Thermoacoustics . Oscillatory gas flow with heat transfer ... · Where innovation starts...

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Thermoacoustics.Oscillatory gas flow with heat transfer tosolid boundaries

Peter in ’t panhuis

April 22, 2009

2/23

/ department of mathematics and computer science

Outline

IntroductionWhat is thermoacoustics?ApplicationsBasic thermoacoustic effect

Overview

Nonlinear oscillations near resonanceNonlinear standing wavesNonlinear standing waves in interaction with a stack

Future work

2/23


Outline


Overview


Future work

2/23


Outline


Overview


Future work

2/23


Outline


Overview


Future work

3/23


Introduction

What is thermoacoustics?

I All effects in acoustics in which heat conduction and entropy variationsplay a role (Rott, 1980)

I Thermoacoustic devices aim to produce refrigeration, heating ormechanical work

Thermoacoustic devices

I Heat power→ acoustic power

• A prime mover converts heat into sound

I Acoustic power→ heat power

• A refrigerator uses sound to produce cooling• A heat pump uses sound to produce heating

I Earliest example is Higgins’ flame (1777)Higgins’ flame

3/23


Introduction

What is thermoacoustics?

I All effects in acoustics in which heat conduction and entropy variationsplay a role (Rott, 1980)

I Thermoacoustic devices aim to produce refrigeration, heating ormechanical work

Thermoacoustic devices

I Heat power→ acoustic power

• A prime mover converts heat into sound

I Acoustic power→ heat power

• A refrigerator uses sound to produce cooling• A heat pump uses sound to produce heating

I Earliest example is Higgins’ flame (1777)Higgins’ flame

4/23


Introduction

ApplicationsI Liquefaction of natural gases

I Upgrading of industrial waste heat

I Downwell power generation

I Food refrigeration

I Airconditioning

I . . .

.

4/23


Introduction





I Airconditioning

I . . .

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4/23


Introduction





I Airconditioning

I . . .

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4/23


Introduction





I Airconditioning

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4/23


Introduction





I Airconditioning

I . . .

.

4/23


Introduction





I Airconditioning

I . . .

.

Benefits

I No moving parts and use of simple materials

I Reliable

I Environmentally friendly

5/23


Geometry

Model: gas-filled tube with porous medium (stack):

Straight tubes Looped tubes

I Standing-wave phasing I Traveling-wave phasing

Prime mover:asdfRefrigerator:asdf

apply high temperature difference across stack;sound is supplied to exhaustspeaker is attached to supply sound;temperature difference arises across stack

5/23


Geometry

Model: gas-filled tube with porous medium (stack):

Straight tubes Looped tubes

I Standing-wave phasing I Traveling-wave phasing

Prime mover:asdfRefrigerator:asdf

apply high temperature difference across stack;sound is supplied to exhaustspeaker is attached to supply sound;temperature difference arises across stack

6/23


Basic thermoacoustic effect

I Thermodynamic cycle of gas parcel near wall in refrigerator

I Bucket brigade: heat is shuttled along the wall

6/23


Basic thermoacoustic effect

I Thermodynamic cycle of gas parcel near wall in refrigerator

I Bucket brigade: heat is shuttled along the wall

7/23


Overview of PhD

ProgressI Weakly nonlinear theory of thermoacoustics

7/23


Overview of PhD


• based on dimensional analysis and small-parameter asymptotics• in wide or narrow tubes• with arbitrary slowly-varying cross-sections

solid

gasrx

Rg ( x, )

Rs ( x, )

g

s

7/23


Overview of PhD



solid

gasrx

Rg ( x, )

Rs ( x, )

g

s

I Standing-wave devices

I Traveling-wave devices

7/23


Overview of PhD



solid

gasrx

Rg ( x, )

Rs ( x, )

g

s



I Nonlinear oscillations near resonance

• Shock formation

7/23


Overview of PhD



solid

gasrx

Rg ( x, )

Rs ( x, )

g

s



I Nonlinear oscillations near resonance

• Shock formation

}Today

8/23


Recap

Small-parameter asymptotics

I Expansion in Mach number Ma with scaled frequency κ

f (x, t ;Ma ) = f0(x)+Ma f1(x)eiκt+M2

a

{f2,0(x)+ f2,2(x)e2iκt

}+ · · · .

