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Czech Technical University in PragueFaculty of Electrical EngineeringDepartment of Control Engineering
bachelorβs thesis
Modelling of acoustic plane waves in gaseswith variable temperature
Hanna Chaika
Supervisor: doc. Dr. Ing. Michal Bednarik
May 2015
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Czech Technical University in Prague Faculty of Electrical Engineering
Department of Control Engineering
BACHELOR PROJECT ASSIGNMENT
Student: Hanna Chaika
Study programme: Cybernetics and Robotics Specialisation: Systems and Control
Title of Bachelor Project: Modelling of acoustic plane waves in gases with variable temperature
Guidelines:
1. Study problems concerning description of acoustic waves in fluids with spatially variable temperature, including derivation of model equations. Take into account one-dimensional temperature distribution. 2. Find analytical solutions of linear model equations for given temperature distributions and realize their analysis. 3. Derive model equation describing nonlinear acoustic plane waves for small temperature gradients. Solve this equation numerically by means of a suitable numerical code in the programming language C. Analyze the numerical results.
Bibliography/Sources:
[1] Press, W. H., Teukolsky, S.A., Vetterling, W. T., Flannerz, B. P.: Numerical Recipes (3ed), Cambridge Univ. Press, New York, 2007, [2] Blackstock, D.T.: Fundamentals of Physical Acoustics, John Eley & Sons, USA, 2000 [3] Rudenko, O V., Shvartsburg, A. B.: Nonlinear and Linear Wave Phenomena in Narrow Pipes, Acoustical Physics, vol. 56, No. 4, pp. 429-434, 2010. [4] Sujith, R. I., Waldherr, G. A. and Zinn, B. T.: An exact solution for one-dimensional acoustic fields in ducts with an axial temperature gradient, Journal of Sound and Vibrafon, vol. 184, No. 3, pp. 389-402, 1995. [5] Subrahmanyam, P. B., Sujith, R. I., Lieuwen, T. C.: A family of exact transient solutions for acoustic wave propagation in inhomogeneous, non-uniform area ducts, Journal of Sound and Vibrafon, vol. 240, No. 4, pp. 705-715, 2001. [6] Brekhovskikh, L. M., Lysanov, Yu.P.: Fundamental of Ocean Acoustics, Sprinter, New York, 1991.
Bachelor Project Supervisor: doc.Dr.Ing. Michal BednaΕΓk
Valid until the summer semester 2014/2015
L.S.
prof. Ing. Michael Ε ebek, DrSc. Head of Department
prof. Ing. Pavel Ripka, CSc. Dean
Prague, January 8, 2014
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AcknowledgementI would like to express my sincere gratitude to my supervisor, doc. Dr. Ing. MichalBednarik, for his immense knowledge, patience and friendly manner.
DeclarationI declare that I completed the presented thesis independently and that all used sourcesare quoted in accordance with the Methodological Instructions that cover the ethicalprinciples for writing an academic thesis.
In Prague on . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Signature
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AbstraktBakalΓ‘ΕskΓ‘ prΓ‘ce se zabΓ½vΓ‘ modelovΓ‘nΓm rovinnΓ½ch akustickΓ½ch vln v plynech s pro-mΔnnou teplotou. Pro Ε‘ΓΕenΓ rovinnΓ½ch vln v teplotnΔ nehomogennΓm prostΕedΓ bylaodvozena vlnovΓ‘ rovnice s promΔnnΓ½mi koeficienty. Pro vybranΓ© teplotnΓ distribucejsou v prΓ‘ci prezentovΓ‘na pΕesnΓ‘ analytickΓ‘ ΕeΕ‘enΓ tΓ©to rovnice. PomocΓ nalezenΓ½chobecnΓ½ch ΕeΕ‘enΓ byly vypoΔteny koeficienty transmise a reflexe. Pro pΕΓpad vln koneΔ-nΓ½ch amplitud (nelineΓ‘rnΓ vlny) Ε‘ΓΕΓcΓch se tekutinou s malΓ½m teplotnΓm gradientembyla odvozena modifikovanΓ‘ Burgersova rovnice. Tato rovnice byla ΕeΕ‘ena numericky vkmitoΔtovΓ© oblasti pomocΓ Runge-Kuttovy metody 4. ΕΓ‘du v programovacΓm jazyce C.
KlΓΔovΓ‘ slovaVlnovΓ‘ rovnice s promΔnnΓ½mi koeficienty; koeficienty transmise a reflexe; Burgersovarovnice.
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AbstractThe thesis presents the modelling of acoustic plane waves in gases with variable temper-ature. The wave equation with variable coefficients for the propagation of plane wavesin a thermally inhomogeneous medium was derived. For chosen temperature distribu-tions the exact analytical solutions of this equation are presented in this thesis. Thecoefficients of transmission and reflection were calculated using found general solutions.For the case of finite amplitude waves (nonlinear waves), propagating in a fluid with alow temperature gradient, the modified Burgers equation was derived. This equationwas solved numerically in the frequency domain using the fourth-order Runge-Kuttamethod written in the C programming language.
KeywordsWave equation with variable coefficients; transmission and reflection coefficients; theBurgers equation.
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Contents
1 Introduction 1
2 Derivation of model equations 22.1 Fundamental equations of fluid mechanics . . . . . . . . . . . . . . . . . 22.2 Derivation of basic model equations for temperature-inhomogeneous region 3
3 Exact analytical solutions for various temperature functions 73.1 Transformation of derivatives . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Analytical solutions for unknown temperature functions . . . . . . . . . 8
3.2.1 First method of finding analytical solution . . . . . . . . . . . . . 83.2.2 Second method of finding an analytical solution . . . . . . . . . . 113.2.3 Third method of finding an analytical solution . . . . . . . . . . 14
3.3 Finding analytical solutions for known temperature distributions . . . . 153.3.1 Linear temperature distribution . . . . . . . . . . . . . . . . . . . 163.3.2 Exponential temperature distribution . . . . . . . . . . . . . . . 18
4 Transmission and reflection coefficients 214.1 Sound velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.2 Sound pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.2.1 Second exact analytical solution . . . . . . . . . . . . . . . . . . 274.2.2 Exact analytical solution for a linear temperature distribution . . 294.2.3 Exact analytical solution for an exponential temperature distri-
bution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
5 Derivation of the Burgers-type equation for temperature-inhomogeneous flu-ids 335.1 Westervelt equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.2 Burgers-type equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.3 Dimensionless Burgers-type equation . . . . . . . . . . . . . . . . . . . . 35
6 Numerical solution of the Burgers-type equation for temperature-inhomogeneousfluids 366.1 Description of the numerical method . . . . . . . . . . . . . . . . . . . . 366.2 Representation of an artificial attenuation filter . . . . . . . . . . . . . . 386.3 Graphical representation of the Burgers-type equation solution for a lin-
ear temperature distribution . . . . . . . . . . . . . . . . . . . . . . . . . 386.3.1 Positive temperature gradient . . . . . . . . . . . . . . . . . . . . 386.3.2 Negative temperature gradient . . . . . . . . . . . . . . . . . . . 396.3.3 Different lengths of a duct with a positive temperature gradient . 40
7 Conclusion 41
Bibliography 43
Appendices
A Attached CD contents I
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Abbreviations
Latin capital lettersπ΄ constant in Eqs. (3.19), (3.24) and (3.71)πΆ1, πΆ2 integration constants in Eqs. (3.34), (3.53), (3.55), (3.88) and (3.116)πΏ characteristic lengthπ acoustic pressure amplitudeπ specific gas constantππ΄, ππ΅ characteristic temperatures of temperature-homogeneous regions A, B
resp.π0 absolute temperature in temperature-inhomogenious regionπ acoustic velocity amplitudeπ dimensionless acoustic velocity
Latin lowercase lettersπ constant in Eq. (3.88)π0 linear sound speed in temperature-inhomogenious regionππ΄ =
βΞΊπ ππ΄, sound speed in temperature-homogeneous region A with
absolute temperature ππ΄ππ΅ =
βΞΊπ ππ΅, sound speed in temperature-homogeneous region B with
absolute temperature ππ΅ππ΄ = π/ππ΄, wave number at π₯ = 0ππ΅ = π/ππ΅, wave number at π₯ = πΏπ pressureπ0 atmospheric pressureπβ² acoustic pressureπ constant in Eq. (3.44)π constant in Eq. (3.17)π‘ timeπ£ acoustic velocity (the particle velocity of the medium)π₯ distance
Greek capital lettersΞ = π2π΄π2 β 4π2, discriminant of Eq. (3.49)Ξ = π0/ππ΄, dimensionless absolute temperatureΞ = πβ²/
(οΈππ΄π
2π΄
)οΈ, dimensionless acoustic pressure
Ο dimensionless acoustic velocity amplitudeΞ¦ dimensionless acoustic pressure amplitude
Greek lowercase lettersπΌ sound diffusivityπ½ coefficient in Eq. (5.2a)πΎ constant in Eq. (3.90)π a shear viscosity
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π a bulk viscosityπ variable, given by Eq. (3.3)π dimensionless timeπ thermal conduction coefficientΞΊ ratio of specific heatsπ small dimensionless parameterπ variable, given by Eq. (3.32)π =
βπ2 β π2, coefficient in Eq. (3.27)
π density of the mediumπ0 density of the medium before sound propagationππ΄, ππ΅ densities of the medium in regions A, B resp.π = π₯/πΏ, dimensionless lengthπ = π‘β π₯/ππ΄, retarded timeπ = 2/
(οΈΞΊπ πΎ2
)οΈ, coefficient in Eq. (3.112)
π angular frequency
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1 Introduction
The description and analysis of acoustic waves in ducts with a region containing tempera-ture-inhomogeneous fluid represents a significant problem of scientific and practical in-terest. This interest is induced by the need to understand how temperature fields affectacoustic processes. This would lead to the possibility of more effective design and con-trol systems in which interactions between acoustic and temperature fields occur. Thisincludes, for instance, thermo-acoustic devices and engines, combustors, automotivemufflers, measuring methods of impedance of high-temperature systems, the investiga-tion of thermo-acoustics and combustion instabilities among other possible applications.
This thesis presents modelling of plane acoustic waves in gases with variable tem-perature. The whole work can be divided into two parts, the first part is devoted tolinear model equations, and the second describes nonlinear acoustic plane waves forsmall temperature gradients.
The analysis consists of seven chapters. Within Chapter 2 the basic linear one-dimensional model equations for fluids in temperature-inhomogeneous regions are de-rived. Chapter 3 is devoted to the exact analytical solutions of linear model equationsfor varoius temperature distributions. Chapter 4 deals with an application of the foundsolutions for calculation of transmission and reflection coefficients. In Chapter 5 is pre-sented a derivation of the Burgers equation for temperature-inhomogeneous fluids. Thenumerical method for the Burgers equation for small temperature gradients is shownin Chapter 6. Chapter 7 states the conclusion.
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2 Derivation of model equations
2.1 Fundamental equations of fluid mechanics
To describe acoustic waves in fluids the following system of equations is considered (seee.g. [1], [2]) :
1. The Navier β Stokes equation (Momentum equation)This equation describes the motion of fluid substances and the general form is
π
[οΈπvππ‘
+ v Β· βv]οΈ
= ββπ+(οΈπ + π3
)οΈβ (β Β· v) + πβ2v , (2.1)
where π is the density of the medium, v is the particle velocity of the medium, π‘is time, π is the pressure, π is a bulk viscosity, π is a shear viscosity.
