Solutions Manual Micro and Nanoscale Fluid Mechanics Transport in Microfluidic Devices

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Micro- and Nanoscale Fluid Mechanics: Transport in Microfluidic Devices Cambridge University Press, 2010 First Edition Solution:

Solution Manual, Version Date: July 21, 2010 Brian J. Kirby Sibley School of Mechanical and Aerospace Engineering Cornell University Ithaca, NY http://www.kirbyresearch.com/textbook July 21, 2010

CHAPTER 1. KINEMATICS, CONSERVATION EQUATIONS, AND BOUNDARY CONDITIONS FOR INCOMPRESSIBLE Soln Manual, FLOW July 21, 2010 http://www.cambridge.org/kirby For moving interfaces with uniform surface tension separating Newtonian fluids, the tangential stress is matched on either side of the interface:  η1

1.11

∂ut,1 ∂un,1 + ∂n ∂t



 = η2

∂ut,2 ∂un,2 + ∂n ∂t

 .

(1.77)

Supplementary reading

Modern introductory texts that cover the basic fluid mechanical equations include Fox, Pritchard, and McDonald [17], Munson, Young, and Okiishi [18], White [19], and Bird, Stewart, and Lightfoot [20]. These texts progress through this material more methodically, and are a good resource for those with minimal fluids training. More advanced treatment can be found in Panton [21], White [22], Kundu and Cohen [23], or Batchelor [24]. Batchelor provides a particularly lucid description of the Newtonian approximation, the fundamental meaning of pressure in this context, and why its form follows naturally from basic assumptions about isotropy of the fluid. Texts on kinetic theory [25, 26] provide a molecular-level description of the foundations of the viscosity and Newtonian model. Although the classical fluids texts are excellent sources for the governing equations, kinematic relations, constitutive relations, and classical boundary conditions, they typically do not treat slip phenomena at liquid–solid interfaces. An excellent and comprehensive review of slip phenomena at liquid–solid interfaces can be found in [27] and the references therein. Slip in gas–solid systems is discussed in [3]. The treatment of surface tension in this chapter is similar to that found in basic fluids texts [21, 24] but omits many critical topics, including surfactants. Reference [28] covers these topics in great detail and is an invaluable resource. References [29, 30] cover flows owing to surface tension gradients, e.g., thermocapillary flows. A detailed discussion of boundary conditions is found in [31]. Although porous media and gels are commonly used in microdevices, especially for chemical separations, this text focuses on bulk fluid flow in micro- and nanochannels and omits consideration of flow through porous media and gels. Reference [29] provides one source to describe these flows. Another fascinating rheological topic that is largely omitted here is the flow of particulate suspensions and granular systems, with blood being a prominent example. Discussions of biorheology can be found in [23], and colloid science texts [29, 32] treat particulate suspensions and their rheology.

1.12

Exercises

1.1 In general, the sum of the extensional strains (εxx + εyy + εzz ) in an incompressible system always has the same value. What is this value? Why is this value known? Ans: The sum of extensional strains is zero, because the sum of extensional strains is c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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proportional to the local change in volume of the flow. Thus for incompressible flow, cons of mass indicates that the sum of these strains is zero.

1.2 For a 2D flow (no z velocity components and all derivatives with respect to z are zero), write the components rate tensor ~~ε in terms of velocity derivatives.  of thestrain 1 ∂u ∂u + ∂v 0   ∂x  2 ∂y ∂x    ∂v Ans:  1 ∂u + ∂v 0  2 ∂y ∂x ∂y 0 0 0

1.3 Given the following strain rate tensors, draw a square-shaped fluid element and then show the shape that fluid element would take after being deformed by the fluid flow.   1 0 0 ~ε =  0 −1 0  . (a) ~ 0 0 0   −1 0 0 ~ε =  0 1 0  . (b) ~ 0 0 0   0 1 0 ~ε =  1 0 0  . (c) ~ 0 0 0 Ans: See Fig. 1.24.

Figure 1.24: Fluid deformation owing to strain rate tensor.

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CHAPTER 1. KINEMATICS, CONSERVATION EQUATIONS, AND BOUNDARY CONDITIONS FOR INCOMPRESSIBLE Soln Manual, FLOW July 21, 2010 http://www.cambridge.org/kirby 1.4 The following strain rate tensor is not valid for incompressible flow. Why?   1 1 1 ~~ε =  1 1 1  . 1 1 1

(1.78)

Ans: The sum of the extensional strains is nonzero; thus this violates conservation of mass

1.5 Could the following tensor be a strain rate tensor? If yes, explain the two properties that this tensor satisfies that make it valid. If no, explain why this tensor could not be a strain rate tensor.   1 1 1 ~~ε =  1 0 −1  . (1.79) −1 1 −1 Ans: No. This tensor is not symmetric.

1.6 Consider an incompressible flow field in cylindrical coordinates with axial symmetry (for example, a laminar jet issuing from a circular orifice). The axial symmetry implies that the flow field is a function of and z but not θ. Can a stream function be derived for this case? If so, what is the relation between the derivatives of the stream function and the and z velocities?

Solution: Yes. conservation of mass in axisymmetric coordinates is: ∇ ·~u = 0

(1.80)

1 ∂ ∂ (1.81) u + uz = 0 ∂ ∂z now we need to define a stream function ψ such that, if u and uz are defined in terms of this stream function, conservation of mass is satisfied automatically. If we define ψ such that ∂ψ = uz (1.82) ∂ and ∂ψ = −u , (1.83) ∂z which is similar to what we use for Cartesian coordinates, this will not work. There is still an inside the ∂∂ derivative for the radial term. Instead, try defining ψ such that c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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∂ψ = u . ∂z

(1.84)

1 ∂ ∂ψ ∂ + uz = 0 . ∂ ∂z ∂z

(1.85)

1 ∂ ∂ ∂ uz = − ψ ∂z ∂ ∂z

(1.86)

  ∂ ∂ 1 ∂ uz = − ψ . ∂z ∂z ∂

(1.87)

∂ψ = − uz , ∂

(1.88)

Substituting this in, find

Rearrange to get

and then

From this, we obtain

so we find that ∂ψ = u ∂z

(1.89)

∂ψ = − uz ∂

(1.90)

satisfies the stream function requirements. Also, both of these relations could be multiplied by any constant, and conservation of mass would still be satisfied. 1.7 Consider the following two velocity gradient tensors:   0 1 0 (a) ∇~u =  1 0 0  . 0 0 0   1 0 0 (b) ∇~u =  0 −1 0  . 0 0 0 Draw the streamlines for each velocity gradient tensor. With respect to the coordinate axes, identify which of these exhibits extensional strain and which exhibits shear strain. Following this, redraw the streamlines that have√been rotated √for each on axes √ √ 45◦ counterclockwise, using x0 = x/ 2 + y/ 2 and y0 = −x/ 2 + y/ 2. Are your conclusions about extensional and shear strain the same for the flow once you have rotated the axes? Do the definitions of extensional and shear strain depend on the coordinate system? http://www.kirbyresearch.com/textbook

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CHAPTER 1. KINEMATICS, CONSERVATION EQUATIONS, AND BOUNDARY CONDITIONS FOR INCOMPRESSIBLE Soln Manual, FLOW July 21, 2010 http://www.cambridge.org/kirby Solution: See Fig. 1.25. The first exhibits shear strain and the second extensional strain with respect to the x and y axes. If we redraw with respect to rotated axes, the first exhibits extensional strain and the second shear strain. For isotropic materials, extensional and shear strain are different only in terms of how the shear is aligned with respect to an arbitrary coordinate system, and thus this distinction is relatively unimportant. For anisotropic materials (complex fluids or crystalline solids), the relation of the shear to the orientation of the material is important. 1.8 Write out the components of ρ~u · ∇~u by using Cartesian derivatives. Is ρ~u · ∇~u a scalar, vector, or second-order tensor? Solution: the solution for this problem is not available 1.9 Give an example of a 3D velocity gradient tensor that corresponds to a purely rotational flow. Solution: A purely rotational flow has an antisymmetric velocity gradient tensor; thus any antisymmetric velocity gradient tensor answers this question. 1.10 Consider a differential Cartesian control volume and examine the convective momentum fluxes into and out of the control volume. Show that, for an incompressible fluid, the net convective outflow of momentum per unit volume is given by ρ~u · ∇~u . Solution: Define a Cartesian control volume with sides of differential length dx, dy, and dz. Define the velocity at the center of the control volume equal to~u =(u,v,w). Velocities at the faces are thus given by a first-order Taylor series expansion, which is exact for vanishingly small dx, dy, dz: left face. At the left face, the velocity components are given by u − 21 ∂u ∂x dx, v − 1 ∂v 1 ∂w 2 ∂x dx, and w − 2 ∂x dx. right face. At the right face, the velocity components are given by u + 12 ∂u ∂x dx, v + 1 ∂v 1 ∂w dx, and w + dx. 2 ∂x 2 ∂x bottom face. At the bottom face, the velocity components are given by u − 21 ∂u ∂y dy, ∂v v − 21 ∂y dy, and w − 12 ∂w ∂y dy. c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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top face. 1 ∂v 2 ∂y dy,

At the top face, the velocity components are given by u + 12 ∂u ∂y dy, v +

and w + 12 ∂w ∂y dy.

back face. At the back face, the velocity components are given by u − 12 ∂u ∂z dz, v − 1 ∂w 1 ∂v 2 ∂z dz, and w − 2 ∂z dz. front face. At the front face, the velocity components are given by u + 12 ∂u ∂z dz, v + 1 ∂v 1 ∂w 2 ∂z dz, and w + 2 ∂z dz. At any surface of a control volume, the outgoing flux density of momentum in a coordinate direction is given by the product of the density, the outward-pointing velocity normal to the surface, and the component of velocity in that coordinate direction. The velocity normal to the surface is given by ~u ·~n , where ~n is the unit outward normal. For example, the momentum flux density of x momentum traveling through a surface normal to the z axis is given by ρuw, and the momentum flux density of y momentum traveling through a surface normal to the y axis is given by ρvv. In all cases, the values of u, v, and w are the values at the surface. So, the outward-pointing velocity for the six faces are: left face: −u, right face: u, bottom face: −v, top face: v, back face: −w, front face: w. The momentum fluxes are given by the product of momentum flux densities with the surface area. The areas for the six faces are: left and right faces: dydz, bottom and top faces: dxdz, back and front faces: dxdy. Thus the momentum fluxes are given as follows. For the left face:      1 ∂u −ρ u − 12 ∂u dx u − dx dydz 2 ∂x  ∂x       1 ∂u 1 ∂v  −ρ u − 2 ∂x dx v − 2 ∂x dx dydz       1 ∂w dx w − dx dydz −ρ u − 12 ∂u 2 ∂x ∂x

(1.91)

For the right face:     

   1 ∂u ρ u + 21 ∂u dx u + dx 2 ∂x  dydz ∂x   1 ∂v ρ u + 12 ∂u ∂x dx  v + 2 ∂x dx dydz  ρ u + 21 ∂u w + 21 ∂w ∂x dx ∂x dx dydz

    

(1.92)

For the bottom face:     

   ∂v −ρ v − 12 ∂x dx u − 21 ∂u dx ∂x  dydz   ∂v ∂v −ρ v − 12 ∂x dx v − 12 ∂x dx dydz    ∂v −ρ v − 12 ∂x dx w − 12 ∂w dx dydz ∂x

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    

(1.93)

c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

CHAPTER 1. KINEMATICS, CONSERVATION EQUATIONS, AND BOUNDARY CONDITIONS FOR INCOMPRESSIBLE Soln Manual, FLOW July 21, 2010 http://www.cambridge.org/kirby For the top face:     

   ∂v ρ v + 12 ∂x dx u + 12 ∂u dx ∂x  dydz   ∂v ∂v ρ v + 12 ∂x dx v + 12 ∂x dx dydz    ∂v ρ v + 12 ∂x dx w + 12 ∂w dx dydz ∂x

    

(1.94)

For the back face:     

   1 ∂u dx u − dx −ρ w − 12 ∂w 2 ∂x  dydz ∂x   1 ∂v −ρ w − 21 ∂w ∂x dx  v − 2 ∂x dx dydz  −ρ w − 12 ∂w w − 21 ∂w ∂x dx ∂x dx dydz

    

(1.95)

For the front face:     

   1 ∂u dx u + dx ρ w + 12 ∂w 2 ∂x  dydz ∂x   1 ∂v ρ w + 12 ∂w ∂x dx  v + 2 ∂x dx dydz  ρ w + 12 ∂w w + 21 ∂w ∂x dx ∂x dx dydz

    

(1.96)

The sum of these six sources of momentum flux gives the net outward convective momentum flux. We neglect all terms with the product of two differential lengths, as these are small compared with terms with only one differential length. The sum of the left and right face fluxes is:     ∂u ρ u ∂u dxdydz + u ∂x    ∂x    ∂v ∂u (1.97)  ρ u ∂x + v ∂x dxdydz      ∂w ∂u ρ u ∂x + w ∂x dxdydz For the bottom and top faces, the sum of the fluxes is:    ∂v ρ v ∂u + u ∂y  dxdydz   ∂y  ∂v + v ∂y dxdydz  ρ v ∂v    ∂y ∂v ρ v ∂w ∂y + w ∂y dxdydz For the back and front faces, the sum of the fluxes is:    ∂w ρ w ∂u + u ∂z  dxdydz   ∂z  + v ∂w  ρ w ∂v ∂z dxdydz   ∂z ∂w ρ w ∂w ∂z + w ∂z dxdydz c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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    

(1.98)

    

(1.99)

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Summing together, we get    ∂u ∂u ∂u ∂v ∂w + v + w + u + u + u ρ u ∂u ∂y ∂z ∂x ∂y ∂z  dxdydz   ∂x  ∂v ∂v ∂u ∂v ∂w + v ∂v dxdydz  ρ u ∂x ∂y + w ∂z + v ∂x + v ∂y + v ∂z    ∂w ∂w ∂u ∂v ∂w ρ u ∂w ∂x + v ∂y + w ∂z + w ∂x + w ∂y + w ∂z dxdydz

    

(1.100)

The last three terms of each momentum flux component sum to zero, because ∂u ∂x + ∂w ∂v ∂y + ∂z = 0. Thus the convective momentum flux is given by     ∂u ∂u ρ u ∂u + v + w dxdydz ∂y ∂z    ∂x    ∂v ∂v ∂v (1.101)  ρ u ∂x + v ∂y + w ∂z dxdydz      ∂w ∂w ρ u ∂w ∂x + v ∂y + w ∂z dxdydz and the convective momentum flux per unit volume is     ∂u ∂u ρ u ∂u + v + w ∂y ∂z     ∂x   ∂v ∂v ∂v ρ u + v + w  ∂y ∂z     ∂x  ∂w ∂w ρ u ∂w + v + w ∂x ∂y ∂z

(1.102)

This expression is equal to ρ~u · ∇~u . 1.11 Consider a differential Cartesian control volume and examine the viscous stresses on the control volume. Do not use a particular model for these viscous stresses; simply ~τ visc is known. Show that the net outflow of momentum from these forces assume that~ ~τ visc . is given by ∇ ·~ Solution: the solution for this problem is not available

~ε = η∇2~u if the viscosity is uniform and the fluid is incompressible. 1.12 Show that ∇ · 2η~ ~ε : Solution: start with 2η~  



 ∂ ∂ ∂  , , · 2η  ∂x ∂y ∂z 

∂u ∂x  1 ∂u ∂v + 2  ∂y ∂x  1 ∂u ∂w 2 ∂z + ∂x

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2η~~ε 



60

1 2

(1.103) 

∂u ∂v ∂y + ∂x ∂v  ∂y  1 ∂v ∂w + 2 ∂z ∂y

1 2 1 2



∂u

 ∂z

+ ∂w ∂x

∂v ∂w ∂z + ∂y ∂w ∂z

      

(1.104)

c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

CHAPTER 1. KINEMATICS, CONSERVATION EQUATIONS, AND BOUNDARY CONDITIONS FOR INCOMPRESSIBLE Soln Manual, FLOW July 21, 2010 http://www.cambridge.org/kirby    η 

 2   2  2 ∂ ∂v ∂ u ∂ ∂w 2 ∂∂xu2 + ∂∂yu2 + ∂y + + 2 ∂z ∂x    ∂z2  ∂x2 ∂ ∂u ∂2 v ∂ v ∂ ∂w ∂ v + 2 + 2 2 + ∂z2 + ∂z ∂y  ∂x ∂y ∂x2  ∂y  ∂ w ∂2 w ∂ ∂u ∂ ∂v ∂2 w ∂x ∂z + ∂x2 + ∂y ∂z + ∂y2 + ∂z2

    

(1.105)

from equality of mixed partials and incompressibility, can subtract away half of the terms:    η 

2 2 ∂2 u + ∂∂yu2 + ∂∂zu2 ∂x2 2 2 ∂2 v + ∂∂yv2 + ∂∂z2v ∂x2 2 2 ∂2 w + ∂∂yw2 + ∂∂zw2 ∂x2

  

(1.106)

1.13 For a one dimensional flow given by ~u = u(y), the strain rate magnitude is given ∂u by 21 ∂u ∂y and the vorticity magnitude is given by ∂y . Are the strain rate and vorticity proportional to each other in general? If not, why are they proportional in this case?

Solution: the solution for this problem is not available

1.14 Write out the Navier–Stokes equations in cylindrical coordinates (see Appendix D). Simplify these equations for the case of plane symmetry.

Solution: the solution for this problem is not available

1.15 Write out the Navier–Stokes equations in cylindrical coordinates (see Appendix D). Simplify these equations for the case of axial symmetry.

Solution: the solution for this problem is not available

1.16 Write out the Navier–Stokes equations in spherical coordinates (see Appendix D). Simplify these equations for the case of axial symmetry. c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Solution: the solution for this problem is not available 1.17 For each of the following Cartesian velocity gradient tensors, (1) calculate the strain rate tensor, (2) calculate the rotation rate tensor, and (3) sketch the streamlines for the flow:   0 1 0 (a) ∇u =  1 0 0  . 0 0 0   −1 0 0 (b) ∇u =  0 1 0  . 0 0 0   0 1 0 (c) ∇u =  −1 0 0  . 0 0 0   0 1 0 (d) ∇u =  0 0 0  . 0 0 0  0 1 0 Solution: For ∇u =  1 0 0 , 0 0 0 

 0 1 0 ~~ε =  1 0 0  0 0 0

(1.107)

~~ω = 0 .

(1.108)

 −1 0 0 ~~ε =  0 1 0  0 0 0

(1.109)



and

See Fig. 1.26.   −1 0 0 For ∇u =  0 1 0 , 0 0 0 

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Figure 1.25: Streamlines for case 1 (left) and case 2 (right) drawn with respect to original (top) and rotated (bottom) axes.

Figure 1.26: Flow streamlines.

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and ~~ω = 0 .

(1.110)

~~ε = 0

(1.111)

 0 1 0 ~~ω =  −1 0 0  . 0 0 0

(1.112)

See Fig. 1.27. 

 0 1 0 For ∇u =  −1 0 0 , 0 0 0

and 

See Fig. 1.28.   0 1 0 For ∇u =  0 0 0 , 0 0 0 

1 2

 0 ~~ε =  1 0 0  2 0 0 0 0

(1.113)

and 

0 ~~ω =  − 1 2 0

1 2

 0 0 0 . 0 0

(1.114)

See Fig. 1.29. 1.18 Consider the 2D flows defined by the following stream functions. The symbols A, B, C, and D denote constants. (a) ψ = Axy . http://www.kirbyresearch.com/textbook

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Figure 1.27: Flow streamlines.

Figure 1.28: Flow streamlines.

Figure 1.29: Flow streamlines.

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(b) ψ = 21 By2 . p  (c) ψ = C ln x2 + y2 .
(d) ψ = −D x2 + y2 . For the flow field denoted by each of the preceding stream functions, execute the following: (a) Show that the flow field satisfies conservation of mass. (b) Derive the four components of the Cartesian strain rate tensor ~~ε . (c) Plot streamlines for these flows in the regions −5 < x < 5 and −5 < y < 5. (d) Assume that the pressure field p(x, y) is known for each flow. Derive the four components of the Cartesian stress tensor ~~τ , assuming that the fluid is Newtonian. (e) Imagine that a 5 × 5 grid of lines (see Fig. 1.30) is visualized by instantaneously making a grid of tiny bubbles in a flow of water. Sketch the result if the grid were convected in the specified flow field starting at time t = 0 and a picture of the deformed grid was taken at a later time.

Figure 1.30: A grid that can be used to visualize how a flow deforms.

Solution: http://www.kirbyresearch.com/textbook

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CHAPTER 1. KINEMATICS, CONSERVATION EQUATIONS, AND BOUNDARY CONDITIONS FOR INCOMPRESSIBLE Soln Manual, FLOW July 21, 2010 http://www.cambridge.org/kirby conservation of mass stream function.

