Final Tar ReportThermoacoustic refrigeration

8/13/2019 Final Tar ReportThermoacoustic refrigeration

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ABSTRACT

Thermoacoustics is a field that combines thermodynamics,fluid dynamics and acoustics. Inthermoacoustics it is possible to construct thermodynamic engines, prime movers and heat

pumps which respectively use heat to create work, and use work to create or move heat.

Thermoacoustics is a relatively new field of science and engineering. The subject is still quite

unknown and not much literature about the subject is available. But there is a positive trend in

the amount of published papers about the subject.

There are two classes of thermoacoustic devices, travelling-wave devices and standing-wave

devices. The first use a standard travelling acoustic wave and the second use a resonator in whichthe acoustic waves interfere causing a standing-wave.

The goal of this project was to develop a virtual model for standing-wave thermoacoustic

refrigerator. The refrigerator should be able to deliver 100 Watts of cooling power at 280 Kelvin.

The device should be at most one meter long.This report gives an introduction in thermoacoustics, and a summary of the results as calculated

it also tries to explain the principle and working of avery basic thermoacoustic model. The

different parts of a thermoacoustic device with their properties are discussed as well.

Several simple devices have been modeled in the deltaEC during the project and these arediscussed. Not all of these devices Were feasible though. The model of the final refrigerator is

presented as well. Because of lack of time and technical difficulties the final device was notconstructed.However the model suggested is feasible . The right amount of resources and funds provided this

project can be modeled and constructed to generate a noticeable temperature drop . However this

project is in its infancy stage and cannot be yet used for commercial purposes.This model is highly desirable over the present day vapour compression system . And also if

constructed they can survive much longer than our present day conventional refrigerators. As it

consists of no reciprocating part , throttle valve and a condenser. But the working models have

relatively low COPs as compared to other V C refrigerators.Even though it has some major drawbacks Thermoacoustic refrigeration is the most potent

answer to the future refrigeration .


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ABBREVIATIONS

p Pressure Pa

V Volume m3

n Amount of substance mol

R Molar gas constant 8.314 JK1mol-1

T Temperature K

.Q Heat flow W

.W Work flow W

Efficiency -COP Coefficient of performance -

I Sound intensity Wm-2

u Particle velocity inx direction ms-1

f Frequency Hz

N Integer value for tones -

a Velocity of sound ms-1

L Length m

x Position inx direction m

u Particle velocity inx direction ms-1k Thermal conductivity Wm-1K-1

cP Specific heat at constant pressure Jkg-1K-1

cV Specific heat at constant volume Jkg-1K-1y Position in y direction m

U Acoustic volume flow m3s-1

A Cross-sectional area m2.

H Total power W.

E Acoustic power W

rh Hydraulic radius, rh A m

Polytropic coefficient, cp cv -

Wavelength m

Density kgm-3

Penetration depth m

Angular frequency, 2f rads-1

Displacement inx direction m

Prandtl number - Dynamic viscosity kgs-1m-1

Thermal expansion coefficient K-1

Perimeter, d m

Temperature gradient operator -

Volumetric porosity / blockage ratio -


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SUBSCRIPTS

osc Oscillating

m Mean valueC Cold side

H Hot side

carnot Maximum achievable Carnot value

Thermal

Viscous

s Stack

crit Critical value

gas Gas parameter

total Total value


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INTRODUCTION

Thermoacoustics is a relatively new field in physics which combines thermodynamics, fluid dynamicsand acoustics. Using heat, acoustic work can be created, or by using acoustic work heat can bemoved or created. The acoustic work is the sound power of a wave.Sound waves require a medium to propagate. In a gas, sound waves are adiabatically compressedand decompressed. During compression pressure increases and so does temperature, and during

decompression pressure and temperature both decrease.

The adiabatic change can be shown using the ideal law for gases:-

pV nRT ........................... (1)

Here p is pressure, V volume, n amount of the substance, R the gas constant and T the

temperature. The following expression (1) can be derived for adiabatic temperature changecaused by pressure change.

p/p (2)

Where is the polytropic coefficient. The formula clearly shows that temperature and pressurechange occur simultaneously. One might think now that these effects would be noticed in daily

life, however, for pressure amplitudes of a typical conversation, the temperature amplitude

would be 10-4

Kelvin . And even at the threshold of pain, 120 dB, temperature oscillates up anddown only about 10

-2Kelvin. It is not surprising then that thermoacoustic effects are unnoticed in

everyday life.

Thermoacoustics is a relatively new topic in science and engineering.


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Figure 1: Sondhauss TubeThe first thermoacoustic device was the Sondhauss tube. Figure (1) shows illustrations of aSondhauss tube In almost all cases where heat is communicated to a body expansion ensues, and

this expansion may be made to do mechanical work. If the phases of the forces thus operative be

favorable, a vibration may be maintained. For the sake of simplicity, a tube, hot at the closed endgetting gradually cooler towards the open end, may be considered. At a quarter of a period before

the phase of the greatest compression the air is moving inwards, i.e., towards the closed end,

and therefore is passing from colder to hotter parts of the tube. But in fact the adjustment of

temperature takes time, and thus the temperature of the air deviates from that of the neighboringparts of the tube, inclining towards the temperature of that part of the tube from which the air has

just come. From this it follows that at the phase of greatest compression heat is received by the

air, and at the phase of greatest rarefaction heat is given up from it, and thus there is a tendencyto maintain the vibrations.

