Mar 2, 2016 - Next, collecting the terms of order Aκ, we find that u10 satisfies iÏ0u10 = â dp11 dx. +. 1. Wo2. â2u10. ây2 . (3.1.6) and applying the boundary ...
Systematic Derivation of the Weakly Non-Linear Theory of Thermoacoustic Devices Version II
P.H.M.W. in ’t panhuis, S.W. Rienstra, J. Molenaar Dep. of Mathematics and Computer Science Technische Universiteit Eindhoven P.O. Box 513, 5600 MB Eindhoven, The Netherlands
Abstract Thermoacoustics is the field concerned with transformations between thermal and acoustic energy. This paper teaches the fundamentals of two kinds of thermoacoustic devices: the thermoacoustic prime mover and the thermoacoustic heat pump or refrigerator. Two technologies, involving standing wave and traveling wave modes, are considered. We will investigate the case of a porous medium and two heat exchangers placed in a gas-filled resonator, in which either a standing or traveling wave is maintained. The central problem is the interaction between the porous medium and the sound field in the tube. The conventional thermoacoustic theory is reexamined and a systematic and consistent weakly non-linear theory is constructed based on dimensional analysis and small parameter asymptotics. The difference with conventional thermoacoustic theory lies in the dimensional analysis. This is a powerful tool in understanding physical effects which are coupled to several dimensionless parameters that appear in the equations, such as the Mach number, the Prandtl number, the Lautrec number and several geometrical quantities. By carefully analyzing limiting situations in which these parameters differ in orders of magnitude, we can study the behavior of the system as a function of parameters connected to geometry, heat transport and viscous effects.
1.
Introduction
The most general interpretation of thermoacoustics, as described by Rott [21], includes all effects in acoustics in which heat conduction and entropy variations of the (gaseous) medium play a role. In this paper, however, we will focus specifically on thermoacoustic devices exploiting thermoacoustic concepts to produce useful refrigeration, heating, or work.
1.1
A brief history
Thermoacoustics has a long history that dates back more than two centuries. For the most part heatdriven oscillations were subject of these investigations. The reverse process, generating temperature differences using acoustic oscillations, is a relatively new phenomenon. The first qualitative explanation was given in 1887 by Lord Rayleigh in his classical work ”The Theory of Sound” [15]. He explains the production of thermoacoustic oscillations as follows: 1
”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 accurate theoretical description of those phenomena was not achieved until much later. The breakthrough came in 1969, when Rott and coworkers started a series of papers [17], [18], [20], [22] in which a successful linear theory of thermoacoustics was given. The first to give a comprehensive picture, was Swift [23] in 1988. He implemented Rott’s theory of thermoacoustic phenomena into creating practical thermoacoustic devices. His work included a short history, experimental results, discussions on how to build these devices, and a coherent development of the theory based on Rott’s work. Since then Swift and others have contributed a lot to the development of thermoacoustic devices. In 2002 Swift [25] also wrote a textbook, providing a complete introduction into thermoacoustics and treating several kinds of thermoacoustic devices. Recently Tijani [26] wrote a Ph.D thesis based on Swift’s work in which he also discusses the linear theory of thermoacoustics. Most of the work on thermoacoustic devices has been been reviewed by Garrett [4] in 2003.
1.2
Classification
We consider thermoacoustic devices of the form shown in Fig. 1, that is, an acoustically resonant tube, containing a fluid (usually a gas) and a porous solid medium. The porous medium is modelled as a stack of parallel plates. Following Garrett [4] we classify the devices based on several criteria: Hot heat exchanger
Cold heat exchanger y gas
x
Stack of parallel plates Figure 1: Thermoacoustic device
I. Prime mover vs. heat pump or refrigerator: A thermoacoustic prime mover absorbs heat at a high temperature and exhausts heat at a lower temperature while producing work as an output. A refrigerator or heat pump absorbs heat at a low temperature and requires the input of mechanical work to exhaust more heat to a higher temperature (Fig. 2). 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, from now on, when we talk about a thermoacoustic refrigerator we mean either a refrigerator or a heat pump. Often 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 the term thermoacoustic engine. 2
TH QH
TH W
QH
W
Device
Device
QC
QC TC
TC
(a) Prime mover
(b) Refrigerator or heat pump
Figure 2: The flows of work and heat inside (a) a thermoacoustic prime mover and (b) a thermoacoustic refrigerator or heat pump
II. 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”. Inside a regenerator the pore size is much smaller than inside a stack. Garrett [4] 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 . If NL & 1 the porous medium is called a stack and if NL ≪ 1 it is called a regenerator. This definition of stacks and regenerators is slightly different from Garret’s2 , but is chosen to stress that the pore size inside a regenerator is very small. III. Standing-wave devices vs. traveling-wave devices: Finally thermoacoustic devices can also be categorized depending on whether there is a traveling or a standing sound wave inside the thermoacoustic device. In section 3.1.7 we show that it is beneficiary to use a stack inside a standing-wave device and a regenerator inside a travelingwave device. Therefore one could also classify thermoacoustic devices as either standing-wave stack-based devices or traveling-wave regenerator-based devices.
1.3
Basic principle of the thermoacoustic effect
The thermoacoustic effect can be understood by following a given parcel of fluid as it moves through the stack or regenerator. Fig. 3 displays the (idealized) cycles a typical fluid parcel goes through as it oscillates alongside the plate. The fluid parcel follows a four-step cycle which depends on the kind of device. • Stack-based devices: The basic thermodynamic cycle in a stack-based acoustic refrigerator or prime mover consists of two reversible adiabatic steps (step 1,3 in Fig. 3(a,b)) and two irreversible isobaric heat-transfer steps (step 2,4 in Fig. 3(a,b)). As NL & 1, there will be an imperfect thermal contact between the fluid and the solid. As a result a phase shift, or time delay, arises between the pressure and the temperature of the gas parcels that are at a distance of a few thermal penetration depths from the stack plate. 1 2
The thermal penetration depth is the distance heat can diffuse through within a characteristic time Garrett defines the porous medium to be a stack if NL ≥ 1, and a regenerator if NL < 1
3
1 - Compression
−→
1 - Compression
−→
1 - Heating
1 - Heating
δQ1
δQ1
−→ δW1
2 - Cooling
2 - Heating
δQ1
δQ1
δW1
δW1
3 - Expansion
3 - Expansion
←
2 - Expansion
3 - Cooling
3 - Cooling
←− δW2
4 - Heating δQ2
4 - Cooling
δW1
2 - Compression
δQ2 ←−
−→
4 - Expansion
δQ2 ← δW2 4 - Compression
δQ2 δW2
δW2
(a) Stack-based refrigerator (b) Stack-based prime mover (c) Regenerator-based refrig- (d) Regenerator-based prime erator (small ∇T ) mover (large ∇T ) (small ∇T ) (large ∇T )
Figure 3: Typical fluid parcels executing the four steps (1-4) of the thermodynamic 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.
4
Parcels that are farther away have no thermal contact and are simply compressed and expanded adiabatically and reversibly by the sound wave. Therefore parcels that are about a thermal penetration depth away from the plate have good enough thermal contact to exchange some heat with the plate, but at the same time are in poor enough contact to produce a time delay between motion and heat transfer. The fact that the operation of stack-based thermoacoustic devices requires pressure and displacement to be primarily in phase, explains why stack-based devices are also called standingwave devices. The difference between the prime-mover and the refrigerator depends on the magnitude of the temperature gradient along the stack plates. During compression (step 1) the fluid parcel is both warmed (adiabatically) and displaced along the plate. Next, if the temperature gradient along the stack is large enough, the plate temperature will be larger than the fluid parcel temperature. Hence heat will flow from the plate to the fluid (step 2). This is the case in Fig. 3(b). Then the parcel expands and moves back to the original position (step 3). There the temperature of the parcel will still be higher than the plate temperature and heat will flow from the fluid to the plate (step 4). As a result heat is transported from a high to a low temperature, thus producing a certain amount of work. In other words the device acts as a prime mover. Similarly if the temperature gradient along the plate is small enough, we find that heat is transported from a low to a high temperature, thus requiring a certain amount of work. Therefore the cycle shown in Fig. 3(a) corresponds to a refrigerator. Thus we find that a low temperature gradient along the plate is the condition for a refrigerator and a high temperature gradient is the condition for a prime mover. The critical temperature gradient is where the temperature change along the plate just matches the adiabatic temperature change of the fluid parcel. • Regenerator-based devices: The basic thermodynamic cycle in a regenerator-based acoustic refrigerator or prime mover consists of two isochoric displacement steps during which heat is exchanged (step 1,3 in Fig. 3(c,d)) and two isothermal compression and expansion steps (step 2,4 in Fig. 3(c,d)). Because the pores in a regenerator are so small compared tot the thermal penetration depth, there will be an almost perfect thermal contact between the fluid and the solid. Therefore during the motion (step 2, 4) the temperature of the wall and the fluid parcel will be the same. As a result there will be a continuous exchange of heat between the gas and the solid, which takes place over a vanishingly small temperature difference and therefore only a negligibly amount of entropy is created. During the compression and expansion (step 1,3), the temperature remains constant. The gas oscillating inside a regenerator requires the same phasing between pressure and velocity as a traveling acoustic wave. Therefore regenerator-based devices are also called traveling-wave devices. The main advantage of regenerator-based devices with respect to stack-based devices is that there are no irreversible processes, so that the ideal efficiency is equal to the Carnot efficiency. On the other hand, because the pores are so narrow, there may be significant viscous dissipation which could lower the efficiency dramatically. 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 2x1 and 5
passing on heat as in a bucket brigade (Fig. 4). 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. TC
TH
L
QC
QH δQ 2x1
|
2x1
W
{z
2x1
}
Figure 4: Work and heat flow 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 Q is transported from the left to the right, using work W . Inside a prime mover the arrows will be reversed, i.e. heat Q is transported from the right to the left and work W is produced.
1.4
Scope
Motivated by the work of Swift [23], [25] and Tijani [26], this paper tries to reconstruct the linear theory of thermoacoustics in a systematic and consistent manner using dimensional analysis and small parameter asymptotics. We will start in section 2 with a detailed description of the model and an overview of the governing equations and boundary conditions. Then in section 3 we try to solve the equations assuming there is no mean steady flow. This is done both for stack- and regenerator-based devices. This is repeated in section 4, but now with a mean steady flow. Finally section 5 shows the different energy flows and their interaction in thermoacoustic devices.
2.
General thermoacoustic theory
We will model the thermoacoustic devices as depicted in Fig. 5 (see also [23] and [26]), where the device is modeled as an acoustically resonant tube, containing a gas and a porous solid medium. For now we will not make any assumptions on whether the tube is open or closed. The porous medium is modeled as a stack of parallel plates, each of thickness 2l and length L. The space between the plates is equal to 2R.
2.1
Governing equations
We will focus on what happens inside the stack. The general equations describing the thermodynamic behavior are [6] ∂v ˜ µ ˜ ˜ ˜ ˜ p + µ∇ ˜ 2 v˜ + ρ˜ ρ˜ ∇(∇ · v˜), (2.1.1) + (˜ v · ∇)˜ v = −∇˜ b+ ξ+ 3 ∂ t˜ ∂ ρ˜ ˜ + ∇ · (˜ ρv˜) = 0, ∂ t˜
(2.1.2) 6
L Cold heat exchanger
Hot heat exchanger
Plate y y′
x
2R Gas
x
2l Plate
gas
Gas Plate etc.