All should obey conservation of mass, that is the point of a

∇ ·~u =

∂u ∂v ∂ ∂ ∂ ∂ + = ψ− ψ=0 ∂x ∂y ∂x ∂y ∂x ∂y

(1.115)

Cartesian strain rate tensor Strain rate tensor is defined in 2D Cartesian coordinates as:     ∂u 1 ∂u ∂v + ~ ~ε =   ∂x  2 ∂y ∂x  (1.116) ∂v ∂v 1 ∂u + 2 ∂y ∂x ∂y In terms of stream func:  ~ ~ε = 

∂ ∂ ψ  2 ∂x ∂y 2  1 ∂ ∂ 2 ∂y2 ψ − ∂x2 ψ

1 2



∂2 ∂2 ψ − ∂x 2ψ ∂y2 ∂ ∂ − ∂x ∂y ψ

  

(1.117)

For ψ = Axy: ~~ε =



~~ε =



A 0 0 −A



1 2B



(1.118)

For ψ = 12 By2 :

For ψ = C ln

0 1 2B

(1.119)

0

p  x 2 + y2 :  ~ ~ε = 

−2Cxy (x2 +y2 )2 C(x2 −y2 ) (x2 +y2 )2

C(x2 −y2 ) (x2 +y2 )2 2Cxy (x2 +y2 )2

 

(1.120)


For ψ = −D x2 + y2 : ~~ε =

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0 0 0 0

 (1.121)

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Figure 1.31: Streamlines for some simple flows. Source: simpleflows.m.

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CHAPTER 1. KINEMATICS, CONSERVATION EQUATIONS, AND BOUNDARY CONDITIONS FOR INCOMPRESSIBLE Soln Manual, FLOW July 21, 2010 http://www.cambridge.org/kirby streamlines. see Fig. 1.31. stress tensor. Stress tensor is given by: ~~τ = 2η~~ε − p~~δ In terms of stream func: 



∂ ∂ 2η ∂x ∂y ψ

~ ~ε = 

 η

(1.122)

η

∂2 ∂2 ψ − ∂x 2ψ ∂y2



∂2 ∂2 ψ − ∂x 2ψ ∂y2 ∂ ∂ −2η ∂x ∂y ψ

  

(1.123)

For ψ = Axy: ~ ~τ =



2ηA − p 0 0 −2ηA − p

 (1.124)

For ψ = 12 By2 : ~~τ =

For ψ = C ln



−p ηB ηB −p

 (1.125)

p  x 2 + y2 :  ~ ~τ = 

−4ηCxy −p (x2 +y2 )2 2ηC(x2 −y2 ) (x2 +y2 )2

2ηC(x2 −y2 ) (x2 +y2 )2 4ηCxy −p (x2 +y2 )2





 

(1.126)


For ψ = −D x2 + y2 : ~~τ =

−p 0 0 −p

(1.127)

grid deformation. See Figures 1.32 through 1.35. 1.19 Create an infinitesimal control volume in cylindrical coordinates with edge lengths d , dθ, and dz. Use the integral equation for conservation of mass: ∂ ∂t

Z

ρdV = −

V

Z

(ρ~u ) · nˆ dA ,

(1.128)

S

where V is the volume of the control volume, to derive the incompressible continuity equation in cylindrical coordinates. c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Figure 1.32: Deformation of a grid by a stagnation flow, ψ = Axy. In general, grid spacing will be closer near origin and larger far from origin; this is not terribly clear from the sketch.

Figure 1.33: Deformation of a grid by a pure shear flow, ψ = 12 By2 .

Figure 1.34: Deformation of a grid by an irrotational vortex, ψ = C ln

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p  x2 + y2 .

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CHAPTER 1. KINEMATICS, CONSERVATION EQUATIONS, AND BOUNDARY CONDITIONS FOR INCOMPRESSIBLE Soln Manual, FLOW July 21, 2010 http://www.cambridge.org/kirby Solution: Evaluate~u ·~n and the area on all six faces: (a) bottom face: ~u ·~n = uz ; dS = dθd ∂ (b) top face: ~u ·~n = −uz − ∂z uz dz; dS = dθd

(c) front face: ~u ·~n = uθ ; dS = d dz ∂ uθ dθ; dS = d dz (d) back face: ~u ·~n = −uθ − ∂θ

(e) left face: ~u ·~n = u ; dS = dθdz (f) right face: ~u ·~n = −u − ∂∂ u d ; dS = ( + d ) dθdz Plug these into the equation: d (ρV ) = 0 = dt

Z

ρ~u ·~n dS

(1.129)

S



      ∂ ∂ ∂ 0 = ρ dθd uz − uz − uz + dθdzu + ( + d )dθdz −u − u + d dz uθ − uθ − uθ dθ ∂z ∂ ∂θ (1.130) divide by ρ d dθdz: 0=−

d d d 1 ∂ ∂ u u − u − uθ uz − − d d ∂θ ∂z

(1.131)

Neglect terms proportional to d / , and divide by -1: 0= Note that

u

d 1 ∂ ∂ u u + uθ uz + + d ∂θ ∂z

+ dd u is equal to 1/ times 0=

d d

(1.132)

( u ):

1 d 1 ∂ ∂ ( u )+ uθ + uz d ∂θ ∂z

(1.133)

1.20 Using thermodynamic arguments, derive the Young–Laplace equation Eq. (1.41). Solution: the solution for this problem is not available

1.21 Use trigonometric and geometric arguments to deriveEq. (1.48). c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Solution: By definition, the line from the center of curvature to the triple point is normal to the line tangent to the interface at that point. Thus α = θ (see Fig. 1.36). From the triangle with (a) the triple point, (b) the center of curvature, and (c) the centerline of the capillary, we can observe that cos θ =

d/2 , R

(1.134)

and thus R = d/2 cos θ. 1.22 Consider a capillary of diameter d oriented along the y axis and inserted into a reservoir of a fluid. Assume the surface tension of the liquid–gas interface is given by γlg . At the interface, the radius of curvature R can be assumed uniform everywhere in the xz plane (i.e., the interface is spherical) if the variations in the local pressure drop across the interface are small compared with the nominal value of the pressure drop across the interface. (a) Write a relation for the pressure drop across the interface as a function of γlg and R. (b) As a function of θ, evaluate the difference in height of the fluid at the center of the capillary with respect to the fluid at the outside edge of the capillary, and thus evaluate the difference in hydrostatic head between the center and edge of the capillary. (c) The criterion for approximating the interface as spherical is that the pressure drop variations from center to edge are small compared with the pressure drop itself. Determine the maximum value for d for which p the interface can be assumed spherical. Your result will be of the order of γlg /ρg. Solution: pressure drop across interface The Young–Laplace equation gives

∆P =

2γlg . R

(1.135)

height difference from center to edge At the center, the distance from the liquid column to the center of curvature of the interface is R. At the edge, the distance from the liquid column to the height of the center of curvature is R sin θ. Thus the difference in height is ∆h = R(1 − sin θ) . (1.136) http://www.kirbyresearch.com/textbook

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Figure 1.35: Deformation of a grid by solid body rotation, ψ = −D x2 + y2 .

Figure 1.36: Geometry of interface in a capillary.

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pressure difference from center to edge The difference between pressures at center and at edge is ∆P = ρgR(1 − sin θ)

(1.137)

criterion for spherical interface shape The hydrostatic head difference between center and edge needs to be small compared with the interfacial pressure drop. Thus ρgR(1 − sin θ)
10a, we obtain ξ
0. Assume the electric field varies linearly along the length L (E = E0 x/L), and complete the following exercises: (a) Describe the cross-section-averaged velocity field if the fluid has viscosity η and the surface has electrokinetic potential ζ. (b) Consider only cross-section-averaged velocities, and describe the equilibrium position, if any, of an electrolyte with an electrophoretic mobility µEP . (c) Consider the dispersion induced by the full flow field, and describe the width of the distribution of an analyte as a function of its valence. (d) Now consider that the electric field is changed as a function of time. Assume that two analytes with different electrophoretic velocities (not controlled by the user) are to be separated and held at x = 0.3L and x = 0.7L and the goal of the system is to minimize the width of the distribution of these analytes. Assume the user can observe the location of the analytes at any time. Design an electric field distribution as a function of time that achieves this goal. Solution: the solution for this problem is not available 12.6 Gradient-elution, moving-boundary electrophoresis causes analytes to be transported through a microchannel with time as a function of the analyte electrophoretic mobility. Consider a 1-mm-long microchannel of circular cross section aligned along the x axis connecting two large reservoirs, the leftmost of which has a set of five cationic electrolytes at 0.1-mM concentration each with electrophoretic mobilities of 0.2, 0.5, 1.0, 1.5, and 2.5×10−8 m2 /V s. An adverse ( dp dx > 0) pressure gradient is applied, c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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generating flow from the rightmost reservoir to the leftmost reservoir. A positive electric field (E = 250 V/cm) is also applied, which generates ion migration in the positive x direction. dp (a) Let dp dx be reduced linearly with time for one minute, expressed by dx = (60 − t) × 1 × 103 Pa. Describe which of the analytes is found in the microchannel as a function of time.

(b) What is the conductivity of the microchannel as a function of time? Solution: the solution for this problem is not available

12.7 In a variable-temperature system, the conductivity of a buffer system varies owing to two primary phenomena: (1) the reduced viscous friction coefficient of the water at high temperatures allows the mobility (and therefore molar conductivity) of the ions to increase with temperature; and (2) increased temperature causes chemical reactions that increase or decrease the number of ions in solution. Assume a buffer system is used to control the pH of the system containing a number of proteins that are to be analyzed. Assume that the concentration of the buffer is large relative to the concentration of the protein analytes. Model ion electrophoretic mobilities as increasing by a factor α per degree away from their nominal value at 0◦ C (µEP /µEP,0 = 1 + αT ). Assume that the conductivity of the buffer system increases an additional β per degree C owing to increased ion concentrations: σ/σ0 = 1 + (α + β)T .

(12.21)

Take σ0 = 2 S/m, where 1 S = 1/Ω. Consider a microchannel of circular cross section and radius a connected to electrical heaters such that a uniform temperature gradient is established in the microchannel from T = T1 at x = 0 to T = T2 at x = L (assume that this temperature gradient is unperturbed by any fluid flow in the system). A current I is applied from x = 0 to x = L. Ignore electroosmosis. See Fig. 12.16.

Solution: See Ross and Locascio Anal. Chem. 2002, 74, 2556-2564 for the original source of a design of a system similar to that described above. (a) Derive the relation for the x variation of electric field in the microchannel as a function of the experimental parameters. http://www.kirbyresearch.com/textbook

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Solution: The relation for the temperature is x x T = T1 + (T2 − T1 ) = T1 + ∆T . L L The current is i = σE =⇒ I = σEA ,

(12.22)

(12.23)

and thus the electric field is I I . = σA σπa2 Writing the conductivity in terms of the temperature, we get E=

σ(T ) = σ0 [1 + (α + β)T ] .

(12.24)

(12.25)

We define γ = α + β, so that  x σ(T ) = σ0 (1 + γT ) =⇒ σ(x) = σ0 1 + γT1 + γ∆T . L Continuity of current then implies that E=

1 I
. πa2 σ0 1 + γT1 + γ∆T Lx

(12.26)

(12.27)

(b) Derive the relation for the x variation of electrophoretic velocity of an analyte uEP as a function of its µEP,0 and experimental parameters. Solution: The electrophoretic mobility of an analyte is uEP = µEP E . In terms of the location x, the electrophoretic mobility is  x µEP = µEP,0 (1 + αT ) = µEP,0 1 + αT1 + α∆T . L The analyte electrophoretic velocity is thus   IµEP,0 1 + αT1 + α∆T Lx uEP = 2 . πa σ0 1 + γT1 + γ∆T Lx

(12.28)

(12.29)

(12.30)

(c) Now consider that a pressure gradient dp dx is applied to the microchannel. For dp a given dx and a given x, calculate the µEP,0 required to lead to a net analyte 4 velocity of zero. Calculate and plot this result for a = 10 µm, dp dx = 1×10 Pa/m, ◦ ◦ ◦ η = 1 mPa s, I = 15 µA, α = 0.02/ C, β = 0.03/ C, T1 = 30 C, T2 = 70◦ C, and L = 5 cm. c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Solution: For Hagen–Poiseuille flow, we have u=

−a2 dp . 8η dx

(12.31)

For zero net analyte velocity, the sum of the mean fluid flow plus the electrophoretic velocity is zero:   −a2 dp IµEP,0 1 + αT1 + α∆T Lx . (12.32) 0= + 8η dx πa2 σ0 1 + γT1 + γ∆T Lx From this, we can write a relation for the reference electrophoretic mobility, πa4 σ0 dp µEP,0 = 8ηI dx



1 + γT1 + γ∆T Lx 1 + αT1 + α∆T Lx

 .

(12.33)

(d) From the previous result, we can assume that analytes are concentrated at an x = x0 such that they stagnate (i.e., their net velocity is zero). Assume that, at equilibrium, the distribution of an analyte around its stagnation point x0 is given by a Gaussian distribution: c(x) = A exp[−B(x − x0 )2 ]. Write the 1D transport equation for the cross-sectional average concentration c, and determine the FWHM width w of the Gaussian distribution at equilibrium. Given the parameters specified previously, calculate and plot w(x). Solution: The 1-D Nernst–Planck equation is d2 d (uc) = Deff 2 c , dx dx

(12.34)

or, expanding using the product rule, u

d d d2 c + c u = Deff 2 c . dx dx dx

(12.35)

d We define δx = x − x0 and linearize u around x0 by defining u = ∆x dx u:

  d d2 d u c + δx c = Deff 2 c . dx dx dx

(12.36)

d We then plug in a Gaussian form for c and show that B = − 12 dx u/Deff , thus

c = A exp

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δx2 d 2Deff / dx u

! .

(12.37)

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We find FWHM by doubling the δx at which c = A/2: s w(x) =

8

Deff 1 ln . d 2 dx u

(12.38)

From the µEP,0 values that can be focused in this device and the Nernst equation: D=

RT µEP , F

(12.39)

we estimate D ' 2e − 12 m2 /s. From u, we can estimate Pe ' ua/D = 600. From Taylor dispersion relations, i.e.,   Pe2 Deff = D 1 + , (12.40) 48 d u varies from -8×10−4 to -2×10−4 ; here we find Deff = 16 × 10−9 . The local dx we use -4×10−4 to approximate the FWHM.

(e) A detector that measures the local concentration of analytes at any location x 4 is placed at x = 0. dp dx is varied linearly with time (from 0 to 5 × 10 Pa/m) and the signal at the detector is recorded. Qualitatively describe how this signal will vary with time and what the meaning of this signal will be in terms of the analytes in the system. Solution: Analytes will “focus” only when µEP,0 is positive. By looking at x = 0 and changing dp dx , we will see a time history of the concentration of analytes with µEP,0 that leads to zero velocity at x = 0. Thus the time history of the signal will be the reverse of an electropherogram—first signal will indicate slow-moving species, then later signal will indicate fast-moving species. 12.8 You are considering making an electrophoretic separation device out of a glass wafer with etched channels. To complete this device you must make some cover and affix it to your etched glass. You are choosing among glass, alumina, and PMMA. Based strictly on transport phenomena (not cost or ease of fabrication or other matters), which of these three cover materials will lead to the best system performance? Explain your answer briefly and qualitatively. Solution: glass. If top and bottom are of the same material, there will be no dispersion.

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12.9 Model a protein as effectively having two charge sites: an amine, which is neutral at high pH but can gain a proton in a reaction with pKa = 8; and a carboxylic acid, which is neutral at low pH but can lose a proton in a reaction with pKa = 4. Assume T = 25◦ C. (a) Using the Henderson–Hasselbach equation, calculate the average total charge of the protein as a function of pH. Plot this relation and determine the isoelectric point, i.e., the pH at which the total charge is zero. Although a molecule cannot have an instantaneous partial charge, molecules will gain and lose charge quickly enough that, from the standpoint of time-averaged molecular motion, they will act as if they had a partial charge. (b) Assume that this protein is inserted into a microchannel used for an IEF separation. Assume the channel is filled with an ampholyte mixture such that the pH ranges linearly in space from pH 3 to pH 10, where pH = 3 is at x = 0 and pH = 10 is at x = 1 cm. An electric field is applied with a magnitude of 100 V/cm. What must the sign of the electric field be to ensure that the protein will concentrate at a specific location in this microchannel? Where will the protein stabilize if the proper sign of electric field is applied? (c) Consider the 1D Nernst–Planck equations for the distribution of the concentration of the protein. Linearize the electrophoretic mobility of the protein around the isoelectric point to simplify the math. Given this approximation, write the equation that the concentration profile must satisfy in steady state. (d) Your equation for the steady-state concentration profile will be satisfied by a distribution with Gaussian form. What is the half-width at half-maximum for the concentration distribution? That is, at what distance from the isoelectric point has the concentration dropped to one half of the concentration at the isoelectric point? (e) How would the half-width at half-maximum change if the electric field were doubled? (f) How would the half-width at half-maximum change if the pKa ’s of the two reactions were 6.5 and 5.5 instead of 8 and 4? Solution: Protein charge. For the amine, the reaction is NH2 + H+ ↔ NH+ 3 .

(12.41)

The Henderson–Hasselbach equation is [NH2 ] pH − pKa = pH − 8 = log10  +  . NH3 http://www.kirbyresearch.com/textbook

(12.42)

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This can be rearranged to show  + NH 1  + 3 = . NH3 + [NH2 ] 1 + 10pH−8

(12.43)

The term on the left is the positive charge (per amine) as a function of pH. For the carboxylic acid, the reaction is COO− + H+ ↔ COOH .

(12.44)

The Henderson–Hasselbach equation is   COO− pH − pKa = log10 . [COOH]

(12.45)

This can be rearranged to show [COOH] 1   − = 1 + 10pH−4 . [COOH] + COO

(12.46)

  COO− 10pH−4   . = 1 + 10pH−4 [COOH] + COO−

(12.47)

or, equivalently,

The left-hand side of the previous equation is the negative charge (per carboxylic acid) as a function of pH. Thus the total charge is 1 10pH−4 − . 1 + 10pH−8 1 + 10pH−4

(12.48)

The charge is zero at pH=6. See Fig. 12.17. Field sign. The electric field must be positive. Positive ions must move toward the high pH for stability. Positive electric field means the potential is high to the left. The pH as a function of space is given by pH = 3 + 700x; the protein stabilizes where the pH is 6, i.e., at x = 4.3 × 10−3 .

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(12.49)

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Figure 12.16: Schematic of a channel with a temperature gradient and an applied current.

Figure 12.17: Charge as a function of pH for a molecule with one amine and one carboxylic acid, with pKa values specified in the legend.

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The profile must satisfy the 1-D Nernst–Planck equations: d d2 (12.50) (uc) = D 2 c , dx dx d d d2 u c+c u = D 2 c. (12.51) dx dx dx If we define ∆x = x−xstagnation and linearize u around xstagnation by defining u = −A∆x, where A is a constant that comes from the 1st order Taylor series expansion around the stagnation point from the previous calculations, we obtain u

d d2 c − Ac = D 2 c . dx dx

(12.52)

We typically find solutions in Gaussian form, so we will use a Gaussian form for c:   x2 (12.53) c = C exp − 2 . x0 Given this Gaussian form, we can calculate the derivatives of c, namely:   −2xC x2 −2xc d c = exp − 2 = 2 dx x0 x0 x0 2

(12.54)

and       d2 −2C x2 x2 −4x2C −2 4x2 c = exp − 2 + = c exp − 2 + + 4 . (12.55) dx2 x0 2 x0 x0 x0 2 x04 x0 Plugging into the steady 1D Nernst–Planck equations, we find   −2 4x2 2Ax2 0 = Dc + Ac − + c, x0 2 x0 2 x04

(12.56)

which simplifies to  0=

   −2D 4D 2A 2 + A − x . + x0 2 x0 2 x04

(12.57)

This is satisfied by 2D , (12.58) A which shows that the width of the equilibrium distribution is related to the ratio of the effect that broadens the distribution (diffusion) to the effect that narrows it (the spatial gradient of the electrophoretic velocity field). Now, recall that u = −Ax = µEP (x)E. Because D/RT = µEP /zF, we can say that x0 2 =

u = −Ax = c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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EFDz(x) , RT

(12.59)

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which means that if we define z(x) = −Bx (we linearized the velocity, which is equivalent to linearizing the charge per ion), we have A = BEFD/RT . Thus x0 2 =

2RT . BEF

(12.60)

The concentration is half the peak when   x2 1 exp − 2 = , x0 2

(12.61)

p which happens at |∆x| = |x0 | ( ln 2). Thus |x0 | HWHM = √ = ln 2

r

4RT . BEF ln 2

(12.62)

B is found by linearizing the charge relation as a function of space. For pKa ’s of 4 and 8, the charge near pH = 6 changes -.045 per pH unit or, for this geometry, -31.6 per meter (evaluated numerically). Thus B = 31.6 m−1 . Plugging in E = 1 × 104 V/m, we get that HWHM = 428 µm .

(12.63)

If the pKa ’s are 5.5 and 6.5, then B = 588 m−1 and the HWHM is 111 µm. The reason why the distribution is tighter is because the charge per ion is changing more strongly with pH and thus the stabilizing√effects are more spatially located. If the field is doubled, the width goes down by 2. . 12.10 Consider an electrophoretic separation. For dispersionless transport, how does the separation resolution depend on the length of the separation channel and the voltage applied across the channel?