2. Equation of continuityThis equation describes the transport of a conserved quantity, i.e. fluid.
ππ
ππ‘+ β Β· (πv) = 0 . (2.2)
3. Energy equation
ππ
(οΈππ
ππ‘+ v Β· βπ
)οΈ= π2
(οΈππ£πππ₯π
+ ππ£πππ₯π
β 23πΏππππ£πππ₯π
)οΈ2+π (β Β· v)2+βΒ·(π βπ ) , (2.3)
where π is entropy per unit mass, π is absolute temperature, π is the thermalconduction coefficient, πΏππ is the Kronecker delta.
4. Equation of stateπ = π (π, π) . (2.4)
β If a perfect gas is considered then the Navier β Stokes equation (2.1) is reduced tothe form
πvππ‘
+ v Β· βv = β1π
βπ , (2.5)
which is called the Euler equation.β Equation of state (2.4) for a perfect gas is (see e.g. [2])
π = π ππ , (2.6)
where π is the specific gas constant, i.e. the (molar) gas constant divided by themolar mass of the gas.
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2.2 Derivation of basic model equations for temperature-inhomogeneous region
2.2 Derivation of basic model equations fortemperature-inhomogeneous region
In order to derive model equations it is necessary to take into consideration four equa-tions from the previous section and make some assumptions, which enable us to ne-glect nonlinear terms. Assuming a perfect, inviscid and non-heat-conducting gas withone-dimensional temperature distribution, Eqs. (2.2), (2.3), (2.5) and (2.6) can betransformed in the following ways (see e.g. [2]) :
β Energy equationIn a perfect gas no energy is dissipated, so dπ/dπ‘ = 0, and hence a specific entropyπ is a constant. From this an isentropic process and an adiabatic gas law takeplace.
β Equation of stateFrom Eq. (2.4) it follows
dπ =(οΈππ
ππ
)οΈπ
dπ+(οΈππ
ππ
)οΈπ
dπ . (2.7)
As the process is isentropic then state equation can be written as π = π(π) and
dπ =(οΈππ
ππ
)οΈπ
dπ . (2.8)
As an adiabatic gas law takes place then
π
π0=(οΈπ
π0
)οΈΞΊ, (2.9)
where π and π0 are different specific pressures, π and π0 are different specific den-sities, ΞΊ is the ratio of specific heats.
By assuming the gas to be inviscid, it is permissible to express the sound speed π
π2 = dπdπ . (2.10)
Taking into account Eqs. (2.9) and (2.6), the equation above can be written as
π2 = ΞΊππ
= ΞΊπ π . (2.11)
Equation (2.10) lets us take into account the following equality
dπdπ‘ = π
2 dπdπ‘ . (2.12)
The total derivative isd(Β·)dπ‘ =
π(Β·)ππ‘
+ ππ(Β·)ππ₯
. (2.13)
Then Eq. (2.12) can be written as
ππ
ππ‘+ π ππ
ππ₯= π2
(οΈππ
ππ‘+ π ππ
ππ₯
)οΈ. (2.14)
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2 Derivation of model equations
β Equation of continuityOne-dimensional equation of continuity is
ππ
ππ‘+ πππ₯
(ππ£) = 0 . (2.15)
From this equation the equality π£ππ/ππ₯ = βππ/ππ‘ β πππ£/ππ₯ can be obtained.After substitution into Eq. (2.14) the resulting equation is
ππ
ππ‘+ π£ ππ
ππ₯= βπ2πππ£
ππ₯. (2.16)
From Eq. (2.11) the above equation can be written as
ππ
ππ‘+ π£ ππ
ππ₯+ ΞΊπππ£
ππ₯= 0 . (2.17)
β Euler equationBy assuming one-dimensional temperature distribution, Eq. (2.5) takes the formof the one-dimensional Euler equation
πππ£
ππ‘+ ππ£ ππ£
ππ₯+ ππππ₯
= 0 . (2.18)
Each of the dependent variables can be expressed in the following way (see e.g. [3])1. the density π(π₯, π‘) = π0(π₯) + πβ²(π₯, π‘) is the sum of the ambient density π0 and an
acoustic density πβ²,2. the velocity π£ = π£(π₯, π‘) is the particle velocity of the medium, resp. acoustic
velocity,3. the pressure π(π₯, π‘) = π0 +πβ²(π₯, π‘) is the sum of the atmospheric pressure π0, which
is supposed to be constant (see e.g. [3]) and the acoustic pressure πβ².The assumptions can be taken into account |πβ²|/π0 βΌ |π£|/π0 βΌ πβ²/π βΌ π βͺ 1, where
π0 is the linear sound speed, π < 0 is a small dimensionless parameter. It is consideredthat π0 = π0(π₯) and temperature π0 = π0(π₯) are dependent on the distance π₯, whereπ0 is the ambient temperature of the fluid.
The substitution of the above dependent variables into the equation of continuity(2.17)
π(π0 + πβ²)ππ‘
+ π£π(π0 + πβ²)
ππ₯+ ΞΊ
(οΈπ0 + πβ²
)οΈ ππ£ππ₯
= 0 (2.19)
and linearization of Eq. (2.19) leads us to the following linear form of the equation ofcontinuity
ππβ²
ππ‘+ ΞΊπ0
ππ£
ππ₯= 0 . (2.20)
The same substitution into the Euler equation (2.18)
(οΈπ0 + πβ²
)οΈ ππ£ππ‘
+(οΈπ0 + πβ²
)οΈπ£ππ£
ππ₯+ π(π0 + π
β²)ππ₯
= 0 (2.21)
and linearization of Eq. (2.21) leads us to the linear form of the Euler equation
π0ππ£
ππ‘+ ππ
β²
ππ₯= 0 . (2.22)
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2.2 Derivation of basic model equations for temperature-inhomogeneous region
Differentiating of Eq. (2.20) with respect to time t and Eq. (2.22) with respect tocoordinate x and eliminating the cross-derivative term by their combination yields
π2πβ²
ππ₯2+ dπ0dπ₯
ππ£
ππ‘β π0
ΞΊπ0π2πβ²
ππ‘2= 0 . (2.23)
Expressing the derivative ππ£/ππ‘ from Eq. (2.22) and substituting it into Eq. (2.23)obtains the equation
π2πβ²
ππ₯2β dπ0dπ₯
1π0
ππβ²
ππ₯β π0
ΞΊπ0π2πβ²
ππ‘2= 0 . (2.24)
From the perfect gas law π0 = π π0π0 the total derivative of π0 with respect to π₯ is
0 = dπ0dπ₯ = π dπ0dπ₯ π0 +π π0
dπ0dπ₯ . (2.25)
After multiplying the above equation by 1/(π π0π0) the resulting expression is
1π0
dπ0dπ₯ +
1π0
dπ0dπ₯ = 0 . (2.26)
From Eq. (2.11) followsπ20 =
ΞΊπ0π0
, (2.27)
and using Eq. (2.26) hence is obtained the wave equation with variable coefficients (seee.g. [3])
π2πβ²
ππ₯2+ 1π0
dπ0dπ₯
ππβ²
ππ₯β 1π20
π2πβ²
ππ‘2= 0 . (2.28)
Assuming a time periodic source of sound, where π is an angular frequency, it ispossible to consider
πβ²(π₯, π‘) = π (π₯)πβjππ‘ . (2.29)
Equation (2.28) can be written now as
d2πdπ₯2 +
1π0
dπ0dπ₯
dπdπ₯ +
π2
π20π = 0 . (2.30)
Now, let us differentiate vice versa - Eq. (2.20) with respect to x and Eq. (2.22) withrespect to t, then the cross-derivative term can be eliminated. This manipulation leadsto the wave equation
π2π£
ππ‘2β π20
π2π£
ππ₯2= 0 . (2.31)
Substitutingπ£(π₯, π‘) = π (π₯)πβjππ‘ , (2.32)
into Eq. (2.31) obtains the equation
d2πdπ₯2 +
π2
π20(π₯)π (π₯) = 0 . (2.33)
This is the Helmholtz equation.
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2 Derivation of model equations
Let us rewrite the model equations (2.28) and (2.31) in dimensionless form (see e.g.[4])
π2Ξ ππ2
+ 1Ξ(π)dΞ(π)
dππΞ ππ
β 1πΆ20 (π)
π2Ξ ππ2
= 0 , (2.34)
π2π
ππ2β πΆ20 (π)
π2π
ππ2= 0 . (2.35)
Here
Ξ = πβ²
ππ΄π2π΄= π
β²
ΞΊπ0, π = π₯
πΏ, π = ππ‘ ,
πΆ20 (π) =π20(π)π2πΏ2
= ΞΊπ π0(π)π2πΏ2
= Ξ(π)β2
, Ξ = π0ππ΄
, π = π£ππ΄
,
(2.36)
where ππ΄ is a characteristic temperature, ππ΄ = π0(ππ΄) =βΞΊπ ππ΄, ππ΄ = π0(ππ΄) =
π0/(π ππ΄), πΏ is a characteristic length and β = ππΏ/ππ΄.Assuming that the solutions of Eqs. (2.34) and (2.35) have a periodic time depen-
dence, i. e.Ξ (π, π) = Ξ¦(π)πβjπ (2.37)
andπ (π, π) = Ο(π)πβjπ , (2.38)
then equations can be written as
d2Ξ¦dπ2 +
1Ξ(π)
dΞ(π)dπ
dΞ¦dπ +
1πΆ20 (π)
Ξ¦(π) = 0 , (2.39)
d2Οdπ2 +
Ο(π)πΆ20 (π)
= 0 . (2.40)
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3 Exact analytical solutions for varioustemperature functions
This chapter is devoted to exact analytical solutions. These solutions can be dividedinto two cases. The first case represents solving equations where the temperature dis-tribution π0(π₯) is unknown, so it is necessary to find both the temperature distributionπ0(π₯) and the solution of the equation. The second case deals with equations whereit is supposed the function π0(π₯) must be known. For this reason only the equationsolution is to be found in this case. Linear and exponential temperature distributionsare assumed.
3.1 Transformation of derivativesSuppose that a function π(π₯) is known then can be derived the following spatial deriva-tives
d(Β·)dπ₯ =
d(Β·)dπ
dπdπ₯ , (3.1)
and
d2(Β·)dπ₯2 =
ddπ₯
(οΈd(Β·)dπ₯
)οΈ= ddπ₯
(οΈd(Β·)dπ
dπdπ₯
)οΈ=[οΈ d
dπ₯
(οΈd(Β·)dπ
)οΈ]οΈ dπdπ₯ +
[οΈ ddπ₯
(οΈdπdπ₯
)οΈ]οΈ d(Β·)dπ
=[οΈ d
dπ
(οΈd(Β·)dπ₯
)οΈ]οΈ dπdπ₯ +
d2πdπ₯2
d(Β·)dπ =
[οΈ ddπ
(οΈd(Β·)dπ
dπdπ₯
)οΈ]οΈ dπdπ₯ +
d2πdπ₯2
d(Β·)dπ
=(οΈdπ
dπ₯
)οΈ2 d2(Β·)dπ2 +
d2πdπ₯2
d(Β·)dπ . (3.2)
Taking into account a new variable (see e.g. [5])
π =β«οΈ π₯
0
1π0(π₯1)
dπ₯1 , (3.3)
thendπdπ₯ =
1π0(π₯)
, (3.4)
d2πdπ₯2 =
ddπ₯
(οΈ 1π0(π₯)
)οΈ. (3.5)
Putting π(π₯) β‘ π(π₯) can be used the transformation relations (3.1) and (3.2) asfollows
d(Β·)dπ₯ =
d(Β·)dπ
dπdπ₯ , (3.6)
d2(Β·)dπ₯2 =
(οΈdπdπ₯
)οΈ2 d2(Β·)dπ2 +
d2πdπ₯2
d(Β·)dπ =
1π20(π₯)
d2(Β·)dπ2 +
ddπ₯
(οΈ 1π0(π₯)
)οΈ d(Β·)dπ . (3.7)
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3 Exact analytical solutions for various temperature functions
3.2 Analytical solutions for unknown temperature functionsIt is assumed that a temperature distribution π0(π₯) in a temperature-inhomogeneousregion of the length πΏ in a waveguide separates two regions of different constant tem-peratures ππ΄ and ππ΅, see Fig. 1 (see e.g. [4]) .
Region Aππ΄
Region Bππ΅
Temperature βinhomogeneous region
0 π₯πΏ
Fig. 1 Temperature regions in a waveguide.
In dimensionless equations the temperature-inhomogeneous region has the lengthequal to 1 and separates two temperature-homogeneous regions with constant dimen-sionless temperatures Ξπ΄ = 1 and Ξπ΅ = ππ΅/ππ΄.
3.2.1 First method of finding analytical solution
To solve the Helmholtz equation (2.33) the first step is to introduce a new functionπΉ (π₯) (see e.g. [6])
π (π₯) =βοΈπ0(π₯)πΉ (π₯) . (3.8)
Its derivatives ared(βπ0πΉ )
dπ₯ = πΉdβπ0
dπ₯ +βπ0
dπΉdπ₯ , (3.9)
d2(βπ0πΉ )dπ₯2 =
dπΉdπ₯
dβπ0dπ₯ + πΉ
d2βπ0dπ₯2 +
dβπ0dπ₯
dπΉdπ₯ +
βπ0
d2πΉdπ₯2 , (3.10)
wheredβπ0
dπ₯ =1
2βπ0dπ0dπ₯ , (3.11)
d2βπ0dπ₯2 = β
14π0
βπ0
(οΈdπ0dπ₯
)οΈ2+ 12βπ0
d2π0dπ₯2 . (3.12)
Substituting into Eq. (2.33)
β πΉ4π20
(οΈdπ0dπ₯
)οΈ2+ πΉ2π0
d2π0dπ₯2 +
12π0
dπ0dπ₯
dπΉdπ₯ +
12π0
dπ0dπ₯
dπΉdπ₯ +
d2πΉdπ₯2 +
π2
π20πΉ = 0 . (3.13)
And rearranging leads to the equation
d2πΉdπ₯2 +
1π0
dπ0dπ₯
dπΉdπ₯ β
14π20
(οΈdπ0dπ₯
)οΈ2πΉ + 12π0
d2π0dπ₯2 πΉ +
π2
π20πΉ = 0 . (3.14)
8
-
3.2 Analytical solutions for unknown temperature functions
Imposing the variable π from Eq. (3.3) then it is possible to write Eq. (3.14) in thefollowing form
1π20
d2πΉdπ2 +
ddπ₯
(οΈ 1π0
)οΈ dπΉdπ +
1π20
dπ0dπ₯
dπΉdπ β
14π20
(οΈdπ0dπ₯
)οΈ2πΉ + 12π0
d2π0dπ₯2 πΉ +
π2
π20πΉ = 0 . (3.15)
After modification Eq. (3.15) takes the form
d2πΉdπ2 + π
2πΉ = πΉ[οΈ
14
(οΈdπ0dπ₯
)οΈ2β π02
d2π0dπ₯2
]οΈ. (3.16)
Equation (3.16) is an ordinary differential equation with varible coefficients. In orderto get an ordinary differential equation with constant coefficients, it is necessary toequalise the expression in the square brackets using a constant π2. After this step isobtained
d2πΉdπ2 + π
2πΉ = π2πΉ (3.17)
andπ2 = 14
(οΈdπ0dπ₯
)οΈ2β π02
d2π0dπ₯2 . (3.18)
In solving the differential equation (3.18) its solution can be written as
π0(π₯) =14
(οΈπ΄2 β 4π2
)οΈπ₯2
π΅+π΄π₯+π΅ , (3.19)
where π΄ and π΅ are integration constants.After some small algebraic manipulation the relation (3.19) can be expressed in the
convenient form
π0(π₯) = ππ΄
[οΈ(οΈπ΄2 β π
2
π2π΄
)οΈπ₯2 + 2π΄π₯+ 1
]οΈ. (3.20)
Now, let us remember some relations with characteristic temperature ππ΄. Fromprevious chapter
π0(ππ΄) = ππ΄ , π0(ππ΄) = ππ΄ . (3.21)
From Eqs. (2.11) and (2.27) it follows that
π0(π₯) =βοΈΞΊπ π0(π₯) . (3.22)
As Eq. (3.22) takes place then with the help of Eq. (3.20) it is possible to write atemperature relation
π0(π₯) =π2π΄ΞΊπ
[οΈ(οΈπ΄2 β π
2
π2π΄
)οΈπ₯2 + 2π΄π₯+ 1
]οΈ2. (3.23)
According to Eqs. (2.36) and (3.23)
π0(π₯) = ππ΄
[οΈ(οΈπ΄2 β π
2
π2π΄
)οΈπ₯2 + 2π΄π₯+ 1
]οΈ2, (3.24)
where ππ΄ = π2π΄/(ΞΊπ ) .
9
-
3 Exact analytical solutions for various temperature functions
Using expressions (2.36) temperature-inhomogeneous region π0 can be expressed indimensionless form (see e.g. [4])
Ξ(π) = π0(π)ππ΄
=[οΈ(οΈπ΄2 β π
2
π2π΄
)οΈπΏ2π2 + 2π΄πΏπ + 1
]οΈ2. (3.25)
Let characteristic length πΏ be equal to 1 m. By choosing different values of constantsπ΄, π and velocity ππ΄ it is possible to see exemplary solutions for temperature Ξ amongmany instances.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
1.5
2
2.5
π
Ξ
ππ΄ = 345 msβ1, |π| = 597.558 sβ1, π΄ = 1 mβ1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
1.2
π
Ξ
ππ΄ = 345 msβ1, |π| = 172.5 sβ1, π΄ = β 0.008 mβ1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
1.5
2
2.5
3
3.5
4
4.5
π
Ξ
ππ΄ = 345 msβ1, |π| = 301.39 sβ1, π΄ = 0.662 mβ1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
1.5
2
2.5
3
3.5
π
Ξ
ππ΄ = 345 msβ1, |π| = 0.97 sβ1, π΄ = 0.328 mβ1
Fig. 2 Dimensionless temperature function Ξ(π) for different coefficients of Eq. (3.25).
To solve Eq. (3.17) let us use the following substitution
π2 = π2 β π2 , (3.26)
where π2βπ2 > 0, because an evanescent wave is not considered. After this substitutionEq. (3.17) can be written as
d2πΉdπ2 + π
2πΉ = 0 . (3.27)
The above equation is a second-order linear ordinary differential equation, where thediscriminant is less than zero. Then the roots of the characteristic equation of Eq.(3.27) are complex
π1,2 = Β±jπ . (3.28)
10
-
3.2 Analytical solutions for unknown temperature functions
The solution of Eq. (3.27) is then
πΉ (π) = πΆ1 cos(ππ) + πΆ2 sin(ππ) , (3.29)
where πΆ1 and πΆ2 are constants.According to Eqs. (3.3) and (3.20) the variable π can be written as
π =β«οΈ π₯
0
1π0(π₯1)
dπ₯1 =β«οΈ π₯
0
1
ππ΄
[οΈ(οΈπ΄2 β π2
π2π΄
)οΈπ₯21 + 2π΄π₯1 + 1
]οΈ dπ₯1 . (3.30)After integration the variable π can be written as
π = β 1|π|
tanh-1[οΈππ΄|π|
(οΈπ΄+
(οΈπ΄2 β π
2
π2π΄
)οΈπ₯1
)οΈ]οΈβββββπ₯
0= β 1
|π|π(π₯) , (3.31)
where
π(π₯) = tanh-1[οΈππ΄|π|
(οΈπ΄+
(οΈπ΄2 β π
2
π2π΄
)οΈπ₯
)οΈ]οΈβ tanh-1
(οΈππ΄|π|π΄
)οΈ. (3.32)
The absolute value of ππ΄ was omitted, because ππ΄ is an indoor temperature whenππ΄ > 0.
Now the function πΉ (π₯) can be written as
πΉ (π₯) = πΆ1 cos(οΈπ
|π|π(π₯)
)οΈβ πΆ2 sin
(οΈπ
|π|π(π₯)
)οΈ. (3.33)
Taking into account Eq. (3.8), the solution of the Helmholtz equation (2.33) is
π (π₯) =βοΈπ0(π₯)πΉ (π₯) =
β―βΈβΈβ·ππ΄[οΈ(οΈπ΄2 β π
2
π2π΄
)οΈπ₯2 + 2π΄π₯+ 1
]οΈ
Γ[οΈπΆ1 cos
(οΈπ
|π|π(π₯)
)οΈβ πΆ2 sin
(οΈπ
|π|π(π₯)
)οΈ]οΈ. (3.34)
The dimensionless form of the velocity is (see e.g. [4])
Ο(π) = π (π)ππ΄
=
β―βΈβΈβ· 1ππ΄
[οΈ(οΈπ΄2 β π
2
π2π΄
)οΈπΏ2π2 + 2π΄πΏπ + 1
]οΈ
Γ[οΈπΆ1 cos
(οΈπ
|π|π(π)
)οΈβ πΆ2 sin
(οΈπ
|π|π(π)
)οΈ]οΈ, (3.35)
where
π(π) = tanh-1[οΈππ΄|π|
(οΈπ΄+
(οΈπ΄2 β π
2
π2π΄
)οΈπΏπ
)οΈ]οΈβ tanh-1
(οΈππ΄|π|π΄
)οΈ. (3.36)
3.2.2 Second method of finding an analytical solution
Now let us solve Eq. (2.30). It can be rewritten once more
d2πdπ₯2 +
1π0
dπ0dπ₯
dπdπ₯ +
π2
π20π = 0 . (3.37)
11
-
3 Exact analytical solutions for various temperature functions
Using Eqs. (3.1)β(3.7) the above equation can be written as
1π20
d2πdπ2 +
ddπ₯
(οΈ 1π0
)οΈ dπdπ +
1π0
dπ0dπ₯
1π0
dπdπ +
π2
π20π = 0 . (3.38)
From Eq. (3.22) can be derived
dπ0dπ₯ =
1ΞΊπ
dπ20dπ₯ . (3.39)
Multiplying Eq. (3.39) by 1/π0 leads to
1π0
dπ0dπ₯ =
2π0π20
dπ0dπ₯ . (3.40)
So, Eq. (3.37) can be modified as follows
1π20
d2πdπ2 β
1π20
dπ0dπ₯
dπdπ +
2π0π30
dπ0dπ₯
dπdπ +
π2
π20π = 0 , (3.41)
and after rearranging and reducing it becomes
d2πdπ2 +
dπ0dπ₯
dπdπ + π
2π = 0 . (3.42)
In order to solve Eq. (3.42) as an ordinary differential equation with constant coeffi-cients, it is necessary take into account that dπ0/dπ₯ = ππππ π‘. Let this constant be π1.Thus, π0 has the form
π0(π₯) = π1π₯+ π2 , (3.43)
where π1 and π2 are integration constants and π2 β₯ 0.By choosing π2 = ππ΄ and denoting π1/π2 = π the form is
π0(π₯) = ππ΄(ππ₯+ 1) . (3.44)
According to Eq. (3.22), function π0(π₯) is equal to
π0(π₯) = ππ΄ (ππ₯+ 1)2 , (3.45)
where
ππ΄ =π2π΄ΞΊπ
. (3.46)
Using expressions (2.36) the dimensionless form of temperature-inhomogeneous re-gion π0 (see e.g. [4])
Ξ(π) = π0(π)ππ΄
= (ππΏπ + 1)2 . (3.47)
By choosing different values of constant π and letting πΏ = 1 m, it is possible to seeexemplary solutions for temperature Ξ among many instances.
