Mathematical aspects of thermoacoustics

Citation for published version (APA): Panhuis, in 't, P. H. M. W. (2009). Mathematical aspects of thermoacoustics. Technische Universiteit Eindhoven. https://doi.org/10.6100/IR642908

DOI: 10.6100/IR642908

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ISBN 978-90-386-1862-3 NUR 919 Subject headings: thermoacoustics; acoustics; acoustic streaming; thermodynamics; per- turbation methods; numerical methods; boundary value problems; nonlinear analysis; shock waves.

This research was financially supported by the Technology Foundation (STW), grant number ETTF.6668. Mathematical Aspects of Thermoacoustics

PROEFSCHRIFT

ter verkrijging van de graad van doctor aan de Technische Universiteit Eindhoven, op gezag van de rector magnificus, prof.dr.ir. C.J. van Duijn, voor een commissie aangewezen door het College voor Promoties in het openbaar te verdedigen op donderdag 25 juni 2009 om 16.00 uur

door

Petrus Hendrikus Maria Wilhelmus in ’t panhuis

geboren te Roermond Dit proefschrift is goedgekeurd door de promotor: prof.dr. J.J.M. Slot

Copromotoren: dr. S.W. Rienstra en prof.dr. J. Molenaar To Ik Cuin

Preface

This project is part of a twin PhD program between the Departments of Mathemat- ics and Computer Science and Applied Physics and is sponsored by the Technology Foundation (STW), Royal Dutch Shell, the Energy Research Centre of the Netherlands (ECN), and Aster Thermoacoustics. I would like to express thanks to all people who participated in this project. First and foremost I would like to thank my mathematical supervisors dr. Sjoerd Rienstra, prof.dr. Han Slot, and prof.dr. Jaap Molenaar for their expert guidance and stimulating support. I am also indebted to my physics colleagues Paul Aben, dr. Jos Zeegers, and prof.dr. Fons de Waele, who helped to broaden and deepen my understanding of the physics involved. I am also grateful for the many useful discussions I have had with the people from ECN, Aster Thermoacoustics, and Shell. My defense committee is formed by prof.dr. Anthony Atchley, prof.dr. Bendiks-Jan Boersma, and prof.dr. Mico Hirschberg, together with my supervisors dr. Sjoerd Rien- stra, prof.dr. Han Slot, and prof.dr. Jaap Molenaar. I would like to thank them for the time invested and their willingness to judge my work. I also want to thank prof.dr. Bob Mattheij for agreeing to be part of the extended defense committee. What made these four years especially enjoyable was the great working atmosphere within CASA and the Low Temperature group. My special thanks go out to my of- fice mate and partner-in-crime Erwin who started and finished his PhD (and Master) at the same time as I did. Many thanks also to all the current and former colleagues that I have had the pleasure to work with, in particular the PhD students and postdocs: Aga, Ali, Andriy, Bart, Berkan, Christina, Darcy, Davit, Dragan, Hans, Jurgen, Kakuba, Kamyar, Kundan, Laura, Marco, Maria ( 2), Mark, Matthias, Maxim, Michiel, Miguel, Mirela, Nico, Oleg, Patricio, Paul, Remko,× Remo, Roxana, Shruti, Sven, Tasnim, Temes- gen, Valeriu, Venkat, Wenqing, Yabin, Yan, Yixin, Yves, Zoran. I fondly think back to our daily lunches at the Kennispoort, the weekly poker games, the road-trips to Den- mark, the regular squash/tennis/football games, and the many nights in town that I have enjoyed with so many of you. Our two secretaries Mar`ese and Enna also deserve a special word of thanks, for making life of a PhD student so much easier by taking care of all administrative details. I am also thankful to the members of the football teams Pusphaira and Old Soccers, and hope they will have more success without me. On a more personal level, I would like to thank all my friends and family, for their continuous love and support. I especially want to show appreciation to my mom for her unbridled enthusiasm and my dad, who I wish could have been here today. I also want to thank my siblings Jos, Hellen, and Dorris and their significant others Marjanne, Joram, and Tonnie. Of course I should not forget to mention my little nephew Sep, who ii Preface is getting so big now. Last, but definitely not least, I would like to thank my girlfriend Jessey, to whom this thesis is dedicated, for her unlimited love and patience in these last few months.

Peter in ’t panhuis Eindhoven, May 2009 Nomenclature

General symbols and variables

A [m2] cross-sectional area [m2] cross-section A 2 b [ms− ] specific body force field 1 c [m s− ] speed of sound 1 1 Cp [J kg− K− ] isobaric specific heat 1 1 Cs [J kg− K− ] specific heat of stack material 1 1 Cv [J kg− K− ] isochoric specific heat d [m] diameter f [Hz] frequency fν viscous Rott function fk thermal Rott function fs solid Rott function Fν viscous Arnott function Fk thermal Arnott function Fs solid Arnott function Fourier transform FG Green’s function 2 g [ms− ] gravitational acceleration H˙ [W] total power 1 h [J kg− ] specific enthalpy Im imaginary part 1 k [m− ] wave number 1 1 K [W K− m− ] thermal conductivity ℓ [m] displacement L [m] typical length 2 1 m˙ [kg m− s− ] time-averaged mass flux 2 1 M˙ [kg m− s− ] time-averaged volumetric mass flux 1 n˙ [mol s− ] molar flow rate P [W] power p [Pa] pressure pA [Pa] pressure oscillation amplitude pamb [Pa] ambient pressure iv Nomenclature

Q˙ [W] heat flow per unit time 2 q [W m− ] heat flux r [m] radial coordinate Re realpart 1 1 Rspec [J kg− K− ] specific gas constant [m] radius R 1 1 s [J kg− K− ] specific entropy 1 S [J kg− ] entropy S surface Su [K] Sutherland’s constant t [s] time T [K] temperature 1 U [m s− ] typical fluid speed U˙ [W] internal energy 1 v =(u, v, w) [m s− ] velocity vector V [m3] volume W˙ [W] acoustic power x =(x, y, z) [m] spatial coordinate Z [N s m3] impedance 1 β [K− ] isobaric volumetric expansion coefficient Γ boundary, interface δ [m] penetration depth 1 ǫ [J kg− ] specific internal energy λ [m] wave length µ [Pa s] dynamic (shear) viscosity ζ [Pa s] second viscosity 3 ρ [kg m− ] density 2 τ [N m− ] viscous stress tensor θ [rad] angular coordinate 1 ω [rad s− ] angular frequency of the acoustic oscillations

Dimensionless numbers

A amplitude Br blockage ratio COP coefficient of performance COPR relative coefficient of performance COPC Carnot coefficient of performance Dr drive ratio Fr Froude number Ma acoustic Mach number NL Lautrec number gas Ns Lautrec number solid Pr Prandtl number R reflection coefficient Nomenclature v

Sk Strouhal number based on δk Wo Womersley number Wζ second Womersley number β coefficient of nonlinearity γ ratio specific heats δ coefficient of stack dissipation ∆ deviation from resonance η coefficient of weak nonlinearity η efficiency ηR relative efficiency ηC Carnot efficiency ε aspect ratio ε driver Mach number εs stack heat capacity ratio κ Helmholtz number σ ratio thermal conductivities φ porosity µ dimensionless viscosity

Sub- and superscripts and special operators a˜ dimensionless a time averaging a transverse averaging ha˙ i per unit time a∗ complex conjugate are f reference value a+ top plate a− bottom plate a0 steady zeroth order a1 first harmonic a2,0 steady second order (streaming) a2,2 second harmonic aC cold aH hot ag gas, fluid ak thermal aL thermal ap isobaric aR right as solid, source, stack center at outer aτ transverse components av isochoric aν viscous vi Nomenclature Contents

Preface i

Nomenclature iii

1 Introduction 1 1.1 Ahistoricalperspective ...... 2 1.2 The basic mechanism of thermoacoustics ...... 6 1.3 Classification...... 8 1.4 Applications...... 10 1.5 Thesisoverview...... 12 1.6 Literaturereview ...... 13

2 Thermodynamics 17 2.1 Lawsofthermodynamics ...... 17 2.2 Thermodynamicperformance ...... 19 2.2.1 Refrigeratororheatpump ...... 19 2.2.2 Primemover ...... 20 2.3 Thethermodynamiccycle ...... 20 2.3.1 Standing-wavephasing ...... 20 2.3.2 Traveling-wavephasing ...... 22 2.3.3 Bucket-brigadeeffect...... 25

3 Modeling 27 3.1 Geometry...... 27 3.2 Governingequations ...... 28 3.3 Scaling ...... 31 3.4 Small-amplitude and long-pore approximation ...... 34

4 Thermoacoustics in two-dimensional pores with variable cross-section 37 4.1 Governingequations ...... 38 4.2 Acoustics...... 39 4.3 Meantemperature ...... 46 4.4 Integration of the generalized Swift equations ...... 51 4.4.1 Exact solution at constant temperature ...... 52 4.4.2 Short-stackapproximation ...... 54 4.4.3 Approximate solution in short wide channels ...... 56 viii Contents

4.5 Acousticstreaming ...... 59 4.6 Secondharmonics...... 61 4.7 Power...... 67 4.7.1 Acousticpower ...... 67 4.7.2 Totalpower ...... 69

5 Thermoacoustics in three-dimensional pores with variablecross-section 73 5.1 Acoustics ...... 74 5.2 Meantemperature ...... 78 5.3 Integration of the generalized Swift equations ...... 83 5.3.1 Idealstack...... 84 5.3.2 Rotationally symmetric pores ...... 84 5.4 Acousticstreaming ...... 85 5.5 Secondharmonics...... 86 5.6 Power...... 90

6 Standing-wave devices 93 6.1 Design ...... 93 6.2 Computations ...... 95 6.3 Athermoacousticcouple...... 96 6.3.1 Acoustically generated temperature differences ...... 97 6.3.2 Acousticpower ...... 99 6.4 Astanding-waverefrigerator ...... 100 6.5 Astanding-waveprimemover ...... 108 6.6 Streaming effects in a thermoacoustic stack ...... 112

7 Traveling-wave devices 115 7.1 Atraveling-waveprimemover ...... 116 7.2 Computations ...... 118 7.2.1 Regeneratorandthermalbuffertube...... 119 7.2.2 Optimizationprocedure ...... 120 7.3 Results ...... 123 7.3.1 Temperature...... 123 7.3.2 Regeneratorefficiency ...... 125 7.3.3 Geometryoptimization ...... 129

8 Nonlinear standing waves 133 8.1 Governingequations ...... 134 8.1.1 Kuznetsov’sequation ...... 134 8.1.2 Bernoulli’sequation ...... 136 8.1.3 Perturbationexpansion ...... 136 8.2 Solutionawayfromresonance...... 139 8.2.1 Arbitraryexcitation...... 140 8.2.2 Harmonicexcitation ...... 141 8.3 Solutionnearresonance ...... 141 8.3.1 Exact solution when δˆ = ∆ˆ = 0 using Mathieu functions ...... 143 8.3.2 Steady-state solution for δˆ = (1) ...... 144 O 1 8.3.3 Steady-state solution for δˆ = (ν− ) ...... 148 O Contents ix

8.4 Results ...... 154 8.4.1 Nonlinear standing waves in a closed tube ...... 155 8.4.2 Nonlinear standing waves in a thermoacoustic refrigerator. . . . . 157

9 Conclusions and discussion 161

Appendices 165

A Thermodynamic constants and relations 165

B Derivations 167 B.1 Total-energyequation...... 167 B.2 Temperatureequation ...... 168

C Green’s functions 169 C.1 Fj-functions ...... 169 C.2 Fj,2-functions ...... 171 C.3 Green’s functions for various stack geometries ...... 172

Bibliography 175

Index 185

Summary 187

Samenvatting 189

Curriculum Vitae 191 x Contents Chapter 1

Introduction

As the name indicates, thermoacoustics combines the fields of thermodynamics and acoustics and describes the interaction between heat and sound. The term was first coined in 1980 by Rott [119] who described its meaning as “rather self-explanatory”. Ac- cording to Rott the most general interpretation of thermoacoustics includes “all effects in acoustics in which heat conduction and entropy variations of the (gaseous) medium play a role”. Ordinarily a sound wave in a fluid is seen as the combined effect of pressure and velocity (displacement) oscillations, but as a response to these (isentropic) pressure os- cillations, temperature oscillations may occur as well. In free space the temperature variations will be small and the gas parcels will expand and compress adiabatically. However, when the fluid is allowed to interact thermally with solid boundaries, heat transfer between the gas and the solid will take place and a wide range of thermoacous- tic effects may occur. It has been realized that under the right operating conditions these thermoacoustic concepts can be harnessed and exploited to create two kinds of thermoacoustic devices: the refrigerator or heat pump that turn sound into useful refrigeration or heating, and the prime mover that turns heat into useful (acoustic) work. Typically, such devices are con- structed from straight or looped tubes with a porous medium suitably located inside. Thermoacoustic devices can be of much practical use, because significant amount of heat and mechanical power can be produced at a reasonable efficiency. Moreover, in contrast to more conventional engines and refrigerators, they can operate without cranks and pistons, and usually have no more than one (mechanically) moving part. Because of this and their inherent simplicity, they are very reliable, require little mainte- nance, and can be produced at relatively low cost. Furthermore, thermoacoustic devices are friendly to the environment, as they use environmentally friendly gases, produce no toxic waste, and are easily adaptable to use solar power [27] or industrial waste heat [127] as a driving source. Currently, research is also being done on the possibil- ities of using biomass to drive a thermoacoustic stove [112], to be used in developing countries. Despite all these advantages there are still some challenges left that need to be re- solved before thermoacoustic devices can be used competitively on a large scale. Firstly, due to the oscillatory nature of the flow and the interaction with solid boundaries all 2 1.1 A historical perspective kinds of complicated flow patterns may arise, such as vortex shedding, turbulence, or asymmetric flow. Furthermore, the heat transfer is far from ideal and entropy is created. Additionally, thermoacoustic devices often operate at high pressure amplitudes, which can give rise to various nonlinear effects, such as the build-up of shock waves. The com- bined effect of all these phenomena can and will degrade the performance and holds back the development of highly efficient devices. Moreover, as long as these effects are not understood and modeled systematically, it will be hard to make accurate theoretical predictions. The mathematical aspects of oscillatory gas flow with heat transfer to solid boundaries in wide or narrow pores will be the topic of this thesis.

1.1 A historical perspective

Thermoacoustics has a long history that dates back more than two centuries. The inter- est in thermoacoustics was first piqued in 1816, when Laplace showed that even Newton was not infallible. Laplace [72] pointed out that Newton’s approximation of the speed of sound [92] was incorrect because he assumed isothermal compression and expansion of the air and did not compensate for variations in temperature. Correcting for these effects, Laplace found a value that was 18% higher.

Generation of sound

Flame Wire screen

Hydrogen Convection supply flow

(a) Higgins’ singing flame (b) Rijke tube

Figure 1.1: (a) Higgins’ singing flame: for suitable positions of the flame the tube will start to produce sound. (b) The Rijke tube: the loudest sound is produced when the heated wire screen is positioned at one-fourth from the bottom of the pipe.

The first records of heat-driven oscillations are the observations of Higgins [54, 102] in 1777, who experimented with an open glass tube in which acoustic oscillations were excited by suitable placement of a hydrogen flame, the so-called “singing flame”. A Introduction 3 similar, but more famous experiment was performed by Rijke [111] who in his efforts to design a new musical instrument from an organ pipe, constructed the so-called “Rijke tube”. As depicted in figure 1.1, he replaced Higgins’ hydrogen flame by a heated wire screen and found that when the screen was positioned in the lower half of the open tube spontaneous oscillations would occur, which were strongest when the screen was located at one fourth of the pipe. The oscillations would stop if the top of the tube was closed, indicating that the convective air current through the pipe was necessary for sound to be produced. Higgins’ and Rijke’s work later led to the birth of combustion science, with applications in rocket science and weapon industry. For a full review on devices related to the Rijke tube we refer to Feldman [41] or more recently Raun et al. [108]. The earliest predecessor of the type of thermoacoustic prime movers considered in this thesis is the Sondhauss tube, depicted in figure 1.2. It was invented in 1850 by Sond- hauss [126] based on an effect often noticed by glass blowers: generation of loud sound when a hot bulb is blown at the end of a cold narrow tube. Sondhauss found that if a steady gas flame was supplied to the closed bulb, the tube would produce a clear sound which was characterized by the length of the tube; the larger the bulb or the longer the tube, the lower the frequency of the sound. Unfortunately, Sondhauss did not manage to explain why the oscillations arose. Feldman [42] also reviewed most of the phenom- ena related to the Sondhauss tube as he did for the Rijke tube. An important difference between these two devices is that the Sondhauss tube does not require a convective air current for oscillations to occur.

Bulb Tube stem

Flame Sound

Figure 1.2: The Sondhauss tube: sound is generated from the tip of the stem, when heat is supplied to the bulb.

Another early example of a thermoacoustic prime mover is the phenomenon known as “Taconis oscillations”, often observed in cryogenic storage vessels. Taconis [139] cooled a gas-filled tube from room temperature to a cryogenic temperature by inserting it into liquid helium and observed spontaneous oscillations of the gas. The conditions for these type of oscillations have been investigated experimentally by Yazaki et al. [156]. The first qualitative explanation for heat-driven oscillations was given in 1887 by Lord Rayleigh. In his classical work “The Theory of Sound” [109], he explains the pro- duction of thermoacoustic oscillations as an interplay between heat fluxes and density variations: 4 1.1 A historical perspective

”If heat be given to the air at the moment of greatest condensation (compression) or taken from it at the moment of greatest rarefaction (expansion), the vibration is encouraged”. Rayleigh’s qualitative understanding turned out to be correct, but a quantitatively ac- curate theoretical description of these phenomena was not achieved until much later. The reverse process, generating temperature differences using acoustic oscillations, is a relatively new phenomenon. In 1964 Gifford and Longsworth [48] invented the pulse-tube refrigerator, by which they managed to cool down to a temperature of 150 K. In their device, depicted in figure 1.3, heat was pumped along the tube wall by sup- plying pressure pulses at low frequencies. Initially it was considered nothing more than an academic curiosity as it was highly inefficient, but current-day pulse-tube cryocool- ers can reach efficiencies up to 20% of the ideal efficiency and temperatures as low as 2 K. In fact, nowadays pulse-tube refrigeration is one of the most favored technologies for cryocooling . For a complete history and review of pulse-tube crycooling we re- fer to Radebaugh [104, 105]. Detailed modeling and numerical analysis of pulse-tube refrigerators can be found in [79–81].

Room temperature Rotary valve

High-pressure Vent Pulse source tube

Regenerator

Cold end

Figure 1.3: The pulse-tube refrigerator of Gifford and Longsworth. The temperature is cooled from room temperature at the hot end to 150 K at the cold end.

Sound-driven cooling was also observed by Merkli and Thomann [88] when they performed experiments on cooling in a simple gas-filled piston-driven resonator. To ex- plain these effects an extended acoustic theory was developed which predicted cooling in the tube where the velocity amplitude was at its maximum, in good agreement with the experiments. A crucial advance in experimental thermoacoustics came in 1962 when Carter et al. [22] realized that the performance of the Sondhauss tube could be improved by suitable placement of a porous medium inside, in the form of a stack of parallel plates. The presence of a “stack”, with heat transfer from one end to the other, makes it much easier to produce a significant temperature difference and will be the essential ingredient for the kind of thermoacoustic devices considered in this thesis. The foundation for theoretical thermoacoustics was laid in 1868 by Kirchhoff [66], Introduction 5 who investigated the acoustic attenuation in a duct due to oscillatory heat transfer be- tween the isothermal tube wall and the gas inside the tube. His results were generalized by Kramers [67] for a tube supporting a temperature gradient. The breakthrough came in 1969 when Rott et al. , inspired by the Taconis oscillations, started an impressive se- ries of articles [91,115–118,120,121], in which a linear theory of thermoacoustics was derived. Rott abandoned the boundary-layer approximation as used by Kirchhoff and Kramers, and formulated the mathematical framework for small-amplitude damped and excited oscillations in wide and narrow tubes with an axial temperature gradient, only assuming that the tube radius was much smaller than the length of the tube. In 1980 Rott summarized his results in a review work [119], which became the cornerstone for the subsequent intensified interest in thermoacoustics. In the eighties a very intensive and successful research program was started at the Los Alamos National Laboratory by Wheatley, Swift, and coworkers [133,151,152]. Us- ing Rott’s theory of thermoacoustic phenomena they started to design and build practi- cal thermoacoustic devices. Important was Hofler’s invention of a standing-wave ther- moacoustic refrigerator [55, 56], which proved that Rott’s theoretical analysis was cor- rect. Hofler’s refrigerator, shown in figure 1.4, used a loudspeaker to drive a closed resonator tube with a stack of plates positioned near the speaker. At the other end of the tube a resonator sphere was attached to simulate an open ending, so that effectively one can speak of a quarter-wave-length resonator. Inside the refrigerator a standing- wave is maintained by the speaker, generating a temperature difference across the stack such that heat is absorbed at the low temperature or waste heat is released at the high temperature.

Hot heat exchanger Cold heat exchanger

Resonator sphere

Driver Stack

Figure 1.4: Hofler’s standing-wave refrigerator. The hot end of the stack is thermally anchored at room temperature and the standing wave generates cooling at the cold end of the stack.

A whole new branch of thermoacoustic devices started in 1979 with Ceperley’s real- ization [23, 24] that thermoacoustic devices based on the Stirling cycle [21] with ideal heat transfer, could reach much higher efficiencies than devices based on standing- wave modes of operation. His idea was to design machines that allow a traveling wave to pass through a dense porous medium (the regenerator) using a toroidal geometry. Yazaki et al. [155] managed to build a traveling-wave prime mover based on these prin- ciples, but at very low efficiency due to large viscous losses. Finally, Backhaus and Swift [14] managed to overcome these problems by designing a traveling-wave prime mover (shown in figure 1.5) that combines the toroidal geometry with a resonator tube to reduce the velocities in the loop. Swift was the first to give a comprehensive analysis of thermoacoustic devices in his 6 1.2 The basic mechanism of thermoacoustics

Jet pump

Load Regenerator

Resonator

Figure 1.5: Schematic drawing of the traveling-wave prime mover of Backhaus and Swift. The sound produced by the regenerator is absorbed by an acoustic load that is attached to the regen- erator. review article [131] based on Rott’s work. He also gives a detailed description of the thermodynamics involved, a complete historical overview, experimental results, and he treats several types of devices. Since then Swift and others have contributed much to the further development and analysis of thermoacoustic devices. Most of the literature has been collected and summarized in Garrett’s review work [45]; in particular we mention Swift’s textbook [135], which provides a clear introduction into thermoacoustics. Lastly, we note that several articles have been written as well [15, 47,133,150,153], aimed at readers new to the field of thermoacoustics, while various educational animations can be found at the website of Los Alamos National Laboratory [130].

1.2 The basic mechanism of thermoacoustics

The thermoacoustic principles can be understood best by following a given parcel of fluid as it oscillates near a solid boundary. We start by considering fluid parcels oscillat- ing far away from the wall in a closed tube supporting a standing wave or in a infinite tube supporting a traveling wave, as depicted in figure 1.6. Note that under isentropic conditions the pressure oscillations are accompanied by temperature oscillations, which is used in figures 1.6(c) and 1.6(d) to draw schematically the temperature-position cycle that a fluid parcel undergoes during one oscillation. Consider a fluid parcel oscillating in the closed tube, far away from the tube wall. In the center of the tube we have a pressure node and a velocity antinode. As a result the parcel will undergo large displacement without temperature variations. On the other hand, near the ends of the tube we have a pressure antinode and a velocity node, and the parcel will almost stand still and undergo large temperature variations, giving a very steep temperature gradient. If the parcel is oscillating away from a velocity or pressure node, then a finite nonzero temperature gradient will arise. Within thermoacoustics the slope of these adiabatic temperature variations is known as the “critical temperature gradient”. In the infinite tube the situation is somewhat different, since the pressure and ve- locity oscillations are in phase for all positions in the tube. As a result the parcels al- ternately move to the right with a high temperature and then to the left with a lower temperature, giving circle-like temperature-position cycles. It follows that when a small traveling-wave component is added to a standing wave, or vice versa, the thermody- Introduction 7

(a) (b) T T x x

(c) (d) T x

(e)

Figure 1.6: Gas parcels oscillating (a) with standing-wave phasing in a closed tube and (b) with traveling-wave phasing in an infinite tube. The pressure oscillations are indicated by dashed lines, the velocity oscillations by solid lines, and the arrows give the direction of oscillation. (c) The temperature of the gas parcel during one oscillation as a function of its position relative to the standing wave. (d) Temperature-position diagram for traveling wave. (e) Temperature-position diagram when a small traveling-wave component is added to the standing wave.

namic cycles will turn into tilted ellipses as shown in figure 1.6(e). Suppose we look at the same parcels, but now oscillating near a wall supporting a temperature gradient in axial direction. We consider two cases:

(a) The tube wall supports a much smaller temperature gradient than the critical tem- perature gradient of the fluid parcel. As a result heat will be transported from the gas parcel into the wall at the hot end and from the wall into the gas parcel at the cold end (figure 1.7(a)). The transport of heat from the low temperature to the hot temperature will require the input of acoustic work. This is the condition for a refrigerator or heat pump.

(b) The tube wall supports a much larger temperature gradient than the critical tem- perature gradient of the fluid parcel. As a result heat will be transported from the wall into the gas parcel at the hot end and from the gas parcel into the wall at the cold end (figure 1.7(b)). The transport of heat from the high temperature to the low temperature will produce acoustic work as output. This is the condition for a prime mover.

This is in a nutshell the basic mechanism of thermoacoustics. In the next chapter we will analyze the combined effect of oscillations and heat transfer in more detail using “Brayton” [21] and “Stirling” cycles [145]. Specifically we will show how the distance to 8 1.3 Classification the solid and the phasing of pressure and velocity affect the heat transfer between gas and solid.

parcel parcel wall wall δQ

δQ T T δQ δQ

position parcel position parcel

(a) Refrigerator (b) Prime mover

Figure 1.7: The temperature-position diagram for the adiabatic parcel-temperature oscillations (—) and the wall temperature (- -). In (a) we consider a refrigerator and apply a small tempera- ture difference and in (b) we consider a prime mover with large temperature difference across the wall. The arrows show the transport of heat δQ from the wall to the gas parcel and vice versa.

1.3 Classification

We consider thermoacoustic devices of the type shown in figure 1.8, that is, a possibly looped duct containing a fluid (usually a gas) and a porous solid medium, if necessary with neighboring heat exchangers. In addition loudspeakers or other sources of sound may be attached to the ends of the tubes. The porous medium is modeled as a collection of narrow arbitrarily shaped pores aligned in the direction of sound propagation. Typ- ical examples are parallel plates and circular or rectangular pores. As will be discussed below, thermoacoustic devices can be divided [45] into several categories: heat-driven versus sound-driven devices, standing-wave versus traveling-wave devices, or stack- based versus regenerator-based devices.

Prime mover vs. refrigerator We distinguish heat-driven and sound-driven devices. A thermoacoustic prime mover absorbs heat at a high temperature and exhausts heat at a lower temperature while producing acoustic work as an output. A refrigerator or heat pump absorbs heat at a low temperature and requires the input of acoustic work to exhaust more heat to a higher temperature. The only difference between a heat pump and a refrigerator is whether the purpose of the device is to extract heat at the lower temperature (refrigeration) or to reject heat at the higher temperature (heating). Therefore, in this thesis we will use the term refrigerator loosely to refer to either of these devices. Introduction 9

(a) straight duct (b) looped duct

Figure 1.8: Schematic view of two possible duct configurations: straight or looped.

In the literature the term thermoacoustic engine is used as well, either to indicate a thermoacoustic prime mover or as a general term to describe all thermoacoustic devices. To avoid confusion we will refrain from using this term.

Stack-based devices vs. regenerator-based devices A second classification depends on whether the porous medium used to exchange heat with the working fluid is a “stack” or a “regenerator”. Typically inside a regenerator the pore size is much smaller than inside a stack. Garrett [45] uses the so-called Lautrec number NL to indicate the difference between a stack and regenerator. The Lautrec number is defined as the ratio between the half pore size and the thermal penetration depth1. Gas parcels that are separated from the wall by a distance much larger than the thermal penetration depth, will have no thermal contact with the wall. If NL & 1 the porous medium is called a stack and the gas parcels in the stack will have imperfect thermal contact with the solid. If NL 1, then the porous medium is called a regen- erator, and the gas parcels inside the regenerator≪ will have perfect thermal contact with the solid. Whenever the pore size is unknown or irrelevant, the porous medium will be referred to as stack.

Standing-wave devices vs. traveling-wave devices Lastly, thermoacoustic devices can also be categorized depending on the phase shift between the pressure and velocity oscillations at the location of the stack. In a closed and empty resonator a pure standing-wave can be maintained and the pressure and velocity oscillations will be exactly 90 degrees out of phase. In an empty infinite tube (or a loop) a pure traveling-wave can be maintained, so that the pressure and velocity oscillations are exactly in phase. As soon as we insert a stack in either of these tubes, the phasing between pressure and velocity will change because of partial reflection at the stack interfaces. Moreover, if we consider a looped tube of the type depicted in

1The thermal penetration depth is the distance heat can diffuse through within a characteristic time 10 1.4 Applications

figure 1.8(b), with a resonator tube attached to it, the phasing will be affected even more. However, in practise both the straight geometry depicted in figure 1.8(a) and the looped geometry depicted in figure 1.8(b) can be chosen such that the sound field at the stack is predominantly a standing wave or a traveling wave. In Section 2.3 we will show that it is beneficial to use a stack inside a standing-wave device and a regenerator inside a traveling-wave device. For this reason thermoacoustic devices are usually classified as either standing-wave stack-based devices or traveling- wave regenerator-based devices.

1.4 Applications

Over the years thermoacoustic devices have found applications in areas like food in- dustry, defense industry, spacecraft, telecommunication, electronics, energy sectors, and consumer products. Some of these devices are heat-driven, some are sound-driven, and others combine the two effects. In this section we will treat a few examples that have been or will be used commercially. Most of these applications are motivated by the quest for reliable, cheap, or environmentally friendly sources of energy.

Down-well power generation The natural-gas industry use sensors to measure properties of the gas that streams through subterranean gas pipes. These sensors are located far below ground and need electrical power, which has to be delivered via batteries or long cables. The reliability of such equipment is usually quite poor, requiring costly repairs or replacements due to difficult accessibility and extreme operating conditions. Since most wells are used for many years, there is a necessity for a cheap and reliable alternative for power genera- tion, which thermoacoustics can provide.

stack

main flow

Figure 1.9: Schematic drawing of the side-branch system. A standing wave is generated in the side branches due to the interaction of the main flow, supplied by the pipeline, with the edges of the side branch. If a stack is suitably positioned in the standing wave, a temperature difference can be generated.

The answer lies in a technique suggested by Slaton and Zeegers [123–125], which avoids the use of moving parts and uses part of the main-flow energy in the gas pipes Introduction 11 to generate aero-acoustic sound. In their experimental set-up, shown in figure 1.9, they attach two side branches to the main pipeline. Due to the interaction of the main flow with the edges of the side branch, vortices will be created. By adjusting the lengths of the side branch to the flow speed, it is possible to match the frequency of the vor- tex shedding with the fundamental standing-wave mode of the side branch, and a standing-wave sound field is created in the side branch. Finally, by suitable placement of a stack a temperature difference can be generated, which can be used to produce electrical power with thermoelectric elements. The flow patterns and vortex shedding in such side-branch systems have been visualized experimentally and numerically by Kriesels et al. [68] and Dequand et al. [36].

Upgrading of industrial waste heat Another important potential application for thermoacoustics is the upgrading of indus- trial waste heat. Huge amounts of heat produced by industry remain unused because of small power outputs or temperatures that are too small. The Energy research Centre of the Netherlands (ECN) has developed a thermoacoustic system that uses part of the waste heat to power a prime mover that drives a heat pump to upgrade the temperature of the remainder. The apparatus is shown in figure 1.10.

Heat Pump Prime Mover

Resonator

Figure 1.10: Upgrading of industrial waste heat using a combination of a prime mover and heat pump. Part of the waste heat is used to power the prime mover, which generates acoustic power to drive the heat pump and heat the remaining part of the waste heat.

Thermoacoustic cryocooling Thermoacoustics is the only technology that can cool to temperatures close to the abso- lute zero without using moving parts and is therefore very interesting for applications requiring crycooling. One such application is the liquefaction of natural gases which requires very low temperatures. At the Los Alamos National Laboratory (LANL) a heat-driven thermoacoustic refrigeration system [134, 138] has been designed capable of liquefying natural gases. Their system, depicted in figure 1.11, uses a toroidal geom- etry attached to a long resonator tube, with a prime mover located in the toroidal part and a refrigerator located near the end of the resonator. Part of the natural gas is burned to supply heat which is converted into acoustic power by the prime mover. The acoustic power is then provided to the refrigerator and subsequently used to cool the remainder of the natural gas until it is liquefied. Thermoacoustic cyrocooling is also applied to the cooling of electronics. The Amer- ican Navy used a Shipboard Electronics ThermoAcoustic Chiller (SETAC) to cool elec- tronics on board of one of their destroyers [86]. A similar application found its way into 12 1.5 Thesis overview

Prime Refrigerator Mover

Figure 1.11: Schematic drawing of a heat-driven thermoacoustic refrigeration system. By burn- ing part of the natural gases at the prime mover, the remainder can be cooled to liquefaction at the refrigerator. spacecraft, when in 1992 the Space ThermoAcoustic Refrigerator (STAR) was launched on the Space Shuttle Discovery. It was the answer [46] to the need for reliable, com- pact, and long-lived spacecraft cryocooling for the cooling of sensors aboard the shut- tle. Other applications of thermoacoustics within spacecraft concern the development of thermoacoustic systems suitable for electricity generation on space missions [16,140].

Food refrigeration and airconditioning

Thermoacoustics refrigerators can also be used to replace conventional food refriger- ators or airconditioning, without the harmful polluting side-effects. One well-known example is the collaboration [101] between Pennsylvania State University and Ben and Jerry’s Homemade Ice Cream, that led to the development of a thermoacoustic in-store ice-cream cabinet, capable of keeping ice cream at -18 ◦C. The aim was to find a cost- competitive and reliable alternative to the use of polluting refrigerants and thereby re- duce the emission of global-warming gases.

Cooking stove

Recently the SCORE (Stove for Cooking, Refrigeration and Electricity) project has started, an international research program led by the University of Nottingham [112]. The aim is to develop a biomass-powered thermoacoustic system, to be used as an affordable, safe, and efficient alternative for the energy needs of third-world countries. Current sources of energy, like open fires, are highly inefficient and can produce serious health hazards. The SCORE stove will serve as a versatile domestic appliance, being a cooker, a refrigerator, and an electricity regenerator all in one.

1.5 Thesis overview

In this thesis we will derive a weakly nonlinear theory of thermoacoustics, applicable to wide and narrow pores with arbitrarily shaped cross-sections that may vary slowly in axial direction. We will use dimensional analysis and small-parameter asymptotics so that (quadratically) nonlinear terms can be systematically included. The use of a dimensionless framework has two main advantages: Introduction 13

Nondimensionalization allows us to analyze limiting situations in which param- • eters differ in orders of magnitude, so that we can study the system as a function of dimensionless parameters connected to geometry, heat transport, and viscous effects.

We can give explicit conditions under which the theory is valid. Furthermore, we • can clarify under which conditions additional assumptions or approximations are justified.

In the end we will apply this theory to model, analyze, and simulate both standing- wave and traveling-wave devices and we will show how this approach can be used as an aid for optimizing their design. The main ingredients of this thesis are therefore the choice of geometry for the stack pores, the use of dimensionless parameters, the inclusion of quadratic nonlinearities, and the modeling and numerical simulation of complete devices. In Section 1.6 we will discuss briefly the relevant literature on each of these topics, but first we will give a chapter-by-chapter overview of the contents of this thesis. In Chapter 2 the thermodynamics of thermoacoustics will be discussed and a mea- sure for the thermodynamic performance is introduced. Then in Chapter 3 we will give the governing equations and we will set the mathematical framework for our small- parameter asymptotics. Chapter 4 is concerned with the systematic development of the theory of thermoacoustics in slowly-varying two-dimensional pores. We will show how the leading-order, first-order, and second-order terms with respect to the acoustic Mach number can be derived, which includes the derivation of the second harmonics and the streaming terms. We will also show how under certain simplifying assumptions analytic expressions can be obtained. In Chapter 5 the results from Chapter 4 will be generalized to arbitrary three-dimensional slowly-varying pores. Next in Chapter 6 the thermoacoustic equations are implemented and validated experimentally for several types of standing-wave devices: a thermoacoustic couple, a thermoacoustic refrigera- tor, and a thermoacoustic prime mover. We will show how the performance is affected by the operating conditions and parameters connected to geometry, stack material, and working gas. Finally we will also show what kind of streaming velocity profiles can occur. In Chapter 7 a specific type of traveling-wave prime mover is modeled and im- plemented numerically. An optimization routine is developed that for given system parameters computes the ideal geometry. Moreover, we we will show how the perfor- mance is affected by variation of various parameters. Then in Chapter 8 an evolution equation will be derived that predicts the development of shock waves near resonance, both in a closed tube and in interaction with a thermoacoustic stack. Lastly, in Chapter 9 we give some conclusions, discuss our results, and give some suggestions for future work.

1.6 Literature review

While Rott’s theory of thermoacoustics [91,115–121] was still limited to two-dimensional pores or three-dimensional cylindrical pores, it was Arnott et al. [5] who first extended Rott’s theory to arbitrary three-dimensional cross-sections, although nonlinear terms were not yet included. We have extended their results by allowing a slow variation in 14 1.6 Literature review the pore cross-section in axial direction. Furthermore, we also allow temperature de- pendence of speed of sound, specific heat, viscosity, and thermal conductivity, and we have eliminated the restriction of steady pore-wall temperature. Additionally we have also incorporated quadratic nonlinearities such as the second harmonics and streaming terms into the analysis. There are numerous articles that have analyzed more specific choices for the stack pore geometry, shown in figure 1.12. The parallel-plate geometry has been investigated most extensively, in particular by Swift [131] and Rott [115], who also investigated cylin- drical cross-sections. Stinson [128] and Roh et al. [113] derived simultaneously expres- sions for rectangular cross-sections and Stinson and Champoux [129] used results of Han [53] to solve the equations for triangular cross-sections. Lastly, we mention Swift and Keolian [136] who calculated the thermoacoustic behavior for so-called pin-array stacks, consisting of a hexagonal array of pins aligned in axial direction.

(a) Parallel plates (b) Circular (c) Rectangular (d) Triangular (e) Pin array

Figure 1.12: Possible stack pore cross-sections.

Previous treatments with variable cross-sections have always been restricted to widely spaced pores [98, 117], whereas in this thesis no such assumption is made. Although variable cross-sections occur mostly within the main resonator tube, they may also occur in the narrow stack pores. A more general formulation would allow stack geometries other than collections of arbitrarily shaped pores. We mention Roh et al. [114], who in- troduce tortuosity and viscous and thermal dynamic shape factors to extend single-pore thermoacoustics to bulk porous medium thermoacoustics. Furthermore, there exists a vast amount of papers on flow through porous media with random or stochastic prop- erties, that could also be applied to thermoacoustic configurations. Auriault [13] gives a clear overview of various techniques that can be used: Statistical modeling [69]; • Self-consistent models [157]; • Volume-averaging techniques [103]; • Method of homogenization for periodic structures [12]. • In addition Auriault [13] gives a short explanation of how the method of homogeniza- tion can be applied to analyze heat and mass transfer in composite materials. A detailed discussion of methods and results from the theory of homogenization and their appli- cations to flow and transport in porous media can be found in [58]. Another approach is demonstrated by Kaminski [63], who combines the homogenization approach with a stochastic description of the physical parameters to analyze viscous incompressible flow with heat transfer. Introduction 15

In our analysis, we assume a steady-state situation in which the variables oscillate at (integer multiples of) the fundamental frequency (cf. [5,119,131]), which is not al- ways the case. Prime movers require a sufficiently high temperature difference before it goes into self-oscillation. Atchley et al. have given a standing-wave analysis of this phe- nomenon and measured the complete evolution of the quality factor of a prime mover from below, through, and above onset of self-oscillation [6–9]. The use of a dimensionless framework is not very wide-spread. Olson and Swift [97] use dimensionless parameters to analyze thermoacoustic devices, but without trying to construct a complete theory of thermoacoustics. Instead dimensionless numbers are used to reduce the number of independent parameters in their experiments and for scaling purposes. There have been many observations [14, 100,132] demonstrating that at high am- plitudes measurements deviate significantly from predictions by linear theory. Stream- ing, turbulence, transition effects, higher harmonics, and shock waves are mentioned as main causes for these deviations. Streaming [75,95] refers to a steady mass-flux density or velocity, usually of second order, that is superimposed on the larger first-order oscil- lations. With the addition of a steady non-zero mean velocity, the gas moves through the tube in a repetitive ”102 steps forward, 98 steps backward” manner as described by Swift [135]. Streaming is important as a nonzero mass flux can seriously affect the performance of thermoacoustic devices. It can cause convective heat transfer, which can be a loss, but it can also be essential to transfer heat to and from the environment. Backhaus and Swift [14], in their analysis of a traveling-wave heat prime mover, show how streaming can cause significant degradation of the efficiency. The concept of mass streaming has been studied by many authors, (see e.g. [17,50, 52,98,117,148]), but restricted to simple geometries such as straight two-dimensional or cylindrical pores, although Olson and Swift [98] do allow slowly varying cross-sections in the tube. Moreover, they show that variable cross-sections can occur in practical geometries and can be used to suppress streaming; a suitable asymmetry in the tube can cause counter-streaming that balances the existing streaming in the tube. Bailliet et al. [17], Rott [117], and Olson and Swift [98] also take into account the temperature dependence of viscosity and thermal conductivity, although the latter two only consider widely spaced pores. Higher harmonics oscillate at integer multiples of the fundamental frequency, and can become quite important at high amplitudes. Atchley and Bass [10], noticed exper- imentally that the generation of higher harmonics can cause highly nonlinear wave- forms that degrade the performance significantly. The impact of the harmonics is great- est when excited near resonance, but Atchley and Gaitan [43] analyzed that this can be suppressed by careful tuning of the resonator. There is no literature that systemat- ically includes the higher harmonics into the analysis, which Swift [135] describes as “a formidable challenge” and Rott [118] as “a hopeless undertaking”. Although a com- plete extension would require going up to fourth order in the asymptotic expansion, we have shown here that the second-harmonic pressure and velocity oscillations can be expressed in terms of the first-harmonic oscillations. It has been noticed that higher harmonics can interact together to form shock waves [59, 60]. Neglecting the nonlinear sound field in the stack, Gusev [49] analyzed shock formations using a nonlinear evolution equation. Modeling the stack by a reflection coefficient, we will derive a new evolution equation that predicts the development of 16 1.6 Literature review shock waves when the tube is excited near resonance. Turbulence arises at high Reynolds numbers, where the assumption of laminar flow is no longer valid. It disrupts boundary layers and can negatively affect the heat trans- fer. Turbulence may also arise due to abrupt changes in the shape or direction of the channels, which leads to the shedding of vortices [2, 89, 158] and can cause significant losses. Swift [44, 135] gave some suggestions on how to include turbulence into the modeling, but they are by no means complete. It is therefore still a challenge to include systematically the effects of turbulence into the analysis. Finally we note that there is a long list of publications on the numerical simulation of heat-driven and sound-driven thermoacoustic devices, focusing on both standing-wave and traveling-wave modes of operation. This list is too long to go into here, but it should be mentioned that a large part of the thermoacoustic community uses the DeltaEC (or DeltaE) code [146,147] which was developed at the Los Alamos National Laboratory based on Swift’s linear theory of thermoacoustics [135]. Chapter 2

Thermodynamics

In this chapter we will explain the basic thermodynamics concepts that are at the basis of thermoacoustics. We will start by giving the fundamental laws of thermodynam- ics and show how they can be used to derive a thermodynamic efficiency and coefficient of performance that indicate how well a thermoacoustic prime mover or refrigerator per- form. Next we will shed some more light on the mechanisms behind the thermoacoustic production of sound or heat by analyzing the thermodynamic cycle a gas parcel experi- ences.