Solution: the solution for this problem is not available

12.11 Consider pressure-driven flow through a porous material for an HPLC separation. Design the geometry of a channel required to fabricate a 1-m pathlength in a 1 cm × 1 cm footprint device. http://www.kirbyresearch.com/textbook

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Solution: the solution for this problem is not available 12.12 Consider flow through a long channel of uniform circular cross section with radius R = 10 µm. The mean velocity of the fluid is measured by use of photobleaching with a focused laser beam to define a fluid bolus of thickness with a Gaussian distribution and a FWHM of w = 10 µm at t = 0 and x = 0, and imaging the concentration of bleached fluid as a function of time. The velocity is inferred by comparing the x locations of the peaks of the -averaged bleached fluid distributions at t = 0 and t = 2 s. This experiment constitutes a 1D SIV measurement of the fluid flow and the resulting images will be reminiscent of the images in Fig. 11.4. The ability to resolve the position of a distribution of a scalar is a function of the signal-to-noise ratio of the experiment. Assume that the signal-to-noise ratio of the experiment is such that the location of the peak of the fluid distribution can be measured with an accuracy of 30% of the FWHM of the Gaussian distribution. What is the accuracy of the velocity measurement when this technique is used? Solution: the solution for this problem is not available

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13.3 An approximate relation for Henry’s function for an infinite cylinder aligned perpendicularly to the applied electric field is given above in relation (13.30). If the cylinder is aligned with the field, Henry’s function is equal to one because there is no distortion of the field. It can be shown (though we do not do so here) that the timeaveraged electrophoretic mobility of an infinite cylinder aligned at randomly varying orientation is equal to the average of the electrophoretic mobilities of the cylinder aligned in each of three orthogonal directions. Given this, derive a relation for the time-averaged Henry’s function of an infinitely long cylindrical particle oriented randomly. Plot approximate results for the Henry function for the following, all on one graph, as a function of a∗ : (a) a cylinder aligned perpendicular to the E field, (b) a cylinder aligned parallel to the E field, (c) a cylinder aligned randomly with respect to the E field, (d) a spherical particle, i.e., relation (13.27). Plot a∗ on a log scale from 1 × 10−2 to 1 × 103 . Solution: See Figure 13.11.

13.4 Plot the electric field lines around an infinite cylinder aligned perpendicular to the field (i.e., field lines around a circle in two dimensions). Recall from the analytical solution that the electric field at the edge of the particle is double the bulk electric field. Draw circles at = 1.1a, = 2a, and = 10a, which nominally denote the outer edge of the double layer for a∗ = 10, 1, and 0.1, respectively. Comment qualitatively on the fraction of the double layer that sees the high-E region around the circumference of the sphere in these three cases, and relate your observations qualitatively to the approximate Henry function for these three a∗ values. Solution: In this case, the field is significantly different from the freestream value at 1.1a and at 2a. In the 1.1a case, the field is approximately equal to that at the surface. At 10a, the field is not significantly different from the freestream. In the first two cases, the double layer sees the increased field, but, in the third case, the double layer for the most part just sees the bulk electric field. Thus we expect that the electrophoretic mobilities in these cases will be different.

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SolnCHAPTER Manual, July 21, 13.2010PARTICLE

ELECTROPHORESIS

Figure 13.10: Electrophoretic mobilities for cylinders and spheres.

Figure 13.11: Electrophoretic mobilities for cylinders and spheres.

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13.5 Here we use a matched asymptotic technique to estimate Henry’s function for a cylinder in electrophoresis. Consider a 2D electroosmotic flow induced by a fixed and motionless glass cylinder with radius a centered at the origin when an electric field of E∞ in the x direction is applied. Assume that the electrokinetic potential ζ is negative and small in magnitude and the electrolyte solution leads to a Debye length of λD . We can calculate the electrical potential field around an insulating cylinder by superposing the electrical potential field for a uniform field, φ = −E∞ x ,

(13.46)

with the electrical potential field for a line dipole, φ = −E∞ x

a2 2

,

(13.47)

to obtain the total potential field:   a2 φ = −E∞ x 1 + 2 .

(13.48)

(a) Recall that the outer solution for the electroosmotic velocity for uniform properties and simple interfaces is given by ~u outer = −

εϕ0 ~ E. η

(13.49)

Derive a relation for φv (x, y) for the outer solution for the electroosmotic flow. (b) For simplicity, consider only the line corresponding to x = 0. Define an outer ∗ variable y∗outer = y−a a and a normalized velocity~u = −~u η/εζE∞ . (c) Recall that the inner solution for the electroosmotic velocity is given by uinner =

εEwall (ϕ − ϕ0 ) . η

(13.50)

Here Ewall is the (assumed uniform) magnitude of the tangent electric field at the wall, obtained by evaluation of the electric field given by Eq. (13.48) at = a. Consider again the line corresponding to x = 0 and define an inner variable ∗ ∗ y∗ inner = y−a λD . Write the expression for uinner (y inner ). (d) Construct a composite asymptotic solution using the multiplicative formula ucomposite (y) = uinner (y) ∗ uouter (y)/uouter (y = a) .

(13.51)

This multiplicative relation does not rigorously satisfy all of the governing equations, but it is a mathematical tool for creating an analytical solution that transitions smoothly from the inner solution (which is rigorously correct to the extent c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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that the local electric field tangent to the wall is uniform) to the outer solution (which is rigorously correct to the extent that the charge density is zero). Whereas an additive composite asymptotic will give good answers only for thin double layers, the multiplicative composite expansion will give good answers in all cases. (e) Consider a∗ = 0.1, a∗ = 10, and a∗ = 1000, where a∗ = a/λD . For each value of a∗ , make a plot of u∗composite , u∗inner , and u∗outer . For your independent variable, use yinner ranging from 1 × 10−3 to 1 × 109 on a log axis. (f) For several values of a∗ ranging from a∗ = 0.1 to a∗ = 1000, numerically evaluate the maximum value that the composite expansion reaches. Normalize your results for the maximum velocity by the value you obtain for a∗ → ∞. Plot these on a semilog axis over the range from a∗ = 0.1 to a∗ = 1000 and compare your result (on the same graph) with the values for f given by relation (13.30). (g) Use your results to explain Henry’s function physically. How do the size of the double layer and the size of the cylinder interact to make the electrophoretic velocity a function of the particle size? What electric field do the ions in the double layer see in the case of (a) thin double layers, (b) thick double layers, or (c) a = λD ? Solution: velocity potential:   εζ a2 φv = − E∞ x 1 + 2 η

(13.52)

outer solution: u∗ = 1 +

1 (y∗

outer + 1)

2

(13.53)

inner solution: u∗inner (y∗ inner ) = −

εEwall ζ [1 − exp (−y∗ inner )] η

(13.54)

composite solution: See Figures 13.12– 13.14. http://www.kirbyresearch.com/textbook

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Figure 13.12: Inner, outer, and composite asymptotics for the x velocity induced by electroosmosis around a cylinder for a∗ = 1000. In the thin-EDL limit, the double layer is entirely in the region where the local electric field is twice the bulk electric field.

Figure 13.13: Inner, outer, and composite asymptotics for the x velocity induced by electroosmosis around a cylinder for a∗ = 10. Here, the double layer is thick enough that much of the double layer samples regions where the local electric field is lower than that at the wall. Hence, the electrostatic force on the net charge density is lower, and the resulting velocity is lower.

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Figure 13.14: Inner, outer, and composite asymptotics for the x velocity induced by electroosmosis around a cylinder for a∗ = 0.1. Here, the double layer is so thick that most of the double layer exists far from the cylinder, where the electric field is equal to the electric field in the bulk. Relative to the thin double layer example, the net charge density here sees an electric field one-half as strong, and thus the particle velocity is one-half as big.

Figure 13.15: Comparison between Eq. (13.30) and the normalized maximum velocity in the composite asymptotic expansion. Although these results differ by 10% or so, they qualitatively match. The differences are caused by errors caused by the multiplicative composite expansion—this is an approximate technique. Also, using the result for x = 0 alone does not completely describe the flow field, leading to a subtle difference.

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comparison to Henry function: See Fig. 13.15. Physical explanation: See captions of Figures 13.12–13.15. 13.6 Survey the literature and identify the approximate range of observed zeta potentials for (a) mammalian cells, (b) bacterial cells, and (c) virions. Solution: the solution for this problem is not available 13.7 Show that the electrophoretic velocity of a long cylinder moving along its axis of symmetry is independent of the Debye length. Explain whether such a cylinder will have a preferred orientation if an electric field is applied in a random direction. Solution: the solution for this problem is not available

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14.3 Presume that you have a strand of DNA with a well-defined persistence length `p . Consider the mean end-to-end length h`e i, which can be written as

 h`e i2 = [~r (s = `c ) −~r (s = 0)] 2 . (14.67) Thus h`e i can be found by integrating the angles:   Z `c Z `c ∂~r (s1 ) ∂~r (s2 ) 2 h`e i = · ds2 . ds1 ∂s ∂s s2 =0 s1 =0

(14.68)

Evaluate the mean end-to-end length h`e i by evaluating this integral for a DNA polymer with a well-defined `p . Simplify your result for h`e i in the limit where `p  `c (a long flexible polymer) and also in the limit where `p  `c (a short, rigid rod). What value of `K must be used in an ideal polymer model to make the end-to-end length predicted by the ideal polymer model match your result for `p  `c ? Solution: The final result is    `c h`e i = 2`p `c − 2`p 1 − exp − . `p 2

2

(14.69)

p 2`p `c .

(14.70)

If `p  `c , we find h`e i =

If `p  `c , we find h`e i = `c .

(14.71)

√ The ideal polymer model ppredicts that `e = `K `c , so for the ideal polymer model to match our result h`e i = 2`p `c above, we must choose `K = 2`p .

(14.72)

14.4 Plot h`e i as a function of Nbp for Y I = 2 × 10−28 N m2 . Use the range Nbp = 100 to Nbp = 10, 000 and plot on a log–log scale. Compare this result with the equivalent plot for a Gaussian chain polymer with `K = 100 nm. http://www.kirbyresearch.com/textbook

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Solution: See Figure 14.16. Note the theories agree at large Nbp but differ at small Nbp .

14.5 In the bead–spring model of DNA, the probability density function for `e is given by Eq. (14.39): " # 32 ! 2 1 ` e

 P (`e ) = 4π`e 2 exp − 2 . (14.73) 4π rg 2 4 rg Evaluate the mean value of `e by integrating P (`e )`e from `e = 0 to `e = ∞. Show that the result corresponds to Eq. (14.38): √  h`e i = 6 rg . (14.74) Solution: the solution for this problem is not available 14.6 In the bead–spring model of DNA, the probability density function for ∆r is given by Eq. (14.41): " # 32 ! 2 ` ` ∆r c

c 

 P (∆r) = 4π∆r2 exp − . (14.75) 4π∆s rg 2 4∆s rg 2 Evaluate the mean value of ∆r by integrating P (∆r)∆r from ∆r = 0 to ∆r = ∞. Show that the result corresponds to Eq. (14.40): r 6∆s  h∆ri = rg . (14.76) `c Solution: the solution for this problem is not available 14.7 Confirm the correctness of the result for the mean value of ∆r for a Gaussian bead– spring chain

 by showing that the mean-square displacement of polymer segments is equal to rg . That is, show that 1 `c 2

Z s2 =`c Z s1 =s2

s2 =0

s1 =0



 |~r (s2 ) −~r (s1 )| 2 ds1 ds2 = rg .

(14.77)

 A general conclusion stemming from this sort of analysis is that rg = h`√e6i for any linear polymer whose probability density function for ∆r is Gaussian. This relationship breaks down, though, for short DNA molecules that are not Gaussian but rather well approximated by a rigid rod. c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Solution: Starting with the definition of the radius of gyration, 1 `c 2

Z s2 =`c Z s1 =s2

s2 =0

1 `c 2

s1 =0



 |~r (s2 ) −~r (s1 )| 2 ds1 ds2 = rg .

 Z s2 =`c Z s1 =s2 6(s2 − s1 ) rg s2 =0

s1 =0

`c

 ds1 ds2 = rg .

Z s2 =`c s 2

 6 rg 2 ds = rg . 2 `c 2 s2 =0 2`c

Z s2 =`c s 2

 6 rg 2 ds2 = rg . 2 `c s2 =0 2`c

 3

 6 rg `c = rg . 3 6`c

  rg = rg .

(14.78) (14.79) (14.80) (14.81) (14.82) (14.83)

 14.8 Evaluate rg for a polymer that is well approximated by an infinitely rigid linear rod

 of length `c (i.e.,h`e i = `c ), and show that rg 6= h`√e6i in this case. Given the same

 h`e i, does a rigid rod have a larger or smaller rg compared with a freely jointed chain? How about for the same `c ? Solution:

Z Z

2  1 s2 =`c s1 =s2

|~r (s2 ) −~r (s1 )| 2 ds1 ds2 . rg = 2 `c s2 =0 s1 =0

(14.84)

Z Z

2 1 s2 =`c s1 =s2 rg = 2 (s2 − s1 )2 ds1 ds2 . `c s2 =0 s1 =0

(14.85)

Z Z

2 1 s2 =`c s1 =s2 2 s2 − 2s1 s2 + s1 2 ds1 ds2 . rg = 2 `c s2 =0 s1 =0

(14.86)

Z

2 1 s2 =`c 1 3 rg = 2 s2 ds2 . `c s2 =0 3

2 1 rg = `c 2 . 12 r

 1 rg = `c . 12 r

 1 h`e i . rg = 12 http://www.kirbyresearch.com/textbook

(14.87) (14.88) (14.89) (14.90)

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This result may be counterintuitive for some—it shows 

that,  for the same h`e i, a rigid rod has a smaller rg . However, for a given `c , the rg for a freely jointed chain is q q

 `c `K `c 2 , whereas the r for an infinitely rigid rod is g 6 12 . Thus if `c /2  `K , the

 rg for the rigid rod is much larger. If `c is not much bigger than `K , then the freely jointed chain model cannot be used. 14.9 Given that the partition function in a dimension i (where i = x, y, z) is given by " #

 ∞ π2 rg 2 (2p + 1)2 8 1 exp − , Zi = 2 Li ∑ π p=0 (2p + 1)2 Li 2

(14.91)

 derive the partition function in the limit where rg  Li .

Solution: We start with the given equation: " #

 ∞ π2 rg 2 (2p + 1)2 8 1 Zi = 2 Li ∑ exp − . π p=0 (2p + 1)2 Li 2

(14.92)

 If rg  Li , then the argument of the exponential function is small until p gets large, at which point the 1/(2p + 1)2 term is small anyway. Thus the exponential functions can be replaced with unity, giving Zi =

∞ 1 8 L . i ∑ π2 p=0 (2p + 1)2

(14.93)

The specified sum is equal to π2 /8, and thus the partition function is given by Zi = Li .

(14.94)

14.10 Given that the partition function in a dimension i (where i = x, y, z) is given by " #

 ∞ π2 rg 2 (2p + 1)2 8 1 Zi = 2 Li ∑ exp − , π p=0 (2p + 1)2 Li 2

(14.95)

 derive the partition function in the limit where rg  Li . c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Solution: We start with the given equation: " #

 ∞ π2 rg 2 (2p + 1)2 1 8 exp − . Zi = 2 Li ∑ π p=0 (2p + 1)2 Li 2

(14.96)

 If rg  Li , then the argument of the exponential function is large, and each term is smaller than the last by a large exponential factor. Thus we retain only the first term, giving "

 # π2 rg 2 8 . Zi = 2 Li exp − π Li 2

(14.97)

14.11 Assume that a linear polymer is confined to and uniformly fills a spherical domain of  radius R. Calculate rg for this polymer configuration. Solution: 3R/5. Solution: the solution for this problem is not available 14.12 Consider a Gaussian bead–spring model. Consider the point on the polymer that is a fraction α of the distance along the polymer backbone. Show that the mean-square distance between this point and the center of mass of the polymer is given by `c `K 2 [1 − 3α(1 − α)] . 3

(14.98)

Solution: the solution for this problem is not available 14.13 Consider a Gaussian bead–spring model of a linear polymer and assume that the two ends are fixed at a distance `e , but the polymer is otherwise free to equilibrate. Consider the component of the mean-square radius of gyration in the direction along the line connecting the two ends of the polymer. Show that this component of the radius of gyration is given by   `2e 1 . (14.99) `c `K 1 + 3 36 `c `K http://www.kirbyresearch.com/textbook

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Solution: the solution for this problem is not available 14.14 Derive the persistence length of a freely rotating chain. Solution: Note that the dot product between the unit tangents of a link and the next link is equal to cos ϑ. Thus, over n links or a distance of n`K , the dot product becomes cosn ϑ or `K 2 exp n ln cos ϑ. Per the definition of persistence length, `p = −`K

1 , cos ϑ

(14.100)

which can also be written as 1 1 + cos ϑ `p = `K . 2 1 − cos ϑ

(14.101)

14.15 Consider a Gaussian bead–spring chain model of a linear polymer in water. Assume that the chain has a positive charge +q on one end and a negative charge −q at the other end. Assume that an electric field of 100 V/cm is applied to the polymer at room temperature. If the polymer has `c = 5 µm and `K = 20 nm, what will h`e i be for this polymer in this field? At this distance, can the Coulomb interaction between the two charges be ignored? Solution: the solution for this problem is not available 14.16 Consider a 50-kbp DNA molecule with a contour length of 22 µm and a radius of gyration (at room temperature) of 0.75 µm that is forced into a long channel with square cross section. Model this DNA molecule as an ideal bead–spring chain. How small must the channel depth/width d be for the free energy of confinement to be 10 kB T ? Solution: Here we use the relation for energy rise upon confinement: "   # π2 rg 2 8 ∆A = 2kB T − ln , 2 d π2 c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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 and find the d/ rg to make the term in brackets equal to 5. This happens at d/ rg = 1.44. Thus the size of the channel must have a depth of

 d = 1.44 rg = 1.08 µm .

(14.103)

14.17 The expression for the probability density function for `e for the freely jointed chain and the freely rotating chain both allow for the possibility that `e will exceed `c . Explain why this apparently contradictory situation exists. Solution: the solution for this problem is not available 14.18 Consider an ideal freely jointed chain. If the ends of the chain are separated from each other with a force F, the resulting mean end-to-end length h`e i is given by h`e i =

F`c `K , 3kB T

(14.104)

if h`e i is small relative to `c . If h`e i cannot be assumed small relative to `c , a more precise relation is     kB T F`K h`e i = `c coth − , (14.105) kB T F`K where coth (x) − 1x is referred to as the Langevin function. Show that lim F`K →0 h`e i = F`c `K 3kB T .

kB T

Plot both relations.

Solution: the solution for this problem is not available 14.19 Derive the entropy of a Gaussian chain polymer as a function of `e . Solution: S = kB ln Ω

(14.106)

Ω P (`e ) = R Ω

(14.107)

Ω is number of states

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Z

Ω

(14.108)

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S = kB ln Ω = kB ln

Z

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 ΩP (`e )

(14.109)

 Z  S = kB ln Ω + ln P (`e )

(14.110)

 Z  S = kB ln Ω + ln P (`e )

(14.111)

S − S(0) = kB ln P (`e )

(14.112)

only P part is a fn of `e . have to get rid of the 4π this with degeneracy argument–scalar vs vector `e .   3 1  S − S(0) = kB ln 4π`e + ln  (14.113) 2 4π rg 2 − `e 2 2 4hrg i `e 2 S − S(0) = −kB 2 4 rg S − S(0) = −kB

3`e 2 2`c `K

(14.114)

(14.115)

14.20 Develop a Rouse (i.e., free-draining) model for DNA diffusion. Model the DNA as an ideal Gaussian chain of beads linked by springs. Assume that the viscous force on the springs is zero, but assume that each bead feels a force described by the Stokes flow relation for drag on an isolated sphere in an infinite medium. (a) Assume all components of the polymer are moving with velocity U and write the force on the DNA molecule as a function of `c , `K , and the bead radius a. (b) Note that the viscous mobility µ relates F and U by U = µF. Write µ for the DNA molecule. (c) Using Einstein’s relation D = µkB T , write D for DNA using the Rouse model. (d) Comment on the dependence of D on `c . How does this fail to match experimental observations for DNA in aqueous solution? What is the physical inaccuracy of the Rouse model for bulk diffusion of DNA? Solution: the solution for this problem is not available

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Figure 14.14: A DNA microarray used for a comparative study of gene expression. Complementary DNA from two sources (typically a control source and an unknown source) are labeled with two different dyes, hybridized to DNA strands immobilized in spots on the microarray, and quantified by reading out the fluorescence from the spots. A key issue for operation of these microarrays is the time required for all DNA from the sample to explore all of the spots on the microarray, an issue discussed in Chapter 4.

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Figure 14.15: The structure of a DNA oligomer (left) and a longer polymer (right). The oligomer appears stiff because its contour length is short compared with its persistence length, and the longer polymer appears flexible because its contour length is long compared with its persistence length.

Figure 14.16: Mean end-to-end length predicted by using the Kratky–Porod and Gaussian chain models.