Thermoacoustic devices can be divided into two classes, standing-wave and travelling-wave

devices. Travelling-wave devices can be described with the Stirling thermodynamic cycle, andstanding-wave devices with the Brayton cycle. These two classes of thermoacoustics devices can

again be divided in two thermodynamic types of engines, a prime mover (or simply heat engine),

and a heat pump. The prime mover creates work using heat and a heat pump creates or moves

heat using work.A thermoacoustic device basically consists of heat exchangers, a resonator, and a stack or

regenerator. With standing-wave devices this part is called a stack, and with travelling-wave

devices this part is the regenerator. The image below shows a typical standing-wave refrigerator.

This refrigerator uses a driver to create sound waves, and the work done by the sound waves isused for cooling.


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ACOUSTICS

Thermoacoustics is based on the principle that sound waves are pressure waves. These soundwaves propagate through the air via molecular collisions. The molecular collisions cause a

disturbance in the air, which in turn creates constructive and destructive interference. The

constructive interference makes the molecules compress, and the destructive interference makesthe molecules expand. This principle is the basis behind the thermoacoustic refrigerator.

One method to control these pressure disturbances is with standing waves. Standing waves are

natural phenomena exhibited by any wave, such as light, sound, or water waves. In a closed tube,columns of air demonstrate these patterns as sound waves reflect back on themselves after

colliding with the end of the tube. When the incident and reflected waves overlap, they interfere

constructively, producing a single waveform. This wave appears to cause the medium to vibrate

in isolated sections as the traveling waves are masked by the interference. Therefore, thesestanding waves seem to vibrate inconstant position and orientation around stationary nodes.

These nodes are located where the two component sound waves interfere to create areas of zero

net displacement. The areas of maximum displacement are located halfway between two nodesand are called antinodes. The maximum compression of the air also occurs at the antinodes. Due

to these node and antinode properties, standing waves are useful because only a small input of

power is needed to create a large amplitude wave. This large amplitude wave then has enoughenergy to cause visible thermoacoustic effects.

Figure 2: Shows the relationship between the phase of the wave, the

pressure, and the actual arrangement of the molecules. The black line

shows the phase of the sound wave, the red shows the pressure and the

dots below represent the actual molecules.


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In physics waves are usually travelling-waves which, as their name suggests, travel through

space. An acoustic travelling-wave has a pressure and velocity component which are in phase.

A special kind of wave is a standing-wave. When having a tube with an acoustic wave, it is

possible to have open and closed ends. At an open end the dynamic pressure will be zero, and at

a closed end the particle velocity will be zero. The particle velocity is the velocity at which aparcel of gas oscillates.An acoustic wave inside a tube reflects at both open and closed ends. Then, depending on the

frequency and amplitude, the reflected wave can interfere with the wave in the original direction,

creating a standing-wave. Figure 7 shows the difference between a travelling-wave and astanding-wave by taking a snapshot in time. When time-averaged, the left picture would consist

of two horizontal lines (RMS values), however, the right picture would have a similar shape as

the snapshot in time.

A pure standing-wave has a 90 phase difference in space between pressure and velocity. Since astanding-wave does not travel, there is no net particle velocity. The acoustic intensity is the

product of pressure and particle velocity. Since a standing-wave has no particle velocity, there is

no acoustic intensity thus no acoustic power.I=*u .(3)

In thermoacoustics however, there exist no pure standing-waves. When a wave passes a stack, a

pressure change occurs due to transfer of heat and losses in the stack. The reflected wave hasnow either more, or less, energy than the incoming wave, and imperfect interference occurs. This

imperfect interference gives rise to a small phase shift; instead of a 90 phase difference between

pressure and velocity, the difference will now be between 85 and 95. For this reason waves inThermoacoustic refrigeration using a standing-wave device

Figure 3: Respective phase difference of fundamental ,First, Second Overtone.


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standing-wave devices always have some real velocity, giving them a small amount of acoustic

power.

For standing-waves, the position at which the wave has a minimum absolute value (zero) iscalled a node, and the position that belongs to a maximum absolute value is called the antinode.

As mentioned there are special frequencies at which interference occurs. These frequencies are

called resonance frequencies.Resonance is the tendency of a system to oscillate at larger amplitude. When damping is small,the resonant frequency is approximately equal to a natural frequency of the system, which is a

frequency of unforced vibrations.


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THERMODYNAMIC BACKGROUND

In thermodynamics different cycles exist. Each cycle is made up of thermodynamic processes.

The most efficient cycle is the ideal Carnot cycle and this cycle exists of two isentropic processesand two isothermal processes. This means the temperature-entropy diagram would be a simple

square like seen in the picture below.

Figure 4: Pressure-volume (left) and temperature-entropy (right) diagrams of ideal Carnot

engine .

Thermodynamic cycles can be run in two directions, positive (clockwise) and negative (counter-

clockwise) direction. A positive cycle gives net work, while a negative cycle actually costs work.Negative cycles are thus not used to create work, but are used to either heat up a system abovethe temperature of the environment (heat pump) or to drain thermal energy from a system to keep

it cooler than its surroundings (refrigerator).

