eece300 –molecules to mechanismscourses.ece.ubc.ca/300/handouts/kmt+diffusion.pdf · measure the...
Post on 19-Oct-2020
4 Views
Preview:
TRANSCRIPT
EECE300 – Molecules to mechanisms
Courtesy Micreon GmbH
2-11-2011 1
Change – Molecules in motions
• We have discussed so far:bits of quantum mechanics, atoms and bonds
Solid state – equilibrium
Transition to macroscopic theories – elasticity
We will discuss more on dynamicsKinetic theory of gases – molecules in motion
Diffusion – the laws of diffusions
Viscosity and noise
Transition toward microfluidics and its applications
Statistical thermodynamics – linking microscopic behavour with bulk properties
2-11-2011 2
States of matter
• Solids – large intermolecular cohesive forces,
small spacing
• Fluids: liquids (small intermolec. forces) +
gases (extremely small intermolec. forces)
• Plasma (significant number of ionized
molecules)
2-11-2011
Q: The dominant state of matter in the universe is:
1. Solid
2. Liquid
3. Gas
4. Plasma
3
Molecules in motion
• Goal: techniques for discussing motion of all kinds of particles in all kinds of fluids
• Simple models for the random motion of gas molecules - accounts for gas pressure + rates of migration for molecules and energy
• Molecular mobility in fluids, ion motions in solutions, in the presence of electric fields
• Diffusion equation – shows how matter and energy spread through various media
2-11-2011 4
Transport properties of substance• Ability to transport matter, energy, or some
other property
– Diffusion = migration of matter down a concentration gradient
– Thermal conduction = migration of temperature down a temperature gradient
– Electric conduction = migration of electric charge under the action of an electrical potential gradient
– Viscosity = migration of linear momentum down a velocity gradient
2-11-2011 5
Molecular motion in gases
• Kinetic model of a perfect gas – the only contribution to the gas energy is given by the kinetic energy of its molecules
• Assumptions:
1. Gas = molecules of mass m in ceaseless random motion
2. The size of molecules is negligible (their diameters much smaller than the average distance travelled between collisions)
3. Interactions – molecules interact only through brief and infrequent elastic collisions (elastic collision = collision in which the total translational kinetic energy of the molecules is conserved)
2-11-2011 6
2-11-2011
Properties of a typical gas and liquid
Property (at STP) Gas (N2) Liquid (H2O)
Molecular diameter 0.3 nm 0.3 nmNumber density 3 × 1025 m-3 2 × 1028 m-3
Intermolecular spacing 3 nm 0.4 nmDisplacement distance 100 nm (“mean free path”) 0.001 nmMolecular velocity 500 m/s 1000 m/s
N.-T. Nguyen and S. T. Wereley, Fundamentals And Applications of Microfluidics7
2-11-2011
Continuum Assumption
A fluid can be modeled as a collection of individual, interacting molecules (Molecular Flow).A fluid can also be modeled as a continuum in which properties are defined to be continuous throughout space (Continuum Model). Macroscale fluid mechanics is based on the continuum model. However, if the molecules are sparsely distributed relative to the length scale of the flow, assuming continuity of fluid and flow properties may lead to incorrect results.
8
2-11-2011
Continuous vs. molecular behaviorMeasure the density of a fluid at a point. (“Point” refers to a small sampling volume of space surrounding the geometric point in which we are interested.)count the number of molecules within the sampling volume, multiply by the molecular mass of each molecule, and divide by the volume of the sampling volume.
N.-T. Nguyen and S. T. Wereley, Fundamentals And Applications of Microfluidics
33
N m N mL
Lρ
ρ⋅ ⋅= ⇒ =
9
2-11-2011
Continuous vs. molecular behaviorAccording to random process theory, in order to get reasonably stationary statistics, less than 1% statistical variations, 104 molecules must be used to compute an average value. Thus, the point quantities (intensive variables in thermodynamics) can be thought of as continuous if the sampling volume is a cube that measures:
493
, 25 3
1070 10
3 10gas ptL mm
−−= = ×
×
493
, 28 3
109 10
2 10liquid ptL mm
−−= = ×
×
10
2-11-2011
Continuous vs. molecular behaviorThe transport quantities such as viscosity and diffusivity must also be continuous in order for the fluid to be treated as a continuum. For the transport quantities to behave continuously, it is important that the fluid molecules interact much more often with themselves than with flow boundaries. As an arbitrary criterion, choose a measurement point to be a cube whose sides are 10 times as large as the molecules’ interaction length scale.Gas: mean free pathLiquid: molecular diameter
6, 10 100 10gas transportL nm m−= × =
9, 10 0.3 3 10liquid transportL nm m−= × = ×
In order to be able to treat a flow as continuous, both its point quantities and its transport quantities must be continuous. Taking the greater of the two length scales, continuous behavior is expected at:
Lgas ~ 1 µmLliquid ~ 10 nm
Liquid: molecules are in a continual state of collision or interaction, so displacement
distance is not a good estimate of how many interactions will be present in the cube.