Stack of parallel plates
(b) Expanded view of stack
(a) Overall view
Figure 5: A thermoacoustic device modelled as an acoustically resonant tube, containing a gas, a stack of parallel plates and heat exchangers at both sides of the stack.
∂ T˜ ˜ T˜ + v˜ · ∇ ∂ t˜
ρcp
!
− β˜T˜
∂ p˜ ˜ ˜ · (∇ ˜ T˜) + Σ: ˜ ∇˜ ˜ v. + v˜ · ∇˜ p = K∇ ∂ t˜
(2.1.3)
Here ρ˜ is the density, v˜ is the velocity, p˜ is the pressure, T˜ is the temperature, s˜ is the entropy per ˜ is the total body force, µ and ξ are the dynamic (shear) and second (bulk) viscosity, unit mass, b respectively; K is the gas thermal conductivity, cp is the specific heat per unit mass, β˜ is the thermal ˜ is the viscous stress tensor, with components expansion coefficient and Σ ∂˜ vj 2 ∂˜ vk ∂˜ vk vi ˜ ij = µ ∂˜ + − δij . (2.1.4) + ξδij Σ ∂x ˜j ∂x ˜i 3 ∂ x ˜k ∂x ˜k Furthermore ρ˜ is related to p˜ and T˜ according to (A.4.9) d˜ ρ=
γ d˜ p − ρ˜β˜ dT˜. c˜2
(2.1.5)
Finally, the temperature T˜s in the plates satisfies the diffusion equation ρ˜s cs
∂ T˜s ˜ 2 T˜s , = Ks ∇ ∂ t˜
(2.1.6)
where Ks , cs and ρs are the thermal conductivity, the specific heat per unit mass and the density of the stack’s material, respectively. These equations will be linearized and simplified using the following assumptions • The theory is linear; second-order effects, such as acoustic streaming and turbulence, are neglected. • The plates are fixed and rigid. • The temperature variations along the stack are much smaller than the absolute temperature. 7
• The temperature dependence of viscosity is neglected. • Oscillating variables have harmonic time dependence at a single angular frequency ω. At the boundaries we impose the no-slip condition v˜ = 0,
y˜ = ±R.
(2.1.7)
The temperatures in the plates and in the gas are coupled at the solid-gas interface where continuity of temperature and heat fluxes is imposed. =: T˜b (x), (2.1.8a) = T˜s ′ T˜ y˜ =∓l
y˜=±R
K
∂ T˜ ∂ y˜
!
∂ T˜s ∂ y˜′
= Ks
y˜=±R
!
,
(2.1.8b)
y˜′ =∓l
We do not impose any conditions at the stack ends, as we are mainly interested about what happens inside the stack, ignoring any entrance effects. The next step is the rescaling of the variables in (2.1.1), (2.1.2) and (2.1.3) such that the equations are dimensionless. We assume a 2D-model where gravity is the only body force and rescale as follows 1 t˜ = t, ω
x ˜ = Lx,
y˜ = Ry,
y˜′ = ly ′ ,
u ˜ = Cu,
v˜ = εCv,
C2 ˜ = gb, p˜ = DC 2 p, ρ˜ = Dρ, T˜ = T, b cp
(2.1.9a) (2.1.9b)
C2 ρ˜s = Ds ρs , T˜s = Ts , cp c˜ = Cc,
(2.1.9c)
cp β˜ = 2 β, C
(2.1.9d)
where C is a typical speed of sound, g is the gravitational acceleration, D and Ds are typical densities for the fluid and solid, respectively, and ε is the aspect ratio of a stack pore defined as ε = R/L ≪ 1.
(2.1.10)
Clearly, ε is a dimensionless parameter. In total there are 15 physical parameters in this problem expressible in 5 independent fundamental physical quantities. Therefore, using the Buckingham π theorem [1], we know that 10 independent dimensionless parameters can be constructed from the original 14 parameters. In addition to ε, we will use the following dimensionless parameters: εp = l/L,
ϑ=
RKs , lK
κ=
U2 , gL
Wo =
√ R 2 , δν
NL =
Fr =
ωL , C R , δk
γ=
cp , cv
A=
Np =
l , δs
Pr =
U , C
(2.1.11a)
2NL2 . Wo2
(2.1.11b)
where cv is the isochoric specific heat, λ = 2π˜ c/ω is the wavelength, ω is the frequency, U is a typical fluid speed, εp is the aspect ratio of the stack plates, κ is a dimensionless wave number (Helmholtz 8
number), A is a Mach number, Fr is the Froude number, P r is the Prandtl number, Wo is the Womersley number and NL and Np are the Lautrec numbers (as defined by Garrett [4]) related to the fluid and plate, respectively. Here the parameters δν , δk and δs are the viscous penetration depth, and the thermal penetration depths for the fluid and solid, respectively. r r r 2ν 2χ 2χs δν = , δk = , δs = . (2.1.12) ω ω ω Here ν = µ/D is the kinematic viscosity and χ = K/(Dcp ) is the thermal diffusivity of the fluid. It can be shown that the first 9 of the dimensionless parameters in (2.1.11) together with ε form 10 independent dimensionless parameters. Obviously the Prandtl number is determined completely by the Womersley and Lautrec numbers. We assume that the velocity is small compared to the speed of sound, i.e. the Mach number A is small. This can be used to linearize the equations mentioned afore. Let q be any of the fluid variables (p, v, T , etc..), then expanding q in powers of A with an equilibrium state q0 we find q(x, y, t) = q0 (x, y) +
∞ X
Ak qk′ (x, y, t),
A ≪ 1.
k=1
(2.1.13a)
where we assume harmonic time-dependence for q1 with dimensionless frequency 1, so that we can write q1′ (x, y, t) = Re q1 (x, y)eit . (2.1.13b)
Since u = O(A), we find that v0 = (u0 , v0 ) ≡ 0. Moreover we demand u1 6= 0,
(2.1.14)
We call (2.1.13) a weakly non-linear expansion to indicate that, although we assume small amplitudes, we still include terms of higher order in A. In dimensionless form we obtain the following system of equations ∂ρ ∂(ρu) ∂(ρv) + + = 0, ∂t ∂x ∂y ρ
∂u ∂u ∂u +u +v ∂t ∂x ∂y
=−
(2.1.15) κ ∂p + ∂x Wo2
A2 ∂2u ∂2u ρbx ε2 2 + 2 + ∂x ∂y Fr εκ + Wo2
2
ε ρ
ρ
∂v ∂v ∂v +u +v ∂t ∂x ∂y
∂T ∂T ∂T +u +v ∂t ∂x ∂y
ε2 κ ∂p + =− ∂y Wo2
− βT
ξ 1 + µ 3
∂2v ∂2u + ∂x2 ∂x∂y
,
(2.1.16)
,
(2.1.17)
2 ∂2v A2 ε 2∂ v + ρby ε + ∂x2 ∂y 2 Fr ε2 κ + Wo2
1 ξ + µ 3
∂p ∂p ∂p +u +v ∂t ∂x ∂y
κ = 2NL2
κ + Wo2 9
∂2v ∂2u + 2 ∂x∂y ∂y
2 ∂2T 2∂ T ε + ∂x2 ∂y 2
"
∂u ∂y
2
#
+ O(ε) .
(2.1.18)
∂Ts κ ρs = ∂t 2Np2
2 ∂ 2 Ts 2 ∂ Ts , + εp ∂x2 ∂y ′2
ρ1 = −ρ0 βT1 +
(2.1.19)
γ p1 . c2
(2.1.20)
subject to v|y=±1 = 0,
(2.1.21a)
T |y=±1 = Ts |y′ =∓1 =: Tb (x),
(2.1.21b)
∂Ts ∂T =ϑ . ∂y y=±1 ∂y ′ y′ =∓1
3.
(2.1.21c)
Thermoacoustic devices without streaming
This chapter discusses both stack-based and regenerator-based thermoacoustic devices in the absence of streaming. With streaming there exists a steady mean velocity, that is superimposed on the larger first-order oscillations. This will be discussed further in the next section. To solve the equations given in the previous section we need to know the magnitude of the dimensionless parameters involved. First note that κ=
ωL c˜L cL L = = ∼ . C Cλ λ λ
(3.0.22)
Thus the dimensionless wave number is a measure for the relative length of the stack. Here we will make the ”short-stack” approximation L ≪ λ (κ ≪ 1), where the stack is considered to be short compared to the engine with the restriction that κ is the largest among the small parameters. The Womersley number and the Lautrec number are related by 2NL2 = P rWo2 . The Prandtl number only depends on material parameters and is usually close to 1. As a result NL and Wo should be of the same order of magnitude. Normally in standing wave machines R ∼ δk (stack) and in traveling wave machines R ≪ δk (regenerator). Also we assume that Np ∼ NL . Furthermore we assume that the amplitudes of the acoustic oscillations can be taken arbitrarily small, with the restriction that ε2 ≪ A, so that ∂ 2 q˜0 /∂ x ˜2 can be neglected with respect to ∂ 2 q˜1 /∂ y˜2 2 (O(ε ) versus O(A)). Swift[23] and Tijani [26] also treated this case, although they did not make these assumptions explicit. Finally, we also assume that ϑ = O(1),
Fr = O(1).
(3.0.23)
Summarizing we have ε2p ≪ ε2 ≪ A ≪ κ2 ≪ 1,
ϑ = O(1),
Fr = O(1),
(3.0.24)
and Wo ∼ 1,
NL ∼ 1,
Np ∼ 1,
for a stack,
(3.0.25)
Wo ≪ 1,
NL ≪ 1,
Np ≪ 1,
for a regenerator.
(3.0.26)
10
The presence of κ as a small parameter suggests the following alternative expansions for the fluid variables q (3.0.27) q = q0 (x, y) + Aκmq Re q10 (x, y) + κq11 (x, y) + κ2 q12 (x, y) eiκt + · · · .
The variables here have either one or two indices. The first index always indicates the corresponding power of A and the second index shows the corresponding power of κ. When necessary we will also use a third index to indicate the power of ε. If any of these indices are left out, it means that the power is zero. As A is a Mach number, we demand that mv = 0. Anticipating that u ∼ p − p0 in the sound field in the main pipe, we also demand mp = 0. Note that the terms second order in A can be neglected since A ≪ κ2 ≪ 1.
3.1
Short stack
In this section we will restrict ourselves to the case of a stack, i.e. Wo ∼ 1,
NL ∼ 1, Np ∼ 1.
(3.1.1)
3.1.1 The horizontal velocity u1 Substitute the expansions given in (3.0.27) in the y-component of the momentum equation (2.1.17). Collecting, the zeroth order terms we find ∂p0 /∂y = 0. Next collecting terms up to order Aκ2 we also find A
∂p10 ∂p11 ∂p12 + Aκ + Aκ2 = 0. ∂y ∂y ∂y
(3.1.2)
This equation can only be satisfied if ∂p10 ∂p11 ∂p12 = = = 0. ∂y ∂y ∂y
(3.1.3)
We do the same for the x-component (2.1.16). The zeroth order equation yields ∂p0 /∂x = 0. Keeping terms of order up to Aκ we obtain iAκρ0 u10 = −A
dp10 dp11 Aκ ∂ 2 u10 − Aκ + , dx dx Wo2 ∂y 2
Collecting terms of order A we find that ∂p10 /∂x = 0. Hence p = p0 + ARe (p10 + κp11 (x)) eit + · · · .