Below are the pressure-volume diagrams of a positive and a negative cycle. The arrow shows thedirection.


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Figure 5 : Pressure-volume diagrams of a positive (left) and a negative (right) cycle . The positive cycle could be a thermoacoustic prime mover and the negative cycle a

thermoacoustic heat pump or refrigerator.

Figure 6: Pressure volume diagram of sound waves.

The First and second laws of the thermodynamics plays an upper bound on the efficiency andCOPs of the prime mover and refrigeratoroperated by the acoustics .

Given that the two devices are working between the temperature boundaries The higher being T h

and the lower being Tctaking heat Qh and Qc respectively.


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Figure 7: Basic diagrams.

Qh= Qc + W .................(4)

For prime mover to create work and for Heat pump to use work by second law

Net entropy of the system should be less than zero

.....................(5)carnot = = .....................(6)

Also the cop of the heat pump-

COPrev carnot= = ............... (7)


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THERMOACOUSTIC REFRIGERATION MODEL USING A

STANDING WAVES DEVICE:-

In a thermoacoustic device the acoustic power is the ability to do work. This would mean

standing-wave devices can do relatively little work; a pure standing-wave device would not evenbe able to do work at all.Thermoacoustic devices have several advantages compared to their vapor compression

counterparts. No environmentally hazardous refrigerants are needed. Instead, a thermoacousticdevice uses air or an inert gas. Another advantage is that no moving parts are used in the device

making the devices reliable and simple, and thus low fabrication costs are expected . Here the

optimal placement of the stack is between the first pressure antinode and the pressure node .

Ideally at a distance of /10 from the hardened end. Further calculation and analysis of thismodel is done using the deltaEC software where the plots of temperature vs distance and the the

temperature vs length and so on can be drawn plotted and then analyzed long with other

parameters.

Figure 8: Schematic of a Thermoacoustic refrigeraton .With variation of pressure and

velocity and Temperature along the resonance tube.


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METHODOLOGY AND PART OF DESCRIPTION OF

THE MODEL

There are several main components involved in a thermoacoustic refrigerator. The main

components are the stack, heat exchangers and resonator. Each of these components has a specificpurpose in thermoacoustic refrigeration.

Stack

The stack of a thermoacoustic refrigerator is a thin walled tube with thin, well-spaced plates

aligned parallel to the tube axis. The addition of more plates to the stack increases the thermal

exchange area, leading to an increased amount of heat flux and thus an increased overall

efficiency of the device.The spacing between the plates in the stack is crucial in a properly functioning device. If the

spacing between the plates is too narrow the good thermal contact between the gas and the stack

keeps the gas at a temperature similar to the stack. If the spacing is too wide much of the gas is inpoor thermal contact with the stack and does not transfer heat effectively to and from the stack.

However, when the temperature difference across the stack is large enough, the air in the tube

oscillates spontaneously.Now the basic thermodynamics and acoustics are explained, the heat pumping along stacks will be

treated. Figure 10 shows an example of a stack which consists of parallel plates along the wavepropagation direction. Figure 11 - A stack consisting of parallel plates with gas flowing through them .

When an acoustically driven parcel of gas moves through the stack, pressure, temperature and

position all oscillate with time. If the gas is enclosed within a tube, interference occurs creatingan acoustic standing wave. Pressure will now be in phase with displacement; the pressure

reaches its maximum or minimum value while at the same time a parcel of gas is at an extreme

of its movement.This simple relation can be put to use-

1. Adiabatic compression of the gas.

When a parcel of gas is displaced from its rightmost position to its leftmost position, the parcel is

adiabatically compressed and thus the temperature increases. At the leftmost position, the parcel

now has higher temperature than the warm plate.

2. Isobaric heat transfer.

The parcel of gas is transferring heat to the plate at constant pressure losing temperature.

3. Adiabatic expansion of the gas.

The parcel of gas is displaced back from its leftmost position to its rightmost position and due to

adiabatic expansion the parcel is cooled to a temperature lower than that of the cold plate.


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4. Isobaric heat transfer.

In the last step the parcel absorbs heat from the cold plate at constant pressure increasing itstemperature back to its original value along the gradient. Figure 9 shows an illustration of this

cycle.

Figure 9: Basic diagram of process of heat exchanger.

The span of movement for an individual parcel is actually very small. However, along the stack

many parcels exist, each following this cycle, passing heat from one parcel to the other whileusing the stack as temporary storage of heat.

The primary constraint in designing the stack is that the layers need to be a few thermal

penetration depths apart, with four thermal penetration depths being the optimum layerseparation. Where thermal penetration depth, k, is defined as the distance that heat can diffuse

through a gas during the time given by

t=51/ f ..................... (8)

Where f is the frequency of the standing wave. k depends on the thermal conductivity, k,

the density of the gas, , and the isobaric specific heat per unit mass, cp, according to

k= ........................(9)

In order to ensure proper thermal interaction between the speaker and the stack, a nonconductivematerial such as Mylar, PVC piping or Kapton, a polyimide film, should be used. If a conductive

material such as copper is used, the temperature difference between the speaker and resonatorwill be very small and thus hard to detect.

