Microphysics Tom Peter, ETH Zurich · Microphysics Tom Peter, ETH Zurich • Observations •...

MicrophysicsTom Peter, ETH Zurich

• Observations• Modelling

1

• Modelling

Thermo-dynamics

AerosolsClouds

Kinetics

Aerosol and Cloud Processes

IN snowice

evaporationdetrainment

nucleation

coagulation

melting

2

precipitation

CCN

H2O molecules

rain

cloud drops

activation

aqueousaerosols

scavenging

Precipitation staircase

Prerequisites for cloud formation:• water• low T• supersaturation• Cloud Condensation Nuclei (CCN)

or Ice Nuclei (IN)

3

Particle size distributions:The result of the interplay of thermodynamics andkinetics in response to outer forcings ( T, hνννν , g)

4

Size distribution of cloud particles near the top of young continental

cumuli (Hobbs et al., 1980)

Size distribution of aerosol particles from photooxidaton of a methylcyclohexane-propane-NOx mixture (Seinfeld,1994)

Example of instrument:

Optical particle counter (OPC):

- “White light counter”- Aerosol forward scattering single particle counter- Mie theory to determine aerosol size [Mie, 1908]- Scattered light amplified by photomultiplier tube

5

- Scattered light amplified by photomultiplier tube- Pulse height detection- Prior to each flight calibration with polystyrene

latex spheres

All OPC channels measure particles larger than a certain size

),,(),,(),,(log

),,(log

trxnrtrxdr

dnrtrx

rd

dntrx

Dd

dnr

rrrr ≡==

The differential particle size distribution

6

7

Size distribution of stratospheric aerosols

Diffusional growth: OCS → SO2 → H2SO4 → aerosolox ox cond

8



9Thomason and Peter, 2006



Background mode

Volcanic mode: Pinatubo, 15°N, June 1991: coagulation of freshly nucleated with largest background particles

10

103

104

105

106

dN/d

(logD

) (

cm-3)

Size distribution in and around Zürich:smallest particles are 10 times more abundant in the citycompared to the country side and 100 times more abundant duirng the day than in the night

11Bukowiecki et al., 2002

10 100 1000 1000010-2

10-1

100

101

102

Urban Area: (Downtown Zürich)

Day (SMPS) Night (SMPS) Day (OPC) Night (OPC)

dN/d

(logD

) (

cm

Dp (nm)10 100 1000 10000

Rural Region:(Zürcher Oberland)

Day (SMPS) Night (SMPS) Day (OPC) Night (OPC)

Thermo-dynamics

AerosolsClouds

Kinetics

Formation of water clouds

12

Clouds

How far do we get with thermodynamics in explaining

cloud formation?

Cloud droplet formationA fierce competition without which precipitation would be massively impeded!

What decides whether the aerosolparticle stays an aerosol particleor becomes a cloud droplet?Köhler Theory

13

RH↑

water moleculeaerosol particle (e.g. NH4

+/SO42--/H2O solution)

diluted aerosol particlecloud droplet

3 big names: Raoult, Kelvin, Köhler

� Raoult’s law (1870)small droplets have higher solute concentrations (salts, acids) and this reduces the H2O vapor pressure� advantage for small droplets

� Kelvin effect (1879)

14

� Kelvin effect (1879)small droplets have a higher H2O vapor pressure(curvature effect)� disadvantage for small droplets

� Köhler equation (1921)balance between Kelvin and Raoult terms� quantitative understanding

Clapeyron’s equation(applies to any phase transition of a pure substance)

∆Vm dp = ∆Sm dT ∆Sm = change of molar entropy during phase transition∆Vm = change of molar volume during phase transition

Clausius-Clapeyron equation(valid for solid-gas and liquid-gas phase transitions)

Gibbs free energy: ∆Gvap = µdn = ∆Hvap - T∆Svap = 0 ∆Hvap: molar enthalpy change during vaporization

