Sound Propagation in Ducts

Transcription

1 Sound Propagation in Ducts Hongbin Ju Department of Mathematics Florida State Universit, Tallahassee, FL Please send comments to: In this section we will discuss sound propagation in ducts. This is the tpical case of sound waves in confinements. Sound wave can onl propagate freel in one direction. It is confined in other dimensions b duct walls. Rectangular Duct with Uniform Mean Flow The scales will be used are: length: duct height h, velocit: sound speed a 0, time: h /a 0, densit: ambient densit r 0, pressure: r 0 a 0, impedance: r 0 a 0. The uniform mean flow has densit and pressure: r = r 0 =1, p = p 0 =1/g (g =1.401). Mean velocit u = M is in +x-axis direction. The linear Euler equations for the fluctuation quantities are: u x + u u x t x = - 1 p r x, u + u u t x = - 1 p r, u z t + u u z x = - 1 p r z, p t + u p x + gp Ê u x x + u + u ˆ z Á = 0. Ë z Variables with bars are mean flow quantities, and other variables are fluctuation quantities. Isentropic process is assumed, therefore densit can be computed directl from pressure: r = p. We will use Fourier Transform and method of separation of variables to solve the equations. In and z directions there are wall boundaries. In x direction the duct extends to infinit. Fourier transforms about x and t can be performed on the Euler equations. It is equivalent to assume the next single Fourier component form of solutions: 1

2 È u x Ï È u ˆ x (,z) u ˆ u (,z) = RealÌ u z u ˆ z (,z) Î p Ó Î p ˆ (,z) e i(k xx-wt ) Substituting (1) into Euler equations, one can obtain the next equations:. (1) d pˆ d pˆ + + b pˆ = 0, b = w - k x, d dz () u ˆ x = k ˆ x p w, 1 dpˆ 1 dpˆ uˆ =, uˆ z =. iw d iw dz (3) where w = w - Mk x. Linear Euler equations with isentropic process have acoustic and vortical waves as the two normal mode solutions. We are onl interested in acoustic solutions, therefore w 0 and Eq.() must have the nontrivial solution (cht1.doc). Rigid Wall Duct If the four sides of the duct are all rigid walls, then boundar conditions are We ma tr the separation of variable: Substitution of it into () leads to: u ˆ = 0, at = 0 and = h ; (4) u ˆ z = 0, at z = 0 and z = d. (5) ˆ p (,z) = R()T(z). 1 d R R d + 1 d T T dz + b = 0. The first term on the left side onl depends on. The second terms onl depends on z. b is constant respect to and z. Therefore there must have: 1 d R R d = k, 1 d T T dz = k z, k + k z = b. One can use trial solution to solve the ODE. The general solution is then: ik -ik ikzz -ikzz pˆ(, z) = ( Ae + Be )( Ce + De ). (6)

3 To satisf boundar conditions (4), (5), we must have A = B, k = mp / h ; (7) C = D, k z = np / d. (8) m and n are integers. Because of the wall confinements, wave numbers k and k z can no longer be an real numbers. The must be some special real numbers, i.e., multiples of p / h and p / d respectivel. For this reason, k and k z are called eigenvalues. p ˆ(, z) is called the eigenfunction which has the next form: ˆ p mn (,z) = cos(mp /h)cos(npz /d), (9) where m = 0,1,L, n = 0,1,L. This is the standing wave solution. Plugging this eigenfunction into Eq.(), we have: For each eigenvalue b mn, we have two k x : b mn = p (m /h) + (n /d) = k + k z. (10) ± k xmn = -wu 0 ± i 1- M ( b -w / 1- M ) 1/ b + w / 1- M ( ) 1/ 1- M. (11) Branch cuts for b -w / 1- M + k xmn ( ) 1/ and b + w / 1- M ( ) 1/ are shown in Fig.1, so that - represents wave propagating in +x direction and k xmn in -x direction. Fig.1, Branch cuts for ( b -w / 1- M ) 1/ and b + w / 1- M ( ) 1/ in b -plane. 3

