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

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

2 Vapor Pressure (e) e < e # e = e # Vapor Pressure e > e #

3 Relative humidity RH = e e # (T) Supersaturation e > e # (T) S RH 100% S 0.5 2% in convective updrafts

4 The Clausius-Clapeyron Equation Used to find the relationship between pressure and temperature along phase boundaries.

5 The Clausius-Clapeyron Equation Used to find the relationship between pressure and temperature along phase boundaries. Need to derive: Latent Heat Gibbs Energy

6 Latent Heat Specific Enthalpy at constant pressure and temperature Specific Enthalpy intensive units h u + pα Internal energy Constant Pressure Mechanical work L = dh du + d pα = = δq = = du + = pdα p = e # L = (u E u F ) + e # (α E α F ) D B D C A B A C

7 Latent Heat Specific Enthalpy at constant pressure and temperature Specific Enthalpy intensive units h u + pα Internal energy Constant Temperature Mechanical work L = T = δq # B T = = ds = T(s E s F ) # C L = T(s E s F )

8 Gibbs Energy L = (u E u F ) + e # (α E α F ) L = T(s E s F ) Combine (u E u F ) + e # (α E α F ) = T(s E s F ) Rearrange u F + e # α F T s F = u E + e # α E T(s E ) Gibbs Energy = G = u + e # α T s G 1 = G 2 Remember Gibbs energy is not constant for temperature and pressure We assumed isothermal, isobaric change in phase to get here.

9 Clausius Clapeyron equation α F de # s F dt = α E de # s E dt α F LM N LO s F = α E LM N LO s E de # dt = s F s E α E α F Recall: L = T(s E s F ) de # dt = s F s E α E α F = L T(α E α F )

10 Clausius Clapeyron equation for the atmosphere de # dt = s F s E α E α F = de # dt = L T(α E ) L T(α E α F ) Substitute in the ideal gas law for water vapor de # dt = Le # R P T E M N = de # M NR e # O = L R P = dt de # e # 1 α E = e # R P T = L R P dt T E Assumption: α E >>>α F Specific Volume of liquid water Specific Volume of water vapor O T E e # T = e #S exp L 1 R R P T S 1 T

11 Saturation Vapor Pressure (Liquid Water) es(t) [Pa] T [ o C ]

12 Vapor Pressure (e) e < e # e = e # Vapor Pressure e > e #

13 The forces on the hydrogen bonding in the liquid give a net inward attractive force to the molecules on the boundary between the liquid and the vapor. The net inward force, divided by the distance along the surface, is called surface tension, σ. Its units are N/m or J/m 2. Sketches of the curvature effect. Left is a flat surface of pure water; right is a curved surface of pure water. Credit: W. Brune

14 If the surface is curved, then the amount of bonding that can go on between any one water molecule on the surface and its neighbors is reduced. As a result, there is a greater probability that any one water molecule can escape from the liquid and enter the vapor phase. Thus, the evaporation rate increases. The greater the curvature, the greater the chance that the surface water molecules can escape. Thus, it takes less energy to remove a molecule from a curved surface than it does from 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

15 If the surface is curved, then the amount of bonding that can go on between any one water molecule on the surface and its neighbors is reduced. As a result, there is a greater probability that any one water molecule can escape from the liquid and enter the vapor phase. Thus, the evaporation rate increases. The greater the curvature, the greater the chance that the surface water molecules can escape. Thus, it takes less energy to remove a molecule from a curved surface than it does from a flat surface.

16 Gibbs Energy L = (u E u F ) + e # (α E α F ) L = T(s E s F ) Combine (u E u F ) + e # (α E α F ) = T(s E s F ) Rearrange u F + e # α F T s F = u E + e # α E T(s E ) Gibbs Energy = G = u + e # α T s G 1 = G 2 No longer true, there is a change in the Gibbs free energy accompanying the formation of a single drop

17 Gibbs Energy for formation of a drop Gibbs Energy = G = u + e # α T s G 1 = G 2 No longer true, there is a change in the Gibbs free energy accompanying the formation of a single drop Drop radius G = G LYZ[\M] G [DYM P^[ZY = n g \ g P + 4πR [ E σ Surface tension where, n = number of molecules; g \ and g P are the Gibbs free energies ofa molecule in the vapor or liquid phases