The stack material should have a high heat capacity and high thermal conductivity in the ydirection. The thermal conductivity in the x direction however, should be very low. Heat

pumping requires the heat storage and this requires high thermal conductivity in the y direction

to be accessible. A low thermal conductivity in x direction is necessary to minimize lossesthrough conduction from hot to cold side. As becomes clear, a material with anisotropic thermal

conductivity would be best.


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Figure 10 Example of a stack used in a device in Los Alamos. Even though the stack looks

solid, it has a volumetric porosity of 83% .

The important dimensions for a stack are its length and the cross-sectional area of the stack. The

length is important for the temperature gradient. The length and cross-sectional area of the stack

also determine how much the sound waves are perturbed. The cross-sectional area available tothe gas compared to the total cross-sectional area is called the volumetric porosity or blockage

ratio of the part. The thickness of walls or plates and the width of gaps also determine the heat

capacity and conductivity. Simply said, all these factors influence the efficiency.

Stacks of different shape exist. Some stacks have parallel plates, some rectangular pores. For a

parallel plate stack, the plate spacing and the thickness of the plates are important dimensions.

The volumetric porosity of a parallel plate stack is given by

=BR==

..................(10)

where yo is half the distance between the pores and l is half the thickness of the plates.

=BR==

.(11)

Finally, Swift suggests that for a good compromise between high power and high efficiency the

optimal position of a stack inside a resonator with two closed ends is

x= /20=L/10 ....................... (12)

wherex is the distance from a pressure antinode to the center of the stack


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Figure 11: Stack Dimensions

Heat Exchanger

The heat exchangers function as a heat pump, driven by the acoustic work produced from the

stack.Heat exchangers are attached to both ends of the stack. The cold heat exchanger removesheat from the cold temperature reservoir Tr and moves that heat to cold side of the stack at a

temperature Tc. The heat exchanger at temperature Th rejects the pumped heat from the coldheat exchanger and the absorbed acoustic work which is at temperature Tc. Without the heat

exchangers, heat would neither be supplied nor extracted from the ends of the stack. The

heat exchanger strips and the nearby stack plates are non-parallel to each other in order toprevent the total blockage of any gaps in the stack by a heat exchanger strip.

Once the hot heat exchanger temperature is high enough for the parcel of gas to oscillate, the

cold heat exchanger can cool to below 0C as the heat is pumped from the cold heat exchanger

to that of the room temperature exchanger.In order to achieve optimum performance, the heat exchanger must be as long as the peak-to-

peak displacement amplitude:

2u1/h ...................................(13)

Where u1is the x-component of the velocity of the longitudinal wave and h is the enthalpy

per unit mass. When a heat exchanger is too long, some parcels of fluid only come into contactwith the ends of the heat exchanger and when it is too short parcels can jump past the heat

exchanger . Both of which serve no purpose and are ineffective ineffective in transporting heat.

Although Equation (13)for the heat exchanger length is ideal for this project it is imprecise byk, which is the distance heat can diffuse longitudinally past the ends of the heat exchanger.

Poor performance of heat exchangers leads to lower efficiencies in thermoacoustic refrigerators.


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Resonator

The resonator is composed of three main parts: the tube, buffer volume, and speaker housing.The resonator needs to be designed in such a way that is compact, light and strong. It must also

impede the dissipation of acoustical energy as much as possible.

The energy dissipation in the resonator can be reduced by a smooth and varying cross-section,preventing harmonics as well.

L=/2 .........................(14)The first consideration is the length of the resonator. The length of the resonator shouldbe a half

that of the wavelength. Another important factor in developing a resonator is safety; since higher

powers can be obtained by increasing the mean pressure the walls of the resonator should have acertain minimal thickness.

Another consideration is the shape and size of the different resonator components.Original

designs simulated an open-end resonator by using a spherical buffer volume.

Working gas

The choice of gas for a thermoacoustic device involves trade-offs between many issues,

including power, efficiency, and convenience.

Thermoacoustic powers generally scale as as pmaA can be seen .A high mean pressure, a high

velocity of sound and a large cross-sectional area would mean more thermoacoustic power. Forthis reason, helium is commonly used in thermoacoustic devices. Hel iums velocity of sound is

much higher than that of air and helium will not condense or freeze at low temperatures.

There are more reasons why a certain gas would be attractive to use. As mentioned the velocity

of sound should be high. The thermal conductivity should be high as well, and the Prandtl

number low, since a low Prandtl number would mean low viscous losses. Mixtures of heliumwith argon, or helium with xenon, are thus also advised since these reduce viscous dissipation as

their Prandtl number is lower. The following table shows for certain gases the values of importantproperties. Conditions are at room temperature and atmospheric pressure. For the model a frequency of174 Hz was chosen (as this was the expected working frequency of the eventual device at time ofwriting).


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DESIGN PARAMETERS

Important length scales :

Before explaining thermoacoustics, it is important to be aware of certain important length scales

An important length scale along the wave-propagation direction is the wavelength of the sound

wave. The wavelength in a tube is determined by the length and whether the tube has open endsor not.

