on Acoustic Jumping Conditions for a Duct with an Area Expansion and a Non-zero Mean Flow

on Acoustic Jumping Conditions for a Duct with an Area Expansion and a Non-zero Mean Flow

Myung-Gon Yoon Department of Precision Mechanical Eng. Gangneung-Wonju National University

Wonju 26403, Republic of Korea

AbstractIn this paper we study the jump conditions for a one-dimensional acoustic model of a duct which has an abrupt area expansion and a non-zero mean flow. We investigate on the compatibility of the three (mass, energy and momentum) jump conditions across an area expansion and then propose to use either the mass-momentum or the energy-momentum jump conditions for a proper one- dimensional acoustic modeling. The combination of mass and energy jump condition is shown to be inappropriate.

KeywordsAcoustic Model, Jump Condition, Resonance Frequency

1. INTRODUCTION

   One-dimensional modeling of a simple duct seems to be the most elementary problem in an acoustics course. Basically for an acoustic model one simply uses a standard wave solution of the Hermholtz equation and then applies two boundary conditions at the inlet and the outlet of the duct, e.g., see [1], [2].

   This simple one-dimensional modeling problem how- ever becomes a bit confusing when there is a sudden area expansion in the duct and the duct is subject to non-zero mean flow. The confusion comes from the fact that there are several options in choosing jump conditions across the area expansion.

   More specifically, a standard one-dimensional solu- tion of the Hermholtz equation leaves us only two undetermined wave functions but there are many jump conditions including the next three conditions ;

   • mass conservation

   • energy conservation

   • momentum relation

   Authors of [3], [4] proposed to model an area expan- sion as a loss of stagnation pressure instead of the energy conservation.

   As will be shown in this paper, we have to choose only two jump conditions among many jump conditions in the course of one-dimensional modeling. This redundancy of jump condition is neither physically appealing nor easily

   justifiable, considering that those jump conditions should be somehow connected.

   A complete resolution comes if some of the jump conditions become dependent and only two independent conditions are left. In fact, this is the case of the zero mean flow and the three (mass, energy and momentum) conditions, as will be shown.

   Another way for a resolution is to justify a certain choice of jump conditions based on physical grounds. One may claim that energy conservation should be discarded as a sudden area expansion does work and thus propose to choose the mass and momentum conditions instead.

   One criticism for this selection is that there is still controversy on a proper momentum relation across an area expansion. For an example, authors of [5] and

   [6] (p. 396) proposed different momentum relations. Furthermore there are important engineering applications in which the energy relation should appear. For an example, in a thermo-acoustic modeling of gas turbines, a flame located at the point of area expansion serves as an energy source and therefore a thermo-acoustic model needs the energy jump condition.

   It seems that the problem of how to choose jump conditions has no definite answers, which motivates our investigation on the relations of three (mass, energy, momentum) jump conditions in this paper.

2. ACOUSTIC MODEL

   Fig. 1: Schematic of a Combustor

   1. Wave functions

      We consider a duct composed of two chambers as

      the perturbations of pressure, velocity and density at each chamber can be represented as combinations

      shown in Fig. 1. Each chamber [x

      k1

      , xk], k [1, 2]

      of wave functions in (1) below. Note that, for each

      k [1, 2], the three functions of pressure, velocity

      has downstream and upstream waves A+ and A waves

      k k and density waves are given as combinations of

      and there exist a sudden area expansion between two

      two independent function A±

      and therefore we

      chambers at x = x1.

      need four unknown

      k

      functions to describe pressure,

      A solution of the Hermholt equation gives that

      velocity and density waves over the full range [x0, x2].

      p (x, t) := p + A+ (t x xk1 \ + A (t xk x \

      k k k

      c + uk

      k c u1

      1 r ( x xk1 \ ( xk x \

      (x, t) := + 1 rA+ (t x xk1 \ + A (t xk x \

      u (x, t) := u + A+ t A t , (k = 1, 2) (1)

      k k c k

      c + uk

      k c uk

      k c2 k

      c + uk

      k c uk

      transform of {Ik, Jk} and A±k (t), respectively, and the

      Define

      + ( +)

      where {I k(s), J-k(s)} and A-±k (s) denotes the Laplace

      I1(t) := A1

      t 1

      J1(t) := A1 (t)

      2

      I2(t) := A+ (t)

      (2)

      quantities 1, 2 are defined

      := + + = (x

      +

      ,

      r 1

      1

      J2(t) := A2 (t 2)

      1 1 1

      1 0 c + u1

      c u1

      (5)

      ± := xi xi1

      i

      x )

      (i = 1, 2)