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15.3

Supplementary reading

Nanofluidic transport has been the subject of several recent books and reviews, for example [52, 190, 192]. The membrane literature, e.g., [193] and parts of [29], is an excellent source for information about nanofluidic issues, although nanoporous membranes typically do not afford the flexibility of geometry and experimental technique required for exploring the physics completely. Nanoscale transport in long straight channels is described in [194, 195, 196], including discussion of valence-dependent electroosmosis. Relevant discussion of the motion of particles in small channels is in [100, 101]. A treatment of current rectification using 1D equilibrium is found in [197], and a number of models are discussed in [52]. Nanochannels also motivate considering EDLs outside the dilute solution limit, because nanoconfinement can change the role of ion size on ion distributions. Some recent work on this includes implementation of modified Poisson–Boltzmann equations stemming from [198, 199, 200] by Liu et al. [201]. Other modified Poisson–Boltzmann and molecular dynamics simulations include [118, 120]. Some descriptions of macromolecular transport in nanochannels can be found in [178, 181]. The matrix formulation of nanochannels in this chapter is reminiscent of the areaaveraged treatment of porous and gel materials [137].

15.4

Exercises

15.1 Define a geometry-dependent effective electroosmotic mobility µEO,eff such that the flow rate per unit length Q0 of an electroosmotic flow between two infinite plates separated by a distance 2d is given by Q0 = 2dµEO,eff~E ,

(15.49)

where d ∗ = d/λD . This effective electroosmotic mobility thus gives the spatiallyaveraged flow rate in this system. µEO,eff is a function of d ∗ . (a) Graph the velocity distribution for several values of d ∗ ranging geometrically from 0.2 to 50. (b) Evaluate µEO,eff as a function of d ∗ and plot this relation from d ∗ = 0.2 to d ∗ = 50. Solution: the solution for this problem is not available

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CHAPTER 15. NANOFLUIDICS: FLUID AND CURRENT FLOW IN MOLECULAR-SCALE AND THICK-EDL Soln Manual,SYSTEMS July 21, 2010 http://www.cambridge.org/kirby 15.2 Consider two infinite parallel plates separated by a distance 2d. Assume the fluid between the plates is water (ε = 80ε0 ; η = 1 mPa s) with a symmetric, monovalent electrolyte with bulk conductivity σbulk = ∑i ci Λi . Assume that both the cation and the anion of the electrolyte have the same molar conductivity Λ. Assume the normalized potential at the wall is given by ϕ∗0 = −0.2. Solution: There is no solution yet, this is just the intro. (a) Calculate and plot the velocity profile between the two plates for the cases in which λD is equal to 5, 0.2, and 0.02 times d. Use E = 100 V/cm. Solution: In general, the 1D Navier–Stokes equations with a Coulombic source term gives εE u= (ϕ − ϕ0 ) , (15.50) η or, adjusting to report in nondimensional terms, u=

εE RT ∗ (ϕ − ϕ∗0 ) . η F

(15.51)

We can safely make the Debye–Hückel approximation here, because sinh(0.2) = 0.201 or so. Thus the potential distribution is known and is given by Eq. (9.23): ϕ∗ = ϕ∗0

cosh y∗ , cosh d ∗

(15.52)

leading to εE RT ∗ u= ϕ η F 0



 cosh y∗ −1 . cosh d ∗

(15.53)

see Fig. 15.5. (b) Define ζeff as the effective electrokinetic potential such that the cross-sectionalarea-averaged velocity u is given by − εζηeff E. Calculate and plot ζeff /ϕ0 vs. d ∗ for 0.1 < d ∗ < 100. Solution: Here we integrate to get the mean velocity. We start with εE RT ∗ u= ϕ η F 0 c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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R d ∗ cosh y∗ ∗ 0 cosh d ∗ − 1 dy , R d∗ 0

dy∗

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which can be evaluated to obtain u=

εE RT ∗ ϕ η F 0



 tanh d ∗ − 1 . d∗

(15.55)

Evaluating the normalized effective interfacial potential, we obtain ζeff uη 1 tanh d ∗ =− . = 1 − ϕ0 εE ϕ∗0 RT /F d∗

(15.56)

See Fig. 15.6, which shows both the effective zeta potential in terms of velocity (which goes down when the double layers are not thin) as well as the effective conductivity of the system (which goes up when the double layers are not thin). (c) Define σeff as the effective conductivity of the channel such that the channel electrical resistance is given by R = L/σeff A. Calculate and plot σeff /σbulk vs. d ∗ for 0.1 < d ∗ < 100. Solution: σbulk = ∑ ci,∞ Λ = 2ci,∞ Λ

(15.57)

i

Spatial variation of concentration: ci = ci,∞ exp(−zi ϕ∗ )

(15.58)

σ(y∗ ) = ci,∞ Λ [exp(−ϕ∗ ) + exp(ϕ∗ )] = 2ci,∞ Λ cosh ϕ∗ R d∗

σ(y∗ ) dy∗ σeff = R0d ∗ = ∗ σbulk 0 σbulk dy

R d∗ 0

cosh ϕ∗ dy∗ R d∗ 0

dy∗

(15.59) (15.60)

Debye–Hückel approx:

cosh y∗ cosh d ∗ Because ϕ∗ small, approximate cosh ϕ∗ = 1 + 12 ϕ∗2 . Integrate, find ϕ∗ = ϕ∗0

∗   ϕ0 tanh d ∗ σeff 1 = 1 + + σbulk 4 d∗ cosh 2 d ∗

(15.61)

(15.62)

see Fig. 15.6. 15.3 Consider two infinite parallel plates separated by a distance 2d. Assume the fluid between the plates is water (ε = 80ε0 ; η = 1 mPa s) with a symmetric, monovalent electrolyte with bulk conductivity σbulk = ∑i ci Λi . Assume that both the cation and the anion of the electrolyte have the same molar conductivity Λ. Assume the normalized potential at the wall is given by ϕ∗0 = 0.3. Assume d ∗ is not large relative to unity. http://www.kirbyresearch.com/textbook

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Figure 15.5: Electroosmotic flow as a function of the double-layer thickness normalized by the half-separation of two plates. When λD is small, the charge on the wall is shielded by the double layer, most of the domain is electroneutral, and the flow approaches a uniform flow. For λD on the same order as d, the charge density is nonzero throughout, and the flow is approximately parabolic. For λD larger than d, the flow remains parabolic, but the velocity magnitude is small. In the limit where λD /d is large, there is little shielding, and the two walls interact electrostatically with minimal influence from the double layer.

Figure 15.6: Electroosmotic flow and effective conductivity as a function of the halfseparation of two plates normalized by the Debye length. When d ∗ is large, the effective zeta potential ζeff is equal to the voltage drop across the double layer ϕ0 , and the double layer plays little role in changing the effective conductivity. d ∗ needs to be about 20 or higher for this to be correct within 5%. When d ∗ = 1, the effective electrokinetic potential is only about 20% of the voltage drop across the double layer, and the net conductivity has increased. Electroosmotic flow is largely suppressed when d ∗ = 0.2 or below, and the normalized effective conductivity approaches 1 + |ϕ∗0 | /2.

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(a) Derive a relation for the net current (i.e., charge flux) I = by flow-induced convection of charge if a pressure gradient system.

R

A uρE dA generated dp dx is applied to this

(b) Presume the reservoirs at both ends of this system are connected to an open circuit (i.e., presume there can be no net current in the system). In this case, there is (at equilibrium) a potential gradient generated in the system. Calculate this potential gradient. Solution: Write I = I 0 h, where h is the depth of channel and I 0 is the current per depth. Determine I 0 that is due to flow: I0 = 2

uρE dy = λD

0

u(y) =

Z d∗

Z d

0

uρE dy∗ ,


−1 dp ∗2
−1 dp 2 d − y2 = d − y∗2 λ2D , 2η dx 2η dx

ρE = ∑ ci zi F = Fc∞ [exp(−ϕ∗ ) − exp(ϕ∗ )] ,

(15.63) (15.64) (15.65)

i

make the Debye–Hückel approximation, ρE = 2Fc∞ sinh(ϕ∗ ) = 2Fc∞ ϕ∗ .

(15.66)

With this, the current becomes Fc∞ λ3D dp I =2 η dx 0

Z d∗

ϕ∗ (d ∗2 − y∗2 ) dy∗ ,

(15.67)

0

and in the Debye–Hückel approximation, ϕ∗ = ϕ∗0 Fc∞ λ3D ϕ∗0 dp I =2 η cosh d ∗ dx 0

Z d∗

cosh y∗ , cosh d ∗


d ∗2 cosh y∗ − y∗2 cosh y∗ dy∗ .

(15.68) (15.69)

0


R Now we integrate by parts, noting that x2 cosh x = x2 + 2 sinh x − 2x cosh x: I0 = 2

Fc∞ λ3D ϕ∗0 dp (2d ∗ cosh d ∗ − 2 sinh d ∗ ) . η cosh d ∗ dx

(15.70)

Because λ2D = εRT /2F 2 c∞ for monovalent symmetric electrolytes, we finally obtain   2hεϕ0 d dp tanh d ∗ I= 1− η dx d∗

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c Brian J. Kirby 596 Micro- and Nanoscale Fluid Mechanics,

CHAPTER 15. NANOFLUIDICS: FLUID AND CURRENT FLOW IN MOLECULAR-SCALE AND THICK-EDL Soln Manual,SYSTEMS July 21, 2010 http://www.cambridge.org/kirby If the net current is zero, there must be another source of current, namely an ohmic current stemming from a potential gradient: dV I 0 = 2σeff Ed = −2σeff λD d ∗ , (15.72) dx    ϕ∗0 tanh d ∗ 1 0 ∗ dV . (15.73) I = −σλD d 1+ dx 4 d ∗ cosh 2 d ∗ Setting the sum of the flow-induced current and the ohmic current to zero, we obtain ! d∗ 1 − tanh dV dp εϕ0 ∗ d (15.74) =
. ∗ dx dx ση 1 + ϕ0 tanh∗ d ∗ + 12 ∗ 4

d

cosh d

15.4 Consider a nanochannel with equal concentrations of NaCl and MgSO4 and a depth 2d = 2λD . For a wall potential of ϕ0 = −3RT /F, estimate the net electromigration of Na+ , Cl− , Mg+2 , and SO−2 4 averaged over the cross section when an electric field E is applied. Solution: the solution for this problem is not available 15.5 Consider a nanochannel of depth 30 nm, width 40 µm, and length 120 µm aligned along the x axis. Assume that the leftmost 60 µm of the channel has a surface charge of 3 mC/m and the rightmost 60 µm of the channel has a surface charge of zero. A voltage is applied at infinite reservoirs at the ends of the nanochannel. Model the electrolyte as symmetric and nonreacting with mobility equal to 7.8×10−8 m2 /V s, and for simplicity assume that the electrolyte concentration is uniform across the cross section and of a magnitude such that the net charge density at the wall is canceled by the net charge density in the fluid. (a) Write the 1D Nernst–Planck equations for transport of the positive and negative electrolytes. (b) Given a surface charge density that is nonuniform in x, use equilibrium electroneutrality to define the difference between the cation and anion concentrations. (c) Integrate the resulting equations to predict the ion distributions in the channel as a function of applied voltage and electrolyte bulk concentration. (d) Compare your results with the experimental data in Fig. 15.7, taken from [197]. Does your model accurately match the experimental results? What does this tell you about the rectification of current in a nanofluidic channel with a surface charge density discontinuity? c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Figure 15.7: Current–voltage response of a nanofluidic channel with two regions of differing net surface charge density. The positive electrode for calculation of the voltage is the rightmost reservoir. Symbols denote experimental results; lines denote model predictions generated by the study’s authors. (From [197].) Solution: the solution for this problem is not available 15.6 Set q00 = 0 in Eq. (15.44) and show that the result is i¯ = σE. Solution: the solution for this problem is not available 15.7 Assume q00 /rh is uniform in Eq. (15.44) and show that, in the Debye–Hückel limit, ε2 ϕ2 the result is i¯ = σE + 2 λD ηr0h . Explain why the convective surface current differs from that predicted in Eq. (9.36) by a factor of two. Solution: the solution for this problem is not available

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CHAPTER 15. NANOFLUIDICS: FLUID AND CURRENT FLOW IN MOLECULAR-SCALE AND THICK-EDL Soln Manual,SYSTEMS July 21, 2010 http://www.cambridge.org/kirby 15.8 Consider flow through an infinite channel of circular cross section and radius R with a dilute suspension of rigid particles of radius a. Assume that the particles are evenly distributed across the channel, except for the steric repulsion that prevents the particle center from approaching any closer than a to the wall. Approximate the particle velocity as the velocity of the fluid at the particle center location if the particle were absent, and calculate the factor by which the particles move faster than the mean fluid velocity. Solution: For flow through an infinite channel of circular cross section, the fluid velocity is given by  
1 dp u= R2 − 2 . (15.75) 4η dx The mean velocity of the fluid is given by RR

  1 dp u= RR = − R2 . 8η dx 2π d =0 =0 u d

(15.76)

If we assume that the particles move with the velocity that the fluid would have at the particle center if the particle were absent, the mean velocity of a particle of radius a is given by R R−a =0 u d u = R R−a . (15.77) =0 2π d Evaluating the integrals, we find that the area of the region that the particles sample is given by Z R−a

2π d = π(R − a)2

(15.78)

=0

The volumetric flow rate of the region that the particles sample is given by   Z R−a
π dp u 2π d = − R4 + 2R3 a − 4R2 a2 − 4Ra3 − a4 8η dx =0 Normalizing this by area to get the mean velocity, we obtain
  4 R + 2R3 a − 4R2 a2 − 4Ra3 − a4 1 dp u= − . 8η dx (R − a) 2

(15.79)

(15.80)

Normalizing this by the mean velocity of the fluid, the velocity increase factor is
R4 + 2R3 a − 4R2 a2 − 4Ra3 − a4 . (R − a) 2 R2

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15.9 Consider a fluid channel of width 100 µm. The channel has a shallow region of depth 500 nm and length 4 mm surrounded by regions of depth 2 µm and length 4 mm. A suspension of 50% by volume, 150-nm diameter particles is introduced at one end of the device. Assume the particles are evenly distributed other than their steric repulsion from the wall. What is the volume fraction of the particles in the 500-nm channel? The Fahraeus effect in blood is the observation that blood hematocrit (i.e., blood cell volume fraction) decreases if a blood sample goes through a narrow tube. Can the Fahraeus effect be explained, in part, by your calculations?

Solution: If we assume that the particles move with a velocity corresponding to the velocity the fluid would have at the particle center if the particle were absent, the particles move faster than pure fluid would by a factor of d 3 − 23 a2 d + 21 a3 , d 2 (d − a)

(15.82)

where d = 250 nm is the half-height of the channel and a = 75 nm is the radius of the particles. This factor evaluates to 1.25, indicating that the particles move 25% faster than pure fluid would. To satisfy conservation of water and conservation of particles, the total rates of water volume and particle volume must be the same. This is satisfied if the particles are 44% of the volume of the system. This calculation assumes that the water still moves with an average velocity equal to the average velocity of pure fluid, which is only correct if the volume fraction of particles is low. For finite particle volume fractions, this is slightly off, as the water is mostly in the slow-moving region of the flow. The Fahraeus effect can partially be explained by this phenomenon–the particles move faster than the fluid in a narrow tube, and thus they must be at a lower concentration in the tube than they are in the bulk. The Fahraeus effect in blood is not totally explained by the preceding calculations, though, because blood cells do not sample all regions of the flow equally. Blood cells are also nonspherical and nonrigid.

15.10 Consider a mixture of 50-nm and 100-nm particles moving through a wide channel of depth 200 nm at mean velocity 10 µm/s. Calculate the different mean velocities of these particles, and determine if a hydrodynamic separation of these particles is possible in this configuration. http://www.kirbyresearch.com/textbook

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CHAPTER 15. NANOFLUIDICS: FLUID AND CURRENT FLOW IN MOLECULAR-SCALE AND THICK-EDL Soln Manual,SYSTEMS July 21, 2010 http://www.cambridge.org/kirby Solution: the solution for this problem is not available 15.11 Consider the data in Fig. 15.8. These data involve measurements of migration time of DNA of varying sizes in nanochannels with periodically varying depths. Given the data, fit the result by using µEPbulk

µEP = 1+

α`c   hrg i2 E 2 exp − 2

,  exp − k∆A BT

(15.83)

d

where α is a fit parameter. Evaluate the efficacy of this model in explaining the DNA transport data. Solution: the solution for this problem is not available

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Figure 15.8: Measurements of (a) DNA migration times and (b) relative mobilities. (From [181].)

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16.6

Supplementary reading

Chang and Yeo [52] cover AC electrokinetic effects in detail and go well beyond the coverage in this chapter. Early derivations of double-layer capacitances can be found in [108, 204]. Bazant and co-workers have studied EDL dynamics in great detail with a view toward managing the fluid flow in nonequilibrium EDLs developed at electrodes or metallic surfaces [111, 112, 202, 205, 206, 207, 208], and this chapter draws directly from that work. Green et al. [203, 209], Ramos et al. [210], and Gonzales et al. [211] have presented descriptions of AC electroosmosis effects, and [212] shows net pumping by asymmetric electrodes. Ref. [213] shows use of traveling waves to generate DC pumping. di Caprio et al. have presented modified Poisson–Boltzmann theories with attention to capacitance effects [214]. Much current attention is focused on the challenges regarding quantitative predictions of electrokinetic phenomena at electrodes—the fluid velocity magnitudes (and even signs) predicted by analysis described in this chapter often do not match experiment, an issue discussed in detail in [208]. A review by Dukhin [167] focuses on the equilibrium assumptions made in Chapter 13 and how departure from equilibrium affects colloidal motion and characterization of surfaces. Although the surfaces discussed in this chapter are all conducting, interfacial charge is also created when an electric field is applied normal to any interface with mismatched permittivity or conductivity—so electric fields applied to any insulating particle whose properties are not matched to its suspending medium creates interfacial charge. The resulting dipole on a particle is the source of dielectrophoretic forces, described in Chapter 17. This interfacial charge also induces fluid flow, with characteristic frequencies similar to those of the ACEO and ICEO flows described in this chapter. In fact, these flows are related to the variation in observed dielectrophoretic responses at kilohertz frequencies described in the dielectric spectroscopy literature as the alpha relaxation. Compared with the flows at conducting surfaces, though, the fluid flows induced at insulating objects are smaller in magnitude.

16.7

Exercises

16.1 Consider the equilibration of the double layer around a small particle or macromolecule with radius a  λD . Explain why the relevant time scale for double-layer equilibration is τ = λ2D /D. Solution: The system size ` is the size of the double layer, so ` = λD . Thus τ = λ2D /D. 16.2 Consider the equilibration of the double layer around a conducting particle with radius a  λD . Explain why the relevant time scale for double-layer equilibration is τ = aλD /D. c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Solution: This problem is the same as the model electrode system, with a resistor (the particle) in between two capacitors (the hemispheres of the double layer). Thus τ = aλD /D. 16.3 For a symmetric electrolyte in the Debye–Hückel limit, evaluate the net charge in the double layer by integrating the charge distribution. Show that the net charge in the double layer is given by q00edl = −εϕ0 /λD . Show that the capacitance per unit area is given by C00 = ε/λD . Solution: integrate the charge distribution. Using the Debye–Hückel approximation, we have: ϕ = ϕ0 exp (−y/λD ) . (16.34) For the charge density, we have: 

zFϕ ρE = ∑ zi ci F = −2zFc0 sinh RT



 = −2zFc0

zFϕ RT

 ,

(16.35)

leading to  ρE = −2zFc0

zFϕ RT

 = −

2z2 F 2 c0 ε ϕ = − 2 ϕ, RT λD

(16.36)

and eventually ρE = −

ε ϕ0 exp (−y/λD ) . λ2D

(16.37)

The total charge is: q00 =

Z ∞ y=0

ρE =

ε ϕ0 (0 − 1) , λD

(16.38)

leading finally to q00 = −

ε ϕ0 . λD

(16.39)

Note that the net charge in the double layer has a sign opposite that of the potential at the wall. The charge density on the wall is given by q00 =

ε ϕ0 . λD

(16.40)

So, the charge density on the wall is given by q00 = http://www.kirbyresearch.com/textbook

ε ϕ0 , λD

(16.41)

c Brian J. Kirby 620 Micro- and Nanoscale Fluid Mechanics,

CHAPTER 16. AC ELECTROKINETICS AND THE DYNAMICS OF DIFFUSE CHARGE Soln Manual, July 21, 2010 http://www.cambridge.org/kirby and the charge density in the double layer is given by q00 = −

ε ϕ0 . λD

(16.42)

. 16.4 For a symmetric electrolyte outside the Debye–Hückel limit, evaluate the net charge in the double layer by evaluating the derivative of the potential at the wall: (a) Start  with  the Poisson–Boltzmann equation (9.16) and multiply both sides by d 2 dy ϕ . (b) Integrate bothsides,  rearranging the derivatives so that on one side you have the d derivative of dy ϕ 2 and on the other side you have dϕ. d (c) To evaluate dy ϕ, integrate over a dummy variable (y0 or ϕ0 , depending on which side of the equation) from the bulk (y0 = ∞; ϕ0 = 0) to some point in the double layer (y0 = y; ϕ0 = ϕ). d (d) Evaluate ε dy ϕ at the wall.