15

∆Hvap: molar enthalpy change during vaporization

� ∆Svap = ∆Hvap/T at equilibrium of two phases

��

��

assumes ∆Hvap to be T-independent

dTRT

∆Hlnpd

pdp

2vap==

−−=

01

vap

0 T1

T1

R

∆H

pp(T)

ln

2vap

vap

vap

RT

p∆H

T∆V

∆H

dTdp ==

Vapor pressure of liquid water and ice

300

400

500

600

700

800

900

1000P

ress

ure

(Pa)

over waterover ice

16

Murphy & Koop 2005, Q. J. R. Meteorol. Soc. 131, 15 39–1565:

p over water in Pa:ln(pwater) ≈ 54.842763− 6763.22/T − 4.210 ln(T) + 0.000367T+ tanh{0.0415(T − 218.8)}

× (53.878− 1331.22/T− 9.44523 ln(T) + 0.014025T)

p over ice in Pa:ln(pice) = 9.550426 − 5723.265/T + 3.53068 ln(T) − 0.00728332T ; T > 110 K

0

100

200

200 210 220 230 240 250 260 270 280

Temperature / K

Vapor pressure of liquid mixtures: Raoult’s law

Mixture of two liquids A und B:

pA = xA pA*, pB = xB pB*

P = pA +pB

P = total pressurep = partial pressurep* = pressure of pure substance

pB*

17

p* = pressure of pure substance

x = mole fraction:

n = mole number∑

=i

11 n

nx

pA*

0

10

20

30

40

50

60

70

0 0.2 0.4 0.6 0.8 1

Vap

or p

ress

ure

(mm

Hg

)

abs VP

VP etOH

VP H2O

Water / ethanol – mixtures at 25.13 °°°°C

total vapor pressure P

partial vapor pressure of etOH = C2H5OH

partial vapor pressure of H2O

Measurement of the vapor-liquid-equilibrium (VLE) ofethanol (etOH) / water mixtures

18

0

1

2

3

4

5

6

7

8

9

0 0.2 0.4 0.6 0.8 1x(etoh)

activ

ity c

oeffi

cien

ts

UNIFAC: g(etOH)

UNIFAC: g(H2O)

VLE g(etoh)

VLE g(h2o)

x(etoh)

γ(etOH) from UNIFAC model calculation

γ(H2O) from UNIFAC model calculation

γ(etOH) from VLE data

γ(H2O) from VLE data

Activity coefficients γ of ethanol / watermixtures

Kelvin effect

rpa

pi

Work required to increase the surface area A of the liquid-vapor interface:dW = σdAσ = surface tensionA = 4πr2 = surface area

dW = σdA4πr2 (p – p ) = A dp = σ d(4π r2) = 8π r σdr

19

� pi = pa + 2 σ / r Laplace equation (valid for bubble or droplet)

4πr2 (pi – pa) = A dp = σ d(4π r2) = 8π r σdr

Chemical potential depends on pressure: (∂µ/∂p)T = Vm,

Vm: molar volume

� Vm,l dpl = Vm,g dpg =

What is the influence of the higher pressure inside a curved surface on the vapor pressure of a droplet?

µg

µl

In equilibrium: µg = µl and dµg = dµl

gg

dpp

RT

∆pRT

VdpV

RTdp

plm,

∆p*p

llm,

p

g ≈= ∫∫+11

⇒

20

For curved surfaces: ∆p = 2σ /r

p(∞) = vapor pressure over flat surface, r = ptcl radius, R = gas constant, Vm.l = liquid molar volume, e.g. of H2O, σ = surface tension

Kelvin Equation/rRTV2 lm,e)p(p(r) σ×∞=

RT

∆pV

)p(p(r) lm,=

∞ln

∆pRT

dpVRT

dpp )p(

llm,)p(

gg

≈= ∫∫∞∞

⇒

⇒

Köhler theory = Raoult’s law + Kelvin Effect

×∞=

×∞+

=

×∞=

×∞=

1e)(p(r)p

e)(pnn

n(r)p

e)(px(r)p

e)p((r)p

/rRTV2

/rRTV2w

sw

ww

/rRTV2www

/rRTV2w

lm,

lm,

wm,

wm,

σ

σ

σ

σ

21

The last two steps assume the solution to be dilute, and make a Taylor expansion; pw(∞) = H2O vapor pressure over pure liquid bulk water, σ = surface tension of solution, r = particle radius, Vm.w = liquid molar volume of H2O, ns = molar density of solute