4 The general solution of p is: p = Real Âcos(mp /h)cos(npz /d)(a + e ikxmn + x + A - e ikxmn - x )e -iwt mn. (1) mn m= 0n= 0 Other variables can be computed from Eq.(3). It is noted that the eigenvalues and eigenfunctions in and z are independent of wave propagating directions in x. And the eigenfunctions are orthogonal. These are not true when there is mean flow and the walls are not rigid. Occasionall one needs to compute eigenvalue b from wave number k x : b = w - k x = -i(k x - k d ) 1/ (k x - k u ) 1/ (1- M ) 1/. (13) where k u = -w /(1- M) and k d = w /(1+ M). The branch cuts for (k x - k d ) 1/ (k x - k u ) 1/ are chosen as in the following figure: and Fig., Branch cuts for (k x - k d ) 1/ and (k x - k u ) 1/ in k x -plane. Soft Wall Duct without Mean Flow We will discuss a simple case in which the opposite boundaries have the same impedance. The boundar conditions are: ˆ p = -iwx p ˆ at = 0 ˆ and p = iwx p ˆ at ˆ p z = -iwx ˆ z p at z = 0 and ˆ p z = iwx ˆ z x and x z are admittance of soft walls. = h ; (13) p at z = d. (14) 4

5 We can still use trial solution Eq.(6). To facilitate the derivation, we can rewrite Eq.(6) as: ˆ p (,z) = cos(k + f )cos(k z z + f z ) (15) b defining A = A'e if, B = A'e -if, C = C'e if z, and D = C'e -if z. To satisf boundar conditions (13) in direction, we must have k tg(k h + f ) = -iwx ; (16) k tg(f ) = iwx. (17) From which we have, ÈÊ iwx ˆ iwx Á -1 tg( k h) = k k ÎË. (18) Eigenvalue k can be solved from this equation b grid search/newton iteration method. Another wa to solve equations (16) and (17) is to note: tg(k h + f ) = -tg(f ). (19) From which one possible solution is k h + f = -f, then: f = - 1 k h, k tg( 1 k h) = -iwx, (0) which corresponds to the even smmetric solution in Morse&Ingard1968, p.504. The other possible solution to Eq.(19) is k h + f = -f + p /, i.e., f = - 1 k h + p, k ctg( 1 k h) = iwx. (1) which corresponds to the odd smmetric solution in Morse&Ingard1968, p

6 k z can be obtained similarl. Eigenvalues k and k z are independent of wave propagation direction in x direction. However, eigenfunctions cos(k + f ) and cos(k z z + f z ) are no longer orthogonal as in the hard wall case. Circular Duct with Uniform Mean Flow Scales used here are: length: duct diameter D, velocit: sound speed a 0, time: D / a0, densit: densit r 0, pressure: r 0 a 0, impedance: r 0 a 0. The mean flow: r = 1 and p =1/ g ( g =1.401) are constant; Mean velocit u = M is in +x-axis direction and constant. The Euler equations for the fluctuation quantities are: u x + u u x t x = - 1 p r x, u r t + u u r x = - 1 p r r, u f + u u f t x = - 1 p r r f, p t + u p x + gp Ê u x x + u r r + u r r + u ˆ f Á = 0. Ë r f (Densit can be computed directl from pressure: r = p.) In radial direction there is wall boundar and in circumferential direction there is periodic boundar. In x direction the duct extends to infinit. Based on Fourier Transform, we can assume the next single Fourier component form of solutions: È u x Ï È u ˆ x (r,f) u r ˆ u r (r,f) = RealÌ u f u ˆ f (r,f) Î p Ó Î p ˆ (r,f) Substituting it into Euler equations, we have: e i(k xx-wt ). () d pˆ dpˆ d pˆ r + r + + r b pˆ = 0, dr dr df (3) u ˆ x = k x r w ˆ 1 dpˆ 1 dpˆ uˆ r =, uˆ f =. irw dr irw rdf (4) 6

7 where w = w - u k x, b = r gp w - k x. To solve p ˆ( r, f) from Eq.(3), assume the separation of variables: Substituting it into Eq.(3), r R d R dr p ˆ( r, f) = R( r) F( f). (5) + r R dr dr + r b = - 1 d F F df. Left side of the equation onl depends on r, while the right side is onl a function of f. Then the must be equal to a constant: The general solution of Eq.(6) is: r R 1 d F F df = -n. (6) dr dr + r b = n, (7) d R dr + r R F(f) = Ae -inf + Be inf. (8) A, B and n are determined b two boundar conditions in f direction. Eq.(7) is the Bessel equation. Its general solution is: R(r) = CJ n (br) + DY n (br), (9) J n (br), Y n (br) are the n th order first and second kinds of Bessel functions. The are the two independent solutions to the Bessel equation (7). C, D and b are determined b two boundar conditions in the radial direction. For a circular duct, periodic boundar condition applies in f direction. The general solution to Eq.(6) is simpl: F ) imf ( f = Be, (30) where n = m is an integer. In the radial direction, the boundar at r = 0 is that pˆ must be finite. Y m (br) is infinitive at r = 0. Therefore the general solution of R (r) is R(r) = CJ m (br). (31) 7

