Köhler Curve. Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics

Size: px

Start display at page:

Download "Köhler Curve. Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics"

Baldric Bryce Weaver
5 years ago
Views:

1 Köhler Curve Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics

2 Review of Kelvin Effect

3 Gibbs Energy for formation of a drop G = G &'()*+, G ).'+ /0)(' = n g * g / + 4πR ) 7 σ where, n = number of molecules; g * and g / are the Gibbs free energies ofa molecule in the vapor or liquid phases n = 4πR ) N 3v * Volume occupied by a molecule in the liquid phase! G = G &'()*+, G ).'+ /0)(' = 4πR ) N 3v * g * g / + 4πR ) 7 σ

4 Gibbs Energy for formation of a drop G = G &'()*+, G ).'+ /0)(' = 4πR ) N NEED TO SOLVE FOR g * g / To put equation into something useful parameters that we can measure 3v * g * g / + 4πR ) 7 σ

5 Enthalpy Sensible Heat Specific Enthalpy intensive units h u + pα Internal energy Mechanical work dh du + d pα = du + αdp + pdα

6 Specific Entropy R 1 δq = 0 ds T

7 Gibbs Energy for formation of a drop L = (u 7 u Z ) + e \ (α 7 α Z ) L = T(s 7 s Z ) Assumption: α 7 >>>α Z When deriving Clausius Clapeyron Equation Change Gibbs Energy = Δg = u + e \ α T s We assume a constant temperature and these two terms become ZERO Assumption: v / >>>v * g * g / = v * v / de \ v / de \ Volume occupied by a molecule in the liquid phase! Volume occupied by a molecule in the particle phase!

8 Gibbs Energy for formation of a drop Assumption: v / >>>v * Change Gibbs Energy = g * g / = v * v ) de \ v / de \ g * g / v / de \ Substitute in the ideal gas law for water vapor v / = R /T e \ g * g / R /T e \ de \

9 Gibbs Energy for formation of a drop Assumption: v / >>>v * Change Gibbs Energy = g * g / = v * v ) de \ v / de \ g * g / v / de \ Substitute in the ideal gas law for water vapor v / = R /T e \ Where: g * g / R /T e \ de \ = R / Tln( e \ e \ ) e s = vapor pressure over the flat surface e \ =vapor pressure at equilibrium

10 Gibbs Energy for formation of a drop Assumption: v / >>>v * Change Gibbs Energy = g * g / = v * v ) de \ v / de \ g * g / v / de \ Substitute in the ideal gas law for water vapor v / = R /T e \ Saturation Ratio g * g / R /T e \ de \ = R / Tln(S)

11 Gibbs Energy for formation of a drop G = G &'()*+, G ).'+ /0)(' = 4πR ) N NEED TO SOLVE FOR g * g / To put equation into something useful parameters that we can measure 3v * g * g / + 4πR ) 7 σ

12 Gibbs Energy for formation of a drop G = G &'()*+, G ).'+ /0)(' = 4πR ) N SOLVED FOR g * g / = R / Tln(S) 3v * g * g / + 4πR ) 7 σ

13 Gibbs Energy for formation of a drop G = G &'()*+, G ).'+ /0)(' = 4πR ) N SOLVED FOR g * g / = R / Tln(S) 3v * g * g / + 4πR ) 7 σ G = G &'()*+, G ).'+ /0)(' = 4πR ) N 3v * R / Tln(S) + 4πR ) 7 σ

14 SURFACE TENSION TERM DOMINATES 1 st TERM DOMINATES

15 Kelvin Equation G R ) = 0 G at R ) = R ) Find point of maximum value= metastable R ) = 2σv * R / Tln(S) e \ = e \ exp 2σv * R / TR ) Where: e s (D p )= water pressure of the droplet of diameter D p e \ = vapor pressure of pure water over flat surface

$Kelvin Equation e \ = e \ exp 2σv * R / TR ) Must be considered in calculations for particles with diameters smaller than about 200 nm Vapor pressure over a curved interface always$

16 Kelvin Equation e \ = e \ exp 2σv * R / TR ) Must be considered in calculations for particles with diameters smaller than about 200 nm Vapor pressure over a curved interface always exceeds that of the same substance over a flat surface. Sketches of the curvature effect. Left is a flat surface of pure water; right is a curved surface of pure water. Credit: W. Brune

17 Kelvin Equation in Terms of Molar Units BECOMES e \ = e \ exp 2σv * R / TR ) e \ = e \ exp 2σM R / Tρ * R ) v * = M mρ * M = Molecular Weight of substance ρ * = liquid phase density

18 Kelvin Equation in Terms of Molar Units e \ = e \ exp 2σM R / Tρ * R ) M = Molecular Weight of substance ρ * = liquid phase density UP TO THIS POINT WE HAVE BEEN THINKING ABOUT A WATER DROPLET SURROUNDED BY WATER VAPOR BUT THAT DOES NOT REPRESENT THE ATMOSPHERE OTHER STUFF IN CLOUDS ---- SOLUTE!! e \ (D ) ) = e \ exp 4v oσ o R / TD ) Where: e s (D p )= water pressure of the droplet of diameter D p e \ = vapor pressure of pure water over flat surface v o = partial molar volume of the water in the solution

20 How cloud drops form Droplets do not form via homogeneous nucleation of pure water Supersaturation needed much greater than that found in the atmosphere Thus form via heterogeneous nucleation Particles that are wettable (measure of the hydroscopicity) can serve as center upon which water vapor condenses.