2.1 Laws of thermodynamics

Thermodynamically speaking a thermoacoustic system is completely characterized by the flows of heat and work, as shown in figure 2.1. Let TH be the temperature of a hot reservoir and TC the temperature of a cold reservoir. In a refrigerator acoustic work W is used to generate a heat flow against the temperature gradient, removing heat QC at the low temperature and releasing heat QH at the high temperature. In a prime mover work W is produced by transporting heat from the high to the low temperature, removing heat QH at the high temperature and releasing heat QC at the low temperature. The energy flows within a thermoacoustic system are governed by the first and sec- ond law of thermodynamics. The first law concerns the conservation of energy and describes the rate of change of the internal energy U˙ of a system with volume V, the heat and enthalpy flows into the system, and the work done by the system [35]:

U˙ = ∑ Q˙ + ∑ nH˙ pV˙ + P. (2.1) m −

Here n˙ represents the molar flow rate entering the system, Hm represents molar en- thalpy, Q˙ represents heat power, and P represents other forms of power done on the system. The summation is used to account for all the different sources of sound and mass that are in contact with the system and plus and minus signs are used to indicate flows into or out of the system, respectively. The second law of thermodynamics states that any process that occurs will tend to increase the total entropy of the system. Mathematically this can be expressed by the 18 2.1 Laws of thermodynamics

QH QH

T T W H H W Refrigerator Prime mover

TC TC

QC QC

(a) Refrigerator or heat pump (b) Prime mover

Figure 2.1: The flows of work W and heat Q inside (a) a thermoacoustic refrigerator or heat pump and (b) a thermoacoustic prime mover. The arrows are used to indicate the exchange of heat and work between the thermoacoustic system and the environment.

inequality S˙ 0, (2.2) i ≥ ˙ where Si represents the entropy production in the system. The second law can also be formulated using an equality [35],

Q˙ S˙ = ∑ + ∑ nS˙ + S˙ . (2.3) T m i It states that the rate of change of entropy S˙ of a thermodynamic system is equal to the sum of the entropy change due to heat flows Q˙ with temperature T, due to mass ˙ flows nS˙ m, and due to the irreversible entropy production in the system Si. Again the summation signs are used to allow for a multitude of sound or mass sources. In the thermoacoustic systems considered there is no mass flow into or out of the system, the volume of the system is constant, and the only work performed on the system is the acoustic work W. As a result, we find that the laws of thermodynamics reduce to ˙ ˙ ˙ QC QH + W, for a refrigerator or heat pump, U˙ = − (2.4) ( Q˙ + Q˙ W˙ , foraprimemover, − C H − and Q˙ Q˙ C H + S˙ , for a refrigerator or heat pump, T − T i s˙ =  C H (2.5)  ˙ ˙  QC QH ˙  + + Si, foraprimemover. − TC TH  In our analysis we consider a (time-averaged) steady-state situation, so that we can put U˙ = 0 and S˙ = 0. It follows that

Q˙ Q˙ + W˙ = 0, (2.6) C − H Thermodynamics 19 for all thermoacoustic devices. In addition, since the entropy generated by the system S˙ 0, we find i ≥ Q˙ Q˙ C H , forarefrigeratororheatpump, (2.7) TC ≤ TH Q˙ Q˙ C H , foraprimemover, (2.8) TC ≥ TH where equality can only be reached in the ideal situation when there are no irreversible processes.

2.2 Thermodynamic performance

2.2.1 Refrigerator or heat pump The performance of a refrigerator or heat pump is measured by the so-called coefficient of performance COP. Since a refrigerator and a heat pump each have a different goal, cooling versus heating, the coefficients of performance are defined differently also. When analyzing a refrigerator we are interested in maximizing the cooling power ˙ QC extracted at temperature TC, while at the same time minimizing the net required acoustic power W˙ . On the other hand, a heat pump aims at maximizing the heating ˙ ˙ power QH at temperature TH, while minimizing W. Therefore, the coefficients of per- formance COPre f (refrigerator) and COPhp (heat pump) are defined as the ratios of these quantities,

Q˙ COP := C , (2.9) re f W˙ Q˙ COP := H . (2.10) hp W˙ The second law of thermodynamics limits the interchange of heat and work. In particular it follows from (2.6) and (2.7) that

Q˙ T COP = C C =: COPC , (2.11) re f Q˙ Q˙ ≤ T T re f H − C H − C Q˙ T COP = H H =: COPC . (2.12) hp Q˙ Q˙ ≤ T T hp H − C H − C The quantity COPC is called the Carnot coefficient of performance , and gives the max- imal performance for all refrigerators or heat pumps. Using COPC we can introduce a relative coefficient of performance COPR as

COP COPR := . (2.13) COPC Note that although COP and COPC may attain any nonnegative value, the relative co- efficient COPR will always be between 0 and 1. 20 2.3 The thermodynamic cycle

2.2.2 Prime mover The performance of a prime mover is measured by the so-called efficiency η. A prime ˙ ˙ mover uses heating power QH to produce as much acoustic power W as possible and thus the efficiency is defined as W˙ η =: . (2.14) ˙ QH Using equation (2.6), we can write

Q˙ Q˙ η = H C . (2.15) ˙− QH Applying equation (2.8), we obtain

TH TC TC η − = 1 =: ηC. (2.16) ≤ TH − TH

By definition both the efficiency η and the Carnot efficiency ηC are between zero and one. The same holds for the relative efficiency ηR which is defined as η ηR := . (2.17) ηC Of course the efficiency is not the only criterion for a good thermoacoustic prime mover. For certain applications one might be primarily interested in maximizing the power output with the efficiency only of minor interest. Other competing criteria can be cost, size, reliability, available materials, safety, and the complexity of the design. Naturally, the same holds for a refrigerator or heat pump.

2.3 The thermodynamic cycle

In Section 1.2 we gave an intuitive explanation for the basic mechanism of thermoa- coustics. In this section we will give a more detailed analysis, describing all the steps a fluid parcel undergoes while it oscillates in a narrow pore. In particular we will discuss the time phasing of the thermodynamic effect and its relation to the pore size, based on the analysis of Swift [131] and Ceperley [23]. Swift showed that for standing-wave devices the fluid-parcel movements are very accurately described by the Brayton cy- cle [21]. Ceperley suggested to design thermoacoustic devices such that the thermody- namic cycle would match the ideal Stirling cycle [145] to reduce irreversible effects to a minimum. Liang [73] analyzed Ceperley’s concepts further using a sinusoidal model that describes the thermodynamic cycles of gas parcels oscillating in a regenerator.

2.3.1 Standing-wave phasing Suppose gas parcels oscillate with standing-wave phasing at a distance y from a solid plate supporting a temperature gradient. As shown in figure 2.2 the parcels will ex- perience a build-up of pressure (compression) and drop of pressure (expansion) while Thermodynamics 21

displacement displacement compression expansion

z }| { z }| {

t ABCDA

Figure 2.2: Velocity (—-) and pressure (– – –) as a function of time in a gas supporting a standing wave. The cycle consists of a compression/displacement step (A-B) and an expan- sion/displacement step (C-D). Depending on the distance between the gas and the solid heating and cooling may occur as well. undergoing displacement. Depending on the size of y relative the thermal penetration depth δk, heat transfer may occur as well. We consider three cases: No thermal contact (y δ ) • k When there is no thermal≫ contact between the gas and the plate the gas parcels will expand and compress adiabatically and reversibly and no heat transfer will take place.

Perfect thermal contact (y δ ) • k In the first step the gas parcels≪ are compressed and displaced towards the hot end of the plate and the parcel. At the same time the parcels will be heated if a large temperature gradient is imposed (prime mover) and cooled if a small temperature gradient is imposed (refrigerator). Because there is perfect thermal contact the compression and heat exchange will take place simultaneously. In the next step the gas parcels are displaced back towards the cold end and the reverse effect takes place. The parcels will be cooled for a prime mover and heated for a refrigerator.

Imperfect thermal contact (y δ ) • k Because of the distance between∼ the gas and the plate, there is a time delay be- tween motion and heat transfer. As a result the gas parcels execute a four-step cycle. In case of a refrigerator the parcels are compressed and displaced (A-B), cooled (B-C), expanded and displaced back (C-D), and heated (D-A). In case of a prime mover the heating and cooling steps are interchanged.

The acoustic power produced or absorbed by a gas parcel can be found from the area p dV in the pressure-volume diagrams. In figure 2.3 we see that heat can only be convertedH in acoustic power or vice versa if there is a time delay between the compres- sion or expansion and the heat exchange. This thermodynamic cycle is what is known as the idealized Brayton cycle. This is the reason why in standing-wave based devices 22 2.3 The thermodynamic cycle

y δ y δ y δ ≫ k ≪ k ∼ k B C pressure pressure pressure A D

volume volume volume (a) prime mover

y δ y δ y δ ≫ k ≪ k ∼ k CB pressure pressure pressure D A

volume volume volume (b) refrigerator

Figure 2.3: Schematic pressure-volume cycle executed by the gas parcel in (a) a standing-wave prime mover and (b) a standing-wave refrigerator. Only when the gas parcels are about a thermal penetration depth away from the plate, pressure and volume variations are out of phase and net work is performed. stacks are used with stack pores that have radii of several penetration depths. The poor thermal contact is crucial for the thermoacoustic effect to occur, but it also gives rise to irreversible heat transfer and friction, which has a negative effect on the efficiency. We will show below that in traveling-wave devices we can have perfect thermal contact and thus potentially higher efficiencies.

2.3.2 Traveling-wave phasing

Suppose now the gas parcels oscillate with traveling-wave phasing at a distance y δk from a solid plate supporting a temperature gradient. As shown in figure 2.4 the parcels≪ will undergo a cycle consisting of four steps:

A-B First the gas parcel is compressed at constant temperature.

B-C Then the parcel is displaced towards the hot end at constant volume and simulta- neously it is heated (prime mover) or cooled (refrigerator).

C-D Next the parcel is expanded at constant temperature. Thermodynamics 23

compression expansion

z }| { z }| {

t ABCDA

cooling/heating heating/cooling displacement| {z } displacement| {z }

Figure 2.4: Velocity (—-) and pressure (– – –) as a function of time in a gas supporting a trav- eling wave in perfect contact with the solid. The gas undergoes a Stirling cycle of compression, cooling/heating, expansion, and heating/cooling.

D-A Finally, the parcel is displaced back towards the cold end at constant volume and simultaneously it is cooled (prime mover) or heated (refrigerator).

Note that there is a continuous exchange of heat between the gas and the solid during the displacement steps. The acoustic power produced or absorbed by the gas parcel can be found from the area p dV in the pressure-volume diagrams. In figure 2.5 we see that heat is con- δ vertedH into acoustic power and vice versa, even though y k (cf. figure 2.3). This thermodynamic cycle is what is known as the idealized Stirli≪ng cycle.

y δ y δ ≪ k ≪ k B C A D C B

pressure D pressure A

volume volume (a) refrigerator (b) prime mover

Figure 2.5: Schematic pressure-volume cycle executed by the gas parcel in (a) a traveling-wave refrigerator and (b) a traveling-wave prime mover at perfect thermal contact with the wall. Only when the gas parcels are about a thermal penetration depth away from the plate, pressure and volume variations are out of phase and net work is performed. 24 2.3 The thermodynamic cycle

This is the reason why in traveling-wave based devices regenerators can be used that have perfect thermal contact between gas and solid. As there is no irreversible heat transfer much higher efficiencies can be obtained. It should be mentioned that because of the very narrow pores, large viscous losses may occur, which is why traveling-wave devices are usually designed such that in the regenerator the velocities are small. In figure 2.6 we have summarized the (idealized) four-step cycles the fluid parcels undergo as they oscillate along the walls of the stack pores of a standing-wave prime mover or refrigerator and the regenerator pores of a traveling-wave prime mover or refrigerator.

1 - Compression 1 - Compression 1 - Compression 1 - Compression

2 - Cooling 2 - Heating 2 - Cooling 2 - Heating

δQ δQ δ δ 1 1 Q1 Q1 δ δ W1 W1 δ δ W1 W1 3 - Expansion 3 - Expansion 3 - Expansion 3 - Expansion

4 - Heating 4 - Cooling 4 - Heating 4 - Cooling

δQ δQ δ δ 2 2 Q2 Q2 δ δ W2 W2 δ δ W2 W2

(a) Standing-wave re- (b) Standing-wave tx (c) Traveling-wave re- (d) Traveling-wave tx frigerator prime mover frigerator prime mover

Figure 2.6: Typical fluid parcels, near a stack plate, executing the four steps (1-4) of the thermo- dynamic cycle in (a) a stack-based standing-wave refrigerator, (b) a stack-based standing-wave prime mover, (c) a regenerator-based traveling-wave refrigerator and (d) a regenerator-based traveling-wave prime mover. Thermodynamics 25

2.3.3 Bucket-brigade effect Usually the displacement of one fluid parcel is small with respect to the length of the plate. Thus there will be an entire train of adjacent fluid parcels, each confined to a short region of length 2δx and passing on heat as in a bucket brigade (figure 2.7). Although a single parcel transports heat δQ˙ over a very small interval, δQ˙ is shuttled along the entire plate because there are many parcels in series. T T ˙ C H ˙ QC QH

δQ˙

δx

W˙ | {z }

Figure 2.7: Work and heat flows inside a thermoacoustic refrigerator. The oscillating fluid parcels work as a bucket brigade, shuttling heat along the stack plate from one parcel of gas to the next. As a result heat is transported from the left to the right, requiring acoustic power W.˙ Inside a prime mover the arrows will be reversed, i.e. heat is transported from the right to the left and acoustic power W˙ is produced. 26 2.3 The thermodynamic cycle Chapter 3

Modeling

3.1 Geometry

As mentioned in the introduction we consider devices of the form shown in figure 3.1. The porous medium is modeled as a collection of narrow arbitrarily shaped pores aligned in the direction of sound propagation as shown in Fig. 3.2. Furthermore, we allow the pore boundary to vary slowly (with respect to the channel radius) in the di- rection of sound propagation, as shown in figure 3.2(b). For the analysis here we focus on what happens inside one narrow channel, which can represent a stack pore, but also the main resonator tube.

(a) straight duct (b) looped duct

Figure 3.1: Schematic view of two possible duct configurations: straight or looped.

In our analysis of the thermoacoustic effect we restrict ourselves to a pore and its neighboring solid, where the system (x, r,θ) forms a cylindrical coordinate system. Let (x) denote the gas cross-section and (x) the half-solid cross-section at position x. Ag As Let Γg denote the gas-solid interface and Γt the outer boundary of the half-solid. We choose Γt and s such that there is no heat flux through Γt (for periodic structures Γt is the centerlineA of the solid). The remaining stack pores and solid can then be modeled 28 3.2 Governing equations

by periodicity. On Γg we write

(x, r,θ) := r (x,θ) = 0, Sg − Rg with (x,θ) the distance of Γ to the pore centerline at position x and angle θ. Similarly Rg g on Γt we write (x, r,θ) := r (x,θ) = 0, St − Rt where t(x,θ) := g(x,θ) + s(x,θ) denotes the distance of Γt to the origin at position x and angleR θ. R R

Γt (x,θ) solid Rs Γg

r gas g(x,θ) R x θ

(a) transverse cut of stack (b) longitudinal cut of one stack pore

Figure 3.2: Porous medium modelled as a collection of arbitrarily shaped pores.

3.2 Governing equations

The general equations describing the thermodynamic behavior in the gas are the well- known conservation laws of mass, momentum and energy [85]. Introducing the con- vective derivative d f ∂ f = + v · f , dt ∂t ∇ these equations can be written in the following form:

dρ mass : = ρ · v, (3.1) dt − ∇ dv momentum : ρ = p + · τ + ρb, (3.2) dt −∇ ∇ dǫ energy: ρ = · q p · v + τ: v. (3.3) dt −∇ − ∇ ∇ Here ρ is the density, v the velocity vector, p the pressure, ρb the body forces (gravity), q the heat flux, ǫ the specific internal energy, and τ the viscous stress tensor. For q and Modeling 29

τ we have the following relations

Fourier’s heat flux model: q = K T, (3.4) − ∇ Newton’s viscous stress tensor: τ = 2µ + ζ( · v) , (3.5) D ∇ I 1 T where T the absolute temperature, K is the thermal conductivity, = 2 ( v +( v) ) the strain rate tensor, and µ and ζ the dynamic (shear) and secondD viscosity.∇ ∇ In general the viscosity and thermal conductivity will depend on temperature. For example, Sutherland’s formula [74] can be used to model the dynamic viscosity as a function of the temperature

1 + S /Tre f T 1/2 µ = µre f u , 1 + S /T re f u  T  re f re f where µ is the value of the dynamic viscosity at reference temperature T and Su is Sutherland’s constant. According to [26] the variation of thermal conductivity is ap- proximately the same as the variation of the viscosity coefficient. The thermodynamic parameters Cp, Cv, β, and c may also depend on temperature (see Appendix A). The temperature dependence of µ and K is particularly important, as it allows investigation of Rayleigh streaming: forced convection as a result of viscous and thermal boundary- layer phenomena. As shown in Appendix B.2 (eq. (B.9)), the equations above can be combined and rewritten to yield the following equation1 for the temperature in the fluid

∂T ∂p ρC + v · T = βT + v · p + · (K T) + τ: v, (3.6) p ∂t ∇ ∂t ∇ ∇ ∇ ∇     where Cp is the isobaric specific heat and β the volumetric expansion coefficient as de- fined in Appendix A. In the solid we only need an equation for the temperature. We have that the temper- ature Ts in the solid satisfies the diffusion equation ∂T ρ C s = · (K T ) , (3.7) s s ∂t ∇ s∇ s where Ks, Cs and ρs are the thermal conductivity, the isobaric specific heat and the den- sity of the solid, respectively. Another useful equation, that can replace (3.3), describes the conservation of energy in the following way:

∂ 1 1 ρ v 2 + ρǫ = · v ρ v 2 + ρh K T v · τ + ρv · b, (3.8) ∂t 2 | | −∇ 2 | | − ∇ −       where h is the specific enthalpy. The details of the derivation leading to this equation can be found in Appendix B.1. The equations (3.1), (3.2), (3.6), and (3.7) can be used to determine v, p, T, and Ts.

1 ∑ The double-dot product results in a scalar: A:B = i, j AijBji. 30 3.2 Governing equations

Additionally one should add an equation of state that relates the density to the pressure and temperature in the gas. For the analysis here it is enough to express the thermody- namic variables ρ, s, ǫ and h in terms of p and T using the thermodynamic equations (A.7)-(A.10) given in Appendix A. For the numerical examples given in Chapters 6 and 7 it is necessary to make a choice and we choose to impose the ideal gas law

p = ρRspecT, where Rspec = Cp Cv is the specific gas constant. However, the analysis presented here is also valid for non-ideal− gases and nowhere will we use this relation. To distinguish between variations along and perpendicular to sound propagation we will use a τ in the index to denote the transverse vector components. For example the transverse gradient and transverse Laplace operator 2 are defined by ∇τ ∇τ ∂ ∂2 2 2 = ex + τ , = + τ . ∇ ∂x ∇ ∇ ∂x2 ∇ The equations will be linearized and simplified using the following assumptions:

the temperature variations along the stack are much smaller than the typical ab- • solute temperature;

time-dependent variables oscillate with fundamental frequency ω. •

At the gas-solid interface we impose the no-slip condition

v = 0 if = 0. (3.9) Sg

The temperatures in the solid and in the gas are coupled at the gas-solid interface Γg by continuity of temperature and heat fluxes:

T = T if = 0, (3.10a) s Sg K T n = K T n if = 0, (3.10b) ∇ · s∇ s · Sg where n is a vector outward normal to the surface. The boundary Γt, with outward surface normal n′, is chosen such that no heat goes through, i.e.

T · n′ = 0 if = 0. (3.11) ∇ s St

Note that since g and t are vectors normal to the surfaces we can also replace the two flux conditions∇S by ∇S

K T · = K T · if = 0, (3.12) ∇ ∇Sg s∇ s ∇Sg Sg T · = 0 if = 0. (3.13) ∇ s ∇St St Modeling 31

3.3 Scaling

To make the equations above dimensionless we apply a straightforward scaling proce- dure. First we scale the spatial coordinate by a typical pore radius Rg:

x = Rgx˜.

Note that because we consider narrow stack pores, the typical length scale L of radius variation is much larger than a radius. We therefore introduce the small parameter ε as the ratio Rg/L between a typical radius and this length scale. At Γg we write x ˜ (X˜ , r˜,θ) = r˜ ˜ (X,θ) = 0, ˜ (X,θ) = R (x,θ), X = εx˜ = . Sg − Rg Rg gRg L

Similarly at Γt we have

˜ (X˜ , r˜,θ) = r˜ ˜ (X,θ) = 0, ˜ (X,θ) = R (x,θ). St − Rt Rt gRt Our formal assumption of slow variation has now been made explicit in the slow vari- able X. As a typical time scale we use L/cre f , and we scale

L t = t˜. cre f Secondly, we rescale the remaining variables as well using characteristic values

re f re f re f re f re f re f 2 τ µ c τ u = c u˜, vτ = εc v˜ τ , p = ρg (c ) p˜, = ˜ , (3.14a) Rg (cre f )2 (cre f )2 ρ ρre f ρ ˜ ρ ρre f ρ ˜ = g ˜, T = re f T, s = s ˜s, Ts = re f Ts, (3.14b) Cp Cp re f 2 ˜ re f 2 re f ˜ h =(c ) h, ǫ =(c ) ǫ˜, s = Cp s˜, b = gb, (3.14c) Cre f p ˜ re f ˜ re f ˜ re f β = β, Cp = Cp Cp, Cs = Cs Cs, c = c c˜, (3.14d) (cre f )2 K = Kre f K˜, µ = µre f µ˜, ζ = ζre fζ˜. (3.14e)

Thirdly, we observe that the system contains 19 parameters, in which 6 physical dimen- sions are involved. The Buckingham π theorem [20] implies that the 19 parameters can be combined into 13 independent dimensionless parameters. In Table 3.1 a possible choice is presented. Here the parameters δν, δζ, δk and δs are the viscous penetration depths based on dynamic and second viscosity and the thermal penetration depths for the fluid and solid, respectively.

re f re f re f re f 2µ 2ζ 2Kg 2K δ = , δ = , δ = , δ = s . ν v re f ζ v re f k v re f re f s re f re f uωρg uωρg uωρg Cp sωρs Cs u u u t t t 32 3.3 Scaling

Symbol Formula Description re f re f γ Cp /Cv specific heats ratio re f 2 2 2 η (c Rg/UL) ε /Ma

ε Rg/L stack pore aspect ratio

φ Rs/Rg porosity re f re f σ Ks /Kg ratio thermal conductivities κ ωL/cre f Helmholtz number

Br Rg/Rt blockage ratio

Dr pA/pamb drive ratio

Ma U/c Mach number 2 Fr U /(gL) Froude number

NL Rg/δk Lautrec number fluid

Ns Rs/δs Lautrec number solid 2 2 Pr δν/δk Prandtl number

Sk ωδk/U Strouhal number based on δk

Wo √2Rg/δν Womersley number based on µ

Wζ √2Rg/δζ Womersley number based on ζ

Table 3.1: Dimensionless parameters

Note that 16 dimensionless numbers are introduced. Since at most 13 can be indepen- dent, we have at least three dependent numbers. For example, Pr, Sk, and η can be expressed in the other parameters as follows

2N2 ε2 κε L η Pr = 2 , = 2 , Sk = . (3.15) Wo Ma NL Ma The remaining dimensionless numbers are independent and can be chosen arbitrarily. In the analysis below we will use explicitly that Ma 1 (small velocity amplitudes) and ε 1 (long pores). Other important parameters in≪ thermoacoustics are κ, S and ≪ k NL. κ is a Helmholtz number and is therefore a measure for the relative length of the stack with respect to the wavelength; short stacks imply small Helmholtz numbers. In Section 4.7.2 it is shown that Sk is a measure for the contribution of the thermoacoustic heat flux to the total heat flux in the stack and can be both small and large depending on the application. As mentioned previously, the Lautrec number distinguishes between the porous medium being a stack (NL 1) or a regenerator (NL 1). These limiting cases are described schematically in Table∼ 3.2. Note also that the gravity≪ is present via the Froude number Fr. The gravitational terms only appear in the momentum equations (3.17) and (3.18), and in the energy equation (3.22), as second or third-order term in Ma. Therefore, as long as the Froude number does not become too small, the gravity will Modeling 33

NL Ma ε κ Sk thermoacoustic heat 1 regenerator small velocities long pores short stack ≪ flow dominates 1 stack large velocities short pores long stack - ∼ heat conduction 1 resonator - - - ≫ dominates

Table 3.2: Parameter ranges for some important dimensionless numbers

have little effect on the thermoacoustic behaviour. It follows from the definition of the Froude number that in large machines, or at very small velocity amplitudes, gravity can become more important. This should be kept in mind when upscaling thermoacoustic devices to a larger size. From here on we will use dimensionless variables, but omit the tildes for conve- nience. Note that ∂ = εe + , v = ue + εv . ∇ x ∂X ∇τ x τ Substituting (3.14) into equations (3.1)-(3.3) we arrive at the following set of dimension- less equations:

∂ρ ∂(ρu) = · (ρv ), (3.16) ∂t − ∂X − ∇τ τ ∂u ∂u ∂p κ M2 ρ + u + v · u = + · (µ u) + a ρb ∂t ∂X τ ∇τ − ∂X 2 ∇ ∇ F x   Wo r ∂ µ ζ ∂u + ε2κ + + · v , (3.17) ∂X 2 2 ∂X ∇τ τ " Wo Wζ !  # ∂ ∂ 2 vτ vτ 1 κ Ma ρ + u + vτ · τ vτ = τ p + · (µ vτ ) + ρbτ ∂t ∂X ∇ − ε2 ∇ 2 ∇ ∇ εF   Wo r µ ζ ∂u + κ + + · v , (3.18) ∇τ 2 2 ∂X ∇τ τ " Wo Wζ !  # ∂ǫ ∂ǫ ∂u κ ρ + u + v · ǫ = p + · v · (K T) + κΣ, (3.19) ∂t ∂X τ ∇τ − ∂X ∇τ τ − 2 ∇ ∇     2NL where

µ ∂u ∂v 2 ∂u ∂w 2 ∂w ∂v 2 Σ = + ε2 + + ε2 + ε2 + 2 ∂y ∂X ∂z ∂X ∂y ∂z Wo "      ∂u 2 ∂v 2 ∂w 2 ε2ζ ∂u 2 +2ε2 + + + + · v . ∂X ∂y ∂z 2 ∂X ∇τ τ       !# Wζ   34 3.4 Small-amplitude and long-pore approximation

Equations (3.16) and (3.17) show that the gravitational term ρb occurs as a second-order effect in the Mach number Ma. The equations (3.6) and (3.7) for the fluid and solid temperature transform into

∂T ∂T ∂p ∂p ρC + u + v T = βT + u + v p p ∂t ∂X τ · ∇τ ∂t ∂X τ · ∇τ     κ · κΣ + 2 (K T) + , (3.20) 2NL ∇ ∇ ∂T κφ2 ρ C s = · (K T ) s s ∂ 2 s s . (3.21) t 2Ns ∇ ∇ Furthermore the conservation of energy (3.8) transforms into

∂ 1 v 1 κ κ M2 ρ v 2 + ρǫ = · ρ v 2 + ρh T v · τ + a ρv · b. ∂t 2 | | −∇ ε 2 | | − 2 ∇ − 2 F   "   2NL Wo # r (3.22) Finally we have the following boundary conditions at the gas-solid interface. First for the velocity we have

v = 0 if = 0. (3.23a) Sg Secondly, for the temperature we can write

T = T if = 0, (3.23b) s Sg K T · = σK T · if = 0, (3.23c) ∇ ∇Sg s∇ s ∇Sg Sg T · = 0 if = 0. (3.23d) ∇ s ∇St St

3.4 Small-amplitude and long-pore approximation

We assume a small maximal amplitude A of acoustic oscillations, i.e. the velocity, pres- sure, density and temperature fluctuations are small relative to their mean value, which can be used to linearize the equations given above. Since all variables are dimension- less, we can use the same A to linearize all variables, e.g. the relative pressure or velocity amplitude. Let f be any of the fluid variables (p, v, T, etc.) with stationary equilibrium profile f0. Expanding f in powers of A using a Fourier series we write

f (x, t) = f (x) + ARe f (x)eiκt + A2 f (x) + A2Re f (x)e2iκt + , A 1. 0 1 2,0 2,2 · · · ≪ h i h i Furthermore, for the velocity we assume v0 = 0. As the velocity was scaled with the speed of sound, the obvious choice for A then becomes the acoustic Mach number Ma, Modeling 35 and we will therefore use the following asymptotic expansion:

f (x, t) = f (x) + M Re f (x)eiκt + M2 f (x) + M2Re f (x)e2iκt + , 0 a 1 a 2,0 a 2,2 · · · h i h i M 1. (3.24) a ≪

The first index is used to indicate the corresponding power of Ma, and the index behind the comma is used to indicate the frequency of the oscillation (cf. [135]). The fluid vari- ables are built up from steady terms and harmonic modes that are integer multiples of the fundamental frequency. We assume that only the first harmonic f1 is excited exter- nally. It then follows from equation (3.26) below that the first-order oscillations excite the second-order fluid variables by the combined effect of the second harmonic f2,2, that oscillates at twice the fundamental frequency, and the steady streaming part f2,0. In the next chapters we will show how both the streaming variables and the second harmonics can be expressed in terms of the first harmonics and the equilibrium values. Note that for the subsequent analysis we will only use the leading-order approxima- tion of the thermodynamic parameters c, β, Cp, and Cv. Their indices will therefore be omitted. Additionally, since we also assume ε 1, we may expand the perturbation variables ≪ fi again to include powers of ε as well. However this would lead to messy derivations. Instead we will assume that ε Ma, so that the geometric and streaming effect can be included at the same order. In∼ the end, when all the analysis has been performed, one can still consider the limits ε Ma or ε Ma and neglect certain terms. We will use an overbar to indicate≪ time-averaging≫ and brackets to indicate transverse averaging in the gas or solid, i.e.

κ 2π/κ f (x) = f (x, t) dt, 2π Z0 1 1 f (X, t) = f (x, t) dA, f s(X, t) = f (x, t) dA, h i A (X) (X) h i A (X) (X) g ZAg s ZAs where Ag and As are the cross-sectional area at position X of the gas and solid, respec- tively. The time-average of a harmonic variable always yields zero, since Re[ae jiκt] 0, for any j N, a C. However the time-average of the product of two harmonic var≡ i- ables is in∈ general∈ not equal to zero, as

iκt iκt 1 1 Re f e Re g e = Re [ f g∗] = Re [ f ∗ g ] . (3.25) 1 1 2 1 1 2 1 1 h i h i Here the superscript “ ” denotes complex conjugation. In addition we can show that the product of two harmonic∗ variables yields a second harmonic plus the constant time- averaged term from (3.25)

iκt iκt 1 2iκt 1 Re f e Re g e = Re f g e + Re [ f g∗] . (3.26) 1 1 2 1 1 2 1 1 h i h i h i From this equation it follows that the first-order acoustics can only excite second har- monics and steady terms, unless externally excited. This motivates why we only in- 36 3.4 Small-amplitude and long-pore approximation

cluded f2,0 and f2,2 as second-order terms in the expansion (3.24). Chapter 4

Thermoacoustics in two-dimensional pores with slowly varying cross-section

For certain types of geometries it suffices to use a two-dimensional representation. Specif- ically we will consider narrow two-dimensional pores with slowly varying cross-sections as shown in figure 4.1, where we use plus-superscripts to denote parameters and vari- ables in the upper plane and minus-superscripts in the bottom plane. In particular + Rg and −g give the distances between the centerline and the top plate and bottom plate, R + respectively, and s and s− denote the thicknesses of the top and bottom plate, re- spectively. AdditionallyR weR define

+ + g + −g s + s− + + + := R R , := R R := + , − := − + −. Rg 2 Rs 2 Rt Rg Rs Rt Rg Rs Two dimensional pores can be used to model a geometry where the three-dimensional pore arises from an extension of a two-dimensional channel into the third-coordinate di- rection. The most common example is the parallel-plate stack, consisting of various par- allel plates stacked on top of each other. Moreover, often a two-dimensional model can describe in good approximation the behavior of a more complicated three-dimensional shape. Due to the assumption of slow variation we will be able to treat the axial varia- tion separately from the transverse variation. In particular we will determine the X- dependence of the variables as solvability conditions for the (higher-order) y-dependence. The method adopted here is also referred to as the method of slow variation [85,143]. 38 4.1 Governing equations

Γ + t Π

Γ + y g + g R X −g Γg− R

Γt− Figure 4.1: Three-dimensional plates as an extension of a narrow two-dimensional channel with slowly varying cross-section. At position X = εx, the plate separation is given by 2 g(X) := + R (X) + −(X) and the cross-sectional area of the gas is given by A (X) = 2Π (X). Rg Rg g Rg

4.1 Governing equations

In two dimensions the equations (3.16)-(3.22) simplify into

∂ρ ∂(ρu) ∂(ρv) + + = 0, (4.1) ∂t ∂X ∂y

∂u ∂u ∂u ∂p κ ∂ ∂u ∂ ∂u M2 ρ + u + v = + ε2 µ + µ + a ρb ∂t ∂X ∂y − ∂X 2 ∂X ∂X ∂y ∂y F x   Wo      r ∂ µ ζ ∂u ∂v + ε2κ + + , (4.2) ∂X 2 2 ∂X ∂y " Wo Wζ !  # ∂v ∂v ∂v 1 ∂p κ ∂ ∂v ∂ ∂v ρ + u + v = + ε2 µ + µ ∂t ∂X ∂y − ε2 ∂y 2 ∂X ∂X ∂y ∂y   Wo      M2 ∂ µ ζ ∂u ∂v + a ρb + κ + + , (4.3) εF y ∂y 2 2 ∂X ∂y r " Wo Wζ !  # ∂T ∂T ∂T ∂p ∂p ∂p ρC + u + v = βT + u + v p ∂t ∂X ∂y ∂t ∂X ∂y     κ ∂ ∂T ∂ ∂T + ε2 K + K + κΣ, (4.4) 2 ∂X ∂X ∂y ∂y 2NL      ∂T κφ2 ∂ ∂T ∂ ∂T ρ C s = ε2 K s + K s , (4.5) s s ∂t 2 ∂X ∂X ∂y ∂y 2Ns      Thermoacoustics in two-dimensional pores with variable cross-section 39

∂ 1 ∂ 1 κK ∂T (ρ u 2 + ε2 v 2) + ρǫ = ρu( u 2 + ε2 v 2) + ρuh ε2 ∂t 2 | | | | − ∂X 2 | | | | − 2 ∂X   " 2NL # ∂ 1 κK ∂T κµ ∂u M2 ρv( u 2 + ε2 v 2) + ρvh u + · + a ρv · b ∂ 2 ∂ 2 ∂ , (4.6) − y " 2 | | | | − 2NL y − Wo y # ∇ T Fr where

µ ∂u ∂v 2 ∂u 2 ∂v 2 ε2ζ ∂u ∂v 2 Σ = + ε2 + 2ε2 + + + , 2 ∂y ∂X ∂X ∂y 2 ∂X ∂y Wo "      !# Wζ   µ ∂u ∂v ζ u + ε2v + v · u + u · v 2 ∂ ∂ 2 W X X ∇ Wζ ∇ = ε2κ  o    . T µ ∂v ζ  v + v · v + v · v  2 ∂y ∇ 2 ∇   Wo   Wζ    The boundary conditions (3.23) reduce to

v = 0 if y = ±, (4.7a) ±Rg T = T if y = ±, (4.7b) s ±Rg ∂ ∂ ∂ ∂ ∂ T Ts 2 ±g T Ts K σK = ε R K σK if y = ±, (4.7c) ∂y − s ∂y ± ∂X ∂X − s ∂X ±Rg   ∂ ∂ ∂ Ts 2 t± Ts K = ε K R if y = ±. (4.7d) s ∂y ± s ∂X ∂X ±Rt

4.2 Acoustics

In this section we will derive two ordinary differential equations in X, with coefficients depending on T , for the acoustic pressure p and the averaged acoustic velocity u . 0 1 h 1i Moreover, it will be shown that p0 is constant and that p1 is independent of y. Finally, explicit expressions will be derived, expressing u1, v1, T1, and ρ1 in terms of T0, p1, and u1 . h i 2 We start from the momentum equation in which we relate ε to Ma by putting ε = 2 ηMa (see equation (3.15)). Expanding the variables in powers of Ma according to (3.24), substituting the expansions into the y-component of the momentum equation (4.3), and keeping terms up to first order in Ma we find ∂p ∂p 0 + M 1 = o(M ). ∂y a ∂y a

0 This equation should be valid for any Ma 1. Therefore, the coefficients of Ma and 1 ∂ ≪∂ ∂ ∂ Ma are equal to zero, so we find that p0/ y = p1/ y = 0, so that p0 and p1 are independent of y. In the same way it can be shown that T0 does not depend on y either. Collecting the 40 4.2 Acoustics leading-order terms from the temperature equations (4.4) and (4.5), we find

∂ ∂T ∂ ∂T K 0 = 0, K s0 = 0. ∂y 0 ∂y ∂y s0 ∂y h i h i The only solution, satisfying the boundary conditions given in (3.23), is given by T = s0 T = T (X). As a result ρ , ρ , K , K , and µ are independent of y as well. The 0 0 0 s0 0 s0 0 exact X-dependence of T0 will be derived in Section 4.3 as a solvability condition for the temperature T2,0. We now turn to the X-component of the momentum equation (4.2). Keeping terms up to first order in Ma, we find that (4.2) reduces to

dp dp κ ∂ ∂u iM κρ u = 0 M 1 + M µ 1 + o(M ). a 0 1 − dX − a dX a 2 ∂y 0 ∂y a Wo  

Thus to leading order we find that dp0/dX = 0 and p0 must be constant. Collecting the terms of first order in Ma, we find that u1 satisfies

i dp 1 ∂2u u = 1 + 1 , (4.8) 1 2 ∂ 2 κρ0 dX αν y where ρ α =(1 + i) 0 W . (4.9) ν 2µ o r 0 Integrating (4.8) subject to v Γ = 0, we find that u and u satisfy | g 1 h 1i

iFν dp1 i(1 fν) dp1 u1 = and u1 = − . (4.10) κρ0 dX h i κρ0 dX

+ where Fν is chosen such that it vanishes on Γg and Γg−,

+ sinh(αν g ) + sinh(αν −g ) Fν := 1 R + R cosh(αν y) − sinh(α ( + −)) ν Rg Rg + cosh(αν g ) cosh(αν −g ) + R −+ R sinh(αν y), (4.11) sinh(α ( + −)) ν Rg Rg and + tanh(αν g) g + −g f := 1 F = R , := R R , (4.12) ν −h νi α Rg 2 νRg consistent with the notation of Arnott et al. [5], Rott [115], and Swift [131]. In particular + when = − = , we find the familiar expression Rg Rg Rg cosh(α y) F = 1 ν . (4.13) ν − cosh(α ) νRg Note that α , f , and are X-dependent. ν ν Rg Thermoacoustics in two-dimensional pores with variable cross-section 41

Next we go back to the temperature equation. Substituting our expansions into (4.4) and (4.5), substituting p0 is constant and T0 = T0(X), and collecting the first-order terms, we find for the gas temperature

dT κK ∂2T M ρ C iκT + u 0 = iM κβT p + M 0 1 + o(M ), (4.14) a 0 p 1 1 dX a 0 1 a 2 ∂ 2 a   2NL y and for the solid temperature

κφK ∂2T iκM ρ C T = M s0 s1 + o(M ). (4.15) a s0 s s1 a 2 ∂ 2 a 2Ns y

Setting the coefficient of Ma equal to zero in (4.14) and (4.15), and substituting (4.10) for u , we find after some manipulation that T and T can be obtained from 1 1 s1

1 ∂2T 1 dT dp βT T 1 = 0 1 F + 0 p , (4.16a) 1 2 ∂ 2 2 ν ρ 1 − αk y −κ ρ0 dX dX 0Cp 2 1 ∂ T T s1 = 0, (4.16b) s1 2 ∂ 2 − αs y where analogous to (4.9) we define

ρ0 αk :=(1 + i) NL, (4.17) s K0Cp ρ α :=(1 + i)φ s0 N . (4.18) s K C s s s0 s

Collecting the first-order terms, the boundary conditions given in (4.7b, 4.7d) can be written as

T = T , y = ±, (4.19a) 1 s1 ±Rg ∂ Ts 1 = 0, y = ±. (4.19b) ∂y ±Rt (4.19c)

Before we can solve (4.16), we first need to introduce some auxiliary functions. For the gas we define

+ + + sinh(αk g ) cosh(αk g ) Fk := 1 +R cosh(αk y) + +R sin(αk y), (4.20) − sinh(α ( + −)) sinh(α ( + −)) k Rg Rg k Rg Rg sinh(αk −g ) cosh(αk −g ) Fk− := 1 +R cosh(αk y) +R sin(αk y), (4.21) − sinh(α ( + −)) − sinh(α ( + −)) k Rg Rg k Rg Rg + F := F + F− 1. (4.22) k k k − 42 4.2 Acoustics

+ + and in the solid (with y = y, y− = − + y) Rt − −Rt + + cosh(αs y ) Fs := 1 , (4.23) − cosh(α +) sRs cosh(αs y−) Fs− := 1 , (4.24) − cosh(α −) sRs + + chosen such that Fk Γ = 0, Fk Γ + = 1, Fk Γ = 0, Fk− Γ + = 0, Fk− Γ = 1, Fs± Γ = 0, and | g± | g | g− | g | g− | g± ∂ ∂ Fs±/ y Γ = 0. Additionally we also define | t±

tanh(αk g) 1 f := 1 F = R , f ± := 1 F± = f , (4.25) k −h ki α k −h k i 2 k kRg + + + + tanh(αs s ) tanh(αs s−) fs := 1 Fs s = +R , fs− := 1 Fs− s− = R . (4.26) −h i α −h i α − sRs sRs Using these auxiliary functions we can integrate (4.16) subject to (4.19), so that

βT0Fk Fk PrFν dT0 dp1 + + T = p + T (1 F ) + T−(1 F−), (4.27a) 1 1 2 − b1 k b1 k ρ0Cp − κ (1 P )ρ dX dX − − − r 0 T± = T± 1 F± , (4.27b) s1 b1 − s +
where Tb = T1 Γ + and Tb− = T1 Γ are yet to be determined. To leading order, the 1 | g 1 | g− remaining boundary condition (4.7c) can be written as

+ ∂ ∂T ∂ ∂T− T1 s1 T1 s1 K0 = Ks σ + , K0 = Ks σ . (4.28) ∂y + − 0 ∂ ∂y 0 ∂ − y= g y + + y= −g y R y = s −R y−= s− R R

Substituting the expressions for T and T , we find the following two coupled equations 1 s1 for T− and T−: b1 b1

+ βT p 1 f / f dT dp T σK +α2 f +T+ = K α2 f 0 1 ν k 0 1 + b1 s0 s s s b1 0 g k k 2 − 2 R R ρ0Cp − κ (1 P )ρ dX dX 2 sinh (α ) − r 0 kRg T− 2 b1 (1 + coth (αk g)) , (4.29) − R 2 ! and

+ 2 2 βT0 p1 1 fν/ fk dT0 dp1 2 Tb σK −α f −T− = K α f (1 + coth (α )) 1 s0 s s s b1 0 g k k 2 − k g R R ρ0Cp − κ (1 P )ρ dX dX − R 2 − r 0 T− b1 + 2 . (4.30) 2 sinh (α ) ! kRg Thermoacoustics in two-dimensional pores with variable cross-section 43

2 2 Here we used that αk = Prαν and ∂F ∂F k = α2 f , ν = α2 f , ∂y ∓Rg k k ∂y ∓Rg ν ν y= ±g y= ±g ±R ±R ∂ α2 ∂ Fk± 1 g k fk Fk± 1 2 2 = R , = (1 + coth (αk g)) gαk fk, ∂y ± 2 sinh2(α ) ∂y ± 2 R R y= ±g k g y= ±g ±R R ∓R ∂ + ∂ Fs + 2 + Fs− 2 + = g αs fs , = −g αs fs−. ∂ −R ∂ − −R y + + y y−= − y = s s R R

Introducing the stack heat-capacity ratios

2 1 K α f ε 0Rg k k s± := 2 , (4.31) σ K ±α f ± s0 Rs s s we can rewrite (4.29) and (4.30) as

εs± εs± 2 1 T± + 1 + coth (α ) T∓ 2 b1 k g b1 " − 2 sinh (α ) # 2 R kRg h i βT0 p1 1 fν/ fk dT0 dp1 ε± = s 2 − . (4.32) ρ0Cp − κ (1 P )ρ dX dX ! − r 0 + Solving for T− and T , we obtain b1 b2

ε+ βT p 1 f / f dT dp T+ = s (1 + +) 0 1 − ν k 0 1 , (4.33a) b1 + ρ 2 1 + εs T 0Cp − κ (1 P )ρ dX dX ! − r 0 εs− βT0 p1 1 fν/ fk dT0 dp1 T− = (1 + −) , (4.33b) b1 2 − 1 + εs− T ρ0Cp − κ (1 P )ρ dX dX ! − r 0 where + + (εs εs−) cosh(2αk g) = + − − R + , (4.34a) T (1 + ε )(1 + ε− +(ε− 1) cosh(2α )+(ε ε−) cosh(2α ) s s s − kRg s − s kRg + (εs− εs ) cosh(2αk g) − = + +− − R + . (4.34b) T (1 + ε−)(1 + ε +(ε 1) cosh(2α )+(ε− ε ) cosh(2α ) s s s − kRg s − s kRg + Note that ± = 0, when s = s−. In an idealT stack the solidR hasR sufficient heat capacity to keep the stack plates isother- mal, so that the stack heat-capacity ratio εs is equal to zero. Figure 4.2 shows the absolute value of εs for a helium-filled parallel-plate stack for various porosities and plate mate- rials and a fixed hydraulic radius. The graphs show that only when the stack plates are thick enough (B 1) the stack can be considered ideal. r ≪ Using relations (A.5) and (A.8) and substituting T1 we can derive the following rela- 44 4.2 Acoustics

1 stainless steel alumina 0.8 glass mylar 0.6 polystyrene |ε | s 0.4

0.2

0 0 0.2 0.4 0.6 0.8 1 B r

Figure 4.2: The absolute value of εs plotted as a function of the blockage ratio Br for various plate materials. We consider a helium-filled parallel-plate stack at room temperature and fix NL = 1. The stack is ideal for small blockage ratios (thick plates). tion for the acoustic density fluctuations:

1 Fk PrFν dT0 dp1 + + ρ1 = [1 +(γ 1)(1 Fk)] p1 + − β ρ0βTb (1 Fk ) c2 − − κ2(1 P ) dX dX − 1 − − r ρ βT−(1 F−) (4.35) − 0 b1 − k + + 1 εs (1 + ) + εs−(1 + −) γ − = 2 1 +( 1) 1 Fk T+ (1 Fk ) T (1 Fk ) p1 c " − − − 1 + εs − − 1 + εs− − !# + Fk Fν + + (1 fν/ fk)(1 Fk ) + Fν + − + εs (1 + ) − −+ " (1 Pr) T (1 Pr)(1 + εs ) − − (1 fν/ fk)(1 Fk−) β dT0 dp1 ε− − + s (1 + ) − − 2 . (4.36) T (1 P )(1 + ε−) κ dX dX − r s  As a result we find 1 γ 1 ρ = 1 + − f p h 1i 2 1 + ε k 1 c  s  fν fk β dT0 dp1 + 1 fν + − . (4.37) − (1 P )(1 + ε ) κ2 dX dX  − r s  where εs is the effective stack heat-capacity ratio, given by

+ + 1 1 1 εs 1 εs− − = − T+ + − T (4.38a) 1 + εs 2 1 + εs 1 + εs− ! + + 1 εs (1 + ) εs−(1 + −) = 1 T+ + T . (4.38b) − 2 1 + εs 1 + εs− !