Molecular diameter is a better estimate.
11
Pressure of a gas• Pressure exerted on the walls
2-11-2011 12
Avogadro constant (NA) = number of molecules per mole
of substance
23 1
total number of constituent particles in a sample
amount of substance in mole
6.022 10
Amole
A
nN
n
N mol−
= =
= ×
Mole = SI unit for amount of substance (unit symbol mol)
1mol = the amount of substance that contains an equal
number of elementary entities as there are atoms in 12g of
C12 isotope = Avogadro’s number = 6.022x1023 (numerical
value of Avogadro’s constant)
Number density of particles = mole AN
n Nn
V Vρ = =
Pressure and molecular speeds• Elastic collisions with the walls => linear
momentum conservation =>
2-11-2011 13
If the wall has area A, then all the particles in the volume
Will reach the wall in the time interval ∆txV Av t∆ = ∆
2 , 0x x y zp mv p p∆ = ∆ = ∆ =
moles AN x
n Nn V v tA
Vρ∆ = ∆ = ∆
Only half particles travelling to the right:
( ) ( )2
12
2
,
moles Ax x x
moles xA
n NP n mv Av t mv
V
n MAv tP where M N m molar mass
V
∆ = ∆ = ∆
∆∆ = = =
Pressure of a gas• The force exerted on the wall:
• The pressure:
• We assumed all the molecules having the same
speed – in reality there is a velocity distribution, and
we consider averaging:
• The root mean square speed of a molecule:
• It results:
2-11-2011 14
2moles xn MAvP
Ft V
∆= =∆2
moles xn MvFp
A V= =
2molesx
n Mp v
V=
22 2 2 2 2 21
3random motion
x y z xc v v v v v c= = + + → =
21
3 molespV n Mc=
Ideal gas law
• Ideal gas - consist of mostly space (dilute gas) with a
few molecules colliding infrequently. In kinetic gas
theory, a gas molecule is considered to move in a
straight line at a constant speed until it strikes
another molecule
• For a dilute gas, the equation of state is the ideal gas
law:
2-11-2011 15
moles N BpV n RT p k Tρ= ⇔ =
3
23
pressure
number of moles, , number density
8.3185 10 universal gas constant
=1.380Boltzmann's co 5 10ns Jtant / K
moles N
BA
p
Nn
VJ
Rmol K
Rk
N
ρ
−
=
= = =
= × =⋅
= = ×
Root mean square speed
• Microscopic analysis:
• Macroscopic: the equation of state
• It results a root mean speed of the molecules in a gas at a temperature T:
• Exm.(standard conditions): T=273.15K, p=101625 Pa• Sound waves are pressure waves => we expect c to
be comparable with the speed of sound in air (340m/s) -> Exm:
2-11-2011 16
21
3 molespV n Mc=
molespV n RT=
3RTc
M=
2 273.15515N T K
mc
s==
2-11-2011
Mean molecular spacing - gases
Using p = n kBT, where n is the number density; at standard conditions n = 2.70 × 1025 m-3.The molecules in a gas are scattered randomly throughout the vessel, so the mean molecular spacing δ = n-⅓ = 3.3 × 10-9 m at standard conditions.Comparing δ to the diameter of a typical gas molecule (N2) gives a measure of effective density. Gases for which δ/d >>1 are dilute gases (as opposed to dense gases). For dilute gases, the most common mode of intermolecular interaction is binary collisions.
9
10
3.3 1010 1
3 10
m
d m
δ −
−
×= ≈ >>×
18
Maxwell speed distribution
• In an actual gas the speeds of individual molecules
span a wide range, and the collisions in the gas
redistribute the speeds among the molecules.