22 Raoult s Law applied to cloud droplet The saturation vapor pressure of water adjacent to a solution droplet (i.e. a water droplet containing some dissolved material, such as sodium chloride or ammonium sulfate) is less than that adjacent to a pure water droplet of the same size.

23 Raoult s Law applied to cloud droplet The saturation vapor pressure of water adjacent to a solution droplet (i.e. a water droplet containing some dissolved material, such as sodium chloride or ammonium sulfate) is less than that adjacent to a pure water droplet of the same size. The fractional reduction in the water vapor pressure is given by Raoult s law. FOR Dilute Solution!

24 Raoult s Law applied to cloud droplet The saturation vapor pressure of water adjacent to a solution droplet (i.e. a water droplet containing some dissolved material, such as sodium chloride or ammonium sulfate) is less than that adjacent to a pure water droplet of the same size. The fractional reduction in the water vapor pressure is given by Raoult s law. Where e \,\(*.,p(q is the saturation vapor pressure of water adjacent to a solution droplet containing a mole fraction x o of pure water e \ is the saturation vapor pressure of water adjacent to a pure water droplet of the same size and at the same temperature. e \,\(*.,p(q e \ FOR Dilute Solution! = x o

25 Raoult s Law applied to cloud droplet The saturation vapor pressure of water adjacent to a solution droplet (i.e. a water droplet containing some dissolved material, such as sodium chloride or ammonium sulfate) is less than that adjacent to a pure water droplet of the same size. The fractional reduction in the water vapor pressure is given by Raoult s law. Where e \,\(*.,p(q is the saturation vapor pressure of water adjacent to a solution droplet containing a mole fraction x o of pure water e \ is the saturation vapor pressure of water adjacent to a pure water droplet of the same size and at the same temperature. e \,\(*.,p(q e \ FOR Dilute FOR Dilute Solution! Solution! = x o

26 Raoult s Law applied to cloud droplet The fractional reduction in the water vapor pressure is given by Raoult s law. Where e \,\(*.,p(q is the saturation vapor pressure of water adjacent to a solution droplet containing a mole fraction x o of pure water e \ is the saturation vapor pressure of water adjacent to a pure water droplet of the same size and at the same temperature. e \,\(*.,p(q = x FOR Dilute/Ideal e o Solution! \ Activity Coefficient, x o = q s q t uq s accounts for non-ideal behavior n w moles of water and n s moles of solute

27 Raoult s Law applied to cloud droplet The fractional reduction in the water vapor pressure is given by Raoult s law. Where e \,\(*.,p(q is the saturation vapor pressure of water adjacent to a solution droplet containing a mole fraction x o of pure water e \ is the saturation vapor pressure of water adjacent to a pure water droplet of the same size and at the same temperature. e \,\(*.,p(q = x FOR Dilute/Ideal e o Solution! \ e \,\(*.,p(q e \ = y o x o Activity Coefficient, accounts for non-ideal behavior y o 1, for dilute solution

28 Kelvin Equation e \ (D ) ) = e \w exp 4v oσ o R / TD ) Where: e s (D p )= water pressure of the droplet of diameter D p e so = vapor pressure of pure water over flat surface v o = partial molar volume of the water in the solution Kelvin Equation and Raoult s Law combined e \ (D ) ) = e \w exp 4v oσ o R / TD ) e \,\(*.,p(q e \ = y o x o

29 Kelvin Equation and Raoult s Law combined e \ (D ) ) = e \ exp 4v oσ o R / TD ) e \,\(*.,p(q e \ = y o x o e \ (D ) ) e \ y o x o = exp 4v oσ o R / TD ) Droplet of Diameter D p containing n w moles of water and n s moles of solute Z x πd ) N = n o v o + n \ v \ v o = partial molar volume of the water in the solution v \ = partial molar volume of the solute in the solution

30 Köhler Curve e \ (D ) ) e \ y o x o = exp 4v oσ o R / TD ) y o 1, for dilute solution ln + t { + t = }/ s~ s { ln( Z s )

31 x o = q s q t uq s Z x πd ) N = n o v o + n \ v \ 1 6 πd ) N n \ v \ = no v o 1 x o = 1 + n \ n o = 1 + n \ v o 1 6 πd ) N n \ v \

32 Köhler Curve e \ (D ) ) e \ y o x o = exp 4v oσ o R / TD ) y o 1, for dilute solution ln + t { + t = }/ s~ s { ln( Z s ) ln e \ D ) e \ = 4v oσ o R / TD ) ln(1 + n \ v o ) 1 6 πd ) N n \ v \