(3.1.4)
(3.1.5)
Next, collecting the terms of order Aκ, we find that u10 satisfies iρ0 u10 = −
dp11 1 ∂ 2 u10 + . dx Wo2 ∂y 2
(3.1.6)
and applying the boundary conditions u1 (x, ±1) = 0, we obtain cosh(αν y) i dp11 1− , u10 (x, y) = ρ0 dx cosh(αν ) 11
(3.1.7)
where r ρ0 √ R αν = (1 + i) ρ0 = (1 + i) Wo. δν 2
(3.1.8)
In the same way it can be shown that cosh(αν y) i dp12 1− . u11 (x, y) = ρ0 dx cosh(αν )
(3.1.9)
This coincides with the solution found by Swift [23] and Tijani [26] in a dimensional form. 3.1.2 The temperature Ts in the plate Using the rescaled energy equation (2.1.18) and the diffusion equation (2.1.19) we will try to find T and Ts . First we solve (2.1.19) subject to (2.1.21b). Substitute the expansions from (3.0.27) into (2.1.19). To leading order we find that ∂ 2 Ts0 /∂y ′2 = 0. That means Ts0 is at most linear in y ′ . In view of symmetry we find that Ts0 is in fact constant in y ′ , i.e.
Ts10 (x, y ′ ) + κTs11 (x, y ′ ) eiκt + · · · ,
(3.1.10)
ρs10 (x, y ′ ) + κρs11 (x, y ′ ) eit + · · · ,
(3.1.11)
Ts (x, y ′ , t) = Ts0 (x) + AκmT Re
Since ρs0 = ρs0 (p0 , Ts0 ), we also assume that ρs (x, y ′ , t) = ρs0 (x) + Aκmρ Re Collecting terms of order A we find ∂ 2 Ts10 . ∂y ′2
(3.1.12)
Ts10 (x, ±1) = Tb10 (x),
(3.1.13)
2iρs0 Np2 Ts10 = Imposing (2.1.21b)
we find that the solution to(3.1.12) is given by Ts10 (x, y ′ ) = Tb10 (x)
cosh(αs y ′ ) , cosh(αs )
(3.1.14)
where αs = (1 + i)ρs0
l = (1 + i)ρs0 Np . δs
(3.1.15)
Similarly Ts11 (x, y ′ ) = Tb11 (x)
cosh(αs y ′ ) , cosh(αs )
(3.1.16)
The exact expression for Tb1 should be determined from matching Ts with T . 12
3.1.3 The temperature T between the plates We substitute the expansions shown in (2.1.13) into equation (2.1.18). To leading order we find ∂ 2 T0 /∂y 2 = 0. That means T0 is at most linear in y. In view of symmetry we find that T0 is in fact constant in y. Therefore we expand T as T (x, y, t) = T0 (x) + AκmT Re (T10 (x, y) + κT11 (x, y)) eit + · · · , (3.1.17) Since ρ0 = ρ0 (p0 , T0 ), we assume that also ρ(x, y, t) = ρ0 (x) + Aκmρ Re (ρ10 (x, y) + κρ11 (x, y)) eit + · · · ,
(3.1.18)
With these results we find to leading order iκ1+mT ρ0 T10 + ρ0 u10
dT0 κ1+mT ∂ 2 T10 = . dx 2NL2 ∂y 2
(3.1.19)
We choose mT = −1 to balance the equation. Then, inserting the expression for u10 given in (3.1.7), we obtain dT0 dp11 cosh(αν y) 1 ∂ 2 T10 iρ0 T10 + i . (3.1.20) 1− = dx dx cosh(αν ) 2NL2 ∂y 2 Next for T11 we find the following differential equations iρ0 T11 − iβT0 p10 + ρ0 u11
1 ∂ 2 T11 dT0 = . dx 2NL2 ∂y 2
(3.1.21)
Inserting (3.1.9), we obtain dT0 dp12 cosh(αν y) 1 ∂ 2 T11 iρ0 T11 + i . 1− − iβT0 p10 = dx dx cosh(αν ) 2NL2 ∂y 2
(3.1.22)
Applying the boundary conditions in (2.1.21), we can solve (3.1.20) and (3.1.22) P r cosh(αν y) 1 dT0 dp11 1− T10 (x, y) = − ρ0 dx dx P r − 1 cosh(αν ) 1 + εs ffνk 1 dT0 dp11 cosh(αk y) , ρ0 (P r − 1)(1 + εs ) dx dx cosh(αk ) 1 dT0 dp12 P r cosh(αν y) βT0 p10 − 1− T11 (x, y) = ρ0 ρ0 dx dx P r − 1 cosh(αν ) # " fν 1 1 + εs fk dT0 dp12 cosh(αk y) βT0 p10 + . − ρ0 ρ0 P r − 1 dx dx (1 + εs ) cosh(αk ) −
(3.1.23a)
(3.1.23b)
where √ R √ αk = (1 + i) ρ0 = (1 + i) ρ0 NL , δk p K ρ˜0 cp tanh(αk ) εs = √ , Ks ρ˜s0 cs tanh(αs )
fν =
(3.1.24)
tanh(αν ) , αν
fk =
tanh(αk ) . αk
(3.1.25)
Note that if we had taken A ≪ ε2 (compare with (3.0.24)), i.e. the amplitudes of the oscillations can be taken arbitrarily small, then the term d2 T0 /dx2 will appear in the equations. Then, for the equations to be satisfied, we must have that T0 = Ts0 is linear in x. 13
3.1.4 The pressure p In order to derive the dimensionless alternative for Rott’s wave equation (cf. [17], [18]), we start with the continuity equation. From the equation of state (2.1.20) we find that mρ = mT = −1. Now to leading order, we obtain iρ10 +
∂ ∂v10 (ρ0 u10 ) + ρ0 = 0, ∂x ∂y
(3.1.26)
Using the x-derivative of equation (3.1.6), the equation of state (2.1.20), and inserting the expression for T10 , we ultimately find 1 + εs ffνk dT0 dp11 cosh(αk y) dT0 dp11 P r cosh(αν y) −β 1− −β dx dx P r − 1 cosh(αν ) (P r − 1)(1 + εs ) dx dx cosh(αk ) −
d2 p11 1 ∂ 3 u10 ∂v10 + + iρ0 = 0. (3.1.27) 2 2 2 dx Wo ∂x∂y ∂y
Next we integrate with respect to y from 0 to 1. Note that v(x, 0) = 0 because of symmetry. Finally, inserting (3.1.7), we obtain Rott’s wave equation (dimensionless) fk − fν dT0 dp11 d 1 − fν dp11 β + ρ0 = 0, (3.1.28) (P r − 1)(1 + εs ) dx dx dx ρ0 dx Here we used 1 dρ0 1 ∂ρ0 dT0 dT0 d 1 =− =− =β . ρ0 dx ρ0 ρ0 dx ρ0 ∂T0 dx dx This result was also found by Rott [18], Swift [23] and Tijani [26]. For an ideal gas and ideal stack with εs = 0 this result was first obtained in [17]. Note that in Rott’s result there is one term more in the equation (namely (1 + (γ − 1)fk /(1 + εs ))p1 ). The dimensional analysis shows that this term can be neglected. Going one order higher we can show, similar to the analysis above, that p12 satisfies the following equation. fk γ−1 i 2 2 f − (c β T − γ + 1) 1 − p10 1 + 0 k c2 1 + εs 1 + εs fk − fν dT0 dp12 d + β + ρ0 (P r − 1)(1 + εs ) dx dx dx
1 − fν dp12 ρ0 dx
= 0. (3.1.29)
Furthermore, if we impose the dimensionless equivalent of relation (A.4.6) c2 β 2 T0 = γ − 1, then (3.1.29) transforms into 1 dT0 dp12 d 1 − fν dp12 fk − fν γ−1 fk p10 + β + ρ0 1+ = 0. (3.1.30) c2 1 + εs (P r − 1)(1 + εs ) dx dx dx ρ0 dx 14
3.1.5 Acoustic power ˜˙ used (or produced in the case of a prime mover) in a segment The time-averaged acoustic power dW of length dx can be found from ˜˙ dW d h ′ ′ i hp1 u1 i , = A˜g d˜ x d˜ x
(3.1.31)
˜˙ = R2 DC 3 W ˙ , W
(3.1.32)
where the overbar indicates time average (over one period), brackets h i indicate averaging in the y˜˜˙ and A˜g are rescaled direction and Ag is the cross-sectional area of the gas within the stack. Next W as A˜g = R2 Ag .
As p is independent of y, we find i ˙ dW d h ′ = A2 Ag p1 hu1 ′ i dx dx
(3.1.33)
1 p′1 hu1 ′ i = Re [p1 hu1 ∗ i] , 2
(3.1.34)
It can be shown that the product p′1 hu1 ′ i satisfies
where the star denotes complex conjugation. Inserting (3.1.34) into (3.1.33) yields ˙ 1 2 dhu1 ∗ i dW ∗ dp1 = A Ag Re p1 + hu1 i . dx 2 dx dx
(3.1.35)
Next, inserting (3.0.27), we find ˙ 1 2 dhu10 ∗ i dW dhu10 ∗ i dhu11 ∗ i ∗ dp11 = A Ag Re p10 + p11 + hu10 i + κRe p10 +··· . dx 2 dx dx dx dx (3.1.36) Combining equations (3.1.9) and (3.1.28), we find iρ0 hu10 i dp11 =− , dx 1 − fν
(3.1.37)
and dhu10 i d =i dx dx =−
1 − fν dp11 ρ0 dx
fk − fν dT0 β hu10 i. (P r − 1)(1 + εs )(1 − fν ) dx
(3.1.38)
Similarly we obtain d dhu11 i =i dx dx
1 − fν dp12 ρ0 dx
fk − fν dT0 i γ−1 = β hu11 i − fk p10 . 1+ (1 − P r)(1 + εs )(1 − fν ) dx ρ 0 c2 1 + εs 15
(3.1.39)
As a result we find ˙ ˙ 20 ˙ 21 dW dW dW = A2 + A2 κ + ··· , dx dx dx
(3.1.40)
where ˙ 20 fk∗ − fν∗ 1 Ag β dT0 dW ∗ = Re p10 hu10 i + · · · , dx 2 1 − P r dx (1 + ε∗s )(1 − fν∗ ) ˙ 21 fk∗ − fν∗ dW 1 Ag β dT0 ∗ ∗ = Re p10 hu11 i + p11 hu10 i dx 2 1 − P r dx (1 + ε∗s )(1 − fν∗ ) Ag γ − 1 1 Im(fν ) − Ag ρ0 |hu10 i|2 − Im (−fk ) |p10 |2 . 2 |1 − fν |2 2ρ0 c2 1 + εs
(3.1.41a)
(3.1.41b)
Three different effects can be observed in (3.1.41). The first term in (3.1.41a), (3.1.41b) is called the source or sink term because it either absorbs (refrigerator, heat pump) or produces acoustic power (prime mover). This term is the unique contribution to thermoacoustics. The second term in (3.1.41b) is the thermal relaxation dissipation term and the third term in (3.1.41b) is the viscous dissipation term. These two terms are always present whenever a wave interacts with a solid surface, and have a dissipative effect in thermoacoustics. One could repeat the previous analysis for a long stack (κ = O(1)) as well, in which case we get ˙2 dW 1 = Ag dx 2
fk∗ κρ0 Im(fν ) γ−1 2 |hu1 i| + κ Im |p1 |2 |1 − fν |2 ρ 0 c2 1 + ε∗s fk∗ − fν∗ β dT0 ∗ − Re p1 hu1 i . (3.1.42) P r − 1 dx (1 − fν∗ )(1 + ε∗s )
3.1.6 The dissipation in acoustic power To see the effect of changing the stack length and reducing the pore size, we test how (3.1.41) behaves for small NL or Wo and small κ. It can be shown that (3.1.41) behaves as ˙2 dW dx
= sink term − thermal dissipation relaxation − viscous dissipation κ . − O = O(1) − O κNL2 Wo2
(3.1.43)
Dissipation is usually undesirable, so the dissipation terms should be smaller than the sink term. For the thermal relaxation dissipation this is already the case, provided the stack is short enough (κ ≪ 1). Furthermore reducing the pore size will reduce thermal relaxation dissipation even more. For the viscous dissipation the situation is different. If the pore size becomes too small (NL , Wo → 0), viscous dissipation will play an increasingly important role. Therefore, from (3.1.43), we find the following inequalities κ ≪ Wo2 ≪ 1.