I mean term + 1st harmonic + streaming term + 2nd harmonic

8/23


Recap




a

{f2,0(x)+ f2,2(x)e2iκt

}+ · · · .


8/23


Recap




a

{f2,0(x)+ f2,2(x)e2iκt

}+ · · · .


8/23


Recap




a

{f2,0(x)+ f2,2(x)e2iκt

}+ · · · .


8/23


Recap




a

{f2,0(x)+ f2,2(x)e2iκt

}+ · · · .


8/23


Recap




a

{f2,0(x)+ f2,2(x)e2iκt

}+ · · · .


I For example, T0, U1, and p1 satisfy

dT0

dX= κ

H2 − M2h0(T0)− Re [a1(T0)p1U∗1 ]

a2(T0)|U1|2 − Kloss

dU1

dX= κa3(T0)p1 + a4(T0)

dT0

dXU1

dp1

dX= κa5(T0)U1

8/23


Recap




a

{f2,0(x)+ f2,2(x)e2iκt

}+ · · · .



dT0

dX= κ

H2 − M2h0(T0)− Re [a1(T0)p1U∗1 ]


dU1

dX= κa3(T0)p1 + a4(T0)

dT0

dXU1

dp1

dX= κa5(T0)U1

I where H is the energy flux

8/23


Recap




a

{f2,0(x)+ f2,2(x)e2iκt

}+ · · · .



dT0

dX= κ

H2 − M2h0(T0)− Re [a1(T0)p1U∗1 ]


dU1

dX= κa3(T0)p1 + a4(T0)

dT0

dXU1

dp1

dX= κa5(T0)U1

I where M is the mass flux and h is the specific enthalpy

8/23


Recap




a

{f2,0(x)+ f2,2(x)e2iκt

}+ · · · .



dT0

dX= κ

H2 − M2h0(T0)− Re [a1(T0)p1U∗1 ]


dU1

dX= κa3(T0)p1 + a4(T0)

dT0

dXU1

dp1

dX= κa5(T0)U1

I where Kloss denotes the losses through heat conduction

9/23


Recap

Standing-wave refrigerator

HW

QHQC

We can

I Compute pressure, velocity, temperature, and energy fluxes

I Test performance:

COP =QC

W, (refrigerator)

η =W

QH, (prime mover)

I Investigate influence of parameters related to geometry, material, and gas.

10/23


Recap

Traveling-wave prime mover

Issues

I More complex geometry

I Streaming becomes important

I More difficult to model and implement numerically

I Better performance

11/23


Nonlinear standing waves

Assumptions

I Closed gas-filled tube with oscillating speaker

sound

I Potential flow

• Viscous interaction with the wall is neglected• One spatial dimension x• u = −∂8/∂x

Kuznetsov equation

∂28

∂t2 − c20 ∇

28 =∂

∂t

[(∇8)2 +

γ − 1

2c20

(∂8

∂t

)2

+bρ0

∇28

],

with b = K(

1cv−

1cp

)+

43η + ζ .

11/23



Assumptions

I Closed gas-filled tube with oscillating speaker

sound

I Potential flow

• Viscous interaction with the wall is neglected• One spatial dimension x• u = −∂8/∂x

Kuznetsov equation

∂28

∂t2 − c20 ∇

28 =∂

∂t

[(∇8)2 +

γ − 1

2c20

(∂8

∂t

)2

+bρ0

∇28

],

with b = K(

1cv−

1cp

)+

43η + ζ .

12/23



Dimensionless parameters

ε =ω`

c0, µ =

ω2Lb

2ρ0c30

, κ =ωLc0, 1 = κ − π.

Equation

∂28

∂t2 −∂28

∂x2 =∂

∂t

[(∂8

∂x

)2

+γ − 1

2

(∂8

∂t

)2

+ 2µ

κ2

∂28

∂x2

].

Boundary conditions

x = 0 : u = −∂8

∂x= 0,

x = 1 : u = −∂8

∂x= εh (κt).

Here we take h (t) = sin(t).

13/23


SolutionAsymptotic expansionWe assume ε � 1 and write

8(x, t) = ε9(x, t)+ ε2ψ(x, t)+ · · · , ε � 1.