33 Köhler Curve e \ (D ) ) e \ y o x o = exp 4v oσ o R / TD ) y o 1, for dilute solution ln + t { + t = }/ s~ s { ln( Z s ) ln e \ D ) e \ = 4v oσ o R / TD ) ln(1 + n \ v o ) 1 6 πd ) N n \ v \ n \ v \ << 1 6 πd ) N, for dilute solution

34 Köhler Curve e \ (D ) ) e \ y o x o = exp 4v oσ o R / TD ) y o 1, for dilute solution ln + t { + t = }/ s~ s { ln( Z s ) n \ v \ << 1 6 πd ) N, for dilute solution ln e \ D ) e \ ln e \ D ) e \ = 4v oσ o R / TD ) ln(1 + n \ v o ) 1 6 πd ) N n \ v \ = 4v oσ o R / TD ) ln(1 + 6n \v o πd ) N )

35 Köhler Curve ln e \ D ) e \ = 4v oσ o R / TD ) ln(1 + 6n \v o πd ) N ) xq t / t { à0, for dilute solution Recall, ln 1 + x x, as x 0 ln e \ D ) e \ = 4v oσ o R / TD ) 6n \v o πd ) N Recall v o = M om ρo ln e \ D ) e \ = 4M oσ o R / Tρ o D ) 6n \M o πρ o D ) N

36 Köhler Curve ln e \ D ) e \ = 4M oσ o R / Tρ o D ) 6n \M o πρ o D ) N It is customary to define A = 4M oσ o R / Tρ o B = 6n \M o πρ o ln e \ D ) e \ = A D ) B D ) N

38 Köhler Curve

39 Koehler curve for two drops: N s =1x10-17 moles (blue solid line), N s =5x10-17 moles (red dashed line). Credit: W. Brune Note that the supersaturation is less than 0.2% for the smaller particle and less than 0.1% for the larger particle. As cooling occurs, which one will activate first?

40 Note that the supersaturation is less than 0.2% for the smaller particle and less than 0.1% for the larger particle. As cooling occurs, which one will activate first? Koehler curve for two drops: N s =1x10-17 moles (blue solid line), N s =5x10-17 moles (red dashed line). Credit: W. Brune ANSWER: The larger particle, because it has a lower critical supersaturation.

41 Problem 3: a) Show that the height of the critical energy barrier ( E ) in Figure 6.1 is given by ΔE = 16πσ N 3 nktln( e e \ 7 HINTS From previous lecture: ΔE = 4 π r 2 σ - 4/3 π r 3 nkt ln S And r= (2 σ)/ nkt ln(s)

42 Problem 10.2 HINTS Use Figure in your book. Study practice problem in section 10.41

43 Form Cloud Drop Credit: W. Brune

44 Form Cloud Drop Interpretation of figure: The ambient supersaturation does not depend on the radius of the particle and so is a straight line on the graph. The basic concept is the particle supersaturation is constrained to the s k curve, the particle grows or shrinks in an effort to have s k = s, that is, to have the particle condensation and evaporation be in equilibrium with the environment s water vapor. If s>s c, then the condensation of ambient water vapor is greater than the evaporation from the drop no matter what the drop radius is. The particle continues to grow, becomes a critical nucleus, and then continues to add water and grow into a cloud drop. The drop can start at any size and it will grow into a cloud drop as long as the ambient supersaturation is greater than the Koehler curve. Credit: W. Brune

45 Form Haze Credit: W. Brune

46 Form Haze Interpretation of figure: If s < s c and the drop radius is very small, then the drop grows to a radius on the left side of s c and stops when s k = s. The drops are called haze drops. They do not nucleate to become cloud drops, but instead stay a smaller size. If a drop finds itself in an environment where s<s k, (as could happen if the drop's air is mixed with drier air), then the evaporation from the drop would exceed the condensation from the environment and the drop would shrink until s=s k. If a drops happens to find itself in an environment where s=s k, the drop can shrink back into a haze drop or grow into a cloud drop, depending on the drop radius. If r < r c, then the drop will shrink until it has a radius for which s=s k. If r > r c and s=s k, then the drop is in unstable equilibrium and can either grow into cloud drops by adding a water molecule or can shrink all the way back to a haze drop with a radius that matches the radius for s=s k. If s = s c (which is a special case of s = s k ) then the drop will grow to radius r = r c and then with the random addition of a little water, will grow a little bit more so that s k < s. At this point, there will be net condensation and the drop will activate and continue to grow. Credit: W. Brune

47 Imagine scenarios with a distribution of cloud condensation nuclei of different sizes and different amounts of solute. Drops with more solute have lower values for the critical supersaturation and therefore are likely to nucleate first because in an updraft, the lower supersaturation is achieved before the greater supersaturation.

48 Imagine scenarios with a distribution of cloud condensation nuclei of different sizes and different amounts of solute. Larger CCN take up the water first and may take up so much water that the ambient supersaturation drops below the critical supersaturation for the smaller CCN. As a result, the larger CCN nucleate cloud drops while the smaller CCN turn into haze. Time 1 Time 2

Kelvin Effect. Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics

Kelvin Effect. Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics Kelvin Effect Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics Vapor Pressure (e) e < e # e = e # Vapor Pressure e > e # Relative humidity RH =