(3.1.44)
If κ and Wo satisfy these inequalities, then viscous and thermal relaxation dissipation can both be neglected with respect to the sink/source term. 16
Summarizing, reducing the pore size decreases thermal relaxation dissipation and increases the viscous dissipation. The latter can be compensated by taking the stack short enough, as indicated in (3.1.44). Note that in the case of small pores viscous dissipation is always more important than thermal relaxation dissipation. 3.1.7 The sink/source term in acoustic power We saw in (3.1.41) that the sink/source term, which we will define as W2s , was of greatest interest s we will assume ε = 0 and neglect in thermoacoustic engines and refrigerators. To interpret W20 s viscosity, so that fν = 0 and P r = 0, ˙ s Ag β dT0 dW 20 = [Re (fk ) Re (p10 hu10 ∗ i) + Im (−fk ) Im (p10 hu10 ∗ i)] . dx 2 dx
(3.1.45)
In a standing wave system Im (p10 hu10 ∗ i) is large and therefore Im (−fk ) is important. Fig. 6 shows that the maximal value is attained for NL ∼ 1, which is exactly the condition assumed for a stack in a standing wave system. In the case of a traveling wave system Re (p10 hu10 ∗ i) is large and Re (fk ) is important. Fig. 6 shows that Re (fk ) reaches its maximal value for NL ≪ 1, which was the condition for a regenerator. 1 real part imaginary part
0.5 fk
0
−0.5 0
1
2
3 NL
4
5
6
Figure 6: Real and imaginary part of fk , plotted as a function of the Lautrec number NL .
3.1.8 Time-averaged total power H˙ ˜˙ in the stack, correct to Finally we will derive an expression for the time-averaged total power H second order in A. We consider the thermoacoustic refrigerator as shown in Fig. 5(a), driven by a loudspeaker. We assume that the refrigerator is thermally insulated from the surroundings, except at the two heat exchangers, so that heat can be exchanged with the outside world only via the two heat exchangers. Work can only be exchanged at the loudspeaker piston. By conservation of energy we have (A.5.6) 1 2 ∂ 1 2 ˜ ˜ ˜ ˜ ˜ ρ˜v˜ + ρ˜ǫ˜ = −∇ · v˜ ρ˜v˜ + ρ˜h − K ∇T − v˜ · Σ , (3.1.46) 2 ∂ t˜ 2 17
˜ are the internal energy and enthalpy per unit mass, respectively and Σ ˜ is the viscous where ǫ˜ and h stress tensor as defined in (2.1.4). The expression on the left represents the rate-of-change of the energy in a unit volume of the fluid, while that on the right is the divergence of the total power density which consists of transfer of mass by the motion of the fluid, transfer of heat and total power due to internal friction, respectively. Integrating (3.1.46) with respect to y˜ (and y˜′ ) from y˜ = 0 to y˜′ = 0 and time-averaging (up to second order in A) yields d d˜ x
"Z
R
ρ˜u ˜v˜2 d˜ y+ 0
Z
R
0
˜ d˜ ρ˜u ˜h y−
Z
R
K 0
∂ T˜ d˜ y− ∂x ˜
Z
Ks
∂ T˜s ′ d˜ y ∂x ˜
Z
˜ 11 + v˜Σ ˜ 21 d˜ u ˜Σ y = 0. (3.1.47)
l
0
−
0
R
˜˙ Π, ˜ the time-averaged total power per unit perimeter The quantity within the square brackets is H/ along x. Thus for a cyclic refrigerator (prime mover) without heat flows to the surroundings, the time˜˙ (correct up to second order) along x averaged total power H ˜ must be independent of x ˜. Rescaling 2 yields Z 1 Z 1 Z Z ϑε2p κ l ∂T s ′ H˙ ε2 κ 1 ∂T 2 2 =ε dy − dy ρuv dy + ρuh dy − Π 2NL2 0 ∂x 2NL2 0 ∂x 0 0 Z 1 εκ ∂u − u 1 + O(ε2 ) dy, (3.1.48) 2 Wo 0 ∂y
˜˙ h ˜ and Π ˜ were rescaled as where H, ˜˙ = R2 DC 3 H, ˙ H
˜ = C 2 h, h
s˜ = cp s,
˜ = RΠ. Π
(3.1.49)
The second integral on the right hand side, can be written up to second order as follows. ρuh = Aρ0 h0 u′1 + A2 ρ0 h0 u′2 + A2 h0 ρ′1 u′1 + A2 ρ0 h′1 u′1 + · · · .
(3.1.50)
The first term disappears because u′1 = Re [u1 eit ] = 0. The second and third term cancel because the second order time-averaged mass flux is assumed zero Z 1 (3.1.51) ρ0 u′2 + ρ′1 u′1 dy = 0. 0
Therefore, inserting the relation
1 dh = dT + (1 − βT )dp, ρ
(3.1.52)
we obtain Z 1 Z 1 H˙ 3 2 2 2 =A ε ρ0 u1 v1 dy + A ρ0 T0 u1 T1 + (1 − βT0 )p1 u1 dy Π 0 0 Z 1 κ dT0 κ ∂u1 2 2 2 −A ε dy + · · · , (3.1.53) − ε + ϑεp u1 2 2 Wo 0 ∂y 2NL dx 18
Clearly the first and last term on the right hand side can be neglected with respect to the second term. As a result we are left with Z 1 Z 1 κ dT0 H˙ 2 2 ′ ′ (1 − βT0 )p′1 u′1 dy − ε2 + ϑε2p =A ρ0 u1 T1 dy + A . (3.1.54) Π 2NL2 dx 0 0 Substituting the expansions given in (3.0.27) we find Z 1 Z A2 Ag 1 1 2 ∗ ˙ ρ0 Re [u∗10 T11 + u∗11 T10 ] dy H= ρ0 Re [u10 T10 ] dy + A Ag 2κ 0 2 0 Z 1 Ag dT0 1 (1 − βT0 )Re [p10 u∗10 ] dy − (ε2 + ϑε2p )κ + ··· + A2 Ag 2 2NL2 dx 0 =: A2 H˙ 2-10 + A2 H˙ 200 + ε2 + ϑε2p κH˙ 012 + · · · .
with H˙ 2-10 , H˙ 200 and H˙ 012 defined as Z Ag 1 H˙ 2-10 = ρ0 Re [u∗10 T10 ] dy, 2 0 Z Z Ag 1 Ag 1 ρ0 Re [u∗10 T11 + u∗11 T10 ] dy + (1 − βT0 )Re [p10 u∗10 ] dy, H˙ 200 = 2 0 2 0 Ag dT0 H˙ 012 = − , 2NL2 dx
(3.1.55)
(3.1.56a) (3.1.56b) (3.1.56c)
where the first index indicates the order of A, the second of κ and the third of ε. Inserting the expressions obtained for T1 and u1 into (3.1.56), we find " # fν ∗) 2 )(f − f (1 + ε s k A ρ |hu i| dT ν fk g 0 10 0 Im fν∗ + , (3.1.57a) H˙ 2-10 = 2(1 − P r)|1 − fν |2 dx (1 + εs )(P r + 1) Ag βT0 (fk − fν∗ ) ∗ ˙ Re p10 hu10 i 1 − H200 = 2 (1 + εs )(1 − fν∗ )(1 + P r) " # (1 + εs ffνk )(fk − fν∗ ) Ag ρ0 Re [hu10 ihu11 ∗ i] dT0 ∗ + Im fν + . (3.1.57b) 2(1 − P r)|1 − fν |2 dx (1 + εs )(P r + 1) Ag dT0 H˙ 012 = − , 2NL2 dx
(3.1.57c)
H˙ 210 represents the contribution of the conduction of heat through gas and stack material to the total total power and H˙ 2-10 and H˙ 200 represent the contribution of the thermoacoustic heat flow. Therefore we find in the case of a short stack with the assumptions given in (3.0.24), that the total power is not so much affected by conduction of heat through gas and stack material.
3.2
Short regenerator
In the previous section we considered the case of a stack, where the Lautrec and Womersley number were of order 1. Here we will discuss the regenerator, for which Wo and NL are very small. Therefore 19
we have to relate Wo and NL to the other small parameters in (3.0.24). In section 3.1.6 we saw that to reduce viscous dissipation Wo should satisfy κ ≪ Wo2 ≪ 1. Therefore we will examine what √ √ happens when Wo approaches its lower bound κ, i.e. we choose Wo = κ. To simplify the equations even further, we also assume NL = Np . The remaining assumptions remain the same, so that ε2p ε2 ≪ ≪ A ≪ κ2 ≪ Wo2 = 2P rNL2 ≪ 1, κ κ
ϑ, P r, Fr = O(1),
We expand the fluid variables in the same way as we did in (3.0.27) for the short stack q = q0 (x, y) + κmq Re (q10 (x, y) + κq11 (x, y)) eit + · · · .
(3.2.1)
(3.2.2)
3.2.1 The horizontal velocity u1
Substitute the expansions given in (3.0.27) in the y-component of the momentum equation (2.1.17). To leading order we find ∂p0 /∂y = 0. Hence p = p0 + ARe (p10 (x) + κp11 (x)) eit + · · · . (3.2.3)
Next, neglecting terms with order higher than Aκ, we obtain A
∂p10 ∂p11 + Aκ = 0. ∂y ∂y
(3.2.4)
This equation can only be satisfied if ∂p10 /∂y = ∂p11 /∂y = 0. We do the same for the x-component (2.1.16). To leading order we find ∂p0 /∂x = 0. Next, we only keep terms of order up to Aκ and Aκ2 /Wo2 and neglect higher order terms. This leads to dp11 Aκ ∂ 2 u10 ∂ 2 u11 dp10 − Aκ + +κ , (3.2.5) iAκρ0 u10 = −A dx dx Wo2 ∂y 2 ∂y 2 Now we insert √ Wo = κ ≪ 1.