Finally, we turn to the continuity equation (4.1). Expanding the variables in powers Thermoacoustics in two-dimensional pores with variable cross-section 45

of Ma and keeping terms up to first order in Ma we find ∂ ∂v iM κρ + M (ρ u ) + M ρ 1 = o(M ), (4.39) a 1 a ∂X 0 1 a 0 ∂y a

We can use equation (4.39) to express v1 in terms of ρ0, u1 and p1,

1 y ∂ κρ ρ v1 = i 1 + ∂ ( 0u1) dyˆ. (4.40) − ρ − X 0 Z−Rg   where we applied the boundary condition that v1 vanishes at Γg−. + Note that v also vanishes at the boundary y = g . Therefore, substituting (4.10) for u , multiplying with iκ, averaging (4.39) over aR cross-section, and using the Leibniz 1 − rule, we obtain the following equation as a consistency relation for v1:

+ 1 d dp iκ d g iκ d −g κ2 ρ 1 R ρ R ρ 1 = g(1 fν) 0u1 + 0u1 h i − g dX R − dX − 2 g dX g − 2 g dX −g R   R R R −R 1 d dp = (1 f ) 1 . (4.41) − dX Rg − ν dX Rg   After substitution of (4.37) we obtain

κ2 γ 1 dT dp f f dT dp β 0 1 ν k β 0 1 2 1 + − fk p1 +(1 fν) + − c 1 + εs − dX dX (1 Pr)(1 + εs) dX dX   − 1 d dp + (1 f ) 1 = 0. (4.42) dX Rg − ν dX Rg   Then using d 1 1 dρ 1 ∂ρ dT dT ρ = 0 = 0 0 = β 0 , (4.43) 0 dX ρ −ρ dX −ρ ∂T dX dX  0  0 0 0 we obtain a reduced wave equation for p1 valid for slowly varying cross-sections,

κ2 γ 1 f f dT dp ν k β 0 1 2 1 + − fk p1 + − c 1 + εs (1 Pr)(1 + εs) dX dX   − ρ d 1 f dp + 0 − ν 1 = 0. (4.44) dX Rg ρ dX Rg  0  Combining (4.10) and (4.44) we can derive the following two coupled ordinary differ- ential equations for p and u 1 h 1i

d( g u1 ) dT R h i = κa p + a 0 u , (4.45a) dX 3 1 4 dX h 1i dp 1 = κa u , (4.45b) dX 5h 1i 46 4.3 Mean temperature where i γ 1 a := 1 + − f , (4.46a) 3 −ρ 2 1 + ε k 0c  s  ( f f )β a := ν − k , (4.46b) 4 − (1 P )(1 f )(1 + ε ) − r − ν s iρ a := 0 . (4.46c) 5 − 1 f − ν

To solve these equations it only remains to determine T0 and apply appropriate bound- ary conditions in X. In the next section the system of equations will be completed, when an ordinary differential equation for the mean temperature T0 is derived. Given T0, p1 and u , we can compute u , T , T , ρ , and v from (4.10), (4.27), (4.35), and (4.40), 1 1 1 s1 1 1 respectively.h i

4.3 Mean temperature

In this section we will the conservation of energy as given by (4.6) to determine T0. When calculating T2,0 we will determine T0 as a solvability condition from the assump- tion of slow variation. 2 2 Putting ε = ηMa , we can rewrite (4.6) as follows:

∂ 1 ∂ 1 κηK ∂T κ ρ( u 2 + ηM2 v 2) + ρǫ = ρu( u 2 + ηM2 v 2) + ρvh M2 ∂t 2 | | a | | − ∂X 2 | | a | | − a 2 ∂X   2NL ! ∂ 1 κK ∂T κµ ∂u M2 ρv( u 2 + ηM2 v 2) + ρvh u + a ρub + o(M4) ∂ a 2 ∂ 2 ∂ x a . − y 2 | | | | − 2NL y − Wo y ! Fr (4.47)

Suppose there is an average heat flux to the boundary given by F˙. Then averaged in time, the left hand side of this equation will reduce to F˙. Consequently, on expanding in powers of Ma and keeping terms up to second order, we find

∂ κηK dT ∂ M2ρ u h + M2h m˙ M2 0 0 + M2ρ v h + M2h n˙ ∂ a 0 1 1 a 0 2 a 2 ∂ a 0 1 1 a 0 2 X " − 2N dX # y L h κK ∂T κ ∂T κµ ∂u M2 0 2,0 M2 K 1 M2 0 u 1 = M2 F˙ + o(M2) a 2 ∂ a 2 1 ∂ a 2 1 ∂ a 2 a , − 2NL y − 2NL y − Wo y # where m˙ 2 and n˙ 2 are the components in X and y directions of the second-order time- averaged mass fluxm ˙ 2,

2iκt iκt iκt 1 m˙ := ρ (v + Re v e ) + Re ρ e Re v e = ρ v + Re [ρ v∗] . 2 0 2,0 2,2 1 1 0 2,0 2 1 1 h i h i h i Thermoacoustics in two-dimensional pores with variable cross-section 47

Using (3.25) and rearranging terms we find

∂2 ∂ ∂ 2 ∂ T2,0 1 T1 d dT0 NL K + Re K∗ + η K = (ρ Re [u∗h ] + 2h m˙ ) 0 ∂ 2 2 ∂y 1 ∂y dX 0 dX κ ∂X 0 1 1 0 2 y     2 ∂ ∂ 2 NL κµ0 u1 2NL + ρ Re [v∗h ] + 2h n˙ Re u∗ F˙ . (4.48) κ ∂y 0 1 1 0 2 − 2 1 ∂y − κ 2 Wo  ! Similarly we can show that (4.5) reduces to

2 ∂ T± ∂ ∂ ± s2,0 1 Ts d dT0 K + Re K∗ 1 + η K = 0. (4.49) s0 ∂ 2 2 ∂ s1 ∂ dX s0 dX y± y± " y± #   We will now use the flux condition (4.7c) and (4.7d) to derive a differential equation for 2 2 T0. Putting ε = σ Ma , averaging in time, expanding in powers of Ma and collecting the second order terms in Ma, we find

∂ ∂ ∂T ∂ ∂ T2,0 1 T1 s2,0 σ Ts g dT0 K + Re K∗ σK Re K∗ 1 = η K σK R , 0 ∂y 2 1 ∂y ± s0 ∂ ± 2 s1 ∂ ± 0 − s0 ∂X dX   y±  y± 
if y = ±, y± = ±, (4.50) ±Rg Rs and ∂T ∂ ∂ s2,0 1 Ts t± dT0 K + Re K∗ 1 = ηK R , if y± = 0. (4.51) s0 ∂ 2 s1 ∂ − s0 ∂X dX y±  y±  We will combine (4.48) and (4.49) with (4.50) and (4.51). First we integrate (4.49) over its cross-section, multiply with σ, and add it to (4.48), again after integrating over its cross-section, while using the Leibniz rule. This leads to

+ + + y = s ∂ ∂ y= g ∂T ∂ R T2,0 1 T1 R s2,0 1 Ts K + Re K∗ + σ K + Re K∗ 1 0 ∂ 1 ∂ s0 ∂ + s1 ∂ + y 2 y y= − y 2 y ! +    −Rg   y =0 y−= − ∂T ∂T Rs s2,0 1 s1 d dT0 d dT0 + σ Ks + + Re Ks∗ + 2η g K0 + 2ησ s Ks 0 ∂y 2 1 ∂y− R dX dX R dX 0 dX  !y−=0     2 ∂ 2 NL 4NL = 2ρ Re [ u∗h ] + 4h M˙ F˙ , (4.52) κ ∂X 0Rg h 1 1i 0 2 − κ Rg 2   ˙ where M2 = g m˙ 2 . After inserting the flux conditions (4.50) and (4.51) and rearrang- ing terms, weR canh simplifyi (4.52) further into

˙ ρ 2 dH2 d ˙ 0 Sk dT0 = h0 M2 + g Re [ u1∗h1 ] (K0 g + σKs s) , (4.53) dX dX " R 2 h i − 2κ R 0 R dX #

˙ dH˙ ˙ where we put g F =: dX . The quantity H is the time-averaged total power (or energy flux) along X. InR steady state, for a cyclic refrigerator or prime mover without heat flows 48 4.3 Mean temperature to the surroundings H˙ will be constant as F˙ = 0. Combining the thermodynamic expressions (A.7) and (A.9) we find

1 1 dh = Tds + dp = C dT + (1 βT)dp. (4.54) ρ p ρ − Moreover, it follows that h = C (T T ), (4.55) 0 p 0 − re f where Tre f is a reference temperature. Substituting (4.54) and (4.55) into (4.53), we find

1 g H˙ = M˙ C (T T ) + ρ C Re [ T u∗ ] + R (1 βT )Re [p u∗ ] 2 2 p 0 − re f 2 0 pRg h 1 1i 2 − 0 1h 1i S2 dT K + σK k 0 , (4.56) − 0Rg s0 Rs 2κ dX   ˙ where the constant of integration has been chosen such that H2 = 0 when there is no exchange of heat with the environment. Before we substitute T1 and u1 we will first prove the following relationships:

1 1 fk± + Pr(1 fν∗) 2 fν∗ Fk±Fν∗ = − − − , (4.57a) h i 1 + Pr

1 fk + Pr(1 fν∗) FkFν∗ = − − , (4.57b) h i 1 + Pr F 2 = 1 Re [ f ] . (4.57c) h| ν| i − ν We will only prove the first statement as it is most general; the other two equalities can be proven in the same way. Define

I := F± F∗ , (4.58) 1 h k ν i ∂F± ∂F∗ I := k ν . (4.59) 2 ∂y ∂y  

We have the following relationships for Fν and Fk±:

1 ∂2 F F = 1 + ν F , (4.60) ν 2 ∂ 2 ν αν y ∂2 1 Fk± F± = 1 + . (4.61) k 2 ∂ 2 αk y

2 2 Integrating by parts, using (4.60), substituting (α∗) = α , and using the boundary ν − ν Thermoacoustics in two-dimensional pores with variable cross-section 49

conditions for Fk± and Fν, we can rewrite I1 as follows:

+ 2 g ∂ 1 R 1 Fν∗ I = F± 1 dy 1 k 2 ∂ 2 2 g −g − α y R Z−R ν ! + 2 g ∂ 1 R Fν∗ = 1 f ± F± dy k 2 k ∂ 2 − − 2 α −g y Rg ν Z−R + ∂ Rg + ∂ ∂ 1 Fν∗ 1 Rg Fk± Fν∗ = f ± F± + y 1 k 2 k ∂ 2 ∂ ∂ d y − y y − − 2 gαν 2 gαν Z g R −g R −R −R 1 I2 = 1 fk± fν∗ + 2 . (4.62) − − 2 αν In the same way, substituting (4.61) rather than (4.60), we find

I2 ∗ I1 = 1 fν 2 . (4.63) − − αk

Eliminating the common term I2 from equations (4.62) and (4.63), we find

1 1 fk± + Pr(1 fν∗) 2 fν∗ I1 = − − − , (4.64) 1 + Pr as claimed in (4.57a). Replacing Fk± by Fk or Fν, we can repeat the same analysis to prove (4.57b) and (4.57c). Substituting expressions (4.10) and (4.27a) for u1 and T1, while using relations (4.57) and (4.38b), we find

2 βT0 FkFν∗ FkFν∗ Pr Fν dT0 2 ∗ ∗ T1u1 = h i p1 u1 + i h i − h| | 2i u1 h i ρ0Cp 1 fν∗ h i κ(1 P ) 1 f dX |h i| − r ν + − | − | + Fν∗ Fk Fν∗ Fν∗ Fk−Fν∗ + Tb u1∗ h i−h i + Tb− u1∗ h i−h i 1 h i 1 f ∗ 1 h i 1 f ∗ − ν − ν βT0 p1 u1∗ fk fν∗ i(1 Pr + PrRe ( fν)) dT0 2 = h i 1 − + − 2 u1 ρ0Cp − (1 + Pr)(1 + εs)(1 fν∗) κ(1 P ) 1 f dX |h i|  −  − r | − ν| 2 i u1 dT0 ( fk fν∗)(1 + εs fν/ fk) ∗ |h i| 2 fν + − . − κ(1 P ) 1 f dX (1 + Pr)(1 + εs) − r | − ν|   Taking the real part, we get

βT0 fk fν∗ Re [ T1u1∗ ] = Re 1 − p1 u1∗ h i ρ C − (1 + P )(1 + ε )(1 f ∗) h i 0 p  r s − ν   2 u1 dT0 ( fk fν∗)(1 + εs fν/ fk) ∗ + |h i| 2 Im fν + − . (4.65) κ(1 P ) 1 f dX (1 + Pr)(1 + εs) − r | − ν|   50 4.3 Mean temperature

After substitution of (4.65) into (4.56) we arrive at

˙ ˙ 1 fk fν∗ H2 = M2Cp T0 Tre f + gRe p1 u1∗ 1 βT0 − − 2 R h i − (1 + P )(1 + ε )(1 f ∗)   r s − ν    2 ρ C u dT (1 + ε f / f )( f f ∗) 0 p|h 1i| 0 s ν k k ν + g 2 Im fν∗ + − R 2κ(1 P ) 1 f dX (1 + Pr)(1 + εs) − r | − ν|   S2 dT K + σK k 0 . (4.66) − 0Rg s0 Rs 2κ dX   ˙ Given H2, we can solve (4.37) for dT0/dX

˙ ˙ dT 2H2 2M2Cp T0 Tre f gRe [a1 p1u1∗] 0 = κ − − − R , (4.67) 2  2 dX a u K + σK S Rg 2|h 1i| − 0Rg s0 Rs k   where

fk fν∗ a1 := 1 βT0 − , (4.68) − (1 + P )(1 + ε )(1 f ∗) r s − ν ρ 2 0Cp u1 (1 + εs fν/ fk)( fk fν∗) |h i| ∗ a2 := 2 Im fν + − . (4.69) (1 P ) 1 f (1 + Pr)(1 + εs) − r | − ν|   Equation (4.67) together with equations (4.46a) and (4.46b) form a complete coupled sys- tem of differential equations for T0, u1 , and p1. This system gives a generalization of the Swift equations [135], extended toh includei the case of slowly varying cross-sections. It is possible to go one step further and improve the expression for the mean tem- perature by determining the correction term T2,0. If we integrate (4.48) and (4.49) twice with respect to y, we can determine T and T up to X-dependent functions T± . In 2,0 s2,0 b2,0 the same way as the differential equation (4.67) for T0 followed as a solvability condition for T and T , we can derive an ordinary differential equation for T as a solvability 2,0 s2,0 b2,0 condition for the fourth-order mean temperatures T and T . Performing this analysis 4,0 s4,0 it is possible to include transverse variations into the mean temperature profile, since T2,0 does depend on y in contrast to T0. In this thesis we will not go this far, as this would require a lengthy derivation of both the second and third harmonics. We will, however, compute the second harmonics in Section 4.6 below. Thermoacoustics in two-dimensional pores with variable cross-section 51

4.4 Integration of the generalized Swift equations

It has been shown in Section 4.2 and 4.3 that all acoustic variables can be determined, if the following system of ordinary differential equations is solved for T , p , and u : 0 1 h 1i ˙ ˙ dT 2H2 2M2Cp T0 Tre f gRe [a1 p1 u1∗ ] 0 = κ − − − R h i , (4.70a)  2  2 dX a u K + σK S Rg 2|h 1i| − 0Rg s0 Rs k   d( g u1 ) dT R h i = κa p + a 0 u , (4.70b) dX 3Rg 1 4Rg dX h 1i dp 1 = κa u , (4.70c) dX 5h 1i where

fk fν∗ a1 := 1 βT0 − , (4.71a) − (1 + P )(1 + ε )(1 f ∗) r s − ν ρ 2 0Cp u1 (1 + εs fν/ fk)( fk fν∗) |h i| ∗ a2 := 2 Im fν + − , (4.71b) (1 P ) 1 f (1 + Pr)(1 + εs) − r | − ν|   i (γ 1) f a := 1 + − k , (4.71c) 3 −ρ 2 1 + ε 0c  s  ( f f )β a := ν − k , (4.71d) 4 − (1 P )(1 + ε )(1 f ) − r s − ν iρ a := 0 . (4.71e) 5 − 1 f − ν With suitable boundary conditions these equations can be integrated numerically. For example, if a stack is positioned in a resonator with the left stack end at distance xL from the closed end, then one can impose

ipL p1(0) = pL cos(kxL), u1 (0) = sin(kxL), T0(0) = TL, (4.72) h i ρLcL where k is the wave number and pL is the pressure amplitude at the closed end. In ˙ addition we still need to impose H2, which is nonzero in general and it may also depend on X. In a steady-state situation, without heat exchange to the surroundings and zero ˙ mass flow, it follows that H2 = 0. In this case the thermoacoustic heat flow is balanced by a return diffusive heat flow in the stack and in the gas, so that the net heat flow is zero. Alternatively, one can also impose a temperature TR on the right stack end and ˙ look for the corresponding H2 that gives the desired temperature difference. In Sections 6 and 7 we will show some practical examples of thermoacoustic systems for which the equations above have been implemented. Under certain conditions, one can even find an exact solution as well. In the next sections we will treat a few of these cases in full detail. We see that the total energy flux through the stack, the cross-sectional variations, the small-amplitude acoustical oscillations and the resulting streaming may cause (1)- O 52 4.4 Integration of the generalized Swift equations

˙ variations in the mean temperature. The energy flux H2 can be changed by external input or extraction of heat at the stack ends and consequently change the mean temper- ature profile. The acoustic oscillations affect the mean temperature via the well-known ˙ shuttling effect (see e.g. [131]). With a nonzero mass flux M2 heat will be carried to the hot or cold heat exchanger and thus affect the mean temperature also. This can either be a loss or a contribution to the heat transfer. 2 Note that the temperature gradient scales with κ and 1/Sk. Thus in the limits κ 0 ∞ → (short-stack limit) or Sk (small velocities; heat conduction is dominating), the tem- perature difference across→ the stack will tend to zero, unless sufficient heat is supplied ˙ or extracted ( H2 1). Furthermore, the velocity and pressure gradients also scale with κ, which| justifies| ≫ the assumption of constant-stack pressure and velocity that is commonly applied in the short-stack approximation (see e.g. [131], [151]). Finally, remember that these equations result from a linearization in both Ma and ε. As a result there is still a dimensionless number Sk appearing in the equations that contains both ε and Ma. In (3.15) we showed that κε Sk = , NL Ma so that the perturbation variables are weakly affected by the linearization parameters Ma and ε. Hence the theory derived here is not exactly linear as it is often described, which is why we prefer to use the term weakly non-linear to indicate that there is still a weak non-linearity involved. If the amplitude Ma becomes large enough such that S 1 then the temperature gradient will arise solely due to the thermoacoustic heat k ≪ flow; if the amplitude Ma becomes very small, so that Sk 1, then the temperature gradient will arise solely due to heat conduction through so≫lid and gas.

4.4.1 Exact solution at constant temperature

If we apply equations (4.70) to a problem involving a channel supporting a constant temperature, then the equations will simplify greatly. This case is particularly inter- esting for the analysis of insulated resonators, with or without variable cross-sections. Baks et al. [18] showed both theoretically and experimentally what kind of cooling pow- ers can be expected (in a pulse tube). In this section we will focus on deriving analytic expressions for the pressure and velocity profiles in arbitrarily shaped channels. Setting dT0/dX = 0, we find we have to solve

d( g u1 ) R h i = κa p , (4.73a) dX 3Rg 1 dp 1 = κa u . (4.73b) dX 5h 1i We can rewrite these equations into a reduced wave equation for the pressure

d2 p dp a d 1 1 2 5 Rg 2 + C + k p1 = 0, C = , (4.74) dX dX g dX a5 R   Thermoacoustics in two-dimensional pores with variable cross-section 53 where the complex wave number k is defined as

κ 1 γ 1 k = κ√ a a = 1 + − f . − 3 5 c 1 f 1 + ε k s − ν  s  The viscous dissipation due to interaction with the wall gives rise to its imaginary part. For wide tubes that have large Lautrec numbers it follows that 0 < Im[k] 1 and there will be little dissipation. When C and k are constant, equation (4.74)− coincides≪ with Webster’s horn equation [110, 149]. Both C and k depend on in a rather complicated manner. However, for straight Rg pores with constant g we find that k is constant and C = 0. It follows that the pressure and velocity can be writtenR as

ikX ikX p1(X) = Ae− + Be , (4.75a) χ ikX ikX u1 (X) = Ae− Be , (4.75b) h i ρ0c −   with A and B integration constants and

1 γ 1 χ = (1 f ) 1 + − f . c − ν 1 + ε k s  s  For variable cross-sections we cannot find an exact solution as above because C and k depend on in a complicated manner. However, in view of the asymptotic behavior Rg of the hyperbolic tangent, it follows that fν, fk, and εs will not vary much in X, especially for wide tubes, so that in good approximation we can replace k by its average value kˆ 1 d and C by dX g. This leads to the following equation for p1 Rg R 2 d p1 1 d dp1 ˆ2 2 + ( g) + k p1 = 0. (4.76) dX g dX R dX R Next, for specific choices of the geometry, we can solve (4.76) analytically. For most practical applications is either constant or linearly changing. If we put Rg = a(X X ), Rg − 0 for some a > 0 and X R, then we need to solve 0 ∈ 2 d p1 1 dp1 ˆ2 2 + + k p1 = 0. (4.77) dX X X0 dX − ˆ After a transformation ξ = k(X X0), we find that p1 solves the Bessel’s differential equation − 2 2 d p1 dp1 2 ξ + ξ + ξ p1 = 0. dξ2 dξ The general solution is given in terms of the zeroth order Bessel functions of the first 54 4.4 Integration of the generalized Swift equations and second kind.

p (X) = AJ [kˆ(X X )] + BY [kˆ(X X )]. (4.78) 1 0 − 0 0 − 0 For several other geometries exact solutions can be obtained as well. For example, hy- perbolic or exponential horns occur frequently in resonators used for thermoacoustic applications (e.g. [142]). Suppose we put

= ae2bX, Rg for some a > 0 and b R, then we need to solve ∈ 2 d p1 dp1 ˆ2 + 2b + k p1 = 0. (4.79) dX2 dX The general solution is given by

aX ˆ ˆ p1(X) = e− A cos(kX) + B sin(kX) . (4.80)   Alternatively, putting 1 = , Rg X X − 0 for some X R, we find an equation 0 ∈ 2 d p1 1 dp1 ˆ2 2 + k p1 = 0, (4.81) dX − X X0 dX − which only differs from (4.77) by the minus-sign in the second term. The general solu- tion is thus given by

p (X) = AJ [ kˆ(X X )] + BY [ kˆ(X X )]. (4.82) 1 0 − − 0 0 − − 0

4.4.2 Short-stack approximation Many practical thermoacoustic devices have stacks of short length relative to the wave- length. One can take advantage of this property to find an approximate solution to (4.70) in the stack. The relevant dimensionless number for this analysis is the Helmholtz num- ber κ based on the length of the stack. We will expand the mean and acoustic variables once again in powers of κ and derive exact solutions for the perturbation variables. Let fi be one of the original perturbation variables, then we expand

f = f + κ f + , κ 1. i i0 i1 · · · ≪ Furthermore, for illustration, we put the following boundary conditions:

X = 0: T = T , u = u , p = p . (4.83) 0 L h 1i L 1 L We consider a steady-state situation H˙ = 0, in which the thermoacoustic heat flow is Thermoacoustics in two-dimensional pores with variable cross-section 55 balanced by a return diffusive heat flow in the stack and in the gas. In addition, we as- sume there is no streaming, so that M˙ = 0. Substituting the expansions into (4.70a), we find to leading order that T00 is constant and (and thus also ρ00, a10, a20, a30, a40 and a50). Subsequently we find from (4.70b) and (4.70c) that p10 and g u10 must be constant as well. It follows from (4.83) that R h i u T = T , u = L , p = p . (4.84) 00 L h 10i 10 L Rg Going one order higher in κ we find that T , p , and u can be found from: 01 11 h 11i

dT gRe [a10 p10 u10∗ ] 01 = R h i , (4.85a) dX (K + σK )S2 a u 2 00Rg s00 Rs k − 20Rg|h 10i| d( g u11 ) dT R h i = a p + a 01 u , (4.85b) dX 30Rg 10 40Rg dX h 10i dp 11 = a u . (4.85c) dX 50h 10i Integrating these equations, we find

gRe [a10 p10 u10∗ ] T01(X) = R h i dX, (4.86a) (K + σK )S2 a u 2 Z 00Rg s00 Rs k − 20Rg|h 10i| 1 dT u (X) = a p + a 01 u dX, (4.86b) h 11i 30 Rg 10 40Rg dX h 10i Rg Z  

p11(X) = a50 u10 dX. (4.86c) Z h i The temperature difference, generated across the stack, is of order κ and thus the tem- perature difference will increase linearly with increasing stack length or frequency. Continuing this way it is possible to determine the higher order terms in κ as well. In general we get for i = 0,1,2...,

2dT ∑ Re a p u ∗ + ∑ a u 0l j+k+l=i 1 j 1k 1l j+k+l=i+1 βj 1k dx h i l i |h i| ≤ T0(i+1) = h i 2 2 " (K + σK )S a u Z 00Rg s00 Rs k − 20|h 10i| 2 (K0 j g + σKs s)Sk dT ∑ R 0j R 0k dX, (4.87a) − (K + σK )S2 a u 2 dx j+k=i+1 00 g s00 s k 20 10 # l i R R − |h i| ≤ 1 dT u = ∑ a p + ∑ a 0k u dX, (4.87b) h 1(i+1)i 3 jRg 1k 4 jRg dX h 1li Rg Z " j+k=i j+k+l=i #

p1(i+1) = ∑ a5 j u1k dX. (4.87c) Z j+k=i h i

Consider a short parallel-plate stack placed at position xs in an acoustic standing wave with wave number k. For constant all three expressions in (4.86) will be linear Rg 56 4.4 Integration of the generalized Swift equations in X. In particular we find that the temperature difference generated across the stack is given by

1 dT0 ∆T0 = dX Z0 dX κ gRe [a10 p10 u10∗ ] d g = R h i + o(κ), if R = 0. (4.88) (K + σK )S2 a u 2 dX 00Rg s00 Rs k − 20Rg|h 10i| Note that a , a , K and K are constant and are found from substitution of T . 10 20 00 s00 00 Putting p = cos(kx ) and u = i cos(kx ), we find that (4.88) transforms into 10 P s h 10i − P s 2 κ gIm [a10] sin(2kxs) ∆T0 = R |P| + o(κ). (4.89) 2(K + σK )S2 a 2 (1 cos(2kx )) 00Rg s00 Rs k − 20Rg|P| | − s This result can be seen as a generalization of the short-stack approximation per- formed by Wheatley et al [151], whose prediction is only valid for wide enough pores. In [151] and slightly modified in [11] a similar analysis was performed for a parallel-plate stack positioned in an acoustic standing-wave. They solved the equations combining a short-stack approximation (κ 1) with a boundary-layer approximation (NL 1). Furthermore, only a leading-order≪ solution was derived assuming constant pressure≫ and velocity in the stack. The result is a similar, but not identical, expression for the temperature difference developed across the stack,

1 p2 L δ (1 + √P ) ∆ A s k r T0 re f re f re f re f sin(2kxs) ∼ 8 ρg c (Ks Rs + Kg Rs)(1 + Pr) 1 2 − 1 pAδk(1 Pr√Pr) 1 + re f re f re f −re f 2 (1 cos(2kxs)) . (4.90) × " 8 ρ T ω(K R + K R )(γ 1)(1 P ) − # s s g s − − r Numerical evaluations show that when N 1 this expression agrees quite well with L ≥ our approximation (4.89). Unsurprisingly, when NL 1 a discrepancy arises because of the underlying assumption of wide pores. ≪

4.4.3 Approximate solution in short wide channels supporting a tem- perature gradient

In the previous sections we have found approximate solutions in the cases where ei- ther ∆T = 0 (section 4.4.1) or H˙ = 0 (section 4.4.2). An approximate solution can also be found if a (large enough) nonzero temperature difference is imposed, provided the channel is short and wide enough. One example of a tube that satisfies such require- ments is the thermal buffer tube that is commonly used in traveling-wave devices (see Chapter 7). We start our analysis from the system of equations (4.70), with boundary conditions Thermoacoustics in two-dimensional pores with variable cross-section 57

X = 0: T = T , p = p , u = u , (4.91a) 0 L 1 L h 1i L X = 1: T0 = TR. (4.91b)

Without loss of generality we may assume TR > TL. Our first assumption is that a T T considerable temperature difference is imposed across the tube, i.e. 0 R − L = (1). ≪ TL O Moreover, we assume the tube is short with respect to the wave length and wide with respect to the penetration depths, so that κ 1 and N 1. We will use the smallness ≪ L ≫ of κ and NL to derive a leading-order approximation of the temperature, pressure, and velocity profiles. Since N 1, we find that to leading order f = f = 0 and L ≫ ν k . . . i . . i a1 = 1, a2 = 0, a3 = 2 , a4 = 0, a5 = 2 . (4.92) −ρ0c −ρ0c ˙ α Moreover, as TR TL, it is suggested by equation (4.70a) that H = (κ ), for some α 1. We therefore≫ expand H˙ , M˙ , T , p , u as O ≤ − 2 2 0 1 h 1i H˙ = κα + o(κα), κ 1, (4.93a) 2 H ≪ M˙ = κβ + o(κβ), κ 1, (4.93b) 2 M ≪ p p = κγ + o(κγ), κ 1, (4.93c) 1 − L P ≪ u u = κδ + o(κδ), κ 1, (4.93d) h 1i − L U ≪ T T = + o(κ0), κ 1, (4.93e) 0 − L T ≪ ρ = + o(κ0), κ 1, (4.93f) 0 D ≪ where α, β, and δ are yet to be determined. Since c, C , K , and K depend only weakly p 0 s0 on T we can assume they are constant. Using the ideal gas law can be expressed in 0 D terms of T0. Substituting (4.92) and (4.93) into (4.70) we find to leading order

α+1 β+1 γ δ d 2κ 2κ Cp κ gRe (pL + κ )(u∗L + κ ∗) T = H − MT − R P U , (4.94a) h 2 i dX − K + σK S 0Rg s0 Rs k   d( g ) 1 δ g γ R U = iκ − R (pL + κ ), (4.94b) dX − c2 P D d 1 γ δ P = iκ − (u + κ ). (4.94c) dX − D L U The exponents α, β, γ, and δ will now be determined by balancing the terms on either sides of the equations in (4.94). It follows from (4.94b) and (4.94c) that the left and right hand side can only be balanced if γ = δ = 1. As a result equation (4.94a) becomes to leading order α+1 β+1 d 2κ 2κ Cp κ gRe [pLu∗L] T = H − MT − R , dX − K + σK S2 0Rg s0 Rs k   58 4.4 Integration of the generalized Swift equations whose left and right hand side can only be balanced if α = 1. Furthermore, it follows that streaming will only affect the mean temperature if β − 1. Putting α = β = 1, γ = δ = 1, we find we have to solve ≤ − −

d 2 2 Cp T = H − M T , (4.95a) dX − K + σK S2 0Rg s0 Rs k   d( g ) i g R U = R pL, (4.95b) dX − c2 d D P = i u , (4.95c) dX − D L subject to

(0) = 0, (0) = 0, (0) = 0, (1) = T T . P U T T R − L Integrating these equations, we find

2 X g(ξ) T (X) = T H dξ, (4.96a) 0 L ( ) 2 − g X Z0 K (ξ) + σK (ξ) S 0Rg s0 Rs k iκp  X (ξ)  L Rg ξ u1 (X) = uL 2 d , (4.96b) h i − c (X) 0 (ξ) Rg Z D X p1(X) = pL iκuL (ξ) dξ, (4.96c) − Z0 D 1 − g(1) 1 g(ξ) =(T T ) dξ , (4.96d) R L  2  H − 2 Z0 K (ξ) + σK (ξ) S 0Rg s0 Rs k     where X 2C pM ξ g(X) = exp 2 d . "− 0 (K (ξ) + σK (ξ))S # Z 0Rg s0 Rs k In particular, when we assume a straight tube, the expression for the temperature profile simplifies into

θX 1 e 2Cp T0(X) = TL +(TR TL) − , θ = M . (4.97) − 1 eθ (K + σK )S2 − 0Rg s0 Rs k We can now distinguish two limiting cases in which there is either a very large mass flux or a very small mass flux. In the limit for ∞ the temperature profile approaches M → ± Thermoacoustics in two-dimensional pores with variable cross-section 59 an almost discontinuous profile,

θ 1 e X T , X = 0, lim T (X; ) = lim T +(T T ) − = L (4.98a) ∞ 0 ∞ L R L θ M θ − 1 e TR, 0 < X 1, M→− →− −  ≤

θX T , 0 X < 1, 1 e L ≤ lim T0(X; ) = lim TL +(TR TL) − θ = (4.98b) ∞ M θ ∞ − 1 e ( T , X = 1. M→ → − R On the other hand, when there is little or no mass streaming we find a linear tempera- ture profile lim T0(X; ) = TL +(TR TL)X. (4.99) 0 M − M→

4.5 Acoustic streaming

This section discusses steady second-order mass flow in the stack driven by first-order acoustic phenomena. The analysis is valid for arbitrarily shaped pores supporting a temperature gradient. Moreover, the temperature dependence of viscosity is taken into account. There are several types of streaming that can occur simultaneously. Three kinds of streaming are shown in Fig. 4.3. Gedeon streaming refers to a nett time-averaged mass flow through a stack pore, i.e. M˙ = 0, typical for looped-tube thermoacoustic devices. Rayleigh streaming refers to a time-averaged6 toroidal circulation within a stack pore driven by boundary-layer effects at the pore walls that can occur even if M˙ = 0. Inner streaming refers to a time-averaged toroidal circulation in the whole stack, so that the nett time-averaged mass flow can differ from pore to pore both in size and direction. Possible causes for inner streaming are inhomogeneities at the stack ends or asymmet- rical pores. Streaming effects are usually undesirable, but it was suggested in [135] that for some applications it can be useful as a substitute for heat exchangers.

(a) Gedeon streaming (b) Rayleigh streaming (c) Inner streaming

Figure 4.3: Three types of mass streaming in stack

We start with the continuity equation (4.1). If we time-average the equation and expand in powers of Ma, then the zeroth and first order terms in Ma will drop out. 60 4.5 Acoustic streaming

Consequently we find to leading order ∂ ∂ ∂ ∂ v2,0 1 ρ u + ρ + Re (ρ u∗) + (ρ v∗) = 0. (4.100) ∂X 0 2,0 0 ∂y 2 ∂X 1 1 ∂y 1 1  
We can use this equation to express v2,0 in terms of u2,0 and known lower-order quanti- ties. Integrating over y, we find

1 y ∂ 1 1 ρ ρ ρ v2,0 = ∂ 0u2,0 + Re ( 1u1∗) dy Re ( 1v1∗) . (4.101) − ρ0 −g X 2 − 2ρ0 Z−R   We can then integrate (4.100) over a cross-section, while noting that v vanishes at the boundary y = and also at the centerline y = 0, to obtain ±Rg

d g ρ u + R Re [ ρ u∗ ] = 0. (4.102) dX Rg 0h 2,0i 2 h 1 1i   ˙ The expression between the brackets is M2 the second-order time-averaged and cross- ˙ sectional-averaged mass flux in the X-direction. It follows that M2 is constant, which is to be expected as there is no mass transport through the stack walls. We can now express u in terms of M˙ and the first order acoustics as follows: h 2,0i 2 ˙ 1 M2 1 u = Re [ ρ u∗ ] . (4.103) h 2,0i ρ − 2 h 1 1i 0 Rg ! Note that even when the average mass flux M˙ = 0, there can still be a nonzero streaming velocity as a result of first-order velocity and density variations. Next we turn to equation (4.3). Expanding in powers of Ma and averaging in time we find to leading order ∂p 2,0 = 0, ∂y so that p2,0 = p2,0(X). Subsequently, time-averaging equation (4.2), we find to leading order ∂2 2 u2,0 Wo dp2,0 2 = f , (4.104) ∂y − κµ0 dX where f is a collection of products of lower-order terms given by

2 ∂ ∂ ∂ ∂ 1 Wo u1 u1 ρ0 1 µ1∗ u1 f := Re iκρ u∗ + ρ u∗ + ρ v∗ b Re . 2 κµ 1 1 0 1 ∂X 0 1 ∂y − F x − 2 ∂y µ ∂y 0  r    0  The first-order acoustics collected in f can be interpreted as a source term for the stream- ing on the left hand side, with the last term being characteristic for Rayleigh streaming. Thermoacoustics in two-dimensional pores with variable cross-section 61

Integrating (4.104) twice with respect to y, we can write

2 + W dp g y u (X, y) = o 2,0 (( +)2 y2) R f (X, yˆ) dyˆ dy C( + y). (4.105) 2,0 κµ g g − 2 0 dX R − − Zy Z0 − R − + where we used that u2,0 vanishes on Γg . C is a constant of integration and will be determined from the no-slip condition on Γg−. Introducing F as the anti-derivative of f ,

y F(X, y) := f (X, yˆ) dyˆ, Z0 we find 2 Wo dp2,0 + C = g −g F . − 2κµ0 dX R − R −h i   Substituting C into (4.105), we find that u2,0 is given by

2 + Wo dp2,0 + Rg + u (X, y) = ( − + y)( y) F(X, y) dy + F ( y). (4.106) 2,0 2κµ dX g g g − 0 R R − − Zy h i R − Computing the cross-sectional average we can relate dp /dX to u as follows: 2,0 h 2,0i + + dp2,0 3κµ0 1 Rg Rg = 2 2 u2,0 + F(X, y) dy dy + g F . (4.107) dX − W h i 2 g −g y R h i! Rg o R Z−R Z Summarizing, given the mass flux M˙ and the first-order acoustics, u , dp /dX, 2 h 2,0i 2,0 u2,0, and v2,0 can be determined consecutively from (4.103), (4.107), (4.105), and (4.101).

