Oscillating particles in fluids: Theory, experiment and ... · Landau and Lifshitz, "Physical...

Oscillating particles in fluids: Theory, experiment and numerics

Oscillating particles in fluids: Theory, experimentand numerics

Victor YakhotBoston University, Boston, MA

May, 2012, Vienna


Boltzmann-BGK Equation in the range 0 ≤ ωτ <∞

Experimental data. Universality.


My collaborators:

Kamil Ekinci (BU);Carlos Colosqui (BU→ Princeton);Devrez Karabacak (BU→Delft).

Hudong Chen, Xiaowen Shan, Ilya Staroselsky (EXA Corp.)


Landau and Lifshitz, ”Physical kinetics”. After presentingderivation of Boltzmann equation they mention in passing:

C ≈ − f − f eq

τ

with τ = const”... is a rough estimate of collision integral”This expression, which is at the foundation of BGK-LBM, hasnever been derived as a consistent approximation


Linear oscillator in liquid/gas.


m0d2x

dt2+ γm0

dx

dt+ κx = R(t) (1)

d2x

dt2+ γ

dx

dt+ ω2

0x =R0

m0cosωt (2)

ω0 =√κ/m0

I (ω − ω0) =R20

8m0

γ

(ω − ω0)2 + γ2

4

The mass m0 = mR + mf where mR is the resonator mass invacuum and the added mass mf is that of the fluid ‘pushed’ by theresonator.The damping γ = γR + γf where γf is FLUIDIC DISSIPATION


For a spherical slowly moving body of radius a, by the Stokesformula :

F = 6πµax =4π

3ρba3γx

F = 16µax

F =32

3µax

For a 2d-disk and ellipsoid, respectively. The mass:.

m0 =4πa3

3(ρb +

1

2ρ)

m0 = πa2(ρb + ρ)

for a sphere and disk, respectively.


m0 → m0 + δm; ω1 ≈ ω0(1− δm

2m0)


Double-Clamped Beam. Schematic.

diagram.pdf

hw

lz(t)

!

~

hw

lz(t)

!

~

h ≈ 0.1µm, w ≈ 0.5µm, l ≈ 10µm .


Array of Nanobeams. Microscope. Dimensions → ω0


Nanobeams as Sensors.

FrequencyAm

plitu

de o

f Mot

ionBiomolecule

NEMS in Fluidic Enviroment


Nanobeams. Biodetection


Nanocantilevers. Biodetection


The modern devices:

m0 ≈ 10−13 − 10−12gr

ω0 ≈ 109Hz

One can work with higher harmonics and achieve higherfrequencies.

Now, detection of the mass of a proton is a reality.


Pressure dependence of resonance curves of NEMS.(Ekinci & Karabacak (2007)


FDT.

(a)(b) (c)

.


FDT

.


In equilibrium if resonator is thermally excited, variance thedisplacement : ∫ ∞

−∞h2P(h)dh =

4kBT

κ

In a confined system this relation is not clear.

Experimental test of FDT


TO DETECT A MASS δm:

ω0δm

2m0 γ

THE MAIN STUMBLING BLOCK IN BIODETECTION ISFLUIDIC DISSIPATION IN WATER (FLUIDS).

FIRST WE HAVE TO UNDERSTAND IT.


This is not so simple. In air

p ≈ 800torr ; τmol−mol ≈ 0.2× 10−9sec

,

τmol−si ≈ 10−9sec

.

p = 100torr ; τmol−mol ≈ 10−9 − 10−8

.

