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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
/ 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
/ 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
/ 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
3/23
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
Introduction
ApplicationsI Liquefaction of natural gases
I Upgrading of industrial waste heat
I Downwell power generation
I Food refrigeration
I Airconditioning
I . . .
.
4/23
/ department of mathematics and computer science
Introduction
ApplicationsI Liquefaction of natural gases
I Upgrading of industrial waste heat
I Downwell power generation
I Food refrigeration
I Airconditioning
I . . .
.aaaa
a
4/23
/ department of mathematics and computer science
Introduction
ApplicationsI Liquefaction of natural gases
I Upgrading of industrial waste heat
I Downwell power generation
I Food refrigeration
I Airconditioning
I . . .
.aaaa
a
4/23
/ department of mathematics and computer science
Introduction
ApplicationsI Liquefaction of natural gases
I Upgrading of industrial waste heat
I Downwell power generation
I Food refrigeration
I Airconditioning
I . . .
.aaa
a
4/23
/ department of mathematics and computer science
Introduction
ApplicationsI Liquefaction of natural gases
I Upgrading of industrial waste heat
I Downwell power generation
I Food refrigeration
I Airconditioning
I . . .
.
4/23
/ department of mathematics and computer science
Introduction
ApplicationsI Liquefaction of natural gases
I Upgrading of industrial waste heat
I Downwell power generation
I Food refrigeration
I Airconditioning
I . . .
.
Benefits
I No moving parts and use of simple materials
I Reliable
I Environmentally friendly
5/23
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
Basic thermoacoustic effect
I Thermodynamic cycle of gas parcel near wall in refrigerator
I Bucket brigade: heat is shuttled along the wall
6/23
/ department of mathematics and computer science
Basic thermoacoustic effect
I Thermodynamic cycle of gas parcel near wall in refrigerator
I Bucket brigade: heat is shuttled along the wall
7/23
/ department of mathematics and computer science
Overview of PhD
ProgressI Weakly nonlinear theory of thermoacoustics
7/23
/ department of mathematics and computer science
Overview of PhD
ProgressI Weakly nonlinear theory of thermoacoustics
• 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
/ department of mathematics and computer science
Overview of PhD
ProgressI Weakly nonlinear theory of thermoacoustics
• 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
I Standing-wave devices
I Traveling-wave devices
7/23
/ department of mathematics and computer science
Overview of PhD
ProgressI Weakly nonlinear theory of thermoacoustics
• 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
I Standing-wave devices
I Traveling-wave devices
I Nonlinear oscillations near resonance
• Shock formation
7/23
/ department of mathematics and computer science
Overview of PhD
ProgressI Weakly nonlinear theory of thermoacoustics
• 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
I Standing-wave devices
I Traveling-wave devices
I Nonlinear oscillations near resonance
• Shock formation
}Today
8/23
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
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
/ department of mathematics and computer science
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
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
I where H is the energy flux
8/23
/ department of mathematics and computer science
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
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
I where M is the mass flux and h is the specific enthalpy
8/23
/ department of mathematics and computer science
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
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
I where Kloss denotes the losses through heat conduction
9/23
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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 .
12/23
/ department of mathematics and computer science
Nonlinear standing waves
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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)
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
/ department of mathematics and computer science
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)
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
/ department of mathematics and computer science
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 =
õ2
2εβκ
When ν � 1, a solution can be obtained using matched asymptotic expansions
15/23
/ department of mathematics and computer science
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 =
õ2
2εβκ
When ν � 1, a solution can be obtained using matched asymptotic expansions
15/23
/ department of mathematics and computer science
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 =
õ2
2εβκ
When ν � 1, a solution can be obtained using matched asymptotic expansions
16/23
/ department of mathematics and computer science
Steady-state solution near resonance
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
/ department of mathematics and computer science
Steady-state solution near resonance
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
/ department of mathematics and computer science
Steady-state solution near resonance
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 :∫ π
2
−π2
V (z) dz = 0
16/23
/ department of mathematics and computer science
Steady-state solution near resonance
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
17/23
/ department of mathematics and computer science
Steady-state solution near resonance
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
I We derive the following equation for V
ν∂2V∂z2 + 2V
∂V∂z− δV = − sin(2z), δ =
2ν(1− R )µ
19/23
/ department of mathematics and computer science
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
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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.
21/23
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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
22/23
/ department of mathematics and computer science
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
/ department of mathematics and computer science
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.