−×∞≈

×∞+

≈

×∞−+

=

3

23

3

4

32exp

)(4311

)/3(411

πr

Vn

rRT

V)(p(r)p

e)(pπr/Vn

(r)p

e)(pVnπr/Vn

(r)p

wm,swm,ww

/rRTVw

wm,sw

/rRTV2w

sswm,sw

wm,

lm,

σ

σ

σ

Köhler theory

−×∞≈ 3

p

wm,s

p

wm,wpw

πD

Vn

RTD

V)(p)(Dp

64exp

σ

Köhler curves for a NaCl particle with a dry diameter of 50 nm

1.004

1.005

22

pw(∞) = H2O vapor pressure over pure liquid bulk water, σ = surface tension of solution, Dp = particle radius, Vm.w = liquid molar volume of H2O, ns = molar density of solute

0.995

0.996

0.997

0.998

0.999

1

1.001

1.002

1.003

1.004

0 1 2 3 4 5

wet diameter, µm

Kelvin term

Raoult term

Köhler curve

p w(r

)/p w

(∞)

0.1

0.2

0.3

0.4

sup

ers

atu

ratio

n, %

Köhler curves for NaCl particles of different dry diameters

−≈

∞= 3

p

wm,s

p

wm,

w

pww

πD

Vn

RTD

V

)(p

)(DpS

64exp

σ

Köhler curves for particles (with dry dia-meters of 100 nm) of different compositions

0.1

0.2

0.3

0.4

sup

ersa

tura

tion

, %

23

-0.3

-0.2

-0.1

0

0.1

0 1 2 3 4 5

wet diameter, µm

sup

ers

atu

ratio

n, %

NaCl 50 nm

NaCl 100 nm

NaCl 200 nm

The Köhler curve describes the equilibrium vapor pre ssure of a droplet with a specified dry diameter as it takes up or los es water.

-0.3

-0.2

-0.1

0

0.1

0 1 2 3 4 5

wet diameter, µm

sup

ersa

tura

tion

, %

NaClAS, full dissociationglycerol1,2-hexanediol

Adiabatic cooling of an air parcelAdiabatic cooling of an air parcel701.57 hPa 701.57 hPa �� 700.00 hPa700.00 hPa

280.18 K 280.18 K �� 280.00 K280.00 K(Lift of ~17 m)(Lift of ~17 m)

χχχχχχχχtottot = 14300 = 14300 ppmppm= 1.43 %= 1.43 %

MonodisperseMonodisperse salt salt particles:particles:

Number density: Number density:

24

Number density: Number density: 2000 cm2000 cm --33

Salt mass:Salt mass:22××××××××1010--1616gg ((NHNH44))22SOSO44

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:






25



33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:






26

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:





χχχχχχχχtottot = = 14300 14300 ppmppm= = 11..43 43 %%



27

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:

Number density: Number density: 2000 2000 cmcm --33







28

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








29

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








30

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








31

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








32

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








33

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








34

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








35

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








36

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








37

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








38

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








39

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








40

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








41

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








42

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:








43



33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Minority withMinority with22××××××××1010--1515gg ((NHNH44))22SOSO44

Mass balance:






44

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









45

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









46

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









47

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









48

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









49

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









50

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









51

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









52

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









53

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









54

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









55

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









56

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









57

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









58

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









59

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:









60

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

Mass balance:






� The minority has the “size ad-vantage”. But whether it loses




For everyone who has will begiven more, and he will have an abundance. Whoever does not have, even what he has will betaken from him.

Denn wer hat, dem wird gegeben, und er wird im Überfluss haben; wer aber

61

33

)()()(34

−==⇒

−==

wvapw

vapw

totw

vapw

w

w

totwww r

r

Tp

NkT

Tp

p

Tp

pSkT

rNnkTnp

υπ

whether it loses or wins the battle depends on the cooling rate, i.e. on kinetics.

Mass balance:




Überfluss haben; wer aber nicht hat, dem wird auch noch weggenommen, was er hat.

Car à celui qui a, on donnera, et il aura encore davantage; mais à celui qui n'a pas, on ôtera même ce qu'il a.

(Mt 25,29)

Thermo-dynamics

AerosolsClouds

Kinetics


Formation of Polar Stratospheric Clouds

62

Clouds


cloud formation?