0 ≤ ωτ ≤ 10− 100


Stokes’ second problem (1851).Infinite plate oscillating in its plane inthe x-direction. Find velocitydistribution u(y , t)

u(0, t) = U cosωt; u(∞, t) = 0

w = v = 0

∂u

∂t+ u · ∇u = −∇p/ρ + ν∇2u


∂u

∂t= ν

∂2u

∂y 2

u(y , t) = Ue−yδ cos(ωt +

y

δ)

δ =

√2ν

ωOverdamped surface wave. No transverse

propagating shear waves. (Definition of fluid)


Force per unit area (viscous stress)

σ = ρν∂u

∂y|y=0 = −U

√ωρµ cos(ωt +

π

4)

Mean dissipation per unit time

−σu =U2

2

√ωµρ

2

Quality factor Q


1

Q=

E

2πEst=γ

ω≈ σ

u(0, t)ω=

√ρµ

ω

γ =√ρµω =

√µω

RTp


Dissipation. x = ωτ

0 20 40 60 80 100x

0.51

510

delta

1 2 5 10 20 50 100x1

1.52

3

5

7

10gamma


Breakdown of Newtonian Hydrodynamics.

∂ui∂t

+ u·∇ui =1

ρ∇jρσij

σij = u′iu′j

Kinetic Theory. Boltzmann Equation.

σij = σ(1)ij + σ

(2)ij + ....

σ(1)ij ≈

ν

2(∂iuj + ∂jui) ≡ νSij


Different geometries. Different scales.


µ = ρν ≈ ρcsλ

σ(1)ij

U2≈ µSij

U2≈ ρλcsU

ρU2L≈ λ

L

csU≈ Kn

Ma

σ(2)ij ≈ λ2 ∂ui

∂xα

∂uj∂xα

Stokes Problems: Si ,αSα,j = SiαSj ,α = 0

and there is no length scales(Landau-Lifshitz)


σ(2)ij ≈ λ2Sij∇ · u

One has to solve dynamic kineticequation.


Stokes’ Second Problem, Revisited. KineticBoltzmann-BGK equation.

VY, Colosqui (2007).

∂f

∂t+ v · ∇f = −f − f eq

τ

f eq =ρ

(2πθ)32

exp(−(v −U(x, t))2

2θ)


Boltzmann’s collision integral:

C =

∫vrel(f ′f ′1 − ff1)dσdp1 ≈

− f

τ (f )+

∫vrel f

′f ′1dσdp1

τ (f ) =

∫vreldσf1dp1 ≈ τ = const???


Relaxation time τ ≈ const.

τ ≈ λ/v ≈ λ/cs

cs-speed of sound. In the air attemperature θ ≈ 300K and pressure

p = 1atm ≈ 1000torr

τ ≈ 200/p × 10−9sec ≈ 0.2× 10−9sec

At this point this relation does notaccount for solid walls, strong shear

etc.


∫C(f )dv =

∫C(f )vdv = 0

ρ =

∫f (v)dv

∂ρ

∂t+∇ · ρU(x) = 0

∂ρUj(x)

∂t+ ∂iρUj(x)Ui(x) + ∂iσij = 0


σij = ρ(vi − Ui(x))(vj − Uj(x))

The goal is to derive the expression forthe stress σij in terms of observables.

This can be done using theChapman-Enskog expansion.Hudong Chen et al (2004).


f = f (0) + εf (1) + ε2f (2) + · · ·

∂t = ε∂t0 + ε2∂t1 + · · ·∇ = ε∇1


In the zeroth order

f = f eq

σij |eq = ρθδij

(ideal gas):


f 1 = −τθ

f 0Sij [(vi − ui)(vj − uj)−(u− v)2

dδij

f (2) = −2τ 2f (0)[(vi−vj)∂j(Sij−∇·uδij .....)+++


∂Uj(x)

∂t+ Ui(x)∂iUj(x) +

1

ρ∇p(x) = 0


Defining θ(x) = 1d (ci − vi(x))2, we

obtain:

∂θ(x)

∂t+∇iρUi(x)θ +

1

d∇iρ(vi − Ui(x))(vj − Uj(x))2

+2

dρ(vi − Ui(x))(vj − Uj(x))Si ,j = 0


In the next order we derive Newtonianapproximation and NS Equations:

σ1 ≈ −2ρν(Sij −1

dδij∇ · u)

