Book2000_course Fluid Mechanics With Vector Theory

7/27/2019 Book2000_course Fluid Mechanics With Vector Theory

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06-703 190 Spring, 2001

Copyright 2000 by Dennis C. Prieve

l= ay (192)

where a is some constant and 0


195/219

06-703 191 Spring, 2001


Near the wall (i.e. fory


196/219

06-703 192 Spring, 2001


which applies fory+>26 (theturbulent core). This coefficient of lny+ corresponds to a=0.4. so (192)

becomes:

l y= 04.

Recall that we reasoned that a had to be in the range of 0 to 1 to be physically realistic.

In the laminar sublayer, Reynolds stress can be totally neglected, leaving just viscous stress. This

close the wall, the total stress is practically a constant equal to the wall shear stress 0:

y


197/219

06-703 193 Spring, 2001


Although the derivation of (199) assumed that

we are very close to the wall (y+


198/219

06-703 194 Spring, 2001


Re 2 v Rz

Thusv R v R v

v

fz

z

f

* * Re

Re

= =

22

2 2123 ;

Relating v* tofand vz to Re, (201) can be written as:

1177 060

ff= . ln Re .d i

or1

4 07 0 6010f

f= . log Re .d i

which fits experimental data remarkable well. A slightly better fit can be obtained by adjusting the

coefficients:

14 0 0 4010

ff= . log Re .d i (202)

which is called Prandtl s (uni versal) l aw of fr iction. It applies virtually over the entire range of

Reynolds numbers normally encountered for turbulent pipe flow: 2100 < Re < 5x106.


199/219

06-703 195 Spring, 2001


Electrohydrodynamics

ORIGIN OF CHARGE

Nearly all particles, bubbles or drops in water are charged.

We know this because when you apply an electric field across

water containing a particle, bubble or drop, you can observe it

to move. An example is the red blood cell experiment at right.

Interface between the two immiscible phases of the particle and

water acquires charge by a number of mechanisms:

adsorption of ions (e.g. SDS and FeOH

+2

)

dissolution of ions from ionic solids (e.g. AgI) dissociation of acidic or basic sites on interface

COOH COO- + H+

NH2 + H+ NH3+

Regardless of how the surface acquires its charge, overall electroneutrality must be obeyed. So the

solution must contain an excess of counterions. For example, if the surface acquires its charged by the

dissociation of COOH groups at the interface, the COO- remaining at the interface gives the interface a

negative charge, but the H+ joins the rest of the ions in the aqueous solution, thereby lending it a net

positive charge.

Counterions would experience an electrostatic attraction for the surface which tends to pile up the ions

next to the interface. On the other hand, the ions undergo diffusion (Brownian motion) which tends to

disperse them uniformly throughout the phase. At equilibrium, a balance is achieved between these two

opposing tendencies. In this equilibrium state, a diffuse

cloud of counter-ions is formed next to the charged

interface.

double layer- layer of charged fixed on interface +

diffuse cloud of counter-ions.

Even though there are equal concentrations of positive

and negative ions far from the surface, this is not true

inside the counterion cloud. Fluid elements inside the


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cloud bear a net charge. Just like we define the mass per volume to be the density of the fluid, we can

define:

e(x,t) = local charge density (charge/volume)

Global electroneutrality requires:

+ =

z e y dyb g0

0

= surface charge density (charge/area)

GOUY-CHAPMAN MODEL OF DOUBLE LAYER

Let's try to quantify this description. In particular, we would like to know how thick this cloud is.Because of the charge on the interface, ions feel a electrostatic force. The force per unit charge is called

the electri c f ield:

E(x) = electric field (force/charge)

Like gravity, this vector field is conservative, thus there must exist an scalar potential, (x) such that:

E = - (203)

Like the gravitational potential, the minus sign is included by convention. is called the electrostatic

potential.