Show that the net charge in the double layer is given by   ∗  zϕ0 εϕ0 2 00 qedl = − sinh . λD zϕ∗0 2 and that the differential capacitance per unit area dq00 /dϕ0 is given by   ∗  zϕ0 ε cosh . C00 = λD 2

(16.43)

(16.44)

Solution: For 1D, we have d 2 ϕ∗ 1 = sinh(zϕ∗ ) . (16.45) dy∗2 z  ∗ 2 . We start by multiplying both We want to integrate the left-hand side to give dϕ dy∗  ∗ sides by 2 dϕ dy∗ , giving dϕ∗ 2 dy∗ 



 ∗ d 2 ϕ∗ dϕ 1 =2 sinh(zϕ∗ ) . dy∗2 dy∗ z

(16.46)

Now we use a dummy variable y0 to integrate from the bulk (y0 = ∞) to a point not infinitely far from the wall (y0 = y∗ ). Note that dy0 = dy∗ : Z y0 =y∗  ∗  2 ∗ Z y0 =y∗  ∗  dϕ d ϕ dϕ 1 0 2 dy = 2 sinh(zϕ∗ )dy0 . (16.47) ∗ ∗2 dy dy dy∗ z y0 =∞ y0 =∞ c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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∗

∗

dϕ 0 ∗ ∗ 0 Now note that dϕ dy∗ dy = dy∗ dy = dϕ = dϕ . Because we are integrating the potential, we must here use a dummy variable ϕ0 and the corresponding values for ϕ∗ in the bulk and in the double layer: Z y0 =y∗  ∗  2 ∗ Z ϕ0 =ϕ∗ dϕ d ϕ 2 0 2 dy = sinh(zϕ∗ )dϕ0 . (16.48) ∗ ∗2 dy dy z y0 =∞ ϕ0 =0

We integrate to find 

dϕ∗ dy∗

Now evaluate; noting that

ϕ0 =ϕ∗  y0 =y∗ 2 2 ∗ = 2 cosh(zϕ ) . 0 z y =∞ ϕ0 =0

(16.49)

dϕ∗ dy∗

and ϕ∗ are both zero in the bulk:  ∗ dϕ 2 2 = 2 [cosh(zϕ∗ ) − 1] . dy∗ z

(16.50)

We note the identity −1 + cosh x = 2 sinh 2 (x/2), giving  ∗  ∗ 4 dϕ 2 2 zϕ = 2 sinh , (16.51) dy∗ z 2  ∗ dϕ∗ 2 zϕ = ± sinh . (16.52) ∗ dy z 2 Of these solutions, only the negative one leads to a bounded result for ϕ, so  ∗ dϕ∗ 2 zϕ = − sinh . (16.53) dy∗ z 2 Returning to dimensional form, we obtain  ∗ RT /F 2 zϕ dϕ =− sinh . dy λD z 2

(16.54)

Now we set q00edl = ε dϕ dy : q00edl

  ∗  zϕ0 εϕ0 2 sinh . =− ∗ λD zϕ0 2

(16.55)

16.5 For a symmetric electrolyte, plot the capacitance (normalized by ε/λD ) versus zϕ∗0 for (a) the nonlinear Poisson–Boltzmann double layer and (b) a Poisson–Boltzmann diffuse double layer in series with a Stern layer with εS = ε/10 and λS = λD /1000. Plot the modified Poisson–Boltzmann model for ξ = 1 × 10−4 , ξ = 1 × 10−4.5 , and ξ = 1 × 10−5 . Plot results over the range 0 < zϕ∗0 < 35 and 0 < C00 λD /ε < 650. Compare the results of the Stern modification with the results of the steric hindrance modification. http://www.kirbyresearch.com/textbook

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CHAPTER 16. AC ELECTROKINETICS AND THE DYNAMICS OF DIFFUSE CHARGE Soln Manual, July 21, 2010 http://www.cambridge.org/kirby Solution: the solution for this problem is not available 16.6 Consider two electrodes separated by a distance 2l and containing a symmetric electrolyte with all ions having mobilities and diffusivities of equal magnitude. Suppose that two different voltages are applied to the two electrodes. How long does it take for the double layers at the electrodes to form? l Treat the bulk fluid as a resistor. Show that the resistance R = σA (where σ is the bulk conductivity and A is the cross-sectional area of the electrolyte linking the two λ2D l . electrodes) can also be written as εDA

Treat each double layer in the Debye–Hückel limit as a Helmholtz capacitor with thickness equal to λD (C = λεAD ). Model the system as a capacitor, resistor, and capacitor in series, and show that RC for this model system is equal to

λD l D .

Given this result, what is the characteristic time for the double layer to equilibrate if you have two microelectrodes separated by 40 µm in a 1-mM NaCl solution? Approximate D for these ions as 1.5×10−9 m2 /s. Solution: For the capacitor, we have C = λεAD , and for two capacitors in series, C = 2l εA 2λD . For the resistor, we have R = σA . To define σ in terms of λD , we write σ =

∑ ci Λi

= 2cΛ = 2czFµEP ,

(16.56)

i

leading to σ = 2czF

zFD 2z2 F 2 c εD = εD = 2 . RT εRT λD

(16.57)

Substituting in for σ, we find R=

2λ2D l , εDA

(16.58)

leading to RC =

τ=

λD l , D

(16.59)

10 × 10−9 m × 20 × 10−6 m = 133 µs . 1.5 × 10−9 m2 /s

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16.7 Reconsider the previous problem by assuming that the double layer has two parts, which can be considered as capacitors in series (Fig. 16.7). The first part is the diffuse double layer, which can be modeled as being of thickness λD with a permittivity of ε. The second is the Stern layer and is modeled as having a capacitance of CS = ε λS , where λS is an effective thickness that accounts for both the thickness of the Stern layer and its electrical permittivity (which is typically below the permittivity of  the bulk  solution). Show that the previous result must be modified by a factor of λS −1 . 1 + λD

Figure 16.7: Equivalent circuit for double-layer charging with Stern layers.

Solution: For two capacitors in series, we find C=

CDCS 1 1 = CD = CD . CS +CD 1 +CD /CS 1 + λS /λD

(16.61)

Thus λD l RC = D

  λS −1 1+ . λD

(16.62)

16.8 Estimate the velocities and time dependence of velocities at the electrode surface during an ACEO cycle. Solution: the solution for this problem is not available 16.9 Given what you know about induced-charge electroosmosis, consider an AC field applied to an infinitely thin metal plate in one of two positions: http://www.kirbyresearch.com/textbook

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CHAPTER 16. AC ELECTROKINETICS AND THE DYNAMICS OF DIFFUSE CHARGE Soln Manual, July 21, 2010 http://www.cambridge.org/kirby (a) With the long edge tangent to the applied electric field. (b) With the long edge normal to the applied electric field. Assume the metal plate is approximately 10 µm long and infinitely thin. Qualitatively predict the flow induced by an AC field applied to this system. Assume the frequency of the AC field is low enough that the field oscillation is slow compared with the double-layer charging time. Include sketches of the electric field lines at various points during the field cycle. Solution: the solution for this problem is not available 16.10 Consider a 20-µm-deep insulating microchannel that is infinitely wide and 1 cm long. Assume that, halfway along the length of the channel, a 200-µm patch of conducting metal is patterned along the wall. Estimate the magnitude of the flow resulting from a 100-Hz signal of 150 V applied at one end of the microchannel while the other end is grounded. Use a Schwarz–Christoffel transform or a numerical simulation to calculate the electric field in the channel. 16.11 Consider the previous problem, but now assume that the conducting metal is covered with a layer of glass that is 1 nm thick. How does this change the flow? How is the flow changed by a layer that is 1 µm thick? Solution: the solution for this problem is not available 16.12 Derive Eq. (16.27) as follows: (a) Write the Poisson equation as ρE = ∇ · ε~E . Expand ~E in terms of the uniformproperty solution ~E 0 (which is large in magnitude and divergence free) plus a perturbation owing to the property variations (which is small in magnitude but of finite divergence). Replace ρE in the electrostatic body force term with an expression using these two electric fields. (b) Write the charge conservation equation and assume that convective charge fluxes are small compared with electrophoretic charge fluxes. Defend this assumption. Derive Eq. (16.26). (c) Assume sinusoidal fields and write the fields as their analytic representations. Solve the resulting equation for the divergence of the perturbation field, and thus derive Eq. (16.27). c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Solution: the solution for this problem is not available 16.13 Describe the governing equations and boundary conditions used to solve for the outer solution for AC electroosmosis between two electrodes. The electrodes are each 40 µm wide and separated by 40 µm. The microchannel depth is 1 µm, and the ionic strength of the medium is 10 mM. The electrode at left is connected to electrical ground, and the electrode at right is energized to a voltage given by V = V0 cos ωt. In particular, the RC time constant for different regions of the electrodes should be different, and thus the boundary conditions on the outer solution for the Laplace equation will be spatially varying. Solution: the solution for this problem is not available 16.14 Consider two infinite parallel plates located at y = ±h, h  λD , connected to an AC power supply that generates a differential voltage of V = V0 cos ωt. The plates are covered with a dielectric thin film of ε = 5ε0 and thickness 50 nm, which is treated as an additional capacitor in parallel with the fluid, which is a 1-mM solution of KCl. A small transverse electric field ~E = xˆ E0 cos (ωt + α) is applied. What is the time-averaged flow field as a function of h and α? Solution: the solution for this problem is not available 16.15 Consider a 1D system of length L with three domains, each of length L/3. Although the system is water throughout, assume that the domains at left and right are at temperature T1 , with permittivity ε1 and conductivity σ1 . The domain in the center is at temperature T2 , with permittivity ε2 and conductivity σ2 . For times t < 0, the applied electric field is zero, as is the net charge density and interfacial charge. (a) At time t = 0, voltage V is applied at left and the voltage at right is held at V = 0. For a time that is long compared with orientational relaxation of the water but slow compared with motion of free charge in the water, what is the net charge density in the water? What is the charge density at the interfaces between the three regions? (b) What is the charge density in the water and at the interfaces for t → ∞? What is the pressure in the central region? http://www.kirbyresearch.com/textbook

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CHAPTER 16. AC ELECTROKINETICS AND THE DYNAMICS OF DIFFUSE CHARGE Soln Manual, July 21, 2010 http://www.cambridge.org/kirby Solution: the solution for this problem is not available 16.16 Assume that a focused laser beam maintains an elevated temperature at a specific point in an otherwise uniform flow. Assume that this temperature is maintained regardless of any fluid flow or other processes. Model this point as a sphere of radius R with a conductivity that is ∆σ different than the surrounding fluid and a permittivity that is −∆ε different than the surrounding fluid. Model the force generated by this hot spot and show that the resulting flow can be approximated by a stresslet whose stress vector is aligned with the applied electric field. Solution: the solution for this problem is not available

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CHAPTER 17. PARTICLE AND DROPLET ACTUATION: DIELECTROPHORESIS, MAGNETOPHORESIS, AND DIGITAL Soln Manual, MICROFLUIDICS July 21, 2010 http://www.cambridge.org/kirby techniques are presented in [215, 265, 266]. A subset of the published work uses “electrodeless” techniques, in which the electric field nonuniformity is created through constrictions in the channel geometry [218, 219, 264]. Many of these techniques are suitable for continuous-flow separation, also implemented with electrodes in [267, 268, 269]. These continuous-flow separations have been applied for polystyrene spheres [219, 270], bacteria [271], yeast [272], and mammalian cells [257]. Individual cells have been addressed in DEP electrode arrays to facilitate single-cell capture, analysis, and release, both in solution [273, 274, 275] and in photopolymerized gels [276]. This chapter omits dielectric spectroscopy, which can be used on colloidal suspensions to infer double-layer and particle properties. The dielectric spectroscopy community describes many of the phenomena described in this chapter using different terminology—for example, the terms alpha and beta relaxations are used to describe the low- and moderatefrequency changes in the dielectric response owing to double-layer asymmetry and fluid flow (alpha dispersion) or Maxwell–Wagner polarization (beta dispersion). Chang and Yeo [52] discuss EWOD theory as well as its limitations. Some examples of digital microfluidics are presented in [221, 277].

17.6

Exercises

17.1 Consider a protein whose permanent dipole moment is 100 D. Model this protein as having one positive charge on one end of the protein and another positive charge on the other end of the protein. How far apart must these charges be to explain the observed permanent dipole moment? Solution: The magnitude of a dipole induced by separated charges is given by p = qδd ,

(17.65)

p . q

(17.66)

so δd =

Here, q = 1.6 × 10−19 C and p = 100 D or p = 3.34 × 10−28 C m. Thus δd = 2e − 9 m .

(17.67)

17.2 Consider a hydrogen atom, consisting of a nucleus with charge q = e and an electron cloud of q = −e, where e is the magnitude of the electron charge. Treat the nucleus and electron cloud as if they were each point charges. By applying an electric field with magnitude E = 100 V/cm in the x direction, the nucleus and electron cloud will c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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be displaced relative to each other until the field caused by one charge on the other that holds them together, q 1 E= , (17.68) 4πε0 ∆x2 balances out the external electric field. Derive a relation for the displacement ∆x in terms of the other given parameters and calculate the magnitude of this displacement. Is this displacement small or large compared with an atomic radius?

Solution: The field caused by one charge on the other has magnitude E=

|q| 1 . 4πε0 ∆x2

(17.69)

Substituting |q| = e, solving this for ∆x, and retaining the positive root, we obtain r ∆x =

e . 4πε0 E

(17.70)

Evaluating this, we obtain ∆x = 3.79 × 10−7 m .

(17.71)

17.3 Derive the Maxwell stress tensor using Gauss’s law. Solution: the solution for this problem is not available

17.4 A polystyrene bead of radius a = 1 µm is suspended in an electrolyte solution. The permittivity of polystyrene is approximately 2ε0 and the conductivity is 2 µS/cm. A silica bead of radius a = 1 µm is also suspended (ε = 4ε0 ; σ = 0.2 µS/cm). The solution is deionized water (ε = 80ε0 ; 0.18 µS/cm). A spatially varying field proportional to E0 cos ωt is applied. Calculate and plot the real part of the Clausius–Mossotti factor for the polystyrene and silica beads as a function of ω. Is there a range of frequencies at which these two beads will feel forces in opposite directions (i.e., one toward regions of high electric field, one towards regions of low electric field)? What is this range? http://www.kirbyresearch.com/textbook

c Brian J. Kirby 658 Micro- and Nanoscale Fluid Mechanics,

CHAPTER 17. PARTICLE AND DROPLET ACTUATION: DIELECTROPHORESIS, MAGNETOPHORESIS, AND DIGITAL Soln Manual, MICROFLUIDICS July 21, 2010 http://www.cambridge.org/kirby Solution: See Fig. 17.13. 17.5 Consider a polystyrene bead of radius a = 1 µm suspended in an electrolyte solution in a microchannel. The permittivity of polystyrene is approximately 2ε0 and the conductivity is 2 µS/cm. The solution is deionized water (ε = 80ε0 ; 0.18 µS/cm). In these conditions, the electrophoretic mobility of this particle is µEP = 1×10−8 m2 /V s. Presume that the geometry of the microchannel is designed such that the potential in the channel (which ranges from x = 0 to x = L) varies according to the following equation: φ = Ax2 . If A is a constant, what value of A will cause the particle to stagnate at x = L/2? Solution: The particle doesn’t stagnate. The electric field is given by E =−

∂φ = −2Ax . ∂x

(17.72)

The gradient of the electric field squared is given by ∇E 2 =


∂ 4A2 x2 = 8A2 x . ∂x

(17.73)

So the relative magnitudes of electrophoresis and dielectrophoresis changes with space in the same way. 17.6 Assume a combined DC plus AC field is applied to a particle: h i ~E = Re ~E DC + ~E AC exp ( jωt)

(17.74)

Where Re indicates “real part of”. Assume the AC and DC fields are aligned in the same direction, and define α = ~E AC /~E DC . Assume that the Clausius–Mossotti factor for the particle and the medium is given for both DC fields ( fCM,DC ) and AC fields ( fCM,AC ). g g (a) Note that the instantaneous effective dipole moment on a sphere is given by   ~p eff = Re 4πεm a3 fCM~E (17.75) g Given this, derive a relation for the effective dipole moment ~peff for the DCoffset AC field in terms of α, Clausius–Mossotti factors, and ~E DC , as well as εm and a. c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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(b) evaluate the instantaneous DEP force. Note that, for a purely AC field with magnitude ~E AC , the time-averaged DEP force is given by   D E ~F = πεm a3 Re fCM ∇ ~E AC 2 (17.76) g Your instantaneous force relation will be more complicated than the time-averaged relation, because it will involve two Clausius–Mossotti factors, as well as timedependent terms. Rearrange your instantaneous force relation so that (a) there are no nonlinear trigonometric terms—i.e., ensure that there are no cos 2 or sin cos terms; and so that (b) the real and imaginary parts of the Clausius– Mossotti factors are listed separately. (c) in the case where fCM,DC = fCM,AC , derive the time-averaged force in terms of g g α, the Clausius–Mossotti factor, and ~E DC , as well as εm and a. Solution: DIPOLE MOMENT: dipole moment is sum of DC and AC terms. DC: h i ~p eff = Re 4πεm a3 fCM,DC~E DC g AC: h i ~p eff = Re 4πεm a3 fCM,AC~E AC exp( jωt) g so total dipole moment is h i ~p eff = 4πεm a3~E DC Re fCM,DC + fCM,AC α exp( jωt) g g

(17.77)

(17.78)

(17.79)

FORCE: Force is given by ~F = ~p · ∇~E

(17.80)

from above recall h i ~p eff = 4πεm a3~E DC Re fCM,DC + fCM,AC α exp( jωt) g g ~ and from definition of E : ∇~E = ∇~E DC Re [1 + α exp ( jωt)]

(17.81)

(17.82)

force is then ~F = ~p · ∇~E

(17.83) i ~F = 4πεm a3~E DC Re fCM,DC + fCM,AC α exp( jωt) · ∇~E DC Re [1 + α exp( jωt)] g g (17.84) h

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c Brian J. Kirby 660 Micro- and Nanoscale Fluid Mechanics,

CHAPTER 17. PARTICLE AND DROPLET ACTUATION: DIELECTROPHORESIS, MAGNETOPHORESIS, AND DIGITAL Soln Manual, MICROFLUIDICS July 21, 2010 http://www.cambridge.org/kirby h i ~F = 4πεm a3~E DC · ∇~E DC Re fCM,DC + fCM,AC α exp( jωt) Re [1 + α exp( jωt)] g g (17.85)   h i 3 ~F = 2πεm a ∇ ~E DC · ~E DC Re fCM,DC + fCM,AC α exp( jωt) Re [1 + α exp( jωt)] g g (17.86) h i ~F = 2πεm a3 ∇ ~E DC 2 Re fCM,DC + fCM,AC α exp( jωt) Re [1 + α exp ( jωt)] g g (17.87) Now evaluate the real parts of the expressions. Elec field component is easy: Re [1 + α exp ( jωt)] = 1 + α cos ωt

(17.88)

Dipole moment part is more complicated. first split up: h i h i h i Re fCM,DC + fCM,AC α exp( jωt) = Re fCM,DC + Re fCM,AC α exp( jωt) (17.89) g g g g Now separate out parts of fCM,AC and exp jωt: g fCM,AC = Re[ fCM,AC ] + jIm[ fCM,AC ] (17.90) g g g α exp jωt = α cos ωt + jα sin ωt (17.91) multiply and take real part: h i Re fCM,AC α exp jωt = Re[ fCM,AC ]α cos ωt − Im[ fCM,AC ]α sin ωt (17.92) g g g add back DC part: h i Re fCM,DC + fCM,AC α exp jωt = Re[ fCM,DC ]+Re[ fCM,AC ]α cos ωt −Im[ fCM,AC ]α sin ωt g g g g g (17.93) now that we have the dipole part, multiply through by 1 + α cos ωt: h i Re fCM,DC + fCM,AC α exp( jωt) g ×Reg [1 + α exp ( jωt)] = Re[ f ] + Re[ f ]α cos ωt CM,DC

CM,AC

g −Im[g fCM,AC ]α sin ωt + Re[ fCM,DC ]α cos ωt g 2 2 +Re[ fCM,AC ]α cos ωt − Im[g fCM,AC ]α2 sin ωt cos ωt g g (17.94) Now note that cos 2 ωt = 12 (1 + cos 2ωt) and sin ωt cos ωt = 21 sin 2ωt: h i Re fCM,DC + fCM,AC α exp( jωt) g ×Reg [1 + α exp ( jωt)] = Re[ fCM,DC ] + Re[ fCM,AC ]α cos ωt g −Im[g fCM,AC ]α sin ωt + Re[ fCM,DC ]α cos ωt g g 1 2 + 2 Re[ fCM,AC ]α (1 + cos 2ωt) − 12 Im[ fCM,AC ]α2 sin 2ωt g g (17.95) c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Plugging this expression into Eq. (17.87) and rearranging to group terms by their time dependence, we find ~F = πεm a3 ∇ ~E DC 2 n × α2 Re[ fCM,AC ] + 2Re[ fCM,DC ]   g g +2α cos ωt Re[ fCM,DC ] + Re[ fCM,AC ] g −2α sin ωtIm[ f g]

(17.96)

CM,AC

+α2 cos 2ωtRe[ g fCM,AC ] g o 2 −α sin 2ωtIm[ fCM,AC ] g TIME AVERAGED FORCE:
~F = α2 + 2 πεm a3 Re[ fCM ]∇ ~E DC 2 . g

(17.97)

17.7 Consider a homogeneous solid sphere of radius a located at the origin of a axisymmetric spherical coordinate system and surrounded by a homogeneous medium. Let the sphere have properties εp and σp and let the medium have properties εm and σm . Assume a sinusoidal electric field is applied, whose analytic representation is ~E = E0 zˆ exp jωt, where zˆ is the unit vector in the z axis direction and e it is understood that the electric field is given by the real part of ~E (specifically, e ~E = E0 zˆ cos jωt). Assume that the analytic representation of the electrical potential φ(r, θ,t) = φ0 (r, θ) exp jωt. e Assume that any net charge density at interfaces in this system is induced by the electric field and is also sinusoidal. Laplace’s equation is the governing equation inside the sphere and in the medium outside the sphere. In axisymmetric spherical coordinates, Laplace’s equation is given by     ∂ 1 ∂ ∂ 2 ∂ r φ0 + sin ϑ φ0 = 0 (17.98) ∂r ∂r sin ϑ ∂ϑ ∂ϑ The general solution of this equation can be written in terms of Legendre polynomials:6  ∞  φ0 (r, ϑ) =

∑

Ak rk + Bk r−k−1 Pk (cos ϑ) ,

(17.99)

k=0 6 This

is a classic example of solution of a differential equation by separation of variables and infinite sums of orthogonal functions. This is analogous to Fourier and Bessel function solutions in Cartesian and cylindrial coordinates, and the Legendre polynomial solution for a dielectric sphere in an infinite medium is solved in most electrodynamics texts. http://www.kirbyresearch.com/textbook

c Brian J. Kirby 662 Micro- and Nanoscale Fluid Mechanics,

CHAPTER 17. PARTICLE AND DROPLET ACTUATION: DIELECTROPHORESIS, MAGNETOPHORESIS, AND DIGITAL Soln Manual, MICROFLUIDICS July 21, 2010 http://www.cambridge.org/kirby where the Legendre polynomials Pk (x) are given by 1 Pk (x) = k 2 k!



d dx

k


k x2 − 1 .

