~x−~y| 0 1 2 v¯ ∗ u − y1 −iω(t− ) 2 c0 p = 2ρ0V S(ω ) t c e dy1dy2. (57) 4πc0|~x| V −b/2 −c/2 2 + y1
7 For simplicity we use the compact source approximation which assumes that the body size is small compared o the wave length. Here, it implies ω∗ << 1 or c << λ. In this case
|~x| 2 v ∗ iω( −t) f = πρ cV S(ω )e c0 , (58) 0 V is the force per unit span b. Using this expression to calculate P, we get
"π v ωc b2 # P = (ρ c3A) M 6( )2( )2 SS. (59) 0 0 6 V 2V A
5 The Quadrupole:
The concept of a dipole as a combination of two monopoles with equal but opposite strength can be generalized to define a quadrupole as a combination of two dipoles of equal but ~0 0 opposite strength separated by ` = {`i}. Following the same procedure, we obtain
2 r fi(t − ) 0 1 0 ∂ c0 p = `j ( ). (60) 4π ∂xi∂xj r If we introduce the tensor
0 Tij = `jfi. (61) The expression for the pressure (60) can be cast as
2 r 1 ∂ Tij(t − ) p0 = ( c0 ) (62) 4π ∂xi∂xj r
Note Tij has the dimension of a shear × volume. The far field expression for a quadrupole is
0 1 ¨ r xixj p = 2 Tij(t − ) 2 (63) 4πrc0 c0 r
If ~`0 is in the same direction as f~, the quadrupole is said to be longitudinal, and we have
2 0 1 d T 2 p = 2 2 cos θ, (64) 4πrc0 dt where T = |`~0|f. On the other hand, if ~`0 is normal to f~, the quadrupole is said to be lateral, and we have 2 0 1 d T p = 2 2 cosθsinθ, (65) 4πrc0 dt
8 6 The Green’s Function of the Wave Equation
For Laplace equation, the Green’s function G(~x,~y) is solution to the inhomogeneous equation
∇2G = −4πδ(~x − ~y).. (66)
Integrate 66 over a sphere centered at ~y of radius r = |~x − ~y|, Z ∇2G d~x = −4π. (67) Σ This can be rewritten using the divergence theorem as Z Z dG ∇2G d~x = ∇G · ~ndΣ = 4πr2 = −4π dr Σ Σ which gives the free-space Green’s function for Laplace equation as 1 G = . (68) r The Green’s function G(~x,~y, t, τ) for the wave equation is solution to
2 ! 2 1 ∂ ∇ − 2 2 G = −4πδ(~x − ~y) δ(t − τ). (69) co ∂t Because of the spherical symmetry, the general solution to 69 is
f(t − τ − r ) G = co ., (70) r For the cases where r 6= 0, t 6= τ, the function satisfies
2 ! 2 1 ∂ ∇ − 2 2 G = 0. co ∂t Near r = 0,
2 ! 2 1 ∂ 2 1 ∇ − 2 2 G ∼ f(t − τ)∇ = −4πδ(~x − ~y) δ(t − τ). co ∂t r Hence, f(t) = δ(t) and we have the following expression for the Green’s function
δ t − τ − |~x−~y | G = co . (71) |~x − ~y |
9 6.1 Harmonic Time-Dependence For a harmonic time-dependence, we seek solutions of the form
φ(~x,t) = ϕ(~x)e−iωt. (72)
The function ϕ satisfies the Helmholtz equation
(∇2 + k2)ϕ = 0, (73)
where k = ω/c0. The Green’s function for 73 is solution to
(∇2 + k2)g(~x,~y) = −4πδ(~x − ~y). (74)
It is readily seen that
e±ikr g(~x,~y) = . (75) r This can be also obtained by taking the Fourier transform in time of 71 and noting that
+∞ e−i(ωt±kr) 1 Z r = δ t − τ ± e−iωτ dτ. (76) r r co −∞
7 Distribution of Sources and Forces
The linearized Euler equations for non-viscous non-heat-conducting fluid with no-mean mo- tion are ∂ ρ ~v = f~ − ∇p0 (77) o ∂t ∂ρ0 + ∇ · ρ ~v = q (78) ∂t o where f~ is a force per unit volume and q is a mass flow rate per unit volume. Combining the two equations gives
2 ! 2 1 ∂ 0 ∂q ~ ∇ − 2 2 p = − + ∇ · f. (79) co ∂t ∂t In order to find solutions to the acoustic wave equation satisfying 79, we consider the inho- mogeneous equation
2 ! 2 1 ∂ 0 ∇ − 2 2 p = −h(~x,t). (80) co ∂t Using the Green’s function 71, and noting that Z h(~x,t) = h(~y, τ)δ(~x − ~y)δ(t − τ)d~ydτ, (81)
10 we get
1 Z h(~y, t − r ) p0 = c0 d~y + p0 , (82) 4π r 0 0 where p0 is a solution to the homogeneous equation and accounts for the effects of boundaries. In what follows we focus on the inhomogeneous solution.
7.1 Source Distribution ∂q In this case h = ∂t . Hence, we have
1 Z ∂q (~y, t − r ) p0 = ∂t c0 d~y, (83) 4π r
7.1.1 Single Source
For a single source at ~y0, q(~y, t) = m(t)δ(~y − ~y0). Substituting into 83 , we get back the pressure expression 30 for a single source.
m˙ (t − r0 ) p0 = c0 , (84) 4πr0
where r0 = |~x − ~y0|.
7.2 Force Distribution
Force a distribution {fi}, we consider the solution to the equation
2 ! 2 1 ∂ ∇ − 2 2 ϕi = fi. (85) co ∂t This gives
r 1 Z fi(~y, t − ) ϕ = − c0 d~y. (86) i 4π r Taking the divergence of (86), we get
r 1 ∂ Z fi(~y, t − ) p0 = − c0 d~y. (87) 4π ∂xi r A more direct expression for the pressure is given by
∂fi r r Z (~y, t − ) fi(~y, t − ) 0 1 ∂t c0 c0 (xi − yi) p = [ + 2 ] d~y. (88) 4π rc0 r r The acoustic field is given by
1 Z ∂fi (~y, t − r ) (x − y ) p0 = [ ∂t c0 ] i i d~y. (89) 4πc0 r r
11 7.2.1 Single Force
For a single force at ~y0, fi(~y, t) = Fi(t)δ(~y − ~y0). Substituting into (88) , we get back the pressure expression (30) for a single source,
dFi r0 r0 (t − ) Fi(t − ) 0 1 dt c0 c0 (xi − y0i) p = [ + 2 ] . (90) 4π r0c0 r0 r0 The acoustic field is 1 dF r p0 = (t − 0 )cosθ, (91) 4πrc0 dt c0 where F is the magnitude of the force F and θ is the angle between F and the observer direction x. Note that as expected these results are identical to those obtained for a dipole. This confirms that a dipole represents the field of a force.
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