9/23/16 1 Atmospheric Dynamics: lecture 3 Moist (cumulus) convection CAPE Relative humidity, mixing ratio Clausius Clapeyron equation Dew point (lapse rate) Lifted condensation level Equivalent potential temperature ([email protected]) (http://www.phys.uu.nl/~nvdelden/dynmeteorology.htm ) Problems 1.7 (p. 37), 1 in Box 1.5 (p.75) and 1.12 + extra problems Moist (cumulus) convection (Espy, 1841) LCL An air parcel will cool as it ascends. At some level it will become saturated. This level is the Lifted Condensation Level (LCL). This is the level of the cloud base.
Transcript
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Atmospheric Dynamics: lecture 3
Moist (cumulus) convection CAPE Relative humidity, mixing ratio Clausius Clapeyron equation Dew point (lapse rate) Lifted condensation level Equivalent potential temperature
Problems 1.7 (p. 37), 1 in Box 1.5 (p.75) and 1.12 + extra problems
Moist (cumulus) convection (Espy, 1841)
LCL
An air parcel will cool as it ascends. At some level it will become saturated. This level is the Lifted Condensation Level (LCL). This is the level of the cloud base.
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Vertical acceleration of an air parcel
€
d2δzdt2
= −gθ0
dθ0dz
δz ≡ −N 2δz
The solution:
€
N 2 ≡ gθ0
dθ0dz
€
δz = exp ±iNt( )
€
N 2 =gθ0
dθ0dz
< 0If Exponential growth instability
€
N 2 =gθ0
dθ0dz
> 0If oscillation stability
Brunt Väisälä-frequency, N: N is about 0.01-0.02 s-1
Section 1.5 & last week
Governed by:
Convective Available Potential Energy (CAPE)
€
dwdt
≈ w dwdz
= Bg.
Assuming a stationary state and horizontal homogeneity we can write:
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B ≡ θ 'θ0
B = buoyancy
or
€
wdw = Bgdz.
Box 1.1
€
dwdt
= g θ 'θ0
≡ BgGoverned by:
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Integrate this equation from a level z1 to a level z2. An air parcel, starting its ascent at a level z1 with vertical velocity w1, will have a velocity w2 at a height z2 given by
€
w22 = w1
2 + 2 ×CAPE,
€
CAPE ≡ g Bdzz1
z2∫ .
€
wdw = Bgdz.
Convective Available Potential Energy (CAPE)
Box 1.1
What is CAPE in your model problem of project 1 (problem 1.6)? What is the associated maximum value of w? Does your model reproduce this value?
€
B ≡ θ 'θ0
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Updraughts in cumulus clouds Figure 1.28
The relatively sharp downdraughts at the edge of the cumulus cloud are a typical feature of cumulus clouds. What effect is responsible for these downdraughts?
The vertical motion in (a) a fair weather cumulus cloud 1.5 km deep, over a track about 250 m below cloud top
Measurement of the vertical profile of (potential) temperature by radiosonde
De Bilt, 26 Sep. 2013, 12 UTC
Launching a radiosonde at KNMI on 30 November 2012
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De Bilt, 26 Sep. 2013, 12 UTC θ
Will an air parcel at the surface, with a temperature of 16.0°C and a dewpoint temperature of 8.0°C, accelerate upwards spontaneously?
Measurement of the vertical profile of (potential) temperature by radiosonde
De Bilt, 26 Sep. 2013, 12 UTC θ
Up to what approximate height will it accelerate upwards?
Measurement of the vertical profile of (potential) temperature by radiosonde
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De Bilt, 26 Sep. 2013, 12 UTC θ
Up to a height between 1348 m and 1481 m
Measurement of the vertical profile of (potential) temperature by radiosonde
De Bilt, 26 Sep. 2013, 12 UTC
Will it reach the “Lifted Condensation Level” (LCL), i.e. will clouds form?
θ
Measurement of the vertical profile of (potential) temperature by radiosonde
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Determining the Lifted condensation level (LCL)
It is the temperature, Td, which an air parcel would have if it were cooled to saturation at constant pressure
What determines saturation?
The LCL can be determined with information of the dewpoint temperature and the temperature at the surface…
Clausius-Clapeyron equation
Equilibrium water vapour pressure as a function of temperature, according to the Clausius-Clapeyron equation assuming Lv is constant (=2.5×106 J K-1) and es=6.1 hPa at T=0°C
€
∂ ln pe∂T
=LvRvT
2
Clausius-Clapeyron equation for the water vapour pressure, pe, which is in equilibrium with the liquid phase:
Lv and Rv are, respectively, the latent heat of evaporation/condensation (2.5×106 J kg-1) and the specific gas constant for water vapour (461.5 J K-1 kg-1).
Clausius-Clapeyron equation for the water vapour pressure, pe, which is in equilibrium with the liquid phase:
Relative humidity is defined as the ratio
€
RH ≡e
es T( )=es Td( )es T( )
e is the actual water vapour pressure:
€
e = ρvRvT
ρv is the water vapour density; Rv is the specifi gas constant for water vapour
(ideal gas law)
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∂ln pe∂T
≡∂lnes∂T
=LvRvT
2
In meteorology:
€
es ≡ pe
Relative humidity in convective layer
FIGURE 2.31. Model simulation of the relative humidity in the atmospheric boundary layer at midday in June in The Netherlands. The hatched regions correspond to clouds (regions where the relative humidity is 100%). The cloud cover in this case is about 20%.