(17.100)

In particular, P1 (x) = x; you can show that this is the only Legendre polynomial required for this problem. The boundary conditions for this problem are • φ0,p = φ0,m at r = a ∂φ

∂φ

• εp dn0,p = εm dn0,m at r = a e e • φ0 (r = ∞) = −E0 z = E0 r cos ϑ • φ0 (r = 0) is bounded Solution: (a) Solve for φ0 inside and outside the sphere. Write both solutions in terms of fCM . g Solution: Write the solution in the medium as ∞

φ0 (r, θ) =

∑

  Ak rk + Bk r−k−1 Pk (cos θ) ,

(17.101)

k=0

and the solution in the particle as ∞

φ0 (r, θ) =

∑

  Ck rk + Dk r−k−1 Pk (cos θ) ,

(17.102)

k=0

BC at r = 0

φ0 must be bounded, so all Dk = 0.

BC at r = ∞ All r−k−1 terms are zero, so ∞

−E0 r cos θ =

∑ Ak rk Pk (cos θ) ,

(17.103)

k=0

from which we can see that A1 = −E0 and all other Ak are zero. The only exception to this is that this problem has an arbitrary electrical potential offset (we could add a uniform potential to the solution everywhere, including the BC at infinity), which can be changed by changing A0 . With the BC as specified in the problem, A0 = 0. c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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BC at r = a ∞

∑

at r = a we can write, owing to continuity of potential,

  Ak rk + Bk r−k−1 Pk (cos θ) =

k=0

∞

∑

  Ck rk + Dk r−k−1 Pk (cos θ) , (17.104)

k=0

and, to satisfy Gauss’s law written at the interface,   ∞  ∞  εm ∑ Ak krk−1 − (k + 1)Bk r−k−2 Pk (cos θ) = εp ∑ Ck krk−1 − (k + 1)Dk r−k−2 Pk (cos θ) . e k=0 e k=0 (17.105) because all theta dependence is in Pk , these relations hold not only for the sums, but also for each term individually, so Ak rk + Bk r−k−1 = Ck rk + Dk r−k−1 ,

(17.106)

and     εm Ak krk−1 − (k + 1)Bk r−k−2 = εp Ck krk−1 − (k + 1)Dk r−k−2 . (17.107) e e for k 6= 1, Ak = Dk = 0. Taking the ratio of the two above equations, we can show that Bk and Ck are nonzero only if (17.108) −(k + 1)εm = kεp , e e which cannot be satisfied for nonnegative k and nonnegative ε. Thus for k 6= 1, e B = C = 0. k

k

As mentioned before, A0 is nonzero if the BC at infinity has a uniform constant (for this problem, we specified this constant to be zero). We set D0 = A0 and B0 = C0 = 0. From above, the only nontrivial solution is for k = 1. The BC at r = a gives A1 a + B1 a−2 = C1 a

(17.109)

and
εm A1 − 2B1 a−3 = εpC1 , (17.110) e e where from the BC at r = ∞ recall A1 = −E0 . The above equations can be solved algebraically (two equations, two unknowns). Rearrange first equation to solve for C1 : A1 a + B1 a−2 = C1 a (17.111) C1 = A1 + B1 a−3 Now substitute into second equation, solve for B1 :

εm A1 − 2B1 a−3 = εp A1 + B1 a−3 , e e http://www.kirbyresearch.com/textbook

(17.112)

(17.113)

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Now write out C1 :



εm − εp A1 = εp + 2εm B1 a−3 e e e e ε − εm p B1 = −A1 a3 e e εp + 2εm e e B1 = E0 a3 fCM g

  C1 = A1 + B1 a−3 = E0 fCM − 1 g From this, we can write the solution for φ0 : φ0,p = −E0 r cos θ + fCM E0 r cos θ g

(17.114) (17.115) (17.116)

(17.117)

(17.118)

and a3 φ0,m = −E0 r cos θ + fCM E0 2 cos θ g r

(17.119)

Thus the complex representation for the electrical potential is   φp = −E0 r cos θ + fCM E0 r cos θ exp jωt g e

(17.120)

and   a3 φm = −E0 r cos θ + fCM E0 2 cos θ exp jωt g r f

(17.121)

The real part of the complex solution gives the physical solution for the electrical potential. Note that fCM is complex, so we must multiply out fCM exp jωt. g g h i fCM exp jωt = Re( fCM ) cos ωt −Im( fCM ) sin ωt + j Im( fCM ) cos ωt + Re( fCM ) sin ωt g g g g g (17.122) It is convenient to put this in the form of a phase lag. If we say that fCM exp jωt = X exp( jωt + α), where X is the magnitude of the response and αg is the phase lag, then by algebra we can show that fCM exp jωt can also be written as g fCM exp jωt = fCM exp( jωt + 6 fCM ) , (17.123) g g g c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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  where fCM = fCM fCM ∗ and 6 fCM = atan2 Im( fCM ), Re( fCM ) . Given this, gcomplex g g in a different gform that gmore obviously we can g rewrite the solutions relates to the final physical solution:

φp = −E0 r cos θ exp jωt + fCM E0 r cos θ exp j(ωt + 6 fCM ) g g e

(17.124)

a3 φm = −E0 r cos θ exp jωt + fCM E0 2 cos θ exp j(ωt + 6 fCM ) r g g f

(17.125)

and

Here we see that the solution has two parts: (a) the applied electric field and (b) the response due to a dipole. This dipole is proportional in magnitude to the magnitude of fCM , and has a phase lag proportional to the angle of fCM . g g Taking the real part of this complex solution, we get the result for the potential in the system φp = −E0 r cos θ cos ωt + fCM E0 r cos θ cos(ωt + 6 fCM ) g g

(17.126)

a3 φm = −E0 r cos θ + fCM E0 2 cos θ cos(ωt + 6 fCM ) r g g

(17.127)

and

(b) Show that the result for the potential outside a sphere can we written as the sum of (i) the sinusoidally-varying applied electric field and (ii) a field that is equivalent to that created by a dipole at the origin with a sinusoidally-varying dipole moment but a different phase. What is the magnitude of this dipole? What is the phase lag? Solution: The potential generated by a dipole ~p = p~z is given by φ= http://www.kirbyresearch.com/textbook

1 p cos θ 4πεm r2

(17.128)

c Brian J. Kirby 666 Micro- and Nanoscale Fluid Mechanics,

CHAPTER 17. PARTICLE AND DROPLET ACTUATION: DIELECTROPHORESIS, MAGNETOPHORESIS, AND DIGITAL Soln Manual, MICROFLUIDICS July 21, 2010 http://www.cambridge.org/kirby The solution for φ in the medium is given by a3 φm = −E0 r cos θ + fCM E0 2 cos θ cos(ωt + 6 fCM ) (17.129) r g g From this, the first term is clearly from the applied electric field, whereas the second term can be written as a dipole: a3 1 p cos θ = f E0 2 cos θ cos(ωt + 6 fCM ) CM 4πεm r2 r g g

(17.130)

and thus p = 4πεm fCM E0 a3 cos(ωt + 6 fCM ) g g

(17.131)

This dipole is proportional in magnitude to the magnitude of fCM , and has a g phase lag proportional to the angle of fCM . g (c) Use Gauss’s law at the interface between particle and medium to calculate the induced charge density at the interface. Solution: Gauss’s law says Z

q= e define q = q0 exp jωt: e

S

~D ·~n dA e

(17.132)

~D ·~n dA

(17.133)

Z

q0 =

S

q0 = ~D m ·~r A − ~D p ·~r A

(17.134)

q0 = εm~E m ·~r A − εp~E p ·~r A

(17.135)

Note above that we are using the permittivity, NOT the complex permittivity. Now note ~E = −∇φ and q000 = q0 /A: q000 = −εm

∂ ∂ φ0,m + εp φ0,p ∂r ∂r

(17.136)

h i h i q000 = −εm −E0 cos θ − 2E0 cos θ fCM + εp −E0 cos θ + E0 cos θ fCM g g (17.137) c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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q000

 
εp − εm = −E0 cos θ εp + 2εm − fCM εp + 2εm g

(17.138)

From this result, we can see that the charge has a cos θ dependence, indicating that it is maximum in magnitude at the two poles (but opposite in sign). This should be intuitive. As written, the charge is also proportional to the difference between the Clausius-Mossoti factor and the infinite-frequency limit of the CM factor. Thus the charge is zero at infinitely high frequency. Because q000 is complex (the CM factor is complex), this means that the charge in general has a phase lag with respect to the applied field. This phase lag is given by   εp − εm 6 − fCM (17.139) εp + 2εm g

and the magnitude of the charge density oscillation is proportional to εp − εm (17.140) εp + 2εm − fCM g and, if we take the real part of q = q0 exp jωt, we get e   
εp − εm εp − εm 6 q = −E0 cos θ εp + 2εm − fCM cos ωt + − fCM εp + 2εm g εp + 2εm g (17.141) 00

(d) Given the result for the time-varying potential induced by a sphere subjected to a sinusoidal electric field, derive the relation for the DEP force as a function of particle size, Clausius–Mossotti factor, and other parameters. Solution: Evaluate the force on the dipole:

where and

~F = ~p · ∇~E ,

(17.142)

  ~p = 4πεm~E 0 a3 fCM cos ωt + 6 fCM g g

(17.143)

~E = ~E 0 cos ωt .

(17.144)

    ~F = 4πεm~E 0 a3 fCM cos ωt + 6 fCM · ∇ ~E 0 cos ωt g  g h  i ~F = 4πεm a3 fCM cos ωt + 6 fCM cos ωt ~E 0 · ∇ ~E 0 g g http://www.kirbyresearch.com/textbook

(17.145) (17.146)

c Brian J. Kirby 668 Micro- and Nanoscale Fluid Mechanics,

CHAPTER 17. PARTICLE AND DROPLET ACTUATION: DIELECTROPHORESIS, MAGNETOPHORESIS, AND DIGITAL Soln Manual, MICROFLUIDICS July 21, 2010 http://www.cambridge.org/kirby   chain rule says that ∇ ~E 0 · ~E 0 = 2~E 0 · ∇~E 0 . also, ~E 0 · ~E 0 = ~E 0 2 ; So   ~F = 2πεm a3 fCM cos ωt + 6 fCM cos ωt∇ ~E 0 2 (17.147) g g writing now the cos and sin components of the dipole:   ~F = 2πεm a3 Re( fCM ) cos ωt − Im( fCM ) sin ωt cos ωt∇ ~E 0 2 (17.148) g g   ~F = 2πεm a3 Re( fCM ) cos 2 ωt − Im( fCM ) sin ωt cos ωt ∇ ~E 0 2 (17.149) g g apply trig identities:      ~F = 2πεm a3 Re( fCM ) 1 + 1 cos 2ωt − Im( fCM ) 1 sin 2ωt ∇ ~E 0 2 2 2 2 g g (17.150) i h  ~F = 2πεm a3 Re( fCM ) + fCM cos 2ωt + 6 fCM ∇ ~E 0 2 (17.151) g g g    ~F = 2πεm a3 Re( fCM ) + fCM cos 2 ωt + 1 6 fCM ∇ ~E 0 2 2 g g g

(17.152)

From this result, we can see two components. The DC response is proportional to Re( fCM ). There is also a response at twice the applied frequency, which is g proportional to fCM and has a phase lag equal to 21 6 fCM . g g 17.8 Show that, in a uniform medium with properties εm and σm and a uniform applied field ~E ext , the induced dipole on a particle of radius a2 consisting of a core of radius a with properties εp and σp surrounded by a shell with properties εm and σm is the same as the induced dipole on a particle with radius a with properties εp and σp . Solution: the solution for this problem is not available 17.9 Consider pressure-driven flow of a particle suspension through a microchannel that has two electrodes as shown in Fig. 17.14. If an AC signal is passed between the two electrodes (labeled V1 and GND), what will happen to the particles? What physical phenomena govern this system? Describe in detail the parameters that govern the device, fluid, and particles, in this system. What parameters must be known to be able to predict the performance of this system? How could you change parameters to: c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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• make particle trajectories independent of the electric field? • make different particles react differently to the electric field? • separate different types of particles? • trap and store particles?

Solution: Comments below highlight key points, and are not necessarily a complete description of the problem. (a) physical phenomena. Key phenomena affecting particles include stokes drag due to the pressure driven flow, gravity, magnetophoresis, dielectrophoresis, electrophoresis, electroosmosis. MAP can be ignored. Gravity can often be ignored for small particles, but not always. electrophoresis/electrophoresis will likely balance out owing to the AC nature of the field, although in some situations AC electroosmosis or electrothermal effects can be observed. dielectrophoresis and Stokes drag are likely the dominant forces. There is also a hydrodynamic lift force on a particle in a nonuniform fluid velocity field, but this is likely negligible in this case owing to the likely modest velocities, as are electrostatic attraction/repulsion between the particle and wall, likely negligible owing to shielding by the Debye layer. (b) parameters. key parameters include the voltage, frequency, electrode size and spacing, channel height, fluid and particle permittivities and conductivities, particle size, pressure gradient, and fluid viscosity. Secondary parameters might include zeta potentials of particle and device. Critical equations are the DEP force equation, the definition of CM factor, Poiseuille flow relations, Laplace equation for the E field, and the Stokes equation for drag on a spherical particle. (c) making particle trajectories independent of E field. Strictly speaking, this requires a CM factor of zero. If we loosen this to imply that the particle trajectories are independent of the applied voltage, we’d separate the electrodes by large distances or make the channel height so large that particles rarely got close to the electrodes. (d) making different particles react differently. Here we will get different particle behavior owing to size for sure. Particles with different permittivities and conductivities will act differently if we make sure the frequency is in a range where the CM factor is changing (i.e.,not pinned at 1 or -0.5). This often requires frequencies between 1×103 Hz and 1×106 Hz. (e) separating different types of particles. Particles experiencing positive DEP will be attracted to the electrodes, and will be motionless (if trapped) or slower (if close to the wall where the poiseuille flow rates are lower). Thus particles with strong positive DEP forces will elute more slowly. By varying the voltages, electrode spacings, or frequencies, this device could be operated in separation http://www.kirbyresearch.com/textbook

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Figure 17.13: Clausius–Mossotti factors (real part). Ideal region for distinguishing between these two particles is 1–50 Hz.

Figure 17.14: A proposed microfluidic device.

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mode (basically an FFF-DEP separation) or in batch mode (trap one type of particles, flush the rest, then turn off the voltage and elute). If in separation mode, a technique to inject a bolus of particle-laden fluid would be required. If gravity is included, the relative effects of DEP and gravity can be evaluated with this technique. (f) trapping and storing particles. Same as last, whereby cranking up the voltage, positive DEP particles can be trapped. Negative DEP particles can be trapped only at the front of the electrode array, not in the array. So this device works poorly for trapping negative DEP particles. (g) Supplementary reading. References that discuss this type of device include Gascoyne PR et al Electrophoresis 23:1973 (2002), Huang Y et al Biophys J 73:1118 (1997), Hughes MP Electrophoresis 23:2569 (2002), Wang XB et al Anal Chem 72:832 (2000). 17.10 Calculate the force on a 10 µm aluminum particle suspended in water if it is in a magnetic field aligned in the x direction with magnitude 1×104 A/m and field gradient (in the same direction) of 1×108 A/m2 . Solution: the solution for this problem is not available 17.11 Consider an idealized 10 µm magnetic particle with a magnetic susceptibility of 3 and an infinitely large linear range, suspended in a solution whose magnetic susceptibility can be approximated as zero. (a) Calculate the force on this particle if it is in a magnetic field with magnetic field in the x direction with magnitude 1×104 A/m and field gradient (in the same direction) of 1×108 A/m2 . (b) Assume that the force described above is applied in a 20 µm microchannel by an external magnet to induce the particles to become trapped on one wall of the channel so that the system may be flushed. Assume that the field and field gradient can be approximated as being uniform for the purposes of this calculation. Of interest is the settling time, i.e., the time required for all magnetic particles in the system to be pulled to one wall. Assuming the solution is water and the particles can be modeled as Stokes spheres, calculate this settling time. Solution: the solution for this problem is not available 17.12 Show how Eq. (17.59) follows from Gauss’s law of magnetism, listed in Eq. (5.5), in the limit where the magnetic susceptibilities of particle and medium are independent of space, magnetic field, and magnetic field history. http://www.kirbyresearch.com/textbook

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CHAPTER 17. PARTICLE AND DROPLET ACTUATION: DIELECTROPHORESIS, MAGNETOPHORESIS, AND DIGITAL Soln Manual, MICROFLUIDICS July 21, 2010 http://www.cambridge.org/kirby Solution: the solution for this problem is not available 17.13 Consider the limit in which the particle magnetic susceptibility varies with the applied magnetic field, but can be assumed constant within the particle because the magnetic field within the particle is approximately constant. Show how the force on a magnetic particle is affected by saturation. Solution: the solution for this problem is not available 17.14 Derive Eq. (??) using geometric arguments and the Young–Laplace equation. Solution: For a spacing of d, we have R=

d . cos θ

(17.153)

This is not R = cosd θ as shown before for a spherical interface, because we are considering a cylindrical interface (i.e.,one of the radii of curvature is infinite). We can generalize using geometric arguments that this can be rewritten as R=

2d , cos θ1 + cos θ2

(17.154)

where θ1 and θ2 are the contact angles on the two sides. Given the Young–Laplace equation (Eq. 1.46), the pressure drop across the unmodified interface is 2γlg cos θ , d and the pressure drop across the modified interface is ∆plg =

γlg (cos θ + cos θw ) . d The net pressure drop is the difference between the two: ∆plg =

∆P =

γlg (cos θ − cos θw ) . d

(17.155)

(17.156)

(17.157)

Here, the sign of the result depends on how we define the pressure differentials and is arbitrary. 17.15 If the contact angle of water on a 50nm-thick layer of Teflon AF (ε/ε0 = 2) is 120◦ and the surface tension of water in contact with air is 73mN/m, calculate the voltage required to reduce the contact angle to 85◦ . c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Solution: The Lippman equation is: cos θw = cos θ +

ε V2. 2wγlg

(17.158)

Substituting in parametric values from the problem statement, we find V = 15.6V .

