Thermoacoustic Devices

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1 Thermoacoustic Devices Peter in t panhuis Sjoerd Rienstra Han Slot 24th April 2008

2 Down-well power generation

3 Vortex shedding in side branch Vortex shedding creates standing wave Porous medium near closed end Temperature difference is generated

4 Vortex shedding in side branch Vortex shedding creates standing wave Porous medium near closed end Temperature difference is generated No moving parts Cheap and reliable energy source

5 Thermoacoustic devices Refrigerator vs. engine (a) Engine: heat power acoustic power. (b) Refrigerator: acoustic power heat power

6 Basic thermoacoustic effect Thermodynamic cycle of gas parcel in refrigerator

7 Basic thermoacoustic effect Thermodynamic cycle of gas parcel in refrigerator Bucket brigade: heat is shuttled along the wall

8 Outline Outline Introduction Modelling Analysis Standing wave devices Traveling wave devices Results Future work

9 Introduction Outline Modelling Analysis Future Work Further reading Geometry Porous medium is modelled as collection of pores: Boundary may vary slowly in x. pore length radius s g gas r Rs ( x, ) solid Rg ( x, ) x

10 Introduction Outline Modelling Analysis Future Work Further reading Geometry Porous medium is modelled as collection of pores: Boundary may vary slowly in x s Rs ( x, ) solid g gas r Rg ( x, ). pore length radius y y Cross-section may have arbitrary shape. e.g. parallel plates, circles, rectangles, triangles z r y z Porous medium is usually called stack or regenerator x

11 Introduction Outline Modelling Analysis Future Work Thermoacoustic devices Porous medium in a gas-filled tube. Two modes of operation: Standing wave Straight tube. Straight tube. Stack has small pores Traveling wave. Looped tube. Stack has tiny pores Looped tube Further reading

12 Equations Conservation of mass, momentum and energy: mass : momentum : energy: ρ + (ρv) = 0, t t (ρv) + (ρvvt ) = p + τ, (ρɛ) + (ρvɛ) = q + ( pvτ + τ v). t

13 Equations Conservation of mass, momentum and energy: mass : momentum : energy: ρ + (ρv) = 0, t t (ρv) + (ρvvt ) = p + τ, (ρɛ) + (ρvɛ) = q + ( pvτ + τ v). t Additional equations: Fourier s law: q = K T, ( ) viscous stress tensor: τ = µ v + ( v) T + ζ( v)i, ideal gas law: p = ρrt. 6 equations, 7 unknowns

14 Equations Thermodynamic relations: dɛ = T ds + p ρ 2 dρ, ds = C p T dt β ρ dp. µ, K, C p, c may depend on temperature. The temperature in the solid is found from ρ s C s T s t = (K s T s ).

15 Boundary conditions No-slip conditions at solid-gas interface v = 0 at Γ g Continuity of temperature and heat fluxes at solid-gas interface r g s T = T s at Γ g, K T n = K s T s n at Γ g, Symmetry of heat flux in solid T s n = 0 at Γ s. Boundary conditions at stack ends depend on application

16 Dimensionless Model Important dimensionless numbers are M a = U/c ε = R/L κ = 2πL/λ acoustic Mach number aspect ratio of stack pore Helmholtz number We assume M a 1 and ε 1

17 Dimensionless Model Important dimensionless numbers are M a = U/c ε = R/L κ = 2πL/λ acoustic Mach number aspect ratio of stack pore Helmholtz number We assume M a 1 and ε 1 We expand variables in powers of M a 1 f (x, t; M a ) = f 0 (x)+m a f 1 (x)e iωt +.+Ma {f 2 2,0 (x) + f 2,2 (x)e 2it} We assume v 0 = 0, so that also p 0 is constant

18 Dimensionless Model Important dimensionless numbers are M a = U/c ε = R/L κ = 2πL/λ acoustic Mach number aspect ratio of stack pore Helmholtz number We assume M a 1 and ε 1 We expand variables in powers of M a 1 { f (x, t; M a ) = f 0 (x)+m a f 1 (x)e iωt +Ma 2 f 2,0 (x) + f 2,2 (x)e 2iωt} +. Including streaming Gas moves in repetitive "101 steps forward, 99 steps backward" manner

19 Method of slow variation We use the method of slow variation X = εx Slender pore assumption: ε 1 Axial and transverse variation can be treated separately T 0 and p 1 only depend on X Define U 1 = 1 u 1 da A g A g System of ODE s for T 0, p 1 and U 1 Green s function for Helmholtz equation on cross-section gives u 1 and T 1. u 2,0 follows from Green s function for Poisson equation on cross-section