8 The general solution of p ˆ( r, f) for a circular duct is: ˆ p (r,f) = A'J m (br)e imf. (3) b is to be determined b the boundar condition at r = Ro ( R o is the radius of the duct). All other variables can be computed from p ˆ (r,f) b Eq.(4). Rigid Wall Circular Duct For a rigid wall duct, the boundar condition at To satisf (33), one has the eigenvalue: r = Ro is ˆ u r = 0. (33) b ' / R, (34) mn = j mn 0 where j' mn is the nth root of J' m (r) = 0. (See Appendix for how to find j' mn.) ± Wave number k xmn can be computed from Eq.(11). Then the time domain solution of p is: p = Real e i(mf-wt ) (A + mn e ik xmn   + x + A - e ikxmn - x )J (rb ). (35) mn m mn m=- n= 0 b mn and J m (rb mn ) are the same for both directions in x. This is not true when there is mean flow the duct wall is not rigid. Annular Duct with Uniform Mean Flow An annular duct has an inner duct with radius R i and an outer duct with radius R o. The similar analsis for the circular duct in the last section applies to the annular duct. The periodic boundar condition and the general solution (30) still appl. In the radial direction, r = 0 is not in the phsical domain as in the circular duct. Therefore Ym ( b mr) should be included in the solution, and the general solution in radial direction is (9) with Bessel function of integer order: R(r) = CJ m (br) + DY m (br). (36) C, D and b are to be determined b two boundar conditions at r = Ri and r = Ro. Rigid Wall Annular Duct 8

9 For a rigid wall annular duct, the boundar conditions are ˆ u r = 0, at r = R i and r = Ro. (37) First of all let's check if b = 0 is the eigenvalue of the problem. Y m (0r) Æ -, then D must be zero and R(r) = CJ m (0r). Since J m (0r) =1 for m = 0, J m (0r) = 0 for m 0, the onl nontrivial solution to Eq.(36) is: This represents a plane wave. R(r) = CJ 0 (b 00 r) = C, b 00 = 0. (38) Now we assume b 0. From boundar condition (37), the dispersion relation is obtained from (36): J' m (br i )Y' m (br o ) = J' m (br o )Y' m (br i ). (39) It is not trivial to solve eigenvalue b from Eq.(39). When the gap of annular duct, (R i - R o )/R o, is small, we can obtain an approximation solution of b. We begin with the Bessel equation (7). B replacing r b the medium radius R c = (R i + R 0 )/, the Bessel equation becomes a second order ODE with constant coefficients: Its general solution is: R c d R dr l ± are both real, or a conjugate pair. To satisf boundar conditions (37), + R c dr dr + (R cb - m )R = 0. R(r) = Ce l + r + De l - r, l ± = -1± i 4(R cb - m ) -1. R c R o - R i R c e -(l + -l - )(R o -R i ) =1, or, 4(R c b - m ) -1 = pn, n=0,1,,... Its solution is b mn = p n (R o - R i ) + 1 R c (m ). (40) 9

10 This is the eigenvalue for annular duct with small gap. If the gap is not small, one ma appl the Newton iteration method to solve Eq.(39) numericall. Define function: f (b) = J' m (br i )Y' m (br o ) - J' m (br o )Y' m (br i ). For initial eigenvalue b (0), the more accurate eigenvalue is b (1) = b (0) + Db, Db = - f (b (0) )/ f '(b (0) ). This process should be repeated until Db is small. A successful Newton method depends on how good initial eigenvalue b (0) is. The next figure shows a tpical f (b). Separate b -axis into intervals,..., b n, b n +1,... If f (b n ) f (b n +1 ) 0, b n is a good choice of b (0). A FORTRAN90 code, modenumber_annularduct.f, using this method to compute eigenvalues in a solid wall annular duct is attached. The time domain solution of p is: p = Real(A + 00 e ik x + 00 x + A - 00 e ik x - 00 x )e -iwt n= 0 if m 0 n=1 if m= 0 m=18, R o =1, R i =11/15. +Real e i(mf-wt ) (A + mn e ik xmn + x + A - mn e ik xmn  m =- Â È - x ) J m (b mn r) - J' m (b mn R o ) Y' m (b mn R o ) Y (b r). (41) m mn Î With solution of pˆ, all other variables can be computed from Eq.(4). C-Duct with Uniform Mean Flow In the outlet of jet engines, the annular duct is often separated b installations, as illustrated b Fig.3. The duct has a C-shape. It is here called C-duct. The C-duct has an inner duct with radius R i and an outer duct with radius R o. The separation plates are at f s and f e. 10