(17.159)

17.16 Consider a sphere in a uniform harmonic applied field with frequency ω. Show that, as ω → 0, the Clausius–Mossotti factor approaches σp − σm fCM = . g σp + 2σm

(17.160)

Solution: the solution for this problem is not available 17.17 Consider a sphere in a uniform harmonic applied field with frequency ω. Show that, as ω → ∞, the Clausius–Mossotti factor approaches εp − εm fCM = . g εp + 2εm

(17.161)

Solution: the solution for this problem is not available 17.18 Two electrodes are located at y = ±50 µm and are driven with a voltage difference of V0 cos ωt, with V0 = 1 V. Two electrodes are located at x = ±50 µm and are driven with a voltage difference of V0 sin ωt. Assume that the field at the origin is well approximated by E0 sin ωt xˆ + E0 cos ωt yˆ , where E0 = 100 V/cm. Derive ~E 0 f and evaluate Re(~E 0 ) × Im(~E 0 ) for this electric field. What is the direction of the f the rotation f pseudovector defining of a particle at the origin if Im( fCM ) < 0? g 17.19 Starting with the general relation for the effective permittivity of a spherical particle with a core and one outer shell:  3  ε1 −ε2 a2 + 2 e e ε1 +2ε2   a3 εp = ε2  13 e (17.162) e , ε 1 −ε2 a2 e e − e e 3 ε1 +2ε2 a1 e e derive a relation for the effective permittivity if the outer shell has a thickness ∆a  a2 . http://www.kirbyresearch.com/textbook

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

a32  a31

εp = ε2  e e

ε1 −ε2 + 2 εe1 +2ε e2



 e , ε1 −εe 2 − εe1 +2ε e2 e e Write a linear expansion for the cubed radius ratio: a32 a31

a32 /a31 = 1 + 3∆a/a2 ,

(17.163)

(17.164)

and find ε1 + ∆a a2 εp = ε2 e e e ε2 + ∆a a2 e


ε1 + 2ε2 e e
. ε1 + 2ε2 e e

(17.165)

17.20 Consider a sphere of radius a with uniform properties εp and σp embedded in a medium with properties εm and σm . Show that, if the interface has a charge density q00 , the solution for the electrical potential outside the sphere if an applied field ~E ext = E0 zˆ cos ωt has a term given by φ=

a2 q00 . εr

(17.166)

Solution: the solution for this problem is not available 17.21 Consider a polystyrene microsphere with εp = 2.5ε0 and σp = 0.05 µS/cm, and an interfacial potential of -50 mV when embedded in a pH = 7, 10 mM KCl solution. A harmonic, uniform electric field is applied. Model the EDL as a conducting layer with a surface conductance owing to the increased ion concentration in the double layer. Ignore surface conductance from electroosmotic convection, and simplify the math by approximating the potential distribution in the EDL as exponential. How does your prediction of the dielectrophoretic force felt by the sphere change depending on whether surface conductance is included? Solution: the solution for this problem is not available ~ 17.22 Integrate ~T · nˆ over the surface of a sphere exposed to a uniform, harmonic applied field, and show that the force felt by the sphere is given by   ~F = πεm a3 ∇(~E 0 · ~E 0 ) Re( fCM ) + Re( fCM ) cos 2ωt + Im( fCM ) sin 2ωt . (17.167) g g g c Brian J. Kirby Micro- and Nanoscale Fluid Mechanics,

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Solution: the solution for this problem is not available 17.23 Consider a spherical particle with radius a2 , with a core with properties ε1 and σ1 and a thin shell with thickness ∆a and properties ε2 and σ2 , where ε2  ε1 and σ2  σ1 . Estimate the effect of this thin shell on the effective properties of the particle by replacing the sphere with a 1D geometry consisting of a cube of side length a2 with properties corresponding to the core, and two thin layers of thickness ∆a on the right and left side with properties corresponding to the shell. Model the electric field as being uniform and applied vertically, and model the system as consisting of three capacitors in parallel: one for the core and one each for the two layers on either side. Using the Helmholtz capacitor relation C = εA/d, determine the capacitances for the components and the system. Using the total capacitance and the geometry of the entire system, derive the effective permittivity of the system and show that it is given by ∆a εp = ε1 + 2ε2 . (17.168) e a2 e e This geometry is a suitable approximation for a sphere with a conductive or highpermittivity shell, because the field lines for such a sphere can for the most part be approximated as being parallel to the shell. Solution: For the core cube, the capacitance is ε1 a2 2 C=e . a2

(17.169)

For each thin side layer, the capacitance is ε2 ∆a a2 C=e . a2

(17.170)

The summed capacitances for capacitors in parallel give a total capacitance of

Now, define εp such that e

Ctot = ε1 a2 + 2ε2 ∆a . e e εp a2 2 Ctot = e , a2

(17.171)

(17.172)

which leads to ∆a εp = ε1 + 2ε2 . e e e a2

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(17.173)

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CHAPTER 17. PARTICLE AND DROPLET ACTUATION: DIELECTROPHORESIS, MAGNETOPHORESIS, AND DIGITAL Soln Manual, MICROFLUIDICS July 21, 2010 http://www.cambridge.org/kirby 17.24 Consider a spherical particle with radius a2 , with a core with properties ε1 and σ1 and a thin shell with thickness ∆a and properties ε2 and σ2 , where ε2  ε1 and σ2  σ1 . Estimate the effect of this thin shell on the effective properties of the particle by replacing the sphere with a 1D geometry consisting of a cube of side length a2 with properties corresponding to the core, and a thin layer of thickness ∆a on the top of the cube with properties corresponding to the shell. Model the electric field as being uniform and applied vertically, and model the system as consisting of two capacitors in series: one for the core and one for the thin layer on top. Using the Helmholtz capacitor relation C = εA/d, determine the capacitances for the components and the system. Using the total capacitance and the geometry of the entire system, derive the effective permittivity of the system and show that it is given by εp = e

ε1 ε2 ∆a e e a2 ε1 + ε2 e

.

(17.174)

e

This geometry is a suitable approximation for a sphere with an insulating or lowpermittivity shell because the field lines for such a sphere can for the most part be approximated as being normal to the shell. Solution: For the core cube, the capacitance is ε1 a2 2 C=e . a2

(17.175)

ε2 a2 2 . C=e ∆a

(17.176)

For the top layer, the capacitance is

The two capacitors in series give a total capacitance of 1 1 ∆a = + . Ctot ε1 a2 ε2 a2 2 e e Now, define εp such that e

εp a2 2 Ctot = e , a2

(17.177)

(17.178)

which leads to εp = e

ε1 ε2 ∆a e e a2 ε1 + ε2 e

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.

(17.179)

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17.25 Consider an ideal dielectric sphere of radius a and uniform permittivity εp located at the origin and embedded in an ideal dielectric medium of permittivity εm . A point charge q is located outside the particle at a distance zq from the origin along the z axis. If the distance between the point charge and a point in the medium is defined as ∆r, the expression 1/∆r can be written (for r < zq ) in terms of Legendre polynomials as  k 1 1 ∞ r = ∑ Pk (cos ϑ) . ∆r zq k=0 zq

(17.180)

(a) Solve for the electric field inside and outside the particle in terms of Legendre polynomials. (b) Define the multipole moment p(k) = 4πεm Ak , where Ak is the Legendre polynomial coefficient from Appendix F, and define p(k) in terms of q, zq , a, and material properties. (c) At the origin, evaluate the electric field and the successive partial derivatives of the electric field with respect to z. Report these functions in terms of material properties, q, and zq . Your result at the origin should give, for the electric field,   q ~E = − zˆ , (17.181) 4πεm zq 2 and for the derivatives, you should obtain ∂k~E =− ∂zk

(k + 1)!q 4πεm zk+2 q

! zˆ .

(17.182)

(d) Write the multipole moment in terms of the material properties, a, and the field and its derivatives at the origin, i.e., show that the multipole moment can be written as 4πεm K (k) a2k+1 ∂k−1 Ez p(k) = , (17.183) (k − 1)! ∂zk−1 where the kth-order polarization coefficient K (k) is written as K (k) =

εp − εm , kεp + (k + 1)εm

(17.184)

and Ez is the z component of the electric field. (e) Given that the problem considers an ideal dielectric sphere in an ideal dielectric medium, evaluate the dipole moment p(1) and show that it is equal to the dipole caused by a uniform electric field; also, evaluate the first-order polarization coefficient K (1) and show that it is identical to the Clausius–Mossotti factor. http://www.kirbyresearch.com/textbook

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CHAPTER 17. PARTICLE AND DROPLET ACTUATION: DIELECTROPHORESIS, MAGNETOPHORESIS, AND DIGITAL Soln Manual, MICROFLUIDICS July 21, 2010 http://www.cambridge.org/kirby (f) By induction, write the multipole moment p(k) and polarization coefficient K (k) e e for a particle with complex permittivity εp embedded in a medium with complex permittivity εm that has a fixed charge eat z = zq with an oscillatory magnitude e q = q0 cos ωt. Solution: the solution for this problem is not available 17.26 It is illustrative to examine the transient response of a uniform, spherical particle’s effective dipole moment to the instantaneous application of a uniform, DC field, that is Eext = 0 for all t < 0 and Eext = E0 zˆ for all t ≥ 0. In this case, the effective dipole moment is given by n h  i σp −σm t ~p (t) = 4πεm a3 E0 zˆ 1 − exp − σp +2σ m   τMW o . (17.185) εp −εm t + εp +2εm exp − τMW Evaluate the effective dipole in the two limits t  τMW and t  τMW , where the Maxwell–Wagner time constant is given by τMW =

εp + 2εm . σp + 2σm

(17.186)

Solution: the solution for this problem is not available

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EXAMPLE PROBLEM B.3 What is the dielectric constant of the following solutions at 25 C? 1. 1-mM NaCl 2. 10-mM NaCl 3. 100-mM NaCl 4. 1-M NaCl Solution: The dielectric increment of Na+ is −8 M−1 , and the dielectric increment of Cl− is −3 M−1 . Thus the total effect of NaCl is −11 M−1 . Assuming that the 25 C dielectric constant of water is 78.4, the dielectric constants of these solutions are 78.4, 78.3, 77.3, and 67.4.

B.6

Supplementary reading

A number of texts [62, 278, 279, 283, 284] cover water properties in detail and serve as useful background for the values cited here. Segel’s text [288] is excellent for solution properties and acid–base chemistry, as it provides many useful definitions, numerous worked exercises, and problems with solutions. It also provides extensive reference pKa and activity data.

B.7

Exercises

Assume dilute solution results throughout. B.1 Derive the Henderson–Hasselbach equation from the rate equations for acid dissociation and recombination. Solution: The dissociation reaction is Ka =

[H+ ][A− ] . [HA]

(B.20)

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[H+ ][A− ] , [HA]

(B.21)

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and use logarithm identities: − log Ka = − log[H+ ] − log

[A− ] . [HA]

(B.22)

Now, we define pKa = − log Ka and pH = − log[H+ ], giving pKa = pH − log

[A− ] , [HA]

(B.23)

or, rearranging: log

[A− ] = pH − pKa . [HA]

(B.24)

B.2 Given the room-temperature Keq for water dissociation, show that relation (B.11) is approximately correct. Solution: The basic reaction is H2 O ←→ H+ + OH− .

(B.25)

For this reaction, the equilibrium constant is given by Keq =

[H+ ][OH− ] . [H2 O]

(B.26)

At 25◦ C, Keq = 1.8 × 10−16 mol/L. If we assume that [H2 O] is constant, then [H2 O] can be calculated from the density of water and its molar mass: [H2 O] =

mol 1000g mol × = 55.6 = 55.6M . L 18g L

(B.27)

From this, at room temperature [H+ ][OH− ] = 1.8 × 10−16 mol/L × 55.6mol/L ' 1 × 10−14 mol2 /L2 .

(B.28)

Thus, pH + pOH ' 14 ,

(B.29)

B.3 Consider two separate systems: one that is 100 mL of pure water at and one with 100 mL containing 100 mM each of KH2 PO4 and K2 HPO4 . What is the resulting pH if 1 mL of 1 N HCl is added to each solution? What does this tell you about the buffering action of the phosphate ions? 706

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Solution: the solution for this problem is not available B.4 Given its molecular weight (32.04 g/mol) and density (791 kg/m3 ), calculate the molarity of pure methanol. Solution: [CH3 OH] =

791 g mol mol × = 24.7 = 24.7M . L 32.04g L

(B.30)

B.5 Calculate the pH of the following solutions: (a) 1-M HCl, (b) 0.02-M H2 SO4 , (c) 10-mM KOH. Solution: 0, 1.4, and 12. B.6 0.1 mol of a weak acid, HA, is added to pure water, and the resulting pH of the system is measured to be 2.2. What is the pKa of the weak acid? Solution: Because the resulting pH is 2.2, we know [H+ ] = 1×10−2.2 and therefore [A− ] = 1 × 10−2.2 . From this, we know [HA] = 0.1 − 1 × 10−2.2 . We can substitute these into [A− ] pKa = pH − log (B.31) [HA] to get pKa = 3.37 .

(B.32)

B.7 1 mol of a weak acid, HA, is added to pure water, and the resulting pH of the system is measured to be 4.15. What is the pKa of the weak acid? http://www.cambridge.org/kirby

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Solution: Because the resulting pH is 4.15, we know [H+ ] = 1 × 10−4.15 and therefore [A− ] = 1 × 10−4.15 . From this, we know [HA] = 1 − 1 × 10−4.15 . We can substitute these into [A− ] (B.33) pKa = pH − log [HA] to get pKa = 8.30 .

(B.34)

B.8 25 mmol of a weak acid, HA, is added to pure water, and the resulting pH of the system is measured to be 5.9. What is the pKa of the weak acid? Solution: Because the resulting pH is 5.9, we know [H+ ] = 1×10−5.9 and therefore [A− ] = 1 × 10−5.9 . From this, we know [HA] = 0.025 − 1 × 10−5.9 . We can substitute these into [A− ] pKa = pH − log (B.35) [HA] to get pKa = 10.2 .

(B.36)

B.9 Phosphoric acid (H3 PO4 ) has three protons that systematically dissociate with pKa ’s −3 −2 of 2.1, 7.2, and 12.3 to form H2 PO− 4 , HPO4 , and PO4 . Thus we can write three equilibrium equations: pKa1

H3 PO4 ←→ H+ + H2 PO− 4 ,

(B.37)

pKa2 −2 + H2 PO− 4 ←→ H + HPO4 , pKa3 −3 + HPO−2 4 ←→ H + PO4 . in the form PO−3 4 , we would say that three in the form HPO−2 4 , we would say that two

(B.39)

−2 − 3[PO−3 4 ] + 2[HPO4 ] + [H2 PO4 ] ,

(B.40)

(B.38)

If a phosphate ion exists protons have dissociated. If it existed protons have dissociated, and so on. At any given pH, there is a statistical distribution of the four possible dissociation states of phosphate and thus an average number of protons that have dissociated. So we can define the concentration of dissociated protons as

and the total concentration of phosphate ions (in all forms) is −2 − [PO−3 4 ] + [HPO4 ] + [H2 PO4 ] + [H3 PO4 ] .

(B.41)

Calculate and plot (as a function of pH) the average number of protons that have dissociated per phosphate ion. 708

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Solution: We must solve for number of protons dissociated per phosphate, given by −2 − 3[PO−3 4 ] + 2[HPO4 ] + 1[H2 PO4 ] . (B.42) −2 − [PO−3 4 ] + [HPO4 ] + [H2 PO4 ] + [H3 PO4 ] from the Henderson–Hasselbach equation, we know that pH − pKa1 = log

[H2 PO− pH−pKa1 4] ⇒ [H2 PO− . 4 ] = [H3 PO4 ] × 10 [H3 PO4 ]

(B.43)

Also,

− pH−pKa2 2pH−pKa1 −pKa2 ⇒ [HPO−2 , (B.44) [HPO−2 4 ] = [H2 PO4 ]×10 4 ] = [H3 PO4 ]×10 −2 pH−pKa3 3pH−pKa1 −pKa2 −pKa3 [PO−3 ⇒ [PO−3 . 4 ] = [HPO4 ] × 10 4 ] = [H3 PO4 ] × 10 (B.45) We substitute these into the formula for dissociation of protons per phosphate (Fig. B.3). It is also illustrative to plot the populations of all four forms of phosphate (Fig. B.4).

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Figure B.3: Dissociated protons as a function of pH

Figure B.4: Populations of forms of phosphate as function of pH.

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C.5

Exercises

C.1 Compare the length of the position vectors for these three cases: (a) x = 1, y = 1, z = 1. (b)

= 1, θ = 1, z = 1.

(c) r = 1, ϑ = 1, ϕ = 1. Compare your results with the magnitude of the velocity vectors for these three cases: (a) u = 1, v = 1, w = 1. (b) u = 1, uθ = 1, uz = 1. (c) ur = 1, uϑ = 1, uϕ = 1. Solution: the solution for this problem is not available C.2 In Cartesian coordinates, show that the operator ∇ · ∇, when operating on the vector ~a = (a , b , c), gives the result  2  ∂ a ∂2 a ∂2 a + + 2 2 2  ∂∂x2 b ∂∂y2 b ∂∂z2 b   2 + 2 + 2 . (C.90) ∂y ∂z   ∂x ∂2 c ∂2 c ∂2 c + ∂y2 + ∂z2 ∂x2 Solution: The ∇ · ∇ operator, when applied to the vector~a , leads to ∇ · ∇~a ,

(C.91)

which implies that we evaluate the gradient of ~a and then evaluate the divergence of the result. Using Cartesian coordinates, we write this as  ∂a ∂a ∂a  ∂x ∂y ∂z    ∂b ∂b ∂b  ∂ ∂ ∂ · (C.92)  ∂x ∂y ∂z ∂x ∂y ∂z  . ∂c ∂x

∂c ∂y

∂c ∂z

Evaluating, we find   

∂ ∂a ∂x ∂x ∂ ∂b ∂x ∂x ∂ ∂c ∂x ∂x

 ∂ ∂a ∂ ∂a + ∂y ∂y + ∂z ∂z ∂ ∂b ∂ ∂b  + ∂y ∂y + ∂z ∂z  , ∂ ∂c ∂ ∂c + ∂y ∂y + ∂z ∂z 730

(C.93)

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which we write as    

∂2 a ∂x2 ∂2 b ∂x2 ∂2 c ∂x2

2

2

+ ∂∂ya2 + ∂∂za2



 2 2 + ∂∂yb2 + ∂∂zb2  . ∂2 c ∂2 c + ∂y2 + ∂z2

(C.94)

Note that the analysis above treats ∇ as a vector, when in fact it is a vector operator. In Cartesian coordinates, we can simply write the ∇ operator as a vector with com∂ ∂ ∂ ponents ( ∂x , ∂y , ∂z ) and treat the divergence as if it is a dot product of the ∇ vector with~a . We cannot do this with any general coordinate systems (and, specifically, we cannot do this in cylindrical or spherical coordinates). So this is convenient shorthand that is correct for Cartesian coordinates, but it is not a sound general approach. Following this approach in cylindrical coordinates will lead to the wrong result.   ~ ~ C.3 Show that ∇ · −pδ = −∇p.

~ Solution: Start with ∇ · −p~δ , and evaluate: ~ ∇ · −p~δ ,    −p 0 ∂ ∂ ∂  0 −p , , · ∂x ∂y ∂z 0 0  ∂ ∂ ∂ ∂x (−p) + ∂y 0 + ∂z 0  ∂ ∂ ∂  ∂x 0 + ∂y (−p) + ∂z 0 ∂ ∂ ∂ ∂x 0 + ∂y 0 + ∂z (−p)

(C.95) 

0 0 , −p   ,

(C.96)

(C.97)



 − ∂p ∂x  ∂p   − ∂y  , − ∂p ∂z

(C.98)

−∇p .

(C.99)

which leads finally to

C.4 By converting both to Cartesian form, show that cylindrical unit vectors can be written in terms of spherical unit vectors as follows: ˆ = rˆ sin ϑ + ϑˆ cos ϑ , θˆ = ϕˆ , zˆ = rˆ cos ϑ − ϑˆ sin ϑ . http://www.cambridge.org/kirby

731

(C.100) (C.101) (C.102)

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Solution: the solution for this problem is not available C.5 By converting both to Cartesian form, show that spherical unit vectors can be written in terms of cylindrical unit vectors as follows: rˆ =

ˆ√

ϑˆ =

ˆ√

2 + z2

z 2 + z2

+ zˆ √ − zˆ √

z , 2 + z2

(C.103)

,

(C.104)

2 + z2

ϕˆ = θˆ .

(C.105)

Solution: the solution for this problem is not available C.6 By converting to Cartesian coordinates, show that the components of a position vector can be converted from spherical to cylindrical with the following relations: = r sin ϑ ,

(C.106)

θ = ϕ,

(C.107)

z = r cos ϑ .

(C.108)

Solution: the solution for this problem is not available C.7 By converting to Cartesian coordinates, show that the components of a position vector can be converted from cylindrical to spherical with the following relations: p 2 + z2 , (C.109) r = ϑ = atan2 ( , z) ,

(C.110)

ϕ = θ.

(C.111)

Solution: the solution for this problem is not available

732

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E.4

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Supplementary reading

A number of texts contain thorough lists of nondimensional parameters with brief descriptions of their meaning; one example can be found in [33]. Nondimensionalization of the Navier–Stokes equations and dynamic similitude is discussed precisely in [24]. For those interested in a presentation of transport that thoroughly stresses nondimensionalization throughout, Ref. [30] makes nondimensionalization a focus. This text neglects many nondimensional parameters, for example, parameters related to buoyancy, such as the Bond, Grashof, and Rayleigh numbers; parameters related to chemical reaction, such as the Damköhler numbers; parameters related to macromolecule relaxation, such as the Weissenberg and Deborah numbers; and parameters related to thermal transport, such as the Brinkman, Biot, and Fourier numbers. A thorough microfluidic review with a focus on the role of nondimensionalization (and mention of many parameters omitted here) can be found in [293].