4.6 Second harmonics

The previous few sections have provided a recipe to determine all acoustic variables, the mean temperature T0, and the streaming variables. In this section we will show how the second harmonics, the variables that oscillate at twice the fundamental frequency, can be computed from the former variables. Before we begin, we generalize the auxiliary functions from Section 4.2,

α j,2 := √2α j, j = ν, k, s. (4.108) 62 4.6 Second harmonics

1 + Similarly we define (with = ( + −)) Rg 2 Rg Rg + sinh(α j,2 g ) + sinh(α j,2 −g ) F := 1 R R cosh(α y), j,2 − sinh(2α ) j,2 j,2Rg + cosh(α j,2 g ) cosh(α j,2 −g ) + R − R sin(α y), j = ν, k, (4.109a) sinh(2α ) j,2 j,2Rg sinh(αk,2 ±g ) cosh(αk,2 ±g ) F± := 1 R cosh(α y) R sin(α y), (4.109b) k,2 − sinh(2α ) k,2 ± sinh(2α ) k,2 k,2Rg k,2Rg cosh(αs,2y±) Fs±,2 := 1 , (4.109c) − cosh(α ±) s,2Rs and we introduce second-harmonic Rott’s functions

f := 1 F , f := 1 F , f ± := 1 F± , f ± := 1 F± ±, (4.110) ν,2 −h ν,2i k,2 −h k,2i k,2 −h k,2i s,2 −h s,2is and the stack heat capacity ratios

2 1 K α f ε 0Rg k,2 k,2 s±,2 := 2 . (4.111) σ K ±α f ± s0 Rs s,2 s,2 Again we start from the momentum equation. Expanding the variables in powers of Ma according to (3.24), substituting the expansions into the y-component of the mo- 2 2 mentum equation (4.3), putting ε = ηMa , and collecting terms of second order in Ma we find ∂p 2,2 = 0, ∂y and we conclude p2,2 must be independent of y. Similarly, collecting all second-order terms in Ma, we find that the X-component of the momentum equation (??) reduces to

1 1 ∂u 1 ∂u dp 2iκρ u + iκρ u + ρ u 1 + ρ v 1 = 2,2 0 2,2 2 1 1 2 0 1 ∂X 2 0 1 ∂y − dX κµ ∂2u 1 ∂ µ ∂u + 0 2,2 + 1 1 , (4.112) 2 ∂ 2 2 ∂y µ ∂y Wo " y  0 # where we substituted (3.26) and (4.104). Rearranging terms we can rewrite (4.112) as

1 ∂2u i dp u 2,2 = 2,2 + A, (4.113) 2,2 2 ∂ 2 κρ − αν,2 y 2 0 dX where A, a source term arising from products of first-order terms, is known and given by i ∂u ∂u κµ ∂ µ ∂u A = iκρ u + ρ u 1 + ρ v 1 0 1 1 . 4κρ 1 1 0 1 ∂X 0 1 ∂y − 2 ∂y µ ∂y 0 ( Wo  0 ) Thermoacoustics in two-dimensional pores with variable cross-section 63

Using the variation-of-constants formula [19], we can write a solution as follows:

iFν,2 dp2,2 i(1 fν,2) dp2,2 u2,2 = + Ψν(A), u2,2 = − + ψν(A), (4.114) 2κρ0 dX h i 2κρ0 dX where Ψ j( f ) (j = ν, k) is the variation-of-constants formula that is chosen such that Ψ j( f ) y= ± = 0, | ±Rg

y y S j Cj Ψ j( f ) := f (X,ζ)Cj(X,ζ) dζ f (X,ζ)S j(X,ζ) dζ α − − α − j,2 Z−Rg j,2 Z−Rg 2 g + + S j S−j Cj/C−j R S j f Cj Cj f S j + − + , (4.115a) − α j,2 h i − h i S S−C /C−   j − j j j C := cosh(α y), C± := cosh(α ±), (4.115b) j j,2 j j,2Rg S := sinh(α y), S± := sinh(α ±), (4.115c) j j,2 j j,2Rg and ψ ( f ) := Ψ ( f ) . Furthermore, for the solid we define j h j i

y± y± Ss Cs Ψ±( f ) := f (X,ζ)C (X,ζ) dζ f (X,ζ)S (X,ζ) dζ s α s α s s,2 Z0 − s,2 Z0 ±C S± ± η d ± dT S± + Rs s f S s C Rt 0 S s C , (4.116a) α s − s − α dX dX s − s s,2   Cs± s s,2  Cs±  C := cosh(α y±), C± := cosh(α ±), (4.116b) s s,2 s s,2Rs S := sinh(α y±), S± := sinh(α ±), (4.116c) s s,2 s s,2Rs

∂Ψ±( f ) d ± dT Ψ± s η Rt 0 satisfying s ( f ) y±= ± = 0 and ∂ y±=0 = dX dX . Additionally we introduce | Rs y± | − the averaged functions ψs±( f ) := Ψs±( f ) s±. Next we turn to the temperatureh equation.i Substituting our expansions into (4.4) and (4.5) and collecting the terms of second order in Ma, we find after some manipulation that T and T can be found from 2,2 s2,2

1 ∂2T 1 dT dp βT T 2,2 = B 0 2,2 F + 0 p , (4.117a) 2,2 2 ∂2 2 ν,2 ρ 2,2 − αk,2 y − 4κ ρ0 dX dX 0Cp ∂2 1 Ts T 2,2 = C, (4.117b) s2,2 2 ∂2 − αs,2 yˆ 64 4.6 Second harmonics where B and C are known and given by

dp dT ∂T dT 4B = iβκT p + T u 1 iκC ρ T 2ρ C Ψ (A) 0 ρ u 1 ρ u 0 1 1 0 1 dX − p 1 1 − 0 p ν dX − 0 1 ∂X − 1 1 dX ∂T κ ∂ ∂T d dT κµ ∂u 2 ρ v 1 + K 1 + 2η K 0 + 0 1 , − 0 1 ∂y 2 ∂y 1 ∂y dX 0 dX 2 ∂y 2NL     Wo   κφ ∂ ∂T d dT 4C = K s1 + 2η K 0 . 2 ∂yˆ s1 ∂yˆ dX s0 dX 2Ns      Using (4.115) and (4.116) and imposing the boundary conditions given in (4.7b) and (4.7d), we can write

βT F F P F dT dp T = Ψ (B) + 0 k,2 p k,2 − r ν,2 0 2,2 + T+ (1 F+ ) 2,2 k 2,2 2 b2,2 k,2 ρ0Cp − 4κ (1 P )ρ dX dX − − r 0 + T− (1 F− ), (4.118a) b2,2 − k,2 T± = Ψ (C) + T± 1 F± , (4.118b) s2,2 s b2,2 − s,2 +
where Tb = T2,2 Γ + and Tb− = T2,2 Γ are yet to be determined. Collecting the second- 2,2 | g 2,2 | g− order terms in the remaining boundary condition (4.7c) and substituting (4.28), we can write

∂ ∂ ∂T Ts K K ∂T ±g dT K 2,2 σK 2,2 = K 1 s1 1 η R (K σK ) 0 0 ∂y s0 ∂ 0 K K ∂y ∂X 0 s dX ± y± − 0 − s0 ! ± −

if y = ± and y± = ±. (4.119) ±Rg Rs Applying these conditions, we find (cf. (4.33))

εs±,2(1 + 2,2± ) βT0 1 1 fν,2/ fk,2 dT0 dp2,2 T± = T D± + p − , (4.120) b2,2 2,2 2 1 + ε± ρ0Cp − 4κ ρ 1 Pr dX dX s,2 0 − ! where D± can be computed from the lower-order terms as

∂ 1 d ±g dT0 1 K1 Ks T1 D± = η(K σK ) R K 1 2 0 s0 0 ∂ K α f ± − dX dX − 2 K0 − Ks y 0 g k,2 k,2 " 0 ! y= ±g R ±R ∂Ψ (B) ∂Ψ± (C) K k σK s 0 ∂ s0 ∂ − y y= ± ∓ y± g y±= s± # ±R R

Thermoacoustics in two-dimensional pores with variable cross-section 65

and ± is given by T2,2 + + (εs,2 εs−,2) cosh(2αk,2 g) 2,2 = + − − R + , T (1 + ε )(1 + ε− +(ε− 1) cosh(2α )+(ε ε− ) cosh(2α ) s,2 s,2 s,2 − k,2Rg s,2 − s,2 k,2Rg + (εs−,2 εs,2) cosh(2αk,2 g) 2,2− = + +− − R + . T (1 + ε− )(1 + ε +(ε 1) cosh(2α )+(ε− ε ) cosh(2α ) s,2 s,2 s,2 − k,2Rg s,2 − s,2 k,2Rg

Using relations (A.5) and (A.8) and substituting T2,2 we can derive the following relation for the second-harmonic density fluctuations:

1 F P F dT dp ρ γ k,2 r ν,2 β 0 2,2 2,2 = 2 1 +( 1)(1 Fk,2) p2,2 + 2 − c − − 4κ (1 Pr) dX dX   − + + 1 ρ βT (1 F ) ρ βT− (1 F− ) ρ βΨ βρ T (4.121) − 0 b2,2 − k,2 − 0 b2,2 − k,2 − 0 k,B − 2 1 1 + + 1 εs,2(1 + 2,2) + εs−,2(1 + 2,2− ) γ − = 2 1 +( 1) 1 Fk,2 +T (1 Fk,2) T (1 Fk,2) p2,2 c " − − − 1 + εs,2 − − 1 + εs−,2 − !# + Fk,2 Fν,2 + + (1 fν,2/ fk,2)(1 Fk,2) + Fν,2 + − + εs,2(1 + 2,2) − −+ " (1 Pr) T (1 Pr)(1 + εs,2) − − (1 fν,2/ fk,2)(1 Fk−,2) β dT0 dp2,2 ε− − + s,2(1 + 2,2) − − 2 T (1 Pr)(1 + εs−,2) # 4κ dX dX − + + ε D ε− D− 1 ρ ρ β s,2 + s,2 + Ψ (B) + 1 T . (4.122) 0 + k ρ 1 − 1 + εs,2 1 + εs−,2 2 0 !

As a result we find

1 γ 1 f f β dT dp ρ ν,2 k,2 0 2,2 2,2 = 2 1 + − fk,2 p2,2 + 1 fν,2 + − 2 h i c 1 + εs,2 − (1 Pr)(1 + εs,2) 4κ dX dX    −  + + εs,2D εs−,2D− 1 ρ0β + + + ψk(B) + ρ1T1 , (4.123) − 1 + εs,2 1 + εs−,2 2ρ0 h i! where εs,2 is given by

+ + 1 1 1 εs,2 2,2 1 εs−,2 2,2− = − +T + − T (4.124a) 1 + εs,2 2 1 + εs,2 1 + εs−,2 ! + + 1 εs,2(1 + 2,2) εs−,2(1 + 2,2− ) = 1 +T + T . (4.124b) − 2 1 + εs,2 1 + εs−,2 !

Finally, we turn to the continuity equation (4.1). Expanding the variables in powers of Ma and collecting terms of second order in Ma, we find ∂ ∂v ∂ ∂ 2iκρ + (ρ u ) + ρ 2,2 = (ρ u ) (ρ v ). (4.125) 2,2 ∂X 0 2,2 0 ∂y − ∂X 1 1 − ∂y 1 1 66 4.6 Second harmonics

We can use this equation to express v2,2 in terms of u2,2, ρ2,2, and the known lower-order quantities,

y ∂ ∂ 1 1 ρ1 v2,2 = 2iκρ2,2 + (ρ0u2,2) + (ρ1u1) dyˆ v1, (4.126) − ρ 0 ∂X 2 ∂X − 2ρ 0 Z   0 where we applied the boundary condition that v2,2 vanishes at Γg−. + Note that v also vanishes at the boundary Γg . Therefore, substituting (4.114) for u2,2, multiplying with 2iκ, averaging over a cross-section, and using the Leibniz rule, we − obtain the following equation as a consistency relation for v2,2:

1 d dp 2iκ ∂ 4κ2 ρ + (1 f ) 2,2 = ( [ρ ψ (A) + ρ u ]). h 2,2i dX Rg − ν,2 dX ∂X Rg 0 ν h 1 1i Rg   Rg After substituting (4.123), we obtain

4κ2 γ 1 f f dT dp dT dp ν,2 k,2 β 0 2,2 β 0 2,2 2 1 + − fk,2 p2,2 + − +(1 fν,2) c 1 + εs,2 (1 Pr)(1 + εs,2) dX dX − dX dX   − 1 d dp + (1 f ) 2,2 = 2iκρ E, (4.127) dX Rg − ν,2 dX − 0 Rg   where E is a source term arising from products of first-order or zeroth-order quantities and is given by

+ + εs,2D εs−,2D− 1 E := 2iκβ + + + ψk,B + ρ1T1 1 + εs,2 1 + εs−,2 2ρ0 h i! 1 ∂ ρ ψ ρ ∂ g [ 0 ν(A) + 1u1 ] . (4.128) − ρ0 g X R h i R   Inserting (4.43), we obtain a wave equation for the second pressure harmonic

2 4κ γ 1 ( fν,2 fk,2)β dT0 dp2,2 2 1 + − fk,2 p2,2 + − c 1 + εs,2 (1 Pr)(1 + εs,2) dX dX   − ρ d 1 f dp + 0 − ν,2 2,2 = 2iκρ E. (4.129) dX Rg ρ dX − 0 Rg  0  Apart from the source term E this equation has a similar structure as the wave equation derived in (4.44) for the first pressure harmonic. Combining (4.114) and (4.129) we derive the following two coupled ordinary differ- Thermoacoustics in two-dimensional pores with variable cross-section 67 ential equations for p and q := u ψ 2,2 2,2 h 2,2i − ν,A

dq dT 1 d g 2,2 = κa p + a 0 R q + E, (4.130a) dX 3,2 2,2 4,2 dX − dX 2,2 Rg ! dp 2,2 = κa q , (4.130b) dX 5,2 2,2 where 2i (γ 1) f a := 1 + − k,2 , (4.131a) 3,2 −ρ 2 1 + ε 0c  s,2  ( f f )β a := ν,2 − k,2 , (4.131b) 4,2 − (1 P )(1 + ε )(1 f ) − r s,2 − ν,2 2iρ a := 0 . (4.131c) 5,2 − 1 f − ν,2 Since all the zeroth-order and first-order terms are known from the previous sections, we can compute subsequently A, B, C, D, and E. Next the system (4.130) can be inte- grated to determine p2,2 and u2,2 , provided appropriate boundary conditions are im- posed. Note that the gravitationalh i effect only affects the streaming terms via the source function f in equation (4.104), and does not appear in the equations for the first and second harmonics.

4.7 Power

To analyze the performance of thermoacoustic systems it is important to have a clear understanding of the energy flows in the system and their interplay. In this section we will elaborate further on the concept of total and acoustic power.

4.7.1 Acoustic power ˙ The time-averaged acoustic power W2 is given by 1 W˙ = Re [p u ∗ ] . (4.132) 2 2 Rg 1h 1 i ˙ Thus it follows from (3.25) that the time-averaged acoustic power dW2 used or produced in a segment of length dX can be found from

dW˙ d 2 = Re p eiκt Re u eiκt . (4.133) dX dX Rgh 1 h 1i i  h i h i  Using (3.25), we find to leading order

˙ dW2 1 d g g d u1∗ dp1 = R Re [p u ∗ ] + R Re p h i + u ∗ . (4.134) dX 2 dX 1h 1 i 2 1 dX h 1 i dX   68 4.7 Power

Substituting (4.45a) and (4.45b) into (4.134) we find

˙ dW2 g β dT0 fk∗ fν∗ g κ(γ 1) fk 2 R ∗ R = Re − p1 u1 −2 Im p1 dX 2 1 Pr dX (1 fν∗)(1 + εs∗) h i − 2 ρ c − 1 + εs | | −  −  0   g κρ0Im [ fν] 2 R − u1 . (4.135) − 2 1 f 2 |h i| | − ν|

The first term contains the temperature gradient dT0/dX and is called the sink or source term. It will either absorb (refrigerator) or produce (prime mover) acoustic power de- pending on the magnitude of the temperature gradient along the stack. This term is the unique contribution to thermoacoustics. The last two terms are the “viscous” and “thermal-relaxation” dissipation terms, respectively. These two terms arise due to the interaction with the wall, and have a dissipative effect in thermoacoustics. ˙ s The sink/source term, which we define as W2, is of greatest interest in thermoacous- tic engines and refrigerators. For interpretation we will neglect viscosity and set σ = ∞, so that fν = Pr = εs = 0, ˙ s dW2 β dT0 = Re [ f ] Re [p u ∗ ] + Im [ f ] Im [p u ∗ ] . (4.136) dX 2 dX k 1h 1 i − k 1h 1 i   In a standing-wave system Im (p u ∗ ) is large and therefore Im ( f ) is important. 1h 1 i − k Fig. 4.4 shows that the maximal value is attained for NL close to 1. In the case of a traveling-wave system Re (p1 u1∗ ) is large and Re ( fk) is important. Fig. 4.4 shows that Re ( f ) reaches its maximalh valuei for N 1. This motivates why commonly k L ≪ stacks (NL 1) are used in standing-wave systems and regenerators (NL 1) in traveling-wave∼ systems. ≪

1 real part imaginary part

0.5 f k

0

−0.5 0 1 2 3 4 5 6 N L

Figure 4.4: Real and imaginary part of fk, plotted as a function of the Lautrec number NL.

To test the significance of the dissipation terms we can investigate how (4.135) be- haves for small NL or Wo. To optimize the thermoacoustic effect one would like to maximize the source term and minimize the dissipation terms. If we consider an ideal Thermoacoustics in two-dimensional pores with variable cross-section 69

parallel-plate stack with εs = 0, then

˙ dW2 g β dT0 fk∗ fν∗ g κ(γ 1) 2 R ∗ R = Re − p1 u1 −2 Im [ fk] p1 dX 2 1 Pr dX (1 fν∗) h i − 2 ρ c − | | −  −  0 g κρ0Im [ fν] 2 R − u1 . (4.137) − 2 1 f 2 |h i| | − ν| 2 For small Wo and NL one can show that ( fk fν)/(1 fν) = (1), Im( fk) = (NL), 2 2 4 − − O O Im( fν) = (Wo ) and 1 fν = (Wo ). Therefore, it follows that the acoustic power behaves asO | − | O

dW˙ dW˙ s dW˙ k dW˙ ν 2 = 2 2 2 dX dX − dX − dX (4.138) 2 1 = (1) NL 2 , Wo, NL 1, O − O − O W ! ≪   o ˙ k ˙ ν where W2 and W2 denote the thermal relaxation dissipation and viscous dissipation, respectively. Unsurprisingly, equation (4.138) shows that the dissipation in a regenerator (N L ≪ 1, Wo 1) is dominated by viscous dissipation and in a stack (NL = (1), Wo = (1)) by thermal≪ relaxation dissipation. In a regenerator there is perfect thermalO contact,O but very small pores and therefore viscous dissipation will be dominant. In a stack, on the other hand, there is imperfect thermal contact, but wider pores. Thus thermal relaxation dissipation is dominant here. Dissipation is usually undesirable, so NL should be chosen carefully.

4.7.2 Total power

˙ In equation (4.66) we derived an expression for the total power H2,

˙ ˙ 1 fk fν∗ H2 = M2Cp T0 Tre f + gRe p1u1∗ 1 βT0 − − 2 R − (1 + Pr)(1 + εs)(1 fν∗)     −  ρ 2 0Cp u1 dT0 (1 + εs fν/ fk)( fk fν∗) |h i| ∗ + g 2 Im fν + − R 2κ(1 P ) 1 f dX (1 + Pr)(1 + εs) − r | − ν|   S2 dT K + σK k 0 . (4.139) − 0Rg s0 Rs 2κ dX   Combining (4.53) and (4.54), we can write the total power as a sum of the acoustic power ˙ ˙ ˙ W, the hydronamic entropy flux Q, the heat flow Qm due to a nett mass flux, and the 70 4.7 Power

˙ heat flow Qloss due to conduction down a temperature gradient,

g 1 H˙ = R Re [p u∗ ] + ρ T Re [ s u∗ ] + M˙ C (T T ) 2 2 1h 1i 2 0 0Rg h 1 1i 2 p 0 − re f S2 dT K + σK k 0 − 0Rg s0 Rs 2κ dX =: W˙ + Q˙ + Q˙ Q˙ . (4.140) 2 2 m,2 − loss,2 To illustrate the behavior of the thermoacoustic devices, idealized energy flows are depicted in figure 4.5. The situation is ideal in the sense that heat conduction and mass ˙ ˙ streaming are neglected, so that Qm,2 = Qloss,2 = 0. Figure 4.5(a) shows a refrigerator that is thermally insulated from the surroundings except at the heat exchangers where heat is exchanged with the environment. On the left acoustic power is supplied, possibly by means of a speaker or some other source of sound. Part of the acoustic power is used to sustain the standing or traveling wave against thermal and viscous dissipation, and part is used for the thermoacoustic effect to transport heat from the cold to the hot heat exchanger. This can be used for cooling at the the cold heat exchanger or heating at the hot heat exchanger. Figure 4.5(b) tells a similar story for the prime mover. A large temperature differ- ence is imposed across the stack, by supplying heat at the hot heat exchanger, or by extracting heat at the cold heat exchanger. As a result acoustic power is generated, part of which is dissipated due to viscous interaction with the resonator wall. The nett result can then be used as a sound source for some external device, possibly even to drive a thermoacoustic refrigerator [78]. If we apply the conservation of energy to a control volume surrounding each heat exchanger, then we can relate the cooling and heating power to the total power and the acoustic power as follows:

˙ ˙ QC = H, refrigerator: − (4.141) ( Q˙ = W˙ H˙ , H − ˙ ˙ ˙ QC = W + H, prime mover: (4.142) ˙ ˙ ( QH = H. − Thermoacoustics in two-dimensional pores with variable cross-section 71

TH TH

TC

TC

TH TC TH TC

W˙ H˙ H˙ W˙

˙ ˙ ˙ ˙ QH QC QH QC

˙ QH ˙ QC

˙ QC ˙ QH

(a) Refrigerator (b) Prime mover

Figure 4.5: Schematic and idealized illustration of the total power H˙ (solid line), the acoustic power W˙ (dashed line) and the hydronamic energy flux Q˙ (dotted line) in and around a parallel- plate stack, insulated everywhere except at the heat exchangers, positioned in (a) a thermoacous- tic refrigerator and (b) a thermoacoustic prime mover. In (a) a small temperature difference is maintained across the stack and acoustic power is absorbed by the stack; in (b) a large temper- ˙ ature difference is maintained and acoustic power is produced. The discontinuities in H2 arise due to heat transfer at the heat exchangers. 72 4.7 Power Chapter 5

Thermoacoustics in three-dimensional pores with slowly varying cross-section

Although a two-dimensional model can give a lot of information, there are many practi- cal devices that require a three-dimensional analysis because its resonator tube or stack pores have non-trivial three-dimensional shapes. These can be cylindrical, rectangular, or triangular pores, pin arrays, wired mesh, or some random porous material. In this chapter we consider shapes of the type shown in figure 5.1: narrow three-dimensional pores with cross-sections that may vary slowly in longitudinal direction [61].

Γt (X,θ) solid Rs Γg

r gas g(X,θ) R X θ

Figure 5.1: Longitudinal cut of a three-dimensional pore with slowly varying cross-section. At position X and angle θ, the radius of the gas and solid is given by g(X,θ) and s(X,θ), respectively. R R

The analysis in this chapter differs from the analysis in chapter 4 in the sense that there is a nontrivial dependence on the third coordinate, which complicates the compu- tation of the transverse variations. It will be necessary to determine Green’s functions for the Poisson’s and modified Helmholtz equation. Moreover, two additional integral equations need to be solved, unless the boundary condition (3.23c) is replaced by the condition of constant wall temperature, which is quite accurate for most purposes. 74 5.1 Acoustics

5.1 Acoustics

We follow an approach similar to that in the previous chapters. First we introduce some auxiliary functions. As in the two-dimensional case, we define αν, αk and αs as follows: ρ α =(1 + i) 0 W , (5.1) ν 2µ o r 0 ρ0 αk =(1 + i) NL, (5.2) s K0Cp

ρs α =(1 + i)φ 0 N . (5.3) s K C s s s0 s

Furthermore, we determine Fj (j = ν, k) from

1 2 Fj 2 τ Fj = 1 in g, (5.4a) − α j ∇ A

Fj = 0 on Γg, (5.4b) and define f j := 1 Fj ( j = ν, k). Note that for two-dimensional pores these defini- tions match the expressions−h i given in the previous chapter. Arnott et al. [5] follow a similar approach, although a slightly different notation is adopted; F(x;α j) in stead of Fj(x). Also there is an additional minus-sign in (5.4a) be- cause they assume a time-dependence of the form eiκt, whereas we follow the conven- tional notation with a positive sign as used by Rott [115] and Swift [131]. We start from the momentum equation and expand the fluid variables in powers of Ma as shown in (3.24). Substituting the expansions into the transverse components of 2 2 the momentum equation (3.18), putting ε = ηMa , and keeping terms up to first order in Ma we find 0 = p M p + o(M ). −∇τ 0 − a∇τ 1 a Collecting the leading-order terms in M we find we find that p = 0, so that p is a a ∇τ 0 0 function of X only. Furthermore, collecting the first-order terms in Ma, we additionally find that τ p1 = 0, so that p1 is also a function of X only. Next we∇ turn to the X-component of the momentum equation (3.17), which neglect- ing higher-order terms in Ma changes into dp dp κ κρ 0 1 · µ iMa 0u1 = Ma + Ma 2 τ ( 0 τ u1) + o(Ma). − dX − dX Wo ∇ ∇

To leading order we find that dp0/dX = 0 and therefore p0 is constant. Next assume that the mean temperature T0 is a function of X only. Below we will show that this is indeed the case. As a result we also find that µ0 = µ0(X) and K = K0(X). Then, collecting the terms of first order in Ma, we find that u1 satisfies

i dp1 1 2 u1 = + 2 τ u1. (5.5) κρ0 dX αν ∇ Thermoacoustics in three-dimensional pores with variable cross-section 75

With the help of equations (5.4), we can integrate (5.5) subject to v Γ = 0 and write u | g 1 and dp1/dX as

iFν dp1 i(1 fν) dp1 u1 = and u1 = − . (5.6) κρ0 dX h i κρ0 dX Next we turn to the temperature equation. Substituting our expansions into (3.20) and (3.21) and keeping terms up to first order in Ma, we find

· κ · Maρ0Cp (iκT1 + v1 T0) = iMaκβT0 p1 + τ K0 τ (T0 + MaT1) , ∇ 2N2 ∇ ∇ L h i φ2 iM ρ C T = · K T + M T . a s0 s s1 2 τ s0 τ s0 a s1 2Ns ∇ ∇ h
i To leading order this reduces into 2T = 2T = 0. An obvious solution, satisfying τ 0 τ s0 the boundary conditions given in (3.23),∇ is∇ that T and T are equal and independent s0 0 of x . In view of the thermodynamic relation (A.8) it also holds that ρ and ρ are τ 0 s0 independent of xτ . Next, collecting the terms of first order in M , we find that T and T can be obtained a 1 s1 from 1 dT dp βT 1 T + 0 1 F 0 p = 2T , (5.7a) 1 2 ν ρ 1 2 τ 1 κ ρ0 dX dX − 0Cp αk ∇ 1 T = 2T , (5.7b) s1 2 τ s1 αs ∇ where we substituted expression (5.6) for u1. To solve the temperature from (5.7) we first need to introduce some additional aux- iliary functions. In [5] a solution is obtained to (5.7) by assuming that the wall temper- ature Ts does not depend on time, which allows a solution as a combination of Fν and Fk-functions. However, with the boundary conditions given in (3.23), this approach will not work here. Assume for now the boundary function g := T Γ is known and choose 1| g gp and gu such that βT0 gu dT0 dp1 g = gp p1 . ρ c2 − κ2(1 P )ρ dX dX 0 − r 0 We can now write for the temperature β T0Fkp Fku PrFν dT0 dp1 T1(x) = p1 2 − , (5.8) ρ0Cp − κ (1 P )ρ dX dX − r 0 βT 1 F dT dp T (x) = 0 (1 F )p su 0 1 , (5.9) s1 sp 1 2 − ρ0Cp − − κ (1 P )ρ dX dX − r 0 76 5.1 Acoustics

where Fk j (j = p, u) satisfies

1 2 Fk j 2 τ Fk j = 1 in g, (5.10a) − αk ∇ A

Fk j = g j on Γg, (5.10b) and Fs j (j = p, u) is found from

1 2 Fs j 2 τ Fs j = 1 in s, (5.11a) − αs ∇ A F = 1 g on Γ , (5.11b) s j − j g F · n′ = 0 on Γ . (5.11c) ∇τ s j τ t The standard way of solving such boundary value problems is making use of the Green’s functions for the given Helmholtz equations on a cross-section with appropriate bound- ary conditions. In Appendix C we will show how the g j and Fi j functions can be deter- mined using Green’s functions. Using relations (A.5) and (A.8) and substituting T1 we can derive the following rela- tion for the acoustic density fluctuations:

1 β(Fku PrFν) dT0 dp1 ρ1 = γ (γ 1)Fkp p1 + − . (5.12) c2 − − κ2(1 P ) dX dX h i − r Finally, we turn to the continuity equation (3.16). Expanding the variables in powers of Ma and keeping terms up to first order in Ma, we find ∂ iM κρ + M (ρ u ) + M ρ τ · vτ = 0. (5.13) a 1 a ∂X 0 1 a 0∇ 1 Next we substitute (5.6). First note that because of the divergence theorem

τ · vτ dS = vτ · n dℓ = 0, ∇ Γ ZAg Z g since v Γ = 0. Therefore, averaging (5.13) over a cross-section and multiplying with | g iκ, we obtain the following equation as a consistency relation for v : − 1

2 1 d dp1 iκ dAg κ ρ1 = Ag(1 fν) ρ0u1 h i − Ag dX − dX − Ag dX Γg   1 d dp = A (1 f ) 1 . (5.14) − A dX g − ν dX g   Thermoacoustics in three-dimensional pores with variable cross-section 77

After substituting (5.12) and putting f = 1 F (j = ν, kp, ku), we obtain j −h ji

κ2 β( f f ) dT dp dT dp γ ν ku 0 1 β 0 1 2 1 +( 1) fkp p1 + − + (1 fν) c − 1 Pr dX dX − dX dX h i − 1 d dp + A (1 f ) 1 = 0. A dX g − ν dX g   Eventually using

d 1 1 dρ 1 ∂ρ dT dT ρ = 0 = 0 0 = β 0 , (5.15) 0 dX ρ −ρ dX −ρ ∂T dX dX  0  0 0 0 we obtain the dimensionless equivalent of Rott’s wave equation for general porous me- dia

κ2 β( f f ) dT dp ρ d 1 f dp γ ν ku 0 1 0 ν 1 2 1 +( 1) fkp p1 + − + Ag − = 0. (5.16) c − 1 Pr dX dX Ag dX ρ0 dX h i −   We can now combine (5.6) and (5.16) to find a coupled system of first order differen- tial equations for p and u . From (5.6) we find 1 h 1i ρ d 1 f dp ρ d 0 A − ν 1 = iκ 0 (A u ). (5.17) A dX g ρ dX − A dX gh 1i g  0  g

Substituting this result into (5.16) and repeating equation (5.6), we find that u1 and p1 are found from h i

d u1 dT0 1 dAg h i = κa3 p1 + a4 u1 , (5.18a) dX dX − Ag dX ! hh ii dp 1 = κa u , (5.18b) dX 5h 1i where i a3 := 1 +(γ 1) fkp , −ρ c2 − 0 h i β( f f ) a := ν − ku , 4 − (1 P )(1 f ) − r − ν iρ a := 0 . 5 − 1 f − ν To complete the system of equations, it still remains to find an equation for the mean temperature T0. The next section explains how this can be done. 78 5.2 Mean temperature

5.2 Mean temperature

As in Section 4.3 we will use the conservation of energy to determine T0 as a consistency condition for the second-order temperature T2,0. 2 2 Putting ε = ηMa , we can rewrite (3.22) as follows:

∂ 1 κ ρ( u 2 + ηM2 v 2) + ρǫ ∂t 2 | | a | τ |   ∂ 1 κηK ∂T = ρu( u 2 + ηM2 v 2) + ρuh M2 ∂ a τ a 2 ∂ − X 2 | | | | − 2NL X ! 1 κK κµ · ρ 2 η 2 2 ρ τ vτ ( u + Ma vτ ) + vτ h 2 τ T 2 u τ u − ∇ 2 | | | | − 2NL ∇ − Wo ∇ ! 2 · Ma · + + ρ (ubx + √ηMavτ bτ ) , (5.19) ∇ T Fr with ∂ ∂ µ u 2 · vτ · ζ · u + ηMa vτ + v u + u v W2 ∂X ∂X ∇ W2 ∇ 2  o   ζ  = Ma ηκ . T µ · ·b b ζ b · b  vτ τ vτ + v vτ + vτ v  W2 ∇ ∇ W2 ∇   o ζ      If there is an average heat flux F˙ to the boundary, thenb theb left hand sideb ofb this equation ˙ will reduce to F after averaging in time. Consequently, on expanding in powers of Ma and keeping terms up to second order, we can neglect gravitational terms and the - term and find T

∂ κηK dT M2ρ u h + M2h m˙ M2 0 0 + · M2ρ v h + M2h m˙ ∂ a 0 1 1 a 0 2 a 2 τ a 0 τ1 1 a 0 τ2 X " − 2N dX # ∇ L h κ κ κµ 2 K0 2 2 0 2 ˙ 2 Ma 2 τ T2,0 Ma 2 K1 τ T1 Ma 2 u1 τ u1 = Ma F2 + o(Ma ), − 2NL ∇ − 2NL ∇ − Wo ∇ # where m˙ andm ˙ are the components in longitudinal and transverse directions of the 2 τ2 second-order time-averaged mass fluxm ˙ 2,

2iκt iκt iκt 1 m˙ := ρ (v + Re v e ) + Re ρ e Re v e = ρ v + Re [ρ v∗] . 2 0 2,0 2,2 1 1 0 2,0 2 1 1 h i h i h i Plugging in relation (3.25) and rearranging terms we find

2 ∂ 2 1 d dT0 NL K T + · Re [K∗ T ] + η K = (ρ Re [u∗h ] + 2h m˙ ) 0∇τ 2,0 2 ∇τ 1 ∇τ 1 dX 0 dX κ ∂X 0 1 1 0 2   2 2 NL κ 2NL + τ · ρ Re vτ∗ h + 2h m˙ τ Re [u∗ τ u ] F˙ . (5.20) κ 0 1 1 0 2 2 1 1 κ 2 ∇ − Wo ∇ ! −   Thermoacoustics in three-dimensional pores with variable cross-section 79

Similarly we can show in the solid that (3.21) reduces to

2 1 d dT0 K τ T + τ · Re K∗ τ T + η K = 0. (5.21) s0 ∇ s2,0 2 ∇ s1 ∇ s1 dX s0 dX     We will now use the flux condition (3.23c) to derive a differential equation for T0. Time-averaging and expanding (3.23c) in powers of Ma and collecting the second order terms in Ma we find

1 1 K τ T + Re [K∗ τ T ] σK τ T σ Re K∗ τ T · τ 0∇ 2,0 2 1 ∇ 1 − s0 ∇ s2,0 − 2 s1 ∇ s1 ∇ Sg     ∂ dT g = η σK K 0 S , = 0. s0 − 0 dX ∂X Sg
This condition can be rewritten as

1 1 K τ T + Re [K∗ τ T ] σK τ T σ Re K∗ τ T · nτ 0∇ 2,0 2 1 ∇ 1 − s0 ∇ s2,0 − 2 s1 ∇ s1   K σK ∂ dT (K σK ) ∂  dT η 0 s0 Rg 0 η 0 − s0 Rg Rg 0 = − ∂ = ∂ , g = 0, (5.22) τ g X dX 2 +(∂ /∂θ)2 X dX S |∇ S | Rg Rg q where n := /( ) is the outward unit normal vector to Γ . Similarly, since τ ∇τ Sg ∇τ Sg g n′ := /( ) is the outward unit normal vector to Γ , we find from (3.23d) τ ∇τ St ∇τ St t ηK ∂ 1 s0 t t dT0 K T · n′ + · Re K∗ T = R R , = 0. (5.23) s0 s2,0 τ τ s1 τ s1 ∂ t ∇ 2 ∇ ∇ 2 +(∂ /∂θ)2 X dX S   Rt Rt q Now on the one hand, by applying the divergence theorem, substituting the flux condi- tions (5.22) and (5.23), and noting that A 1 2π 2 dθ and dℓ =( 2 +(∂ /∂θ)2)1/2dθ, α= 2 0 R R R R 80 5.2 Mean temperature we find

2 1 d dT0 K T + · Re [K∗ T ] + η K dS 0∇τ 2,0 2 ∇τ 1 ∇τ 1 dX 0 dX ZAg    2 1 d dT0 + σ K τ T + τ · Re K∗ τ T + η K dS s0 ∇ s2,0 2 ∇ s1 ∇ s1 dX s0 dX ZAs      1 σ · = K0 τ T2,0 + Re [K1∗ τ T1] σKs τ Ts Re Ks∗ τ Ts nτ dℓ Γ ∇ 2 ∇ − 0 ∇ 2,0 − 2 1 ∇ 1 Z g     1 · + σ Ks Ts + Re Ks∗ τ Ts nτ′ dℓ Γ 0 ∇ 2,0 2 1 ∇ 1 Z t   d dT  d  dT + ηA K 0 + ησ A K 0 g dX 0 dX s dX s0 dX     2π ∂ 2π dT g dT ∂ = η(K σK ) 0 R dθ + ησK 0 Rt dθ 0 s0 g ∂ s0 t ∂ − dX Z0 R X dX Z0 R X d dT d dT + ηA K 0 + ησ A K 0 g dX 0 dX s dX s0 dX     d dT = η K A + σK A 0 . (5.24) dX 0 g s0 s dX    On the other hand, combining (5.20) and (5.21), applying the divergence theorem and using v Γ = 0, we also have | g

2 1 d dT0 K T + · Re [K∗ T ] + η K dS 0∇τ 2,0 2 ∇τ 1 ∇τ 1 dX 0 dX ZAg    2 1 d dT0 + σ K τ T + τ · Re K∗ τ T + η K dS s0 ∇ s2,0 2 ∇ s1 ∇ s1 dX s0 dX ZAs    2   NL d = ρ0Re [u1∗h1] + 2h0m˙ 2 dS κ g dX ZA   2 2 NL κµ0 2NL + τ · ρ Re [v∗h ] + 2h m˙ τ Re [u∗ τ u ] F˙ dS κ ∇ 0 1 1 0 2 − 2 1∇ 1 − κ 2 ZAg " Wo ! # 2 ˙ NL d dH2 = A ρ Re [ u∗h ] + 2h M˙ 2 , (5.25) κ dX g 0 h 1 1i 0 2 − dX   ˙ ˙ 1 dH˙ where we put M2 := Ag m˙ 2 and F =: , . Finally, equating the right hand sides of h i Ag dX equations (5.24) and (5.25), we get

2 ˙ d ˙ Sk dT0 dH2 2h0 M2 + Agρ0Re [ h1u1∗ ] (K0 Ag + σKs As) = . (5.26) dX h i − 0 κ dX ! dX

˙ The quantity within the big brackets is H2, the time-averaged total power (or energy flux) along X. In steady state, for a cyclic refrigerator or prime mover without heat flows to the surroundings (F˙ = 0), H˙ will be constant. Thermoacoustics in three-dimensional pores with variable cross-section 81

Substituting the thermodynamic expressions (A.7) and (A.9),

1 1 dh = Tds + dp = C dT + (1 βT)dp, ρ p ρ − into (5.26), we find ˙ ˙ H2 M2 1 1 = h0 + ρ0CpRe [ T1u1∗ ] + (1 βT0)Re [p1 u1∗ ] Ag Ag 2 h i 2 − h i A S2 dT K + σK s k 0 . (5.27) 0 s0 − Ag ! 2κ dX

Next we will show that

2 Fk j + Pr Fν∗ fb j Fν = Re [ Fν ] , Fk jFν∗ = h i h i − , j = u, p, (5.28) h| | i h i h i 1 + Pr where 1 · fb j := 2 Fk j τ Fν∗ nτ dℓ, j = u, p. Γ ∇ Agαν Z g Define I := F F∗ , I := F · F∗ . (5.29) 1 h k j ν i 2 h∇τ k j ∇τ ν i It follows from (5.10) and (5.4) that

1 2 Fν = 1 + 2 τ Fν, (5.30) αν ∇ 1 2 Fk j = 1 + 2 τ Fk j, j = u, p. (5.31) αk ∇

Using (5.31), the divergence theorem, and the boundary condition Fν = 0 on Γg, we can rewrite I1 as follows:

1 1 2 I = Fν∗ 1 + τ F dS 1 A α2 ∇ k j g ZAg k ! 1 1 ∗ · ∗ ∗ · = Fν + 2 τ Fν τ Fk j dS 2 τ Fν τ Fk j dS h i A α g ∇ ∇ − A α g ∇ ∇ g k ZA   g k ZA 1 I2 ∗ ∗ · = Fν + 2 Fν τ Fk j nτ dℓ 2 h i A α Γg ∇ − α g k Z   k I2 = Fν∗ 2 . (5.32) h i − αk 82 5.2 Mean temperature

2 2 In the same way, substituting (5.30) rather than (5.31) and noting (α∗) = α , we find ν − ν

I2 I1 = Fk j fb j + 2 , (5.33) h i − αν where the extra term fb j arises because Fk j does not necessarily vanish on the interface Γg. Eliminating the common term I2 from equations (5.32) and (5.33), we find

Fk j + Pr Fν∗ fb j I1 = h i h i − , (5.34) 1 + Pr as claimed in (5.28). Replacing Fk j by Fν, we can repeat the same analysis to prove the first claim of (5.28). Integrating (A.9), we also find that

h = C (T T ), 0 p 0 − re f where Tre f is some reference temperature. Substituting (5.8) into (5.27) and using (5.28) we find after some manipulation

˙ f f ∗ + f H2 1 ˙ 1 kp ν bp = M2Cp T0 Tre f + Re p1u1∗ 1 βT0 − Ag Ag − 2 " − (1 + Pr)(1 fν∗) !#   − ρ 2 0Cp u1 dT0 fku fν∗ + fbu |h i| ∗ + 2 Im fν + − 2κ(1 P ) 1 f dX 1 + Pr − r | − ν|   A S2 dT K + σK s k 0 . (5.35) 0 s0 − Ag ! 2κ dX

This expression represents the total power along the X-direction (wave direction) in terms of T , p , u , M˙ , material properties and geometry. Given H˙ , independent of 0 1 h 1i 2 2 X, we can solve (5.35) for dT0/dX,

˙ ˙ dT 2H2 2M2Cp T0 Tre f AgRe [a1 p1u1∗] 0 = κ − − − , (5.36) 2  2 dX A a u K A + σK A S g 2|h 1i| − 0 g s0 s k   where

fkp fν∗ + fbp a1 := 1 βT0 − , (5.37) − (1 + P )(1 f ∗) r − ν ρ 0Cp fku fν∗ + fbu ∗ a2 := 2 Im fν + − . (5.38) (1 P ) 1 f 1 + Pr − r | − ν|   Equation (5.36) together with equations (5.18b) and (5.18a) form a complete coupled sys- tem of differential equations for T0, u1 , and p1. This system gives a generalization of the Swift equations [135], for arbitraryh three-dimensionai l slowly varying cross-sections. It is possible to go a step further and improve the expression for the mean temper- Thermoacoustics in three-dimensional pores with variable cross-section 83

ature by determining the correction term T2,0. Integrating (5.20) and (5.21) using an appropriate Green’s function, we can determine T and T up to some X-dependent 2,0 s2,0 function T (= T (X)). However, before we can determine T , we need to com- b2,0 b2,0 b2,0 pute the second and third harmonics, as T follows as a solvability condition for the b2,0 fourth-order mean temperatures T and T . 4,0 s4,0

5.3 Integration of the generalized Swift equations

In equations (5.18a), (5.18b) and (5.36) we have derived a coupled system of differential equations for the mean temperature T , the acoustic velocity u , and the acoustic pres- 0 h 1i sure p1. It follows that that all acoustic variables can be determined after integration of

˙ ˙ dT 2H2 2M2Cp T0 Tre f AgRe [a1 p1 u1 ∗] 0 = κ − − − h i , (5.39a) dX A a u 2 K A + σK A S2 g 2|h 1i| − 0 g s0 s k   d u1 dT0 1 dAg h i = κa3 p1 + a4 u1 , (5.39b) dX dX − Ag dX ! h i dp 1 = κa u , (5.39c) dX 5h 1i with

fkp fν∗ + fbp a1 := 1 βT0 − , (5.40a) − (1 + P )(1 f ∗) r − ν ρ 0Cp fku fν∗ + fbu a2 := 2 Im fν∗ + − , (5.40b) (1 P ) 1 f 1 + Pr − r | − ν|   i a3 := 1 +(γ 1) fkp , (5.40c) −ρ c2 − 0 h i β( f f ) a := ν − ku , (5.40d) 4 − (1 P )(1 f ) − r − ν iρ a := 0 . (5.40e) 5 − 1 f − ν Equations (5.39)form a system of five coupled equations, determining the five real vari- ˙ ables: Re(p1), Im(p1), Re( u1 ), Im( u1 ), and T0. Given the total energy flux H2, the ˙ h i h i mass flux M2, the geometry, and appropriate boundary conditions in X, these equations can be integrated numerically. This system of equations has a similar structure as the one we obtained for two- dimensional pores in (4.70). The difference lies in the a j functions that are defined in a slightly different way. The analytic solutions obtained in Sections 4.4.1 and 4.4.2 can therefore be generalized to three-dimensional pores straightforwardly, by replacing the a j functions by those given above. It is true that (5.40) is more difficult to compute than (4.71), but below we will discuss two cases for which the a j functions simplify greatly. 84 5.3 Integration of the generalized Swift equations

5.3.1 Ideal stack

A stack is considered ideal if the pore walls are of sufficiently high heat capacity and thermal conductivity, so that the wall temperature is locally unaffected by the acoustic temperature variations in the gas. In short, an ideal stack has T = 0. Although in s1 reality stacks are almost never ideal, they can get close under certain conditions. For example in figure 4.2 we showed that a parallel-plate stack can approach an ideal stack (with εs = 0) provided the stack plates are thick enough. If we put the assumption of T = 0 into our boundary conditions, then it follows s1 that gu = gp = 0, so that Fsu = Fsp = 1, Fkp = Fku = Fk and fbu = fbp = 0. As a result the a j functions can be expressed as

fk fν∗ a1 := 1 βT0 − , − (1 + P )(1 f ∗) r − ν ρ 0Cp fk fν∗ a2 := 2 Im fν∗ + − , (1 P ) 1 f 1 + Pr − r | − ν|   i γ a3 := 2 [1 +( 1) fk] , −ρ0c − β( f f ) a := ν − k , 4 − (1 P )(1 f ) − r − ν iρ a := 0 . 5 − 1 f − ν

These expressions match the a j functions obtained for two-dimensional pores in (4.71) in case εs = 0.