      := + + = (x

      x ) r 1 + 1 .

      ci ± ui

      2 2 2 2 1 c + u2 c u2

      variables : x (space) and t (time). However, as we already

      Each wave function A±k (x, t) has two independent

      The relations I = R e1sJ and J = R e2sI in

      know how those two variables {x, t} are combined to

      appear as a single variable in the wave function A±k in

      { }

      (4) can be written in a matrix form as

      1

      1

      1

      2

      2

      2

      r1 R1e1s 0 0 r0

      X =

      (1), we have only to know Ik(t), Jk(t) in (2) which have one time variable only, for a complete construction of A±(x, t). In other words, by fixing one variable

      0 0 R2e2s 1 0

      where

      (6)

      k

      x = x1, we lose nothing. The same is true with the

      pressure, velocity and density functions. Note that, with x = x1 in (1), we have

      pk(x1, t) pk = pIk(x1, t) = Ik + Jk c[uk(x1, t) uk] = cuIk(x1, t) = Ik Jk (3) c2[k(x1, t) ] = c2Ik(x1, t) = Ik + Jk

   2. Boundary Conditions

      { }

      Acoustic boundary conditions at x x0, x2 are characterized by reflection coefficients which are transfer functions between incident and reflected waves at both boundaries.

      Two reflection coefficients are defined as

      X := I1 J1 I2 J2 . (7)

   3. Jump Conditions

   { }

   As we have four unknown functions Ik, Jk (k = 1, 2) for a description of acoustic model, from two boundary conditions (6), we need two additional rela- tions between functions. Therefore we need to have two jump conditions across the area expansion at x = x1 in Fig. 1.

   We assume incompressive flows and thus the sound speed c and mean density are constants over the full range [x0, x2].

   The mass, momentum, energy rate (denoted m, f, e, respectively) are given

   R1 :=

   1

   =

   1

   1

   =

   1

   m = Au

   A-+ A-+e +s I

   A1 e

   1

   A1 e

   1

   1

   1

   (4)

   (8)

   – s

   – ( ++ )s

   J e1s

   f = A(p + u2)

   A2

   A2

   e

   A2 e2s J 2

   2

   A

   2

   e

   2

   2

   2

   2

   e = A(pu + u3/2), := .

   where A denotes cross sectional area.

   R2 :=

   -+- +s =

   -+- ( ++ )s = I e s

   1

   Remark 1: Throughout this paper, by saying mass, momentum and eergy without the term rate explicitly, we mean the rates of those quantities across x = x1.

   The perturbations of quantities in (8), denoted by

   mI, fI, eI, become

   By substituting the representation of a momentum perturbation in (10) to a perturbation of the equality (14), it follows that

   f

   A1( 1)p1I + f1I f2I

   = + 2M1 2M1 (1 + 2M2)

   + 2 uuI + u2I

   mI/A = uI

   fI/A = pI

   + uI

   (1 2M ) 1 X = 0. (15)

   (9)

   2

   k

   f

   eI/A = (p + 3u2/2)uI + upI + u3I/2

   By dividing both sides with and using M2 = M1/, the momentum relation can be written as

   where we used the facts p = c2 and c2I = pI.

   Let us introduce a new notation. From now on two

   + 2M1 2M1

   2M1 + 2M1

   1 X = 0. (16)

   subscripts {1, 2} of functions {m, f, e, A} of x rep- resent their magnitude at x x1 (left limit) and x x1 (right limit), respectively.

   We assume that the mean flow is much slower than the sound speed. That is, Mk := uk/c « 1.

   Up to now we have found four jump conditions corresponding to the mass, energy, and momentum re- lations across a sudden area expansion, which can be summarized in a 3 × 4 matrix form as

   Making use of (3), the perturbation form (9) can be rewritten as, for k = 1, 2,

   cmIk

   (mass) (energy) (momentum)

   1 + M1 1 + M1

   1 + M1 1 + M1

   + 2M1 2M1

   k

   A = (Ik Jk) + Mk(Ik + Jk)

   M1 M1

   0

   fkI

   Ak

   k

   k

   k

   k

   k

   = (I + J ) + 2M (I

   J )

   (10)

   M1 M1 X = 0 (17)

   2M1 + 2M1

   0

   eIk

   1

   = (I

   J ) + M (I

   k

   + J )

   Note that in the case of zero mean flow, i.e., M1 = 0,

   cAk

   1

   k k k

   this matrix has a form

   k

   1 I

   where for simplicity we assumed M 2 = 0. This expres- sion can be written in a matrix form as

   k

   c mIk 1 + Mk 1 + Mk

   A

   k

   eIk/c

   Mk +1

   1

   Mk 1 1

   Jk

   fI = 1 + 2Mk 1 2Mk (11)

   k

   k

   1 1 0

   1 1 X = 0 . (18)

   01

   Thus the first (mass) and second (energy) rows become identical, leaving us only two independent relations. An

   implication of this result is that, when there is no mean

   implication of this result is that, when there is no mean

   From this representation, the mass conservation con- dition mI mI = 0 can be written as

   flow, we have only two independent jump conditions and there is no ambiguity in choosing jump conditions.