We obtain with F ′ = f and ϒ quadratically nonlinear

9(x, t) = F1(t − x)− F2(t + x),

ψ(x, t) = G1(t − x)+ G2(t + x)+ ϒ(F1,2, f1,2, f ′1,2

),

13/23


SolutionAsymptotic expansionWe assume ε � 1 and write

8(x, t) = ε9(x, t)+ ε2ψ(x, t)+ · · · , ε � 1.

We obtain with F ′ = f and ϒ quadratically nonlinear

9(x, t) = F1(t − x)+ F2(t + x),

ψ(x, t) = G1(t − x)+ G2(t + x)+ ϒ(F1,2, f1,2, f ′1,2

),

satisfying u(0, t) = 0.

f and g follow from u(1, t) = ε sin(κt).

14/23


Solution

We need to solve:

f (t − 1)− f (t + 1)+ εϒ(F , f , f ′, f ′′

)+ ε

{g(t − 1)− g(t + 1)

}= ε sin(κt)+ o(ε)

14/23


Solution

We need to solve:

f (t − 1)− f (t + 1)+ εϒ(F , f , f ′, f ′′

)+ ε

{g(t − 1)− g(t + 1)


We can distinguish two cases:

1. Away from resonance (1 6= 0)

⇒ Linear solution using Fourier transform

f (t) =cos(κt)2 sin(κ)

g(t) =βκ

4cos(2κt)

sin2(κ)−µ

2εcos(κ) sin(κt)

sin2(κ)

⇒ To leading order u(x, t) = ε sin(κx)sin(κ) sin(κt)+ o(ε, µ)

14/23


Solution

We need to solve:

f (t − 1)− f (t + 1)+ εϒ(F , f , f ′, f ′′

)+ ε

{g(t − 1)− g(t + 1)


We can distinguish two cases:

2. Near resonance (1� 1).

⇒ Nonlinear solution using a multiple-scales approach⇒ Let v(ζ, T ) = f (t)

with “slow” time T = εκt and “fast” time ζ = κt + π⇒ f will be almost periodic

f (t − 1)− f (t + 1) = O (ε,1).

15/23


Steady-state solution near resonance

For ε, µ,1� 1, we have

πε∂v∂T+1

∂v∂ζ− εβκv

∂v∂ζ−µ

κ2

∂2v∂ζ 2 = ε sin(ζ )+ o(ε, µ,1)

Putting

v(ζ, T ) =1

εβκ+

√2εβκ

V (z, T ), z =ζ

2

We find

in steady state

−2πεµ

∂V∂T+ν

∂2V∂z2 + 2V

∂V∂z= − sin(2z), ν =

√µ2

2εβκ

When ν � 1, a solution can be obtained using matched asymptotic expansions

15/23




πε∂v∂T+1


∂v∂ζ−µ

κ2

∂2v∂ζ 2 = ε sin(ζ )+ o(ε, µ,1)

Putting

v(ζ, T ) =1

εβκ+

√2εβκ

V (z, T ), z =ζ

2

We find

in steady state

−2πεµ

∂V∂T+ν

∂2V∂z2 + 2V

∂V∂z= − sin(2z), ν =

√µ2

2εβκ


15/23




πε∂v∂T+1


∂v∂ζ−µ

κ2

∂2v∂ζ 2 = ε sin(ζ )+ o(ε, µ,1)

