Simplified Microphysics. condensation evaporation. evaporation

Transcription

1 Simplified Microphysics water vapor condensation evaporation cloud droplets evaporation condensation collection rain drops fall out (precipitation) = 0 (reversible) = (irreversible)

2 Simplified Microphysics dθ dw dl dr = L c p π C = C + E r = C A r = P r + A r E r P r = 0 (reversible) P r = (irreversible)

3 More Simplified Microphysics dθ dw dl = L c p π C = C = C A r A r = 0 (reversible) A r = (irreversible)

4 p n+1 θ θ w θ n+1 w n+1 w p n θ n w n saturation mixing ratio saturation adiabat dry adiabat

5 1. Adiabatic. No phase changes involving cloud droplets (C=0): θ n, w n θ, w p n+1 θ θ n+1 w n+1 θ w w 2. Isobaric. Only phase changes involving cloud droplets operate ( C > 0): θ, w θ n+1, w n+1 p n θ n saturation mixing ratio saturation adiabat w n dry adiabat

6 1. Adiabatic. No phase changes involving cloud droplets (C=0): θ n, w n θ, w p n+1 θ θ n+1 w n+1 θ w w (θ, w )=(θ n, w n ) 2. Isobaric. Only phase changes involving cloud droplets operate ( C > 0): θ, w θ n+1, w n+1 p n θ n saturation mixing ratio saturation adiabat w n dry adiabat θ n+1, w n+1 =(θ + θ, w + w)

7 2. Isobaric. Only phase changes involving cloud droplets operate ( C > 0): θ n+1, w n+1 = (θ + θ, w + w) Conservation of energy p n+1 p n (First Law of Thermodynamics): 0=c p π θ + L w or 0=c p T + L w Adjusted state is exactly saturated: w n+1 = w s (πθ n+1,p n+1 ) or w n+1 = w s (T n+1,p n+1 ) θ θ n+1 w n+1 θ w θ n saturation mixing ratio saturation adiabat w n w dry adiabat

8 0=c p π θ + L w w n+1 = w s (πθ n+1,p n+1 ) which can be written as 0=c p π(θ n+1 θ )+L(w n+1 w ) w n+1 = w s (πθ n+1,p n+1 )

9 θ n+1 + γw n+1 = θ + γw (5) where γ L/(c p π). se two equations express This conservation set governsof energy (first law of thermodynam conservation of suspended water mixing ratio (vapor and cloud droplets), ctively. They also form w n+1 a set + l n+1 of two = w equations + l in three unknowns: θ n+1, w n. (6) We first assume that the air will be exactly saturated after adjustment, so that where w s (T,p) is the saturation mixing ratio, w n+1 = w s (T n+1,p n+1 ), (7) w s (T,p)=0.622 e s(t ) p e s (T ), (8) and e s (T ) is the saturation vapor pressure. One may use Bolton s (1980) formula for e s (T ): 17.67Tc e s (T )=6.112 exp, (9) T c where e s is in mb, T c = T T 0, and T 0 = K.

10 Saturation Adjustment Algorithm Equation (7) closes the set (5), (6), and (7). However, this set must be solved iteratively because w s is a non-linear function of T. To obtain a direct (non-iterative) solution, expand w s in a Taylor series in T about w s (T,p n+1 ) and neglect all terms of second and higher order: w n+1 w s (T,p n+1 ws )+ (T n+1 T ). (10) T T =T,p=p n+1 The set (5), (6), and (10) can now be solved algebraically for θ n+1, w n+1, and l n+1.

11 To solve the set, we first write (10) in terms of θ instead of T : w n+1 = w s + α (θ n+1 θ ), (11) where ws w s (T,p n+1 ), α α(t,p n+1 ), and πp α(t,p) (p e s (T )) 2 des dt T. (12) For de s /dt, one may use the Clausius-Clapeyron equation: de s dt = Le s R v T 2, (13) where L = J/kg and R v = J/(kg K). Now use (11) in (5) to eliminate w n+1. Then solve for θ n+1 : θ n+1 = θ + γ 1+γα (w w s). (14) Once θ n+1 is known from (14), we can immediately obtain w n+1 from (11), and l n+1 from (6).

12 If w n+1 w +l, then (6) implies that l n+1 0. ies that l 0. This means that our assumption If w n+1 >w +l, (6) implies that l n+1 < 0, which n+1 means that our assumption of saturation is incorrect. Therefore, l n+1 = 0 replaces (10). Then (5) and (6) become w n+1 = w + l, θ n+1 = θ γ(w n+1 w ).

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