   1 2

3. COMPATIBILITY OF JUMP CONDITIONS

   (12)

   f1 + M1 1 + M1 M1 M11 X = 0

   where := A2/A1 denotes the area ratio and we used

   M1 = M2 from the mass continuity u1A1 = u2A2.

   In a similar way, the energy conservation condition

   1. Mass and Energy Relations

      Suppose we take the following mass and momentum relations in (17) for granted. Then a rewriting of (17) gives the next equality

      J1

      eI1 eI2 = 0 can be written as

      f1 + M1 1 + M1 M1 M11

      X = 0.

      1 + M1 1 + M1

      1 + M1 1 + M1

      I1 =

      (13)

      Momentum balance across an area expansion can be modeled as

      + M1 + M1

      + M1 + M1

      I2 . (19)

      J2

      f2 f1 = (A2 A1)p1 = A1( 1)p1, (14) e.g., see [6](p. 396 Eq. (13.67c)).

      Our aim is to compare the momentum relation in (17)

      with a new momentum relation which comes from the mass and energy conditions.

      { } { }

      Firstly we express I2, J2 in terms of I1, J1 from and the corresponding unperturbed form the matrix relation (19) assuming M 2 = 0 and obtain

      k

      e e

      = A

      ( 1)(3 1) u3

      (25)

      fl fl

      2 1 1 6 2

      A

      1

      2 1 = ( 1)(I1 + J1) 2(M1 M2)(I1 J1)

      l l

      ( 1)

      (pl l \

      = ( 1)p1 2 M1cu1

      can be obtained.

      As the term in the right hand side of (25) is always negative for all 1, it follows that a sudden area

      = ( 1)

      2

      1 u1u1

      .

      (20)

      expansion causes an energy loss.

      { }

      In addition, we represent the energy terms e1, e2 of

      (24) explicitly to have

      Then, by recovering an unperturbed form from (20) to

      have

      ( \ A u

      p2

      1 2 A u

      p1

      1

      + 2 u1

      1 2

      1 2

      2

      2

      + 2 u2

      1

      f2 f1 = A1( 1) p1 u1 . (21)

      ( 1)(3 1)

      A comparison of this result with the momentum relation (14) reveals that our new momentum relation

      which is equal to

      = A2u2

      u2, (26)

      2

      6

      which is a pure consequence of the mass and energy

      p2 1 2 1 2 p1 1 2

      conservation in (19), has an additional term.

      + 2 u2 + 2 Keu2 =

      + 2 u1 (27)

      In order words, the mass conservation condition being taken for granted, the energy conservation condition is equivalent to having a smaller momentum increase than expected from the momentum relation (14).

      This difference however is not significant. To see this, note firstly that the additional term in (20) can be written

      1 1 1

      2 u ul = 2M1 cul . (22)

      2

      1

      1

      «

      Secondly, we know from (3) that c ul1 has the same order of pl1. Hence if M1 then the momentum difference fl fl in (21) is dominated by the term pl , recovering the momentum relation (14).

   2. Mass and Momentum Relations

   In this section we assume that the mass and momen- tum relations hold and, under that assumption, develop a new energy relation and then compare it with the energy conservation relation in (17).

   The mass and momentum conditions (the first and third rows of the matrix (17) can be written as

   + 2M1 2M1 J1

   1 + M1 1 + M1 I1

   By expressing {I2, J2} in terms of {I1, J1} from this

   = + M1 + M1 I2 . (23)

   after a cancellation of the mass rate where

   e

   K := ( 1)(3 1) > 0. (28)

   3

   The stagnation enthalpy is defined

   k k

   H := pk + 1 u2 (29)

   2

   and thus the energy loss in (27) across a sudden area expansion can be seen as a loss of stagnation enthalpy as follows ;

   2 e 2 1

   H + 1 K u2 = H . (30)

   2

   The quantity Ke can be called as a stagnation enthalpy loss coefficient associated with an area expansion.

   In conclusion the combination of mass and momentum relations automatically results in an energy loss which contradicts with the energy conservation in (17).