• The precise form of speed distribution for molecules
of gas at temperature T:
2-11-2011 19
( )23/2
2 242
Mv
RTM
f v v eRT
ππ
− =
Fraction of molecules in the range v1 to v2:
( ) ( )2
1
1 2
v
v
p v v v f v dv< < = ∫
Maxwell distribution of speeds
• Derived from Boltzmann distribution – the fraction of molecules with velocity (vx,vy,vz) is proportional to the exponential of their kinetic energy:
2-11-2011 20
( )
( ) ( ) ( ) ( )
22 2
2 2 2
3 3 31 1 1 1
1 1 1
1 1 1
2 2 2
, ,
, ,
yx z
B B B B
x y z
mvmv mvE
k T k T k T k Tx y z
x y z x y z
E mv mv mv
f v v v C e C e C e C e
f v v v f v f v f v
− − − −
= + +
= =
=Normalization condition:
( ) 21/2
/23 31 1 1
3/2 3/2
1
21 1
2 2
x Bmv k T Bx x x
B
k Tf v dv C e dv C
m
m MC
k T RT
π
π π
∞+ ∞
∞ ∞
+−
− −
= ⇔ = =
= =
∫ ∫
2axe dxa
π∞
−∞
+−
= ∫
Velocity distribution (cont.)
• Probability that a molecule has a velocity in
the range:
2-11-2011 21
( ) ( ), , , ,x y z x x y y z zv v v v v dv v dv v dv≤ < + + +�
( )( )
23/2
2
2 2 2 2
, ,2
where
Mv
RTx y z x y z x y z
x y z
Mf v v v dv dv dv e dv dv dv
RT
v v v v
π− =
= + +
The probability that a molecule has a speed in the range
v to v+dv is:
( )23/2
2 242
Mv
RTM
f v v eRT
ππ
− =
Consequences• Due to the exponential term, very few
molecules have very high speeds
• Heavy molecules are unlikely to be
found with very high speeds
• When T increases, a larger fraction of
molecules is expected to have high speeds
• The fraction of molecules with very low
speeds is also low
2-11-2011 22
Exm: Mean speed of molecules in gas
• Compute the mean speed of N2 molecules in
air at 250C.
2-11-2011 23
( )2
0
3/2 3/2 23 2
0
1 24 4
2 2 2
8475
Mv
RT
c v vf v dv
M M RTv e dv
RT RT M
RT mc v
M s
π ππ π
π
−
∞
∞
= =
= =
= = =
∫
∫
Most probable speed: * 2RTc
M=
Root mean speed: 3RTc
M=
Relative mean speed and collisions
• The mean speed with which one molecule
approaches another:
2-11-2011 24
2relc c=Collision frequency: we count a ‘hit’ whenever the
centres of two molecules come within a distance d
of each other, where d= the collision diameter
(the order of the actual molecule diameter)
Collision frequency
• Use the kinetic model:
2-11-2011 25
• Use the relative mean speed (all molecules
except one are frozen)
• “Collision tube” swept in ∆t interval:
2Cross-sectional area:
Length: rel
d
L c t
σ π== ∆
Collision volume:
Number of stationary molecules within: rel
N rel
moles A AN
B
V c t
c t
n N pNN p
V V RT k T
σρ σ
ρ
∆ = ∆∆
= = = =
Collision frequency: c rel N rel
B
pf c c
k Tσ ρ σ= =
2 "collision cross-section" of the moleculedσ π= =
Mean free path• Collision frequency exm: N2 molecule in a
sample at p=1atm, T=250C => fc=5x109s-1
• Mean free path (λλλλ)=the average distance a
molecule travels between collisions
• Exm: typical gas (N2, O2) at 1atm, 250C:
λ=70nm (~103 molecular diameters)
2-11-2011 26
2B
cc
k TccT
f pλ
σ= = =c rel
B
pf c
k Tσ= ⇒
Generic summary
• A typical gas (N2, O2) in normal conditions (1atm, 250C) can be thought as a collection of molecules travelling with a mean speed ~500m/s. Each molecule makes a collision every 1ns, and between collisions it travels about 103 molecular diameters.
• The kinetic model of gases is valid if the diameter of the molecules is much smaller than the mean free path (the molec. spend most of their time without interacting)
2-11-2011 27
2-11-2011
Knudsen number
The Knudsen number Kn is the ratio of the mean free path of the atoms or molecules in a fluid (gas or liquid) to the characteristic length of the system.Kn = λ/L where L is the characteristic length of the system.
Different orders of magnitude for Kn define different flow regimesKn < 0.001 continuum flow, no-slip boundary conditions0.001 < Kn < 0.1 continuum flow, slip boundary conditions0.1 < Kn < 10 transition flow regimeKn > 10 free molecular flow (mean free path >>
characteristic sizes)The Knudsen number is a measure of how rarefied a flow is, or how low the density is, relative to the length scale of the flow.