Relative humidity is reasonably constant at the surface, but not in the cloud layer!!
Simulation with cloud model (labels %):
Van Delden & Oerlemans, 1982
Convective boundary layer
Lifted condensation level
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Mixing ratio, r, in terms of partial pressures
€
rv =ρvρd
Water vapour mixing ratio:
€
rv =ρvρd
=eRv
Rdpd
=εepd
≈εep
€
with ε ≡ RdRv
€
pd = ρdRdTDry air:
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e = ρvRvTWater vapour:
Ideal gases and Dalton’s law*:
*Dalton’s law: total pressure is sum of partial pressures
Water vapour mixing ratio:
Dew point lapse rate
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dedTd
=LeRvTd
2
Previous slide:
Substitute es=e and T=Td in the Clausius Clapeyron equation:
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e ≈ rv pε
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dedTd
≈Lrv pRvεTd
2
With (1) and (2):
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dedTd
≈rvεdpdTd
€
dTddp
=RvTd
2
Lp
With the hydrostatic equation:
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dTddz
= −ρgRvTd
2
Lp≈ −
gRvTd2
LRdT≈ −
gTdLε
Finding the lifted condensation level
rv is constant (see problem 1.8)!
(1)
(2)
€
Lv ≡ Lc ≡ L
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dTddz
≈ −gTdLε
≈ −10× 285
2.5×106 × 0.62≈ −0.0018 K m-1
Figure 1, Box 1.4 €
Γdew ≡ −dTddz
€
Γd ≡gcp
Derive this expression for the dry-adiabatic lapse rate
Dew point lapse rate Finding the lifted condensation level
Dew point lapse rate
Figure 1, Box 1.4
Extra problem (see box 1.5)
When is
Investigate the consequences for conditional instability, cloud formation, precipitation and global water cycle if this is indeed the case.
€
−dTddz
≡ Γdew >Γd?€
dTddz
≡ −Γdew ≈gTdLε
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Radiosonde measurements De Bilt, 26 Sep. 2013, 12 UTC
Determine the lifted condensation level (LCL). Will clouds form?
θ
Radiosonde measurements De Bilt, 26 Sep. 2013, 12 UTC θ
The lifting condensation level is found by solving:
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Ts −ΓdzLCL =Tds −ΓdewzLCL
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16.0 − gcpzLCL
⎛
⎝ ⎜
⎞
⎠ ⎟ = 8.0 − 1.8
103zLCL
⎛ ⎝ ⎜
⎞ ⎠ ⎟
€
zLCL =1 km Yes: clouds will form!
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What happens above the LCL?
If the air parcel continues its ascent after reaching the LCL, condensation of water vapour will occur, which will be accompanied by release of latent heat…..
Latent heat release in updraught
rs is saturation mixing ratio
€
mJ = −L dmv
dt
L (=2.5×106 J kg-1) is the latent heat of condensation€
J = −L drsdt
Section 1.14
? The rate of heating due to condensation is mJ (m is mass of air parcel):
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Latent heat release in updraught
The rate of heating due to condensation is mJ (m is mass of air parcel):
rs is saturation mixing ratio
€
mJ = −L dmv
dt
L (=2.5×106 J kg-1) is the latent heat of condensation
Change in rs following the motion is primarily due to ascent:
€
drsdt
≅ w drsdz
for w > 0;
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drsdt
≅ 0 for w ≤ 0.
€
J = −L drsdt
Section 1.14
Conditional instability Assume that θ=θ0(z)+θ’, with θ’<<θ0. Then:
€
dθ 'dt
=−θ0
gN 2w if w ≤ 0;{
€
dθ 'dt
≈−θ0gNm
2w if w > 0,
Latent heat release only in the updraught!
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dθdt
≈dθ 'dt
+w dθ0dz
=JΠ.
J=0 if w≤0 and J=-Lwdrs/dz if w>0
Section 1.14
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Conditional instability
If Nm<0 and w>0 then
€
dθ 'dt
> 0: positive buoyancy and upward acceleration
Assume that θ=θ0(z)+θ’, with θ’<<θ0. Then:
Section 1.14
€
dθ 'dt
=−θ0
gN 2w if w ≤ 0;{
€
dθ 'dt
≈−θ0gNm
2w if w > 0,
€
dθdt
≈dθ 'dt
+w dθ0dz
=JΠ.
J=0 if w≤0 and J=-Lwdrs/dz if w>0
Nm is the "moist" Brunt Väisälä frequency
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Nm2 ≡ N 2 +
gLθ0Π0
drsdz,
Conditional instability Assume that θ=θ0(z)+θ’, with θ’<<θ0. Then:
Nm is the "moist" Brunt Väisälä frequency
€
Nm2 ≡ N 2 +
gLθ0Π0
drsdz,
Frequently: Nm2 <0 and N2>0. In these circumstances the atmosphere is
statically or buoyantly unstable only with respect to saturated upward motion. This is called conditional instability.
Section 1.14
€
dθ 'dt
=−θ0
gN 2w if w ≤ 0;{
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dθ 'dt
≈−θ0gNm
2w if w > 0,
€
dθdt
≈dθ 'dt
+w dθ0dz
=JΠ.
J=0 if w≤0 and J=-Lwdrs/dz if w>0
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Equivalent potential temperature Previous slide:
Define a pseudo- or moist adiabatic process in which a “equivalent potential temperature”, θe, is constant. That is, θe is constant following saturated ascent.