3 Types of Polar Stratospheric Clouds (PSCs)

(1) Ice (2) NAT (nitic acid trihydrate

= HNO3 ⋅⋅⋅⋅ 3H2O, crystalline)(3) STS (supercooled ternary

solutions, H2SO4/HNO3/H2O)

Lidar observations

63

Lidar observationsPSCs over the Norwegian Alps Wirth et al. (1994)

Hanson and Mauersberger (1986):

Mass spectrometric measurements Formation of NAT PSCs

64

Carslaw et al. (1994):

Thermodynamic modeling of STS (supercooled ternarysolutions, H2SO4/HNO3/H2O)

65

Thermodynamics of Electrolytes. I. Theoretical Basis and General Equations

Kenneth S. Pitzer, J. Phys. Chem., 77, 268 - 277, 1973:

66

Pitzer Ion-Interaction Model:

long range ionic short range short range neglectinteraction potential 2-body interactions 3-body interactions higher order( I = ion strength) of species (i, j, k, w) of species (i, j, k, w) terms

Activities:

.....11

)( 2∑ ∑ +++=ij ijk

kjiijkw

jiijw

w

ex

nnnn

nnn

IfnRTG µλ

ii n

GRT

a∂∂×= 1

ln

67

Theoretical decription for f(I): Debye-Hückel theory (1928)

Data used to parameterize the ineraction potentials λij , µijk :

(1) pvap measurements: for water

for solutes, e.g. HCl ↔ H+ + Cl-

(2) Electromotive force measurements:

(3) Enthalpy measurements: (4) Measurements of heat capacities.(5) Measurements of solubilities.(6) Measurements of dissociation coefficients

)ln(ClH −+×+= aa

RTF

EE 0

)/()/(

TTG

H1∂

∆∂=∆

wwwww n

GRT

aapp∂∂×=×= 10 ln;

iH nG

RTa

K

aap

∂∂×== −+ 1

HClClH

HCl ln;

Pitzer Ion-Interaction Model:

Activities:

pvap for H2O

∑ ∑++=ij ijk

kjiijkw

jiijw

w

ex

nnnn

nnn

IfnRTG µλ 2

11)(

ii n

GRT

a∂∂×= 1

ln

G∂1

68

�

pvap for HNO3

�

wwwww n

GRT

aapp∂∂×=×= 10 ln;

iH nG

RTa

K

aap

∂∂×==

+ 1ln; H

HH 3

-3

3 NO

NON

O

Alti

tude

[km

]S

urfa

ce a

rea

cm3 ]

Ice

Ice

Bac

ksca

tter

Rat

io

Bac

ksca

tter

Rat

io

Box modeling of PSCs

(a) Lidar backscatter ratio

(b) Backscatter along trajectory of ice, NAT and STSRed: measured, black: calc.

(c) Calculated surface area densities

69

Sur

face

are

a [ µµ µµ

m2 /

cmM

ixin

g R

atio

[p

pbv]

Time [h]

Distance [km]

densities

(d) Calculated chemical effects, mainly due to HCl + ClONO2 → Cl2 + HNO3

Thermo-dynamics

AerosolsClouds

Kinetics



Deliquescence of aerosols

70

Clouds


cloud formation?

Deliquescence RH of organic mixtures

T = 25°C Marcolli et al. (2004)


Eutonic mixtures:

M2: malic + malonic



Eutonic mixtures:

M2: malic + malonic

M3: malic + malonic + maleic



Eutonic mixtures:

M2: malic + malonic


M4: malic + malonic + maleic + glutaric



Eutonic mixtures:

M2: malic + malonic



M5: malic + malonic + maleic + glutaric + methylsuccinic



Eutonic mixtures:

M2: malic + malonic






The more complex the solution the lower its deliquescence (and hence

Eutonic mixtures:

M2: malic + malonic





lower its deliquescence (and hence efflorescence) RH (entropic effect). Aerosols with such complex com-positions stay liquid to very low RH

Thermo-dynamics

AerosolsClouds

Kinetics



Deliquescence of aerosols

Homogeneous nucleation of water droplets

78

Clouds


cloud formation?When do we need a kinetic

treatment?