12
-
3.2 Analytical solutions for unknown temperature functions
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
1.5
2
2.5
3
3.5
4
π
Ξ
π = 1 mβ1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
1.2
π
Ξ
π = β 0.5 mβ1
Fig. 3 Dimensionless temperature function Ξ(π) for different coefficients of Eq. (3.47).
To solve Eq. (3.42) let us substitute the expression of sound velocity (3.44). ThenEq. (3.42) takes the form
d2πdπ2 + ππ΄π
dπdπ + π
2π = 0 . (3.48)
The solution of the above equation can be found through the characteristic equationof an appropriate equation.
π2 + ππ΄ππ+ π2 = 0 . (3.49)
The discriminant of Eq. (3.49) is
Ξ = π2π΄π2 β 4π2 . (3.50)
There are different possible solutions according to the sign of the discriminant. Thesolutions are represented below in the table, where πΆ1 and πΆ2 are constants.
Ξ > 0 Ξ = 0 Ξ < 0
π1,2 = βππ΄πΒ±β
Ξ2 π1 = π2 = β
12ππ΄π π1,2 =
βππ΄πΒ±iβ
Ξ2
π (π) = πΆ1π(β12 ππ΄π+
12
βΞ)π
+πΆ2π(β12 ππ΄πβ
12
βΞ)π
π (π) = πβ12 ππ΄ππ
Γ [πΆ1 + πΆ2π]π (π) = πβ
12 ππ΄ππ
[οΈπΆ1 cos
(οΈ12β
Ξπ)οΈ
+ πΆ2 sin(οΈ
12β
Ξπ)οΈ]οΈ
Tab. 1 The solution of Eq. (3.48).
According to Eqs. (3.3) and (3.44) the variable π can be written as
π =β«οΈ π₯
0
1π0(π₯1)
dπ₯1 =β«οΈ π₯
0
1ππ΄(ππ₯1 + 1)
dπ₯1 . (3.51)
Notice that π₯1 cannot be equal to β1/π.After integration the variable π can be written as
π = ln (ππ₯1 + 1)ππ΄π
ββββπ₯0
= 1ππ΄π
ln (ππ₯+ 1) . (3.52)
Now it is possible to write the solution of Eq. (2.30) according to the discriminantof Eq. (3.49)
13
-
3 Exact analytical solutions for various temperature functions
β If Ξ > 0, an evanescent wave is not considered.β If Ξ = 0,
π (π₯) = 1βππ₯+ 1 [πΆ1 + πΆ2 ln(ππ₯+ 1)] , (3.53)
where πΆ1 and πΆ2 are integration constants. According to Eqs. (2.36) and (3.21)the dimensionless form of the pressure is (see e.g. [4])
Ξ¦(π) = 1βππΏπ + 1
[πΆ1 + πΆ2 ln(ππΏπ + 1)] , (3.54)
where πΆ1 and πΆ2 are new constants.β If Ξ < 0,
π (π₯) = 1βππ₯+ 1
[οΈπΆ1 cos
(οΈβοΈ14 β
π2
π2π΄π2 ln(ππ₯+ 1)
)οΈ
+πΆ2 sin(οΈβοΈ
14 β
π2
π2π΄π2 ln(ππ₯+ 1)
)οΈ]οΈ, (3.55)
where πΆ1 and πΆ2 are integration constants. The dimensionless form of this solutionis (see e.g. [4])
Ξ¦(π) = 1βππΏπ + 1
[οΈπΆ1 cos
(οΈβοΈ14 β
π2
π2π΄π2 ln (ππΏπ + 1)
)οΈ
+πΆ2 sin(οΈβοΈ
14 β
π2
π2π΄π2 ln (ππΏπ + 1)
)οΈ]οΈ, (3.56)
where πΆ1 and πΆ2 are new constants.
3.2.3 Third method of finding an analytical solutionThere is one more possible way to solve Eq. (2.28). Let us start with multiplying thisequation by π20
π2πβ²
ππ‘2β π
20π0
dπ0dπ₯
ππβ²
ππ₯β π20
π2πβ²
ππ₯2= 0 . (3.57)
With reference to the following expression
π20π0
π
ππ₯
(οΈπ0ππβ²
ππ₯
)οΈ= π
20π0
dπ0dπ₯
ππβ²
ππ₯+ π20
π2πβ²
ππ₯2, (3.58)
Eq. (3.57) can be written as (see e.g. [6])
π2πβ²
ππ‘2β π
20π0
π
ππ₯
(οΈπ0ππβ²
ππ₯
)οΈ= 0 . (3.59)
Now let us substitute a new function πΉ (π₯, π‘) instead of the pressure
πβ²(π₯, π‘) = πΉ (π₯, π‘)/βοΈπ0(π₯) . (3.60)
Equation (3.59) takes the form
π2(οΈπΉ/
βπ0)οΈ
ππ‘2β π
20π0
π
ππ₯
(οΈπ0π(οΈπΉ/
βπ0)οΈ
ππ₯
)οΈ= 0 . (3.61)
14
-
3.3 Finding analytical solutions for known temperature distributions
Applying all simplifications, the above equation takes the form
π2πΉ
ππ₯2β 1π20
π2πΉ
ππ‘2= 1β
π0
d2βπ0
dπ₯2 πΉ . (3.62)
Let us remove from Eq. (3.62) derivatives dependent on t. The function F canbe represented as πΉ = πΉ (π₯, π‘) = πΉ (π₯)πβjππ‘ for this purpose. After substitution andrearranging the resulting equation is
d2πΉdπ₯2 =
(οΈ1βπ0
d2βπ0
dπ₯2 βπ2
π20
)οΈπΉ . (3.63)
Concerning Eq. (3.22) and the right part in brackets of the equation above, which isa constant π, it is possible to derive the following equation
1π0
d2π0dπ₯2 β
π2
π20= π . (3.64)
This is a nonlinear second-order ordinary differential equation, that can be solvedby an appropriate numerical method, but unfortunately its analytical solution is notknown and for this reason Eq. (2.28) cannot be solved analytically.
3.3 Finding analytical solutions for known temperaturedistributions
Let us write Eq. (3.37) once more
d2πdπ₯2 +
1π0
dπ0dπ₯
dπdπ₯ +
π2
π20π = 0 . (3.65)
According to the transformations (3.1) and (3.7) it is possible to express the followingderivatives
dπdπ₯ =
dπdπ0
dπ0dπ₯ , (3.66)
d2πdπ₯2 =
(οΈdπ0dπ₯
)οΈ2 d2πdπ 20
+ d2π0
dπ₯2dπdπ0
. (3.67)
Substituting expressions (3.66) and (3.67) into Eq. (3.65), using the equality π20 =ΞΊπ π0 and rearranging the terms leads to the equation(οΈdπ0
dπ₯
)οΈ2 d2πdπ 20
+[οΈ
1π0
(οΈdπ0dπ₯
)οΈ2+ d
2π0dπ₯2
]οΈdπdπ0
+ π2
ΞΊπ π
π0= 0 . (3.68)
Keeping in mind the equality
ddπ₯
(οΈπ0
dπ0dπ₯
)οΈ= dπ0dπ₯
dπ0dπ₯ + π0
d2π0dπ₯2 =
(οΈdπ0dπ₯
)οΈ2+ π0
d2π0dπ₯2 , (3.69)
Eq. (3.68) can be written as (see e.g. [3])(οΈdπ0dπ₯
)οΈ2 d2πdπ 20
+ 1π0
ddπ₯
(οΈπ0
dπ0dπ₯
)οΈ dπdπ0
+ π2
ΞΊπ π
π0= 0 . (3.70)
15
-
3 Exact analytical solutions for various temperature functions
3.3.1 Linear temperature distributionIn this section an acoustic pressure of a duct with a linear temperature distribution isstudied. The linear temperature distribution can be given by the expression (see e.g.[3])
π0(π₯) = ππ΄(π΄π₯+ 1) , (3.71)where π΄ is constant.
The dimensionless form of a temperature function Ξ is according to Eq. (2.36) (seee.g. [4])
Ξ(π) = π0(π)ππ΄
= π΄πΏπ + 1 . (3.72)
Some dimensionless linear temperature distributions are made possible by choosingdifferent values of constant π΄ and letting πΏ = 1 m are shown in Fig. 4.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
π
Ξ
π΄ = 1 mβ1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
1.2
π
Ξ
π΄ = β 0.5 mβ1
Fig. 4 Some possible linear dimensionless temperature distributions Ξ.
As derivatives of the temperature function (3.71) are
dπ0dπ₯ = ππ΄π΄ and
ddπ₯
(οΈπ0
dπ0dπ₯
)οΈ= ddπ₯
(οΈπ 2π΄π΄
2π₯+ π 2π΄π΄)οΈ
= π 2π΄π΄2 , (3.73)
then multiplying Eq. (3.70) by 1/π 2π΄π΄2 and substituting the above derivatives into thisequation obtains the following equation (see e.g. [3])
d2πdπ 20
+ 1π0
dπdπ0
+ π2/(οΈπ 2π΄π΄
2)οΈΞΊπ
π
π0= 0 . (3.74)
To simplify Eq. (3.74) a new independent variable π is introduced
π 2 = ππ0 , (3.75)
where the constant π is given by
π = 4π2
π 2π΄π΄2ΞΊπ
. (3.76)
It is necessary to know the first and the second derivatives of a new variable π withrespect to π0
2π dπ = πdπ0 βdπ dπ0
= π2π , (3.77)
16
-
3.3 Finding analytical solutions for known temperature distributions
2(οΈ dπ
dπ0
)οΈ2+ 2π d
2π
dπ 20= 0 β d
2π
dπ 20= β π
2
4π 3 . (3.78)
According to transformation of derivatives the derivatives of Eq. (3.74) then
dπdπ0
= dπdπ0dπ dπ0
= π2π dπdπ0
, (3.79)
d2πdπ 20
=(οΈ dπ
dπ0
)οΈ2 d2πdπ 2 +
d2π dπ 20
dπdπ =
π2
4π 2d2πdπ 2 β
π2
4π 3dπdπ , (3.80)
and Eq. (3.74) transforms to Eq. (3.83). Some steps of transforming are included
π2
4π 2d2πdπ 2 β
(οΈπ2
4π 3 βπ2
2π 3
)οΈdπdπ +
π2
4π 2π = 0 , (3.81)
π2
4π 2d2πdπ 2 +
π2
4π 3dπdπ +
π2
4π 2π = 0 . (3.82)
By multiplying Eq. (3.82) by 4π 2/π2 the final equation after the transformation is
d2πdπ 2 +
1π
dπdπ + π = 0 . (3.83)
Equation (3.83) is the zeroth order Bessel equation. The solution to the equation iswell known and given by
π (π ) = πΆ1J0(π ) + πΆ2Y0(π ) , (3.84)
where πΆ1 and πΆ2 are complex integration constants, J0 and Y0 are the Bessel andNeumann functions of the order zero.
Let us express variable π
π =βοΈππ0 =
βοΈ4π2
π 2π΄π΄2ΞΊπ
βοΈπ0 =
π
π
βοΈπ0 , (3.85)
whereπ = ππ΄|π΄|2
βΞΊπ . (3.86)
Then acoustic pressure can be rewritten as
π (π0) = πΆ1J0(οΈπ
π
βοΈπ0
)οΈ+ πΆ2Y0
(οΈπ
π
βοΈπ0
)οΈ. (3.87)
Let us substitute temperature distribution (3.71)
π (π₯) = πΆ1J0(οΈπ
π
βοΈππ΄(π΄π₯+ 1)
)οΈ+ πΆ2Y0
(οΈπ
π
βοΈππ΄(π΄π₯+ 1)
)οΈ. (3.88)
The dimensionless form of the pressure is (see e.g. [4])
Ξ¦(π) = πΆ1J0(οΈπ΅
βπ΄πΏπ + 1
)οΈ+ πΆ2Y0
(οΈπ΅
βπ΄πΏπ + 1
)οΈ, (3.89)
where πΆ1, πΆ2 are new constants, π΅ =(οΈπ
βππ΄)οΈ/π.