Putting

v(ζ, T ) =1

εβκ+

√2εβκ

V (z, T ), z =ζ

2

We find in steady state

−2πεµ

∂V∂T+ν

∂2V∂z2 + 2V

∂V∂z= − sin(2z), ν =

√µ2

2εβκ


16/23



Matched asymptotic expansionsWe solve for ν � 1

ν∂2V∂z2 + 2V

∂V∂z= − sin(2z), −

π2 < z < π

2

We introduce a boundary-layer coordinate s by z = zi + νs

V (z) = W (s), near z =zi

Boundary conditions

matching : limz↑zi

V = lims→−∞

W , limz↓zi

V = lims→+∞

W

periodicity : V(π2

)= V

(−π2

),

zero average : W (0) = 0

16/23




ν∂2V∂z2 + 2V

∂V∂z= − sin(2z), −

π2 < z < π

2



Boundary conditions


V = lims→−∞

W , limz↓zi

V = lims→+∞

W

periodicity : V(π2

)= V

(−π2

),


16/23




ν∂2V∂z2 + 2V

∂V∂z= − sin(2z), −

π2 < z < π

2



Boundary conditions


V = lims→−∞

W , limz↓zi

V = lims→+∞

W

periodicity : V(π2

)= V

(−π2

),

zero average :∫ π

2

−π2

V (z) dz = 0

16/23




ν∂2V∂z2 + 2V

∂V∂z= − sin(2z), −

π2 < z < π

2



Boundary conditions


V = lims→−∞

W , limz↓zi

V = lims→+∞

W

periodicity : V(π2

)= V

(−π2

),


17/23



Asymptotic expansion

V = V0 + νV1 + · · · ,

W = W0 + νW1 + · · · ,

We find zi = 0 and

V0 = −sgn(z) cos(z), W0 = tanh(s),

V1 =sin(z)− sgn(z)

2 cos(z), W1 =

12

(s

2 cosh2(s)+ tanh(s)

).

18/23


Velocity at exact resonance (1 = 0)

During one period a shock travels back and forth in the tube.

— velocity for µ = 2.7 · 10−4 and ε = 6.8 · 10−6

--- velocity for µ = 0 and ε = 6.8 · 10−6

0 0.2 0.4 0.6 0.8 1−1.5

−1

−0.5

0

0.5

1

1.5

u (m

/s)

x/L

— ωt = −π

— ωt = −3π/4

— ωt = −π/2

— ωt = −π/4

— ωt = 0

— ωt = −3π/4

0 0.2 0.4 0.6 0.8 1−1.5

−1

−0.5

0

0.5

1

1.5

x/L

u (m

/s)

— ωt = 0

— ωt = π/4

— ωt = π/2

— ωt = 3π/4

— ωt = π

— ωt = π/4

19/23


Coupling with stack

I With closed end at x = 0 and x = x − x0

u(x, t) = ε{f (t − x)− f (t + x)

}+ ε2ϒ

(F , f , f ′, f ′′

)+ ε2

{g(t − x)− g(t + x)

}

19/23


Coupling with stack

I With closed end at x = 0 and x = x − x0

u(x, t) = ε{f (t − x)− f (t + x)

}+ ε2ϒ

(F , f , f ′, f ′′

)+ ε2

{g(t − x)− g(t + x)

}⇒ The wave reflects at x = 0 with amplitude 1

19/23


Coupling with stack

I With stack at x = 0 and x = x − x0

u(x, t) = ε{Rf (t − x)− f (t + x)

}+ ε2ϒ

(F , f , f ′, f ′′,R , x0

)+ ε2

{Rg(t − x)− g(t + x)

}⇒ The wave effectively reflects at some x0 with a slight loss, modeled

by the factor R ≤ 1

19/23


Coupling with stack


u(x, t) = ε{Rf (t − x)− f (t + x)

}+ ε2ϒ

(F , f , f ′, f ′′,R , x0

)+ ε2

{Rg(t − x)− g(t + x)



I We derive the following equation for V

ν∂2V∂z2 + 2V

∂V∂z− δV = − sin(2z), δ =

2ν(1− R )µ

19/23


Coupling with stack


u(x, t) = ε{Rf (t − x)− f (t + x)

}+ ε2ϒ

(F , f , f ′, f ′′,R , x0

)+ ε2

{Rg(t − x)− g(t + x)



I We derive the following equation for V

ν∂2V∂z2 + 2V

∂V∂z− δV = − sin(2z), δ =

2ν(1− R )µ

I Due to the extra term we cannot find analytical expressions for the outersolution

20/23


Coupling with stack

Outer solution

I We solve numerically

V0

(2∂V0

∂z− δ

)= − sin(2z), −

π2 < z < π

2 (1)

V0(−π2 ) = V0(

π2 ) (2)

.