ν = θτ

Sij =1

2(∂ui∂xj

+∂uj∂ui

)


σ(2)ij = 2ρν ˙(τSij) +

4ρν2

θ(SikSkj − δijSklSkl/d)−

−2ρν2

θ(SikΩkj + SjkΩki)

A = (∂t + U · ∇)A

Ωij =1

2(U j

i − U ij )


σ(1)ij + σ

(2)ij = −2ρν(1− τ∂t)Sij

∂u

∂t+ u · ∇u = 2ρν(1− τ∂t)

∂2u

∂y 2


Two dimensionless expansionparameters appear:

σij = σ(1)ij +σ

(2)ij ≈ 2ρν(1+τ

∂

∂t+O(

ν|S |θ

))Sij

Wi = τ∂t → τω

S ≈ U/L; ν = λcs ; θ ≈ c2s

MaKn = νS/θ ≈ U

cs

λ

L


VISCO-ELASTIC TRANSITION. (VY,Colosqui, JFM 2007).

For a problem of oscillating plate:

SikSkj = 0

∇ ·U = 0

SiαSαβ.....Sθ,j = 0


This is correct to all orders and asωτ →∞ (VY, Colosqui)

τ∂2u

∂t2+∂u

∂t= ν

∂2u

∂y 2

u(0, t) = U cosωt; u(∞, t) = w = v = 0

As ωτ → 0, we recover Stokes problemWhen ωτ →∞, wave equation.


All this can be exactly derived in anon-perturbative way (Chen,

Staroselsky, Orszag, Shan, VY...).

∂f

∂t+ v · ∇f = −f − f eq

τ

Interested in a unidirectional flow,which is incompressible, we, withoutloss of generality, set ρ = const = 1.


f (x, v, t) =

∫ tτ

0

e−sf eq(x− vτ s, v, t − τ s)ds +

f0(x− vt, v)e−tτ

t τ

f (y , z , v, t) =

∫ ∞0

e−sf eq(x−vτ s, v, t−τ s)ds

where x = y j + zk, and j and k are unitvectors along the y and z axis.


Spatial shift operator:F (x + a) = ea∂x F (x)

f (x, v, t) =

∫ ∞0

e−se−τsv·∇f eq(x, v, t−τ s)ds

Unidirectional flow: ρ = const = 1:

u(x, t) =

∫vdv

∫ ∞0

e−se−τsv·∇f eq(x, v, t−τ s)ds

(3)


Simple Gaussian integration gives:

U(y , t) =

∫ ∞0

e−ses2λ2∇2/2U(y , t − τ s)

λ2 = τ 2θ

sec2 ·cm2/sec2 = cm2


Differentiate twice over time; useidentity

∂

∂s

[e−se

s2λ2∇2

2

]= [sλ2∇2 − 1]

[e−se

s2λ2∇2

2

]

τ∂2u

∂t2+∂u

∂t=

ν∇2u + νλ2

∫ ∞0

s2e−ses2λ2∇2

2 ∇4u(y , t − τ s)ds


This hydrodynamic equation is exact.

τ∂2u

∂t2+∂u

∂t= ν(1 + φ)∇2u (4)

φ = λ2

∫ ∞0

dss2e−ses2λ2∇2

2 e−sτ∂t∇2 (5)

φ = −Wi

2

∫ ∞0

dss2e−s(1−iWi)− s2Wi2

2


Wi = ωτ → 0

ν = θτ ;∇2 ≈ 1/δ2 = ω/2ν =ω

2θτ

φ ≈ ωτ

2

∫ ∞0

dss2e−se−s2ωτ

4 e isωτ = ωτ → 0

∇2 = 1/δ2 = ω/2ν


Diffusion equation; Stokes’ problem;


Large-Weissenberg Limit Wi = ωτ 1:

φ ∼ −Wi

∫ 1√Wi

0

s2ds ∼ −Wi−1/2 → 0

∂2u

∂t2+

1

τ

∂u

∂t− c2∂

2u

∂y 2≡ Θu = 0 (6)

The TE derived from the CEexpansion before (VY, Colosqui. JFM

(2007).


