Modeling of a centrifuge experiment: the Lamm equations

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1 Modeling of a centrifuge experiment: the Lamm equations O. Gonzalez and J. Li UT-Austin 29 October 2008

2 Problem statement

3 Setup Consider an experiment in which a 2-component solution (solute + solvent) is separated in a centrifuge. rotor h spin ω cell air r a solute + solvent r b elapsed time air solvent solute t = 0 (cloudy mixture) t >> 0 (separated mixture)

4 Goals Derive model for concentration of solute particles. Use model to estimate shape from concentration data. Begin with simple case of spherical particles. Generalize to particles of arbitrary shape.

5 Simplifying assumptions

6 Assumption 1 In a coord system attached to and rotating with the rotor, assume: (time-averaged) velocity v of a solute particle due to centrifugal force is small and in radial direction, solvent is macroscopically stationary. rotor solute particle solvent v

7 Assumption 2 Assume solute concentration ρ and velocity v depend only on radial coordinate r and time t. z φ h (r, θ,z) ρ = ρ(r, t), v = v(r, t). x y h << 1 φ << 1

8 Assumption 3 Assume particles are spherical and independent, with velocity proportional to centrifugal force (Stokes law). ω solute particle v r solvent f γ, m γ = radius, m = bouyant mass µ = viscosity, T = abs temperature v µ, T v = f 6πγµ, f = m ω 2 r = v = Sω 2 r, S = m 6πγµ.

9 Assumption 4 Assume mass flux J of solute particles across radial surface at r can be decomposed into two parts (diffusion + convection). r J J(r, t) = D ρ + ρv [ Mass Area Time ]. For spherical particles in a fluid D = kt 6πγµ (Stokes-Einstein).

10 Governing equations

11 Setup Consider the region R between two radial surfaces Ω r0 where r 0 < r 1 are arbitrary. and Ω r1 φ Ω r 0 h ra R r0 r1 Ωr 1 rb

12 Conservation of mass For the region R we have ( ) mass of solute t inside of R = ( ) mass flow in to R ( ) mass flow out of R

13 Conservation of mass For the region R we have ( ) mass of solute t inside of R = ( ) mass flow in to R ( ) mass flow out of R r1 t r 0 ρ(r, t) Area(Ω r ) dr = J(r 0, t) Area(Ω r0 ) J(r 1, t) Area(Ω r1 )

14 Conservation of mass For the region R we have ( ) mass of solute t inside of R = ( ) mass flow in to R ( ) mass flow out of R r1 t r 0 ρ(r, t) Area(Ω r ) dr = J(r 0, t) Area(Ω r0 ) J(r 1, t) Area(Ω r1 ) r1 t r 0 ρ(r, t) Area(Ω r ) dr = r 1 r 0 [ ] J(r, t) Area(Ω r ) dr.

15 Conservation of mass For the region R we have ( ) mass of solute t inside of R = ( ) mass flow in to R ( ) mass flow out of R r1 t r 0 ρ(r, t) Area(Ω r ) dr = J(r 0, t) Area(Ω r0 ) J(r 1, t) Area(Ω r1 ) r1 t r 0 ρ(r, t) Area(Ω r ) dr = r 1 r 0 [ ] J(r, t) Area(Ω r ) dr. Substituting Area(Ω r ) = hφr and rearranging gives r1 [ hφr ρ t + ( hφrj) ] dr = 0, r 0 < r 1. r 0

16 Localization Assuming the integrand is continuous, we deduce by the arbitrariness of r 0 and r 1 φhr ρ t + ( ) φhrj = 0, r a < r < r b.

17 Localization Assuming the integrand is continuous, we deduce by the arbitrariness of r 0 and r 1 φhr ρ t + ( ) φhrj = 0, r a < r < r b. Substituting J = D ρ + ρv and v = Sω2 r gives, after cancelling factors ρ t + 1 r ( Sω 2 r 2 ρ rd ρ ) = 0, r a < r < r b.

18 Boundary conditions Solute particles cannot cross surfaces at r a and r b. air r a solute + solvent rb J(r, t) = D ρ + ρv = 0, r = r a, r b.

19 Summary of model For spherical particles in dilute solution we obtain the Lamm equations ( ) ρ t = 1 r rd ρ Sω2 r 2 ρ, r a < r < r b, t > 0 D ρ = Sω2 rρ, r = r a, t > 0 D ρ = Sω2 rρ, r = r b, t > 0 ρ(r, 0) = ρ 0 (r), r a r r b, t = 0 D = kt /6πγµ, S = m /6πγµ. ω r a v r b f v γ, m µ, T

20 Analysis of experiment

21 Problem Deduce particle radius γ from experimental data on solute concentration ρ. rotor air solute + solvent spin ω cell air solvent solute r a r b elapsed time ρ t = 0 ρ t > 0 ρ t >> 0 r a r b r a r f (t) r b r a r b

22 Simplification of Lamm equations In experiments, D Sω 2 ra 2 (ω is large). Thus it is reasonable to ignore D ρ in regions where ρ is moderate (away from r = r b ): ( ) ρ t = 1 r rd ρ Sω2 r 2 ρ, r a < r < r b, t > 0 D ρ = Sω2 rρ, r = r a, t > 0 D ρ = Sω2 rρ, r = r b, t > 0 ρ(r, 0) = ρ 0 (r), r a r r b, t = 0.