For a tube with two closed ends, the length of the tube can be calculated by dividing the

wavelength by two

L =/ 2 ........................(15)The wavelength is related to the frequency by the speed of sound

= a/ f ...................... (16)

Another important length scale in the direction of motion of the gas is the gas displacementamplitude. The distance across which a parcel of gas can move is twice the displacement

amplitude. The displacement amplitude at a certain position along the wave-propagation

direction is given by dividing the velocity amplitude with the angular frequency of the wave

sc = uosc/

Figure 12: This illustration shows some of the length scales. The length of the tube

determines the wavelength. The spacing between the stack plates is around a penetration

depth and a parcel of gas is able to move across twice the displacement amplitude.


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The displacement amplitude is often a very small fraction of the stack length. The figure above

shows this length scale. As can be seen, the displacement amplitude varies ; the amplitude is

largest in the middle (velocity antinode) and smallest at the hard ends (velocity nodes).

The velocity of the gas parcel as function of the positionx in the tube is given by

Uosc= sin (

) .................. (18)

and depends on the amplitude of the pressure p, the mean density of the gas , the velocity of

sound a. and the reduced wavelength.This velocity can be calculated as well when both the volume flow and the cross-sectional area

are known.

uosc =Uosc/Atotal ........................ (19)

Perpendicular to the direction of the motion of gas there are two more important length scalescalled penetration depths.

The thermal penetration depth is defined as

k= .................(20)

and is roughly the distance across which the heat can diffuse through the fluid during a time

interval in the order of 2/.

The viscous penetration depth is defined as

v=

....................(21)

and is roughly the distance across which momentum is lost during a time interval in the order of

2 /.

In the above equations the following gas properties are mentioned; k is the thermal conductivity,

the dynamic viscosity, the mean density, cp the heat capacity and the angular frequency.At distances much greater than these penetration depths from the nearest solid boundary, the gas


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has neither thermal contact nor viscous contact with the solid boundaries. In parts of the device

whose lateral dimensions are of the order of the viscous and thermal penetration depths, the gas

does feel both thermal and viscous effects from the boundaries.

The heat exchange components in thermoacoustic systems must have dimensions in the order of

k in order to exchange heat with the working gas. The viscous penetration depth v ,howevershould beas small as possible in order to reduce viscous losses.

The Prandtl number describes the ratio between the viscous and thermal penetration depth

and should be as small as possible in thermoacoustics.

=2 = ......................(22)

For typical gases this value is usually around 1.

SWIFT THERMOACOUSTIC MODEL

In 1988 Swift published a paper called Thermoacoustic engines in which he describes thefundamentals of thermoacoustic engines. Swift uses the thermoacoustic theory provided by Rott

which is based on the Navier-Stokes equation.

Swift starts in his paper with a single plate approximation for which he makes many assumptions

and then continues with a numeric model with only few assumptions. He also shows that thisfinal model, with certain additional assumptions, yields the same results as his single plate

approximation. In the following paragraphs summaries will follow of both models.

3.1. Single plate approximationIn his paper Swift starts with a stack consisting of a single plate and makes many assumptions:

1) The system is in steady state.

2) Short-stack approximation: the stack length Lsis much smaller than the reduced wavelength

rThe stack will now have no effect on the sound wave, and no pressure drop would occur.

3) The plate is far enough from pressure and velocity nodes that p osc and uosc can be assumed

uniform over the entire plate.

4) The fluid has zero viscosity so that uoscdoes not depend ony direction.

5) The plate has a large enough heat capacity per unit area that its temperature does not change

appreciably.

6) The plate has a given mean temperature gradient in thex direction called Tm.

7) The plates thermal conductivity in thex direction is neglected.

8) The fluids thermal conductivity in thex direction is neglected as well.

9) The mean fluid temperature as function of its position in the x direction is independent of ydirection and is the same as that of the plate.


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Additionally, Swift adds certain boundary conditions. The temperature in the y direction should

be between 0 Kelvin at y=0 and a finite amount of Kelvins at y= . The solution for the

oscillating temperature Swift finds is given by Tm mCp

Tosc = ( ) ................(23)

where is the thermal expansion coefficient.

The mean-temperature gradient along a stack Tm is obtained by dividing the temperature

difference over the stack Tmwith the length of the stackLs

Tm= Tm/Ls ................(24)The fluid far from the plate iny direction y>>k makes negligible thermal contact with the plate.

Then setting equation (18) equal to zero gives a critical mean-temperature gradient

Tcrit = .......................(25)

The critical temperature gradient is important because, as we will see later, it is the boundary

between a prime mover and a heat pump when Tm~ T crit. .

For an ideal gas, the coefficient of thermal expansion is given by =1/Tm and the velocity isgiven by the volume flow divided by the cross-sectional area uosc= Uosc /A. The critical

temperature gradient along a stack can then be written as

Tcrit= .............................(26)

Equation (18) shows that the factor with the exponential power reaches a magnitude of 1 for

y>>k and 0 for y k(no contact with solid).


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Figure 11 : The real and imaginary part of the temperature as well as the magnitude

(modulus).