NucleationFormation of a critical cluster from a sequence of bimolecular additions:

A + A ↔↔↔↔ A2

A2 + A ↔↔↔↔ A3

…Ai-1 + A ↔↔↔↔ Ai (critical cluster)

- supersaturation: necessary but not sufficient- need to form new surface (Kelvin equation)

Clathrate structure

79

- need to form new surface (Kelvin equation)- need to overcome energy barrier by a critical cluster

What energy is needed to form a critical cluster?What is the size of the critical cluster?

Excess free energy for cluster formation: ∆G = ∆GS + ∆GV

∆GS > 0 � excess free energy required to form cluster surface (expense) ∆GV < 0 � excess free energy released from volume transformation (gain)

Ice –Ih structure(Hale & Plummer, 1974)

Nucleation: Classical Theory – a poor man’s approach

where

∞=−=

)p(p

kTlnµµµ lg∆

µνπσπ

µσ

∆−=

∆−=∆+∆=∆

m

vs

rr

nA

GGG

34

43

2

∆G

∆Gcrit

- G∆ V ∆GS

rcrit

80

� Critical radius:

� Critical energy (barrier): Assumptions?

∆µ

σνr mc

2=

( )222

23

2

23

2

23

2

23

)(/ln(3

16)(3

16

)(332

)(16

∞=

∆=

∆−

∆=∆

ppTk

vv

vvG

mm

mmcrit

πσµ

πσµ

πσµ

πσ

∆ ∆G + GS V

rcrit

Obtain prefactor K from kinetic collision frequency:

mm : molecular massNv : vapor concentration

If we assume an Arrhenius reaction velocity equation commonly used for the rate of a thermally activated process, the rate of nucleation, J, is given as:

SNv

mK vm

m

22/12

=

πσ

( )

∞−=

∆−= 222

23

)(/ln(3

16expexp

ppTk

vK

kTG

KJ mcrit πσ

81

Nv : vapor concentrationSmm π

Example: critical radius, number, and nucleation rate for water droplets at 298 K

p/p(∞) rc (Å) ic J (cm-3s-1)

1 ∞ ∞ - ∞

2 15.1 482 1.3 x 10-47

3 9.5 121 8.9 x 10-4

4 7.6 60 6.4 x 10+7

5 6.5 39 3.8 x 10+11

σ = 0.072 N/m ; νm = 2.99 x 10-29 m3 ; mm =2.99 x 10-26 kg

Experiments on homogeneous nucleation from vapors b y a nucleation pulse method

82

• Premixed vapor and carrier gas (Xe, Kr, Ar, Ne or He)

• Expansion is held for texp (a few ms)

• Small recompression to quench further nucleation� nuclei can grow into droplets of observable size (µm)

• Number density C of the droplets obtained from scattered light� J = C / texp Wagner & Strey, JPC 1983

Strey et al., JPC, 1994

83Strey et al., JPC, 1994, 98, 7748 – 7758.

Homogeneous nucleation exp.:

ptot and scattered light flux, θ = 15°

Significant light scattering occurs only after the nucleation pulse

Nucleation pulse:

Obtain experimental pressure drop ∆pexpt and duration ∆texpt of the nucleation pulse

Homogeneous nucleation rates for water: gas-to-liqu id and liquid-to-solid

84

Nucleation rate measurements and classical (Becker-Döring) theory (Viisanen et al., J. Chem. Phys., 1993).

Solid lines belong to solid points.

Variation of the rate of homogeneous ice nucleation in supercooled water (from Pruppacher and Klett, Kluwer, 1997).

Thermo-dynamics

AerosolsClouds

Kinetics




Homogeneous nucleation of ice in aqueous solutionsDeliquescence of

aerosols

85

Clouds



treatment?

A quantitative thermodynamic ice freezing theory Koop et al., Nature 2000

Unifies freezing nucleation temperatures and rates for 18 different aqueous solutions:

H2SO4

HNO

Homogeneous ice nucleation

86

H2SO4

HNO3

HNO3/H2SO4

NH4HSO4

(NH4)2SO4

NH4F

LiCl

NaCl

KCl

NH4Cl

CaCl2MnCl2

Ca(NO3)2H2O2

urea

ethylene

glycol

glucose Relative humidity

∆aw

∆aw

Thermo-dynamics

AerosolsClouds

Kinetics





aerosolsHom. efflorescence and liquid-liquid separation

87

Clouds



treatment?

separation

Deliquescence/efflorescence of aerosol particles

88

Electrodynamic particle trap

� store single micron-sized particle� air flow with controlled RH (5-95 %)� air flow with controlled T (160-310 K)� multiple spectroscopic analysis methods

Efflorescence measurements

� here (NH4)2SO4 (sulfate) particle� also NH4HSO4 (bisulfate)� and (NH4)3H(SO4)2 (letovicite)

hysteresis(measured in trap)

Trajectory-based global analysis (Colberg et al., 2003):left: full account for hysteresisright: allow efflorescence w/o supersaturation

O/H/SO/HSONH 22444

−−+

89

Thermodynamic properties of aerosols• evidence for prevalence of liquid organic aerosols• evidence for liquid-liquid phase separations

90

Raman microscope:

Thermo-dynamics

AerosolsClouds

Kinetics






91

Clouds



treatment?

Heterogeneous nucleation of ice on mineral dust

Heterogeneous nucleation on Arizona test dust (ATD):A first active-site-attribution approach

Hea

t flo

w (a

.u.)

1 wt% ATD

DSC expt on emulsified aqueous suspension of ATD (1 wt%)

Marcolli et al., ACP, 2007.

92

230 235 240 245 250 255 260

Temperature (K)

Hea

t flo

w (a

.u.)

Homogeneousfreezing of dust-

free droplets

Hetero-geneous

freezing on ATD

(a) DSC experiments on an emulsified ATD suspension �

(b) Assuming all ATD particles to have the same active site(c) Attributing stochastically a single active site per ATD ptcl(d) Active site distribution on each ATD particleMore realistic dusts than ATD: Pinti et al.

Where does mineral dust come from?

93

Location of preferential dust sources

W. African

Bodélé

Taklamakan

Gobi

94

Percentage of model grid box that is a preferential dust source, calculated from the extent of potential lake areas, excluding areas of actual lakes [Tegen et al., 2003, Quat. Sci. Rev].

Bodélé

Lagrangian tracing of RH ice

RHice,Lagr > 100 %

RHwater,Lagr < 100 %

“cold cirrus”T = -40°C

“warm thin cirrus”

altitude

9595

Transported specific humidity (Q) at beginning of each trajectory (t = 0)No condensation and mixing � trace Q only up to RHw ≤ 100%Checked ECMWF’s T(t = 0) and Q(t = 0) are realistic

T < 0°C

Q (t = 0)

distance

40

60

80

100

120

140

160

180

200

avgdry wet

RH

ice

[%] Liquid Liquid

water water cloudcloud

MPCMPC

MPCMPC’’

Warm thin

cirrus

Cold thin

cirrus

ClasscalClasscalcirruscirrus Mixed phase Mixed phase

cloudscloudsClassical Classical

cirruscirrus’’

Cloud formation processes200

150

100

50

96

210 220 230 240 250 260 270 280 2900

20

40 avgdry wet

Temperature [K]220 230 240 250 260 270 280 290

# sa

tura

ting

traj

ecto

ries

[K-1

] 0

2000

1000

0210

Distribution of trajectories from Taklamakan

10

12

14

16

18Mixed phase

clouds

Results by Cloud Type%

of a

ll tr

ajec

tory

poi

nts

(1.7

Mio

in e

ach

regi

on)

Dust gets

Potentially big effect

herenegligible

97

0

2

4

6

8

10

Classical cirrus

Cold thin cirrrus

Liquid water clouds

Warm thin cirrrus

% o

f all

traj

ecto

ry p

oint

s (1

.7 M

io in

eac

h re

gion

)

No effect on cold cirrus

Do “warm thin cirrus” exist at all?