Solution:

u = Ueyδ− cos(ωt − y

δ+)

δ

δ±=

(1 + ω2τ 2)14 [cos(

tan−1 ωτ

2)± sin(

tan−1 ωτ

2)]


We have three scales and three Knudsen

numbers:

δ; δ−; δ+

Kn =λ

δKn− =

λ

δ−; Kn+ =

λ

δ+


λ ≈ τcs =1

σn= τ√θ ∝ τ

√p

Newtonian regime; Stokes’ scale:

δ =

√2ν

ω≈√λcsω≈ λ

√1

ωτ

Knδ =√ωτ =

√Wi


as ωτ →∞,

Kn− →1

2; Kn+ → ωτ = Wi

u(y , t)→ U0e−2yλ cos(ωt − ωτ y

λ)

In the Knudsen layer y ≤ λ we havePROPAGATING SHEAR(TRANSVERSE) wave.

LIKE IN SOLIDS



Velocity distribution vs τω. Colosqui, VY. LBM-BGKequation.

widthwidth



σ(0, t) =

ρν exp(− t

τ)

∫ t

−∞

∂u(0, λ)

∂yexp(

λ

τ)dλ

τ(7)

ν∂u(0)

∂y= U(−cosωt

δ−+

sinωt

δ+)



Ddissipation rate per unit time perunit area of the plate:

W = −u(0, t)σ(0, t)

W (τ, ω) =1

2

µU2

1 + ω2τ 2(

1

δ−+ωτ

δ+) (8)

(instead of√

p in Newtonian regime.



Dissipation rate per cycle vs τω.

2 4 6 8 10

0.2

0.4

0.6

0.8

1

1.2

1.4



Plane oscillator.

M∂ttx + Sγ∂tx + ω20x = MR(t)

∂ttx +γ

ρphp∂tx + ω2

0x = R(t)



γxt = γu(0, t) = 2ρσ(0, t)/ms

ms = ρph

γ = gρ√ωpν

ρph(1 + ω2τ 2p )

34

×

[(1 + ωτp) cos(1

2tan−1ωτp)−

(1− ωτp) sin(1

2tan−1ωτp)]



In general:

E ≡ W =SU2

2

√ωρµ

2f (ωτ )



f (ωτ ) =1

(1 + ω2τ 2)34

[(1+ωτ ) cos(tan−1 ωτ

2)−

(1− ωτ ) sin(tan−1 ωτ

2)]



We predict:

γ ∝√ω;

1

Q=γ

ω∝ 1/

√ω; γ ∝ √p

γ = const;1

Q∝ ω−1; γ ∝ p



Dissipation. VY, Colosqui (2007); Ekinci, Karabacak(2007).



EXPERIMENTAL DATA.KARABACAK, VY, EKINCI (PRL,

2007).



Normalized fluidic dissipation γn vs pressure. a. Cantilever(53× 2× 460µm); b. beams (230nm × 200nm × 9.6µm);c. (240nm × 200nm × 3.6µm); d. relaxation time τ vs p.



Quality factor Q vs pressure and frequency.



Universality. Large quartz resonator. Ekinci,Karabacak,VY; PRL (2008)

Base

Mountingclips

Bondingarea

ElectrodesQuartzblank

Seal

Pins

D ≈ 1cm



1/Q vs pressure for different ω



1/Q vs 1 ≤ p ≤ 103 ; ≤ 101S/m ≤ 104; nems, mems,macro).



1/Q vs 10 ≤ p ≤ 103 ; ≤ 101S/m ≤ 104;NEMS, MEMS, Quartz (macro).


Summary.0. Solving the BGK equation we predicted a transition fromviscous to visco-elastic behavior at ωτ ≈ 1 in a simple fluid. (VY;Colosqui).1. This effect has been demonstrated on experimental data onresonators and numerically using LBGK simulations (Ekinci;Karabacak;).