11 Fig.3, C-duct. The first major difference of C-duct from the annular duct is the periodic boundar condition is no longer applicable in f direction. Similar to the argument for Eq.(9) in rectangular duct, the general solution of F (f ) is: where F(f) = Bcos[n m (f - f s )], (4) n m = mp /(f e - f s ), m = 0,1,,L. (43) m is an nonnegative integer. n m ma not be an integer as in the circular or annular duct unless when f e - f s = p. (When f e - f s = p, the solution in a C duct is the same as that in an annular duct.) In the radial direction the general solution is: R(r) = CJ n m (br) + DY n m (br). (44) The second major difference of the C-duct from annular duct is the Bessel functions here ma have non-integer orders. Rigid Wall C-Duct For the rigid wall C-duct, the boundar conditions in radial direction are ˆ u r = 0, at r = R i and r = Ro. (45) 11

12 The similar argument as for Eq.(38) leads to the plane wave solution: R(r) = J 0 (b 00 r), b 00 = 0. (46) If b 0, the next dispersion relation can be derived from (44) and (45): J' n m (br i )Y' n m (br o ) = J' n m (br o )Y' n m (br i ). (47) Eigenvalue b is to be determined b this equation. The difficult involved here is the Bessel functions with noninteger order. From the similar procedures for Eq.(40), when the gap of annular duct, (R i - R o )/R o, is small, the eigenvalue is approximated b: b mn = p n (R o - R i ) + 1 R (n m + 1 ), n=0,1,,... (48) c 4 If the gap of the annular duct is not small, we can use (48) as an initial value in a Newton iteration method. The final results need to be reorganized. The time domain solution of p is: Ï p = Real (A + 00 e ik x + 00 x + A - 00 e ik x - 00 Ì x )e -iwt + Ó È cos[n m (f - f s )] J n m (b mn r) - J' n m (b mn R o ) Y' n m (b mn R o ) Y n m (b mn r) Î (A + mne ik xmn + x + A - mn e ik xmn   - x )e -iwt m= 0n= 0 if m 0 n=1 if m= 0 With solution of pˆ, all other variables can be computed from Eq.(4).. (49) Circular Duct with Nonuniform Mean Flow (Boundar Laer) Scales used will be: length: duct radius R, velocit: sound speed a 0, densit: ambient densit r 0, pressure: r 0 a 0, impedance: r 0 a 0. The mean flow: r 0 =1 and p 0 =1/g (g =1.401) are constant; Shear mean velocit u 0 (r) is in +x-axis direction. Assume the next form of solutions: 1

13 È u x Ï È u ˆ x (r) u r ˆ u r (r) = RealÌ u f u ˆ f (r) Î p Ó Î p ˆ (r) e i(ax+mf-wt ). (50) (Densit can be computed directl from pressure: r = p.) From Euler equations, one can obtain the next equations for shear flow: dˆ u r dr - i È w - 1 Ê w a ˆ Á + m Ê p ˆ + a du 0 Î Ë r w dr + 1 ˆ Á u ˆ Ë r r = 0, (51) dˆ p dr - iw u ˆ r = 0, (5) where w = w - u 0 a. For a wall with impedance Z, the boundar condition is p ˆ = Zˆ u r, at r =1. (53) Equations (51) and (5) and boundar condition (53) form an eigenvalue problem. Onl some special values of a can satisf these equations. These a are eigenvalues, and the respective functions p ˆ and u ˆ r are eigenfunctions. An important propert of eigenvalues of Eqs.(51)&(5) is given here. Suppose a, p ˆ, u ˆ r are a set of eigenvalue and eigenfunctions of this problem, then a *, p ˆ *, u ˆ * r form a set of eigenvalue and eigenfunctions for Eqs.(51)&(5) satisfing the next B.C.: p ˆ * = Z * u ˆ * r, at r =1. (54) Here superscript * represents conjugate. Under some circumstances B.C.(53) is exactl the same as B.C.(54), such as for hard wall, u ˆ r = u ˆ * r = 0, or the impedance with onl resistance, Z = Z *. Then a *, p ˆ *, u ˆ * r are also the set of eigenvalue and eigenfunctions for the original problem [Eqs.(51)&(5) with B.C.(53)]. Newton iteration method can be used to solve the eigenvalues and eigenfunctions. For an initial guess of the eigenvalue a 0, integrate Eqs.(51)&(5) numericall from the outside of the boundar laer to r =1 to get u ˆ r and p ˆ at r =1. u ˆ r and p ˆ are functions of a 0 : u ˆ r (a 0 ), p ˆ (a 0 ), which ma not satisf B.C. (53) at r =1. Similarl we can have another solution u ˆ r (a 1 ) and p ˆ (a 1 ) for initial eigenvalue a 1 near a 0. Suppose the exact eigenvalue is a 0 + Da, i.e., ˆ p (a 0 + Da) = Zˆ u r (a 0 + Da). 13