20 System of ODE s ODE s: dt 0 dx = F 1(T 0, p 1, U 1, V; Ṁ, Ḣ, geometry, material) du 1 dx = F 2(T 0, p 1, U 1 ; geometry, material) dp 1 dx = F 3(T 0, U 1 ; geometry, material) ( ) dv dx = F dt0 4 ; geometry, material dx Ḣ and Ṁ are the time-averaged energy flux and mass flux V is an auxiliary variable that disappears when da g /dx = 0

21 System of ODE s ODE s (with variable cross-section): dt 0 dx = F 1(T 0, p 1, U 1, V; Ṁ, Ḣ, geometry, material) du 1 dx = F 2(T 0, p 1, U 1 ; geometry, material) dp 1 dx = F 3(T 0, U 1 ; geometry, material) ( ) dv dx = F dt0 4 ; geometry, material dx Ḣ and Ṁ are the time-averaged energy flux and mass flux V is an auxiliary variable that disappears when da g /dx = 0

22 Standing-wave refrigerator Boundary conditions: Impose U 1 = 0 at closed end Impose p 1 = Dp 0 at closed end D is drive ratio Impose continuity of mass and momentum T H T C TH TC < Tcrit Ẇ H x s Impose T C and T H Shoot in Ḣ to obtain desired T H Q C Q H

23 Standing-wave refrigerator Boundary conditions: Impose U 1 = 0 at closed end Impose p 1 = Dp 0 at closed end D is drive ratio Impose continuity of mass and momentum T H T C TH TC < Tcrit Ẇ H x s Impose T C and T H Shoot in Ḣ to obtain desired T H Energy balance: Q C = Ḣ Ẇ Q H = Ḣ Efficiency: COP = Q C Ẇ COPC = Q C T C T H T C COPR = COP COPR Q H

24 High-amplitude calculations COPR D=5% D=10% D=20% D=30% D=40% temperature difference (K) COPR vs. T H T C : at optimal stack position COPR stabilizes for large drive ratios critical temperature is reached

25 High-amplitude calculations COPR D=5% D=10% D=20% D=30% D=40% temperature difference (K) COPR vs. T H T C : at optimal stack position COPR stabilizes for large drive ratios critical temperature is reached COPR opt COPR opt vs. stack position: D = 30% COPR opt is maximal near the closed end kx s

26 Introduction Outline Modelling Analysis Future Work Further reading Traveling-wave devices Traveling wave refrigerator: Looped geometry Stack TC IT TH TBT TC sound Looped geometry TBT Th Thin pores Real impedance Z = p1 /U1 TH TC arises Acoustic power is dissipated Thermal buffer tube (TBT). Temperature is reduced to TC Resonator sound Tc.... IT Tc Unrolled Geometry. Supplies lost acoustic power Inertance tube (IT). Ensures p1 and U1 fit

27 Introduction Outline Modelling Analysis Future Work Further reading Traveling-wave devices Traveling wave engine: Looped geometry Stack TC IT TH.... TBT TC sound Looped geometry TBT Th Thermal buffer tube (TBT). Temperature is reduced to TC Resonator sound Tc Thin pores Real impedance Z = p1 /U1 TH TC is imposed Acoustic power is produced IT Tc Unrolled Geometry. Absorbs excess acoustic power Inertance tube (IT). Ensures p1 and U1 fit

28 Introduction Outline Modelling Analysis Future Work Traveling-wave devices sound sound TBT Tc Th TBT IT Tc Tc Th IT Tc Computations: On unrolled geometry In regenerator and TBT we solve system of ODE s. Yields T0, p1 and U1 Everywhere else we have an analytic solution. Wave equation for p1 and U1 (T0 is constant). p1 = Ae ikx + Be ikx 1. U1 = (Ae ikx Be ikx ) ρ0 c Further reading

29 Introduction Outline Modelling Analysis Future Work Traveling-wave devices sound sound TBT Tc Th TBT IT Tc Tc Th IT Tc Computations: Continuity of mass and momentum at interfaces At x = 0 impose p1 and U1 such that Z is real. Overdetermined system of equations. Fit geometry parameters. Minimize least squares solution Further reading

30 Goals Traveling wave devices Find geometry parameters Compute efficiencies Investigate effect of streaming Higher amplitudes Include higher harmonics Research shock waves Include end effects Validate against measurements

31 Further reading N. Rott Thermoacoustics Adv. in Appl. Mech. (20), G.W. Swift Thermoacoustic engines JASA (84), 1988.

Thermoacoustic Devices

Thermoacoustic Devices Peter in t panhuis Sjoerd Rienstra Han Slot 9th May 2007 Introduction Thermoacoustics All effects in acoustics in which heat conduction and entropy variations play a role. (Rott,