Not shown in this report however, is that both the heat flux and the acoustic power depend on the

imaginary part of Tosc . When the real part is largest and reaches the absolute value (maximum

temperature, maximum pressure), there will be a significant imaginary part around y k , andhere the heat transfer takes place. Now this phase shift or delay, caused by the time it takes theheat to diffuse through the fluid, provides a natural ability to produce the proper phasing for heat

pumping along the stack.

There is one drawback to the imperfect thermal contact between the gas and the solid material ofthe stack: heat transfer over a non-zero temperature span must create entropy. This means even

an idealized stack-based thermoacoustic device cannot achieve ideal Carnot thermodynamic

performance if power is produced. Standing-wave devices are thus intrinsic irreversible devices.

Swift derived the following expressions for the heat flux along the single plate stack

H=Q= -.. k .Tm..posc.uosc.(-1) .........................(27)

In this equation is the temperature gradient factor which is the mean temperature gradient

divided by the critical temperature gradient


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=

..........(28)

When the temperature gradient factor equals one, there is no heat flux. When this factor is larger

than one, Tm>Tcrit heat flux is towards the pressure antinode and heat is consumed by thestack. If the factor is smaller than one, Tm


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By making the mentioned assumptions, Swift arrived at the following central thermoacoustic

engine effects:

1) The thermal boundary condition imposed on the fluid by the plate causes a phase shift in time

of the oscillating temperature in the fluid about a thermal penetration depth away from the

plate.2) Heat flows hydrodynamically, parallel to the plate in the fluid about a thermal penetration

depth away from the plate.

3) Acoustic power is absorbed or produced by the fluid about a thermal penetration depth away

from the plate.

4) Whether heat flows up or down the temperature gradient, and whether power is absorbed or

produced, depends on the magnitude of the mean-temperature gradient relative to a critical

temperature gradient.

Swifts Estimations

The following are a few simple order-of-magnitude estimates based on Swifts approximations.

For standing-wave devices the total power in the stack is in the order of

H .|posc|.|Uosc| ...............................(32)Swift also showed that the critical temperature gradient is in the same order as the mean

temperature in the stack divided by the reduced wavelength

TcritTm/rThe size or frequency can be estimated using

a = .f ........................(33)

with the length of the device in the range of /4 to /2.


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GENERAL CALCULATIONS

Calculations are in general done for a pressure of 5 bars and ambient temperature of 293 Kelvin.

1. General values

The length of the device is one meter and the wavelength will be=2.L=2.1= 2m

The value k of the wave is

k= = = 3.14 m-1

The reduced wavelength is

r==.318 m

With room temperature the velocity of sound is 343 m/s and the frequency becomes

f== 172 Hz

Angular frequency becomes

=2f=1080 rad/s

The cross section area of the tube is acircle with adia of 0.10 m

Atotal=

..d

2

= 79 cm

2

Even though the penetration depths are not same in all the parts it is calculated as per the ambient

conditions.

The thermal penetration depth is given by

k= =

= .089 mm

The viscous penetration depth is given by

v= = =.075 mm


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The prandtl number will be

=2 = =0.71

Stacks

The stack should preferably consist of small rectangular channels. A ceramic called Celcor 400is used:

Type: Celcor Length 2.5 cm stack L =2.5 cm Half pore width a=b=.660 mm

Half thickness of walls l=.070 mm Thermal conductivity k= 2.5 Wm-1K-1

The width of the gaps in the stack should be a few times the thermal penetration depth, since theimaginary part of the temperature is largest here. Both stacks are of the material Celcor and have

similar dimensions; half the width of the pores is a=b= .660 mm and half the thickness of the

walls is l= .070 mm. The volumetric porosity then becomes

= BR ==

=

=0.82

The area available for the gas .Total cross sectional area multiplied with volumetric porosity

Agas=.Atotal = 0.82*79= 64cm2

For adecent compromise in acoustic power andhigh efficiency The optimal position of the stack

is given by:-

x= /20 = 2/20 = 0.10 m

This is the distance from pressure antinode. Both stacks are at the same distance from the pressureantinodes. The lengths of the stack can be found by optimization. This was done in DeltaE and will not be

shown here.A final length of 2.5 cm for both stacks was chosen since the stack material was only available in

this thickness. A slightly thicker one for the refrigerator is recommended, although it depends on

the wanted temperature difference across the stack.

The mean temperature difference divided by across the length for refrigerator will be

Tm= =

= 520 K/m


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27

The estimated critical temperature gradient is

Tcrit= Tm/r= 293/.318= 920 K/m

The approximated critical temperature gradient for the refrigerator becomes

Tcrit==

=1060 K/m

The temperature gradient operatorfor the refrigerator is

=

= =0.54

COP of the designed model is given by

COPrev carnot= = = = 21.5And actual COP of the model is

COPactual=.COPrev,carnot= .54*21.5= 11.63


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29

TITLE Model for thermoacoustic refrigeration

!->C:\Users\a\Desktop\google drive\Thermoacoustic refrigeration

model.out

!Created@22:36:51 06-Nov-2013 with DeltaEC version 6.3b11.12!under

win32,

using Win 6.1.7600 () under Python DeltaEC.!--------------------------------- 0 --------------------------------

-

BEGIN Initial Parameters

!All Pressures from 1 to 25 bar give realistic results.

