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Transcript of 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
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06-703 191 Spring, 2001
Copyright 2000 by Dennis C. Prieve
Near the wall (i.e. fory
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06-703 192 Spring, 2001
Copyright 2000 by Dennis C. Prieve
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
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Although the derivation of (199) assumed that
we are very close to the wall (y+
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06-703 194 Spring, 2001
Copyright 2000 by Dennis C. Prieve
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.
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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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06-703 199 Spring, 2001
Copyright 2000 by Dennis C. Prieve
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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06-703 200 Spring, 2001
Copyright 2000 by Dennis C. Prieve
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