23 Simplification of Lamm equations In experiments, D Sω 2 ra 2 (ω is large). Thus it is reasonable to ignore D ρ in regions where ρ is moderate (away from r = r b ): ( ) ρ t = 1 r rd ρ Sω2 r 2 ρ, r a < r < r b, t > 0 D ρ = Sω2 rρ, r = r a, t > 0 D ρ = Sω2 rρ, r = r b, t > 0 ρ(r, 0) = ρ 0 (r), r a r r b, t = 0. The leading-order equations away from r = r b are: ( ) ρ t = 1 r Sω 2 r 2 ρ, r > r a, t > 0 0 = Sω 2 rρ, r = r a, t > 0 ρ(r, 0) = ρ 0 (r), r r a, t = 0.

24 Solution of leading-order equations Using the method of characteristics we obtain, for a general initial condition ρ 0 (r) { 0, r < ra e Sω2 t ρ(r, t) = ) e 2Sω2t ρ 0 (re Sω2 t, r r a e Sω2t. ρ t = 0 ρ t > 0 r a r b r a r f (t) r b

25 Estimation of radius The leading-order solution implies that the location of the moving front is r f (t) = r a e Sω2t. Thus S can be determined from data on the front location r f. ln( r f / r a ) 1 S ω 2 t Thus: experiment S = m 6πγµ particle radius.

26 Generalization

27 Stokes law f ext τ ext fluid particle ω v [ ] v = ω [ ] M1 M 3 M 2 M 4 }{{} M [ ] ext f. τ M R 6 6 Stokes mobility matrix determined by shape of solute particle.

28 Conservation of mass ext f τ ext (r,q) (q,η) Dom Dom x SO 3 Ω x A σ t + g 1 (gj) = 0 in Ω A, t > 0 J n = 0 on ( Ω) A, t 0 J n periodic on Ω ( A), t 0 σ = σ 0 in Ω A, t = 0.

29 Conservation of mass ext f τ ext (r,q) (q,η) Dom Dom x SO 3 Ω x A σ t + g 1 (gj) = 0 in Ω A, t > 0 J n = 0 on ( Ω) A, t 0 J n periodic on Ω ( A), t 0 σ = σ 0 in Ω A, t = 0. σ = mass concentration in config space. J = D σ + σch ext, D = βbmb T, C = bmc. β = Boltzmann factor. g, b, c = geometric factors. M = Stokes mobility matrix. h ext = (f, τ) ext = external loads.

30 Averaging result Result. Due to a time-scale separation, the general balance of mass equation σ t + g 1 (gj) = 0 reduces to the Lamm equation ρ t + 1 r ( Sω 2 r 2 ρ rd ρ ) = 0, where ρ = SO 3 σ dv, D = kt 3 tr(m 1), S = m 3 tr(m 1).

31 Averaging result Result. Due to a time-scale separation, the general balance of mass equation σ t + g 1 (gj) = 0 reduces to the Lamm equation ρ t + 1 r ( Sω 2 r 2 ρ rd ρ ) = 0, where ρ = SO 3 σ dv, D = kt 3 tr(m 1), S = m 3 tr(m 1). Remarks. ρ is usual concentration per unit volume of physical domain.

32 Averaging result Result. Due to a time-scale separation, the general balance of mass equation σ t + g 1 (gj) = 0 reduces to the Lamm equation ρ t + 1 r ( Sω 2 r 2 ρ rd ρ ) = 0, where ρ = SO 3 σ dv, D = kt 3 tr(m 1), S = m 3 tr(m 1). Remarks. ρ is usual concentration per unit volume of physical domain. Result follows by averaging over fast rotational time-scale.

33 Averaging result Result. Due to a time-scale separation, the general balance of mass equation σ t + g 1 (gj) = 0 reduces to the Lamm equation ρ t + 1 r ( Sω 2 r 2 ρ rd ρ ) = 0, where ρ = SO 3 σ dv, D = kt 3 tr(m 1), S = m 3 tr(m 1). Remarks. ρ is usual concentration per unit volume of physical domain. Result follows by averaging over fast rotational time-scale. Recover classic results in case of spherical particles.

34 Averaging result Result. Due to a time-scale separation, the general balance of mass equation σ t + g 1 (gj) = 0 reduces to the Lamm equation ρ t + 1 r ( Sω 2 r 2 ρ rd ρ ) = 0, where ρ = SO 3 σ dv, D = kt 3 tr(m 1), S = m 3 tr(m 1). Remarks. ρ is usual concentration per unit volume of physical domain. Result follows by averaging over fast rotational time-scale. Recover classic results in case of spherical particles. As before, experiment S info on shape.

35 Intuitive explanation In terms of components in the molecular frame, assuming τ ext = 0, we have [ ] [ ] [ ] ext v M1 M = 3 f v = M ω M 2 M 4 τ 1 f ext. centrifuge frame x z y molecular frame

36 Intuitive explanation In terms of components in the molecular frame, assuming τ ext = 0, we have [ ] [ ] [ ] ext v M1 M = 3 f v = M ω M 2 M 4 τ 1 f ext. centrifuge frame x z y molecular frame Let [v] c and [f ext ] c be components in the centrifuge frame, and let Q SO 3 be the rotation from the centrifuge to molecular frame. Then [v] c = [M 1 ] c [f ext ] c, [M 1 ] c = QM 1 Q T.

37 Intuitive explanation Due to fast rotational diffusion, Q wanders uniformly over SO 3. Averaging the previous relation, assuming [f ext ] c is fixed, yields where Since [v] avg c [M 1 ] avg c = [v] avg c = [M 1 ] avg c [f ext ] c, SO 3 QM 1 Q T dv = 1 dv 3 tr(m 1)I. SO 3 [f ext ] c, the molecule behaves as if it were spherical.

38 The End

On the hydrodynamic diffusion of rigid particles

On the hydrodynamic diffusion of rigid particles O. Gonzalez Introduction Basic problem. Characterize how the diffusion and sedimentation properties of particles depend on their shape. Diffusion: Sedimentation: