FB
Ptop
Buoyant force
The volume of gas (liquid) is
at rest with respect to its
surrounding gas (liquid): the
force of gravity is balanced
by the buoyant force
Vg
Pbottom
FB  ( Pbottom  Ptop ) A
h
Note that the buoyant force does not care what’s
inside this volume (a brick, a gas, or vacuum): it
depends only on the volume and the density of the
outside gas (liquid).
Ptop Pbottom
P
Lecture 3. Transport Phenomena (Ch.1)
Lecture 2 – various processes in macro systems near the state of
equilibrium can be described by a handful of macro parameters. Quasistatic processes – sufficiently slow processes, at any moment the system
is almost in equilibrium.
It is important to know how much time it takes for a system to approach an
equilibrium state. A system is not in equilibrium when the macroscopic
parameters (T, P, etc.) are not constant throughout the system. To
approach equilibrium, these non-uniformities have to be ironed out through
the transport of energy, momentum, and mass from one part of the
system to another. The mechanism of transport is molecular collisions. Our
goal - to estimate the characteristic rates of approaching equilibrium, and,
thus, to impose limitations on the rates of “quasi-static” processes.
1.
Transfers of Q (“Heat” Conduction)
2.
Transfers of Mass (Diffusion)
One-dimensional (1D) case:
n(x,t)
T(x,t)
x
The Mean Free Path of Molecules
Transports energy, momentum, mass – due to random
thermal motion of molecules in gases and liquids.
The mean free path l - the average distance traveled
by a molecule btw two successive collisions.
An estimate: one molecule is moving with a constant
speed v, the other molecules are fixed. Model of hard
spheres, the radius of molecule r ~ 110-10 m.
The av. distance traversed by a molecule until the 1st collision is the
distance in which the av. # of molecules in this cylinder is 1.
 2 r 2 l 
N
1
V

l
n = N/V
– the density of molecules
1 V
1

4 r 2 N  n
Maxwell:
l
1
2 n
 = 4r2 – the cross section
The average time interval between successive collisions - the collision time:

l
v
v
- the most probable speed of a molecule
Some Numbers:
l
1
n

for an ideal gas:
PV  NkBT
P  nkBT

1 T
l 
n P
1 V k B T 1.38 10 23 J/K  300K
 26
3
air at norm. conditions:
 


4

10
m
n N
P
105 Pa
V
3
d
~ 3 109 m
the intermol. distance
N
P = 105 Pa: l ~ 10-7 m - 30 times greater than d
P = 10-2 Pa (10-4mbar):
l ~ 1 m (size of a typical vacuum chamber)
- at this P, there are still ~2.5
1012 molecule/cm3 (!)
l
 n 2 / 3  P 2 / 3
d
The collision time at norm. conditions:  ~ 10-7m / 500m/s = 2·10-10 s
For H2 gas in interstellar space, where the density is ~ 1 molecule/ cm3,
l ~ 1013 m -
~ 100 times greater than the Sun-Earth distance (1.5 1011 m)
l
Transport in Gases (Liquids)
Box 1
x=-l
Box 2
x=0
x=l
Simplified approach: consider the “ballistic”
molecule exchange between two “boxes” within the
gas (thickness of each box should be comparable to
the mean free path of molecules, l). During the
average time between molecular collisions, ,
roughly half the molecules in Box 1 will move to the
right in Box 2, while roughly half the molecules in
Box 2 will move to the left in Box 1.
Each molecule “carries” some quantity  (mass, kin. energy, etc.), within
each box -  = N  = A l n . E.g., the flux of the number of molecules
across the border per unit area of the border, Jx:
Jn 
N 1
1 
n 
1 n
n
 v nx  l   n x  l   v   2l    v l   D
At 6
6 
x 
3 x
x
in a 3D case, on average 1/6 of the
molecules have a velocity along +x or –x
the diffusion constant
“-” - if n/x is negative, the flux is in the positive x direction
(the current flows from high density to low density)


In a 3D case, J n   Dn JU   KthT
n(x,t)
Jx
x
n1
J
Diffusion
n2
J
Fick’s Law:
Diffusion – the flow of randomly moving particles
caused by variations of the concentration of
particles. Example: a mixture of two gases, the total
concentration n = n1+n2 =const over the volume (P
= const).
1 n
n
Jx   l v
 D
3 x
x
1
D  lv
3
- the diffusion coefficient
(numerical pre-factor depends on the dimensionality: 3D – 1/3; 2D – 1/2)
1
D l v
3
its dimensions [L]2 [t]-1, its units m2 s-1
Typically, at normal conditions, l ~10-7 m, v ~300 m/s  D ~ 10-5 m2 s-1
(in liquids, D is much smaller, ~10-10 m2 s-1)
For electrons in well-ordered semiconductor heterostructures at low T:
l ~10-5 m, v ~105 m/s  D ~ 1 m2 s-1
Diffusion Coefficient of an Ideal Gas
for an ideal gas:
1 k BT
l 
n
P
from the equipartition theorem:
and at a const. pressure:
3/ 2
T
T
D  T 1/ 2 
P
P
v  T 1/ 2
therefore, at a const. temperature:
( Pr. 1.70 )
1
D
P
D  T 3/ 2
The Diffusion Equation
flow in
flow out
change of n inside:
J x x At
J x x  x At
n(x,t)
x+x
x
J
n
 x
t
x
combining with
J x  D
n
x
we’ll get the equation that describes one-dimensional diffusion:
n
 2n
D 2
t
x
t1 =0
the solution which corresponds to an initial condition
that all particles are at x =0 at t =0:
t2 =t
x =0
the diffusion equation
 x2 
C