1.0133E+05 a Mean P Pa

176.71 b Freq Hz

298.00 c TBeg K

1.3458E+04 d |p| Pa

0.0000 e Ph(p) deg

0.0000 f |U| m^3/s

0.0000 g Ph(U) deg

0.0000 h Htot Wair Gas type

!--------------------------------- 1 --------------------------------

SURFACE Top

!Area belongs to dia of .10 m

7.8540E-03 a Area m^2 1.3458E+04 A |p| Pa

0.0000 B Ph(p) deg

3.2852E-05 C |U| m^3/s

180.00 D Ph(U) deg

0.0000 E Htot W

stainless Solid type -0.22106 F Edot W

!--------------------------------- 2 --------------------------------

DUCT Top Cylinder

sameas 1a a Area m^2 Mstr 1.3047E+04 A |p| Pa

0.31416 b Perim m 2a 1.0708E-02 B Ph(p) deg

7.7000E-02 c Length m 6.3154E-02 C |U| m^3/s

5.0000E-04 d Srough -90.117 D Ph(U) deg

0.0000 E Htot W


!--------------------------------- 3 --------------------------------

HX Heat input - warm side - power input

!!Heat xchangers are fin types.

7.8000E-03 a Area m^2 1.3001E+04 A |p| Pa

0.5000 b GasA/A 2.9854E-02 B Ph(p) deg

2.0000E-03 c Length m 6.3981E-02 C |U| m^3/s

1.0000E-03 d y0 m -90.143 D Ph(U) deg95.151 e HeatIn W G 95.151 E Htot W

0.0000 f SolidT K -1.2575 F Edot W

298.00 G GasT K

copper Solid type 390.93 H SolidT K

!--------------------------------- 4 --------------------------------

STKRECT Warm Stack

!!The gaps are rectangular so STKRECT is used.


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30

sameas 1a a Area m^2 1.2509E+04 A |p| Pa

0.81741 b GasA/A 0.61514 B Ph(p) deg

2.5000E-02 c Length m 8.0101E-02 C |U| m^3/s

6.6000E-04 d aa m -89.712 D Ph(U) deg

7.0000E-05 e Lplate m 95.151 E Htot W

sameas 4d f bb m -2.8635 F Edot W

298.00 G TBeg Kcelcor Solid type 254.60 H TEnd K

!--------------------------------- 5 --------------------------------

HX Heat input-Cold side-ambient temp


0.5000 b GasA/A 0.72379 B Ph(p) deg

8.0000E-03 c Length m 8.3249E-02 C |U| m^3/s

1.0000E-03 d y0 m -89.77 D Ph(U) deg

17.879 e HeatIn W G 113.03 E Htot W


254.60 G GasT K


!--------------------------------- 6 --------------------------------DUCT Resonator


0.31416 b Perim m 6a -179.43 B Ph(p) deg

0.7500 c Length m 3.8335E-02 C |U| m^3/s

5.0000E-04 d Srough -86.317 D Ph(U) deg

113.03 E Htot W


!--------------------------------- 7 --------------------------------

HX Cold output - cold side -> power output


sameas 5b b GasA/A -179.44 B Ph(p) deg

sameas 5c c Length m 3.5043E-02 C |U| m^3/s

sameas 5d d y0 m -85.845 D Ph(U) deg

-100.0 e HeatIn W 13.030 E Htot W


254.60 G GasT K


!--------------------------------- 8 --------------------------------

-

STKRECT Cold stack


0.81741 b GasA/A -179.56 B Ph(p) deg

2.5000E-02 c Length m 1.7349E-02 C |U| m^3/s

sameas 4d d aa m -78.262 D Ph(U) deg

sameas 4e e Lplate m 13.030 E Htot Wsameas 8d f bb m -22.496 F Edot W

254.60 G TBeg K

celcor Solid type 276.07 H TEnd K

!--------------------------------- 9 --------------------------------

-

HX Cold output - warm side -> ambient temp


sameas 5b b GasA/A -179.53 B Ph(p) deg


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31

sameas 5c c Length m 1.4077E-02 C |U| m^3/s

sameas 5d d y0 m -75.086 D Ph(U) deg

-13.394 e HeatIn W G -0.36367 E Htot W


276.07 G GasT K


!--------------------------------- 10 ---------------------------------

DUCT Bottom Cylinder


0.31416 b Perim m 10a -179.29 B Ph(p) deg

8.2000E-02 c Length m 5.3237E-02 C |U| m^3/s

5.0000E-04 d Srough 86.727 D Ph(U) deg

-0.36367 E Htot W


!--------------------------------- 11 --------------------------------

-

SURFACE Bottom

sameas 1a a Area m^2 1.2993E+04 A |p| Pa-179.29 B Ph(p) deg

5.3239E-02 C |U| m^3/s

86.695 D Ph(U) deg

-0.36367 E Htot W

ideal Solid type -24.19 F Edot W

!--------------------------------- 12 --------------------------------

-

HARDEND End

sameas 12G a R(1/z) =12G 1.2993E+04 A |p| Pa

sameas 12H b I(1/z) =12H -179.29 B Ph(p) deg

sameas 12E c Htot W =12E 5.3239E-02 C |U| m^3/s

86.695 D Ph(U) deg

-0.36367 E Htot W

-24.19 F Edot W

-1.5539E-02 G R(1/z)

-0.22164 H I(1/z)

! The restart information below was generated by a previous run

! and will be used by DeltaEC the next time it opens this file.

guessz 3e 5e 9e

xprecn 1.0339E-04 7.6451E-05 1.5290E-04

targs 12a 12b 12c

hilite 0c 0d 3e

mstr-slave 3 2 -2 6 -2 10 -2

! Plot start, end, and step values. May be edited if you wish.