17
-
3 Exact analytical solutions for various temperature functions
3.3.2 Exponential temperature distributionThis section investigates the acoustic pressure of a duct with an exponential tempera-ture distribution that is given by the expression (see e.g. [3])
π0(π₯) = ππ΄πβπΎπ₯ , (3.90)
where ππ΄ and πΎ are constants.The dimensionless form Ξ of the temperature-inhomogeneous region π0 is (see e.g.
[4])
Ξ(π) = π0(π)ππ΄
= πβπΎπΏπ . (3.91)
Some dimensionless exponential temperature distributions made possible by choosingdifferent values of constant πΎ and letting πΏ = 1 m are shown in Fig. 5.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
1.5
2
2.5
3
π
Ξ
πΎ = β ln 3 mβ1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
1.2
π
Ξ
πΎ = β ln 0.5 mβ1
Fig. 5 Some possible exponential dimensionless temperature distributions Ξ.
Let us find derivatives to substitute into Eq. (3.70)
dπ0dπ₯ = βππ΄πΎπ
βπΎπ₯ and ddπ₯
(οΈπ0
dπ0dπ₯
)οΈ= ddπ₯
(οΈβπ 2π΄πΎπβ2πΎπ₯
)οΈ= 2π 2π΄πΎ2πβ2πΎπ₯ .
(3.92)Substituting into Eq. (3.70) and reducing gives the equation
(ππ΄πΎπβπΎπ₯)2d2πdπ 20
+ 2ππ΄πΎ2πβπΎπ₯dπdπ0
+ π2
ΞΊπ π
ππ΄πβπΎπ₯= 0 . (3.93)
Multiplying Eq. (3.93) by 1/πΎ2
(ππ΄πβπΎπ₯)2d2πdπ 20
+ 2ππ΄πβπΎπ₯dπdπ0
+ π2
ΞΊπ πΎ2π
ππ΄πβπΎπ₯= 0 (3.94)
and using backward substitition ππ΄πβπΎπ₯ = π0 leads to the equation (see e.g. [3])
π 20d2πdπ 20
+ 2π0dπdπ0
+ π2
ΞΊπ πΎ2π
π0= 0 . (3.95)
To simplify the previous equation two new variables π€ and π§ are introduced
π€ = πβοΈπ0 and π§2 = π
1π0
, (3.96)
18
-
3.3 Finding analytical solutions for known temperature distributions
where constant π isπ = 4π
2
ΞΊπ πΎ2. (3.97)
1. Let us start to work with variable π€ in the following three steps:β expressing the derivatives to substitute into Eq. (3.95)
dπdπ0
= ddπ0
(οΈπ€
π1/20
)οΈ= 1π
1/20
dπ€dπ0
β 12π€
π3/20
. (3.98)
d2πdπ 20
= β121
π3/20
dπ€dπ0
+ 1π
1/20
d2π€dπ 20
β 12
[οΈπ
3/20 dπ€/dπ0 β 3/2π€π
1/20
π 30
]οΈ
= β 1π
3/20
dπ€dπ0
+ 1π
1/20
d2π€dπ 20
+ 34π€
π5/20
. (3.99)
β substitution into Eq. (3.95)
π 20
[οΈβ 1π
3/20
dπ€dπ0
+ 1π
1/20
d2π€dπ 20
+ 34π€
π5/20
]οΈ+ 2π0
[οΈ1
π1/20
dπ€dπ0
β 12π€
π3/20
]οΈ
+ π2
ΞΊπ πΎ2π€
π3/20
= 0 . (3.100)
β multiplying and rearranging terms in the previous equation gives the equationbelow
π3/20
d2π€dπ 20
+ π 1/20dπ€dπ0
β 14π€
π1/20
+ π2
ΞΊπ πΎ2π€
π3/20
= 0 . (3.101)
2. Now let us work with the second variable π§:β finding the first and the second derivatives of a new variable π§ with respect
to the temperature π0
2π§dπ§ = βπ 1π 20
dπ0 βdπ§dπ0
= βπ2π§1π 20
, (3.102)
d2π§dπ 20
= βπ2
(οΈβ 1π§2
1π 20
dπ§dπ0
β 2 1π 30
1π§
)οΈ= π2
(οΈ21π§
1π 30
β π2π§31π 40
)οΈ= π
π§
1π 30
β π2
4π§31π 40
dπ§dπ0
. (3.103)
β according to the transformation of derivatives the derivatives of Eq. (3.101)then
dπ€dπ0
= dπ€dπ§dπ§dπ0
= βπ2π§1π 20
dπ€dπ§ , (3.104)
d2π€dπ 20
=(οΈ dπ§
dπ0
)οΈ2 d2π€dπ§2 +
d2π§dπ 20
dπ€dπ§ =
π2
4π§21π 40
d2π€dπ§2 +
(οΈπ
π§
1π 30
β π2
4π§31π 40
)οΈdπ€dπ§ .
(3.105)
19
-
3 Exact analytical solutions for various temperature functions
β substituting the above derivatives into Eq. (3.101)
π3/20
[οΈπ2
4π§21π 40
d2π€dπ§2 +
(οΈπ
π§
1π 30
β π2
4π§31π 40
)οΈdπ€dπ§
]οΈβ π 1/20
π
2π§1π 20
dπ€dπ§ β
14
π€
π1/20
+ π4π€
π3/20
= 0 . (3.106)
β reducing and rearranging of terms
π2
4π§21
π5/20
d2π€dπ§2 +
(οΈπ
2π§1
π3/20
β π2
4π§31
π5/20
)οΈdπ€dπ§ β
14
π€
π1/20
+ π4π€
π3/20
= 0 . (3.107)
β from Eqs. (3.96) it follows that π0 = π/π§2, substitution of this equality intoEq. (3.107)
π2
4π§2π§5
π5/2d2π€dπ§2 +
(οΈπ
2π§π§3
π3/2β π
2
4π§3π§5
π5/2
)οΈdπ€dπ§ β
14π€π§
π1/2+π4
π€π§3
π3/2= 0 . (3.108)
β after some reductions and rearranging some termsπ§3
4π1/2d2π€dπ§2 +
14π§2
π1/2dπ€dπ§ β
14π€π§
π1/2+ 14
π€π§3
π1/2= 0 . (3.109)
β multiplying the above equation by 4π1/2/π§3
d2π€dπ§2 +
1π§
dπ€dπ§ +
(οΈ1 β 1
π§2
)οΈπ€ = 0 . (3.110)
Equation (3.110) is the first order Bessel differential equation. The solution is givenas
π€(π§) = πΆ1J1(π§) + πΆ2Y1(π§) , (3.111)
where πΆ1 and πΆ2 are integration complex constants and J1 and Y1 are the Bessel andNeumann functions of the first order. Let us express variable π§
π§ =βοΈπ
π0=βοΈ
4π2ΞΊπ πΎ2
βοΈ1π0
= ππ 1βπ0
, (3.112)
whereπ = 2βοΈ
ΞΊπ πΎ2. (3.113)
So, the solution of Eq. (3.110) can be written in the form
π€(π0) = πΆ1J1(οΈππ
1βπ0
)οΈ+ πΆ2Y1
(οΈππ
1βπ0
)οΈ. (3.114)
According to Eqs. (3.96) the acoustic pressure then is
π (π0) =π€(π0)βπ0
= 1βπ0
[οΈπΆ1J1
(οΈππ
1βπ0
)οΈ+ πΆ2Y1
(οΈππ
1βπ0
)οΈ]οΈ. (3.115)
Let us substitute temperature distribution (3.90)
π (π₯) = 1βππ΄πβπΎπ₯
[οΈπΆ1J1
(οΈππ
1βππ΄πβπΎπ₯
)οΈ+ πΆ2Y1
(οΈππ
1βππ΄πβπΎπ₯
)οΈ]οΈ. (3.116)
The dimensionless form of the pressure is (see e.g. [4])
Ξ¦(π) =βππΎπΏπ
[οΈπΆ1J1
(οΈπ΅
βππΎπΏπ
)οΈ+ πΆ2Y1
(οΈπ΅
βππΎπΏπ
)οΈ]οΈ, (3.117)
where πΆ1, πΆ2 are new constants, π΅ = (ππ)/βππ΄ .
20
-
4 Transmission and reflection coefficients
This chapter deals with the calculation of transmission and reflection coefficients forexact analytical solutions derived in the previous chapter.
4.1 Sound velocity
The first exact analytical solution was derived for an acoustic velocity. Consider thereflection and transmission problem through the temperature-inhomogeneous regionfor an incident plane wave sketched in Fig. 6. The wave is partly reflected and partlytransmitted (see e.g. [2]).
Region A
ππ
ππ
ππ΄
ππ΄(π₯)
Region B
ππ‘
ππ΅
ππ΅(π₯)
Temperature βinhomogeneous region
π (π₯)
0 π₯πΏ
Fig. 6 Reflection and transmission of a sound velocity in a waveguide (see e.g. [2]).
It is essential to know the amplitudes ππ and ππ‘ of the reflected and transmittedwaves respectively and their integral constants.
In the region A can be written
ππ΄ = πππjππ΄π₯ + πππβjππ΄π₯ , (4.1)
where the quantities ππ and ππ are the complex velocity amplitudes of the incidentand reflected waves, and
ππ΄ =π
ππ΄. (4.2)
In the temperature-inhomogeneous region the velocity amplitude is given by Eq.(3.34). Let us write it once more
π (π₯) =βοΈπ0(π₯)πΉ (π₯) =
β―βΈβΈβ·ππ΄[οΈ(οΈπ΄2 β π
2
π2π΄
)οΈπ₯2 + 2π΄π₯+ 1
]οΈ
Γ[οΈπΆ1 cos
(οΈπ
|π|π(π₯)
)οΈβ πΆ2 sin
(οΈπ
|π|π(π₯)
)οΈ]οΈ, (4.3)
21
-
4 Transmission and reflection coefficients
where
π(π₯) = tanh-1[οΈππ΄|π|
(οΈπ΄+
(οΈπ΄2 β π
2
π2π΄
)οΈπ₯
)οΈ]οΈβ tanh-1
(οΈππ΄|π|π΄
)οΈ, π2 = π2 β π2 . (4.4)
In the region B can be written
ππ΅ = ππ‘πjππ΅(π₯βπΏ) , (4.5)
where ππ‘ is the complex acoustic velocity amplitude of the transmitted wave and
ππ΅ =βοΈΞΊπ ππ΅ , (4.6a) ππ΅ =
π
ππ΅. (4.6b)
The acoustic velocities must be the same on the interface of temperature-homogeneousand temperature-inhomogeneous regions. This is true for the both interfaces that havecoordinates π₯ = 0 and π₯ = πΏ. Consequently, there are two conditions in the waveguide
1. conditionππ΄(0) = π (0) , (4.7)
2. conditionππ΅(πΏ) = π (πΏ) . (4.8)
But these conditions contain four unknown variables ππ, ππ‘, πΆ1 and πΆ2. So, it isnecessary to impose two more conditions.