Dust gets into cirrus

only via MPC Affects cloud

properties and precip.?

negligible

40

60

80

100

120

140

160

180

200

avgdry wet

RH

ice

[%] Liquid Liquid

water water cloudcloud

MPCMPC

MPCMPC’’

Warm thin cirrus

Cold thin cirrus

Classical Classical cirruscirrus Mixed phase Mixed phase

cloudscloudsClassical Classical cirruscirrus’’

Cloud formation processes200

150

100

50?!AIDA chamber

experiments[Möhler et al., 2006]↑↑↑↑ Saharan

IN efficiency drops by > 1 order of magnitude upon coating with 1-10 nm of H 2SO4, (NH4)2SO4, C2H4(COOH)2[Wex et al., 2009]

CFDC measure-ments[Salam et al., 2006]

Thermal dif-fusion chamber[Schaller & Fukuta, 1979]↑↑↑↑ kaolinite

AIDA warm measurements [Field et al., 2006]

Saharan

?!Microscope cold stage [Roberts & Hallett, 1968]

Montmorillonite

98

210 220 230 240 250 260 270 280 2900

20

40 avgdry wet

Temperature [K]220 230 240 250 260 270 280 290

# sa

tura

ting

traj

ecto

ries

[K-1

]

0

2000

1000

0

210

Distribution of trajectories from Taklamakan

↑↑↑↑ Saharan↑↑↑↑ Taklimakan↓↓↓↓ ATD

↑↑↑↑ kaolinite↓↓↓↓ montmorillon.

↑↑↑↑ kaolinite– local soil↓↓↓↓ silver iodide

SaharanAsian

Montmorillonite↑↑↑↑ unprocessed↓↓↓↓ preactivated

� Availability of bare dust for cold cirrus is negligible� Mineral dust unlikely competitor to homogeneous nucleation� Availability for mixed-phase clouds higher from Asian deserts

Thermo-dynamics

AerosolsClouds

Kinetics






99

Clouds



treatment?


Diffusion growth of droplet or ice particle

Diffusion Growth of Flat Surface or Water Droplet

1-D 3-D

(1) Continuity equationn = H2O molecule number density (in the gas phase)j = molecules per area per time = molecular flux

=∂

∂t

txn ),(

100

(2) Fick’s LawDiffusive flux of molecules is a result of number density gradients:(diffusion coefficient [D] = cm2 s-1) j =

(3) Diffusion EquationCombine (1) and (2) :

∂t

The droplet growth problem:

r = radial coordinatea = droplet radiusn = number density of H2O moleculesn∞ = number density far from dropletna = number density just above

= droplet surface

Transform from Cartesian coordinates to spherical c oordinates:

101

(e.g. formulae in Jackson’sbook on electrodynamics)

The droplet growth problem is in good approximation spherically symmetric. Therefore, the diffusion of H2O molecules towards a small water droplet of radius a can be described by the radial diffusion equation:

where Dg is the diffusion constant of H2O molecules in air.

n)(rrr

DnDtn

2

2

g2

g ∂∂×=∇=

∂∂ 1

Task: For stationary conditions (∂/∂t = 0), derive n from this equation.

Need two boundary conditions:n(r → ∞) = n∞ = const far awayn(r → a) = na = const above droplet surface.

n)(rrr

DnDtn

2

2

g2

g ∂∂×=∇=

∂∂ 1

102

2)()(and)()(ra

nnDrn

Drjnr

nnrn gg ∞∞∞ −=∂∂−=+−= aa

a

Interpretation:

• p∞ = n∞ kT is the H2O partial pressure.• pa = na kT is the H2O vapor pressure. • (na – n∞) < 0 � water uptake• (na – n∞) > 0 � water loss• (na – n∞) = 0 � equilibrium

Droplet growth:

103

S = droplet surface areaN = # H2O molecules in dropletV = droplet volumeVm = H2O molecular volume

)( ∞−−= nna

DV

dtda

ag

m

Growth/evaporation law for a droplet

⇒⇒⇒⇒

pa = na kT is the H2O vapor pressure

But there is a serious problem with our solution:

⇒⇒

104

⇒⇒⇒⇒ Violation of flux limitation

This result diverges for . But this is physical non-sense!

The flux cannot become arbitrarily large but is limited by molecular bombardment.

Continuum Theory Molecular Theorydiffusion equation statistical thermodynamics

In mathematical terms, we need to change our boundary condition:

Need a flux boundary condition, not a concentration boundary condition!

Molecular bombardment on surface: (v = mean molecular thermal velocity)

⇒⇒⇒⇒ Hertz-KnudsenEquation

nvj41=

)(4

!1

)()( avapag nnva

nnDaj −=−= ∞α

105

v = mean molecular thermal velocityα = mass accommodation coefficient

� 1 - α is the fraction of colliding molecules that is reflected by surface

From this equation determine na :

No divergence!Finite growth andevaporation rates!