5.3.2 Rotationally symmetric pores

Another simplification arises if we consider pores that are rotationally symmetric, such as cylindrical or conical pores. Then the acoustic temperature fluctuations at the solid/gas interface will only depend on X, so that the gi functions can be computed as indicated in (C.11). Putting 1 K0 Agαk tanh(αk) ε := , s σ K A α tanh(α ) s0 s s s Thermoacoustics in three-dimensional pores with variable cross-section 85

we can show that the a j functions can be expressed in the familiar form

fk fν∗ a1 := 1 βT0 − , − (1 + P )(1 + ε )(1 f ∗) r s − ν ρ 2 0Cp u1 (1 + εs fν/ fk)( fk fν∗) |h i| ∗ a2 := 2 Im fν + − , (1 P ) 1 f (1 + Pr)(1 + εs) − r | − ν|   i (γ 1) f a := 1 + − k , 3 −ρ 2 1 + ε 0c  s  ( f f )β a := ν − k , 4 − (1 P )(1 + ε )(1 f ) − r s − ν iρ a := 0 . 5 − 1 f − ν

5.4 Acoustic streaming

In this section we generalize the analysis, derived first in Section 4.5 for two-dimensional pores, to three-dimensional pores. We discuss steady second-order mass flow in the stack driven by first-order acoustic phenomena. The analysis is valid for arbitrary slowly varying pores supporting a temperature gradient. Moreover, the temperature dependence of viscosity is taken into account. We start with the continuity equation (3.16). If we time-average the equation and expand in powers of Ma, then the zeroth and first order terms in Ma will drop out. Consequently we find to leading order

∂ 1 ∂ ρ u + ρ τ · vτ + Re (ρ u∗) + τ · ρ vτ∗ = 0. ∂X 0 2,0 0∇ 2,0 2 ∂X 1 1 ∇ 1 1  

Again applying the divergence theorem and noting that vτ Γ = 0, we can average over | g a cross-section to find

d Ag A ρ u + Re [ ρ u∗ ] = 0. dX g 0h 2,0i 2 h 1 1i   ˙ The expression between the brackets is M2 the time-averaged and cross-sectional-averaged ˙ mass flux in the X-direction. It follows that M2 is constant, which is to be expected as there is no mass transport through the stack walls. We can now express u2,0 in terms ˙ h i of M2 and the first order acoustics as follows:

˙ 1 M2 1 u2,0 = Re [ ρ1u1∗ ] . (5.41) h i ρ0 Ag − 2 h i !

Next we turn to equation (3.18). Expanding in powers of Ma and ε and averaging in time we find to leading order in Ma and ε that

p = 0, ∇τ 2,0 86 5.5 Second harmonics

so that p2,0 = p2,0(X). Subsequently, time-averaging equation (3.17), we find to leading order

2 2 Wo dp2,0 τ u2,0 = f , (5.42) ∇ − κµ0 dX where f is given by

2 1 Wo ρ0 1 µ1∗ f := Re iκρ u∗ + ρ v∗ · u b Re · u . 2 κµ 1 1 0 1 ∇ 1 − F x − 2 ∇τ µ ∇τ 1 0  r    0  The first-order acoustics collected in f can be interpreted as a source term for the stream- ing on the left hand side, with the last term being characteristic for Rayleigh streaming. We can also see this as a Poisson’s equation for the streaming velocity u2,0, which may be solved using a Green’s function. Introducing the Green’s function Gm that, for fixed ˆx (X), satisfies τ ∈ Ag G (x; ˆx) = δ(x ˆx ), x (X), (5.43) ∇τ m τ − τ τ ∈ Ag G (x; ˆx) = 0, (x) = 0, (5.44) m Sg we can write

2 p Wo d 2,0 ˆ ˆ u2,0(x) = Gm(x; ˆx) dS + Gm(x; ˆx) f (ˆx) dS. (5.45) κµ0 dX g(X) g(X) ZA ZA Computing the cross-sectional average we can relate dp /dX to u as follows: m20 h m,20i ˆ dp κµ u2,0 Gm(x; ˆx) f (ˆx) dS 2,0 = 0 h i−h Ag i . (5.46) 2 ˆ dX Wo R Gm(x; ˆx) dS h Ag i R ˙ Summarizing, given the mass flux M2 and the first-order acoustics, it only remains to compute the Green’s function G for the desired geometry. Then u , dp /dX m h 2,0i 2,0 and u2,0 can be determined consecutively from (5.41), (5.46) and (5.45).

5.5 Second harmonics

We start with generalizing the auxiliary functions given in (4.9), (4.17), and (4.18)

α j,2 := √2α j, j = ν, k, s. (5.47)

Furthermore, for given source function f , we introduce functions Ψ j( f ) that satisfy

1 Ψ 2Ψ j,2 2 τ j,2 = f in g, (5.48a) − α j,2 ∇ A

Ψ j,2 = 0 on Γg. (5.48b) Thermoacoustics in three-dimensional pores with variable cross-section 87

We expand all variables in powers of Ma according to (3.24), and substitute the expansions into the transverse component of the momentum equation (3.18). Putting 2 2 ε = ηMa , and collecting terms of second order in Ma we find

p = 0, ∇τ 2,2 and we conclude p2,2 must be independent of xτ . Similarly, collecting all second-order terms in Ma, we find that the X-component of the momentum equation (3.17) reduces into ∂ 1 1 u1 1 dp2,2 2iκρ u + iκρ u + ρ u + ρ vτ · τ u = 0 2,2 2 1 1 2 0 1 ∂X 2 0 1 ∇ 1 − dX κµ 1 µ + 0 2u + · 1 u , (5.49) 2 ∇τ 2,2 2 ∇τ µ ∇τ 1 Wo   0  where we substituted (3.26). Note that the steady component from this equation was chosen such that it vanishes. Rearranging terms we can rewrite (5.49) as

1 i dp u 2u = 2,2 + A, (5.50) 2,2 2 τ 2,2 κρ − αν,2 ∇ 2 0 dX where A, a source term arising from products of first-order terms, is known and given by ∂ i u1 κµ0 µ1 A = iκρ u + ρ u + ρ vτ · τ u τ · τ u . 4κρ 1 1 0 1 ∂X 0 1 ∇ 1 − 2 ∇ µ ∇ 1 0 ( Wo  0 ) The solution, satisfying the no-slip condition, can be written as

iFν,2 dp2,2 i(1 fν,2) dp2,2 u2,2 = + Ψν,2(A), u2,2 = − + ψν,2(A). (5.51) 2κρ0 dX h i 2κρ0 dX Next we turn to the temperature equation. Substituting our expansions into (3.21) and (3.22) and collecting the terms of second order in Ma, we find after some manipula- tion that T and T can be found from 2,2 s2,2

1 1 dT dp βT T 2T = B 0 2,2 F + 0 p , (5.52a) 2,2 2 τ 2,2 2 ν,2 ρ 2,2 − αk,2 ∇ − 4κ ρ0 dX dX 0Cp 1 T 2T = C, (5.52b) s2,2 2 τ s2,2 − αs,2 ∇ 88 5.5 Second harmonics where B and C are known and given by

dp dT ∂T dT 4B = iβκT p + T u 1 iκC ρ T 2ρ C Ψ (A) 0 ρ u 1 ρ u 0 1 1 0 1 dX − p 1 1 − 0 p ν,2 dX − 0 1 ∂X − 1 1 dX κ d dT0 κµ0 2 ρ vτ · τ T + τ · (K τ T ) + 2η K + τ u , − 0 1 ∇ 1 2 ∇ 1∇ 1 dX 0 dX 2 |∇ 1| 2NL   Wo κφ d dT0 4C = τ · K τ T + 2η K . 2 ∇ s1 ∇ s1 dX s0 dX 2Ns   
As in Section 5.1 we denote the temperature on the boundary by g, in particular g2,2 := T Γ , and we write 2,2| g

βT0 gu,2 dT0 dp2,2 g2,2 = gp,2 p2,2 . ρ c2 − 4κ2(1 P )ρ dX dX 0 − r 0 Using (5.48) and imposing the boundary conditions given in (3.23), we find

βT F P F dT dp Ψ 0 ku,2 r ν,2 0 2,2 T2,2 = k,2(B) + Fkp,2 p2,2 2 − , (5.53a) ρ0Cp − 4κ (1 P )ρ dX dX − r 0 βT 1 F dT dp T = Ψ (C) + 0 (1 F )p − su,2 0 2,2 , (5.53b) s2,2 s,2 sp,2 2,2 2 ρ0Cp − − 4κ (1 P )ρ dX dX − r 0 where Fk j,2 (j = p, u) satisfies

1 2 Fk j,2 2 τ Fk j,2 = 1 in g, (5.54a) − α j,2 ∇ A

Fk j,2 = g j,2 on Γg, (5.54b) and Fs j,2 (j = p, u) is found from

1 2 Fs j,2 2 τ Fs j,2 = 1 in s, (5.55a) − αs,2 ∇ A F = 1 g on Γ , (5.55b) s j,2 − j,2 g F · n′ = 0 on Γ . (5.55c) ∇τ s j,2 τ t

Using relations (A.5) and (A.8) and substituting T2,2 we can derive the following relation for the second-harmonic density fluctuations:

1 1 ρ ρ γ ρ β Ψ 1 2,2 = 2 1 +( 1)(1 Fkp,2) p2,2 0 k,B + T1 c − − − 2 ρ0 h i   Fku,2 Fν,2 β dT0 dp2,2 + Fν,2 + − 2 . (5.56) 1 Pr 4κ dX dX  −  Thermoacoustics in three-dimensional pores with variable cross-section 89

As a result we find

1 f f β dT dp ρ γ ν,2 ku,2 0 2,2 2,2 = 2 1 +( 1) fkp,2 p2,2 + 1 fν,2 + − 2 h i c − − 1 Pr 4κ dX dX h i  −  1 ρ β ψ + ρ T . (5.57) − 0 k,B 2ρ h 1 1i  0  Finally, we turn to the continuity equation (3.16). Expanding the variables in powers of Ma and collecting terms of second order in Ma we find ∂ ∂ 2iκρ + (ρ u ) + ρ τ · vτ = (ρ u ) τ · (ρ vτ ). (5.58) 2,2 ∂X 0 2,2 0∇ 2,2 − ∂X 1 1 − ∇ 1 1

We know that v vanishes at Γg. Therefore, substituting (5.51) for u2,2, multiplying with iκ, and averaging over a cross-section, we obtain the following equation as a consis- − tency relation for v2,2:

1 d dp 2iκ ∂ 4κ2 ρ + A (1 f ) 2,2 = (A ρ u + ρ ψ (A) ). h 2,2i A dX g − ν,2 dX A ∂X g h 1 1i 0 ν,2 g   g   After substituting (5.57), we obtain

4κ2 f f dT dp dT dp γ ν,2 ku,2 β 0 2,2 β 0 2,2 2 1 +( 1) fkp,2 p2,2 + − +(1 fν,2) c − 1 Pr dX dX − dX dX h i − 1 d dp + A (1 f ) 2,2 = 2iκρ E, (5.59) A dX g − ν,2 dX − 0 g   where E is a source term arising from products of first-order or zeroth-order quantities and is given by

1 1 ∂ E := 2iκβ ψ + ρ T A ρ ψ (A) + ρ u . k,B 2ρ h 1 1i − ρ A ∂X g 0 ν,2 h 1 1i  0  0 g    Inserting (4.43), we obtain a wave equation for the second pressure harmonic

4κ2 ( f f )β dT dp γ ν,2 ku,2 0 2,2 2 1 +( 1) fkp,2 p2,2 + − c − 1 Pr dX dX h i − ρ d 1 f dp + 0 A − ν,2 2,2 = 2iκρ E. (5.60) A dX g ρ dX − 0 g  0  Apart from the source term E this equation has a similar structure as the wave equation derived in (4.44) for the first pressure harmonic. Combining (5.51) and (5.60) we derive the following two coupled ordinary differen- 90 5.6 Power tial equations for p and q := u ψ (A) 2,2 2,2 h 2,2i − ν,2

dq dT 1 d g 2,2 = κa p + a 0 R q + E, (5.61a) dX 3,2 2,2 4,2 dX − dX 2,2 Rg ! dp 2,2 = κa q , (5.61b) dX 5,2 2,2 where 2i a3,2 := 1 +(γ 1) fkp,2 , (5.62a) −ρ c2 − 0 h i ( f f )β a := ν,2 − ku,2 , (5.62b) 4,2 − (1 P )(1 f ) − r − ν,2 2iρ a := 0 . (5.62c) 5,2 − 1 f − ν,2 Since all the zeroth-order and first-order terms are given by the equations in the pre- vious sections, we can compute subsequently A, B, C, and E. Moreover, Appendix C shows how the Fj and Ψ j functions can be computed for a given cross-sectional geom- etry. Having done this, we can integrate the system (5.61) to determine p2,2 and u2,2 , provided appropriate boundary conditions are imposed. h i

5.6 Power

In Section 5.2 we derived the following expression for the total power H˙ :

˙ f f ∗ + f H2 1 ˙ 1 kp ν bp = M2Cp T0 Tre f + Re p1u1∗ 1 βT0 − Ag Ag − 2 " − (1 + Pr)(1 fν∗) !#   − ρ 2 0Cp u1 dT0 fku fν∗ + fbu |h i| ∗ + 2 Im fν + − 2κ(1 P ) 1 f dX 1 + Pr − r | − ν|   A S2 dT K + σK s k 0 . (5.63) 0 s0 − Ag ! 2κ dX

As mentioned in Section 4.7.2, the total power can be written as a sum of the acoustic ˙ ˙ ˙ power W, the hydrodynamic entropy flux Q, the heat flow Qm due to a nett mass flux, ˙ and the heat flow Qloss due to conduction down a temperature gradient. For the time- ˙ averaged acoustic power dW2 used or produced in a segment of length dX, second order in Ma, we can write

dW˙ d 2 = A Re p eiκt Re u eiκt . (5.64) dX dX gh 1 h 1i i  h i h i  Thermoacoustics in three-dimensional pores with variable cross-section 91

Using (3.25), we find to leading order

˙ dW2 1 dAg Ag d u1∗ dp1 = Re [p u ∗ ] + Re p h i + u ∗ . (5.65) dX 2 dX 1h 1 i 2 1 dX h 1 i dX   Substituting (5.18b) and (5.18a) into (5.65) we find

˙ dW2 Ag β dT0 fku∗ fν∗ Ag κ(γ 1) 2 ∗ = Re − p1 u1 −2 Im fkp p1 dX 2 1 Pr dX (1 fν∗) h i − 2 ρ c − | | −  −  0 h i Ag κρ0Im [ fν] 2 − u1 . (5.66) − 2 1 f 2 |h i| | − ν| 92 5.6 Power Chapter 6

Standing-wave devices

In this chapter we present the results from our computations on standing-wave devices. We will simulate both refrigerators and prime movers and where possible we will com- pare our numerical results to analytic approximations and experimental data. The in- fluence of geometry parameters, material and other parameters such as drive ratio and mean pressure is investigated. The performance of the machines is tested in terms of the temperature difference, power-output, and efficiencies.

6.1 Design

We consider two kinds of standing-wave devices: the thermoacoustic refrigerator and the thermoacoustic prime mover. As depicted in figure 6.1, they are modeled in three parts: an acoustically resonant tube, containing a gas, a parallel-plate stack, and two heat exchangers. In case of a thermoacoustic refrigerator a driver is attached to one end, and the other end is closed. In case of a prime mover, one end will be closed and one will be open. When designing a thermoacoustic system several parameters are important, related to the choice of material, working gas, geometry, and operating conditions. Below we will discuss a few of these parameters that are important in our computations.

Stack position

Perhaps the most important design parameter is the position of the stack. Without a stack, a standing-wave would be maintained in the resonator, with velocity nodes at the closed ends and pressure nodes at the open ends. With a stack, the standing-wave will be altered, but if the stack is short enough, then the standing wave will not be perturbed appreciably, as we saw in our derivation of the short stack approximation in Section 4.4.2. The distance between the stack center and the closed end is denoted by xs. Using the wave number k = 2π/λ, we can introduce a dimensionless number kxs that relates the stack to its position in the acoustic wave. Since the power output is proportional to the product p1 u1 , it follows that the stack has to be positioned somewhere between a pressure and ah velocityi node. 94 6.1 Design

xs xs

TH TC TH TC sound sound

˙ ˙ ˙ ˙ QH QC QH QC

(a) Standing-wave refrigerator (b) Standing-wave prime mover

Figure 6.1: Schematic model of (a) thermoacoustic refrigerator and (b) standing-wave ther- moacoustic prime mover, illustrated with the standing-wave pressure and velocity profiles. The model consists of an acoustically resonant tube filled with gas, a stack of parallel plates positioned at distance xs from the closed end, and two heat exchangers. In (a) we have a half-wave-length resonator attached to a speaker and in (b) we have a quarter-wave-length resonator that supplies sound to the exhaust.

Stack length

As mentioned previously the stack length is vital in the design of a thermoacoustic stack. First of all, the stack should be kept short with respect to the wave length, so that the acoustic field is not significantly altered. Furthermore, as the stack length is increased, the viscous and thermal dissipation will increase as well, reducing the efficiency sig- nificantly. On the other hand, if the stack length is increased then more heat can be pumped by the stack, so that higher temperature differences and larger power-output can be obtained. The optimal stack length is obtained by balancing these effects.

Plate separation

In standing-wave systems it is beneficial to have pores with a radius of one or more thermal penetration depths, i.e. NL 1. This is necessary to create the optimal phasing between pressure and velocity and create≥ an optimal heat-shuttling effect. It was shown in equations (4.136) and (4.138) that (in an idealized situation) the highest acoustic pow- ers are obtained if N 1. L ∼ Standing-wave devices 95

Plate material ˙ The term Qloss in the expression for the total power (4.140) represents the heat conduc- tion through gas and plate material and it is a loss term; it has a negative effect on the performance. To reduce the effect of this term a material must be chosen with a low thermal conductivity. On the other hand, we must take a heat capacity Cs larger than the heat capacity Cp of the working gas, so that the plate temperature can be considered steady (εs = 0).

Drive ratio

The drive ratio Dr is defined as the ratio between the dynamic pressure amplitude pA (at the closed end) and the mean pressure p0,

pA Dr = , p0 and is a measure for the amplitude of the sound field. In general one is interested in high drive ratios because this leads to larger power outputs and larger temperature dif- ferences. However, at high drive ratios nonlinear effects will start to become important that may degrade the performance. Turbulence may occur, but also shock waves due to interaction of the higher harmonics.

6.2 Computations

We have implemented the system (4.70) for the configuration shown in figure 6.1(a). There are three regions for which a solution has to be computed: the resonator, the heat exchangers, and the stack. Each region requires a different approach.

Resonator We consider an insulated resonator, that is kept at a constant temperature, without heat leaks to the environment. Although in reality the resonator will be cylindrical we will model it as straight two-dimensional channel to simplify the calculations. In Section 4.4.1 it was shown that for straight two-dimensional pores the exact solution looks as follows:

ikrX ikrX p1(X) = Ae− + Be , (6.1a) χ r ikrX ikrX u1 (X) = Ae− Be , (6.1b) h i ρ0c −   with A and B integration constants and

1 γ 1 χ = (1 f ) 1 + − f , (6.2) r c − ν 1 + ε k s  s  κ 1 γ 1 k = κ √ a a = 1 + − f , (6.3) r r − 3 5 c 1 f 1 + ε k s − ν  s  96 6.3 A thermoacoustic couple

where χr and kr are computed from the resonator properties. The viscous dissipation in the resonator is accounted for as the wave number kr will have a small imaginary part.

Heat exchangers The heat exchangers are used to exchange heat with the environment. In our compu- tations they are modeled as infinitely short with temperature TH at the hot end and temperature TC at the cold end.

Stack Inside the stack there will be a nontrivial temperature difference and the full system of equations (4.70) needs to be solved. For this standard ode-solvers from MatLab can be used. The result can be compared to the short-stack approximation derived in Section 4.4.2, provided the stack is short enough and H˙ = 0.

Boundary and interface conditions To fix all the integration constants, boundary conditions have to be imposed. We start our computations at the closed end, where we give the pressure amplitude and impose zero velocity. At the interface with the stack we impose continuity of mass and momen- tum, which gives continuity of pressure p1 and volumetric velocity Ag u1 . We still need boundary conditions for the temperature. There areh twoi constants of integration and so we can choose to fix the temperature at the heat exchangers on either side of the stack. Alternatively, one can also impose the temperature on one side and a cooling (or heating) power, and the temperature on the other side will follow. Lastly, we mention that typically there is no time-averaged mass flux M˙ in the sys- tem. With a nonzero M˙ there would be an accumulation or depletion of mass at the closed end, which is not physically realistic. In the looped geometries that we consider in Chapter 7, nonzero mass fluxes may be present.

6.3 A thermoacoustic couple

The simplest type of thermoacoustic devices is the so-called thermoacoustic couple (TAC), a stack of short parallel plates without heat exchangers positioned in a half- wave-length resonator, with a driver providing a standing-wave. It is of the type shown in figure 6.1(a), but without heat exchangers. Due to its simplicity a thermoacoustic couple is perfect for a quantitative under- standing of the basic thermoacoustic effect. There are no heat exchangers, but a steady- state temperature profile is developed across the stack from a balance of the thermoa- ˙ ˙ ˙ coustic heat flow (W + Q) by a return diffusive heat flow (Qloss) in the stack and in the gas. The resulting energy flux through a stack pore will be equal to zero (H˙ = 0). The first measurements of thermoacoustic couples were performed by Wheatley et al. [151], who measured the temperature difference developed across the stack as a function of its position kxs for drive ratios of approximately 0.3%. Additionally, they derived a theoretical prediction (see (4.90)) for the temperature difference, using a boundary-layer and short-stack approximation, which was later slightly modified by Standing-wave devices 97

Atchley et al. [11]. Atchley et al also extended the measurements to higher driver ratios of up to 2%. We will validate our numerical code by comparison against these measure- ments.

6.3.1 Acoustically generated temperature differences We consider the thermoacoustic couple TAC #3 as described in [11]. It comprises a parallel-plate stack placed in helium-filled resonator with plates that are a lamination of 302 stainless steel and G-10 fiberglass, epoxied together. All relevant parameters are given in Table 6.1.

Parameter Symbol Value Unit total power H˙ 0 W mass flux M˙ 0 kg m/s speed of sound c 1020 m/s isobaric specific heat Cp 5190 J/(kg K) isochoric specific heat Cv 3110 J/(kg K) 3 plate separation dg 1.52 10− m · 4 thickness fiberglass d f g 1.02 10− m · 5 thickness stainless steel dss 8.89 10− m · 4 thickness plate d = d + d 1.91 10− m s f g ss · frequency f 696 Hz thermal conductivity Helium Kg 0.16 W/(m K) thermal conductivity fiberglass K f g 0.48 W/(m K) thermal conductivity stainless steel Kss 11.8 W/(m K) d K +d K thermal conductivity plate K = fg fg ss ss 5.76 W/(m K) s ds 3 stack length L 6.85 10− m s · average pressure p 1.14 105 Pa 0 · Prandtl number Pr 0.68 - ambient temperature TL 298 K

Table 6.1: Specifications for TAC #3

Using the procedure given in Section 6.2, we have calculated numerically the ex- pected temperature differences developed across the stack in case H˙ = 0 for various drive ratios. In figure 6.2 we plot the temperature difference across the stack as a func- tion of kxs, the relative position of the stack center for drive ratios of 0.28 and 1.99%. The numerical outcome is compared with the measurements of [11], the short-stack ap- proximation (4.89) and the short-stack/boundary-layer approximation (4.90). For this configuration we have that κ 0.015 1 and NL 2.75. It is therefore no surprise that the short-stack approximation∼ and the≪ numerics agree∼ very well. Further- more, the match with (4.90) is remarkably good, even though NL is not much greater than 1. The match with the measurements is not bad either, but for high drive ratios the fit seems to become worse; both the numerics and the theoretical predictions over- predict the measured values. In [11] this is attributed to uncertainties in the thermal conductivity of the stack material or possible measurement errors. 98 6.3 A thermoacoustic couple

0.3 6 numerics numerics 0.2 short stack approx. 4 short stack approx. Wheatley et al. Wheatley et al. Atchley et al. 0.1 2 Atchley et al.

0 0

−0.1 −2

temperature difference (K) −0.2 temperature difference (K) −4

−0.3 −6 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0.5 1 1.5 2 2.5 3 3.5 4 4.5 kxs kxs (a) Dr = 0.28% (b) Dr = 1.99%

Figure 6.2: The temperature difference across the stack as a function of the relative position of the stack center. The solid line shows the results from the numerics, the dashed line shows the short-stack approximation (4.89), the squares shows the values predicted by Wheatley et al. , and the triangles show the measurements of [11].

In an article by Piccolo et al. [99], a simplified numerical model was derived for de- scribing time-average transverse heat transfer near the edges of a thermally isolated thermoacoustic stack at low Mach numbers. The difference with the approach used here lies in the inclusion of a y-dependence in the mean temperature and various simplify- ing assumptions such as constant pressure and velocity inside the stack. This model was compared to the experimental data of Atchley et al. [11] and to what Picollo et al. call standard thermoacoustic theory. The latter is an approximation similar to the ap- proximation of Wheatley et al. [151]. Figure 6.3 show a close-up of figure 6.2(b) near the maximal temperature difference and includes the results of [99].

6 7 rence (K) 4 iffe e d

2 peratur tem

0 1,5 2 2,5 3 kxs

Figure 6.3: Close-up of figure 6.2(b) near the maximal value for Dr = 1.99%. The results are compared to those of Piccolo et al [99]. Standing-wave devices 99

To show the effect of increasing drive ratio we can compute for various drive ra- tios the temperature difference numerically. In figure 6.4 the results are compared to those obtained experimentally in [11] for various drive ratios between 0.17% and 1.99%. Comparing the results, we again observe reasonable agreement, with larger deviations for high drive ratios. Both in figure 6.4(a) and figure 6.4(b) the temperature difference progresses from a perfect sinusoid for small amplitudes to a sawtooth curve for large drive ratios with extremes shifting towards the pressure antinodes.

6

4

2

0

−2

−4 temperature difference (K) temperature difference (K)

−6 0.5 1.5 2.5 3.5 4.5 kx kxs s (a) Numerics (b) Measurements

Figure 6.4: The (a) computed and (b) measured temperature difference across the stack as a function of the relative position of the stack for drive ratios ranging between 0.17 and 1.99%.

Finally we also look at the impact of the stack on the velocity and pressure field in the tube. Fig. 6.5 shows the pressure and velocity amplitude throughout the tube with the stack positioned at kxs = π/4 together with the pressure and velocity amplitudes in the absence of the stack. The plot shows that the pressure and velocity are hardly effected by the stack because it is so short. Note the discontinuities in the velocity at the stack ends which arise to maintain the same volumetric mass flux, as the cross-sectional area of the gas is smaller in the stack than in the resonator.

6.3.2 Acoustic power

Since a thermoacoustic couple does not have heat exchangers, the heating and cooling power will be equal to zero and the coefficient of performance has no meaning. How- ever, the acoustic power will be nonzero and is necessary to obtain the steady-state temperature profile. In Section 4.7.1 we showed that the acoustic power produced or absorbed in a segment of length dX is composed of a source/sink term, a viscous dissi- pation term, and a thermal relaxation term,

dW˙ dW˙ s dW˙ k dW˙ ν 2 = 2 2 2 dX dX − dX − dX (6.4) 2 1 = (1) NL 2 , Wo, NL 1. O − O − O W ! ≪   o 100 6.4 A standing-wave refrigerator

2500 14 with stack with stack without stack 12 without stack 2000 10

1500 8 (Pa) (Pa) 1 1 p 1000 p 6 4 500 2

0 0 0 π/4 π/2 3π/4 π 0 π/4 π/2 3π/4 π kxs kxs (a) pressure amplitude (b) velocity amplitude

Figure 6.5: (a) The absolute value of the acoustic pressure and (b) the absolute value of the crosswise-averaged velocity throughout the tube both without a stack and with a stack positioned π at kxs = 4 . The drive ratio is 1.99 %.

To test this equation for a real application, we consider the thermoacoustic couple as introduced in the previous section. We put the stack at 5 cm from the closed end, apply a drive ratio of 2%, and vary NL. The remaining parameters are chosen as in Table 6.1. ˙ In figure 6.6 we have plotted the acoustic power ∆W2 absorbed by the stack together with its source term and dissipation components. ˙ k Looking at the graph of the thermal relaxation dissipation ∆W2 , we indeed observe ˙ k ˙ k that ∆W2 tends to zero for decreasing NL, but only until NL 0.1. Below this value ∆W2 ˙ k 2∼ starts to grow rapidly again because ∆W2 scales with p1 . As the pore size becomes smaller and smaller, the pressure drop (and also the velocit| | y) in the stack will become larger and larger, canceling the effect of the prefactor Im( fk). For the viscous dissipation ˙ ν 2 ∆W2 the situation is simpler. Both u1 and its prefactor will explode for small NL, and ˙ ν | | therefore ∆W2 too, as the graph clearly shows. The graph also shows that the source term is maximal for NL close to 1. Below this value the viscous dissipation increases dramatically and therefore, for the case considered here, NL should not be taken smaller than 1. This confirms the choice of NL 1 that is commonly applied in standing-wave devices. ≥

6.4 A standing-wave refrigerator

The aim of this section is to study how a standard thermoacoustic refrigerator as de- picted in 6.1(a) behaves in terms of parameters such as cooling power, drive ratio, tem- perature difference, mean pressure, stack position, stack length, and coefficients of per- formance. In particular we are interested in what happens at large amplitudes, going up to drive ratios of 40%. In all likelihood drive ratios of 40% are too high to obtain very accurate predictions of the performance of the refrigerator, since nonlinearities will se- riously start to affect the performance. Furthermore, there are all kinds of practical considerations and limitations that prevent us from achieving these high amplitudes. Still these computations can give some quantitative insight into the behavior of the re- frigerator and it shows how the refrigerator would work under ideal circumstances. Standing-wave devices 101

400 acoustic power source ) 2 300 therm. dissipation visc. dissipation 200

100

acoustic power (W/m 0

−100 0 0.5 1 1.5 2 N L

Figure 6.6: The acoustic power per unit area absorbed by the thermoacoustic couple as a func- tion of the Lautrec number for H˙ = 0 and a drive ratio of 2%. The stack is positioned at 5cm from the closed end.

The cooling and heating power depend on the position of the stack inside the res- onator. In figure 6.7 we show a detailed description of all the energy flows in and around the stack depending on whether the stack is placed near the closed end or near the driver. In both cases the hot end will face the nearest end of the resonator and the total power H˙ through the stack will be constant and directed from the cold to the hot end. When the stack is positioned near the driver, we assume that the driver supplies ˙ ˙ acoustic power Wdriver which is reduced to WH due to viscous interaction with the res- ˙ ˙ onator wall. Most of WH is then absorbed by the stack, and the remaining amount WC will be dissipated by the resonator wall left of the cold end. Alternatively, when the stack is positioned near the closed end, we again assume that the driver supplies acous- ˙ ˙ ˙ tic power Wdriver which is reduced to WC by dissipation in the resonator. Most of WC ˙ is then absorbed by the stack, and the remaining amount WH will be dissipated by the resonator left of the hot end. In either case, we have the following expressions for the cooling power extracted at the cold end and the heating power supplied to the hot end,

Q˙ = H˙ W˙ , Q˙ = H˙ W˙ . (6.5) C − C H − H The performance of the refrigerator will be measured by the (relative) coefficient of per- formance Q˙ COP T COP = C , COPR = 1, COPC = C . (6.6) W˙ COPC ≤ T T H − C ˙ ˙ ˙ To compute the stack efficiency we can substitute W = WC WH and to compute the ˙ ˙ − refrigerator efficiency we can put W = Wdriver. Due to wall attenuation and the second law of thermodynamics we have that 0 COPstack COPdriver COPC. Our numerical calculations are motivated≤ by experiments≤ ≤ performed at Shell by Araujo [4], where a simple air-filled standing-wave refrigerator with a ceramic stack 102 6.4 A standing-wave refrigerator

xs xs

TH TC TC TH ˙ ˙ ˙ ˙ WH H˙ WC WC H˙ WH

˙ ˙ ˙ ˙ QH QC QC QH

(a) Stack near closed end (b) Stack near driver

Figure 6.7: Schematic model of the energy flows in a thermoacoustic refrigerator with a stack positioned (a) near the closed end or (b) near the driver. The hot end always faces the nearest end ˙ of the resonator. Heating power QH is supplied to the hot end and cooling power is extracted from the cold end. Acoustic power W˙ W˙ is absorbed by the stack. C − H

was tested. To simplify calculations we assume a parallel-plate structure for the stack as opposed to the honey-comb stack tested by Araujo, but with the same porosity. For most of our computations we have kept the stack center fixed at 2.5 cm from the closed end, to maximize the COPR. All other relevant parameters are given in Table 6.2.

Parameter Symbol Value Unit mass flux M˙ 0 kg m/s speed of sound c 375 m/s

isobaric specific heat Cp 1000 J/(kg K)

isochoric specific heat Cv 714 J/(kg K)

tube diameter db 0.073 m · 4 pore diameter dg 3.7 10− m · 5 plate diameter ds 6 10− m frequency f 393 Hz

thermal conductivity gas Kg 0.0262 W/(m K)

thermal conductivity plate Ks 2 W/(m K)

stack length Ls 0.03 m · 5 average pressure p0 1 10 Pa

temperature hot heat exchanger TH 370 K

distance between stack centre and closed end xs 0.025 m

Table 6.2: Parameters for the standing-wave refrigerator tested in [4].

In all figures shown below the left column shows the result for a mean pressure of 1 bar and the right column shows the result when the mean pressure is increased to 20 bar, while keeping NL and Br constant. Standing-wave devices 103

1.2 1.2

1 1

0.8 0.8

0.6 0.6 COP COP D=5% D=5% 0.4 0.4 D=10% D=10% D=20% D=20% 0.2 D=30% 0.2 D=30% D=40% D=40% 0 0 0 0.5 1 1.5 2 0 1 2 3 4 2 4 2 5 cooling power (W/m ) x 10 cooling power (W/m ) x 10

(a) p0 = 1 bar (b) p0 = 20 bar

1.5 1.5

1 1 COP COP

0.5 D=5% 0.5 D=5% D=10% D=10% D=20% D=20% D=30% D=30% D=40% D=40% 0 0 0 25 50 75 100 125 0 25 50 75 100 125 temperature difference (K) temperature difference (K)

(c) p0 = 1 bar (d) p0 = 20 bar

Figure 6.8: The coefficient of power COP as a function of the temperature difference TH TC or ˙ − cooling power QC. The calculations are repeated for various drive ratios and for mean pressures equal to 1 or 20 bar.

In figure 6.8 we have plotted the COP as a function of both the cooling power and the temperature difference for various drive ratios. The plots show that for increasing cooling power, the COP becomes larger. With the cooling power increasing, the temper- ature difference across the stack naturally becomes smaller, and a peak COP is attained for a temperature difference equal to zero. Furthermore, it is noticed that for increasing drive ratio the COP increases as well. However, plots 6.8(c) and 6.8(d) show that beyond 30% for p0 = 1 bar and beyond 10% for p0 = 20 bar, increasing the drive ratio has little effect, because the temperature drop approaches the critical temperature difference. It follows that the critical temperature difference is approximately 115 K for both p0 = 1 bar and p0 = 20 bar. Swift [135] predicts for the critical temperature gradient, ω p T = 1 , (6.7) ∇ crit ρ c u 0 p h 1i which gives a temperature difference of approximately 247 K. We conclude that this prediction, based on an inviscid approach, overpredicts the numerical result. Finally, we note that going from 1 bar to 20 bar, the cooling power needs to be increased by roughly a factor 20 as well, to obtain the same temperature difference. In figure 6.9 we have repeated the previous calculations for the relative coefficient of performance COPR and the same conclusions can be drawn. However, in contrast 104 6.4 A standing-wave refrigerator to the COP, the COPR has a peak value, which appears to converge to a fixed value for increasing drive ratio. The peak for 20 bar is only slightly higher than for 1 bar. However, at a mean pressure of 1 bar a higher drive ratio are required to approach the peak. Note also that the two zeros of the COPR occur where the cooling power and temperature difference equal zero.

0.3 0.3 D=5% D=5% 0.25 D=10% 0.25 D=10% D=20% D=20% D=30% D=30% 0.2 0.2 D=40% D=40%

0.15 0.15 COPR COPR 0.1 0.1

0.05 0.05

0 0 0 2 4 6 8 10 0 0.5 1 1.5 2 2 4 2 6 cooling power (W/m ) x 10 cooling power (W/m ) x 10

(a) p0 = 1 bar (b) p0 = 20 bar

0.3 0.3 D=5% D=5% 0.25 D=10% 0.25 D=10% D=20% D=20% D=30% D=30% 0.2 D=40% 0.2 D=40%

0.15 0.15 COPR COPR 0.1 0.1

0.05 0.05

0 0 0 25 50 75 100 125 0 25 50 75 100 125 temperature difference (K) temperature difference (K)

(c) p0 = 1 bar (d) p0 = 20 bar

Figure 6.9: The relative coefficient of power COPR as a function of the temperature difference ˙ TH TC or cooling power QC. The calculations are repeated for various drive ratios and for mean− pressures equal to 1 or 20 bar.

In figure 6.9 we observed that the peak COPR-value appears to converge to a fixed value. In figure 6.10 we examine this by plotting for each drive ratio the optimal value of the coefficients of performance, i.e. the values of the COP and COPR for which the COPR attains its peak value. Both graphs seem to converge to the same fixed value for high drive ratios. Particularly for 20 bar the convergence is quite fast. In the previous calculations the stack was fixed at 2.5 cm from the closed end. The location of the stack can be optimized to maximize the efficiency. In figure 6.11 we vary the position of the stack for a drive ratio of 30% and mean pressures of 1 and 20 bar. We choose this drive ratio to ensure that the COPRopt approaches its maximal value (see figure 6.10). The resulting graphs for 1 and 20 bar are almost overlapping. Furthermore both reach their peak value near kxs = 0.16, which corresponds with a stack center positioned at 2.5 cm from the closed end. This is in fact the case considered in all other plots in this section. Standing-wave devices 105

0.96 0.3

0.25 0.94

0.2 0.92 opt opt 0.15 COP

0.9 COPR 0.1

0.88 p = 1 bar p = 1 bar 0 0.05 0 p = 20 bar p = 20 bar 0 0 0.86 0 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 drive ratio drive ratio (a) (b)

Figure 6.10: The optimal coefficients of performance COPopt and COPRopt, corresponding to the peak COPR-values, as a function of the drive ratio. The calculations are performed for mean pressures equal to 1 or 20 bar.

2 0.3 p = 1 bar p = 1 bar 0 0 p = 20 bar 0.25 p = 20 bar 0 0 1.5 0.2 opt opt 1 0.15 COP COPR 0.1 0.5 0.05

0 0 0 0.5 1 1.5 0 0.5 1 1.5 kX kX s s (a) (b)

Figure 6.11: The optimal coefficients of performance COPopt and COPRopt, corresponding to the peak COPR-values, as a function of the scaled stack position. The calculations are performed for a drive ratio of 30% and mean pressures equal to 1 and 20 bar.

Figure 6.12 shows how the cooling power, temperature difference and drive ratio in the stack depend on each other. We again observe the same critical temperature differ- ences of approximately 115 K, where the effect of the drive ratio is minimal, which is reflected in the nodes where all the lines cross. In both cases the corresponding cooling power is approximately -1800 W/m2, which is caused almost completely by the conduc- tion through gas and stack material. For high drive ratios, the temperature difference will approach the critical temperature difference more and more. Left of the node, the device will act as a prime mover, while right of the node, for positive cooling powers, the device will act as a refrigerator. Shown in figure 6.13, are the acoustic and cooling power as a function of the (scaled) position of the stack center, while keeping the temperature difference over the stack equal to zero. The peak acoustic powers are obtained near the tube center, which cor- responds to 1/4 of the wave length. The peak cooling powers on the other hand are obtained at 1/4 and 3/4 of the tube length, which is at about 1/8 and 3/8 of the wave 106 6.4 A standing-wave refrigerator

150 150

125 125

100 100

75 75

D = 5% D = 5% 50 50 D = 10% D = 10% D = 20% D = 20%

temperature difference (K) 25 D = 30% temperature difference (K) 25 D = 30% D = 40% D = 40% 0 0 −5000 0 5000 10000 −1 0 1 2 2 2 5 cooling power (W/m ) cooling power (W/m ) x 10

(a) p0 = 1 bar (b) p0 = 20 bar

Figure 6.12: Mutual dependence of cooling power, temperature difference and drive ratio. The calculations are performed for mean pressures equal to 1 or 20 bar. length. In the right part of the graph, the cooling power turns negative and the device starts to work as a heater. Note also that the power is amplified by a factor 20, as the mean pressure is increased from 1 to 20 bar.

4 x 10 2500 5 D=1% D=1% D=2% D=2% ) ) 2 2000 D=3% 2 4 D=3%

1500 3

1000 2 acoustic power (W/m 500 acoustic power (W/m 1

0 0 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 kX kX s s

(a) p0 = 1 bar (b) p0 = 20 bar

4 x 10 4000 8 D=1% D=1% D=2% D=2% ) ) 2 3000 D=3% 2 6 D=3%

2000 4

1000 2 cooling power (W/m cooling power (W/m

0 0 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 kX kX s s

(c) p0 = 1 bar (d) p0 = 20 bar

Figure 6.13: Supplied cooling power and absorbed acoustic power as a function of the scaled stack position. The temperature difference across the stack is kept at 0 K, and the drive ratio is varied between 1 and 3%. The calculations are repeated for mean pressures equal to 1 and 20 bar.

So far, we have kept the stack length fixed at 3 cm in all our computations. The stack Standing-wave devices 107 length is, however, an important parameter in the design of a stack that should be taken neither too small nor too large for optimal performance. In figure 6.14 we show the COPR as a function of both the cooling power and the temperature difference across the stack for various stack lengths and a drive ratio of 30%.