32
2-11-2011
Flow Regimes
molecular flow: λ/L > 10gas-wall collisions predominategas molecules do not collide with one another
viscous flow: λ/L < 0.01character of flow is determined by gas-gas interactions
33
2-11-2011
N.-T. Nguyen and S. T. Wereley, Fundamentals And Applications of Microfluidics34
2-11-2011
Viscosity
35
Transport properties of a perfect gas• Commonly expressed in terms of ‘phenomenological’
equations (macroscopic/empirical laws)
• Rate of migration for a property – its flux (J)
• Phenomenological equation: the flux is proportional
with a gradient of a physical variable
2-11-2011 40
( ) quantity passing through the area
Area x TimeJ quantity =
.matter
no moleculesJ
Area Time=
×
3, where n=number density of particles [m]matter
dnJ
dz−
∼
energy
dTJ
dz∼
Fick’s first law of diffusion
• Fick’s 1st law:
• D- the diffusion coefficient [m2/s]
• Heat conduction (energy flux):
2-11-2011 41
matterJ D n= − ∇�
( ' )energyJ T Fourier s lawκ= − ⋅∇�
κ=coefficient of thermal conductivity[J/(K⋅m⋅s)
Momentum flux
• Connection between the flux of momentum and viscosity - Newtonian flow (layers moving past one another)
2-11-2011 42
x
xp
dvJ
dzη= −
η=coefficient of viscosity (‘the viscosity’)
[kg/(m⋅s)]
• The layer close to the wall is stationary, while the
velocity of successive layers varies with the distance
from the wall
• Due to their vy, vz velocity components, molecules move
between layers and bring with them their m⋅vx momentum
• A layer is retarded by molecules arriving from a more
slowly moving layer, and accelerated by molec. arriving
from a more rapidly moving layer
• Fluid viscosity is the net retarding effect
Transport parameters
• Transport properties of perfect gases:
2-11-2011 43
Flux of matter – collisions with surfaces
• To account for gas transport ->the rate at which molecules strike an area (imaginary surface/wall)
2-11-2011 44
( ) #Collision flux =
Collision frequency [# hits/s] = J
p
p
collisionsJ
Area tArea
× ∆⋅
• A molecule with vx>0 will strike the wall in ∆t, if it lies
within a distance vx ∆t of the wall
• Uniform velocity (vx=ct):
collisions N x p N xN A v t J vρ ρ= ⋅ ⋅ ∆ ⇒ =
Flux of matter (2)
• For a general probability distribution of the
velocities:
2-11-2011 45
( ) ( )0 0
collisions N x x p N x xN A t v f v dv J v f v dvρ ρ∞ ∞
= ⋅ ⋅∆ ⇒ =∫ ∫
• Apply to Maxwell velocity distribution:
( )2
2
2
0 0
0
1
2 2 4
1
2
x
B
mv
k T Bx x x x x
B
ax
k Tmv f v dv v e dv c
k T m
xe dxa
π π
∞ ∞
∞
−
−
= = =
=
∫ ∫
∫
1
4 2 2 2
1
4 2
moles AB B Bp N N
B
p N
B
n Nk T k T k TpJ c
m V m k T m
pJ c
mk T
ρ ρπ π π
ρπ
= = = =
= =
Exm: p=100kPa, T=300K => Jp = 3x1023 cm-2 s-1
Derivation of the diffusion coefficient
• On average, molecules passing through the area A at z=0 have travelled about one mean free path since their last collision [notation: n(z)=ρN(z)]
2-11-2011 46
( ) ( ) ( )
( ) ( ) ( )
0 0
0 0
dnn n
dzdn
n ndz
λ λ
λ λ
− ≈ −
≈ +
Average number of particles crossing A0 from the left during ∆t:
( ) ( )0
1 1
4 4L R L RN A t n c J n cλ λ→ →∆ = ∆ − ⇒ = −
Average number of particles crossing from the right:
( ) ( )0
1 1
4 4R L R LN A t n c J n cλ λ→ →∆ = ∆ ⇒ =
Net flux: ( ) ( )10 0
2z L R R L
dnJ J J c
dzλ→ →= − = −
Refinement• We found a net flux:
• We have not taken into account that some particles
might make a long flight to the plane => they have a
higher chance of colliding before
2-11-2011 47
�� 0 = ��→� − ��→� = −1
2�� ��
��0
The final result: 1
3D cλ=
J D n= − ∇�
Fick’s first law of diffusion:
Commentaries
• The mean free path λ~T/p => D decreases as p increases (T=ct) => gas molecules diffuse more slowly as pressure increases
• Both mean speed and λ increase with temperature => molecules in a hot sample diffuse faster than in a cool sample (for given dn/dz)