4. RESONANT FREQUENCY

By substituting the boundary conditions (6) to the mass, energy and momentum relations across an area expansion, we obtain a complex valued 4 × 2 matrix

(mass) (1 + M )R e1s 1 + M

(1 + M1)R1e

1 + M1

+ 2M1 2M1 J2

(energy)

(momentum)

1 1 1

1s

( + 2M1)R1e1s + 2M1

equality, the energy difference can be written as a linear

M1 + ( M1)R2e2s

1 l

J1

0

combination of {I2, J2} as

e2 e1 = cA1

2 M2 (I2 J2)

M1 + ( M1)R2e2s

= 0 .

(31)

l l ( 1)(3 1)

2 2M1 ( 2M1)R2e2s

(24)

I 2 0

= A

( 1)(3 1) u 2ul

Depending on which jump conditions in (31) we

1 2 2 2

choose, a homogeneous 2×2 square matrix, call it (s),

– –

is determined and the corresponding matrix quation al- lows non-zero solutions J1, I2 only when the determinant

| | | |

(s) is zero. As the determinant (s) is nothing but the characteristic equation of a differential equation associated with the matrix equation (31), let us call the roots of |(s)| = 0 as characteristic roots.

Characteristic roots are complex numbers in general but typically they are close to the imaginary axis. Hence the image of a real-valued function |(jw)| defined on

the real axis w R, has a local minimum at w = w

such that s = + jw is a characteristic root for some real number . The frequency w or f := w/2 is commonly called as a resonance frequency.

There are three possible ways in choosing two jump conditions among the three rows of the matrix (31). However, it turns out that the mass-energy combina- tion (first and second rows) should be avoided. An obvious reason is that, as we have already observed, the first and second row becomes identical when M1 = 0 and hence a singularity occurs. A hidden but more fundamental drawback of the mass-energy combination is that characteristic roots are independent of mean flow. In fact, elementary calculations with (31) can reveal that the equation |(s)| = 0 of this case is given

1

2

+ 1

1

2

R R e(1+2)s 1 (R e1s R e2s 1 = 0

(32)

2500

2000

Imaginary

1500

1000

500

0

-20 -10 0 10 20 30 40 50 60 70 80

Real

Fig. 2: Characteristic Roots (0 M1 0.5)

V. CONCLUSION

We considered a one-dimensional acoustic model of

a duct which has an abrupt area expansion and sub- ject to a non-zero mean flow. We investigated on the

relations between three (mass, energy and momentum) jump conditions across an area expansion and found

that the combination of mass-energy jump condition is not a proper choice for an acoustic modeling. It was

also shown that the combination of either the mass- momentum or the energy-momentum jump condition

which is independent of the mean flow M1.

With the mass-energy combination excluded, there are only two cases ; mass-momentum and energy-momentum combinations.

Note that the only difference between the mass (first) and energy (second) row of the matrix (31) is that M1 of the first row is replaced with M1 in the second. As a result, the two choices do not make much difference as long as the mean flow is small. Roughly speaking, with a fixed momentum relation, the energy relation emphasizes the effect of mean flow times than the mass relation.

Finally, let us consider s simple numerical example with parameters

L1 = 1.0 (m), L2 = 1.5 (m), = 5,

c = 443 (m/s), = 1.4, R1 = R2 = 1. (33)

For a set of mean flow 0 M1 0.5, the charac- teristic roots of both the mass-momentum and energy- momentum combinations are numerically found and shown in Fig. 2. In overall the two combinations have similar characteristic roots but show some differences as

result in a similar model for reasonably small mean flow.

REFERENCES

1. L. E. Kinsler, A. R. Frey, A. B. Coppens, and J. V. Sanders,

   Fundamentals of Acoustics, 4th ed. John Wiley & Sons, 2000.

2. S. W. Rienstra and A. Hirschberg, An Introduction to Acoustics. Eindhoven University of Technology, 2017.

3. V. B. Panicker and M. L. Munjal, Aero-acoustic analysis of straight-through mufflers with simple and extended tube ex- pansion chambers, Journal of the Indian Institute of Science, vol. 63, no. 1, pp. 119, 1981.

4. M. L. Munjal, Acoustics of Ducts and Mufflers, 2nd ed. Wiley, 2014.

5. P. Davies, Practical flow duct acoustics, Journal of Sound and Vibration, vol. 124, no. 1, pp. 91 115, 1988.

6. T. Lieuwen and V. Yang, Combustion instabilities in gas turbine engines: operational experience, fundamental mechanisms and modeling, ser. Progress in astronautics and aeronautics. Amer- ican Institute of Aeronautics and Astronautics, 2005.

mean flow increases.