0.5 0.7 L = 0.1 cm L = 0.1 cm s s L = 0.5 cm 0.6 L = 0.5 cm 0.4 s s L = 1 cm L = 1 cm s 0.5 s L = 2 cm L = 2 cm 0.3 s 0.4 s L = 3 cm L = 3 cm s s

COPR L = 4 cm COPR 0.3 L = 4 cm 0.2 s s 0.2 0.1 0.1

0 0 0 1 2 3 4 5 6 0 2 4 6 8 10 12 2 4 2 5 cooling power (W/m ) x 10 cooling power (W/m ) x 10

(a) p0 = 1 bar (b) p0 = 20 bar

0.5 0.7 L = 0.1 cm L = 0.1 cm s s L = 0.5 cm 0.6 L = 0.5 cm 0.4 s s L = 1 cm L = 1 cm s 0.5 s L = 2 cm L = 2 cm 0.3 s 0.4 s L = 3 cm L = 3 cm s s

COPR L = 4 cm COPR 0.3 L = 4 cm 0.2 s s 0.2 0.1 0.1

0 0 0 50 100 150 0 50 100 150 temperature difference (K) temperature difference (K)

(c) p0 = 1 bar (d) p0 = 20 bar

Figure 6.14: The relative coefficient of power COPR as a function of the temperature difference ˙ TH TC or cooling power QC. The calculations are repeated for various stack lengths, a drive ratio− of 30% and for mean pressures equal to 1 and 20 bar.

On the one hand, we see that in order to obtain a large COPR we need to take the stack as short as possible. On the other hand, we see that these large COPR are only achieved at very small temperature differences and cooling powers. Therefore, if one is satisfied with a small temperature difference across the stack, then very large COPR can be obtained. In figure 6.15 we have shown this more clearly by plotting the peak COPR value as a function of the stack length. For mean pressures of both 1 and 20 bar we see that the maximal COPR is obtained when the stack length Ls approaches zero. The other extreme is when the stack length approaches 0.5 cm and the refrigerator becomes highly inefficient. For longer stacks the refrigerator even stops working altogether. If one is more interested in large cooling powers and temperature differences and less in a large COPR, then one can try to maximize the achievable cooling power and temperature differences that is obtained from figure 6.15 at zero COPR. In figure 6.16 we plot these values as a function of the stack length. It turns out that the maximal cooling power is obtained for stack lengths close to 2 cm and the maximal temperature difference is obtained for stack lengths close to 4 cm. Therefore, the choice of Ls = 3 cm, 108 6.5 A standing-wave prime mover that was made in all previous calculations, seems to be a reasonable choice.

0.7 5 p = 1 bar p = 1 bar 0 0 0.6 p = 20 bar p = 20 bar 0 4 0 0.5 3 opt 0.4 opt

0.3 COP

COPR 2 0.2 1 0.1

0 0 0 2 4 6 8 0 2 4 6 8 L (cm) L (cm) s s (a) (b)

Figure 6.15: (a) COPRopt and (b) COPopt as a function of the stack length for a 30% drive ratio and mean pressures of 1 and 20 bar.

5 x 10 12 100 p = 1 bar p = 1 bar 0 0 10 p = 20 bar p = 20 bar 0 80 0

8 60 max max 6 T Qc ∆ 40 4

20 2

0 0 0 2 4 6 8 0 2 4 6 8 L (cm) L (cm) s s (a) (b)

Figure 6.16: The maximal achievable cooling power and temperature difference as a function of the stack length for a 30% drive ratio and mean pressures of 1 and 20 bar.

6.5 A standing-wave prime mover

In this section we aim to analyze a thermoacoustic prime mover of the type depicted in 6.1(b). A higher goal is a comparison with the traveling-wave prime mover that will be discussed in Chapter 7, which is based on the “Swift”-type looped geometry [14]. To ensure an accurate and meaningful comparison we will choose a configuration that resembles this traveling-wave configuration as close as possible. In [14] a stainless steel regenerator is positioned in a helium-filled looped resonator with parameters as given in 6.5. Instead of the regenerator we will model a parallel-plate stack placed in a straight helium-filled resonator with the same operating conditions. The stack position, porosity, and Lautrec number will be optimized to maximize the efficiency. In the end we will also examine how the choice of plate material affects the performance. ˙ ˙ Given the acoustic power WH and WC at the hot and cold end, and the total power Standing-wave devices 109

parameter symbol value unit mass flux M˙ 0 kg m/s speed of sound c 1020 m/s isobaric specific heat Cp 5193.2 J/(kg K) isochoric specific heat Cv 3115.9 J/(kg K) thermal conductivity gas Kg 0.16 W/(m K) stack length Ls 7.3 cm gas radius Rg 4.2 cm solid radius Rs 3.0 cm tube radius Rb 7.3 cm frequency f 84 Hz · 6 mean pressure p0 3.1 10 Pa drive ratio Dr 5 % low temperature TC 300 K high temperature TH 600 K

Table 6.3: Parameters: a helium-filled parallel-plate stack through the stack H˙ , we can compute

Q˙ = H˙ W˙ , Q˙ = H˙ W˙ , (6.8) C − C H − H ˙ ˙ ˙ with the sign of H, WC, and WH chosen such that they are positive. The performance of the prime mover will be measured by the (relative) efficiency

W˙ W˙ T η = C − H , η = H COP. (6.9) Q˙ R T T H H − C

Note that that 0 ηR 1 due to the second law of thermodynamics. We start with≤ the parameter≤ values given in Table 6.5. We vary the position of the stack in the resonator and for each position we compute the relative efficiency ηR. Figure 6.17 shows the results of these computations for different temperature differences across the stack. It turns out that for all temperature differences the optimal stack position lies near the closed end with kxs = 0.045, and as the temperature difference increases the optimal stack position shifts more and more to the closed end. The plots also show that not every stack position is suitable for a prime mover operating at a given temperature difference. The temperature difference has to be larger than the critical temperature gradient which is maximal near the closed end. The larger the imposed temperature difference, the bigger the region where the stack produces sound. Next we fix the temperature difference at 300 K and kxs at 0.045 and we optimize the Lautrec number NL. Figure 6.18 shows the relative efficiency as a function of the Lautrec number. The optimal value is obtained for NL 0.795. The last parameter we wish to optimize is the porosity≈ of the stack. We vary the blockage ratio Br between 0 and 1 and compute the relative efficiency. It should be noted that for these computations εs cannot be neglected. When Br approaches 1, εs starts to grow and the wall temperature is no longer a steady variable. Figure 6.19 shows ηR as a function of B and it turns out that η is optimal when B 0.887. r R r ≈ 110 6.5 A standing-wave prime mover

0.2 T −T = 200 K H C T −T = 250 K H C T −T = 300 K 0.15 H C T −T = 350 K η H C R T −T = 400 K 0.1 H C

0.05

0 0 0.1 0.2 0.3 0.4 kx s

Figure 6.17: The relative stack efficiency ηR as a function of the relative stack position kxs for different temperature differences across the stack. The optimal stack position is kxs 0.03 if the blockage ratio is equal to 0.72 and the Lautrec number is equal to 0.275. ≈

0.4

0.795 0.3

η R0.2

0.1

0 0 0.5 1 1.5 2 2.5 3 N L

Figure 6.18: The relative stack efficiency ηR as a function of the Lautrec number NL. The optimal Lautrec number is N 0.795 when T T = 300 K, kx = 0.045,and B = 0.72. L ≈ H − C s r

0.5

0.4 0.887

η 0.3 R

0.2

0.1

0 0 0.2 0.4 0.6 0.8 1 B r

Figure 6.19: The relative stack efficiency ηR as a function of the blockage ratio Br. The optimal efficiency is obtained for B 0.887,when T T = 300 K, kx = 0.045,and N = 0.795. r ≈ H − C s L Standing-wave devices 111

So far, following [14], we used stainless steel for the stack plates in all our calcu- lations. However, there is a wide range of materials possible, ranging from metal-like materials to polystyrene that is used to manufacture plastic straws. In figure 6.20 we test several options by repeating the calculations of figure 6.19 for different plate materi- als. Furthermore in figure 6.21 we show the corresponding acoustic and cooling power, again as a function of the blockage ratio Br. It turns out that for all materials, similar efficiencies and acoustic powers are obtained , although mylar seems to work best. Fur- thermore all materials seem to achieve their maximal efficiency close to Br = 0.83. The highest acoustic powers are obtained for small porosities, but at a very low efficiency, which requires an enormous input of heating power,. We will show in Chapter 7 that for a traveling-wave geometry, the choice of material has a much greater impact on both the efficiency and the heating and acoustic power.

0.5 stainless steel alumina 0.4 glass mylar polystyrene η 0.3 R

0.2

0.1

0 0 0.2 0.4 0.6 0.8 1 B r

Figure 6.20: The relative stack efficiency ηR as a function of the blockage ratio Br for several materials, when T T = 300 K, kx = 0.045,and N = 0.795. H − C s L

6 6 x 10 x 10 2 2 stainless steel stainless steel alumina alumina glass glass 1.5 1.5 mylar mylar polystyrene polystyrene

1 1 heating power (W) 0.5 acoustic power (W) 0.5

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 φ B r (a) (b)

Figure 6.21: The (a) supplied heating power and (b) produced acoustic power at the stack as a function of the blockage ratio B for several materials, when T T = 300 K, kx = 0.045, r H − C s and NL = 0.795. 112 6.6 Streaming effects in a thermoacoustic stack

6.6 Streaming effects in a thermoacoustic stack

In this section we will try to investigate how the streaming velocity u2,0 behaves as a ˙ function of NL when M = 0. Waxler [148] examined the streaming velocity within straight two-dimensional pores using a comparable formal perturbation expansion. Waxler considered a thermoacoustic stack filled with dry air at room temperature positioned inside a resonator at λ/8 from a closed end. Across the stack a temperature difference of 20K was imposed with the highest temperature at the side of the closed end. The exact value for the drive ratio and mean pressure were not given and neither were the stack plate material and thickness. As a result his results cannot be reproduced exactly . Qualitatively, however, we can compare our results. The parameter values for our computations are shown in Table 6.4.

Parameter Symbol Value Unit drive ratio Dr 2 % mass flux M˙ 0 kg m/s speed of sound c 347 m/s isobaric specific heat Cp 1000 J/(kg K) isochoric specific heat Cv 714 J/(kg K) isobaric specific heat solid Cs 1300 J/(kg K) gas radius Rg NLδk m · 5 plate radius Rs 1.5 10− m thermal conductivity air Kg 0.0264 W/(m K) thermal conductivity plate Ks 0.08 W/(m K) stack length Ls 0.05λ m · 5 mean pressure p0 1.14 10 Pa Prandtl number Pr 0.68 - temperature hot heat exchanger TH 290 K temperature cold heat exchanger TC 270 K

Table 6.4: Parameters: an air-filled parallel-plate stack

Figure 6.22 shows the streaming velocities u2,0 together with the mass flux m˙ as a ∞ function of X and y in the stack, neglecting gravitational effects (Fr ). In figure 6.23 we zoom in at the center of the stack and show the streaming velocity→ as a function of y only. The computations are repeated for various Lautrec numbers ranging between 0.1 and 8. Note that the area in the graph below m˙ is equal to zero so that for every X the average mass flux is zero. Just like Waxler we observe several transitions as NL increases. For NL Pr, the velocity is mostly towards the right, where the acoustic wave originates. This≪ is the regime where the acoustic Bernouilli effect is unimportant. Due to the acoustic source mass is driven in the direction that acoustic intensity must flow. To maintain the zero mass flux, the mean velocity must drive the mass back to the left. As NL increases, the Bernoulli effect becomes more important. For N P there is a transition and as L ∼ r NL increases more, boundary layer flow starts to develop. Finally for sufficiently large NL, in addition to the boundary-layer flow, the parabolic profile starts to appear that is typical for wide channels. Standing-wave devices 113

(a) NL = 0.1 (b) NL = 0.1

(c) NL = 0.7 (d) NL = 0.7

(e) NL = 2 (f) NL = 2

(g) NL = 8 (h) NL = 8

Figure 6.22: The streaming velocity u2,0 (left column) and the mass flux m˙ (right column) as a function of X and y inside the stack. The profiles are computed for NL varying between 0.1 and 8. 114 6.6 Streaming effects in a thermoacoustic stack

−6 x 10 2

1

0 (m/s)

2,0 N = 0.1 u −1 L N = 0.7 L −2 N = 2 L N = 8 L −3 −1 −0.5 0 0.5 1 y/R g

Figure 6.23: The streaming velocity profile u2,0 as a function of y in the center of the stack. The profiles are computed for NL varying between 0.1 and 8. Chapter 7

Traveling-wave devices

Traveling-wave devices come in all sorts and sizes. These can be refrigerators [84,142], prime movers [14, 78, 141, 155], or combinations of the two [34,134,154]. Ceperley [23] was the first to realize that a traveling acoustic wave propagating through a regenerator undergoes a thermodynamic cycle similar to the Stirling cycle, so that potentially higher efficiencies could be obtained. These concepts were first transformed into a working prime mover by Yazaki et al. [155], who managed to design a machine consisting of a regenerator placed in a looped tube, generating spontaneous traveling-wave gas oscil- lations around the loop. However, due to the high acoustic velocities there were large viscous losses, reducing the efficiency significantly. Backhaus and Swift [14] improved on these ideas by combining the loop with a resonance tube, so that in the loop the traveling-wave phasing could be obtained, but at low acoustic velocities, thereby ob- taining efficiencies up to 30%. In this chapter we will present the results from our simulations of a traveling-wave prime mover similar to that of Backhaus and Swift [14], consisting of a looped tube with a regenerator and thermal buffer tube, connected to a resonator tube with an acous- tic load. The usual approach in simulating such devices [14, 34,76,78,141] consists of providing a geometry and looking for the right values of system parameters such as frequency, power input and imposed temperature difference that give a stable system. Here we will employ the opposite approach. We only fix the geometry of the regenerator and thermal buffer tube, and look for an appropriate configuration of the loop and res- onator that gives us the desired system parameters. Such an approach can be used as a useful guide for the design of practical devices. Given certain specifications, like power- input, temperature impedance, frequency, drive ratio, and regenerator impedance, the optimal geometry can be computed. Moreover, the machine can be designed to prevent the occurrence of mass streaming, which is known to have a negative impact on the performance. We will derive an optimization procedure that finds the traveling-wave geometry that fits given system parameters. In addition we will compute the efficiency of the prime mover, and its sensitivity to the choice of parameters related to the regenerator. It will follow that this kind of traveling-wave prime mover works potentially much better than an equivalent standing-wave prime mover. 116 7.1 A traveling-wave prime mover

7.1 A traveling-wave prime mover

We consider a traveling-wave prime mover of the type shown in figure 7.1 (cf. [14]): a short looped tube attached to a half-wave-length resonance tube with a variable acoustic load. Within the loop several segments can be distinguished: the regenerator, thermal buffer tube, T-junction, inertance tube, and compliance. Below we will discuss each of these segments.

4

1 λ ˙ 2 WC TC A 3 ˙ TH (a) Schematic scale drawing of traveling-wave prime mover WH

B

T 2 C ˙ 1 Wload

5 6

(b) Detailed drawing of torus section

Figure 7.1: (a) Schematic scale drawing of a traveling-wave prime mover consisting of a short looped tube attached to a half-wave-length resonator with variable acoustic load. (b) Detailed view of torus section that provides sound to the resonator. The system is divided into several different segments: regenerator (A), thermal buffer tube (B), tube with T-junction (1,2), iner- tance tube (3), compliance (4), and the resonator (5,6) with acoustic load. The (parallel-plate) ˙ ˙ ˙ regenerator amplifies the sound from WC to WH and Wload is supplied to the acoustic load.

Regenerator (A) Standing-wave devices require stacks that have pores of several penetration depths wide. Due to the resulting imperfect thermal contact between gas and solid, entropy is created, so that the ideal Carnot efficiency can never be achieved. In traveling-wave devices regenerators can be used that have pores that are very small compared to the penetration depth, so that we can make use of the reversible Stirling cycles [23], which requires the pressure and velocity to be in phase at the regenerator. Furthermore, small gas velocities are preferred to prevent large amounts of viscous dissipation in the nar- row regenerator pores. Additionally, we want to maximize the product of pressure and Traveling-wave devices 117 velocity to maximize the produced acoustic power. It is therefore advantageous to re- duce the velocity oscillations and increase the pressure oscillations at the regenerator. For this purpose a resonator is attached in such a way that the loop lies at a velocity node (and pressure antinode) of the resonator. We will simulate a helium-filled parallel-plate regenerator made of stainless steel with heat exchangers on either side. The heat exchangers are used to supply heating ˙ ˙ power and amplify the acoustic power from WC at the cold end to WH at the hot end. ˙ ˙ The difference between WH and WC is the acoustic power produced by the regenerator ˙ and will be denoted by WA. As in [14] we will start with a porosity Br = 0.72 and Lautrec number NL = 0.275. Later we will optimize the geometry and material of the regenerator to maximize the efficiency.

Thermal buffer tube (B) The thermal buffer tube is used as a thermal buffer between the hot heat exchanger and the room temperature. Usually its tapered to reduce Rayleigh streaming, but in our analysis we consider a straight tube for simplicity.

T-junction (1,2,5) The T-junction connects the loop to the resonator. Part of the sound produced by the regenerator will be supplied to the resonator and part will be fed back to the loop to compensate for the wall attenuation. Due to the presence of sharp edges flow sepa- ration may occur, resulting in vortex shedding and subsequent turbulent dissipation. Disselhorst and Van Wijngaarden [37] showed that vortex shedding will only take place if the Strouhal number, pertaining to the radius of curvature of the edges and the acous- tic velocity amplitude, is smaller than one. In our modeling and simulations we will ignore such effects.

Inertance tube and compliance (3,4) Clockwise from the T-junction we have the inertance tube and the compliance that give a contraction and expansion of the tube. Their lengths and diameters have to be cho- sen carefully so that the pressure and velocity match periodically across the loop. In lumped-elements models the inertance tube is modeled as a piston with a certain mass and the compliance as a large volume at constant pressure.

Resonance tube with acoustic load (5,6) For the resonance tube we use a straight tube with a variable acoustic load to which ˙ ˙ ˙ acoustic power Wload is supplied. The difference between Wload and WA is caused by wall attenuation in the loop and resonator. In [14] a quarter-wave length resonator is used with a conical shape to reduce the acoustic velocities, but here we have used a straight half-wave length resonator to simplify calculations. The acoustic load is modeled as a narrow channel connected to a large reservoir that is kept at constant pressure. The highest load efficiencies will be obtained when the load is positioned near the T-junction, but in practice it is necessary to keep some distance due to the complicated flow patterns around the T-junction. In practical applications the 118 7.2 Computations

T

x

A B 1 2 3 4

˙ ˙ HA WH ˙ ˙ WC HB

Q˙ Q˙ C H 5

Figure 7.2: Torus section unrolled. We distinguish the regenerator (A), thermal buffer tube (B), T-junction (1,2,5), inertance tube (3), and the compliance (4). The top graph shows a schematic temperature profile. In the regenerator the temperature increases and in the thermal buffer tube the temperature drops back to its original value. Everywhere else the temperature is assumed ˙ ˙ constant. Heating power QH is supplied to the hot end and cooling power QC is extracted from the cold end of the regenerator. As a result the acoustic power is amplified by the regenerator ˙ ˙ ˙ from WC to WH. The total power in the regenerator and thermal buffer tube is denoted by HA ˙ and HB, respectively. load may be replaced by a traveling-wave refrigerator that is driven thermoacoustically by the prime mover (see e.g. [34]).

7.2 Computations

If we cut the torus in figure 7.1 at the regenerator and unroll the geometry we get a configuration as shown in figure 7.2. In our computations we will work on the unrolled geometry and model the ends by periodicity; influence of edge effects is not taken into account. Figure 7.2 also shows the energy flows in and around the regenerator. Heating ˙ ˙ power QH is supplied to the hot end and cooling power QC is extracted from the cold ˙ end of the regenerator. This leads to a total power HA in the regenerator directed to its ˙ cold heat exchanger and a total power HB inside the thermal buffer directed to its cold ˙ heat exchanger. Moreover, the acoustic power is amplified from WC at the cold end to ˙ WH at the hot end. Putting control volumes around the hot and cold heat exchangers and applying conservation of energy, we find

Q˙ = W˙ H˙ , Q˙ = H˙ H˙ , W˙ = W˙ W˙ , (7.1) C C − A H B − A A H − C Traveling-wave devices 119

˙ where WA denotes the acoustic power produced by the regenerator. We can now intro- duce efficiencies for the regenerator and acoustic load as

W˙ W˙ η = A , η = load . (7.2) A ˙ load ˙ QH QH

The relative efficiencies are obtained by dividing by the Carnot efficiency ηC,

η TH TC ηR = , ηC = − . (7.3) ηC TH

The Carnot efficiency gives an upper bound for the efficiency, so that 0 ηR 1. As ˙ ˙ ≤ ≤ WA will be larger than Wload because of wall attenuation, we obtain ηload ηA ηC. The computations consist of two parts. First we give the geometry≤ for the≤ regen- erator and the thermal buffer tube, we impose a impedance and pressure amplitude at the regenerator, and we impose a temperature difference across the regenerator and thermal buffer tube. Integration of the system (4.70) will then give the complete veloc- ity, pressure and temperature profile in the regenerator and thermal buffer tube. From this we get boundary conditions for the pressure and velocity in the loop and resonator. As a result of this approach we impose more conditions than available unknowns. We circumvent this problem, by adapting the geometry parameters of the loop and res- onator until all boundary conditions are satisfied. In particular we will vary the radii and lengths of the various segments.

7.2.1 Regenerator and thermal buffer tube

The first part of the geometry consists of the regenerator and thermal buffer tube as depicted in figure 7.2.1. The variables in the regenerator will be indicated by an index A and in the thermal buffer tube by an an index B.

TC TH TC

pA pB A B uA uB

Figure 7.3: Regenerator (A) and thermal buffer tube (B).

Left of the regenerator a pressure pA and an impedance ZA is imposed, which leads to the following initial conditions:

p1A(0) = pA, (7.4a) u (0) = u = p /Z . (7.4b) h 1Ai A A A At the interface between the regenerator and thermal buffer tube we impose continuity of mass and momentum, which results in continuity of pressure and volumetric velocity, 120 7.2 Computations so that

p1A(1) = p1B(0), (7.5a) 1 u1A (1) = u1B (0), (7.5b) h i Br h i where Br is the blockage ratio as given in Table 3.1. Additionally, we impose a temper- ature difference T T across the regenerator and thermal buffer tube, H − C

T0A(0) = TC, T0A(1) = TH, (7.6a)

T0B(0) = TH, T0B(1) = TC. (7.6b)

Integrating the system of ODE’s given in (4.70), subject to these boundary conditions, we can compute numerically T , p and u throughout the regenerator and thermal 0 1 h 1i buffer tube. This will yield a pressure pB and velocity uB at the end of the thermal buffer tube. Note that only pA and uA are imposed; pB and uB are an outcome of the computations.

7.2.2 Optimization procedure Having solved for the regenerator and thermal buffer tube, we still need to model and simulate the remaining part of the system that is shown in figure 7.4. The lengths (L j) and radii (R j) of each segment will be adapted so that all boundary conditions are satis- fied. The regenerator and thermal buffer tube are kept fixed throughout this procedure.

(pB, uB) ˙ 1 Wload 5 6

2

3

4 Figure 7.4: Tube with T-junction, inertance tube, compliance and reso- (pA, uA) nance tube with acoustic load.

Since each segment is a straight tube at constant temperature, we can use the expres- sions derived in equations (4.75), which express the pressure and velocity in segment j (j = 1,...6)as

ikjX ikjX p1 j(X) = A je− + Bje , 0 X 1, (7.7a) χ ≤ ≤ j ikjX ikjX u1 j (X) = A je− Bje , 0 X 1, (7.7b) h i ρ0c − ≤ ≤   Traveling-wave devices 121

with A j and Bj integration constants and

1 γ 1 χ = (1 f ) 1 + − f , (7.8) j v ν j ε k j c u − 1 + s j ! u t κ j 1 γ 1 ωL j k = κ a a = 1 + − f , κ = . (7.9) j j 3 j 5 j v ε k j j re f − c u 1 fν j 1 + s j ! c q u − t The unknowns A j and Bj differ in each section and have to be determined from the boundary and interface conditions. The left and right end of each segment correspond with X = 0 and X = 1, respectively. It still remains to model the side branch with acoustic load. The acoustic load is dissipated via a thin channel leading to a reservoir which is kept at constant pressure. We can apply Darcy’s law to relate the pressure pload and the velocity uload at the load,

uload = Zload pload, (7.10) where Zload is an impedance depending on the area, viscosity, and length of the channel. We will use pA, uA, pB and uB as boundary conditions for the ends of the loop as indicated in figure 7.4. In addition we will impose continuity of mass and momentum across every interface. Moreover, at the end of the resonator we have a closed ending, which results in a zero velocity. This leads to the following set of conditions:

u = u (0), p = p (0), (7.11a) B h 11i B 11 u (1) = b u (0) + b u (0), p (1) = p (0) = p (0), (7.11b) h 11i 2h 12i 5h 15i 11 12 15 u (1) = b u (0), p (1) = p (0), (7.11c) h 12i 3h 13i 12 13 b u (1) = b u (0), p (1) = p (0), (7.11d) 3h 13i 4h 14i 13 14 b u (1) = B u , p (1) = p , (7.11e) 4h 14i r A 14 A b u (1) = b u + b u (0), p (1) = p (0) = p , (7.11f) 5h 15i load load 6h 16i 15 16 load u (1) = 0. (7.11g) h 16i 2 2 where b j = R j /RA gives the ratio between the cross-sectional area of segments j and A. Substituting (7.7), we find we have to solve the following equation for the constants A j and Bj: M− 0 A = F (7.12) 0 M+ B     with vectors

A = A1 A2 A3 A4 A5 A6 ,

B =  B1 B2 B3 B4 B5 B6 ,  F =  pB ρ0cuB 0000000 pA ρ0cBruA 0 pload ρ0cbloaduload 0 ,   122 7.2 Computations and matrices

1

χ1  ik  e− 1 1 ik −  e− 1 1  −  ik1   χ1e− b2χ2 b5χ5   − ik −   e− 2 1  −  ik2   χ2e− b3χ3   − ik  M− =  e− 3 1 , −  ik3   χ3e− b4χ4   − ik   e− 4   ik4   χ4e−   ik   e− 5 1   ik −   e− 5   ik5   χ5e− b6χ6   − ik   e− 6      and

1 χ  − 1  eik1 1 −  eik1 1  −  ik1   χ1e b2χ2 b5χ5   −   eik2 1  −  ik2   χ2e b3χ3  +  −  M =  eik3 1 1  . − −  ik3   χ3e b4χ4 b4χ4   − −   eik4     χ eik4   − 4   eik5 1   −   eik5     χ eik5 b χ   − 5 6 6   eik6   −    + 15 6 6 15 Note that M−, M C C , A, B C , F C . As there are more equations (15) than unknowns (12)∈ this is× an overdetermined∈ ∈ system of equations. By variation of the geometry parameters we will try to minimize the distance between the least-squares solution and the real solution. In particular we will do this by minimizing the relative error

1 (M(M∗ M)− M∗ I)F M− 0 RELTOL := k − k , with M := , F 0 M+ k k   subject to 8 (real) geometry parameters L1, L2, L3, L4, b3, b4, b5, Zload. Traveling-wave devices 123

7.3 Results

We have implemented the optimization procedure above using the parameter values given in Table 7.1. Following Backhaus and Swift [14] we choose 30-bar helium as the working gas and a stainless-steel regenerator that we model with a parallel-plate ge- ometry. The regenerator parameters are chosen such that the regenerator has the same porosity, length, hydraulic radius as in [14]. Additionally the thermal buffer tube has also been given the same length as in [14]. The geometry parameters of the remaining segments will be determined from the optimization procedure.

variable value unit Dr 5 % 3 ZA 30ρAcA N s m− LA 7.3 cm LB 24 cm Rg 0.42 mm Rs 0.30 mm RA 4.45 cm RB 4.45 cm f 84 Hz · 6 p0 3.1 10 Pa TC 300 K TH 600 K 1 M˙ 0 kg s−

Table 7.1: Input parameters for numerical simulation of regenerator and thermal buffer tube

In the sections below we will first show what kinds of temperature profiles one can expect in the regenerator and thermal buffer tube, and how they are affected by the lengths of the components and the presence of mass streaming. Next, we will focus on the regenerator and try to look for its optimal design that maximizes its efficiency. Finally, we will give the machine geometry that follows from the minimization routine for the optimal regenerator and thermal-buffer-tube design, and we show the pressure, velocity, and acoustic-power profiles that can be expected throughout the prime mover.

7.3.1 Temperature In figure 7.5 we show the temperature profiles in the regenerator and thermal buffer tube computed using the parameter values given in Table 7.1. The temperature profile increases and decreases almost linearly as is commonly assumed. If the length of the regenerator or thermal buffer tube is changed then a deviation from the linear temperature profile can occur. This is shown in figure 7.6 where the temperature is plotted for different lengths of the regenerator or thermal buffer tube. For very long thermal buffer tubes the temperature even increases first before dropping to the low temperature TC. So far we have only considered the case with no streaming, i.e. M˙ = 0. To show the effect of streaming on the temperature profile we have plotted for some choices of M˙ the 124 7.3 Results

600

550

500 A B 450

400 mean temperature (K) 350

300 0 0.05 0.1 0.15 0.2 0.25 0.3 x(m)

Figure 7.5: Mean temperature profile in regenerator (A) and thermal buffer tube (B).

600 650 L = 10 cm s 550 600 L = 20 cm s L = 2 cm 550 L = 30 cm 500 s s L = 4 cm L = 40 cm s 500 s 450 L = 6 cm L = 50 cm s s L = 8 cm 450 L = 60 cm s s 400 L = 10 cm s 400 mean temperature (K) L = 12 cm mean temperature (K) 350 s 350 L = 14 cm s 300 300 0 0.05 0.1 0.15 0.1 0.2 0.3 0.4 0.5 0.6 0.7 x(m) x(m) (a) regenerator (b) thermal buffer tube

Figure 7.6: Mean temperature profiles in (a) regenerator and (b) thermal buffer tube. In (a) LB was fixed at 24 cm and LA was varied and in (b) LA was fixed at 7.3 cm and LB was varied.

corresponding temperature distribution in 7.7. As M˙ increases the temperature profile in the thermal buffer tube changes from linear to exponentia| | l to almost boundary-layer- like behavior, whereas the regenerator-temperature profile is not affected so much. This shows that the presence of a mass flux can distort the mean-temperature profile signif- icantly. This behavior was in fact predicted in Section 4.4.3. It was shown in equations (4.97)-(4.99) that for a relatively short and wide tube (as we have here) with little or no mass streaming a linear temperature profile is expected and as the mass flux increases the temperature distribution will start to look more and more like an exponential profile Traveling-wave devices 125 until eventually we get

T , X = 0, ˙ H lim T0(X; M2) = ˙ ∞ M2 ( T , 0 < X 1, →− C ≤

T , 0 X < 1, ˙ H ≤ lim T0(X; M2) = M˙ ∞ 2 → ( TC, X = 1. The same limiting behavior is observed in figure 7.7. In addition we see that for positive mass fluxes the temperature first increases before dropping to the low temperature. It is possible that due to the high positive mass flux, high-temperature gas parcels are transported away from the hot end whose heat cannot be pumped fast enough to the cold end, and heat is accumulated near the hot heat exchanger which gives rise to a temperature increase.

600 650

M˙ /A = 0.025 550 g − 600 M˙ /A = 0.005 g − 550 500  M˙ /A = 0.001 g − 500 ˙ 450  M/Ag = 0 450 N M˙ /A = 0.001 400 g 400 × M˙ /A = 0.005

mean temperature (K) g mean temperature (K) 350 + ˙ 350 M/Ag = 0.025 300 300 0 0.02 0.04 0.06 0.1 0.15 0.2 0.25 0.3 x(m) x(m) (a) regenerator (b) thermal buffer tube

Figure 7.7: Temperature distributions in (a) regenerator and (b) thermal buffer tube, for different ˙ 1 2 average mass fluxes M/Ag [kg s− m− ] through the regenerator.

7.3.2 Regenerator efficiency So far we have used the design parameters for the regenerator and thermal buffer tube that were also used in [14]. In this section we will look a bit more closely on what the optimal design is for the regenerator in terms of power-output and efficiency. The efficiency in this section will therefore be defined as the relative regenerator efficiency

ηR = ηA/ηC.

Several parameters will affect the performance of the prime mover. The most im- portant geometry parameters are the Lautrec number NL, the blockage ratio Br and the regenerator length LA. Important operating conditions are the impedance ZA, the drive ratio Dr and the temperature difference TH TC. In figure 7.8 we show how variation of these parameters degrades or improves the− performance. In all these computations we have used the parameter values as given in Table 7.1, unless otherwise specified. 126 7.3 Results

0.7 0.8 0.275 0.6 0.917 0.6 0.5

η 0.4 η R R0.4 0.3

0.2 0.2 0.1

0 0 0 0.2 0.4 0.6 0.8 1 1.2 0 0.2 0.4 0.6 0.8 1 N B L r (a) (b)

0.7 0.8 6.8 0.6 16.5 0.6 0.5

η 0.4 η R R0.4 0.3

0.2 0.2 0.1

0 0 0 3 6 9 12 15 18 0 10 20 30 40 50 L Z /(ρ c) A A (c) (d)

1 0.8

0.8 0.6 233

η 0.6 η R R0.4 0.4

0.2 0.2

0 0 0 0.05 0.1 0.15 0.2 0 200 400 600 800 1000 drive ratio temperature difference (K) (e) (f)

Figure 7.8: The relative regenerator efficiency as a function of (a) the Lautrec number NL (com- puted at room temperature), (b) the blockage ratio Br, (c) the regenerator length LA, (d) the impedance ZA, (e) the drive ratio Dr, and (f) the temperature difference across the stack.

It is well-known that for the optimal working of a traveling-wave device the Lautrec number should be small, so that there is a perfect thermal contact between the gas and the wall. In figure 7.8(a) we see that reality is a bit more subtle. Indeed, NL should be smaller than 1, but the optimal value is reached for NL 0.275 (computed at room temperature) which corresponds with a hydraulic radius of∼ 42 µm, exactly the value as used in [14]. When NL becomes too small the viscous dissipation becomes important and it starts affecting the performance. For a given hydraulic radius we can also change the porosity Br, which is effectively Traveling-wave devices 127 the same as changing the thickness of the solid. We see in figure 7.8(b) that the more gas the better the efficiency. However, for high porosities, as the plates become infinitely thin, the efficiencies drop back to zero because there is not enough regenerator material to transport the heat. The effect of the regenerator length is shown in figure 7.8(c). When the stack is too short, only a small temperature difference can be achieved and therefore the efficiency will drop also. On the other hand, when the stack is too long, there will be lots of dissipation, which also reduces the efficiency. We see from the graph that the optimal value is obtained near LA = 6.8 cm, which is quite close to the choice of 7.3 cm used in [14]. We mentioned earlier that for the gas in the regenerator to undergo the ideal Stirling cycle, it is necessary that the pressure and velocity are in phase. Moreover, the velocity oscillations should be small to reduce viscous dissipation and the pressure oscillations should be large to maximize the acoustic power. A trade-off is therefore expected, and it is shown in figure 7.8(d) that an impedance of ZA 16.5 ρc gives the highest efficien- cies. ≈ Next, in figure 7.8(e), we demonstrate how the efficiency is influenced by the drive ratio. When the drive ratio is increased at a fixed temperature difference, the power output will increase, but the losses stay the same. As a result we observe higher effi- ciencies for increasing drive ratios, but as the driver ratios increases more and more, the efficiency profiles converge due to the restriction of the Carnot efficiency. Lastly, we mention the impact of the imposed temperature difference. We first note that it should be large enough, otherwise the device cannot act as a prime mover; it follows from 7.8(f) that a minimal temperature difference of approximately 24 K is nec- essary. The optimal efficiency is achieved near 233 K and for larger temperature differ- ences the efficiency starts to decrease slowly.

0.7 0 0.6

η 0.5 R

0.4

0.3

0.2 −0.3 −0.25 −0.2 −0.15 −0.1 −0.05 0 0.05 mass flux (kg/m2/s)

˙ Figure 7.9: The relative regenerator efficiency as a function of (d) the average mass flux M/Ag 1 2 [kg s− m− ] through the regenerator.

So far we have neglected the influence of streaming. If a nonzero mass flux is in- ˙ cluded the efficiencies ηR should decrease because when M = 0 a large time-averaged convective enthalpy flux can arise, mixing gas of different t6 emperatures and degrad- ing the performance. If we plot the relative efficiency as a function of the average 128 7.3 Results

˙ mass flux M/Ag through the regenerator (figure 7.9), then we see that this is indeed the case. As expected, the maximal efficiency is obtained when M˙ = 0 and decreases rapidly for increasing or decreasing mass flux. The graph does not show values beyond 1 2 M˙ /A 0.03 kg s− m− because of numerical difficulties; a very steep temperature g ≈ profile is necessary (see figure 7.7) to reach the desired low temperature TC, which re- ˙ quires a very precise guess of the total power HB. In all the previous calculations we have used stainless steel for the regenerator ma- terial. If we want to make a study of what material is best we can repeat the previous calculations for different materials. In particular we are interested in the effect of the porosity on the performance. Apart from the material and porosity, the remaining pa- rameters are chosen as in Table 7.1. In figures 7.10 and 7.11 we have plotted the relative efficiency of the regenerator, the supplied heating power, and the produced acoustic power as a function the blockage ratio Br for stainless steel, alumina ceramics, glass, mylar, and plastic.

1

0.8

η 0.6 R

0.4 stainless steel alumina glass 0.2 mylar polystyrene 0 0 0.2 0.4 0.6 0.8 1 B r

Figure 7.10: The relative regenerator efficiency as a function of the blockage ratio Br for several materials.

1000 300 stainless steel stainless steel alumina 250 alumina 800 glass glass mylar mylar 200 600 polystyrene polystyrene 150 400 100 heating power (W) acoustic power (W) 200 50

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 B B r r (a) (b)

Figure 7.11: The (a) supplied heating power and (b) produced acoustic power at the regenerator as a function of the blockage ratio Br for several materials. Traveling-wave devices 129

We see that for most materials the optimal efficiency is obtained for high porosities, i.e. very thin stack plates. However, when the plates become too thin (Br 1), the efficiency drops back to zero again because there is not enough plate material→ to carry all the heat. For some materials, such as mylar or plastic straws, a high efficiency is reached for almost all porosities, provided the stack plates are neither too thick (B 0) r → or too thin (Br 1). However, although the efficiency may be high, low values of Br are not practical,→ as almost no acoustic power is produced (see figure 7.11). We observe that for all materials the most acoustic power is produced for Br between 0.75 and 0.9 with still a relatively high efficiency. Depending on whether one is interested more in high efficiencies or high power output it follows that either glass or stainless steel are the best options for the regenerator material. We can compare these results to stack efficiency of the standing-wave prime mover simulated in Section 6.5, whose configuration has also been optimized to obtain maxi- mal efficiencies under the same operating conditions. Comparing figures 6.20 and 7.10 it follows that the efficiencies for the traveling-wave configuration are about twice as high as those of the standing-wave configuration. Furthermore, as opposed to the standing- wave configuration, the choice of material is much more critical for the regenerator, especially at low porosities.

7.3.3 Geometry optimization In this section we will determine the geometry for the loop and resonator that fits the regenerator and thermal buffer tube given by the input parameters in Table 7.2. We used the results from figure 7.8 to choose the optimal design that for a temperature difference of 300 K and drive ratio of 5% maximizes the regenerator efficiency, i.e. Br = 0.917, ˙ NL = 0.275, LA = 6.8cm, ZA = 16.5ρc, and M = 0. Using MATLAB’s lsqnonlin we have minimized RELTOL by careful choice of the parameters L1, L2, L3, L4, b3, b4, b5, and Zload, subject to the the constraints that all parameters be positive, b3 < 1 (contraction), and b4 > 1 (expansion). 8 The routine converged with a relative error of approximately 4 10− and yielded the geometry given in figure 7.12 and Table 7.2. This choice of loop and· resonator gives a machine (load) efficiency of 0.35. Note that viscous dissipation due to interaction with the wall is included in all segments. Although the addition of viscous dissipations affects the wave number only slightly (e.g. k1 = 0.5172 vs. k1 = 0.5185 0.0007i),it has a significant impact on the performance of the device. Most importantly− the efficiency η is affected, which decreases from 0.39 at the regenerator to 0.35 at the load. Also the ˙ ˙ acoustic power Wload provided to the load is reduced significantly from WA that was generated by the regenerator. It should be noted that the minimization procedure is quite sensitive to the initial guess. Because we have so many geometry parameters to choose (more than available unknowns), different geometries might work. Small changes in the initial guess can lead to very different geometries and some of these can be quite unrealistic, e.g. a very short and wide compliance or a very thin and long inertance tube. In figure 7.13 we show the acoustic power as a function of its position in the prime mover. Within the regenerator there is a sharp amplification of the sound and there is an ˙ increase in acoustic power of about WA = 211 W. Some of this is lost in the loop due to viscous dissipation, and 206 W is supplied to the resonator, which explains the drop in 130 7.3 Results

Input Output variable value unit variable value unit Dr 5 % L1 14.1 cm ZA/(ρAcA) 16.5 - L2 15.9 cm LA 6.8 cm L3 18.0 cm LB 24 cm L4 20.3 cm L5/λ 1/8 - R3 2.93 cm L6/λ 1/2 - R4 4.96 cm Rg 42.1 µm R5 3.23 cm ˙ Rs 3.81 µm HA 123 W ˙ RA 4.45 cm HB 661 W ˙ RB 4.45 cm QH 538 W ˙ R1 4.45 cm WA 211 W ˙ Rload 9.94 mm Wload 185 W f 84 Hz ηA 0.39 - · 6 p0 3.1 10 Pa ηload 0.35 - TC 300 K ηA,R 0.79 - TH 600 K ηload,R 0.69 - 1 8 M˙ 0 kg s− RELTOL 4.14 · 10− -

Table 7.2: Input and output parameters of minimization routine.

A B load 1 56 2 3 1 m 4

Figure 7.12: Scaled version of geometry obtained from minimization routine. From top to bot- tom and left to right we have: regenerator (A), thermal buffer tube (B), tube with T-junction (1,2), inertance tube (3), compliance (4), resonator with load (5,6). The input and output param- eters are given in Tables 7.1 and 7.2.

W˙ at the T-junction. Inside the resonator there is dissipation as well, so that eventually ˙ an amount of Wload = 185 W can be provided to the acoustic load. Next, in figures 7.14-7.17 we show the volumetric velocity A u and the pressure gh 1i p1 all along the system. At the regenerator we have a small velocity, a large pressure and a real impedance. The length of the whole loop is much smaller than a wave length, and it is positioned near a pressure antinode of the resonator, so that the pressure is approximately constant in the loop and the velocity close to zero. At the T-junction there is a drop in A u due to the flow into the resonator. gh 1i Traveling-wave devices 131

700 250 B 1 650 200 5 600 150 A 550 100 500 acoustic power (W) acoustic power (W) 2 3 4 50 450 6 400 0 0 0.2 0.4 0.6 0.8 1 0 1 2 3 4 5 6 x(m) x(m) (a) in loop (b) in resonator

Figure 7.13: Acoustic power as a function of the position in the prime mover.

−3 x 10 10 0.06 B 1 0.04 9 2 3 /s) /s) 3 3 0.02 8 4 >) (m >) (m 1 A 1 0 A

−3 x 10 6 0 6 −0.02 4 5 −0.04 /s) /s) 3 3 2 −0.06 ) (m ) (m 1 1 U U

g −0.08 g 0

Im(A −0.1 Re(A 6 −2 5 −0.12

−4 −0.14 0 2 4 6 8 0 1 2 3 4 5 6 x(m) x(m) (a) Re A u (b) Im A u gh 1i gh 1i

Figure 7.15: Real and imaginary part of volumetric velocity as a function of the position in the resonance tube. 132 7.3 Results

5 x 10 1.56 3000 4 B 2500 1.54 2000 A 1 1.52 2 3 1500 ) (Pa)

) (Pa) A 1 1 1.5 1000 Im(p Re(p B 3 500 2 1.48 4 1 0 1.46 −500 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 x(m) x(m)

(a) Re p1 (b) Im p1

Figure 7.16: Real and imaginary part of acoustic pressure as a function of the position in the loop.