2. Size of the device in a vast range 10−4 ≤ L ≤ 1cm does notenter the results represented as a function of x = ωτ . This pointsto the universality which is a much broader class than thewell-known SIMILARITY.3. The transition is due to formation of propagating shear wave inthe Knudsen layer. Is it a new physics emerged from LBM?(Mandelshtam/ Leontovich (chem reactions), Boon (relaxation dueto polymers etc)...

I do not know any work showing thisphenomenon in simple flows. Are you?


3. We have found a UNIVERSALITY in the entire range0 ≤ ωτ ≤ ∞

u = Uφ(r

L,λ

L, ωτ,Re)

with the scaling function f (x) derived from kinetic equation.


4. LBGK simulations agree with experimental results. (non-trivialoutcome).

5. The transition is general U(x , 0) = U sin(ky). (Colosqui et alPhys. Pluids. (January))


6. Bodies of finite size (Colosqui)7. Boundary conditions; slip -no-slip.


Different geometries. Different scales.


Bluff Body. Linear Dimension L. Fig.1a

σ(2)ij ≈ ρλ

2U2/L2

σ(1) + σ(2) ≈ ρνU

L+ ρλ2

U2

L2≈

ρνU

L(1 +

U

cs

λ

L) ≈ ρU

L(1 +

τ

T)

Kn = λ/L; Wi = τ/T

T is a shedding period.


Boundary Layer.

σ(1) ≈ ρν ∂u

∂y≈ ρcsλ

U

δ(x)

σ(2) ∝ ∂ui

∂xα

∂uj

∂xα= 0

σ(2) ≈ ρλ2∂uu∇ · u ≈ ρλ2 U2

xδ

σ(1) + σ(2) ≈ ρcsλU

δ(1 +

λ2

δ2)


In the Stokes problem


Future: 1. This may be a huge area with applications in chemicalengineering, combustion, bio/medical engineering etc. Based onwhat we know LBM may become the most important tool......


TIME The parameter τ i: s a time scale characterizing return of aninitially perturbed system to thermodynamic equilibrium. If, forexample, at time t = 0, a small volume fraction of a gas is locallyperturbed from thermodynamic equilibrium by an excess energy∆E (local heat release, for example), due to intermolecularcollisions or collisions with the walls of the vessel, this localperturbation (inhomogeneity) must disappear on a time-scale τ .The first-principle derivation of this parameter, requiring fulldescription of all microscopic details of interactions is very difficultand, in general, is impossible. It is especially hard if relaxationprocess involves interactions with the solid bodies and excitationsof phonons and other internal degrees of freedom by a gasmolecule impacting the surface. It is clear that in the case of aflow generated by an oscillating solid surface, it is the gas -surfaceinteraction which is responsible for the dissipation process. Forideal gas of the number density n colliding with the wall:


τ ≈ λcs ∝csσn∝ kBθ

32

σp≡ I

p

where the cross-section σ includes all microscopic information. It iseasy to estimate the relaxation time in a bulk of a gas of hardspheres of diameter d where σ ∝ d2. For a nitrogen gas it givesτ ≈ 180× 10−9sec . The relaxation time can experimentally beestablished from the relation ωτ ≈ 1 marking the bending in adissipation curve γ(τω (See Fig. [XX]. The observed dependenceτ ∝ 1/p is shown on Fig. [XX]. Our experimental data giveI ≈ 1850− 1500 for nanocantilevers and nanobeams andI ≈ 400− 500 for the quartz resonators with aluminum - coveredsurface. The fact that all silicon -based nanodevices can becharcterized by the same magnitude of coefficientI ≈ I = 1850× 10−9sec points to the microscopic details ofgas-surface interaction as an important factor.

Oscillating particles in fluids: Theory, experiment and ... · Landau and Lifshitz, "Physical...

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Transcript of Oscillating particles in fluids: Theory, experiment and ... · Landau and Lifshitz, "Physical...