(3.2.6)
Collecting the terms of order A, we find that u10 satisfies dp10 ∂ 2 u10 = , 2 ∂y dx
(3.2.7)
and applying the boundary conditions u10 (x, ±1) = 0, we obtain u10 (x, y) = −
1 dp10 (1 − y 2 ). 2 dx
(3.2.8)
Finally, collecting the terms of order Aκ, we find −
dp11 ∂ 2 u11 i dp10 ρ0 (1 − y 2 ) = − + , 2 dx dx ∂y 2
(3.2.9)
where we substituted (3.2.8) for u10 . Applying the boundary conditions u11 (x, ±1) = 0, we obtain u11 (x, y) =
i dp10 1 dp11 ρ0 (1 − y 2 )(5 − y 2 ) − (1 − y 2 ). 24 dx 2 dx 20
(3.2.10)
3.2.2 The temperature Ts in the plate Using the rescaled energy equation (2.1.18) and the diffusion equation (2.1.19) we will try to find T and Ts . First we solve (2.1.19) subject to (2.1.21b). Substitute the expansions from (3.0.27) into (2.1.19). To leading order we obtain ∂ 2 Ts0 /∂y ′2 = 0. That means Ts0 is at most linear in y ′ . In view of symmetry we find that Ts0 is in fact constant in y ′ , i.e. Ts (x, y ′ , t) = Ts0 (x)+AκmT Re Ts10 (x, y ′ ) + κTs11 (x, y ′ ) + κ2 Ts12 (x, y ′ ) eiκt +· · · , (3.2.11) Collecting terms up to order AκmT +2 we find iκP rρs0 (Ts10 + κTs11 ) =
2 ∂ 2 Ts10 ∂ 2 Ts11 2 ∂ Ts12 + κ + κ = 0. ∂y ′2 ∂y ′2 ∂y ′2
Then to leading order we find ∂ 2 Ts10 = 0. ∂y ′2 Imposing (2.1.21b) we find Ts10 (x, y ′ ) = Tb10 (x),
(3.2.12)
Next we find for Ts11 ∂ 2 Ts11 = iP rρs0 Tb10 . ∂y ′2 Applying the boundary conditions in (2.1.21b) we find i Ts11 (x, y ′ ) = − P rρs0 Tb10 (x) 1 − y ′2 + Tb11 (x). 2
(3.2.13)
Finally Ts12 satisfies
1 ∂ 2 Ts12 = iP rρs0 Ts11 = iP rρs0 Tb11 + P r2 ρ2s0 Tb10 1 − y ′2 . ′2 ∂y 2
Applying the boundary conditions in (2.1.21b) we find Ts12 (x, y ′ ) = −
i 1 P r2 ρ2s0 Tb10 y ′4 − 6y ′2 + 5 − P rρs0 Tb11 1 − y ′2 + Tb12 (x). (3.2.14) 24 2
The exact expression for Tb1 should be determined from matching Ts with T . 3.2.3 The temperature T between the plates
We substitute the expansions shown in (2.1.13) into equation (2.1.18). To leading order we find ∂ 2 T0 /∂y 2 = 0. That means T0 is at most linear in y. In view of symmetry we find that T0 is in fact constant in y. Therefore we expand T as T (x, y, t) = T0 (x) + AκmT Re T10 (x, y) + κT11 (x, y) + κ2 T12 (x, y) eit + · · · , (3.2.15) 21
Since ρ0 = ρ0 (p0 , T0 ), we assume that also ρ(x, y, t) = ρ0 (x) + AκmT Re ρ10 (x, y) + κρ11 (x, y) + κ2 ρ12 (x, y) eit + · · · ,
(3.2.16)
Collecting terms up to order Aκ we find
dT0 κmT Aρ0 u10 + iAκ1+mT ρ0 T10 − iAκβT0 p10 = dx Pr
2 ∂ 2 T11 ∂ T10 + Aκ A . ∂y 2 ∂y 2
(3.2.17)
Obviously mT cannot be positive, otherwise u10 ≡ 0. Two possible choices remain: mT = 0, mT = −1. With mT = 0 the first and last term in (3.2.17) are balanced, whereas with mT = −1 the first two terms are balanced. Let’s first examine the case mT = 0. To leading order we find ρ0
∂ 2 T10 P r dp10 dT0 2 (y − 1) = . 2 dx dx ∂y 2
where we substituted the expression for u10 given in (3.1.7). Applying the boundary conditions in (2.1.21b), we find T10 (x, y) = ρ0
P r dp10 dT0 4 y − 6y 2 + 5 + Tb10 . 24 dx dx
Now Tb10 still needs to be determined. The remaining boundary conditions in (2.1.21c), however, cannot be satisfied, unless dp10 =0 dx
⇒
u10 (x, y) = 0,
Again we find u10 ≡ 0, contradicting (2.1.14) and thus mT 6= 0. Next consider the case mT = −1. To leading order we find ∂ 2 T10 /∂y 2 = 0, hence T10 (x) = Ts10 (x) = Tb10 (x),
(3.2.18)
where Tb10 is yet to be determined. Tb10 will be fixed by solving the equation for T11 . Note how this agrees with our intuition that for very small pores the temperature of the plate and gas parcels will be practically the same. Collecting the terms of order A, we find −
ρ0 dT0 dp10 1 ∂ 2 T11 (1 − y 2 ) + iρ0 Tb10 = . 2 dx dx P r ∂y 2
Applying the boundary conditions in (2.1.21b), we find T11 (x, y) =
P rρ0 dT0 dp10 4 i (y − 6y 2 + 5) − P rρ0 Tb10 1 − y 2 + Tb11 . 24 dx dx 2
The boundary conditions in (2.1.21c) lead to Tb10 = T10 = Ts10 = − =
ρ0 dT0 dp10 i , 3 ρ0 + ϑρs0 dx dx
iρ0 dT0 hu10 i. ρ0 + ϑρs0 dx 22
(3.2.19)
As a result we find P rρ0 dT0 dp10 4 ρ20 Pr dT0 dp10 (y − 6y 2 + 5) − 1 − y 2 + Tb11 , 24 dx dx 6 ρ0 + ϑρs0 dx dx (3.2.20)
T11 (x, y) =
Ts11 (x, y ′ ) = −
P r ρ0 ρs0 dT0 dp10 1 − y ′2 + Tb11 , 6 ρ0 + ϑρs0 dx dx
(3.2.21)
where Tb11 is yet to be determined. To find Tb11 we have to go one order higher. Collecting terms of order Aκ, we find ρ0 u11
dT0 1 ∂ 2 T12 + iρ0 T11 − iβT0 p10 = . dx P r ∂y 2
Applying the boundary conditions in (2.1.21b), we find iP r(1 + P r) dT0 dp10 2 6 ρ (y − 15y 4 + 75y 2 − 61) 720 dx dx 0 i ρ0 P r dT0 dp11 4 + y − 6y 2 + 5 + P rβT0 p10 1 − y 2 24 dx dx 2
T12 (x, y) =
−
iP rρ0 P r2 2 ρ0 Tb10 y 4 − 6y 2 + 5 − Tb11 1 − y 2 + Tb12 , 24 2
with Tb12 still unknown. The boundary conditions in (2.1.21c) lead to
Tb11
1 ρ20 + ϑρ2s0 2 27 dT0 ρ0 = − Pr Pr + hu10 i ρ0 + 3 ρ0 + ϑρs0 5 40 ρ0 + ϑρs0 dx +
iρ0 dT0 βT0 p10 hu11 i + . (3.2.22) ρ0 + ϑρs0 dx ρ0 + ϑρs0
3.2.4 The pressure p In order to derive Rott’s wave equation, we start with the continuity equation. From the equation of state (2.1.20) we find that mρ = mT = −1. Thus up to terms of order Aκ we find − iAρ0 βT10 − iAκρ0 βT11 + iAκ
∂ ∂ γ p10 + A (ρ0 u10 ) + Aκ (ρ0 u11 ) 2 c ∂x ∂x + ρ0
∂v11 ∂v10 + Aκρ0 = 0, (3.2.23) ∂y ∂y
where we inserted the equation of state (2.1.20). Collecting terms of order A and Aκ, we obtain −iρ0 βT10 +
∂ ∂v10 (ρ0 u10 ) + ρ0 = 0, ∂x ∂y
−iρ0 βT11 + i
(3.2.24)
γ ∂ ∂v11 p10 + (ρ0 u11 ) + ρ0 = 0. 2 c ∂x ∂y 23
(3.2.25)
Inserting the expressions for T1 and u1 , we ultimately find −iρ0 βTb10
1 d − 2 dx
∂v10 dp10 1 − y 2 + ρ0 ρ0 = 0, dx ∂y
and ∂v11 γ dp11 − iρ0 βTb11 1 − y 2 + ρ0 + i 2 p10 ρ0 dx ∂y c dT0 dp10 1 4 ρ0 1 − iP rρ20 β (y − 6y 2 + 5) − 1 − y2 dx dx 24 6 ρ0 + ϑρs0 i d 2 dp10 + ρ0 (y 4 − 6y 2 + 5) = 0. 24 dx dx 1 d − 2 dx
Next we integrate with respect to y from 0 to 1. Note that v(x, 0) = 0 because of symmetry and v(x, 1) = 0 because of the boundary conditions. We obtain d dx
dT0 dp10 dp10 ρ20 β , ρ0 = −3iρ0 βTb10 = − dx ρ0 + ϑρs0 dx dx =
d dx
(3.2.26a)
3ρ20 β dT0 hu10 i, ρ0 + ϑρs0 dx
(3.2.26b)
11 1 ρ0 dp11 γ 2 dT0 dp10 − ρ0 = −3iρ0 βTb11 + 3i 2 p10 − iP rρ0 β dx c dx dx 40 3 ρ0 + ϑρs0 11 d 2 dp10 +i ρ0 40 dx dx 33 γ − P r ρ20 βTb10 = −3iρ0 βTb11 + 3i 2 p10 + c 40 +i
33 dT0 (1 + P r)ρ20 β hu10 i. 40 dx
(3.2.26c)
Imposing two boundary conditions for both p10 and p11 will give a complete boundary value problem for p10 and p11 . During the derivation leading to this result, the following assumptions were used ε2p ε2 ≪ ≪ A ≪ κ2 ≪ 1, κ κ
ϑ = O(1),
κ = Wo2 =
2Np2 2NL2 = , Pr Pr
(3.2.27)
3.2.5 Acoustic power ˜˙ is given by Remember from the previous section that the time-averaged acoustic power dW ˙ dW 1 d = A2 Ag (Re [p1 hu1 ∗ i]) dx 2 dx
(3.2.28) 24
Inserting (3.2.2) into (3.2.28) yields ˙ dW 1 2 dhu10 ∗ i ∗ dp10 = A Ag Re p10 + hu10 i dx 2 dx dx dhu11 ∗ i dhu10 ∗ i ∗ dp11 ∗ dp10 +κRe p10 + p11 + hu10 i + hu11 i + ··· dx dx dx dx =: A2
˙ 20 ˙ 21 dW dW + A2 κ + ··· . dx dx
(3.2.29)
Combining equations (3.2.8), (3.2.10) and (3.2.26), we find after a lengthy calculation 1 d2 p10 1 d dp10 1 dρ0 dp10 dhu10 i =− = − ρ + 0 dx 3 dx2 3ρ0 dx dx 3ρ0 dx dx =
ϑρs0 dT0 hu10 i. β ρ0 + ϑρs0 dx
(3.2.30)
11i dT0 11i dhu10 i 1 d2 p11 dhu11 i = ρ0 β hu10 i − ρ0 − dx 40 dx 40 dx 3 dx2 11i ρ20 β 1 d dT0 dp11 1 dT0 dp11 = hu10 i − ρ0 − β 40 ρ0 + ϑρs0 dx 3ρ0 dx dx 3 dx dx ρ20 + ϑρ2s0 dT0 i 27 11P r dT0 ρ20 β = − P rρ0 β hu i + i − hu10 i 10 2 3 (ρ0 + ϑρs0 ) dx 40 15 ρ0 + ϑρs0 dx −
11i dT0 β 2 T0 p10 ϑρs0 β dT0 γ P rρ0 β hu10 i + i p10 + hu11 i. −i 2 40 dx ρ0 + ϑρs0 ρ0 c ρ0 + ϑρs0 dx
(3.2.31)
Inserting this result into (3.2.29), we find ˙ 20 1 dhu10 ∗ i dp10 dW = Ag Re p10 + hu10 ∗ i dx 2 dx dx 1 ϑρs0 dT0 3 = Ag β Re [p10 hu10 ∗ i] − Ag |hu10 i|2 , (3.2.32a) 2 ρ0 + ϑρs0 dx 2 ˙ 21 1 dhu11 ∗ i dhu10 ∗ i dW ∗ dp11 ∗ dp10 = Ag Re p10 + p11 + hu10 i + hu11 i dx 2 dx dx dx dx ρ20 + ϑρ2s0 1 27 11P r 11 dT0 1 ρ0 = Ag − − P r ρ0 β Im [p10 hu10 ∗ i] − Pr 2 2 40 15 ρ0 + ϑρs0 3 (ρ0 + ϑρs0 ) 40 dx ϑρs0 β dT0 1 Re [p10 hu11 ∗ i + p11 hu10 ∗ i] − 3Ag Re [hu10 ihu11 ∗ i] + Ag 2 ρ0 + ϑρs0 dx
(3.2.32b)
˙ 20 and the first two terms in W ˙ 21 are the sink/source terms. They either absorb (reThe first term in W frigerator) or produce acoustic power (prime mover) depending on the magnitude of the temperature ˙ 20 and W ˙ 21 is the viscous dissipation. To find the gradient along the stack. The last term in both W thermal relaxation dissipation we would have to go one order higher in κ. Thus, as expected, viscous dissipation plays a key role in regenerators and thermal relaxation dissipation only a minor role. This 25
is to be expected because if the pores become smaller then viscous dissipation will increase. Furthermore due to the perfect thermal contact between the gas parcels and the plate, thermal relaxation dissipation will decrease. Finally, if we look at (3.2.32a), then we can see that this almost exactly coincides with the sink term in (3.1.45), for NL ≪ 1 (Im(−fk ) → 0 and Re(fk ) → 1). 3.2.6 Time-averaged total power H˙ In section 3.1.8 we saw that the time-averaged total power H˙ in the stack is given by Z 1 Z 1 ε2 + ϑε2p dT0 2 2 ′ ′ ˙ ρ0 u1 T1 dy + A H = A Ag . (1 − βT0 )p′1 u′1 dy − Pr dx 0 0 Substituting the expansions given in (3.2.2) we find Z 1 Z 1 1 2 −1 1 2 ∗ ˙ H = A κ Ag ρ0 Re [u10 T10 ] dy + A Ag ρ0 Re [u∗10 T11 + u∗11 T10 ] dy 2 2 0 0 Z 1 2 ε + ϑε2p dT0 1 (1 − βT0 )Re [p10 u∗10 ] dy − Ag + ··· + A2 Ag 2 Pr dx 0 =: A2 κ−1 H˙ 2-10 + A2 H˙ 200 + ε2 + ϑε2p H˙ 002 + · · · .