4 x 10 15 4000

3000

10 2000 1000 ) (Pa) ) (Pa) 6 1 5 1 0

5 Im(p Re(p 5 −1000

−2000 6 0 −3000 0 1 2 3 4 5 6 0 1 2 3 4 5 6 x(m) x(m)

(a) Re p1 (b) Im p1

Figure 7.17: Real and imaginary part of acoustic pressure as a function of the position in the resonance tube. Chapter 8

Nonlinear standing waves

It is well-known [28, 32, 33, 39] that disturbances can arise in a closed tube, when a pis- ton supplies oscillations at or near a resonance frequency. In particular shock waves may occur, or deformed sinusoidal profiles. Nonlinear standing waves may also occur in practical thermoacoustic devices, where high amplitudes are quite common. Highly nonlinear acoustic-wave forms have been observed experimentally [10], arising due to generation of higher harmonics and generating losses that degrade the performance. Dequand et al. [36] also observed shock formation in side-branch systems of the type shown in Figure 1.9. Replacing the thermoacoustic stack by a piston, Gaitan and Atch- ley [43] analyzed the energy transfer into the higher harmonics for a prime-mover con- figuration and showed that it can be partly suppressed by varying the cross-section of the prime mover. Hofler’s refrigerator (Figure 1.4) also uses a variable cross-section with a horn and reservoir to avoid the formation of shock waves. An elaborate theoretical model for calculating nonlinear disturbances (both shock and pre-shock) in closed tubes was derived by Chester [28] and improved upon by var- ious others [62,65,93,94,144]. Chester’s model includes effects of compressive viscosity and of shear viscosity in the boundary layers near the walls, and starts from the fun- damental one-dimensional hydrodynamic equations with a correction for wall friction using boundary-layer equations. Coppens and Sanders [32] derive a one-dimensional nonlinear acoustic-wave equa- tion describing finite-amplitude (pre-shock) standing waves in closed tubes. Using a perturbation expansion they arrive at a set of linear equations that need to be solved it- eratively. This was later extended [33] to a three-dimensional model for acoustic waves in “lossy” cavities. Menguy and Gilbert [87] have also analyzed nonlinear oscillations in gas-filled tubes, both theoretically and experimentally. In their approach several small parameters are identified and a multiple-scales perturbation solution is attempted. Starting from the Navier-Stokes equations, two generalized Burgers equations are derived that describe the propagation of a weakly nonlinear acoustic wave propagating in a cylindrical tube before shock formation. These equations are solved numerically to second order using truncated Fourier series, while allowing for arbitrary boundary conditions. Finally, the results are successfully validated against experimental results for standing-wave and traveling-wave configurations. 134 8.1 Governing equations

Much of the later work on nonlinear standing waves [1, 31, 39, 51,60,94,122] starts with the Kuznetsov’s equation [70], a single second-order nonlinear wave equation that describes the nonlinear propagation of sound in a dissipative and irrotational medium. Neglecting wall friction, Enflo and Hedberg [40] used Kuznetsov’s equation to derive a one-dimensional evolution equation for the velocity potential and found a steady-state solution also using a multiple-scales approach. The main difference with the analysis of Menguy and Gilbert [87] is the assumption of irrotational flow, which allows for the computation of an analytical solution. Makarov and Ochmann [82, 83, 96] have written an extensive three-part review on all modern investigations of nonlinear and thermoviscous phenomena in acoustic fields in fluids. They treat [83] various nonlinear evolution equations, most of which start from the simple wave equation with nonlinear source terms. In particular they also go into the Kuznetsov equation and the Khokhlov-Zabolotskaya-Kuznetsov (KZK) equa- tion which is a generalization of the Kuznetsov equation and includes the effect of weak diffraction. Ochmann and Makarov [96] also address the topic of thermoacoustic vibra- tions in a closed pipe with an infinitesimally thin heater located in the center of the pipe. By relating the heat release of the heater to the pressure fluctuations, shock waves can be predicted [29,30]. In this chapter we will try to predict the wave forms that may arise when a ther- moacoustic refrigerator or prime mover is excited near its resonance frequency. Only wide tubes are considered with uniform cross-sections in which the viscous interaction with the wall can be neglected. We generalize the approach of Enflo and Hedberg [40] to predict the nonlinear wave profiles when a thermoacoustic stack is placed inside the tube. As in [40] we will use Kuznetsov’s equation to determine the sound field in the tube, but with a reflection condition to simulate the presence of a thermoacoustic stack. Gusev et al. [49] also use a reflection condition to model a prime mover, but assume that the sound field inside the stack is not affected by the nonlinearities.

8.1 Governing equations

To simplify analysis we assume irrotational flow, i.e. v = 0. As a result the flow velocity can be described as the gradient of a velocity poten∇ × tial,

v = Φ. −∇ The assumption of irrotational flow is correct, provided the viscous interaction with the wall can be neglected. This is indeed the case, since we consider wide and straight tubes with negligible wall dissipation. Moreover, we consider such cases in which the wall dissipation can be neglected with respect to the dissipation in the shock.

8.1.1 Kuznetsov’s equation We start from the Kuznetsov equation [70], which is given by

∂2Φ ∂ 1 ∂Φ 2 b c2 2Φ = ( Φ)2 + (γ 1) + 2Φ , (8.1) ∂ 2 − 0∇ ∂t ∇ 2 − ∂t ρ ∇ t " 2c0   0 # Nonlinear standing waves 135

where c0 and ρ0 are the equilibrium values of the speed of sound and density of the fluid, independent of x, and b is a constant that represents the total effect of viscosity and heat conduction, 1 1 4 b = K + µ + ζ. Cv − Cp ! 3 As before γ is defined as the ratio of the specific heats

Cp γ = . Cv Let the ends of the tube be denoted by x = 0 and x = L. From here on we will restrict ourselves to one space dimension with the velocity u = ∂Φ/∂x as dependent variable, −

∂2Φ ∂2Φ ∂ ∂Φ 2 1 ∂Φ 2 b ∂2Φ c2 = + (γ 1) + . (8.2) ∂ 2 − 0 ∂ 2 ∂t ∂x 2 − ∂t ρ ∂ 2 t x "  2c0   0 x # If the left end of the tube is closed then a zero velocity is imposed

u(0, t) = 0.

It follows that the sound reflects at x = 0 with reflection coefficient R = 1. We will now model the presence of a stack at x = 0 by requiring the wave to reflect effectively at a point τL < 0 with reflection coefficient R , − 0 R( τL, t) = R . − 0

Depending on how much sound is dissipated in the stack R0 will vary between 0 (very dense stack) and 1 (no stack). Assuming a short stack with relatively wide pores one can expect that R0 will be close to 1. At x = L we impose an excitation h with angular frequency ω and amplitude U.

u(L, t) = Uh(ωt) = ℓωh(ωt). where ℓ is the maximal displacement of the excitation. The sound source will be mod- eled by putting h(t) = sin(t). We next rescale

L ˜ 2 x = Lx˜, t = t˜, u = c0u˜, Φ = Lc0Φ, p = ρ0c0 p˜. c0 The relevant dimensionless numbers for this problem are

2 U ω Lb ωL γ + 1 Cp ε µ κ β γ = , = 3 , = , = , = , (8.3a) c0 2ρ0c0 c0 2 Cv δ = 1 R , µ = µ(1 + τ), κ = κ(1 + τ), ∆ = κ π. (8.3b) − 0 0 0 0 − 136 8.1 Governing equations

Dropping the tildes we arrive at the following boundary value problem:

∂2Φ ∂2Φ ∂ ∂Φ 2 γ 1 ∂Φ 2 2µ ∂2Φ = + − + , (8.4) ∂ 2 − ∂ 2 ∂t ∂x 2 ∂t 2 ∂ 2 t x "    κ x # subject to

R( τ, t) = R , (8.5a) − 0 ∂Φ u(1, t) = (1, t) = εh(κt). (8.5b) − ∂x

8.1.2 Bernoulli’s equation

If the density is assumed constant and gravity is neglected, the momentum balance as part of the Euler equations can be integrated to Bernoulli’s equation for unsteady potential flow ∂ Φ 1 2 p ∂ + u + = C, (8.5c) t 2 ρ0 where C only depends on time. Therefore the pressure p can be expressed in Φ and u as

1 ∂Φ p = ρ C u2 . (8.5d) 0 − 2 − ∂t   With the rescaling given above, we find that the dimensionless pressure is given by

∂Φ 1 p(x, t) = C(t) (x, t) u2(x, t). (8.6) − ∂t − 2

8.1.3 Perturbation expansion

We will use the smallness of Φ, ε, µ, δ, and ∆ to find an approximate solution correct up to second order in these small parameters. To make a small-parameter analysis possible we will use ε as the small parameter and relate µ, δ and ∆ to ε. First, assuming Φ is small, we expand

Φ(x, t;ε) = εΨ(x, t) + ε2ψ(x, t) + (ε3), ε 1. (8.7) O ≪ Substituting the expansion into (8.4) we find that to leading order

∂2Ψ ∂2Ψ = 0, ε 1, µ 1. ∂t2 − ∂x2 ≪ ≪ The general solution is given as a sum of left and right-running waves

Ψ(x, t) = F (t x) + F (t + x), (8.8) 1 − 2 Nonlinear standing waves 137

for still arbitrary functions F1 and F2. Next assuming µ = (ε) and collecting the first- order terms in ε and µ, we find O

∂2ψ ∂2ψ =(γ + 1) f1(t x) f1′ (t x) + f2(t + x) f2′ (t + x) ∂t2 − ∂x2 − − h 2µ i +(γ 3) f (t x) f ′ (t + x) + f ′ (t x) f (t + x) + f ′′(t x) + f ′′(t + x) . − 1 − 2 1 − 2 εκ2 1 − 2 h i h i where f F′ . If we change to the variables 1,2 ≡ 1,2 ξ = t x, η = t + x, (8.9) − then we find

∂2ψ 4 =(γ + 1) f (ξ) f ′ (ξ) + f (η) f ′ (η) +(γ 3) f (ξ) f ′(η) + f ′ (ξ) f (η) ∂ξ∂η 1 1 2 2 − 1 2 1 2 h i h 2µ i + f ′′(ξ) + f ′′(η) . εκ2 1 2 h i Integrating with respect to ξ and η we find

γ + 1 4ψ(ξ, η) = η f 2(ξ) + ξ f 2(η) +(γ 3) F (ξ) f (η) + f (ξ)F (η) 2 1 2 − 1 2 1 2 µ h i h i + 2 η f ′ (ξ) + ξ f ′ (η) + A(ξ) + B(η). εκ2 1 2 h i A suitable redefinition of the integration constants

γ + 1 2 2µ A(ξ) := ξ f (ξ) ξ f ′ (ξ) + G (ξ), − 2 1 − εκ2 1 1 γ + 1 2 2µ B(η) := η f (η) ξ f ′ (η) + G (η), − 2 2 − εκ2 2 2 leads to γ + 1 ψ(x, t) = G (t x) + G (t + x) + x f 2(t x) + f 2(t + x) 1 − 2 4 1 − 2 γ 3 h i + − F (t x) f (t + x) + f (t x)F (t + x) 4 1 − 2 1 − 2 µ h i + x f ′ (t x) + f ′ (t + x) . (8.10) εκ2 1 − 2 h i 138 8.1 Governing equations and

Φ(x, t) = ε F (t x) + F (t + x) + ε2 G (t x) + G (t + x) 1 − 2 1 − 2 h γ + 1 i h i + ε2 x f 2(t x) f 2(t + x) 4 1 − − 2 γ 3 h i + ε2 − F (t x) f (t + x) + f (t x)F (t + x) 4 1 − 2 1 − 2 εµ h 3 i + x f ′ (t x) f ′ (t + x) + (ε ), (8.11) κ2 1 − − 2 O h i so that

u(x, t) = ε f (t x) f (t + x) + ε2 g (t x) g (t + x) 1 − − 2 1 − − 2 n γ + 1 o n o ε2 f 2(t x) f 2(t + x) − 4 1 − − 2 nh i 2x f (t x) f ′ (t x) + f (t + x) f ′ (t + x) − 1 − 1 − 2 2 2 γh 3 io ε − F (t x) f ′ (t + x) f ′ (t x)F (t + x) − 4 1 − 2 − 1 − 2 εµ n o 3 f ′ (t x) f ′ (t + x) x f ′′(t x) + f ′′(t + x) + (ε ). (8.12) − κ2 1 − − 2 − 1 − 2 O n h io Note that some of the second-order terms on the right hand side are of the same form. We can therefore redefine gi as

γ + 1 2 µ g˜ (t) := g (t) f (t) + f ′(t) , i = 1,2, i i − 4 i ε i   and find

u(x, t) = ε f (t x) f (t + x) + ε2 g˜ (t x) g˜ (t + x) 1 − − 2 1 − − 2 n 2 γ + 1 o n o + ε x f (t x) f ′ (t x) + f (t + x) f ′(t + x) 2 1 − 1 − 2 2 2 γ 3 h i + ε − f ′ (t x)F (t + x) F (t x) f ′ (t + x) 4 1 − 2 − 1 − 2 εµ n 3 o + x f ′′(t x) + f ′′(t + x) + (ε ). (8.13) κ2 1 − 2 O h i In [39] a zero velocity was imposed at x = 0, so that f1 = f2 and g˜1 = g˜2. Here the situation is a bit different due to condition (8.5a). It assumes that because of the presence of the stack the left-running wave is effectively reflected at some point τ 0 − ≤ with reduced amplitude R0. As a result we can put

f (t x) = f (t x), f (t + x) = f (t + x), x = x + τ, (8.14) 1 − − 2 g˜ (t x) = g(t x), g˜ (t + x) = g(t + x), (8.15) 1 − − 2 Nonlinear standing waves 139 where f and g are yet to be determined. Thus we can write for u

2 2 u(x, t) = ε R f (t x) f (t + x) + ε βx R f (t x) f ′(t x) + f (t + x) f ′(t + x) 0 − − 0 − − n 2 γ 3 o h i + ε − R f ′(t x)F(t + x) F(t x) f ′(t + x) (8.16) 4 0 − − − εµ n 2 o 3 + x R f ′′(t x) + f ′′(t + x) + ε R g(t x) g(t + x) + (ε ). κ2 0 − 0 − − O h i n o The remaining boundary condition (8.5b) can be used to derive an equation for the unknown functions f and g,

2 h(κt) = R f (t t ) f (t + t ) + εβt R f (t t ) f ′(t t ) + f (t + t ) f ′(t + t ) 0 − 0 − 0 0 0 − 0 − 0 0 0 nγ 3 o h i + ε − R f ′(t t )F(t + t ) F(t t ) f ′(t + t ) 4 0 − 0 0 − − 0 0 εµ0 n o 2 + R f ′′(t t ) + f ′′(t + t ) + ε R g(t t ) g(t + t ) + (ε ), (8.17) κ2 0 − 0 0 0 − 0 − 0 O h i n o where t0 = 1 + τ. In the sections below we will solve this equation for two cases; we first look for a solution away from resonance for an arbitrary excitation h and R0 = t0 = 1, and next we look for a solution near resonance with h(t) = sin(t) and arbitrary R0 and t0. Having done that we can use the second solution to determine the sound field between the right stack end and the piston when the tube is near resonance and the first solution can be used to determine the sound field between the closed end (with R0 = t0 = 1) and the left stack end.

8.2 Solution away from resonance

Depending on the order of magnitude of ε, µ, and δ, different approximations to (8.17) are possible. We assume µ, δ = (ε) and introduce O µ δ µˆ = , δˆ = . (8.18) ε ε Moreover, for the subsequent analysis we also need

κ = nπ, n Z. (8.19) 0 6 ∈ The last condition implies that we have to stay away from resonance (∆ = 0) and is necessary to avoid division by zero. Below we will consider two cases; first for6 arbitrary excitation h and then for harmonic excitation h(t) = sin(t). 140 8.2 Solution away from resonance

8.2.1 Arbitrary excitation

Substituting (8.18), we find that (8.17) transforms into

h(κt) = f (t t ) f (t + t ) + εβt f (t t ) f ′(t t ) + f (t + t ) f ′(t + t ) − 0 − 0 0 − 0 − 0 0 0 n γ 3 o h i 2εδˆ f (t t ) + ε − f ′(t t )F(t + t ) F(t t ) f ′(t + t ) − − 0 4 − 0 0 − − 0 0 µˆ0 n o2 + ε f ′′(t t ) + f ′′(t + t ) + ε g(t t ) g(t + t ) + (ε ). (8.20) κ2 − 0 0 − 0 − 0 O h i n o Setting ε = 0 we find to leading order

f (t t ) f (t + t ) = h(κt). − 0 − 0 We will look for a solution to this equation in the frequency domain. Taking the Fourier transform we find F 1 ω 2i sin(ωt ) [ f ] (ω) = [ h] . (8.21) − 0 F κ F κ   1 Rearranging terms and taking the inverse Fourier transform − , we obtain F ∞ (ω) 1 inπt/t0 i [ h] f (t) = − A (t) + ∑ ane , A(ω) = F , (8.22) F n= ∞ 2 sin(ωt0) h i − where the second term is a solution of the homogeneous problem. From causality argu- ments we conclude that an = 0 for all n. Next, collecting the second-order terms, we find

g(t t ) g(t + t ) = 2δˆ f (t t ) βt f (t t ) f ′(t t ) + f (t + t ) f ′(t + t ) − 0 − 0 − 0 − 0 − 0 − 0 0 0 h i µˆ0 β 2 f ′′(t t ) + f ′′(t + t ) − f ′(t t )F(t + t ) F(t t ) f ′(t + t ) . − κ2 − 0 0 − 2 − 0 0 − − 0 0 h i n o Taking the Fourier transform on either side, we find

2i sin(ωt ) [ g] (ω) = [ k] (ω), (8.23) − 0 F F with

k(t) = 2δˆ f (t t ) βt f (t t ) f ′(t t ) + f (t + t ) f ′(t + t ) − 0 − 0 − 0 − 0 0 0 h i µˆ0 β 2 f ′′(t t ) + f ′′(t + t ) − f ′(t t )F(t + t ) F(t t ) f ′(t + t ) . − κ2 − 0 0 − 2 − 0 0 − − 0 0 h i n o Finally, applying the inverse Fourier transform we obtain

1 i [ k] (ω) g(t) = − B (t), B(ω) = F . (8.24) F 2 sin(ωt0) h i Nonlinear standing waves 141

8.2.2 Harmonic excitation

Substituting h(t) = sin(t) into (8.21), we find

√2π √2π [ f ] (ω) = [δ(ω + κ) δ(ω κ)] = [δ(ω + κ) + δ(ω κ)] . F − 4 sin(ωt0) − − 4 sin(κ0) −

Here and below δ( ) is used to denote the δ-function and is not to be confused with the constant δ defined· in (8.3). We can take the inverse Fourier transform and find

cos(κt) f (t) = . (8.25) 2 sin(κ0) Next, substituting f into (8.23), we obtain

βκ √2π [ g] (ω) = 0 [δ(ω + 2κ) + δ(ω 2κ)] F 8 sin(2κ0) tan(κ0) 2 − (µˆ + δˆ) √2πi 0 [δ(ω + κ) δ(ω κ)] − 2 sin(κ0) tan(κ0) 2 − − δˆ √2π + [δ(ω + κ) + δ(ω κ)] . 2 sin(κ0) 2 − Finally, applying the inverse Fourier transform we obtain

βκ cos(2κt) µˆ + δˆ cos(κ ) sin(κt) δˆ cos(κt) g(t) = 0 0 0 + . (8.26) 2 2 κ 4 sin (κ0) − 2 sin (κ0) 2 sin( 0) The full solution, correct up to second order in ε, can be found by substituting f and g into (8.16). For example, if we use only the leading-order terms, we find

u(x, t) = ε [ f (κt κx) f (κt + κx)] + o(ε) − − sin(κx) = ε sin(κt) + o(ε). (8.27) sin(κ0)

For t0 = 0 this coincides with the standard solution for the sound field in closed tubes without wall friction.

8.3 Solution near resonance

In the previous section we showed how equation (8.4) can be solved using Fourier trans- formations, provided one stays away from resonance. In this section we will show that under certain conditions we can find a solution near resonance using matched asymp- totic expansions. With the substitution

f (t) = v(κt), g(t) = w(κt), vˆ′ = v, ζ = κt κ , ζ = κt + κ , 1 − 0 2 0 142 8.3 Solution near resonance we arrive at the following equation

2 h(κt) = R v(ζ ) v(ζ ) + εβκ R v(ζ )v′(ζ ) + v(ζ )v′(ζ ) 0 1 − 2 0 0 1 1 2 2 n o h β 2 i + µ R v′′(ζ ) + v′′(ζ ) + εR − v′(ζ )vˆ(ζ ) vˆ(ζ )v′(ζ ) 0 0 1 2 0 2 1 2 − 1 2 h i n o + ε R w(ζ ) w(ζ ) + o(ε). (8.28) 0 1 − 2 n o When ω approaches the first resonance frequency, in the sense that ∆ 1, we can find an approximate solution to (8.28). Since ∆ = κ π, we have ≪ 0 − ζ = κt π ∆, ζ = κt + π + ∆. 1 − − 2 Moreover, we can substitute for v(ζ ) v(ζ ) the difference of their series expansion 1 − 2 v(ζ ) v(ζ ) = v(κt π) v(κt + π) ∆ v′(κt π) + v′(κt + π) , ∆ 1. 1 − 2 − − − − ≪ If h is a periodic excitation near resonance, we can expect the function v to be almost periodic. We model this by introducing two time scales,

ζ = κt + π, T = εκt, v˜(ζ, T) = v(ζ), where ζ is the “fast” time and T is the “slow” time. We then assume that the deviation from periodicity occurs in the slow time scale and not in the fast time scale, so that v˜(ζ, T) = v˜(ζ 2π, T). Dropping the tildes, we can now write − ∂v v(ζ 2π, T 2πε) = v(ζ 2π, T) 2πε (ζ 2π, T) − − − − ∂T − ∂v = v(ζ, T) 2πε (ζ, T), ε 1. − ∂T ≪ As a result we have ∂v v(κt + π) v(κt π) = v(ζ, T) v(ζ 2π, T 2πε) = 2πε (ζ, T), ε 1, − − − − − ∂T ≪ and ∂v ∂v v(ζ ) v(ζ ) = 2πε (ζ, T) 2∆ (ζ, T), ∆,ε 1. (8.29) 1 − 2 − ∂T − ∂ζ ≪ Inserting (8.29) into (8.28) and putting R = 1 δ, we find 0 − ∂v ∂v ∂v ∂2v h(ζ π) = 2πε 2∆ δv + 2εβκ v + 2µ + o(ε, ∆, µ , δ). (8.30) − − ∂T − ∂ζ − 0 ∂ζ 0 ∂ζ2 0

It follows that to match the left and right hand side of the equation, we need to rescale Nonlinear standing waves 143 v. We introduce V by

∆ 2 ζ v(ζ, T) = + V(z, T), z = , εβκ0 sεβκ0 2

This rescaling does not violate the asymptotic expansion given in (8.7) since ∆ 1. We find that (8.30) transforms into ≪

∂V 1 ∂2V δ δ∆ 2πε ∂V 1 V + ν V ν = h(2z π), (8.31) ∂ ∂ 2 µ 2 µ ∂ z " 2 z − 0 − µ0 − 0 T # 2 − where µ2 ν = 0 . s 2εβκ0 In Section 8.3.1 we will give an analytical (z, T)-dependent solution for the case δˆ = ∆ˆ = 0, derived by Rudenko et al. [122]. We conclude from (8.31) that the effects of the stack (δ), resonance (∆), viscosity (µ0), and nonlinearity (ε) are balanced with the energy inflow (h), when

δ ∆ µ √ε. ∼ ∼ 0 ∼ However, with this ordering of dimensionless parameters it is not possible to derive an analytic solution to (8.31). If instead we assume

δ ∆ µ , ν 1, ∼ ∼ 0 ≪ then a steady-state solution (T ∞) can be determined using the smallness of ν. At T ∞ a steady-state field is reached→ due to the interaction of dissipation, nonlinear → ∂ losses and energy inflow from the the source. Setting ∂T = 0, putting h(ζ) = sin(π), and introducing δ ∆ δˆ = , ∆ˆ = , µ0 µ0 we find dV d2V V + ν δˆV ν2δˆ∆ˆ = ( z) 2 ∂ 2 2 2 sin 2 . (8.32) z " ∂z − # − − In Section 8.3.2 we will derive a solution using matched asymptotic expansions, when δˆ, ∆ˆ = (1). Next in Section 8.3.3 we show numerically how the solution changes when O 1 the dissipation due to the stack is increased such that δˆ = (ν− ). O

8.3.1 Exact solution when δˆ = ∆ˆ = 0 using Mathieu functions

Consider a closed tube at resonance (∆ = 0), periodically excited (h(t) = sin(t)), and without a stack ( τ = δ = 0). For this case it was shown by Rudenko et al. [64,122] that the solution to (8.31) with zero initial condition v(T = 0,ζ) = 0 can be written in terms 144 8.3 Solution near resonance of periodic Mathieu functions:

∞ λ µ V(z, T) = ∑ a exp 2n T CE (z, q) , (8.33) n − 4πε 2n n=0   where π 2 CE ζ , q dζ επβ 1 a = 0 0 2 , q = = , 2n 2π 2 ζ ζ µ2 ν2 R0 CE2n 2 , q
d 2 4 and λ2n is the characteristicR eigenvalue
corresponding to the periodic Mathieu eigen- function CE2n in the notation of Abramowitz and Stegun [3]. In particular when q 1 we can write ≫ 1 1 λ = 2q + 2√q + . 0 − − 4 − 32√q · · · The solution above simplifies in a few special cases. In the limit for T ∞ a steady- state solution is approached and →

2µ′ d ζ v(ζ) = ln CE , q . (8.34) πβ dζ 0 2   Since we assume ν 1 we also have q 1. As a result v can be approximated by [3] ≪ ≫ 2 ζ 2exp 2√qζ v(ζ) = cos − , 0 ζ π, (8.35) sεπβ 2 − 1 + 2exp 2√qζ ≤ ≤ "   −
# so that for q ∞ (ν 0) we find
→ → 2 ζ v(ζ) = cos sgn(ζ), π ζ π. (8.36) επβ 2 − ≤ ≤ s  

8.3.2 Steady-state solution for δˆ = (1) O Assuming δˆ, ∆ˆ = (1), we can use the smallness of ν to find an approximate solution using the methodO of matched asymptotic expansions [57]. First we look for an outer ˆ solution V that is valid outside a narrow region around some point zi ( π/2, π/2). Then we look for an inner solution Vˇ that is valid within this small region.∈ − A combina- tion of these two solution will give the full composite expansion. We need some boundary conditions to guarantee a unique solution. A natural con- dition would be V( π/2) = V(π/2), (8.37) − so that in resonance, when ∆ = 0, v wil be periodic as well. Moreover, in the transi- tion point we will assume V(zi) = 0 for reasons of symmetry. The location zi will be determined from the matching conditions for Vˆ and Vˇ in the intermediate region. We expand the outer solution in powers of ν,

Vˆ = Vˆ + νVˆ + , ν 1. 0 1 · · · ≪ Nonlinear standing waves 145

Inserting the expansions into (8.32) and collecting the terms of order ν0 and ν1, we find

dVˆ 2Vˆ 0 = sin(2z), (8.38) 0 dz − 2 ˆ ˆ ˆ d V0 dV0V1 ˆ + 2 2δV0 = 0. (8.39) dz2 dz − Integrating (8.38) we find

Vˆ = cos2(z) + C . (8.40) 0 ± 0 q where C0 is a constant of integration. Applying condition (8.37), we find that C0 = 0 and

Vˆ = cos(z). (8.41) 0 ± ˆ ˆ We see that two possible solutions for V0 exist. Now before we go to V1, we will first turn to the inner solution. We expect that in the narrow region around zi there will be a transition between the “-” and the “+” solution. Therefore an inner solution Vˇ is introduced as z z V(z) = Vˇ (s), s = − i . ν Substitution into equation (8.32) yields

2 ˇ d V dV 2 ˆ ˆ ˆ 3 + 2V 2ν δV = ν sin(2zi + 2νs) + 2δ∆ν , ν 1. (8.42) ds2 ds − − ≪ Again we expand in powers of ν,

Vˇ (s) = Vˇ (s) + νVˇ (s) + , ν 1. 0 1 · · · ≪

It turns out that to find a solution we need to let zi depend on ν and we expand

z = z + νz + , ν 1. i i0 i1 · · · ≪ Next we substitute the expansion and equate the coefficients of ν0 and ν1 to find

2 ˇ ˇ d V0 ˇ dV0 + 2V0 = 0, (8.43) ds2 ds 2 ˇ ˇ ˇ d V1 dV0V1 + 2 = sin(2zi ). (8.44) ds2 ds − 0

We integrate (8.43) subject to Vˇ (0) = 0. By separation of variables we find ˇ V0 = a0 tanh (a0s) , (8.45) where a0 is yet to be determined. We can determine a0 by matching the inner solution 146 8.3 Solution near resonance

Vˇ to the outer solution Vˆ found above. Taking the limit for s ∞, we observe that 0 0 → ± ˇ ˇ lim V0 = a0 , lim V0 = a0 , s ∞ s +∞ →− −| | → | | ˇ ˆ and we see that the inner solution V0 and outer solution V0 match, provided the change a = (z ) is from the “-” solution to the “+” solution and 0 cos i0 0. Summarizing, we find ≥ cos(z), ζ < z , ˆ i V0(z) = − (8.46) ( cos(z), ζ > zi, provided z stays away from π/2. Furthermore i ± Vˇ (s) = a [a s] a = (z ) 0 0 tanh 0 , 0 cos i0 . (8.47) ˆ Next we will compute the first-order terms. Substituting V0 into (8.39), we find after integration

1 C 1 Vˆ = 1 + 1 + 2δˆ tan(z). (8.48) 1 ± 2 cos(z) 2
Applying (8.37), we obtain C = 1 2δˆ and 1 − − ˆ sin(z) + 1 1 + 2δ , ζ < zi, ˆ 2cos(z) V1(z) =  (8.49) 
sin(z) 1  1 + 2δˆ − , ζ > z . 2cos(z) i 
 Note that only this choice for C1 cancels the singularity that would otherwise arise in ˇ z = π/2. It still remains to determine V1. We integrate (8.44) once with respect to s and apply± the integrating factor

ˇ 2 exp 2 V0 ds = cosh (a0s).  Z  We can then rewrite (8.44) as

d cosh2(a s)Vˇ = s sin(2z ) cosh2(a s) B cosh2(a s), ds 0 1 − i0 0 − 1 0 h i where B1 is a constant of integration. Integrating once again with respect to s, we get

2 1 s s sin(2zi ) Vˇ = sin(2z ) B + 0 [cosh (2a s) 1] 1 2 i0 1 2 0 cosh (a0s) (− 4 − 2 8a0 − 1 B + sin(2z )s sinh (2a s) . − 4a 1 i0 0 0    However, the first-order outer and inner solution do not necessarily attain a common limit as is the case for the zeroth-order terms. We will show that they only match when Nonlinear standing waves 147 sin(2z ) is close to zero, thereby fixing z . For this purpose we introduce an intermedi- i0 i0 ate variable z z ν σ = − i = s, η(ν) 1, η(ν) η(ν) ≪ that is positioned between the (1) coordinate of the outer layer and the (ν) coordi- nate of the inner layer. We thereforeO assume O

ν η(ν) 1, ≪ ≪ and substitute σ in the outer and inner solution in some overlapping region where both solutions should hold. For this we choose η1(ν) and η2(ν) such that ν η1(ν) η(ν) η (ν) 1 where the outer solution is valid for η that satisfies η (ν) ≤ η(ν) ≪1 ≪ 2 ≤ 1 ≪ ≤ and the inner solution for η that satisfies ν η(ν) η2(ν). Without loss of generality we may assume σ > 0, so that ≤ ≪

sin(z + ησ) 1 sin(zi ) 1 Vˆ = 1 + 2δˆ i − = 1 + 2δˆ 0 − + o(1), 1 2cos(z + ησ) 2cos(z ) i i0

and since η/ν 1 ≫

1 ση 2 ση sin(2zi ) η Vˇ = sin(2z ) B + 0 cosh 2a σ 1 1 2 η i0 1 2 0 cosh (a σ ) (− 2ν − 2ν 8a ν − 0 ν   0 h   i 1 η η B + ( z )σ a σ 1 sin 2 i0 sinh 2 0 − 4a0 ν ν h i   δˆ sin(zi ) 1 = 1 + 4 0 − + o(1), µˆ 2cos(z )  0  i0 provided z = 0, B = 1 + 2δˆ. (8.50) i0 1 With these restrictions Vˆ and Vˇ will attain a common limit for z z = (η(ν)). Thus 1 1 − i O the matching condition fixes the position of the boundary layer. If necessary, zi can be determined up to higher-order accuracy by computing and matching the higher-order z V inner and outer solutions. However, knowledge of i0 is enough to determine up to first-order accuracy. Summarizing we find that the outer solution Vˆ and the inner solution Vˇ are given by sin(z) + 1 cos(z) + ν 1 + 2δˆ + o(ν), ζ < z , − 2cos(z) i Vˆ (z) = 
(8.51)  ˆ sin(z) 1  cos(z) + ν 1 + 2δ − + o(ν), ζ > zi, 2cos(z) 
and since a0 = 1,

ν s ˇ δˆ V(s) = tanh (s) 1 + 2 2 + tanh(s) . (8.52) − 2 ( cosh (s) )
148 8.3 Solution near resonance

The composite solution can be found by adding both solutions and subtracting the com- mon part. We find

z z V(z) = sgn(z z ) [cos(z) 1] + tanh − i − i − ν   ν sin(z) sgn(z z ) + 1 + 2δˆ − − i + sgn(z z ) 2 cos(z) − i 
(z z )/ν z z − i tanh − i + o(ν). (8.53) − 2 z zi − ν cosh −ν  ) As a result we obtain for ν 1
≪ ∆ 2 ζ ζ v(ζ) = + sgn (ζ) cos 1 + tanh (8.54) εβκ εβκ 2 − 2ν 0 s 0        µ + 2δ sin ζ sgn (ζ) ζ ζ + 0 2 − + sgn (ζ) tanh + , 2εβκ cos ζ − ν 2 ζ − 2ν · · · 0 (
2 2 cosh 2ν  )

∞ where we put zi = 0. Since tanh(x) approaches sgn(x) if x and cos(x) ap- proaches 1 if x 0, we can rewrite the solution as → ± → ∆ 2 ζ ζ v(ζ) = + cos tanh εβκ εβκ 2 2ν 0 s 0     µ + 2δ ζ tanh ζ ζ + 0 tan 2ν + . (8.55) 2εβκ 2 − cos ζ − ν 2 ζ · · · 0 (   2
2 cosh 2ν ) Letting ∆, δ, ν 0 this expression coincides with
the one derived
in equation (8.36) using Mathieu functions.→ Finally note that if we compute the average v, while omitting the odd parts of the integrand, we find to leading order

1 π ∆ v = v(ζ) dζ = + o(1), ν 1. (8.56) 2π π βκ ε ≪ Z− 0 We conclude that in exact resonance, when ∆ = 0, v will have zero average.

1 8.3.3 Steady-state solution for δˆ = (ν− ) O Previously we assumed δˆ = (1) with ε. In this section we will investigate what hap- pens when the dissipation byO the stack is increased by assuming νδˆ = (1). More precisely we will define O 2νδ δˇ := 2νδˆ = . (8.57) µ0 Nonlinear standing waves 149

With this rescaling, equation (8.32) transforms into

d2V dV ν + 2V δˇV = sin(2z) + νδˇ∆ˆ . (8.58) dz2 dz − − Again we will use the smallness of ν to find an approximate solution via the method of matched asymptotic expansions using an outer solution Vˆ and an inner solution Vˇ that is valid inside a narrow region around some point zi ( π/2, π/2). Remember the boundary conditions are periodicity for the outer solution,∈ −

V( π/2) = V(π/2), (8.59) − and symmetry for the inner solution,

V(zi) = 0. (8.60)

Before we turn to the outer solution we will first determine the inner solution. As before we introduce Vˇ by z z V(z) = Vˇ (s), s = − i . ν

Then Vˇ satisfies 2 ˇ ˇ d V ˇ dV ˇ ˇ ˇ ˆ 2 + 2V δνV = ν sin(2zi + 2νs) + δ∆ν , ν 1. (8.61) ds2 ds − − ≪ We expand in powers of ν 1, ≪ Vˇ (s) = Vˇ (s) + νVˇ (s) + , 0 1 · · · z = z + νz + . i i0 i1 · · · Next we substitute the expansions into (8.58) and equate the coefficients of ν0 and ν1 to find 2 ˇ ˇ d V0 ˇ dV0 + 2V0 = 0, (8.62) ds2 ds 2 ˇ ˇ ˇ d V1 dV0V1 ˇ ˇ + 2 δV0 = sin(2zi ). (8.63) ds2 ds − − 0

Then we integrate (8.62) subject to Vˇ (0) = 0. By separation of variables, we find ˇ V0 = a0 tanh [a0s] , (8.64) where a0 is yet to be determined. Without loss of generality we may assume a0 0. It ˇ ≥ still remains to determine V1. We integrate (8.63) once with respect to s and apply the integrating factor ˇ 2 exp 2 V0 ds = cosh (a0s).  Z  150 8.3 Solution near resonance

We can then rewrite (8.63) as

d cosh2(a s)Vˇ = sin(2z )s + B cosh2(a s) + δˇ log [cosh(a z)] , ds 0 1 − i0 1 0 0 h i
where B1 is a constant of integration. Integrating once again with respect to s, we get

2 1 s s sin(2zi ) Vˇ = sin(2z ) B + 0 [cosh (2a s) 1] 1 2 i0 1 2 0 cosh (a0s) (− 4 − 2 8a0 − 1 B + sin(2z )s sinh (2a s) + δˇL(s) . − 4a 1 i0 0 0    where s L(s) := log [cosh(a0σ)] dσ Z0 2 1 2 π 1 2s = s s log 2 + + dilog(1 + e− ) 2 − 24 2 2 ∞ n 2ns 1 2 π 1 ∑ ( 1) e− s s log 2 + + − 2 s 0, 2 − 24 2 n=1 n ≥ =  ∞  1 π 2 1 ( 1)ne2ns  2 ∑  s s log2 − 2 s 0. − 2 − − 24 − 2 n=1 n ≤  It follows that to avoid blow-up of the solution for large s we need to have sin(2z ) = 0, i0 z = i.e. i0 0, and we obtain

1 s B Vˇ = B 1 sinh (2a s) + δˇL(s) . (8.65) 1 2 − 1 2 − 4a 0 cosh (a0s)  0 

The constants a0 and B1 will be determined from the matching condition between the inner and outer solution. Taking the limits for s ∞, we find the following matching conditions for Vˆ : → ± ˆ ˆ lim V0 = a0 0, lim V0 = a0 0, (8.66) z 0 − ≤ z 0 ≥ ↑ ↓ ˆ B1 ˆ B1 lim V1 = , lim V1 = . (8.67) z 0 2a z 0 − 2a ↑ 0 ↓ 0 Next we turn to the outer solution. Substituting the expansion

Vˆ = Vˆ + νVˆ + , ν 1, 0 1 · · · ≪ Nonlinear standing waves 151 into (8.58) and collecting the terms of order ν0 and ν1, we find

dVˆ 2Vˆ 0 δˇVˆ = sin(2z), (8.68) 0 dz − 0 − d2Vˆ dVˆ dVˆ 0 + 2Vˆ 1 + 2 0 δˇ Vˆ = δˇ∆ˆ . (8.69) 2 0 dz dz − 1 dz   ˆ The equation for V0 cannot be solved in closed form and will be solved numerically. ˆ ˆ Therefore we will first turn to V1, which can be expressed in V0. With the integrating factor z ˆ z δˇ 1 dV0(s) ˇ ˆ exp 2 δ ds = E(z)V0(z), E(z) = exp ds , 0 ˆ ds − 0 ˆ Z 2V0(s)    Z 2V0(s)  we can rewrite (8.69) as

dEVˆ Vˆ E d2Vˆ 0 1 δˇ∆ˆ 0 = 2 . dz 2 " − dz #

One more integration yields

z E(s) d2Vˆ (s) Vˆ (z) = δˇ∆ˆ 0 ds. (8.70) 1 ˆ 2 Z 2E(z)V0(z) " − ds # ˆ ˆ Therefore once V0 is known we can apply (8.70) to determine V1. However, before we ˆ ˆ compute V0 numerically we will first examine some properties of V0. We consider the nonlinear differential equation

dY 2Y δˇY = sin(2z), (8.71) dz − − subject to either Y(π/2) = α, or Y( π/2) = β. Since 2Y′ = δˇ sin(2z)/Y it is clear that if Y(z) = 0 for any z away from 0− then the solution will break− down. This is shown in figure 8.1 in which we plot Y for various α or β. For α or β too close to zero, the solution may cross the the horizontal axis and break down. Theorem 1 below shows that, given α = 0, Y can only cross zero in the right part of the domain and for given β = 0 in the left6 part of the domain. 6

Theorem 1 Suppose Y satisfies (8.71), then for any ǫ with 0 < ǫ < π/2 it holds that (i) if α > 0, then Y(z) > 0 for all z (ǫ, π/2]; ∈ (ii) if β < 0, then Y(z) > 0 for all z [ π/2, ǫ); ∈ − − (iii) if α < 0, then Y(z) < 0 for all z (ǫ, π/2]; ∈ (iv) if β > 0, then Y(z) > 0 for all z [ π/2, ǫ). ∈ − − Proof. 152 8.3 Solution near resonance

(i) We will prove by contradiction. Suppose 0 < ǫ < π/2, α > 0 and z [ǫ, π/2) 0 ∈ such that Y(z0) = 0 and Y(z0) > 0 for all z [0, z0). Then Y′(z0) > 0. However, from (8.71) we also find ∈

δˇ sin(2z) lim Y′(z) = lim = ∞, z z0 z z0 2 − 2Y(z) − ↓ ↓ and we arrive at a contradiction. (ii-iv) The proof goes analogous to that of (i). 2

3 3 α = −3 α 2 = −2 2 α = −1 α 1 = −0.1 1 α = 0.1 Y α = 1 Y 0 0 β = 3 β = 2 −1 −1 β = 1 β = 0.1 −2 −2 β = −0.1 β = −1 −3 −3 −1.5 −1 −0.5 0 0.5 1 1.5 −1.5 −1 −0.5 0 0.5 1 1.5 z z (a) (b)

Figure 8.1: Y as a function of z for δˇ = 1 and (a) Y( π/2) = α or (b) Y(π/2) = β. If 0 α 2.5 or 0 β 2.5, then the solution will break− down when it crosses the horizontal axis.≤ ≤ ≥ ≥ −

It now follows from Theorem 1 and the matching condition (8.66) that a necessary ˆ condition for the inner and outer solutions to match is a change from a “-” solution V0− ˆ + (with β < 0) to a “+” solution V0 (with α > 0). Moreover, Theorem 1 ensures that such a negative solution always exists for z < 0 and such a positive solution for z > 0. Since α > 0 and β < 0 it is clear that (8.59) can only be satisfied when α 0 and β 0, although neither can be exactly equal to zero. Finally, there are still the↓ two matching↑ ˆ + conditions in (8.66) that need to be satisfied. If we take a0 := V0 (0), then the second of the two matching conditions will be satisfied. Next, if we also choose β = α, then it ˆ ˆ + − follows from Theorem 2 that V0−(0) = V0 (0) = a0 and the first condition in (8.66) will be satisfied as well. − −

Theorem 2 Suppose Y satisfies (8.71) subject to Y(π/2) = α, then Y: z Y( z) satisfies (8.71) with Y( π/2) = α. 7→ − − Proof. − − The statement follows immediately from substitution of Y into (8.71). 2

As an example of what V0 looks like we have plot in figure 8.2 the composite solution that is obtained by combining the leading order inner and outer solutions. The plots 5 have been computed for various values of ν and δˇ, α = 10− and ∆ = 0. We observe Nonlinear standing waves 153 that as the dissipation in the stack is increased and δˇ gets bigger, the shock wave gets less steep, until it completely disappears for δˇ 1. Similarly, when ν increases, the viscous effects in the tube become more dominant,≫ and the shock wave becomes less steep.