14 To the first order of Talor expansion, we have: p ˆ (a Da = 0 ) - Zˆ u r (a 0 ) Z dˆ u r(a 0 ) da - dˆ p (a. 0) da If we use dˆ u (a ) r 0 ª u ˆ (a ) - u ˆ (a ) r 1 r 0, dˆ p (a ) 0 da a 1 -a 0 da ª p ˆ (a ) - p ˆ (a ) 1 0, we can get Da to have a a 1 -a 0 better approximation to the accurate eigenvalue. The process can be repeated until certain accurac is achieved. This is the Newton s iteration method. Choosing appropriate initial value a 0 is critical for a successful Newton s iteration method. Eigenvalues of plug flow without boundar laer, Eq.(34) and Eq.(11), are good choice of the initial eigenvalues. In most of the time these choices work well. But in some cases Newton s method fail when using these initial eigenvalues. Another wa for choosing initial eigenvalues is the grid search. Compute function: f (a) = ˆ p (a)/z - ˆ u r (a) (55) for a matrix in a-plane, draw the contours of real(f)=0 and imag(f)=0 in the a-plane using a commercial software. The intersection points in the a-plane should be the solution of the eigenvalues. Read in these data and use them as the initial eigenvalues in the Newton iteration. As an example, let's find eigenvalues for mean flow with turbulent boundar laer in a circular rigid wall duct with radius: The velocit profile of the turbulent boundar laer is shown in Fig.4. Mach number: 0.45, boundar laer thickness: 1, Fig.4, Velocit profile of turbulent boundar laer, Mach number: 0.45, boundar laer thickness: 1. 14

15 The grid search method is used to get the initial guess of eigenvalues. Since the wall is rigid, both a and a * are the eigenvalues according to the propert discussed above. Therefore in Fig.5 onl the up half of the a-plane is needed. Read in the data of the intersection points of solid and dashed lines as the initial values in the Newton s method. Fig.5, Contours of real(f)=0 (solid lines) and imag(f)=0 (dashed lines) in the a-plane The next figures show the results. Fig.6 shows the eigenvalues when frequenc is just above the cuton frequenc. Fig.7 is for frequenc is well above the cuton frequenc. Fig.8 is for frequenc well below the cuton frequenc. In most of the cases, the effect of shear flow is to move real parts of plug flow eigenvalues to right while the imaginar parts keep nearl the same. This change is ver small, and using plug flow eigenvalues as initial values in Newton method will work. The onl exception is when the frequenc is just above the cut-on frequenc (Fig6). There are two cuton modes in the plug flow. The boundar laer makes these propagating modes into cut-off modes. This explains wh a cut-on mode wave was damped in experiments. In this case the plug flow eigenvalues as the initial guess will fail. Fig.6, Eigenvalues at 741Hz for plug flow (square) and shear flow with turbulent boundar laer (triangle). Mach number: 0.45, boundar laer thickness: 1, duct 15

16 radius: 6.33, azimuthal mode number m=, cut-on frequenc for the 1 st radial mode of the plug flow: 740.6Hz. [Real(a) in the right figure is rescaled to show the difference.] Fig.7, Eigenvalues at 800Hz for plug flow (square) and shear flow with turbulent boundar laer (triangle). Mach number: 0.45, boundar laer thickness: 1, duct radius: 6.33, azimuthal mode number m=, cut-on frequenc for the 1 st radial mode of the plug flow: 740.6Hz. Fig.8, Eigenvalues at 730Hz for plug flow (square) and shear flow with turbulent boundar laer (triangle). Mach number: 0.45, boundar laer thickness: 1, duct radius: 6.33, azimuthal mode number m=, cut-on frequenc for the 1 st radial mode of the plug flow: 740.6Hz. [Real(a) in the right figure is rescaled to show the difference.] A numerical integration method can also be used. 16