From the equation of continuity (2.20) and Eq. (2.27) the following equation can bewritten
ππβ²
ππ‘+ π0π20
ππ£
ππ₯= 0 . (4.9)
Substitution of the equalities (2.29) and (2.32) into Eq. (4.9)
π (π₯) = π0π20
jπdπdπ₯ . (4.10)
Let us distinguish densities and velocities in temperature-homogeneous regions π΄ andπ΅
π0(π₯ = 0) = ππ΄ , π0(π₯ = 0) =βοΈΞΊπ ππ΄ = ππ΄ , (4.11)
π0(π₯ = πΏ) = ππ΅ , π0(π₯ = πΏ) =βοΈΞΊπ ππ΅ = ππ΅ . (4.12)
From Eqs. (4.1), (4.5), (4.10), (4.11), (4.12) the acoustic pressures in regions A andB are
ππ΄ =ππ΄π
2π΄
jπ jππ΄(οΈπππ
jππ΄π₯ β πππβjππ΄π₯)οΈ, (4.13)
ππ΅ =ππ΅π
2π΅
jπ jππ΅ππ‘πjππ΅(π₯βπΏ) . (4.14)
The pressures also must be the same on the interface of temperature-homogeneousand temperature-inhomogeneous regions. For the interface at coordinate π₯ = 0 thefollowing steps are introduced
ππ΄|π₯=0 = π (π₯)|π₯=0 , (4.15)
22
-
4.1 Sound velocity
ππ΄π2π΄
jπ jππ΄(οΈπππ
jππ΄π₯ β πππβjππ΄π₯)οΈβββββ
π₯=0= ππ΄π
2π΄
jπdπdπ₯
βββββπ₯=0
. (4.16)
After reducing the term ππ΄π2π΄/jπ in Eq. (4.16) the left part jππ΄(οΈπππ
jππ΄π₯ β πππβjππ΄π₯)οΈ
then is the acoustic velocity derivative dππ΄/dπ₯. So,dππ΄dπ₯
ββββπ₯=0
= jππ΄ (ππ β ππ) =dπdπ₯
ββββπ₯=0
. (4.17)
Similarly, the acoustic velocity derivative at coordinate π₯ = πΏ isdππ΅dπ₯
ββββπ₯=πΏ
= jππ΅ππ‘ =dπdπ₯
ββββπ₯=πΏ
. (4.18)
As a consequence, the acoustic velocity derivatives must be the same on the inter-face of temperature-homogeneous and temperature-inhomogeneous regions. For bothinterfaces at coordinates π₯ = 0 and π₯ = πΏ the two following boundary conditions aredefined
3. conditiondππ΄dπ₯
ββββπ₯=0
= dπdπ₯
ββββπ₯=0
, (4.19)
4. conditiondππ΅dπ₯
ββββπ₯=πΏ
= dπdπ₯
ββββπ₯=πΏ
. (4.20)
Substituting Eqs. (4.1), (4.3), (4.5) into the four derived boundary conditions (4.7),(4.8), (4.19), (4.20) produces the following system of equations
ππ + ππ = πΆ1πΊ1 , (4.21)ππ‘ = πΆ1π»1 + πΆ2π»2 , (4.22)
jππ΄ (ππ β ππ) = πΆ1π1 + πΆ2π2 , (4.23)jππ΅ππ‘ = πΆ1π1 + πΆ2π2 , (4.24)
where
πΊ1 =βππ΄ , π(πΏ) = π΄+
(οΈπ΄2 β π
2
π2π΄
)οΈπΏ , π =
βοΈπ2 β π2 , (4.25)
π(πΏ) =
β―βΈβΈβ·ππ΄[οΈ(οΈπ΄2 β π
2
π2π΄
)οΈπΏ2 + 2π΄πΏ+ 1
]οΈ, (4.26)
π(πΏ) = tanh-1[οΈππ΄|π|π(πΏ)
]οΈβ tanh-1
(οΈππ΄|π|π΄
)οΈ, (4.27)
π»1 = π(πΏ) cos(οΈπ
|π|π(πΏ)
)οΈ, π»2 = βπ(πΏ) sin
(οΈπ
|π|π(πΏ)
)οΈ, (4.28)
π1 =βππ΄π΄ , π2 =
πβππ΄
, (4.29)
π(πΏ) =πππ΄
(οΈπ΄2 β π2
π2π΄
)οΈπ(πΏ)
π2π΄π2(πΏ) β π2
, π(πΏ) = ππ΄π(πΏ)π(πΏ) , (4.30)
π1 = π(πΏ) sin(οΈπ
|π|π(πΏ)
)οΈ+ π(πΏ) cos
(οΈπ
|π|π(πΏ)
)οΈ, (4.31)
π2 = π(πΏ) cos(οΈπ
|π|π(πΏ)
)οΈβ π(πΏ) sin
(οΈπ
|π|π(πΏ)
)οΈ. (4.32)
23
-
4 Transmission and reflection coefficients
Solving the system of Eqs. (4.21)β(4.24) obtains
πΆ1 =2ππ΄ (ππ΅π»2 + jπ2)ππ
ππ΄ππ΅πΊ1π»2 +π1π2 βπ2π1 + j [ππ΄πΊ1π2 + ππ΅ (π»1π2 βπ»2π1)], (4.33)
πΆ2 = β2ππ΄ (ππ΅π»1 + jπ1)ππ
ππ΄ππ΅πΊ1π»2 +π1π2 βπ2π1 + j [ππ΄πΊ1π2 + ππ΅ (π»1π2 βπ»2π1)], (4.34)
ππ =(ππ΄ππ΅πΊ1π»2 βπ1π2 +π2π1 + j [ππ΄πΊ1π2 β ππ΅ (π»1π2 βπ»2π1)])ππππ΄ππ΅πΊ1π»2 +π1π2 βπ2π1 + j [ππ΄πΊ1π2 + ππ΅ (π»1π2 βπ»2π1)]
, (4.35)
ππ‘ =j2ππ΄ (π»1π2 βπ»2π1)ππ
ππ΄ππ΅πΊ1π»2 +π1π2 βπ2π1 + j [ππ΄πΊ1π2 + ππ΅ (π»1π2 βπ»2π1)]. (4.36)
As ππ is optional, the reflection coefficient π (π£) and transmission coefficient ππ(π£) canbe calculated on the basis of solutions of the system equations. The coefficients (seee.g. [2], [4]) are defined by
π (π£) = ππππ
(4.37)
and
ππ(π£) = ππ‘ππ. (4.38)
By supposing values of air constants as ΞΊ = 7/5 and π = 287.058 Jkgβ1Kβ1, settingthe characteristic length πΏ to 1 m and choosing different values of constants π΄, π andvelocity ππ΄ it is possible to see a frequency dependence of the wave reflection andtransmission coefficients of the exact analytical solutions given by Eq. (4.3). Hereattention must be paid to one serious condition
π2 = π2 β π2 β₯ 0 . (4.39)
That is why the graphs start at point π = |π|.
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000
0.05
0.1
0.15
0.2
0.25
0.3
0.35
π (rad/s)
|R(v
) |(-
)
Reflection coefficient
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000.94
0.95
0.96
0.97
0.98
0.99
1
π (rad/s)
|Tr(v
) |(-
)
Transmission coefficient
Fig. 7 Dependence of modulii of reflection and transmission coefficients for ππ΄ = 345 msβ1,|π| = 597.558 sβ1, π΄ = 1 mβ1 in Eq. (4.3) on angular frequency.
24
-
4.1 Sound velocity
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
π (rad/s)
|R(v
) |(-
)
Reflection coefficient
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000.84
0.85
0.85
0.85
0.85
0.85
0.86
0.86
0.86
π (rad/s)
|Tr(v
) |(-
)
Transmission coefficient
Fig. 8 Dependence of modulii of reflection and transmission coefficients for ππ΄ = 345 msβ1,|π| = 172.5 sβ1, π΄ = β0.008 mβ1 in Eq. (4.3) on angular frequency.
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000
0.05
0.1
0.15
0.2
0.25
0.3
0.35
π (rad/s)
|R(v
) |(-
)
Reflection coefficient
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0001.34
1.36
1.38
1.4
1.42
1.44
1.46
π (rad/s)
|Tr(v
) |(-
)Transmission coefficient
Fig. 9 Dependence of modulii of reflection and transmission coefficients for ππ΄ = 345 msβ1,|π| = 301.39 sβ1, π΄ = β0.662 mβ1 in Eq. (4.3) on angular frequency.
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000
0.05
0.1
0.15
0.2
0.25
0.3
0.35
π (rad/s)
|R(v
) |(-
)
Reflection coefficient
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0001.27
1.28
1.29
1.3
1.31
1.32
1.33
1.34
1.35
1.36
π (rad/s)
|Tr(v
) |(-
)
Transmission coefficient
Fig. 10 Dependence of modulii of reflection and transmission coefficients for ππ΄ = 345 msβ1,|π| = 0.97 sβ1, π΄ = 0.328 mβ1 in Eq. (4.3) on angular frequency.
25
-
4 Transmission and reflection coefficients
4.2 Sound pressure
The following exact analytical solutions are derived for sound pressure.Consider the reflection and transmission problem through the temperature-inhomoge-
neous region for an incident plane wave sketched in Fig. 11. The wave is partly reflectedand partly transmitted as it was for sound velocity, but now the object of interest issound pressure (see e.g. [2], [4]).
Region A
ππ
ππ
ππ΄
ππ΄(π₯)
Region B
ππ‘
ππ΅
ππ΅(π₯)
Temperature βinhomogeneous region
π (π₯)
0 π₯πΏ
Fig. 11 Reflection and transmission of a sound pressure in a waveguide (see e.g. [2], [4]).
Now, it is necessary to know the amplitudes ππ and ππ‘ of reflected and transmittedwaves respectively and also their integral constants.
In the region A can be written
ππ΄ = πππjππ΄π₯ + πππβjππ΄π₯ , (4.40)
where the quantities ππ and ππ are the complex velocity amplitudes of the incidentand reflected waves, and the same Eq. (4.2) is applied.
In the region B can be written
ππ΅ = ππ‘πjππ΅(π₯βπΏ) , (4.41)
where ππ‘ is the complex velocity amplitude of the transmitted wave, and Eqs. (4.6)are true.
In the temperature-inhomogeneous region pressure amplitude is given by appropriateequalities, derived in the previous chapter, and all of these equalities are discussedbelow.
To calculate ππ, ππ‘, πΆ1 and πΆ2 it is necessary to deduce boundary conditions. Theseconditions can be imposed in a similar manner to boundary conditions of sound velocity.
From the linear form of the Euler equation (2.22) it follows that
ππ£
ππ‘= β 1
π0
ππβ²
ππ₯. (4.42)
After substitution of equalities (2.29) and (2.32) into Eq. (4.42)
π
ππ‘
(οΈπ (π₯)πβjππ‘
)οΈ= β 1
π0
π
ππ₯
(οΈπ (π₯)πβjππ‘
)οΈ(4.43)
26
-
4.2 Sound pressure
the amplitude of sound velocity then is
π (π₯) = 1jπ0πdπdπ₯ . (4.44)
From Eqs. (4.11), (4.12), (4.40), (4.41), (4.44) the acoustic velocities at regions Aand B are
ππ΄ =jππ΄
jππ΄π(οΈπππ
jππ΄π₯ β πππβjππ΄π₯)οΈ, (4.45)
ππ΅ =jππ΅
jππ΅πππ‘π
jππ΅(π₯βπΏ) . (4.46)
The velocities also must be the same on the interface of temperature-homogeneousand temperature-inhomogeneous regions. For an interface at coordinate π₯ = 0 thefollowing steps are introduced
ππ΄|π₯=0 = π (π₯)|π₯=0 , (4.47)
jππ΄jππ΄π
(οΈπππ
jππ΄π₯ β πππβjππ΄π₯)οΈββββ
π₯=0= 1jππ΄π
dπdπ₯
ββββπ₯=0
. (4.48)
After reducing the term 1/jππ΄π in Eq. (4.48) the left part jππ΄(οΈπππ
jππ΄π₯ β πππβjππ΄π₯)οΈ
isthen an acoustic pressure derivative dππ΄/dπ₯. So
dππ΄dπ₯
ββββπ₯=0
= jππ΄ (ππ β ππ) =dπdπ₯
ββββπ₯=0
. (4.49)
Similarly, the acoustic pressure derivative at coordinate π₯ = πΏ isdππ΅dπ₯
ββββπ₯=πΏ
= jππ΅ππ‘ =dπdπ₯
ββββπ₯=πΏ
. (4.50)
Finally, the four boundary conditions for the rest analytical solutions can be intro-duced
1. conditionππ΄(0) = π (0) , (4.51)
2. conditionππ΅(πΏ) = π (πΏ) , (4.52)
3. conditiondππ΄dπ₯
ββββπ₯=0
= dπdπ₯
ββββπ₯=0
, (4.53)
4. conditiondππ΅dπ₯
ββββπ₯=πΏ
= dπdπ₯
ββββπ₯=πΏ
. (4.54)
4.2.1 Second exact analytical solutionIn the temperature-inhomogeneous region the pressure amplitude is given by Eqs. (3.53)and (3.55).