0.75 1 ν = 0.01 δˇ = 0 ˇ 0.5 ν = 0.05 δ = 1 δˇ ν = 0.1 0.5 = 2 δˇ = 5 0.25 ν = 0.2 V V 0 0 0 0

−0.25 −0.5 −0.5

−0.75 −1 −1.5 −1 −0.5 0 0.5 1 1.5 −1.5 −1 −0.5 0 0.5 1 1.5 z z (a) (b)

5 ˇ Figure 8.2: V0 as a function of z for α = 10− and ∆ = 0. In (a) ν is varied for δ = 1 and in (b) δˇ is varied for ν = 0.05.

ˆ ˆ + Once V0− and V0 have been computed, we can use (8.70) to compute

π/2 E+(s) d2Vˆ +(s) Vˆ +(z) = 0 δˇ∆ˆ ds + C , (8.72) 1 + ˆ + 2 1 Zz 2E (z)V0 (z) " ds − # z 2 ˆ E−(s) d V0−(s) ˆ − δˇ∆ˆ V1 (z) = 2 ds + C1, (8.73) π/2 2E−(z)Vˆ −(z) − Z− 0 " ds # ˆ + ˆ ˆ ˆ + where we imposed V1 (π/2) = V1−( π/2). Substituting V0−(z) = V0 ( z), we can write − − −

z ˆ + ˇ ˆ E−(s) V1−(z) = V1 ( z) δ∆ ds + C1. (8.74) − − − π/2 E−(z)Vˆ −(z) Z− 0 Furthermore the two matching conditions in (8.67) can be reduced to one condition if + we impose Vˆ −(0) = Vˆ (0). This leads to the following expression for C : 1 − 1 1 0 π/2 + ˇ ˆ E−(s) ˇ ˆ E (s) C1 = δ∆ ds = δ∆ + + ds. π/2 E−(0)Vˆ −(0) − 0 E (0)Vˆ (0) Z− 0 Z 0

Substituting C1 back into (8.72) and (8.74) we find

π/2 E+(s) d2Vˆ +(s) π/2 E+(s) Vˆ +(z) = 0 δˇ∆ˆ ds δˇ∆ˆ ds, (8.75) 1 + ˆ + 2 + ˆ + Zz 2E (z)V0 (z) " ds − # − Z0 E (0)V0 (0) z 0 ˆ + ˇ ˆ E−(s) ˇ ˆ E−(s) V1−(z) = V1 ( z) + δ∆ ds + δ∆ ds. (8.76) − − π/2 E−(z)Vˆ −(z) π/2 E−(0)Vˆ −(0) Z− 0 Z− 0 154 8.4 Results

Lastly we determine B1 from the remaining matching condition in (8.67). We find ˆ + B1 = 2a0 lim V1 (z) − z 0 ↓ π/2 d2Vˆ +(s) + 0 δˇ∆ˆ = E (s) 2 + ds, − Z0 " ds # ˆ + + where we used a0 = V0 (0) and E (0) = 1. The full composite solution is now found by adding the inner and outer solutions and subtracting the common part.

ˆ ˇ z ˆ ˇ z B1 V0−(z) + V0− + a0 + ν V1−(z) + V1− z 0. ν ν − 2a0 ≤ V(z) =         + + z + + z B1  Vˆ (z) + Vˇ a + ν Vˆ (z) + Vˇ + z 0. 0 0 ν − 0 1 1 ν 2a ≥  0        Note that since V0(z) = V0( z), it follows that V = 0 + (ν), so that to leading order V has zero average.− If in− addition ∆ˆ = 0, then it followsO from (8.74) that even V = 0 + (ν2). O ˇ Figure 8.3 below shows the V1 profiles for various ν and δ. For the calculations we used the V0 profiles shown in figure 8.2 . Special precaution has to be taken near z = π/2 where there is a removable singularity similar to what we observed in the previous± section in (8.49) ; the numerical integration converges if z = π/2 is excluded. ±

6 ν = 0.01 δˇ = 0 2 ˇ ν = 0.05 4 δ = 0.5 ˇ ν = 0.1 δ = 1 1 δˇ = 1.5 ν = 0.2 2 V V 1 0 1 0

−2 −1

−4 −2 −6 −1.5 −1 −0.5 0 0.5 1 1.5 −1.5 −1 −0.5 0 0.5 1 1.5 z z (a) (b)

5 ˇ Figure 8.3: V1 as a function of z for α = 10− and ∆ = 0. In (a) ν is varied for δ = 1 and in (b) δˇ is varied for ν = 0.05.

8.4 Results

We have tested the equations derived in the previous sections for two kinds of configu- rations: Nonlinear standing waves 155

(a) A closed tube excited near resonance by a speaker generating velocity oscillations at a relative amplitude ε 1 with h(t) = sin(t) (figure 8.4(a)). This configuration can be used to model a thermoacoustic≪ prime mover, with the speaker simulating the presence of the sound-producing stack.

(b) A thermoacoustic refrigerator, consisting of a closed tube with a stack of parallel plates positioned near the closed end. The speaker provides sound at a relative amplitude ε 1 with h(t) = sin(t) (figure 8.4(b)). ≪

L L

sound sound

LA LB LC

(a) Closed tube (b) Closed tube with thermoacoustic stack

Figure 8.4: We consider two kinds of configurations: (a) a closed tube driven near resonance and (b) a thermoacoustic refrigerator driven near resonance with a stack near the closed end.

8.4.1 Nonlinear standing waves in a closed tube

We have simulated a closed tube at exact resonance with the parameters as given in Table 8.1. For this configuration we can compare the results of the matched asymptotics in Section 8.3.2 for δ = 0 with the solution computed in Section 8.3.1 using Mathieu functions. First in figure 8.5 we give the nonlinear pressure and velocity profiles as a function of both time and the position in the tube. Then in figure 8.6 we show how the nonlin- earity travels inside the tube at given times and we compare the result of the matched asymptotics (solid line) with the zero-viscosity solution (dashed line) given in (8.36). From figures 8.5 and 8.6 we conclude that as time changes the nonlinearity moves back and forth through the tube and enters and leaves the tube at the piston (x = L). Note that the pressure is an odd function of time and velocity is an even function of time. The zero-viscosity solution shows a discontinuous shock-like profile. We expect that as ν decreases the matched asymptotics will converge to the same shock profile. Indeed it follows from figure 8.7 that as ν decreases and the effect of viscosity becomes smaller the profile will look more and more like the zero-viscosity shock wave. Note also that 6 although the sound waves were excited with a Mach number of only ε 6 · 10− , we 3 ≈ find peak Mach numbers close to 6 · 10− . 156 8.4 Results

(a) velocity (b) pressure

Figure 8.5: (a) The velocity u and (b) the pressure p as a function of place and time in the closed tube. The profiles were computed at exact resonance using the parameter values from Table 8.1.

1.5 400

300 1 200 0.5 ωt = 0 100 ωt = π/4 − 0 ωt = π/2 0 p (Pa) u (m/s) ωt = −3π/4 ωt = −π −100 −0.5 − −200 −1 −300

−1.5 −400 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 x/L x/L (a)

1.5 400

300 1 200 0.5 ωt = 0 100 ωt = π/4 0 ωt = π/2 0 p (Pa) u (m/s) ωt = 3π/4 −100 −0.5 ωt = π −200 −1 −300

−1.5 −400 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 x/L x/L (b)

Figure 8.6: The velocity and pressure inside the closed tube at various times. The computations were performed at exact resonance (∆ = 0), and there is a good agreement between the matched asymptotics (-) and the zero-viscosity solution (- -). Nonlinear standing waves 157

Physical parameters Dimensionless parameters symbol value unit symbol value 3 1 U 2.1 · 10− m s− κ π 1 c0 330 m s− ∆ 0 L 0.5 m δ 0 2 f 330 - ν 4.4 · 10− 2 1 2 b 1 · 10− kg m− s− γ 1.4 3 · 6 ρ0 1 kg m− ε 6.3 10− 6 ℓ 1 · 10− m τ 0

Table 8.1: Parameter values for simulation of closed tube.

1 200 ν = 0.1, ν = 0.1, ν = 0.05 ν = 0.05 0.5 ν = 0.01 100 ν = 0.01

0 0 p (Pa) u (m/s)

−0.5 −100

−1 −200 -π -π/2 0 π/2 π -π -π/2 0 π/2 π ω t ω t (a) (b)

Figure 8.7: (a) The velocity u and (b) the pressure p as a function of time in the center of the tube. As we vary ν from 0.1 to 0.01 (and viscosity decreases) we see that the nonlinear wave profile starts to look more and more like a shock wave.

8.4.2 Nonlinear standing waves in a thermoacoustic refrigerator In this section we will simulate a thermoacoustic refrigerator of the type shown in figure 8.4(b). In particular we will simulate the thermoacoustic couple of Atchley et al. [11] that was described in Section 6.3. In addition to the stack specifications given in Table 6.1, we will use the parameter values given in Table 8.2 to describe segments A and C. The value of τC depends on the frequency f and the length of the third segment LC and is chosen such that

ωLC(1 + τC) ∆C = π = 0, c0 − and a nonlinearity will arise in the right part of the tube. In the left part of the tube we have τA = 0 and ωLA ∆A = π = 0. c0 − 6

We have implemented this configuration in three steps: 158 8.4 Results

Physical parameters Dimensionless parameters symbol value unit symbol value symbol value · 3 1 U 4.37 10− m s− κA 0.215 κC π · 3 1 c0 1.02 10 m s− ∆A -2.93 ∆C 0 LA 5 cm δA 0 δC 5e-3 · 2 LB 6.85 cm τA 0 τC 8.44 10− · 3 LC 73.1 cm γ 1.67 νC 1.58 10− · 4 ℓ 1e-6 m εC 4 10− f 696 Hz 4 1 2 b 2.87 · 10− kg m− s− 3 ρ0 0.184 kg m−

Table 8.2: Parameter values for thermoacoustic refrigerator.

1. First we simulate the segment C, between the stack and the speaker. At x = 0 we impose a reflection coefficient by fixing δ and at x = LC we assume harmonic excitation with h(t) = sin(t). Implementing the equations given in Section 8.3.3 we can compute the pressure pC and the velocity uC.

2. We then compute the discrete Fourier transform of pC and uC at x = 0 and use it as a boundary condition for the sound field in the stack by applying continuity of mass and momentum for each mode. For each mode the pressure, velocity, and temperature profiles are computed by implementing the system of ode’s given in (4.70). Finally the full nonlinear sound field in the stack is computed using the inverse of the discrete Fourier transform. 3. Segment A is not at resonance and thus we can compute the sound field by imple- menting the off-resonance solution given in Section 8.3.1. The nonlinearity comes in via the right velocity condition which follows from the nonlinear velocity field in the stack by continuity of mass across the interface. In figure 8.8 we give the nonlinear pressure and velocity profiles as a function of both ˇ time and the position for a reflection coefficient R0 = 0.995 (δ 1.67). Then in fig- ure 8.9 we show how the nonlinearity travels inside the thermoacoustic≈ refrigerator at given times. As for the closed tube without stack we conclude that as time changes the nonlinearity moves back and forth through the tube. Inside the stack the nonlinear- ity is damped due to dissipation in the narrow pores. Moreover, it follows from from ˇ figure 8.10 that as δ is increased (and R0 becomes smaller), the sharp nonlinearity is smoothened, and the velocity and pressure profiles change from the nonlinear shock waves into a more familiar linear sinusoid profile. Nonlinear standing waves 159

(a) velocity in stack (b) velocity in tube

(c) pressure in stack (d) pressure in tube

Figure 8.8: The velocity u and pressure p as a function of place and time in (a,c) the stack and (b,d) the tube. The computations were performed at resonance with a reflection coefficient of R = 0.995 and at a frequency of 696 Hz. 160 8.4 Results

−3 x 10 2 300

200 1 ωt = 0 100

/s) ω = π/ 3 t 4 − 0 · · · ωt = π/2 0 p (Pa) u (m

g −

A ωt = 3π/4 − −100 −1 ωt = π − −200

−2 −300 0 0.2 0.4 0.6 0.8 0 0.2 0.4 0.6 0.8 x(m) x(m) (a) volumetric velocity (b) pressure

−3 x 10 2 300

200 1 ωt = 0 100

/s) ω π 3 t = /4 0 · · · ωt = π/2 0 p (Pa) u (m g A ωt = 3π/4 −100 −1 ωt = π −200

−2 −300 0 0.2 0.4 0.6 0.8 0 0.2 0.4 0.6 0.8 x(m) x(m) (c) volumetric velocity (d) pressure

Figure 8.9: The velocity and pressure as a function of the position in the refrigerator at various times.

0.5 600 δˇ = 0 0 400 δˇ = 1 δˇ = 2 −0.5 200 −1 0 p (Pa)

u (m/s) −1.5 −200 −2 δˇ = 0 −2.5 δˇ = 1 −400 δˇ = 2 −3 −600 -π -π/2 0 π/2 π -π -π/2 0 π/2 π ω t ω t (a) velocity (b) pressure

Figure 8.10: (a) The velocity u and (b) the pressure p as a function of time at the right stack end. As we vary δˇ from 0 to 2 the nonlinearity is smoothened. Chapter 9

Conclusions and discussion

A weakly non-linear theory of thermoacoustics for arbitrary and slowly varying pore cross-sections, applicable for stacks, regenerators, and resonators, has been developed based on systematic dimensional analysis and using small-parameter asymptotics. Cru- cial assumptions for the asymptotic expansions are small amplitudes (Dr, Ma 1) and slow longitudinal variation (ε 1). The theory is weakly non-linear in the sense≪ that al- ≪ though the equations are linearized, terms up to second order in Ma and ε are included; in addition to the first and second harmonics this also includes streaming, second-order time-averaged mass flow. Using the smallness of ε it is possible to decouple the transverse and longitudinal variation of the fluid variables. It follows that a coupled system of ordinary differential equations has to be solved for the mean temperature, the acoustic pressure, and the av- erage acoustic velocity, which can easily be implemented numerically. The problem of determining the transverse velocity variation is reduced to finding Green’s functions for a modified Helmholtz equation on the given cross-section and solving two additional integral equations. Next the streaming velocity can be determined using a Green’s func- tion for the Poisson equation. Finally it is shown that the second-harmonic pressure and velocity can also be determined from a system of ordinary differential equations simi- lar to the one found for the first harmonics. The asymptotic theory can in principle be extended to include higher-order terms, such as the higher harmonics and higher-order streaming, but this is avoided as it would require a lengthy and messy derivation and would add little more understanding. In addition to Ma and ε we have identified several other dimensionless parame- ters (Sk,κ, NL, Br) that are crucial in thermoacoustics. For these parameters we have indicated various parameter regimes, each signifying specific geometrical or physical constraints. For specific parameter regimes an approximate analytic solution can be ob- tained. For example, at constant temperature the acoustic velocity and pressure in a straight tube can be written as a linear combination of sines and cosines with a complex wave number; for linearly changing cross-sections this becomes a combination of Bessel functions of the first and second kind. The wave number will have a small imaginary part to account for viscous dissipation, depending on the width of the tube. Another approach predicts the steady-state temperature difference across the stack using a short- stack approximation (κ 1), which shows a change from a sinusoid profile for small ≪ 162

Heat Pump Prime Mover

Resonator

Figure 9.1: Combination of traveling-wave heat pump and prime mover using a double- Helmholtz resonator. The regenerators are located near velocity nodes of the tube to reduces viscous dissipation and the gas loop surrounding the regenerator gives the traveling-wave phas- ing. The sound produced by the prime mover is used to drive the refrigerator. amplitudes to a sawtooth profile for large amplitudes. A third approximation assumes wide tubes (NL 1) and large temperature differences. It turns out that a tube that satisfies these conditions≫ will have a temperature profile that changes from a linear pro- file in the absence of streaming to an exponential profile when the average mass flux is increased. The thermoacoustic equations have been applied to two kinds of thermoacoustic systems: standing-wave (straight-tube) devices and traveling-wave (looped-tube) de- vices. The standing-wave devices are modeled as (closed) straight tubes with a stack and heat exchangers placed inside and, in case of a refrigerator, driven by a speaker. To validate our equations a numerical simulation of a thermoacoustic couple has been per- formed and compared to experimental and analytic results with very good agreement. Subsequently it has been shown numerically how a standing-wave prime mover and refrigerator perform under various operating conditions. In particular we have investi- gated how parameters like drive ratio, temperature difference, mass flux, stack length, stack position, hydraulic radius, porosity, plate material, mean pressure, power-input, and power-output affect the relative coefficient of performance and efficiency. A traveling-wave prime mover has been modeled as a resonance tube with variable acoustic load connected to a short looped tube containing a regenerator and thermal buffer tube. The resonator is there to create the optimal phasing between pressure and velocity at the location of the loop. In addition the loop has a contraction and expansion to create the proper phasing at the regenerator. In our numerical simulations we fixed the regenerator and thermal buffer tube and performed an optimization routine that determines a loop and resonator configuration that gives a stable system for given sys- tem parameters such as drive ratio, regenerator impedance, temperature difference, and frequency. This optimization routine can be a very useful tool in the design of practi- cal traveling-wave systems. In addition we have optimized several of the regenerator’s properties, such as impedance, drive ratio, temperature difference, regenerator length, porosity, hydraulic radius, mass flux, and regenerator material, to maximize the regen- erator efficiency. A comparison with an equivalent standing-wave prime mover shows that the traveling-wave prime mover can potentially yield efficiencies that are twice as high. The logical next step in the analysis of traveling-wave devices would combine the traveling-wave prime mover with a traveling-wave refrigerator. One could for example remove the acoustic load and attach a looped refrigerator in its place, a so-called ther- moacoustically driven refrigerator (cf. [77]). Another possible approach uses a double- Conclusions and discussion 163

Helmholtz resonator (figure 9.1), i.e. a closed resonator tube that expands towards the tube ends, with regenerators placed near both ends. If sufficient heat is supplied to one of the regenerators it is possible to produce enough sound to generate a significant temperature difference across the other regenerator (cf. [78]). A third configuration, as proposed by Yazaki et al. [154], assumes the looped geometry given in figure 9.2, where the first regenerator generates spontaneous gas oscillations which are absorbed by the secondary regenerator that is placed further along the loop. This is a much simpler geometry which ensures the gas in the regenerator to execute the ideal Stirling cycle. However, large velocities may occur near the regenerators, causing large viscous losses and degrading the performance. The first two approaches are therefore recommended.

Refrigerator

Prime mover

Figure 9.2: Looped tube equipped with prime mover and refrigerator. If enough heat is supplied to the first regenerator, then spontaneous gas oscillations will arise around the loop, generating a temperature difference across the secondary regenerator.

Lastly, we have investigated the nonlinear disturbances that can arise in a thermoa- coustic refrigerator or prime mover, when it is excited near resonance. It is shown that nonlinear shock-like profiles may arise, due to a balance of energy inflow from the source with nonlinear absorption and viscous dissipation in the gas and stack. The presence of a stack in a closed tube is modeled by means of a reflection condition, as a result of which we can compute the nonlinear wave profiles both in the stack in the tube. 164 Appendix A

Thermodynamic constants and relations

In dimensional form we have the following thermodynamic constants and relations (taken from [25]).

∂p c2 = , (A.1) ∂ρ  s ∂s ∂h C = T = , (A.2) p ∂T ∂T  p  p ∂s ∂ǫ C = T = , (A.3) v ∂T ∂T  ρ  ρ 1 ∂ρ β = , (A.4) −ρ ∂T  p c2β2T = C (γ 1), (A.5) p − p = ρh ρǫ, (A.6) − Cp β Cp β ds = dT dp, s1 = T1 p1, (A.7) T − ρ ⇒ T0 − ρ0 γ γ dρ = dp ρβ dT, ρ1 = p1 ρ0βT1, (A.8) c2 − ⇒ c2 −

1 p1 dh = Tds + dp, h1 = T0s1 + , (A.9) ρ ⇒ ρ0 p p ǫ ρ ǫ 0 ρ d = Tds + 2 d , 1 = T0s1 2 1. (A.10) ρ ⇒ − ρ0 166 Appendix B

Derivations

B.1 Total-energy equation

From the conservation of mass (3.1), momentum (3.2), and energy (3.3), we can derive the following expressions

∂ρ = · (ρv) , (B.1) ∂t −∇ ∂v ρ = ρ(v · )v p + · τ + ρb, (B.2) ∂t − ∇ − ∇ ∇ ∂ǫ ρ = ρ(v · )ǫ + · (K T) p · v + τ: v. (B.3) ∂t − ∇ ∇ ∇ − ∇ ∇ We can use these relations to derive an equation for the change in total energy per unit mass 1 ρ v 2 + ρǫ, 2 | | ∂ ∂ρ 1 ∂v ∂ǫ 1 ρ v 2 + ρǫ = v 2 + ǫ + ρv · + ρ ∂t 2 | | ∂t 2 | | ∂t ∂t     1 = v 2 · (ρv) ρv · (v · )v ǫ · (ρv) ρv · ǫ + ρv · b − 2 | | ∇ − ∇ − ∇ − ∇ v · p p · v + v · ( · τ) + τ: v + · (K T) . (B.4) − ∇ − ∇ ∇ ∇ ∇ ∇ Since · (v · τ) = v · ( · τ) + τ: v, (B.5) ∇ ∇ ∇ we can write ∂ 1 1 ρ v 2 + ρǫ = · ρ v 2v + ρǫv + pv K T v · τ + ρv · b, ∂t 2 | | − ∇ 2 | | − ∇ −     1 = · ρv v 2 + h K T v · τ + ρv · b, (B.6) − ∇ 2 | | − ∇ −     where we used that h = ǫ + p/ρ. The same equation appears in [71], but without the gravitational term. It describes the rate of change of the total specific energy of the fluid 168 B.2 Temperature equation due to an energy flux arising from four phenomena: transfer of mass by the motion of the fluid, transfer of heat, internal friction, and gravitational effects.

B.2 Temperature equation

Starting from equation (3.3), we can derive an equation for the fluid temperature T. After substitution of (3.5) and ǫ = h p/ρ, we obtain − dh d p ρ = ρ + · (K T) p · v + τ: v dt dt ρ ∇ ∇ − ∇ ∇   dp p dρ = + · (K T) p · v + τ: v dt ∇ ∇ − ρ dt − ∇ ∇ dp = + · (K T) + τ: v, (B.7) dt ∇ ∇ ∇ where we substituted the continuity equation (3.1) into the last equality. It follows from the thermodynamic relations (A.9) and (A.7) that the total derivative of h can be written as follows: dh ds dp ρ = ρT + dt dt dt dT dp = ρC +(1 βT) . (B.8) p dt − dt If we substitute this back into (B.7), then we find

dT dp ρC = βT + · (K T) + τ: v. (B.9) p dt dt ∇ ∇ ∇ Appendix C

Green’s functions

In this section we will first show how the Fj-functions given in Section 5.1 and the Fj,2- functions given in Section 5.5 can be computed using Green’s functions [38]. Then we will show how the Green’s functions can be computed for specific geometries.

C.1 Fj-functions

First we introduce the Green’s functions G and G . For every X we fix ˆx (X), set ν k τ ∈ Ag ˆx := Xex + ˆxτ , and solve for j = ν, k 1 2 δ Gj(x; ˆx) 2 τ Gj(x; ˆx) = (xτ ˆxτ ), xτ g(X), (C.1a) − α j ∇ − ∈ A G (x; ˆx) = 0, (x) = 0. (C.1b) j Sg

Using the Green’s identities, the Fj-functions introduced in (5.4) can be expressed in Gj as follows: ˆ Fj(x) = Gj(x; ˆx) dS, j = ν, k. (C.2) ZAg

In the same way we can also introduce the Green’s functions Gs. For fixed ˆxτ s(X) we solve ∈ A 1 2 δ Gs(x; ˆx) 2 τ Gs(x; ˆx) = (xτ ˆxτ ), xτ s(X), (C.3a) − αs ∇ − ∈ A G (x; ˆx) = 0, (x) = 0, (C.3b) s Sg G (x; ˆx) · n′ = 0, (x) = 0. (C.3c) ∇τ s τ St 170 C.1 Fj-functions

Given g j it can be shown that

ˆ ˆ · ˆ Fk j(x) = Gk(x; ˆx) dS g j(ˆx) τ Gk(x; ˆx) nτ dℓ, − Γ ∇ ZAg Z g ˆ ˆ · ˆ Fs j(x) = Gs(x; ˆx) dS + (1 g j(ˆx)) τ Gs(x; ˆx) nτ dℓ. Γ − ∇ ZAs Z g The hats in the gradients and integrals are used to indicate that the differentiation or integration is with respect to ˆx. It only remains to determine the unknown boundary functions gu and gp for which we will use the boundary condition (3.23c). If we impose

F · n = σ F · n , = 0, ∇τ kp τ − ∇τ sp τ Sg (F P F ) · n = σ F · n , = 0, ∇τ ku − r ν τ − ∇τ su τ Sg then (3.23c) is satisfied to leading order. We now find that gu and gp are found from the following integral equations:

ˆ Kb(x; ˆx)g j(ˆx) dℓ = Φ j(x), g(x) = 0, j = u, p, (C.4) Γ S Z g where Φu, Φp and Kb are defined as

Φ (x) = G (x; ˆx) · n dSˆ, p ∇τ k τ ZAg Φ (x) = (G (x; ˆx) P G (x; ˆx)) · n dSˆ, u ∇τ k − r ν τ ZAg K (x; ˆx) = ˆ G (x; ˆx) σφG (x; ˆx) · n · n . b ∇τ ∇τ k − s τ τ     In most cases these integral equations are not trivially solved. In [5] this problem was avoided by neglecting the acoustic fluctuations in the wall, i.e. gu = gp = 0. In that case Fsu = Fsp = 1 and Fkp = Fku = Fk. The second case for which the solution is simple is when is rotationally symmetric, i.e. circular or conical pores. Then K and Ag b the g j-functions will be constant on Γg(X) and we find

Φ j g j = , j = u, p. (C.5) Kb These two cases are discussed in more detail in Sections 5.3.1 and 5.3.2. For the general case one has to resort to a numerical implementation or to the general theory of integral equations. For example a solution can be attempted in the form of a sum of orthogonal basis functions. Green’s functions 171

C.2 Fj,2-functions

The Fj,2-functions can be determined in the same way as the Fj-functions. We first de- termine Green’s functions Gj,2 (ν, k) from

1 2 δ Gj,2(x; ˆx) 2 τ Gj,2(x; ˆx) = (xτ ˆxτ ), xτ g(X), (C.6a) − α j,2 ∇ − ∈ A G (x; ˆx) = 0, (x) = 0, (C.6b) j,2 Sg and Gs,2 from 1 2 δ Gs,2(x; ˆx) 2 τ Gs,2(x; ˆx) = (xτ ˆxτ ), xτ s(X), (C.7a) − αs,2 ∇ − ∈ A G (x; ˆx) = 0, (x) = 0, (C.7b) s,2 Sg G (x; ˆx) · n′ = 0, (x) = 0. (C.7c) ∇τ s,2 τ St

It follows that the Ψ j,2-functions as introduced in (5.48) are given by

ˆ Ψ j,2( f ) (x) = f (x)Gj,2(x; ˆx) dS, j = ν, k. (C.8) g h i ZA In particular F follows from Ψ by substitution of f 1, j,2 j,2 ≡ ˆ Fj,2(x) = Gj,2(x; ˆx) dS, j = ν, k. (C.9) ZAg

Furthermore Fk j,2 (k = u, p) is given by

ˆ ˆ · ˆ Fk j,2(x) = Gk,2(x; ˆx) dS g j,2(ˆx) τ Gk,2(x; ˆx) nτ dℓ, − Γ ∇ ZAg Z g ˆ ˆ · ˆ Fs j,2(x) = Gs,2(x; ˆx) dS + (1 g j,2(ˆx)) τ Gs,2(x; ˆx) nτ dℓ, Γ − ∇ ZAs Z g where the boundary functions gu,2 and gp,2 are found from the following integral equa- tions:

ˆ Kb,2(x; ˆx)g j,2(ˆx) dℓ = Φ j,2(x), g(x) = 0, j = u, p, (C.10) Γ S Z g where Φu,2, Φp,2 and K are defined as

Φ (x) = G (x; ˆx) · n dSˆ, p,2 ∇τ k,2 τ ZAg Φ (x) = G (x; ˆx) P G (x; ˆx) · n dSˆ, u,2 ∇τ k,2 − r ν,2 τ ZAg
K (x; ˆx) = ˆ G (x; ˆx) σφG (x; ˆx) · n · n . b,2 ∇τ ∇τ k,2 − s,2 τ τ     172 C.3 Green’s functions for various stack geometries

n Laplace: 2 Modified Helmholtz: 2 1/α2 ∇τ ∇τ − j 1 1 1 xτ ˆxτ exp α j xτ ˆxτ − 2 | − | 2α j − | − |   1 1 2 log x ˆx K ( α x ˆx ) − 2π | τ − τ | 2π 0 − j| τ − τ | Table C.1: Free-space Green’s functions on Rn.

As above we note that if we assume a steady wall temperature then gu,2 = gp,2 = 0, so that Fsu = Fsp = 1 and Fkp = Fku = Fk. Moreover, for rotationally symmetric pores we find that Kb,2 and the g j,2-functions will be constant on Γg(X), so that

Φ j,2 g j,2 = , j = u, p. (C.11) Kb,2

C.3 Green’s functions for various stack geometries

There is more than one way to determine the Green’s functions Gj. One way is using the method of images [38]. The method of images adds homogeneous solutions to the free-space Green’s function in such a way that their sum satisfies the right boundary conditions. The free-space Green’s functions are given in Table C.1 and are fundamental solutions of the Laplace and modified Helmholtz equations that have suitable behavior at infinity. As an example we consider the case n = 1 where we have a geometry as shown in Fig. C.1(a), so that xτ = y. Define 1 Φ j(x, ˆx) := exp α j y yˆ , j = ν, k, s. 2α j − | − |   We now want to add a homogeneous function such that the resulting function vanishes at Γ . Introducing sources at the reflection points 2 yˆ and 2 yˆ, we can cancel g Rs − − Rs − the contribution of yˆ on Γg. However, to eliminate the contributions of 2 s yˆ and 2 yˆ we have to introduce even more sources. Continuing this way wRe can− write − Rs − the Green’s functions Gj (j = ν, k) in the form of an infinite sum,

∞

Gj(x, ˆx) = ∑ Φ j [x; ak] Φ j [x; bk] , j = m, ν, k, (C.12) k= ∞ − −   where a =(X, yˆ + 4k ), b =(X, yˆ +(4k 2) ). k Rs k − − Rs Similarly we can show that in the solid Gs is given by the sum ∞ k Gs(x, ˆx) = ∑ ( 1) Φs [x; ask] Φs [x; bsk] , (C.13) k= ∞ − − −   Green’s functions 173

2bg y y θ r 2 t 2 2a 2ag R Rs t z Rs Rt

2bt (a) Parallel plates (b) Circular cross-sections (c) Rectangular cross-sections

Figure C.1: Various stack geometries.

where

a =(X, yˆ + 2k( )), b =(X, yˆ + 2k( ) + 2 ). sk Rt − Rs sk − Rt − Rs Rs

For n = 2 we will employ a different approach that is also used by [38] and that solves for the Green’s functions by expanding in eigenfunctions. For circular pores this leads to the following expressions:

∞ ∞ 1 J (k rˆ)J (k r) ∑ ∑ n ni n ni θ θˆ Gm(x; ˆx) = 2 2 2 cos n( ) , (C.14) π s n= ∞ i=1 kni Jn′ (kni s) − R − R ∞ ∞ 2   1 α J (k rˆ)J (k r) ∑ ∑ j n ni n ni θ θˆ ν Gj(x; ˆx) = 2 2 2 2 cos n( ) , j = k, , (C.15) π s n= ∞ i=1 (kni α j )Jn′ (kni s) − R − − R   ∞ ∞ 2 ˆ 2 α j ni(lnirˆ) ni(lnir) cos[n(θ θ)] G (x; ˆx) = ∑ ∑ J J − , (C.16) s π 2 2 2 2 2 2 n= ∞ i=1 ǫn(kni α j )[ t ni′ (lni t) s ni′ (lni s)] − − R J R − R J R where the prime denotes differentiation and

2, n = 0, ni(r) = Yn(lni s)Jn(r) Jn(lni s)Yn(r), ǫn = J R − R ( 1, n > 0, and Jn and Yn are the Bessel functions of the first and second kind, respectively. Fur- thermore, the eigenvalues kni and lni are computed from d J (k ) = 0, Jn (l ) = 0. n niRs dr niRt 174 C.3 Green’s functions for various stack geometries

0 1

−0.05 0.5 −0.1 0 −0.15

−0.2 −0.5 1 1 1 1 0 0.5 0 0 0 −0.5 −1 −1 −1 z y z −1 y

(a) Re Gj (b) Im Gj    

Figure C.2: The Greens function Gj for the Helmholtz equation over a rectangular region with Dirichlet boundary conditions on the sides when a = b = 1, α = 1 + i, and yˆ = zˆ = 0.3. g g j −

For rectangular pores we obtain the following Green’s functions:

∞ 4 ∑ gin Gm = 2 2 2 2 2 2 , (C.17) agbg i,n=1 i π /ag + n π /bg ∞ 2 4 α g ∑ j in ν Gj = 2 2 2 2 2 2 2 , j = k, , (C.18) agbg i π /a + n π /b 4α i,n=1 g g − j ∞ α2 4 ∑ s sin Gs = 2 2 2 2 2 2 2 , (C.19) (as)(bs) i π /(a ) + n π /(b ) 4α i,n odd s s − s with eigenfunctions

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A Stirling ...... 7,20 acoustics ...... 1,39,74 thermodynamic ...... 7,20 adiabatic ...... 6 airconditioning ...... 12 D approximation Darcy’slaw ...... 121 boundary-layer ...... 96 DeltaE ...... 16 short-stack ...... 54,96, 161 DeltaEC ...... 16 wide-channel ...... 56 devices asymptotics ...... 12,34,161 heat-driven ...... 8 attenuation ...... 5 sound-driven ...... 8 dimensionalanalysis ...... 12 B dimensionless parameters ...... 13, 15, 31 Bernoulli dissipation ...... 68 effect ...... 112 thermal-relaxation ...... 68, 100 equation ...... 136 viscous ...... 68,100 bucket-brigadeeffect ...... 25 down-well power generation ...... 10 Buckingham π theorem ...... 31 E C efficiency ...... 1,20,119,125 coefficient of performance ....19, 101, 103 Carnot ...... 20,119 Carnot ...... 19 relative ...... 20,119 relative ...... 19,103 entropy ...... 18 composite expansion ...... 144 conservation ...... 28 F energy ...... 28 foodrefrigerators ...... 12 mass ...... 28 Fourier transform momentum ...... 28 continuous ...... 140 convectivederivative ...... 28 discrete ...... 158 critical temperature gradient ...... 103 Fourier’slaw ...... 29 cross-sections cylindrical ...... 14,73 G parallel-plate ...... 14, 37,96 gasliquefaction ...... 11 pin-array ...... 14,73 Green’sfunction ...... 73,169 rectangular ...... 14,73 triangular ...... 14,73 H wired-mesh ...... 73 harmonics cryocooling ...... 4,11 first ...... 35 cycle higher ...... 15,95,133,161 Brayton ...... 7,20 second ...... 14,35,61,86 186 Index

I sidebranch ...... 11 industrialwasteheat ...... 11 singingflame ...... 2 innersolution ...... 144,148 Sondhausstube ...... 3 integralequations ...... 73, 170 speedofsound ...... 2,165 integrating factor ...... 149 stack ...... 4,9,22 irrotationalflow ...... 134 standingwave ...... 6,9,20,162 nonlinear ...... 133 K standing-wave ...... 93 Kuznetsov’s equation ...... 134 statisticalmodeling ...... 14 streaming . 14, 15, 35, 59, 85, 112, 123, 127, L 161 least-squares solution ...... 122 Gedeon ...... 59 Los Alamos National Laboratory . . . 11, 16 inner ...... 59 Rayleigh ...... 29,59 M Sutherland’s formula ...... 29 Mathieufunctions ...... 143 method T homogenization ...... 14 Taconicoscillations ...... 3 matched asymptotic expansions . 144 thermoacoustic slowvariation ...... 37 couple ...... 96,157 devices ...... 1 N heatpump ...... 1,7,19 nondimensionalization ...... 13 prime mover ...... 1, 7, 8, 108, 133 nonlinearities ...... 13, 100, 134 refrigerator ...... 1, 7, 8, 19, 100, 157 O thermoacoustics ...... 1 oscillations spacecraft ...... 12 heat-driven ...... 8 weaklynonlinear ...... 12 sound-driven ...... 8 thermodynamics ...... 1,6,17 outersolution ...... 144,148 laws ...... 17 transitioneffects ...... 15 P travelingwave ...... 6,9,22,162 pulse-tube refrigerator ...... 4 turbulence ...... 2,15,95

R V reflectioncondition ...... 134 volume-averaging techniques ...... 14 regenerator ...... 5,9,24 vortexshedding ...... 2 resonance ...... 139,141 frequency ...... 134 Rijketube ...... 3

S SCOREstove ...... 12 self-consistent models ...... 14 self-oscillation ...... 15 above ...... 15 below ...... 15 through ...... 15 shock waves ...... 15,95, 133, 155, 158 Summary

This thesis addresses the mathematical aspects of thermoacoustics, a subfield within physical acoustics that comprises all effects in which heat conduction and entropy vari- ations of the gaseous medium play a role. We focus specifically on the theoretical basis of two kinds of devices: the thermoacoustic prime mover, that uses heat to produce sound, and the thermoacoustic heat pump or refrigerator, that use sound to produce heating or cooling. Two kinds of geometry are considered. The first one is the so-called standing-wave geometry that consists of a closed straight tube (the resonator) with a porous medium (the stack) placed inside. The second one is the so-called traveling-wave geometry that consists of a resonator attached to a looped tube with a porous medium (regenerator) placed inside. The stack and the regenerator differ in the sense that the regenerator uses thinner pores to ensure perfect thermal contact. The stack or regenerator can in principle have any arbitrary shape, but are modeled as a collecting of long narrow arbitrarily shaped pores. If the purpose of the device is to generate cooling or heating, then usually a speaker is attached to the regenerator to generate the necessary sound. By means of a systematic approach based on small-parameter asymptotics and di- mensional analysis, we have derived a general theory for the thermal and acoustic be- havior in a pore. First a linear theory is derived, predicting the thermoacoustic behavior between two closely placed parallel plates. Then the theory is extended by consider- ing arbitrarily shaped pores with the only restriction that the pore cross-sections vary slowly in longitudinal direction. Finally, the theory is completed by the inclusion of nonlinear second-order effects such as streaming, higher harmonics, and shock-waves. It is shown how the presence of any of these nonlinear phenomena (negatively) affects the performance of the device. The final step in the analysis is the linking of the sound field in the stack or regenera- tor to that of the main tube. For the standing-wave device this is rather straightforward, but for the traveling-wave device all sorts of complications arise due to the complicated geometry. A numerical optimization routine has been developed that chooses the right geometry to ensure that all variables match continuously across every interface and the right flow behavior is attained at the position of the regenerator. Doing so, we can pre- dict the flow behavior throughout the device and validate it against experimental data. The numerical routine can be a valuable aid in the design of traveling-wave devices; by variation of the relevant problem parameters one can look for the optimal traveling- wave geometry in terms of power output or efficiency. 188 Summary Samenvatting

Dit proefschrift is gewijd aan de wiskundige aspecten van thermoakoestiek, het vakge- bied dat bestaat uit alle akoestische verschijnselen waarin de warmtegeleiding en en- tropie variaties van het gas een rol spelen. In het bijzonder richten we ons op de the- oretische fundamenten van twee soorten thermoakoestische apparaten: de motor, die warmte omzet in geluid, en de warmtepomp of koelkast, die geluid gebruiken om ver- warming of koeling te genereren. We onderscheiden staande-golf en lopende-golf systemen. De staande-golf systemen maken maken gebruik van een gesloten rechte buis (de resonator) waarin een poreus medium, de stack, geplaatst wordt om warmte om te zetten in geluid of geluid in warmte. De stack is opgebouwd uit nauwe kanaaltjes met willekeurige doorsneden. Als het doel van het apparaat koelen of verwarmen is, dan is het gebruikelijk om de buis met een luidspreker te verbinden om het noodzakelijke geluid te genereren. In de lopende-golf systemen wordt de resonator gecombineerd met een lusvormige geometrie en wordt de stack vervangen door een regenerator. De lusvormige geometrie is nodig om het geluidsveld in de regenerator het lopende golf karakter te geven. De regenerator is ver- gelijkbaar met de stack, maar heeft veel nauwere kanaaltjes om perfect warmte-contact met het gas te garanderen. Gebruik makend van een systematische aanpak, gebaseerd op dimensie analyse en kleine-parameter asymptotiek, hebben we getracht een algemene thermoakoestische theorie af te leiden. We beginnen met een lineaire theorie, die het thermoakoestische gedrag voorspelt tussen twee parallelle platen. Vervolgens wordt de theorie uitge- breid door willekeurige drie-dimensionale stack kanalen te beschouwen met de enige beperking dat de dwarsdoorsneden langzaam veranderen in axiale richting. Uitein- delijk wordt de theorie voltooid door het toevoegen van kwadratische niet-lineaire ef- fecten zoals hogere harmonischen, schok golven, of stromingseffecten. Bovendien laten we zien hoe de aanwezigheid van deze niet-lineaire effecten de prestaties van de ther- moakoestische apparaten (negatief) be¨ınvloedt. De laatste stap in de analyse is de koppeling tussen het geluidsveld in de stack of regenerator en het geluidsveld van de hoofdbuis. Dit is relatief eenvoudig voor de staande-golf apparaten, maar de lopende-golf apparaten vereisen een speciale aanpak vanwege hun complexe geometrie. We hebben een numerieke optimalisatie routine ont- wikkeld die voor elke sectie de juiste afmetingen uitrekent zodat alle variabelen continu aansluiten over elke interface en het gewenste geluidsveld bereikt wordt bij de regene- rator. Op deze manier kan men het hele apparaat doorrekenen en valideren met met experimentele data. De numerieke routine kan een waardevolle houvast zijn voor het ontwerpen van praktische lopende-golf apparaten. 190 Samenvatting Curriculum Vitae

Peter in ’t panhuis was born in Roermond, The Netherlands, on July 10th 1981. After fin- ishing his pre-university education in Sittard at the Serviam College (now Trevianum) in 1999, he started his studies in Technical Mathematics at the Eindhoven University of Technology that same year. During his studies he did an internship at the Marcus Wal- lenberg Laboratory in Stockholm, Sweden, entitled “Calculations of the acoustic end correction of a semi-infinite circular pipe issuing a subsonic cold or hot het with co- flow”. In April 2005 he obtained his master’s degree in Industrial and Applied Math- ematics after writing a master’s thesis entitled “Li-ion battery modelling” under the supervision of prof.dr. Hans van Duijn and dr. Evgeniy Verbitskiy. From May 2005 till June 2009 he has been working as a PhD student at the Eind- hoven University of Technology within the Applied Analysis group, which is part of the Department of Mathematics and Computer Science, and the Low Temperature group, which is part of the Department of Applied Physics. This project, entitled “High-amplitude oscillatory gas flow in interaction with solid boundaries”, was sponsored by the Technl- ogy Foundation (STW) and was performed under the supervision of dr. Sjoerd Rienstra, prof.dr. Han Slot, prof.dr. Jaap Molenaar, and dr. Jos Zeegers. In addition to writing this thesis, he has taught classes to students of various faculties at the Eindhoven University of Technology. Moreover, he has participated in four study groups “Mathematics with Industry” at universities in the Netherlands (Eindhoven, Utrecht) and Denmark (Odense, Lyngby).