18 Gibbs Energy for formation of a drop G = G LYZ[\M] G [DYM P^[ZY = n g \ g P + 4πR [ E σ where, n = number of molecules; g \ and g P are the Gibbs free energies ofa molecule in the vapor or liquid phases n = 4πR [ r 3v \ G = G LYZ[\M] G [DYM P^[ZY = 4πR [ r 3v \ g \ g P + 4πR [ E σ

19 Gibbs Energy for formation of a drop G = G LYZ[\M] G [DYM P^[ZY = n g \ g P + 4πR [ E σ where, n = number of molecules; g \ and g P are the Gibbs free energies ofa molecule in the vapor or liquid phases n = 4πR [ r 3v \ Volume occupied by a molecule in the liquid phase! G = G LYZ[\M] G [DYM P^[ZY = 4πR [ r 3v \ g \ g P + 4πR [ E σ

20 Gibbs Energy for formation of a drop G = G LYZ[\M] G [DYM P^[ZY = n g \ g P + 4πR [ E σ where, n = number of molecules; g \ and g P are the Gibbs free energies ofa molecule in the vapor or liquid phases n = 4πR [ r 3v \ Volume occupied by a molecule in the liquid phase! G = G LYZ[\M] G [DYM P^[ZY = 4πR [ r 3v \ g \ g P + 4πR [ E σ

21 Gibbs Energy for formation of a drop G = G LYZ[\M] G [DYM P^[ZY = 4πR [ r NEED TO SOLVE FOR g \ g P To put equation into something useful parameters that we can measure 3v \ g \ g P + 4πR [ E σ

22 Gibbs Energy for formation of a drop L = (u E u F ) + e # (α E α F ) L = T(s E s F ) Assumption: α E >>>α F 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 P >>>v \ g \ g P = v \ v P de # v P de # Volume occupied by a molecule in the liquid phase! Volume occupied by a molecule in the particle phase!

23 Gibbs Energy for formation of a drop Assumption: v P >>>v \ Change Gibbs Energy = g \ g P = v \ v [ de # v P de # g \ g P v P de #

24 Gibbs Energy for formation of a drop Assumption: v P >>>v \ Change Gibbs Energy = g \ g P = v \ v [ de # v P de # g \ g P v P de # Substitute in the ideal gas law for water vapor v P = R PT e # g \ g P R PT e # de #

25 Gibbs Energy for formation of a drop Assumption: v P >>>v \ Change Gibbs Energy = g \ g P = v \ v [ de # v P de # g \ g P v P de # Substitute in the ideal gas law for water vapor v P = R PT e # Where: g \ g P R PT e # de # = R P Tln( e # e #S ) e s = vapor pressure over the flat surface e s0 =vapor pressure at equilibrium

26 Gibbs Energy for formation of a drop Assumption: v P >>>v \ Change Gibbs Energy = g \ g P = v \ v [ de # v P de # g \ g P v P de # Substitute in the ideal gas law for water vapor v P = R PT e # Saturation Ratio g \ g P R PT e # de # = R P Tln(S)

27 Gibbs Energy for formation of a drop G = G LYZ[\M] G [DYM P^[ZY = 4πR [ r NEED TO SOLVE FOR g \ g P To put equation into something useful parameters that we can measure 3v \ g \ g P + 4πR [ E σ

28 Gibbs Energy for formation of a drop G = G LYZ[\M] G [DYM P^[ZY = 4πR [ r SOLVED FOR g \ g P = R P Tln(S) 3v \ g \ g P + 4πR [ E σ

29 Gibbs Energy for formation of a drop G = G LYZ[\M] G [DYM P^[ZY = 4πR [ r SOLVED FOR g \ g P = R P Tln(S) 3v \ g \ g P + 4πR [ E σ G = G LYZ[\M] G [DYM P^[ZY = 4πR [ r 3v \ R P Tln(S) + 4πR [ E σ

31 SURFACE TENSION TERM DOMINATES 1 st TERM DOMINATES

32 Kelvin Equation G R [ = 0 Find point of maximum value= metastable G at R [ = R [ R [ = 2σv \ R P Tln(S) e # = e #S exp 2σv \ R P TR [

33 Kelvin Equation e # = e #S exp 2σv \ R P TR [ 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

34 Kelvin Equation e # = e #S exp 2σv \ R P 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

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

Köhler Curve. Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics Köhler Curve Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics Review of Kelvin Effect Gibbs Energy for formation of a drop G = G &'()*+, G ).'+