Inserting T1 and u1 into (3.2.34), we find Ag iρ20 dT0 2 ˙ H2-10 = Re |hu10 i| = 0, 2 ρ0 + ϑρs0 dx
(3.2.33)
(3.2.34)
(3.2.35)
Ag ρ0 + (1 − βT0 )ϑρs0 Ag ρ20 dT0 H˙ 200 = Im [hu10 ihu11 ∗ i] Re [p10 hu10 ∗ i] − 2 ρ0 + ϑρs0 2 ρ0 + ϑρs0 dx (3.2.36) −
Ag 2
ρ2 + ϑρ2s0 1 Pr 0 − 3 (ρ0 + ϑρs0 )2
4 27 Pr + 15 40
dT0 17 ρ0 P r ρ20 |hu10 i|2 , − ρ0 + ϑρs0 105 dx
Ag dT0 . H˙ 002 = − P r dx
(3.2.37)
H˙ 002 represents the contribution of the conduction of heat through gas and stack material to the total total power and H˙ 2-10 and H˙ 200 represent the contribution of the thermoacoustic heat flow. The term H˙ 002 is an undesirable nuisance, but can normally be neglected provided ε and εp are small enough.
4.
Thermoacoustic devices with streaming
This chapter discusses thermoacoustic devices with streaming. Streaming refers to a steady mass-flux density or velocity, usually of second order, that is superimposed on the larger first-order oscillations. With the addition of a steady non-zero mean velocity along x, the gas moves through the tube in a repetitive ”102 steps forward, 98 steps backward” manner. As a result we will have to adapt our expansions in (2.1.13) to include a steady mean flow u ˜m . However, we will still assume that um = u ˜m /C ≪ 1. We include this in our expansion by demanding um = Aα um,α + o (Aα ) ,
(4.0.38) 26
for some constant α > 0 and with um,α 6= 0. This suggests the following alternative expansion for v and the remaining fluid variables q v(x, y, t) =
∞ X
αj
A vm,αj (x, y) +
j=1
∞ X
Ak vk′ (x, y, t),
(4.0.39a)
k=1
q(x, y, t) = q0 (x, y) +
∞ X
Aαj qm,αj (x, y) +
j=1
∞ X
Ak qk′ (x, y, t),
(4.0.39b)
k=1
where αj = jα and q1′ is assumed to be of the form q1′ (x, y, t) = Re q1 (x, y)eiκt .
(4.0.40)
We will insert these expansions in the equations and boundary conditions given in (2.1.15)-(2.1.21).
4.1
Estimating α
So far we have not made any comments on the exact value of α other than that it should be positive. It turns out that with the choice of dimensionless parameters made thusfar, α has to be equal to 2. We will show this here only for the case of a ”long” stack (κ = O(1), NL , = O(1)) to simplify the calculations. A similar analysis can be performed for the short stack or regenerator. We start with the same ordering for the dimensionless parameters as in (3.0.24) for the case without a mean flow except that now κ = O(1). κ = O(1), Wo = O(1),
NL = O(1),
Np = O(1),
ϑ = O(1),
ε2p ≪ ε2 ≪ A ≪ 1,
Fr = O(1), (4.1.1a) (4.1.1b)
We can distinguish three cases: 0 < α < 2, α = 2 and α > 2. 0 < α < 2: Let us first assume 0 < α < 2 and insert the expansions (4.0.39) into the x-component of the timeaveraged momentum equation. Taking terms up to order Aα and using p0 is constant, we find Aα
dpm,α κ ∂ 2 um,α = Aα . 2 2 Wo ∂y dx
(4.1.2)
This leads to um,α = −
Wo2 dpm,α (1 − y 2 ). 2κ dx
(4.1.3)
Next we insert the expansions into the time-averaged heat equation for the plate temperature. Collecting terms of order Aα we find d2 Tsm,α dy ′ 2
= 0,
⇒
Tsm,α = Tsm,α (x).
Furthermore collecting the Aα -terms in the time-averaged temperature equations we also find dT0 Wo2 ρ0 dT0 dpm,α κ ∂ 2 Tm,α = ρ u = − (1 − y 2 ). 0 m,α dx 2κ dx dx 2NL2 ∂y 2 27
Integrating twice with respect to y and applying the boundary conditions given in (2.1.21b), we obtain Tm,α =
Wo2 NL2 ρ0 dT0 dpm,α 4 (y − 6y 2 + 5) + Tsm,α (x). 12κ2 dx dx
The remaining boundary conditions given in (2.1.21c) say dTsm,α ∂Tm,α (x, ±1) = ϑ (x, ∓1), ∂y dy ′ The right hand side of this condition is zero, so we find Wo2 NL2 ρ0 dT0 dpm,α 3 2Wo2 NL2 ρ0 dT0 dpm,α = ± (y − 3y) = 0. 3κ2 dx dx 3κ2 dx dx y=±1
This equation can only be satisfied if dpm,α /dx = 0 and thus also um,α = 0. This contradicts our initial assumption given in (4.0.38), that um,α 6= 0. We conclude that α ≥ 2. α > 2: Suppose now that α > 2, then to leading order the time-averaged momentum equation becomes 2
A
κRe [iρ1 u∗1 ]
A2 ∗ ∂u1 ∗ ∂u1 + v1 ρ 0 bx . = + A ρ0 Re u1 ∂x ∂y Fr 2
(4.1.4)
However, the first-order variables occurring in this equation have already been determined and therefore this equation will not be satisfied in general. The same can be said about the heat equation and the energy equation. We conclude α cannot be larger than two. Summarizing, it is clear that α = 2. That means if there is a mean velocity then it is essentially second order in the amplitude A. This was also suggested by Swift [25]. The reason that the analysis does not go wrong for α = 2 is that then products of first order terms balance the second order timeindependent terms (e.g. the terms in equation (4.1.2) and (4.1.4)). The next two sections will show how this can be done for the short stack and regenerator.