Dg =

)/4(1 vaD

nn

a

DV

dtda

g

vapgm α+

−×−= ∞

=+

−×−= ∞nnD

Vda vapg

m αλ

106

=+

×−=aa

Vdt m αλ /1

Thermo-dynamics

AerosolsClouds

Kinetics






107

Clouds



treatment?



Microphysics and micrometeorological interplay

Importance of meso-scale temperature fluctuations for ice

nucleation

0 10 20 30time (hours)

0.8

0.4

frac

tion

of o

ccur

renc

e

MASP (0.3-40 µm)

T (K)

245

240

235

108

SUCCESS campaign, cirrus cloud at ~ 7 km altitude

time (hours)0

0.8

0.4

0

0.0001 0.01 0 100 10000

frac

tion

of o

ccur

renc

e

Model

0.0001 0.01 0 100 10000ice particle number density (cm-3)

Hoyle et al. (JAS, 2005)

Thermo-dynamics

AerosolsClouds

Kinetics






109

Clouds



treatment?




Glass formation

Example:

H2O uptake impedance of glassy aerosols: sucrose particle

deli-que-scence

no efflorescence

impeded H2O uptake

Particle mass(UDC signal)

110

particle at 291 K

(5-day experiment, each leg ~ 1 day)

RMSD (intensity)

crystal

crystal (aspherical)

droplet or glass bead (perfectly spherical)

Light scattering

Thermo-dynamics

AerosolsClouds

Kinetics






111

Clouds



treatment?




Glass formation

Combined frost point / backscatter measurements

Very last slide:

Low clouds, cirrus and stratospheric aerosol above Zurich, 16 September 2009

112

Highlight:New Instrument development

COBALDLightweightBackscatter Sonde

Feature Specification Remark

wavelengths 455 nm & 870 nm color index 1-15

Backscatter 104 from unperturbed dynamic range stratosph. aerosol to

thick anvil outflowthick anvil outflow

time resolution 1 s 0.05-3 s selectable

dimensions 17 × 14 × 12 cm3 incl. 3 cm insulation

total weight 540 g suited for piggyback

power supply 8 × LR61 (1.5V AA) for > 3 h of operation2 × 6LR61 (9V)

data interface 19.2 kbit/s, settings for SRS-C34,logic level RS232 adaptable to telemetry

altitude range ground to > 30 km cp. weather sondes

More on this: Cirisan et al., future work

Aerosol and Cloud Processes

INsnow

ice

evaporation

detrainmentnucleation

coagulation

melting

precipitation

CCN

H2O molecules

rain

cloud drops

activation

aqueous

aerosols

scavenging

Thanks!

115

Last slide:The differential particle size distribution

and its corresponding transport equation:

),,(),,(log

trxnrtrxrd

dnr

rr ≡

116

),)(,(),,())(,(

),,(),,(),(

),,(

),,(),,(),,(),,(),,(

3/1333/133

0

0

2

trrxntrxnrrrKrd

trxntrxnrrKrd

trxJ

trxnDtrxrnv

trxzn

vtrxnvtrxtn

rr

r

rr

nuc

rrrrr

srr

′−′′−′′+

′′′−

=

∇+∂

∂+∂∂+∇⋅+

∂∂

∫

∫∞

rr

rr

r

rrrrrr

aerosol

What decides whether the aerosolparticle stays an aerosol particle




moister

117cloud droplet

particle stays an aerosol particleor becomes a cloud droplet?

� Köhler Theory

moister

Cloud droplet formationA fierce competition without which precipitation would be massively impeded!

What decides whether the aerosolparticle stays an aerosol particleor becomes a cloud droplet?Köhler Theory

118

cooling




Different pathways for ice nucleation, e.g.:

Homogeneous nucleation of solution droplets

Heterogeneous nucleation on solid particles

119

Cloud condensation nucleiinsoluble aerosol

120

Depending on surface, composition,temperature, relative humidity ...

Hier Colberg

Hoyle paper

Do we need a kinetic treatment for H2o transport

121

treatment for H2o transport in strat?

Raoult lived when?

Snowwhite / Cobald

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