[=] volts (energy/charge)

In thermodynamics, we learn that the criteria for phase equilibrium is equality of chemical potenti alifor each chemical species i in each phase. For an ideal solution, recall that changes in chemical potential

at constant temperature and pressure is given by

i

i T P ic

kT

c

FHG

IKJ

=,

or d kT d ci T P ib g , ln= (204)

where kTis the thermal energy possessed by every ion or molecule and i is the chemical energy perion. When the species is charged, we must add an electrical contribution to the chemical energy to

obtain the electrochemical potentialof each species.

electrical potential energy: zie


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wherezi is the number of elemental charges per unit ion (including sign) and e (>0) is the magnitude of

the elemental charge (on proton); thus zie is the charge on one ion. At equilibrium, ionic species

distribute themselves in an electric field such that their electrochemical potential is constant everywhere:

ie

i iz ex x xb g b g b g= + = const (205)

For example, diffusion of nonelectrolytes (i.e. zi = 0) takes place so that i is constant throughout thesolution, which means that the concentration ci is uniform at equilibrium. If the electrostatic potential

varies from one location to the next, the concentration of charged species will also not be uniform.

(205) can be re-written in differential form as

d d z e d ie

i i = = +0

After substituting (204), we have kT d c z e d i iln + = 0

which relates changes in ion concentration to changes in potential. This can be integrated to obtainBol tzmann' s Equati onrelating the local ion concentration to the local electrostatic potential

c Az e

kTi

ix

xb g

b g=

L

NM

O

QPexp

whereA is some integration constant. In our problem, we expect the ion concentration to depend only

ony and we expect that the ionic concentration will approach the bulk value ci at some distance from

the charged interface. For this to happen, the electrostatic potential must tend to some constant (say

0 as y) orE0:

asy : cici, 0

To satisfy these boundary conditions, Boltzmanns equation can be rewritten as

c y cz e y

kTi i

ib gb g

= L

NM

O

QP exp

(206)

So if I knew the (y), I could calculate the concentration profile. But alas (y) is still unknown. I needanother equation. The extra equation is provided by Coulomb s lawof electrostatics. If I knew the

distribution of charges within a system, I could calculate the force on them using Columbs law. For a

continuum, Coulombs law can be written as:

=.E 4

e or = 2 4

e (207)

where is another material property called the electri c permi ttivityof the medium between the ions(e.g. the water):


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vacuum2

2

Coul

N - m= 1113 10 10.

water vacuum= 78

This equation is called Poisson' s Equation. The charge density of the fluid arises from the charge

born by each ion. Adding up the charges from each species i, then substituting Boltzmans equation,

gives

e i i

i

i ii

i

y z ec y z ecz e y

kTa f a f

a f= =

LNM

OQP exp

Special Case: symmetric binary electrolyte, which means there are only two different species of ions and

they have the same magnitude of charge:

z+ = -z- =z

c+ = c- = c

Then Boltzmann's equation can be used to calculate the charge density:

e y z ecz e

kTz ec

z e

kT

zecze

kT

ze

kTzec

ze

kT

( ) exp exp

exp exp sinh

= LNM

OQP

+ LNM

OQP

= FHGIKJ +

FHG

IKJ

LNM

OQP

= FHGIKJ

+ ++

2

(208)

Substituting this into Poisson's equation, we get:

d

dy

ec ze

kT

2

2

8

= FHGIKJ

sinh

The argument of the sinh must be dimensionless, so lets use this combination to define a new

dimensionless potential

ze

kT

We can make the on the left-hand side dimensionless by multiplying both sides by ze/kT. Then wehave

d

dy

2

2

2 = sinh (209)


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where 2

2 28 1= =z e c

kT[ ]

cm2

Eq. (209) is a nonlinear ODE. Despite its innoccuous appearance, it is much more difficult to solve than

linear equations. However, a particular solution is possible:

subject to: = 0 aty=0

= 0 aty=

tanh tanh y

eyb g

4 4

0L

NM

O

QP =

FHG

IKJ

(210)

which is called the Gouy-Chapman modelfor the double layer. In the special case in which the

surface potential is small, these expressions simplify to:

0


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which is analogous to the body force exerted by gravity:

gravitational body force/volume = g

Including electrostatics, but neglecting gravity, the Navier-Stokes equation becomes:

Dv/Dt= 2v - p + eE

In the Gouy-Chapman model described in the previous section, the fluid is stagnant. At hydrostatic

equilibrium, any electric fiel

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