nx, t  
exp  
Dt
 4 Dt 
C is a normalization factor
the rms displacement of particles:
x 2  Dt
Brownian Motion (self-diffusion)
Historical background:
The experiment by the botanist R. Brown concerning the
drifting of tiny (~ 1m) specks in liquids and gases, had been
known since 1827.
Brownian motion was in focus of the struggle for and against
the atomic structure of matter, which went on during the
second half of the 19th century and involved many prominent
physicists.
Ernst Mach: “If the belief in the existence of atoms is so crucial in your eyes, I
hereby withdraw from the physicist’s way of thought...”
Albert Einstein explained the phenomenon on the basis of the kinetic theory
(1905), connected in a quantitative manner the Brownian motion and such
macroscopic quantities as the coefficients of mobility and viscosity – and
brought the debate to a conclusion in a short time.
Observing the Brownian motion under a microscope, Jean Perrin measured
the Boltzman constant and Avogadro number in 1908 (Nobel 1926).
Gaussian
distribution
x
Brownian
Motion
(cont.)
 x2 
C

nx, t  
exp  
Dt
 4 Dt 
x2
a 1D random walk
of a drunk
t
 Dt
the rms displacement
t
A body that participates in a random walk, or a subject of random collisions with the
gas molecules. Its average displacement is zero. However, the average square
distance grows linearly with time:




n - a randomly

after N steps, the position is RN 1
RN 1  RN 1  l n
oriented unit
2

vector
 2 2 2
 


RN 1  RN  l n  RN  l  2l n  RN

2
2
after averaging ( RN 1  0 ):
RN 1  RN  l 2

2
RN  N l 2  t
For air at normal conditions ( l  107 m v  500 m/s D  1.7 105 m 2 /s ) , it takes
L2
t 
~ 105 s for a molecule to “diffuse” over 1m: odor spreads by convection
D
For electrons in metals at 300K ( l  107 m v  106 m/s D  3 102 m 2 /s ), it takes
L2
t 
~ 30 s to “diffuse” over 1m. For the electron gas in metals, convection can be ignored:
D
the electron velocities are randomized by impurity/phonon scattering.
Static Energy Flow by “Heat” Conduction
In general, the energy transport due
to molecular motion is described by
the equation of heat conduction:
T
1 JU K th  2T


t
C x
C x 2
JU   K th
T
x
Thus, in principle, if you know the initial conditions, e.g. T(x,t=t0), you can
describe the process by solving the equation. Often, you are asked to consider a
different situation: a static flow of energy from a “hot” object to a “cold” one. (At
what rate the internal energy is transferred between two systems with T1  T2 or
between parts of a non-equilibrium system (if one can introduce Ti) ?) The
temperature distribution in this formulation is time-independent, and we need to
calculate the flux of thermal energy JU due to the heat conduction
(diffusion/intermixing of particles with different energies, interactions between the
particles that vibrate but do not move “translationally”).
area A
T1
T2
x
Heat conduction ( static heat flow, T = const)
T(x)
T1
T2
JU
x
Fourier Heat Conduction Law
Q 
Tt
A 
x
JU 
Q
t
  K th A
T
x
“-” - if T increases from left to
right, energy flows from right to left
Kth [W/K·m] – the thermal conductivity (material-specific)
Pr. 1.56
For a window glass (Kth =0.8W/mK, 3 mm thick, A=1m2) and T = 20K:
Q
20 K
 (0.8 W/m  K )(1 m )
 5300 W
t
0.003 m
2
T1
What “flows”
Flux
Driving “force”
“Resistance”
G
~ 10 times greater than in reality, a thin
layer of still air
must contribute to thermal insulation.
J U power   GT ,
T2
A
x
G – the thermal conductivity [W/K]
R =1/G – the thermal resistivity
Electricity
Thermal Physics
Charge Q
Th. Energy, Q
Currant dQ/dt
El.-stat. pot.
difference
Power Q/dt
Temperature
difference
El. resistance R
G  K th
Th. resistance R
Connection in series (Pr. 1.57):
T1
T2
Rtot = R1 + R2
Connection in parallel:
Rtot-1 = R1-1 + R2-1
T1
T2
Relaxation Time due to Thermal Conductivity
the heat capacity (specific heat)
(a rough estimate)
U
CT
C
~