4.2
Short stack
We will assume the same ordering for the dimensionless parameters as in (3.0.24) for the case without a mean flow Wo = O(1),
NL = O(1),
Np = O(1),
ϑ = O(1),
Fr = O(1),
ε2p ≪ ε2 ≪ A ≪ κ2 ≪ 1,
(4.2.1a) (4.2.1b)
and expand the fluid perturbation variables q1 and qm,α as q1 (x, y) = κmq q10 (x, y) + κq11 (x, y) + κ2 q12 (x, y) + · · · , qm,α (x, y) = κmqm qm,α0 (x, y) + κqm,α1 (x, y) + κ2 qm,α2 (x, y) + · · · . 28
(4.2.2a) (4.2.2b)
4.2.1 The steady horizontal velocity component um,α Take α = 2 as shown in the previous section. As α > 1 the calculation of the first order perturbation variables T1 , Ts1 , u1 and p1 in the previous chapter remains unaffected. Substituting the expansions (4.0.39) into the time-averaged momentum equation, collecting the A2 terms and expanding in κ, we find 2
A Re
−iρ0 βT10 u∗10
+
∂u10 ρ0 u∗10 ∂x
+
∗ ∂u10 ρ0 v10
∂y
= −A2 κmpm +
dpm,20 dx
A2 1 ∂ 2 um,20 ρ0 bx + A2 κ1+mum . Fr Wo2 ∂y 2
Taking mum = −1 and mpm = 0 balances the equation. As a result we find that um,20 and pm,20 satisfy 1 1 ∂ 2 um,20 dpm,20 ∗ ∗ ∂u10 ∗ ∂u10 − = Re −iρ0 βT10 u10 + ρ0 u10 + ρ0 v10 − ρ0 bx . (4.2.3) 2 2 Wo ∂y dx ∂x ∂y Fr Thus the mean velocity and pressure are driven by first order oscillations and body forces (gravity). 4.2.2 The steady temperature Tsm,α in the plate The time-averaged part of the diffusion equation for Ts reads A2 κ ∂ 2 Tsm,α 1 2 A κRe iρs1 Ts∗1 + = 0. 2 2Np2 ∂y ′2
(4.2.4)
Here we only included terms up to order A2 . Expanding in κ, we find 1 −1 1 ∂ 2 Tsm,20 κ Re iρs10 Ts∗10 = κ1+mTm = 0. 2 2Np2 ∂y ′2
(4.2.5)
The choice mTm = −2 balances the equation and thus Tsm,20 satisfies 1 ∂ 2 Tsm,20 = 0. Re iρs10 Ts∗10 = 2 Np ∂y ′2
(4.2.6)
4.2.3 The steady temperature Tm,α Substituting the expansions (4.0.39) in the time-averaged temperature equation, we obtain 1 2 1 2 ∗ ∗ ∗ ∂T1 ∗ ∂T1 ∗ dp1 A κRe [iρ1 T1 − iβp1 T1 ] + A Re ρ0 u1 + ρ 0 v1 − βT0 u1 + 2 2 ∂x ∂y dx dT0 A2 κ ∂ 2 Tm,2 1 A2 κ ∂u1 2 2 . (4.2.7) A ρ0 um,2 = + dx 2 Wo2 ∂y 2NL2 ∂y 2
Again we only included terms up to order A2 . Expanding in κ, we find to leading order dT0 1 1 ∂ 2 Tm,20 ∗ ∂T10 ∗ ∂T10 − ρ0 um,20 = ρ0 Re u10 + v10 . dx 2 ∂x ∂y 2NL2 ∂y 2 29
(4.2.8)
Remember that the variables p1 , T1 , T s1 and u1 are given by the expressions in the previous chapter, as α > 1. Solving (4.2.6) and (4.2.8) subject to the boundary conditions Tm,20 (x, ±1) = Tsm,20 (x, ±1),
(4.2.9a)
∂Tsm,20 ∂Tm,20 (x, ±1) = ϑ (x, ∓1). ∂y ∂y
(4.2.9b)
we will find Tm,20 and Tsm,20 . 4.2.4 The continuity equation Substituting the expansions (4.0.39) in the time-averaged continuity equation, we obtain ∂(ρ0 um,2 ) ∂vm,2 1 ∂(ρ1 u∗1 ) ∂(ρ1 v1∗ ) + ρ0 + Re + = 0. ∂x ∂y 2 ∂x ∂y Integrating with respect to y from 0 to 1 and expanding in κ, we find to leading order Z 1 Z 1 1 d d ∗ (ρ0 βT10 u10 ) dy = 0. ρ0 um,20 dy − Re dx 0 2 dx 0 Hence dM˙ 2-1 = 0, dx
(4.2.10)
where M˙ 2-1 = ρ0
Z
0
1
1 um,20 dy − ρ0 βRe 2
Z
1
T10 u∗10
0
dy .
(4.2.11)
4.2.5 The unsteady terms p1 , u1 and T1 As α > 1 the calculation of the perturbation variables T1 , u1 , p1 in the previous chapter remains unaffected. Thus for p1 we have p10 = constant, fk − fν dT0 dp11 d 1 − fν dp11 β + ρ0 = 0, (P r − 1)(1 + εs ) dx dx dx ρ0 dx fk − fν 1 dT0 dp12 d 1 − fν dp12 γ−1 fk p10 + β + ρ0 1+ = 0. c2 1 + εs (P r − 1)(1 + εs ) dx dx dx ρ0 dx For u1 we have i dp11 u10 (x, y) = 1− ρ0 dx i dp12 1− u11 (x, y) = ρ0 dx
cosh(αν y) , cosh(αν ) cosh(αν y) . cosh(αν ) 30
Finally, for T1 we have 1 dT0 dp11 P r cosh(αν y) T10 (x, y) = − 1− ρ0 dx dx P r − 1 cosh(αν ) 1 + εs ffνk dT0 dp11 cosh(αk y) 1 , ρ0 (P r − 1)(1 + εs ) dx dx cosh(αk ) 1 dT0 dp12 P r cosh(αν y) βT0 p10 − 1− T11 (x, y) = ρ0 ρ0 dx dx P r − 1 cosh(αν ) " # fν βT0 1 1 + εs fk dT0 dp12 cosh(αk y) − p10 + ,. ρ0 ρ0 P r − 1 dx dx (1 + εs ) cosh(αk ) −
4.2.6 Time-averaged total power H˙ In the same way as we did in equation (3.1.55) in the absence of a mean flow, we can derive an ˙ In the case of a mean flow, we get an additional term equation for the time-averaged total power H. in equation (3.1.50) involving um,2 , that is ρuh = Aρ0 h0 u′1 + A2 ρ0 h0 u′2 + A2 h0 ρ′1 u′1 + A2 ρ0 h′1 u′1 + A2 ρ0 h0 um,2 + · · · = A2 h0 ρ′1 u′1 + A2 ρ0 h′1 u′1 + A2 ρ0 h0 um,2 + · · · . With this change, we also get an additional steady flow term H˙ m,2-10 in equation (3.1.55) H˙ = A2 κ−1 H˙ m,2-10 + A2 κ−1 H˙ 2-10 + A2 H˙ 200 + A2 H˙ m,200 + ε2 + ϑε2p κH˙ 012 + · · · , (4.2.12)
where
H˙ m,2-10 = ρ0 h0
Z
1
Z
1
0
= ρ0 h0
0
h0 um,20 dy + Re 2
Z
1
ρ10 u∗10
0
1 um,20 dy − ρ0 h0 βRe 2
Z
dy
1
T10 u∗10 0
dy
= h0 M˙ 2-1 ,
(4.2.13)
and similarly H˙ m,200 = h0 M˙ 20 .
4.3
Short regenerator
We will assume the same ordering for the dimensionless parameters as in (3.2.27) for the case without a mean flow ε2p ε2 ≪ ≪ A ≪ κ2 ≪ 1, κ κ
ϑ = O(1),
κ = Wo2 =
2Np2 2NL2 = . Pr Pr
and again expand the fluid perturbation variables q1 and qm,α as q1 (x, y) = κmq q10 (x, y) + κq11 (x, y) + κ2 q12 (x, y) + · · · , 31
(4.3.1)
(4.3.2a)
qm,2 (x, y) = κmqm qm,20 (x, y) + κqm,21 (x, y) + κ2 qm,22 (x, y) + · · · .
(4.3.2b)
In the same way as we did for the stack, the mean time-independent terms can be obtained. We will not repeat this analysis and only mention the final outcome. 4.3.1 The steady flow terms For um,2 and pm,2 we find mum = mpm = 0, hence um,2 = um,20 + · · · ,
pm,2 = pm,20 + · · · ,
with um,20 and pm,20 satisfying ∂ 2 um,20 dpm,20 1 ∗ ∗ ∂u10 ∗ ∂u10 − = Re −iρ βT u + ρ u + ρ v ρ 0 bx , − 0 10 10 0 10 0 10 ∂y 2 dx ∂x ∂y Fr
(4.3.3)
dM˙ 20 = 0, dx
(4.3.4)
subject to um,20 (x, ±1) = um,21 (x, ±1) = 0 with Z 1 i 1 1 hγ ∗ ∗ ˙ M20 = ρ0 hum,20 i − ρ0 βRe (T10 u11 + T11 u10 ) dy + Re 2 p10 hu10 ∗ i . 2 2 c 0
(4.3.5)
For the steady temperature terms we find mTm = −1 and thus Tm,2 = κ−1 Tm,20 + · · · ,
Tsm,2 = κ−1 Tsm,20 + · · · ,
with Tm,20 and Tsm,20 satisfying 1 ∂ 2 Tm,20 1 ∗ ∂T10 ∗ ∂T10 = ρ Re u + v , 0 10 10 P r ∂y 2 2 ∂x ∂y 1 ∂ 2 Tsm,20 ∗ = Re iρ T , s s 10 10 P r ∂y ′2
(4.3.6) (4.3.7)
subject to
Tm,20 (x, ±1) = Tsm,20 (x, ±1),
(4.3.8a)
∂Tsm,20 ∂Tm,20 (x, ±1) = ϑ (x, ∓1). ∂y ∂y
(4.3.8b)
4.3.2 The unsteady flow terms As α > 1 the calculation of the perturbation variables T1 , u1 , p1 in the previous chapter remains unaffected. Thus for p1 we have d dT0 dp10 dp10 ρ20 β , ρ0 − dx dx ρ0 + ϑρs0 dx dx d 33 dp11 γ − P r ρ20 βTb10 ρ0 = −3iρ0 βTb11 + 3i 2 p10 + dx dx c 40 +i
dT0 33 (1 + P r)ρ20 β hu10 i. 40 dx 32
For u1 we have u10 (x, y) = − u11 (x, y) =
1 dp10 (1 − y 2 ), 2 dx
1 dp11 i dp10 ρ0 (1 − y 2 )(5 − y 2 ) − (1 − y 2 ). 24 dx 2 dx
Finally, for T1 we have T10 (x) = T11 (x, y) =
dT0 iρ0 hu10 i, ρ0 + ϑρs0 dx ρ20 Pr dT0 dp10 P rρ0 dT0 dp10 4 1 − y 2 + Tb11 , (y − 6y 2 + 5) − 24 dx dx 6 ρ0 + ϑρs0 dx dx
4.3.3 Time-averaged total power H˙ In the same way as we did for the stack we find an additional steady flow term H˙ m,210 , so that H˙ = A2 H˙ m,200 + A2 H˙ 200 + ε2 + ϑε2p H˙ 002 + · · · , (4.3.9)
where
H˙ m,200 = h0 M˙ 20 .
5.
(4.3.10)
Energy fluxes in thermoacoustic devices
The preceding sections show the derivation of expressions for the total power flow H˙ 2 and the acoustic ˙ 2 absorbed in a thermoacoustic refrigerator or heat pump or produced in a prime mover. As power W noted earlier in equation (3.1.55), the total power is the sum of the acoustic power, the hydrodynamic heat flux and the conduction heat flux. This section will give an idealized illustration of the different flows and their interaction in a thermoacoustic refrigerator. Fig. 7(a) shows a standing wave refrigerator thermally insulated from the surroundings except at the heat exchangers left and right of the stack, where heat can be exchanged with the surroundings. The arrows show the different energy flows into or out of the system except for the conductive heat flow which is neglected for ease of discussion. A loudspeaker (driver) sustains a standing acoustic ˙ 2 in the form of a traveling acoustic wave. wave in the resonator by supplying acoustic power W Equation (3.1.40) shows that part of this wave will be used to sustain the standing wave against the thermal and viscous dissipations, and the rest of the power will be used by the thermoacoustic effect to transport heat from the cold heat exchanger to the hot heat exchanger. Fig. 7(b) illustrates the behavior of the different energy flows as a function of the position in the tube. A part of the acoustic power delivered by the speaker is dissipated at the resonator wall (first and second term in (3.1.40), to the left and right of the stack. The part dissipated at the right of the stack shows up as heat at the cold heat exchanger, decreasing the effective cooling power of the refrigerator. The acoustic power in the stack decreases monotonously, as it is used to transport heat from the cold heat exchanger to the hot heat exchanger, and to overcome viscous forces inside the stack. This power used shows up as heat at the hot heat exchanger. The power dissipated at the tube wall to the left of the stack also shows up as heat at the hot heat exchanger. 33
Control volume Q˙ H TH ˙2 W
TC H˙ 2
Driver Q˙ C
Thermally insulated tube
(a)
˙2 Acoustic power W
Q˙ C Q˙ H Total power H˙ 2 Heat power Q˙ 2 (b)
Figure 7: (a) A standing wave refrigerator, insulated everywhere except at the heat exchangers. (b) Illustration ˙ 2 , Q˙ 2 and H˙ 2 in the refrigerator. The discontinuities in H˙ 2 are due to heat transfers at the heat of W exchangers.