E.5

Exercises

E.1 Derive Eq. (E.8) from the dimensional Navier–Stokes equations by substituting in the relations in Eqs. (E.2)–(E.6). Solution: the solution for this problem is not available

E.2 We wish to study the flow of water through a circular microchannel with radius R = 4 µm and mean velocity 100 µm/s. We expect experimental resolution of the velocity profile to 1/100 of the channel. Unfortunately, most experimental techniques for measuring fluid velocity cannot localize the measurement to a region as small as 40 nm. Describe how you might select the channel size and fluid such that you could make a dynamically similar measurement in a larger channel and achieve the desired resolution. Solution: Flows are similar if the Reynolds number is the same in each case. We need to increase ` and keep Re the same as the microchannel-water case. We could simply decrease the velocity and increase the size of the channel, although in a larger channel it will be difficult to reproduce such a small velocity with water. The best result will involve a higher-viscosity fluid such as glycerin, in which case the channel could be made roughly on the order of 1 m with the same velocity. There are many possible solutions. http://www.cambridge.org/kirby

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This leads us to a nondimensional del operator, giving ∇∗ =

∇ 1/`

(E.32)

∇ . 1/`2

(E.33)

~u , U

(E.34)

and ∇∗2 = We nondimensionalize the velocity by ~u ∗ =

and the time by the characteristic time of the boundary conditions: t∗ =

t . tc

(E.35)

Returning to the dimensional equation: ∂ci +~u · ∇ci = Di ∇2 ci , ∂t

(E.36)

1 1 ∂ci 1 +U~u ∗ · ∇ci = Di 2 ∇2 ci . ∗ tc ∂t ` `

(E.37)

we substitute and find

Multiplying both sides by `/U, we get 1 ∂ci Di 2 +~u ∗ · ∇ci = ∇ ci . ∗ St ∂t U`

(E.38)

From this we see that Pe appears whether we nondimensionalize ci or not. This is because the passive scalar mass transfer equation is linear in the concentration. To match Eq. (E.11), we divide through by a reference concentration to get 1 ∂c∗i 1 ∗2 ∗ +~u ∗ · ∇∗ c∗i = ∇ ci . ∗ St ∂t Pei

(E.39)

For steady boundary conditions, St = 1 and drops out of the equation. E.5 From the standpoint of dimensional analysis, define the physical parameters and the fundamental physical quantities for the Poisson–Boltzmann equation, and explain why two of the nondimensional quantities that govern the equation are the nondimensional voltage and nondimensional length.

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and ~ ~G 0 =

~ ~ 1 ~~δ + ∆ r ∆ r 8πη∆r ∆r2

! .

(F.38)

The impact of these analytical solutions is that they both provide critical insight into the distance dependence of the velocity fields, and enable solutions in simplified cases.

F.5

Supplementary reading

Griffiths [55] and Jones [61] are both excellent sources for Laplace equation solutions. Griffiths [55] is expository and provides an accessible introduction. It also covers the separation of variables solution for axisymmetric coordinates. Jones [61] covers the general multipolar theory in considerable detail in an appendix, with specific attention to the solution of electrostatic and electrodynamic problems involving particles in a medium. Jackson [56] provides the most thorough treatment of separation of variables and the Green’s function solution techniques. Derivations of the Green’s function solutions for Stokes flow can be found that use Fourier transform techniques [32] or a limiting approach in which a sphere’s radius approaches zero in a limiting process. Kim and Karrila [89] present a detailed description of the multipolar solutions for Stokes flow, which focuses on mathematical formulation. Russel et al. [32] include a detailed discussion of this material, including a general formulation for Stokes multipoles. Happel and Brenner [88] also provide a useful resource.

F.6

Exercises

F.1 By substitution, show that each of the first three Legendre polynomials, defined by Eq. (F.3), satisfy Legendre’s differential equation:  
d 2 d 1−x Pk (x) + k(k + 1)Pk (x) = 0 . (F.39) dx dx Solution: For k = 0, d dx



1 − x2


d (1) + 0(1)(1) = 0 , dx

(F.40)

which leads to 0+0 = 0.

766

(F.41)

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For k = 1, d dx




d 1−x x + 1(2)(x) = 0 , dx

(F.42)


 d  1 − x2 + 2x = 0 , dx

(F.43)

2

which leads to −2x + 2x = 0 .

(F.44)

For k = 2, d dx




d 1−x dx 2



   1 2 1 2 (3x − 1) + (3x − 1) = 0 , 2 2


 1 d  1 − x2 6x + 9x2 − 3 = 0 , 2 dx
d 6x − 6x3 + 18x2 − 6 = 0 , dx

(F.45)

(F.46) (F.47)

which leads to −18x2 + 6 + 18x2 − 6 = 0 .

(F.48)

F.2 Consider the axisymmetric Laplace equation, Eq. (F.1). Find a general solution using separation of variables. (a) Start by assuming that the solution for φ can be written as the product of two functions, one which is a function of r only and one which is a function of ϑ only: φ = R(r)Θ(ϑ) . (F.49) (b) Insert this relation into the Laplace equation and rearrange the equation so that all terms involving r are on one side and all terms involving ϑ are on the other. (c) Because each side of this equation must be a constant, each side can be rewritten as an ordinary differential equation with an arbitrary constant. This arbitrary constant is typically written in a form proportional to k(k + 1) in anticipation of the boundary conditions that the physical problem will need to satisfy. The R side of the equation can be rearranged to obtain an Euler differential equation: r2 http://www.cambridge.org/kirby

d2 d R + 2r R = k(k + 1)R , 2 dr dr 767

(F.50)

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which is satisfied by polynomial solutions in r. The ϑ side of this equation is most easily handled by rewriting the equation by applying the transform x = cos ϑ (the variable x here is not a distance or coordinate but simply a transform variable). With this transform, the equation can be rearranged into Legendre’s differential equation:  
d d 1 − x2 Pk (x) + k(k + 1)Pk (x) = 0 , (F.51) dx dx which allows for Legendre polynomial solutions. (d) Substitute cos ϑ back in for x, and write the general solution as a product of the R and Θ functions, and identify the allowable values of k by noting that the solution for φ = 0 must be equal to the solution for φ = 2π. With the allowable values of k specified, write the full infinite sum solution as written in Eq. (F.69). Solution: Start with the axisymmetric Laplace equation:     ∂ ∂φ 1 ∂ ∂φ r2 + sin ϑ = 0. ∂r ∂r sin ϑ ∂ϑ ∂ϑ

(F.52)

Now assume φ = R(r)Θ(ϑ) and substitute into the previous equation, giving     1 ∂ ∂RΘ ∂ 2 ∂RΘ r + sin ϑ = 0. (F.53) ∂r ∂r sin ϑ ∂ϑ ∂ϑ Evaluate the derivatives to obtain
1 ∂ ∂ 2 0
r RΘ + RΘ0 sin ϑ = 0 , ∂r sin ϑ ∂ϑ and then 2rR0 Θ + r2 R00 Θ +

RΘ00 sin ϑ + RΘ0 cos ϑ = 0, sin ϑ

(F.54)

(F.55)

finally leading to 2rR0 Θ + r2 R00 Θ + RΘ00 + RΘ0 cot ϑ = 0 ,

(F.56)

Now multiply through by 1/RΘ and rearrange to put all R terms on one side and all Θ terms on the other, giving 2rR0 r2 R00 Θ00 Θ0 cot(ϑ) + =− − . R R Θ Θ

(F.57)

Both sides of this equation must be a constant, which we will set to be k(k + 1). The Θ equation is thus given by Θ00 + Θ0 cot(ϑ) = −k(k + 1)Θ . 768

(F.58)

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Now we want to make the transform x = cos ϑ. this substitution, sin 2 ϑ

dx dϑ

= sin ϑ, so dϑ = csc ϑ dx. Making

d2 Θ + cot(ϑ) sin ϑΘ = −k(k + 1)Θ , dx2

(F.59)

d2 Θ + cos(ϑ)Θ = −k(k + 1)Θ , dx2

(F.60)

or sin 2 ϑ

We replace cos ϑ with x, and replace sin 2 ϑ with 1 − cos 2 ϑ which is 1 − x2 , giving 1 − x2


d2 Θ + xΘ = −k(k + 1)Θ , dx2

This, then, is equal to the Legendre differential equation:  
d 2 d 1−x Θ + k(k + 1)Θ = 0 , dx dx

(F.61)

(F.62)

and thus the solutions for Θ are proportional to the Legendre polynomials: Θ = Pk (x) = FPk (cos ϑ) .

(F.63)

Because Θ(0) must equal Θ(2π), k must be an integer. The R equation is given by 2rR0 r2 R00 + = k(k + 1) . R R

(F.64)

This Euler equation is solved by a polynomial form. Assuming R(r) = Hrλ , we get 2λ + λ(λ − 1) = k(k + 1) ,

(F.65)

λ(λ + 1) = k(k + 1) ,

(F.66)

which simplifies to and therefore λ = k or λ = −k − 1. Thus, for any nonzero integer k, the solution is given by   φ = RΘ = Ck rk + Dk r−k−1 Ek Pk (cos ϑ) .

(F.67)

The nature of the R solution is such that the solution for k is the same as the one for −k − 1. Thus, to put the solution in the notation of the given result, we set Ek (Ck + D−k−1 ) = Ak and Ek (Dk +C−k−1 ) = Bk , giving   φ = RΘ = Ak rk + Bk r−k−1 Pk (cos ϑ) . (F.68) http://www.cambridge.org/kirby

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Now we need sum only nonnegative values for k. The final solution is thus ∞

φ(r, ϑ) =

  k −k−1 A r + B r Pk (cos ϑ) , k ∑ k

(F.69)

k=0

F.3 Consider the 2D cylindrical Laplace equation, Eq. (F.18). Find a general solution using separation of variables. (a) Start by assuming that the solution for φ can be written as the product of two functions, one that is a function of only and one that is a function of θ only: φ = R( )Θ(θ) .

(F.70)

(b) Insert this relation into the Laplace equation and rearrange the equation so that all terms involving are on one side and all terms involving θ are on the other. (c) Because each side of this equation must be a constant, each side can be rewritten as an ordinary differential equation with an arbitrary constant. This arbitrary constant is typically written in a form proportional to k2 in anticipation of the boundary conditions that the physical problem will need to satisfy. The R side of the equation can be rearranged to obtain an Euler differential equation, 2 2∂ R ∂ 2

+

∂R = k2 R , ∂

(F.71)

which is satisfied by polynomial solutions in . The θ side of this equation can be rearranged to obtain ∂Θ + k2 Θ = 0 , (F.72) ∂θ which leads to sine and cosine solutions. (d) Write the general solution as a product of the R and Θ functions, and identify the allowable values of k by noting that the solution for φ = 0 must be equal to the solution for φ = 2π. With the allowable values of k specified, write the full infinite sum solution as written in Eq. (F.19). Solution: Start with the 2D cylindrical Laplace equation:   1 ∂ ∂φ 1 ∂2 φ + 2 2 = 0. ∂ ∂r ∂θ

(F.73)

Now assume φ = R( )Θ(θ) and substitute into the previous equation, giving   1 ∂ ∂RΘ 1 ∂2 RΘ + 2 = 0. (F.74) ∂ ∂r ∂θ2 770

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Evaluate the derivatives to obtain 1 ∂ ∂ and then

1


1 R0 Θ + 2 RΘ00 = 0 ,

R0 Θ + R00 Θ +

1 2

(F.75)

RΘ00 = 0 .

(F.76)

Now multiply through by 2 /RΘ and rearrange to put all R terms on one side and all Θ terms on the other, giving 2 R00 R0 Θ00 + =− . R R Θ

(F.77)

Both sides of this equation must be a constant, which we will set to be k2 . The Θ equation is thus given by Θ00 = −k2 , (F.78) Θ or, rearranging, Θ00 + k2 Θ = 0 , (F.79) which leads to the general solution Θ(θ) = F cos(kθ) + G sin(kθ), or, equivalently, Θ(θ) = E cos k(θ + α), where α is a constant. Because Θ(0) must equal Θ(2π), k must be an integer. The R equation is given by 2 R00 R0 + = k2 . R R

This Euler equation is solved by a polynomial form. Assuming R( ) = H

(F.80) λ,

we get

λ + λ(λ − 1) = k2 ,

(F.81)

λ = ±k .

(F.82)

which simplifies to

Thus, for any nonzero integer k, the solution is given by   φ = RΘ = Ck k + Dk −k Ek cos k(θ + α) .

(F.83)

The nature of the R solution is such that the solution for k is the same as the one for −k. Thus, to put the solution in the notation of the given result, we set Ek (Ck +C−k ) = Ak and Ek (Dk + D−k ) = Bk , giving   φ = RΘ = Ak k + Bk −k cos k(θ + α) . (F.84) Now we need sum only positive values for k. http://www.cambridge.org/kirby

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For k=0, the theta equation is simple (Θ = 1). For the R equation, the polynomial solution above (R constant) solves the equation, but it is not the only solution. For k=0, the R equation can also be simplified to R00 + R0 = 0 ,

(F.85)

which we could not do for k 6= 0. Because this equation has no R, it is an Euler equation for R0 and is satisfied by solutions of the form R0 = I λ . Plugging this in, we find that λ = −1, so R0 = I/ and thus R = I ln . For k = 0, then, we set A0 = D(C1 +C2 ) and set B0 = DI, and thus the solution is given by φ = A0 + B0 ln .

(F.86)

The final solution is thus ∞  φ( , θ) = A0 + B0 ln + ∑ Ak

k

+ Bk

−k



cos(kθ + αk ) .

(F.87)

k=1

F.4 List the positions and strengths of the dipoles required for approximating a linear axisymmetric octupole at the origin aligned along the z axis. Solution: the solution for this problem is not available F.5 List the positions and strengths of the quadrupoles required for approximating a linear axisymmetric octupole at the origin aligned along the z axis. Solution: the solution for this problem is not available F.6 List the positions and strengths of the monopoles required for approximating a linear axisymmetric quadrupole at the origin aligned along the z axis. Solution: the solution for this problem is not available

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F.7 Consider two linear axisymmetric monopoles of strength ±B0 , located at finite z locations ± 21 d. Along the z axis, compare the resulting values of φ calculated in two ways: (a) By superposing the two monopole solutions; (b) By approximating these two monopoles as a dipole with strength B1 = B0 d. Plot the two results. At what values of z/d is the dipole solution within 5% of the exact (two-monopole) solution? Solution: the solution for this problem is not available F.8 Show that Eq. (F.11) is a solution of the axisymmetric spherical Laplace equation (F.1). Solution: the solution for this problem is not available F.9 Show that Eq. (F.22) is a solution of the 2D cylindrical Laplace equation (F.18). Solution: the solution for this problem is not available F.10 Calculate and plot the velocity induced by a stresslet with ~S =

√ √ 2 2 ˆ x + 2 2 zˆ .

F.11 Calculate and plot the velocity induced by a rotlet with ~T = zˆ . F.12 Show that the velocity fields of a stresslet and a Stokeslet can be combined to give the velocity field around a sphere in Stokes flow. F.13 Consider a sphere that is rotating, perhaps a cell rotating because of a rotating electric field. Calculate the velocity field induced by this sphere. F.14 Consider five spherical, 1-µm-diameter particles at (0,0,0), (10,10,10), (0,−10, 0), (5,−5,5), and (−20,0,0), in which all locations are specified in units of µm. Assume uniform flow in the x direction. What is the viscous force on the particle at the origin owing to the flow?

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The description of water models here is restricted to static charge models and ignores charge-on-spring models, inducible dipole models, and fluctuating charge models, as well as any aspect of quantum treatment of the water molecule. Sources that discuss water models, a description of water properties, or the relation between these include [304, 305, 306, 307].

H.7

Exercises

H.1 Consider a pair potential given by e2 = ∞

if

∆r < a

(H.29)

e2 = kB T

if

a < ∆r < 2a

(H.30)

e2 = 0

if

2a < ∆r

(H.31)

Calculate and plot the Mayer f function fM for this potential. Solution: the solution for this problem is not available H.2 Write a numerical routine to solve the Ornstein–Zernike equation with hypernettedchain closure to find the radial distribution function for a homogeneous Lennard– Jones fluid. Proceed as follows: (a) Use an iterative technique that, in turn, uses the hypernetted-chain closure in Eq. (H.17) to solve for ftc and the Ornstein–Zernike equation (H.16) to solve for fdc . (b) Start by setting ftc (∆r) = fdc (∆r) = 0 on a domain that ranges from ∆r = 0 to ∆r = 512σ. (c) In each step, define a new ftc by using the hypernetted-chain relation: ftc,new (∆r) = −1 + exp [−e1 (∆r)/kB T + ftc,old (∆r) − fdc (∆r)] .

(H.32)

Note that e1 (∆r) in this case is the Lennard–Jones potential. (d) In each step, define a new fdc by Fourier-transforming ftc and fdc , applying the Fourier-transformed Ornstein–Zernike equation to get a new fˆdc , and inverse Fourier-transforming fˆdc to get a new fdc . We do this because the Fouriertransformed Ornstein–Zernike equation is much easier to deal with (the spatial integral becomes a product when Fourier transformed): fˆtc (k) = fˆdc (k) + ρ fˆdc (k) fˆtc (k) . http://www.cambridge.org/kirby

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(H.33)

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Here k is the frequency variable and fˆ is the Fourier transform of f . This can be rearranged to give fˆdc,new (k) =

fˆtc,new (k) 1 + ρ fˆtc,new (k)

(H.34)

So we Fourier-transform ftc and fdc to get fˆtc and fˆdc , apply Eq. (H.34), and then transform fˆdc back.9 (e) The two previous steps are repeated until the solutions for ftc and fdc are no longer changing. Some attention to numerical stability is needed, especially if ρ∗ is high and T ∗ is low. Plot your results for nine cases as follows: three values of ρ∗ (0.1, 0.4, 0.8) and three values of T ∗ (0.5, 1.0, 1.5).

Solution: See plots in Fig. H.6.

H.3 Given the results from Exercise H.2, calculate the potential of mean force that atoms see in this case. Plot the emf for the nine cases from Exercise H.2.

Solution: See plots in Fig. H.7.

H.4 Calculate the magnitude of the dipole moment (in debyes) for a water molecule given the geometry of the SPC model. 9 Note that one must be careful about how one transforms fˆ and f ˆ . The FFT and IFFT algorithms in this tc dc case should be symmetric with regards to the number of points. That is, for Eq. (H.34) to apply, the definition of the FFT must be

fˆ(k) = N −1/2

N

f (n) exp(2πi/N)(n−1)(k−1) ,

(H.35)

∑ fˆ(k) exp(2πi/N)−(n−1)(k−1) .

(H.36)

∑

n=1

and the IFFT must be f (n) = N −1/2

N

k=1

many software packages define the fast Fourier transform function slightly differently.

808

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Figure H.6: Solutions of the Ornstein–Zernike equation for the radial distribution function of Lennard–Jones atoms for several different ρ∗ and T ∗ values. The hypernetted chain closure (Eq. (H.17)) was used for these calculations.

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Figure H.7: Potentials of mean force, given solutions of the Ornstein–Zernike equation for the radial distribution function of Lennard–Jones atoms for several different ρ∗ and T ∗ values. The hypernetted chain closure (Eq. (H.17)) was used for these calculations. Note that at low density and high temperature, the effective potential is the same as the Lennard– Jones potential. At higher densities, especially at low temperature, the effect of multiple atoms is to make the potential of mean force oscillatory.

810

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Solution: The magnitude of the dipole moment is given by the charge multiplied by the distance between charges along the axis of symmetry. For the SPC model, the distance in angstroms along the axis of symmetry is given by   109.47◦ d = cos , (H.37) 2 which gives d = 0.577 Å or d = 0.577 × 10−10 m. The charge magnitude is 0.82e or 1.31×10−19 C. Thus the dipole moment is given by p = 7.57 × 10−30 C m or, because 1 D = 3.33 × 10−30 C m, p = 2.27 D .

(H.38)

H.5 Consider hard spheres of radius a. (a) What is the closest approach of the centers of two spheres? (b) What is the doublet potential for the interaction between the two spheres? (c) Calculate the excluded volume. How does the excluded volume compare with the volume of one of the spheres? Solution: 2a; e = ∞ if ∆r12 < 2a e = 0 if ∆r12 > 2a;

4 3 3 π(2a) 4 3 3 πa

=8

H.6 Show that the nonintegrability of the Coulomb pair potential guarantees that physical systems must be overall electroneutral. Solution: the solution for this problem is not available H.7 Using the equation for monopole interaction potentials, explain why sodium chloride might be expected to be a crystalline solid when dry but dissolves when exposed to water. Solution: The electrical permittivity of water is about 80 times higher than air or space. Thus the forces holding ions together (or equivalently, the potential energies gained when oppositely-charged ions are brought in proximity) are reduced by approximately 80. This makes kB T large relative to e1 when the salt is in water, whereas e1 in air or space is large relative to kB T .

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