β Ξ = π2π΄π2 β 4π2 = 0Let us write Eq. (3.53) once more
π (π₯) = 1βππ₯+ 1 [πΆ1 + πΆ2 ln(ππ₯+ 1)] . (4.55)
For this solution only one frequency can be found, that is why we are not interestedin this solution.
27
-
4 Transmission and reflection coefficients
β Ξ = π2π΄π2 β 4π2 < 0According to Eq. (4.2) the exact analytical solution (3.55) can be rewritten as
π (π₯) = 1βππ₯+ 1
β‘β£πΆ1 cosβββοΈ1
4 βπ2π΄π2
ln(ππ₯+ 1)
ββ +πΆ2 sin
βββοΈ14 β
π2π΄π2
ln(ππ₯+ 1)
ββ β€β¦ . (4.56)According to the system of Eqs. (4.51)β(4.54), the system of equations for theexact analytical solution (4.56) is then
ππ + ππ = πΆ1 , (4.57)ππ‘ = πΆ1π»1 + πΆ2π»2 , (4.58)
jππ΄ (ππ β ππ) = πΆ1π1 + πΆ2π2 , (4.59)jππ΅ππ‘ = πΆ1π1 + πΆ2π2 , (4.60)
where
π(πΏ) =
βοΈ14 β
π2π΄π2
ln(ππΏ+ 1) , (4.61)
π»1 =1β
ππΏ+ 1cos (π(πΏ)) , π»2 =
1βππΏ+ 1
sin (π(πΏ)) , (4.62)
π1 = β12π , π2 = π
βοΈ14 β
π2π΄π2
, (4.63)
π1 = βπ(ππΏ+ 1)β32
β‘β£12 cos (π(πΏ)) +
βοΈ14 β
π2π΄π2
sin (π(πΏ))
β€β¦ , (4.64)π2 = βπ(ππΏ+ 1)β
32
β‘β£12 sin (π(πΏ)) β
βοΈ14 β
π2π΄π2
cos (π(πΏ))
β€β¦ . (4.65)The solution of the system of Eqs. (4.57)β(4.60) is
πΆ1 =2ππ΄ (ππ΅π»2 + jπ2)ππ
ππ΄ππ΅π»2 +π1π2 βπ2π1 + j [ππ΄π2 β ππ΅ (π»2π1 βπ»1π2)], (4.66)
πΆ2 = β2ππ΄ (ππ΅π»1 + jπ1)ππ
ππ΄ππ΅π»2 +π1π2 βπ2π1 + j [ππ΄π2 β ππ΅ (π»2π1 βπ»1π2)], (4.67)
ππ =(ππ΄ππ΅π»2 βπ1π2 +π2π1 + j [ππ΄π2 + ππ΅ (π»2π1 βπ»1π2)])ππππ΄ππ΅π»2 +π1π2 βπ2π1 + j [ππ΄π2 β ππ΅ (π»2π1 βπ»1π2)]
, (4.68)
ππ‘ =j2ππ΄ (π»1π2 βπ»2π1)ππ
ππ΄ππ΅π»2 +π1π2 βπ2π1 + j [ππ΄π2 β ππ΅ (π»2π1 βπ»1π2)]. (4.69)
The calculation of reflection and transmission coefficients (ππ is optional) is thesame as for sound velocity (see e.g. [2], [4])
π (π ) = ππππ
(4.70)
andππ(π ) = ππ‘
ππ. (4.71)
28
-
4.2 Sound pressure
By supposing values of air constants as ΞΊ = 7/5 and π = 287.058 Jkgβ1Kβ1, settingthe characteristic length πΏ to 1 m and choosing different values of constants π, ππ΄it is possible to see a frequency dependence of the wave reflection and transmissioncoefficients of the exact analytical solutions given by Eq. (4.56). Ξ = π2π΄π2 β 4π2 < 0,ππ΄ > 0 and from Eq. (4.56) it can be concluded that π > β1/π, then only π > 0 canbe examined. Let us choose ππ΄ = 296 K and some different values of π to see thesefrequency dependencies.
0 500 1,000 1,500 2,000 2,500 3,000 3,500 4,000 4,500 5,0000.3
0.4
0.5
0.6
0.7
0.8
0.9
1
π (rad/s)
|R(p
) |(-
)
Reflection coefficient
π = 1 mβ1π = 2 mβ1π = 3 mβ1
0 500 1,000 1,500 2,000 2,500 3,000 3,500 4,000 4,500 5,0000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
π (rad/s)
|Tr(p
) |(-
)
Transmission coefficient
π = 1 mβ1π = 2 mβ1π = 3 mβ1
Fig. 12 Dependence of modulii of reflection and transmission coefficients for ππ΄ = 296 K anddifferent values of π in Eq. (4.56) on angular frequency.
4.2.2 Exact analytical solution for a linear temperature distributionThe exact analytical solution for a linear temperature distribution from Chapter 3 is
π (π₯) = πΆ1J0(οΈπ
π
βοΈππ΄(π΄π₯+ 1)
)οΈ+ πΆ2Y0
(οΈπ
π
βοΈππ΄(π΄π₯+ 1)
)οΈ, (4.72)
where π = ππ΄|π΄|βΞΊπ /2 .
After all calculations according to the system of Eqs. (4.51)β(4.54) the followingsystem of equations is produced
ππ + ππ = πΆ1πΊ1 + πΆ2πΊ2 , (4.73)ππ‘ = πΆ1π»1 + πΆ2π»2 , (4.74)
jππ΄ (ππ β ππ) = πΆ1π1 + πΆ2π2 , (4.75)jππ΅ππ‘ = πΆ1π1 + πΆ2π2 , (4.76)
where
πΊ1 = J0(οΈπ
π
βοΈππ΄
)οΈ, πΊ2 = Y0
(οΈπ
π
βοΈππ΄
)οΈ, (4.77)
π»1 = J0(οΈπ
π
βοΈππ΄(π΄πΏ+ 1)
)οΈ, π»2 = Y0
(οΈπ
π
βοΈππ΄(π΄πΏ+ 1)
)οΈ, (4.78)
π1 = βππ΄
2πβοΈππ΄J1
(οΈπ
π
βοΈππ΄
)οΈ, π2 = β
ππ΄
2πβοΈππ΄Y1
(οΈπ
π
βοΈππ΄
)οΈ, (4.79)
π1 = βππ΄
2π
βοΈππ΄
π΄πΏ+ 1J1(οΈπ
π
βοΈππ΄(π΄πΏ+ 1)
)οΈ, (4.80)
π2 = βππ΄
2π
βοΈππ΄
π΄πΏ+ 1Y1(οΈπ
π
βοΈππ΄(π΄πΏ+ 1)
)οΈ. (4.81)
29
-
4 Transmission and reflection coefficients
Solution of the system of Eqs. (4.73)β(4.76) is
πΆ1 =2ππ΄ (ππ΅π»2 + jπ2)ππ
ππ΄ππ΅ (πΊ1π»2 βπΊ2π»1) +π1π2 βπ2π1 + j [ππ΄ (πΊ1π2 βπΊ2π1) + ππ΅ (π»1π2 βπ»2π1)],
(4.82)πΆ2 = β1Γ
2ππ΄ (ππ΅π»1 + jπ1)ππππ΄ππ΅ (πΊ1π»2 βπΊ2π»1) +π1π2 βπ2π1 + j [ππ΄ (πΊ1π2 βπΊ2π1) + ππ΅ (π»1π2 βπ»2π1)]
,
(4.83)ππ = ππΓππ΄ππ΅ (πΊ1π»2 βπΊ2π»1) βπ1π2 +π2π1 + j [ππ΄ (πΊ1π2 βπΊ2π1) β ππ΅ (π»1π2 βπ»2π1)]ππ΄ππ΅ (πΊ1π»2 βπΊ2π»1) +π1π2 βπ2π1 + j [ππ΄ (πΊ1π2 βπΊ2π1) + ππ΅ (π»1π2 βπ»2π1)]
,
(4.84)ππ‘ = ππΓ
j2ππ΄ (π»1π2 βπ»2π1)ππ΄ππ΅ (πΊ1π»2 βπΊ2π»1) +π1π2 βπ2π1 + j [ππ΄ (πΊ1π2 βπΊ2π1) + ππ΅ (π»1π2 βπ»2π1)]
.
(4.85)
The reflection and transmission coefficients can be calculated according to Eqs.(4.70), (4.71) (see e.g. [2], [4]).
By supposing values of constants for air as ΞΊ = 7/5 and π = 287.058 Jkgβ1Kβ1,setting the characteristic length πΏ to 1 m and choosing different values of constants π΄,ππ΄ it is possible to see a frequency dependence of the wave reflection and transmissioncoefficients of the exact analytical solution given by Eq. (4.72). Let us choose ππ΄ = 296K for a positive temperature gradient, ππ΄ = 752 K for a negative temperature gradientand some different values of π΄ to see these frequency dependencies.
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000
0.05
0.1
0.15
0.2
0.25
π (rad/s)
|R(p
) |(-
)
Reflection coefficient
π΄ = 456/ππ΄ mβ1π΄ = 238/ππ΄ mβ1π΄ = 110/ππ΄ mβ1
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000.76
0.78
0.8
0.82
0.84
0.86
0.88
0.9
0.92
0.94
π (rad/s)
|Tr(p
) |(-
)
Transmission coefficient
π΄ = 456/ππ΄ mβ1π΄ = 238/ππ΄ mβ1π΄ = 110/ππ΄ mβ1
Fig. 13 Dependence of modulii of reflection and transmission coefficients for ππ΄ = 296 K anddifferent values of π΄ (positive gradient) in Eq. (4.72) on angular frequency.
30
-
4.2 Sound pressure
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000
0.05
0.1
0.15
0.2
0.25
π (rad/s)
|R(p
) |(-
)
Reflection coefficient
π΄ = β218/ππ΄ mβ1π΄ = β346/ππ΄ mβ1π΄ = β456/ππ΄ mβ1
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000.9
0.95
1
1.05
1.1
1.15
1.2
1.25
1.3
π (rad/s)
|Tr(p
) |(-
)
Transmission coefficient
π΄ = β218/ππ΄ mβ1π΄ = β346/ππ΄ mβ1π΄ = β456/ππ΄ mβ1
Fig. 14 Dependence of modulii of reflection and transmission coefficients for ππ΄ = 752 K anddifferent values of π΄ (negative gradient) in Eq. (4.72) on angular frequency.
4.2.3 Exact analytical solution for an exponential temperature distribution
The exact analytical solution for an exponential temperature distribution was derivedin the previous chapter
π (π₯) = 1βππ΄πβπΎπ₯
[οΈπΆ1J1
(οΈππ
1βππ΄πβπΎπ₯
)οΈ+ πΆ2Y1
(οΈππ
1βππ΄πβπΎπ₯
)οΈ]οΈ, (4.86)
where π = 2/βοΈΞΊπ πΎ2 .
According to the system of Eqs. (4.51)β(4.54), the system of equations for an exactanalytical solution (4.86) is the same as the system of Eqs. (4.73)β(4.76), therefore thesolution has the form of Eqs. (4.82)β(4.85), but notations πΊ1, πΊ2, π»1, π»2, π1, π2, π1and π2 have different meanings
πΊ1 =1βππ΄
J1(οΈππ
1βππ΄
)οΈ, πΊ2 =
1βππ΄
Y1(οΈππ
1βππ΄
)οΈ, (4.87)
π»1 =1βοΈ
ππ΄πβπΎπΏJ1
(οΈππ
1βοΈππ΄πβπΎπΏ
)οΈ, π»2 =
1βοΈππ΄πβπΎπΏ
Y1
(οΈππ
1βοΈππ΄πβπΎπΏ
)οΈ, (4.88)
π1 =πΎππ
2ππ΄J0(οΈππ
1βππ΄
)οΈ, π2 =
πΎππ
2ππ΄Y0(οΈππ
1βππ΄
)οΈ, (4.89)
π1 =πΎππ
2ππ΄πβπΎπΏJ0
(οΈππ
1βοΈππ΄πβπΎπΏ
)οΈ, π2 =
πΎππ
2ππ΄πβπΎπΏY0
(οΈππ
1βοΈππ΄πβπΎπΏ
)οΈ. (4.90)
Reflection and transmission coefficients can be calculated according to Eqs. (4.70),(4.71) (see e.g. [2], [4]).
By supposing values of air constants as ΞΊ = 7/5 and π = 287.058 Jkgβ1Kβ1, settingthe characteristic length πΏ to 1 m and choosing different values of constants πΎ, ππ΄it is possible to see a frequency dependence of the wave reflection and transmissioncoefficients of the exact analytical solution given by Eq. (4.86). Let us choose ππ΄ = 296K for both positive and negative temperature gradients and some different values of πΎto see frequency dependencies.
31
-
4 Transmission and reflection coefficients
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000.92
0.93
0.94
0.95
0.96
0.97
0.98
π (rad/s)
|R(p
) |(-
)Reflection coefficient
πΎ = β ln 1.1 mβ1πΎ = β ln 2.0 mβ1πΎ = β ln 3.0 mβ1
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0001.4
1.5
1.6
1.7
1.8
1.9
2
π (rad/s)
|Tr(p
) |(-
)
Transmission coefficient
πΎ = β ln 1.1 mβ1πΎ = β ln 2.0 mβ1πΎ = β ln 3.0 mβ1
Fig. 15 Dependence of modulii of reflection and transmission coefficients for ππ΄ = 296 K anddifferent values of πΎ (positive gradient) in Eq. (4.86) on angular frequency.
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0000.65
0.7
0.75
0.8
0.85
0.9
0.95
1
π (rad/s)
|R(p
) |(-
)
Reflection coefficient
πΎ = β ln 0.8 mβ1πΎ = β ln 0.5 mβ1πΎ = β ln 0.1 mβ1
0 1,000 2,000 3,000 4,000 5,000 6,000 7,0001.5
2
2.5
3
3.5
4
π (rad/s)
|Tr(p
) |(-
)Transmission coefficient
πΎ = β ln 0.8 mβ1πΎ = β ln 0.5 mβ1πΎ = β ln 0.1 mβ1
Fig. 16 Dependence of modulii of reflection and transmission coefficients for ππ΄ = 296 K anddifferent values of πΎ (negative gradient) in Eq. (4.86) on angular frequency.
32
-
5 Derivation of the Burgers-type equationfor temperature-inhomogeneous fluids
In this section the Burgers-type equation for nonlinear acoustic waves in the temperature-inhomogeneous fluids and its dimensionless form is derivated. The classical Burgersequation is the most widely used model equation for studying the combined effects ofdissipation and nonlinearity on progressive plane waves (see e.g. [7]).
5.1 Westervelt equationLet us begin with the nonlinear acoustic wave equation for progressive waves, which iscalled the Westervelt equation (see e.g. [1], [8], [9])
οΏ½2πβ² = β πΌπ0π40
π3πβ²
ππ‘3β π½π0π40
π2πβ²2
ππ‘2, (5.1)
where π½ is the nonlinearity coefficient, πΌ is the sound diffusivity andοΏ½2 is the dβAlembertionoperator for plane waves. These notations mean
π½ = ΞΊ + 12 ,(5.2a)
οΏ½2(Β·) = π2(Β·)ππ₯2
β 1π20
π(Β·)ππ‘
,
(5.2b)
πΌ = 43π+ π + π (οΈ
1ππ
β 1ππ
)οΈβΌ π ,
(5.2c)where ππ and ππ are specific heats at constant volume and pressure respectively andπ < 1 a small dimensionless parameter.
The left hand side of Eq. (5.1) represents the canonical wave equation for acousticwaves in homogeneous fluids. For temperature-inhomogeneous media it was derived thewave equation (2.24) which can be expressed as
οΏ½2π πβ² = 0 , (5.3)
where οΏ½2π (Β·) =π2(Β·)ππ₯2 β
1π0
dπ0dπ₯
π(Β·)ππ₯ β
1π20
π2(Β·)ππ‘2 is the dβAlembertion operator for plane waves
in temperature-inhomogeneous media. On the basis of this operator the Westerveltequation (5.1) can be modified for a temperature inhomogeneous media
οΏ½2π πβ² = β πΌ
π0π40
π3πβ²
ππ‘3β π½π0π40
π2πβ²2
ππ‘2. (5.4)
The validity of this equation is restricted to regions with low temperature gradients.
5.2 Burgers-type equationGiven that the spatial variation of a plane progressive acoustic wave is small enoughin proportion to one wavelength it is possible to apply the multiple-scale method tosimplify Eq. (5.4). The retarded time π = π‘β π₯/ππ΄ is introduced.
33
-
5 Derivation of the Burgers-type equation for temperature-inhomogeneous fluids
Let us find the solution of Eq. (5.4) in the form (see e.g. [7])
π = π(π₯1, π) , (5.5)
where
π₯1 = ππ₯ , (5.6a) π = π‘βπ₯
ππ΄. (5.6b)
In the retarded time frame (i.e., for an observer in a reference frame that movesat speed ππ΄), nonlinearity and absorption separately produce only slow variations asfunctions of distance. Moreover, the relative order of the variations due to each effectis the same, i.e., it is π(π). Thus, it can be anticipated that the combined effects ofnonlinearity and absorbtion will introduce variations of the same order. The coordinateπ₯1 is referred to as the slow scale corresponding to the retarded time frame π .
To derive a simplified progressive-wave equation that accounts for both absorptionand nonlinearity, let us first rewrite Eq. (5.4) in the new coordinate system (π₯1, π).Transformations of the partial derivatives are
π(Β·)ππ₯
= ππ(Β·)ππ₯1
β 1ππ΄
π(Β·)ππ
, (5.7)
π2(Β·)ππ₯2
= βπ 2ππ΄
π2(Β·)ππππ₯1
+ 1π2π΄
π2(Β·)ππ2
, (5.8)
π(Β·)ππ‘
= π(Β·)ππ
,π2(Β·)ππ‘2
= π2(Β·)ππ2
,π3(Β·)ππ‘3
= π3(Β·)ππ3
, (5.9)
applying the introduced notations
π2πβ²
ππ₯2= βπ 2
ππ΄
π2πβ²
ππππ₯1+ 1π2π΄
π2πβ²
ππ2,
dπ0dπ₯ = π
dπ0dπ₯1
,ππβ²
ππ₯= π ππ
β²
ππ₯1β 1ππ΄
ππβ²
ππ,
π2πβ²
ππ‘2= π
2πβ²
ππ2,
π3πβ²
ππ‘3= π
3πβ²
ππ3.
(5.10)
Substitution of derivatives (5.10) into Eq. (5.4) gives
βπ 2ππ΄
π2πβ²
ππππ₯1+ 1π2π΄
π2πβ²
ππ2βπ
2
π0
dπ0dπ₯1
ππβ²
ππ₯1+ ππ0ππ΄
dπ0dπ₯1
ππβ²
ππβ 1π20
π2πβ²
ππ2= β πΌ
π0π40
π3πβ²
ππ3β 2π½π0π40
πβ²π2πβ²
ππ2.
(5.11)We are interested in the second approximation of the above equation. The third term
in Eq. (5.11) is π(π3) and is therefore discarded. After that and small rearranging Eq.(5.11) can be written as
βπ 2ππ΄
π2πβ²
ππππ₯1+(οΈ
1π2π΄
β 1π20
)οΈπ2πβ²
ππ2+ ππ0ππ΄
dπ0dπ₯1
ππβ²
ππ= β πΌ
π0π40
π3πβ²
ππ3β 2π½π0π40
πβ²π2πβ²
ππ2. (5.12)
Integration of Eq. (5.12) with respect to π , multiplication of the resulting equationby βππ΄/2 and having noted
βππ΄2
(οΈ1π2π΄
β 1π20
)οΈ= 12ππ΄
π2π΄ β π20π20
= 12ππ΄ππ΄ β π0π0
= 1 β Ξ (π)2ππ΄Ξ (π)(5.13)
34
-
5.3 Dimensionless Burgers-type equation
leads to equation
πππβ²
ππ₯1+ 1 β Ξ (π)2ππ΄Ξ (π)
ππβ²
ππβ π2π0
dπ0dπ₯1
πβ² = πΌππ΄2π0π40
π2πβ²
ππ2+ π½ππ΄π0π40
πβ²ππβ²
ππ. (5.14)
Applying Eq. (2.26) and returning to the physical coordinate π₯ in place of π₯1 theabove equation can be written as
ππβ²
ππ₯+ 1 β Ξ (π)2ππ΄Ξ (π)
ππβ²
ππ+ 12π0
dπ0dπ₯ π
β² β πΌππ΄2π0π40
π2πβ²
ππ2β π½ππ΄π0π40
πβ²ππβ²
ππ= 0 , (5.15)
and this is the Burgers-type equation for a temperature-inhomogeneous fluid.
5.3 Dimensionless Burgers-type equationUsing expressions (2.36) the Burgers-type equation (5.15) can be written as
πΞ ππ
+ [1 β Ξ (π)]ππΏ2ππ΄Ξ (π)πΞ ππ
+ 12Ξ (π)dΞ (π)
dπ Ξ βπΌπ2πΏ
2Ξ (π) ππ΄π3π΄π2Ξ ππ2
β π½ππΏΞ (π) ππ΄Ξ πΞ ππ
= 0 .
(5.16)Introducing new constants
π = ππΏ2ππ΄, πΊ = πΌπ
2πΏ
2ππ΄π3π΄, π = π½ππΏ
ππ΄(5.17)
and remembering
π = π₯πΏ, π = ππ , (5.18)
the Burgers-type equation (5.16) can be rewritten into the form
πΞ ππ
+ 1 β Ξ(π)Ξ(π) ππΞ ππ
+ 12Ξ(π)Ξ dΞ(π)
dπ βπΊ
Ξ(π)π2Ξ ππ2
β πΞ(π)Ξ πΞ ππ
= 0 . (5.19)
Equation (5.19) is the dimensionless Burgers-type equation for plane progressive non-linear waves in temperature-inhomogeneous fluids.
35
-
6 N