• λ~collision cross-section (σ) => smaller molecules diffuse faster than larger ones
2-11-2011 48
The diffusion equation
• Time-dependent diffusion processes ->
spreading of inhomogeneities with time
• The diffusion equation (Fick’s second law of
diffusion) – relates the rate of change in
concentration at a point to the spatial
variation of the concentration at that point
• For 1D case:
2-11-2011 49
2
2
n nD
t z
∂ ∂=∂ ∂
Equation of continuity
• The diffusion equation is related to the
conservation of the total quantity of
substance -> the related conservation law is
the eqn. of continuity
2-11-2011 50
• Assume a fluid with properties dependent on z-axis (1D case)
• Consider a thin slab in the fluid, of area A
• Assume the substance in diffusing in the positive z-direction
• Net amount of substance entering the slab per second
from below: ( )1 1
1
2zInflux z J z z A = − ∆
• Net amount of substance leaving the slab per second
through the top surface:
( )1 1
1
2zEfflux z J z z A = + ∆
Equation of continuity
• The rate of change of the amount of
substance in the thin slab:
• Equation of continuity in 1D (conservation of
matter):
2-11-2011 51
( )
( ) ( ) ( ) ( ) ( )
( ) ( )
1 1
1 1 1 1 1
1 1
2 2
2 2
z z
z zz z
z
d nA z z zAJ z AJ z
dt
J Jdn z zz z J z z J z z
dt z z
Jdnz z
dt z
∆ ∆ ∆ = − − +
∂ ∂∆ ∆ ∆ = − − + ∂ ∂
∂= −∂
0zJn
t z
∂∂ + =∂ ∂
Equation of continuity in 3D
• The 3D version:
• ∇�� = the divergence of the flux J (a measure of
the rate at which ‘stream lines’ of a vector
quantity diverge from each other
• The equation expresses the conservation of
matter
2-11-2011 52
0 0yx zJJ Jn n
Jt t x y z
∂∂ ∂∂ ∂+ ∇ = ⇔ + + + =∂ ∂ ∂ ∂ ∂
�
Fick’s second law of diffusion
• 1D case:
• If the diffusion coefficient is independent of position
=> diffusion equation for 1D:
• If the concentration n(x,y,z) and if D is ct, then we get
the diffusion equation in 3D:
2-11-2011 53
zJn nD
t z z z
∂∂ ∂ ∂ = − = ∂ ∂ ∂ ∂
2
2
n nD
t z
∂ ∂=∂ ∂
2 2 22
2 2 2
n n n n nD n D
t t x y z
∂ ∂ ∂ ∂ ∂= ∇ ⇔ = + + ∂ ∂ ∂ ∂ ∂
Interpretation
• The rate of change in concentration depends
on the curvature (second derivative) of the
concentration with respect to distance ->
there is a natural tendency for the
nonuniformity in distribution to disappear.
2-11-2011 54
Quiz
• What is the equation of continuity for electric
charges (particles with charge q), if an electric
field might be present?
– A.
– B.
– C. Both
2-11-2011 55
0, electric current densityc
q j jt
∂ + ∇ = =∂
� �
2 0c
D ct
∂ − ∇ =∂
Solutions for the diffusion equation• It is a PDE – 2nd order w.r.t. space coordinates, and 1st
order w.r.t. time => we must specify two boundary
conditions for the spatial dependence and a single
initial condition for the time-dependence
• Exm (1D): a layer of sugar on the bottom of a deep
baker of water
– IC: at t=0 all N0 particles are at x=0
– BC: (1) the concentration must be everywhere finite
(2) the total amount of particles is ct (nmoles=N0/NA)
2-11-2011 56
( ) ( )2
2, ,
c cx t D x t
t x
∂ ∂=∂ ∂
Exm: Diffusion from finite source
• The solution is given by:
2-11-2011 57
( )2
4,x
moles Dtn
c x t eA Dtπ
−=
Exm: ion implantation in semiconductors
2-11-2011 58
Exm2: diffusion from infinite source
• Constant surface
concentration as boundary
condition:
2-11-2011 59
Example: Si oxidation
Example: Si oxidation
2-11-2011 61
Diffusion with convection
• Convection = transport of particles arising from the motion of a streaming fluid
• If we ignore diffusion, the convective flux when fluid is flowing at velocity v is:
• The rate of change in concentration:
• Generalized diffusion equation (both diffusion and convection occur):
2-11-2011 62
( ) ( ),,
c x t Av tJ c x t v
A t
∆= =
∆
in outJ Jc c v cc c x v
t x x x x
−∂ ∂ ∂ = = − + ∆ = − ∂ ∆ ∂ ∆ ∂
2
2
c c cD v
t x x
∂ ∂ ∂= −∂ ∂ ∂
top related