! Outer Loop: | Inner Loop .


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32

RESULTS AND CONCLUSIONS

During all this time a literature study on thermoacoustics has been done, as well as modeling of a

device, for testing purposes was done The goal of the project was to develop a refrigerator. A

DeltaE model and calculations were made for a refrigerator however, the device itself was notconstructed because of lack of time and the other complications.The main point of TAR studysuggests that despite a very low efficiency, due to experimental conditions in very limited budget

academic environment and at atmospheric pressure, the system already exhibits interesting

cooling capacity. Significant temperature difference has been measured over a period of time,providing a proof of principle for modest size systems. If it is not possible to act on time

variation, it is possible to influence the magnitude of temperature curve. As shown by theoretical

study, thermal power varies linearly with pressure, once all other variables are fixed. So the

interest of the device is increased when it is pressurized as evident from experimental andtheoretical efficiencies. The results suggest a forthcoming more complete theoretical study to be

done in continuity of present project, taking into account physical imperfections of fluids,

introduction of more appropriate heat sources and with a more relevant experimental model.The only moving part in a thermoacoustic heat pump is the vibrating loudspeaker, so the

technology should prove to be reliable as well as low in cost. Thermoacoustic heat pumps also

use environmentally benign noble gases or mixtures there of another advantage is that

thermoacoustic devices are well suited for proportional control, i.e., the ability to adjust thecooling output provided by the cooler to the heat load. This should result in an efficiency

advantage compared to smaller vapor compression cooling systems that are usually operated

only at full capacity and are cycled on and off to match the cooling load.Thermoacoustic coolers can also be easily powered directly from a heat source. This is

accomplished by coupling a thermoacoustic engine and cooler into one devicethe heat powers

the engine, which generates the acoustic power used by the cooler. Unfortunately,

thermoacoustic cooling is currently less efficient than vapor compression cooling.. Power densitycould also be an issue for some applications where space is limited. However, there are no

apparent barriers limiting improvement upon these deficiencies with further research and

development.Currently, no thermoacoustic air-conditioning technology has been commercialized. Cool Sound

Industries, Inc. and ThermoAcoustics Corp. have launched websites, but at this point, they serve

only to promote thermoacoustic air conditioning along with other applications. Anothercommercial entity, Clever Fellows Innovation Consortium, Inc., has already begun selling

thermoacoustic cryocoolers. According to the research group at Pennsylvania State University,

the largest hurdle to commercialization is the lack of individuals with the necessary background

in acoustics, transduction, heat exchanger design, and instrumentation. Additionally, theymention the lack of suppliers for the specialized components (acoustic drivers, stacks and

regenerators, and heat exchangers) that are needed to make a thermoacoustic device. At this

point, most of the components must be custom designed and manufactured. The current focus,

specifically in air-conditioning applications, is to improve efficiency as well as to increase powerdensity. Stack-based coolers currently have an efficiency 20-30% below equivalent vapor

compression coolers. Traveling-wave coolers, being intrinsically more efficient than stack-

based coolers, have potentially better performance, yet no theoretical or experimentalperformance data could be found for this type of device.


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REFERENCES

1. J. W. Strutt (Lord Rayleigh), The Theory of Sound,

2.T. Yazaki, A. Tominaga, and Y.Narahara, "Experiments on Thermally Driven Acoustic

Oscillations of Gaseous Helium," J. Low Temp. Phys, 41,45 (1980).3.J. C. Wheatley, T. Hofler, G. W. Swift, and A. Migliori, "Experiments with an Intrinsically

4.Irreversible Acoustic Heat Engine," Phys. Rev. Lett. 50, 499 (1983); "An Intrinsically

5. Irreversible Thermoacoustic Heat Engine," J. Acoust. SOC. Am. 74, 153 (1983); "6. Heat Pumping Engine," U.S. Patent No. 4,398,398 (Aug. 16, 1983); "Intrinsically

7. Irreversible Heat Engine," U.S. Patent No. 4,489,553 @ec. 25, 1984).

8. T. J. Hofler, "Thermoacoustic Refrigerator Design and Performance," Ph.D. dissertation,9. Physics Dept., Univ. Calif. San Diego (1986); "Concepts for Thermoacoustic

Refrigeration

10.G. W. Swift, "Thermoacoustic Engines," J. Acoust. Soc. Am. 84(4), 1145-1 180 (1988).11. "Acoustic Cooling Engine," U. S. PatentNo. 4,722,201 (Feb. 2, 1988).

12 S. L. Garrett, J. A. Adeff, and T. J. Hofler, ThermoAcoustic Refrigeration for Space

Final Tar ReportThermoacoustic refrigeration

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Transcript of Final Tar ReportThermoacoustic refrigeration