 T1  T2  T1  
Q / dt G T
G
the thermal conductivity
G
U = CT1
environment
T2
the thermal
conductivity
G  K th
A
x
Problem 1.60: A frying pan is quickly heated on the stovetop to 2000C. It has
an iron handle that is 20 cm long. Estimate how much time should pass before
the end of the handle is too hot to grab (the density of iron  = 7.9 g/cm3, its
specific heat c = 0.45 J/g·K, the thermal conductivity Kth=80 W/m·K).
C  c A L  c L2 7900 kg  m 3  450 J  kg 1  K 1  0.1m 
 


~ 400 s
1
1
1
G K A
K th
80 J  s  m  K
th
L
2
Thermal Conductivity of an Ideal Gas
Energy “flow”, t ~  :
the time between two
consecutive collisions

Q
T
  K th A

x
l
Box 1
T1
l
v
Q 1 U1  U 2  1 CV T1  T2  1 CV l dT 1
dT



 CV v

2

2

2  dx 2
dx
1 CV v 1 CV l v 1 CV


lv
2 A
2 Al
2 V
f
Nk B
CV
f
2
cV 

 n kB
V
V
2
K th 
Box 2
T2
T
the specific
heat capacity
K th 
f
n kB l v
4
The thermal conductivity of air at norm. conditions:
K th 
f
5
W
n k B l v   2.4 10 25 m3 1.38 10  23 J/K 10 7 m  500m/s  0.02
4
4
m K
(exp. value – 0.026 W/m·K)
Thermal Conductivity of Gases (cont.)
K ht 
f
n kB l v
4
 l   n  , v 
1
T
m
 Kth 
1. Kth  1 / m
- an argon-filled window helps to reduce Q
2. Thermal conductivity of an ideal gas
is independent of the gas density!
Dewar
(at higher densities, more
molecules participate in the
energy transfer, but they
carry the energy over a
shorter distance)
This conclusion holds only if L >> l .
For L < l , Kth  n
T
m
Sate-of-the-art Bolometers
(direct detectors of e.-m. radiation)
Q
dt
power   RI 2  Ge ph T T  Ge ph T Te  Tph 
specific heat of electrons
9
10
HEDDs
meanders
Nb
G, W/Km
3
8
  VTc Ge ph
10
7
10
6
10
Ti
1000
5
10

4
10
0.1
Te
electrons
Ge-ph= Ce/e-ph
Tph phonons
G = Cph/es
T
T, K
heat sink
Ge  ph  G
 ~  e ph
1
Time constant (µs)
ħ
100
BT85-4
10
BT87-1
BT100-1
BT121-1
BT121-4
1
10
100
Temperature (mK)
1000
Momentum Transfer, Viscosity
Drag – transfer of the momentum in the
direction perpendicular to velocity.
A  u x , top  u x , bottom 
 px
 Fx 
t
z
Fx
d ux

A
z
ux
Laminar flow of a gas (fluid) between two
surfaces moving with respect to each other.
Fx – the viscous drag force,  - the coefficient of viscosity
Fx/A – shear stress
Viscosity of an ideal gas ( Pr. 1.66 ):
p z 
z
z ~ l
Box 2
ux (z2)
Box 1
ux (z1)
1
1
1
Nmux ( z1 )  Nmux ( z2 )   Nm u x
2
2
2
Fx 1 pz Nmu x v


A A 
2 Al

Fx  1
 du
   v l x
A 2
dz

1
vl
2
 T1/2
Effusion of an Ideal Gas
- the process of a gas escaping through a small hole (a << l)
into a vacuum (Pr. 1.22) – the collisionless regime.
The opposite limit of a very large hole ( a >> l ) – the hydrodynamic regime.
The number of molecules that escape through
a hole of area A in 1 sec, Nh, in terms of P(t ), T
(how is T changing in the process?)
P  Nh
2m v x 1
p 1
 Nh
t A
t A
Nh 
P A t
2m v x
vx

1
1
2
m vx  k BT ,
2
2
vx 
vx
2

k BT
m
Nh = - N, where N is the total # of molecules in a system
 N 
P A t
2m
m
NkBT At

k BT
V 2m
m
ANt

k BT
2V
k BT
m
2V
 t
N (t )  N (0) exp    ,  
A
 
Depressurizing of a space ship,
V - 50m3, A of a hole in a wall – 10-4 m2
(clearly, a << l does not apply)

N
A

t
2V
k BT
1
N  N
m

m
k BT
2  50 m3
30 1.7 10 27 kg
 106 102  0.3 s  3000 s
4
2
 23
10 m
1.38 10 J/K  30 K