Inside the stack, the heat flux grows from Q˙ C at the right end of the stack to Q˙ H at the left end of the stack. The two discontinuities in the heat flux arise as a result of the heat exchangers at TC and TH , supplying heat Q˙ C and removing heat Q˙ H . The total power flow is everywhere the sum of acoustic work and heat fluxes. Within the stack the total power remains constant. Conservation of energy holds everywhere. Applying the principle of energy conservation to the control volume shown in Fig. 7(a), in steady state over a cycle, the energy inside the control volume cannot change. Hence the rate H˙ 2 at which energy flows out is equal to the rate Q˙ C at which energy ˙ 2 is equal to Q˙ H − Q˙ C . flows in. Similarly we also find that W
6.
Numerical integration of Rott’s acoustic approximation
The previous sections have shown how the acoustic pressure p1 , velocity u1 , density ρ1 and temperature T1 can be calculated based on a given mean temperature T0 . Furthermore the total power H˙ can be calculated given p1 , hu1 i, M˙ 2 , T0 and dT0 /dx. Thus the question still remains how to determine the mean temperature T0 . In section 3.1.8 we have shown that for a cyclic refrigerator (prime mover) without heat flows to 34
the surroundings, the time-averaged total power E˙ 2 along x must be independent of x, i.e. E˙ 2 should satisfy, dT 0 H˙ 2 T0 , (6.0.11) , hu1 i, p1 ; M˙ 2 = constant. dx where we assume M˙ 2 is a given constant. If a constant total power is imposed, then (6.0.12) yields a first order differential equation for T0 of the following kind: dT0 = F T0 , hu1 i, p1 ; M˙ 2 , H˙ 2 , dx
(6.0.12)
where F : R × C2 × R2 → R is a known function. Additionally, e.g. for a stack, the equations derived for u1 and p1 can be rewritten into the following equations for hu1 i and p1 , dhu10 i dT0 = G 3 T0 , hu10 i, (6.0.13a) dx dx dhu11 i dT0 = G 3 T0 , hu11 i − G1 (T0 )p10 , (6.0.13b) dx dx dp10 = 0, dx
(6.0.13c)
dp11 = G2 (T0 )hu10 i, dx
(6.0.13d)
where G1 , G2 and G3 are given by i γ−1 fk , G1 (T0 ) = 1+ ρ 0 c2 1 + εs G2 (T0 ) = −
iρ0 , 1 − fν
dT0 β(fk − fν ) dT0 . = G 3 T0 , dx (1 − P r)(1 + εs )(1 − fν ) dx Note that ρ0 , fν and fk can all be expressed in T0 only. Summarizing, it is clear that T0 , p1 and hu1 i have to be solved simultaneously from a coupled system of three ordinary differential equation with suitable boundary conditions, given the constant time-averaged mass-flux M˙ 2 and total power H˙ 2 . Since T0 is real-valued and p1 and hu1 i are complexvalued, there are in fact 5 ordinary differential equations for the five real quantities T0 , Re(p1 ), Im(p1 ), Re(hu1 i) and Im(hu1 i). Hence 5 real conditions are necessary to fix T0 , p1 and hu1 i.
7.
Conclusions
A linear theory of thermoacoustics has been developed based on a dimensional analysis and using small parameter asymptotics. The validity domain of this linear theory can be expressed in the dimensionless parameters. For the stack this is ε2p ≪ ε2 ≪ A ≪ κ2 ≪ 1,
(7.0.14a) 35
Wo = O(1),
NL = O(1),
Np = O(1),
ϑ = O(1),
Fr = O(1),
(7.0.14b)
and for the regenerator we have ε2p ε2 ≪ ≪ A ≪ κ2 = Wo4 ≪ 1, κ κ Wo ∼ NL ,
Np ∼ N L ,
(7.0.15a)
ϑ = O(1)
Fr = O(1).
(7.0.15b)
Using this linear theory all the relevant variables could be determined. Similar results to those of Swift [23], [25] were obtained. The effect of the dimensionless parameters is most clear in the acoustic power. Viscous and thermal relaxation dissipation is reduced by taking the stack short (κ ≪ 1). On the other hand, reducing the pore size (NL ≪ 1) increases the viscous dissipation and reduces the thermal relaxation dissipation. However, the former can be prevented by taking the stack short enough, such that κ ≪ Wo2 ≪ 1. The linearization was first performed assuming no streaming. Repeating the analysis with streaming by adding a (small) steady mean velocity showed that the mean velocity had to be second order in the (dimensionless) amplitude A. In fact streaming is always present and arises due to body forces and first-order oscillations.
Acknowledgements This project3 is part of a twin PhD program with the Physics Low Temperature Group. The authors would like to thank the Technology Foundation (STW), Shell, the Netherlands Energy Research Foundation (ECN) and Aster Thermoacoustics for funding this project.
3
Project number: 10002154
36
A
Nomenclature
Note that the tildes are used to indicate dimensional variables
A.1
General constants and variables
C
typical speed of sound
c˜
speed of sound
cp
isobaric specific heat
cv
isochoric specific heat
D
typical density
˜˙ H
total power
˜ h
specific enthalpy
K
thermal conductivity
l
plate half-thickness
L
length of the stack plates
p˜
pressure
˜˙ Q s˜
heat flux specific entropy
t˜ T˜
temperature
U
typical fluid speed
u ˜
velocity component in direction of sound propagation
v˜
velocity component perpendicular to the direction of sound propagation
˜˙ W
acoustic power
x ˜
space coordinate along sound propagation
y˜
space coordinate perpendicular to sound propagation
R
plate half-separation
β˜
isobaric volumetric expansion coefficient
δ
penetration depth
ǫ˜
specific internal energy
χ
thermal diffusivity
λ
wave length
µ
dynamic (shear) viscosity
ν
kinematic viscosity
ξ
second (bulk) viscosity
ρ˜
density
˜ Σ
viscous stress tensor
time
37
ω
A.2
angular frequency of the acoustic oscillations
Dimensionless Numbers U C
A=
(A.2.1)
NL =
R δk
Lautrec number for the fluid
(A.2.2)
Np =
l δs
Lautrec number for the solid
(A.2.3)
Pr =
2NL2 Wo2
Prandtl number
(A.2.4)
Wo =
√ R 2 δν
Womersley Number
(A.2.5)
aspect ratio of the space between two stack plates
(A.2.6)
aspect ratio of the stack plate
(A.2.7)
ε=
A.3
amplitude of the acoustic oscillations (Mach number)
R L
εp =
l L
γ=
cp cv
ratio isobaric and isochoric specific heat
(A.2.8)
ϑ=
RKs lK
ratio of the heat fluxes leaving the gas and the plate
(A.2.9)
κ=
ωL C
dimensionless wavenumber (Helmholtz number)
(A.2.10)
Thermoacoustic auxiliary functions and variables fk =
tanh(αk ) αk
fν =
tanh(αν ) , αν
(A.3.1)
Rott’s function
(A.3.2)
αν = (1 + i)
R δν
(A.3.3)
αk = (1 + i)
R δk
(A.3.4)
αs = (1 + i)
l δs
(A.3.5) 38
p K ρ˜0 cp tanh(αk ) εs = √ Ks ρ˜s0 cs tanh(αs )
A.4
stack heat capacity ratio
(A.3.6)
Thermodynamic constants and relations
These relations were taken from [3]. ∂ p˜ 2 , c˜ = ∂ ρ˜ s˜ cp = T˜
cv = T˜
∂˜ s ∂ T˜
∂˜ s ∂ T˜
1 β˜ = − ρ˜ γ=
=
p˜
∂ ρ˜ ∂ T˜
= ρ˜
˜ ∂h ∂ T˜
!
∂˜ ǫ ∂ T˜
(A.4.1)
,
(A.4.2)
,
(A.4.3)
p˜
ρ˜
,
(A.4.4)
p˜
cp , cv
(A.4.5)
c˜2 β˜2 T˜ = cp (γ − 1),
(A.4.6)
˜ − ρ˜ǫ˜, p˜ = ρ˜h
(A.4.7)
d˜ s=
cp ˜ β˜ dT − d˜ p, ρ˜ T˜
⇒
s˜1 =
cp ˜ β˜ T1 − p˜1 , ρ˜0 T˜0
(A.4.8)
d˜ ρ=
γ d˜ p − ρ˜β˜ dT˜, c˜2
⇒
ρ˜1 =
γ p˜1 − ρ˜0 β˜T˜1 , c˜2
(A.4.9)
1 d˜ p, ρ˜
⇒
˜ 1 = T˜0 s˜1 + p˜1 , h ρ˜0
(A.4.10)
p˜ ρ, d˜ ǫ = T˜d˜ s + 2 d˜ ρ˜
⇒
p˜0 ǫ˜1 = T˜0 s˜1 − 2 ρ˜1 , ρ˜0
(A.4.11)
˜ = T˜d˜ dh s+
For an ideal gas some of these equations simplify. An ideal gas is a gas for which p˜/(˜ ρT˜) = constant. The value of this constant is denoted by R. p˜ = ρ˜RT˜,
(A.4.12)
R = cp − cv ,
(A.4.13)
β˜T˜ = 1,
(A.4.14) 39
c˜2 = γRT˜,
(A.4.15)
Some ideal gases have the special property that their specific heats cp and cv are constant. These gases are called perfect gases4
A.5
Fundamental equations
These equations were taken from [6]. ∂ ρ˜ ˜ + ∇ · (˜ ρv˜) = 0, ∂ t˜ ∂v ˜ ˜ v = −∇˜ ˜ p + µ∇ ˜ 2 v˜ + ξ + µ ∇( ˜ ∇ ˜ · v˜), + (˜ v · ∇)˜ ρ˜ 3 ∂ t˜ ρ˜T˜
ρ˜s
∂˜ s ˜ ˜ · (K ∇ ˜ T˜) + Σ: ˜ ∇˜ ˜ v, ˜ + v · ∇˜ s =∇ ∂ t˜
∂ T˜s ˜ 2 T˜s , = χs ∇ ∂ t˜
(A.5.1)
(A.5.2)
(A.5.3)
(A.5.4)
Using the thermodynamic relations in (A.4.8) - (A.4.11) we can also add the following equations
ρ˜cp
∂ T˜ ˜ T˜ + v˜ · ∇ ∂ t˜
!
− β˜T˜
∂ p˜ ˜ ˜ · (∇ ˜ T˜) + Σ: ˜ ∇˜ ˜ v, + v˜ · ∇˜ p = K∇ ∂ t˜
1 2 ∂ 1 2 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ρ˜v˜ + ρ˜ǫ˜ = −∇ · v ρ˜v˜ + ρ˜h − K ∇T − v · Σ , 2 ∂ t˜ 2
A.6
(A.5.5)
(A.5.6)
Sub- and superscripts and special symbols
∼ −
dimensional time-averaging
∗
complex conjugate
hi
y-averaging
· C H
time-differentiation cold hot
k m p
thermal mean isobaric, plate
r
real-valued
4 Sometimes in literature different definitions are used for ideal and perfect gases. Chapman [3] for example makes no distinction between ideal and perfect gases.
40
s v
stack, sink, source isochoric
δ
boundary layer
ν
viscous
41
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