I If V0(−π2 ) = α

I If V0(π2 ) = β

−1.5 −1 −0.5 0 0.5 1 1.5−3

−2

−1

0

1

2

3

z

Y

α = −3α = −2α = −1α = −0.1α = 0.1α = 1

−1.5 −1 −0.5 0 0.5 1 1.5−3

−2

−1

0

1

2

3

z

Y

β = 3β = 2β = 1β = 0.1β = −0.1β = −1

⇒ We combine a “left”-solution V− with a “right”-solution V+

⇒ Only when V+0 (π2 ) = −V

−

0 (−π2 ) = β ↓ 0 , we can satisfy (2)

⇒ The inner solution will connect V− and V+ in z = 0.

20/23


Coupling with stack

Outer solution


V0

(2∂V0

∂z− δ

)= − sin(2z), −

π2 < z < π

2 (1)

V0(−π2 ) = V0(

π2 ) (2)

.

I If V0(−π2 ) = α

I If V0(π2 ) = β

−1.5 −1 −0.5 0 0.5 1 1.5−3

−2

−1

0

1

2

3

z

Y

α = −3α = −2α = −1α = −0.1α = 0.1α = 1

−1.5 −1 −0.5 0 0.5 1 1.5−3

−2

−1

0

1

2

3

z

Y

β = 3β = 2β = 1β = 0.1β = −0.1β = −1



−

0 (−π2 ) = β ↓ 0 , we can satisfy (2)


20/23


Coupling with stack

Outer solution


V0

(2∂V0

∂z− δ

)= − sin(2z), −

π2 < z < π

2 (1)

V0(−π2 ) = V0(

π2 ) (2)

.

I If V0(−π2 ) = α

I If V0(π2 ) = β

−1.5 −1 −0.5 0 0.5 1 1.5−3

−2

−1

0

1

2

3

z

Y

α = −3α = −2α = −1α = −0.1α = 0.1α = 1

−1.5 −1 −0.5 0 0.5 1 1.5−3

−2

−1

0

1

2

3

z

Y

β = 3β = 2β = 1β = 0.1β = −0.1β = −1



−

0 (−π2 ) = β ↓ 0 , we can satisfy (2)


20/23


Coupling with stack

Outer solution


V0

(2∂V0

∂z− δ

)= − sin(2z), −

π2 < z < π

2 (1)

V0(−π2 ) = V0(

π2 ) (2)

.

I If V0(−π2 ) = α

I If V0(π2 ) = β

−1.5 −1 −0.5 0 0.5 1 1.5−3

−2

−1

0

1

2

3

z

Y

α = −3α = −2α = −1α = −0.1α = 0.1α = 1

−1.5 −1 −0.5 0 0.5 1 1.5−3

−2

−1

0

1

2

3

z

Y

β = 3β = 2β = 1β = 0.1β = −0.1β = −1



−

0 (−π2 ) = β ↓ 0 , we can satisfy (2)


21/23


A thermoacoustic refrigerator

stack

�

tube

-

⇐

I For given R and x0 we compute the velocity in the tube

I We apply a discrete Fourier transform to the velocity field in the tube

I For each harmonic mode we compute the velocity in the stack

I Full velocity field follows from the inverse Fourier transform

21/23



stack

�

tube

-

⇐





21/23



stack

�

tube

-

⇐





21/23



stack

�

tube

-

⇐





21/23



stack

�

tube

-⇐





22/23


Challenges

What is next?

I Combine traveling-wave prime mover with traveling wave refrigerator

I Include turbulence into the modeling

I Include other nonlinear effects

I Generalize single-pore model to bulk porous media

⇒ Homogenization, statistical modeling, . . .

I . . .

23/23


Further reading

S. Backhaus and G.W. Swift

A thermoacoustic-Stirling heat engine: Detailed study

JASA (107), 2000.

B.O. Enflo and C.M. Hedberg

Theory of nonlinear acoustics in fluids

Kluwer Academic Publishers, 2002.

P.H.M.W. in ’t panhuis, S.W. Rienstra, J. Molenaar, J.J.M. Slot

Weakly non-linear thermoacoustics for stacks with slowly varying pore cross-sections

JFM (618), 2009.

N. Rott

Thermoacoustics

Adv. in Appl. Mech. (20), 1980.

G.W. Swift

Thermoacoustic engines

JASA (84), 1988.

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