Download Atmospheric Dynamics: lecture 4

9/25/15 Atmospheric Dynamics: lecture 4
Moist (cumulus) convection
Latent heat release in the updraught
Conditional Instability
Equivalent potential temperature
Convective Available Potential Energy
Potential instability
Tephigram
Problem 1.12
([email protected])
(http://www.phys.uu.nl/~nvdelden/dynmeteorology.htm)
Section 1.14
Latent heat release in updraught
The rate of heating due to condensation is mJ (m is mass of air parcel):
mJ = −L
dr
dmv m ≈ md = constant
J ≈ −L s
dt
dt
rs is saturation mixing ratio
L (=2.5×106 J kg-1) is€
the latent heat of condensation
€
€
Change in rs following the motion is primarily due to ascent:
drs
dr
≅ w s for w > 0;
dt
dz
drs
≅ 0 for w ≤ 0.
dt
€
€
1 9/25/15 Section 1.14
Conditional instability
Assume that θ=θ0(z)+θ’, with θ’<<θ0. We have (eq 1.23/1.33),
{
dθ dθ '
dθ
J
≈
+w 0 = .
dt dt
dz Π
€
J=0 if w≤0 and J=-Lwdrs/dz if w>0
€
dθ ' −θ 0 2
=
N w if w ≤ 0;
dt
g
dθ ' −θ 0
≈
N m2 w if w > 0,
dt
g
(latent heat release only in the updraught!!!)
N m2 ≡ N 2 +
gL drs
,
θ 0Π 0 dz
€
Nm is the "moist" Brunt Väisälä frequency
dθ '
> 0: positive buoyancy and upward acceleration
If Nm<0 and w>0 then
€
dt
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
Equivalent potential temperature
Previous slide:
N m2 ≡ N 2 +
gL drs
,
θ 0Π 0 dz
Define a pseudo- or moist adiabatic process in which a
“equivalent potential temperature”, θe, is constant.
That is, θe €
is constant following saturated ascent.
Simply define:
2
Nm
=
then
€
g d (θ e )0
(θ e )0 dz
⎛ Lr ⎞
θ e ≈ θ exp⎜ s ⎟
⎝θΠ ⎠
Extra problem for
exercise session:
Show that equivalent
potential temperature is
approximately
conserved in saturated
upward motion.
€
2 9/25/15 cold and dry air
Warm conveyor belt
cold and dry air
Warm conveyor belt
3 9/25/15 Tropical cyclone “Nadine”: warm/moist core
4 9/25/15 cold and dry air
Cyclone with cold/dry core
Section 1.14
Equivalent potential temperature
⎛ Lr ⎞
θ e ≈ θ exp⎜ s ⎟
⎝θΠ ⎠
Equivalent potential
temperature unsaturated air parcel:
€
actual mixing ratio
⎛ Lr ⎞
⎛ Lr ⎞
θ e = θexp⎜
=
θ
exp
⎟
⎜
⎟
⎝θΠ LCL ⎠
⎝ c pTLCL ⎠
(LCL: lifting condensation level)
€
5 9/25/15 “Tephigram”:
Explanation:
⎛ p ⎞R /c p
θ = T ⎜ ref ⎟
⎝ p ⎠
constant θ
constant pressure
T
€
constant θe
⎛ Lr ⎞
θ e ≈ θ exp⎜ s ⎟
⎝θΠ ⎠
€
constant saturation mixing ratio
Re
rs ≈ d s ?
Rv p
€
isotherm
€
Tephigram
Parcel of air is lifted
from the ground. What
“temperature-profile”
does it follow?
temperature
LNB
LNB=level of no
buoyancy
dew point temperature
diluted parcel
undiluted parcel
LCL=lifted
condensation level
LCL
6 9/25/15 0°C
Tephigram
Close-up
diluted parcel
undiluted parcel
Temperature (environment)
Dew point temperature (environment)
700 hPa
LCL
800 hPa
dT/dz=9.8 K/km
900 hPa
dTd/dz=1.8 K/km
1000 hPa
10°C
20°C
30°C
A case study of convection on 10 July 2013
x
Paris
Barcelona x
where will cumulus convection occur?
7 9/25/15 Paris
Compute the height
of the LCL
07145 Trappes Observations at 00Z 10 Jul 2013
----------------------------------------------------------------------------PRES
HGHT
TEMP
DWPT
RELH
MIXR
DRCT
SKNT
THTA
THTE
THTV
hPa
m
C
C
%
g/kg
deg
knot
K
K
K
----------------------------------------------------------------------------1002.0
168
19.4
14.9
75 10.74
10
6 292.4 323.1 294.3
1000.0
184
19.2
14.7
75 10.62
10
7 292.4 322.7 294.2
973.0
419
18.4
14.3
77 10.63
37
16 293.8 324.5 295.7
955.0
580
20.0
16.2
79 12.27
56
22 297.0 332.8 299.2
946.0
662
19.4
15.7
79 11.99
65
25 297.2 332.1 299.3
925.0
855
17.8
14.5
81 11.34
60
21 297.5 330.6 299.5
875.0
1326
14.1
11.7
86
9.99
60
11 298.4 327.8 300.2
850.0
1572
12.2
10.3
88
9.33
60
13 298.9 326.5 300.6
802.0
2057
8.4
7.2
92
8.00
57
18 299.9 323.7 301.3
778.0
2309
10.2 -13.8
17
1.70
56
21 304.4 310.0 304.7
770.0
2394
9.8 -14.3
17
1.64
55
22 304.8 310.3 305.1
700.0
3179
5.6 -19.4
15
1.18
50
21 308.6 312.7 308.9
660.0
3657
3.0 -24.0
12
0.84
49
20 311.0 313.9 311.1
611.0
4277
-1.3 -29.3
10
0.56
48
20 312.9 315.0 313.1
500.0
5840 -12.7 -21.7
47
1.36
45
18 317.5 322.2 317.8
483.0
6103 -14.5 -21.5
55
1.43
44
18 318.4 323.4 318.7
400.0
7500 -25.3 -41.3
21
0.26
40
16 322.0 323.0 322.1
394.0
7609 -26.3 -43.3
19
0.21
40
16 322.1 322.9 322.1
330.0
8865 -36.1 -48.1
28
0.15
37
14 325.4 326.0 325.4
300.0
9520 -40.7
35
13 327.9
327.9
250.0 10740 -49.5
30
11 332.3
332.3
Will cumulus clouds (and thunderstorms) form over Paris?
(Paris) 00 UTC 10 July 2013
Temperature
Dewpoint temperature
LCL at 550 m
8 9/25/15 (Paris) 00 UTC 10 July 2013
Potential instability if
∂θ e
<0
∂z
Temperature
€
Dewpoint temperature
∂θ e
<0
∂z
€
∂θ e
<0
∂z
€
(Paris) 12 UTC 10 July 2013
Temperature
Dewpoint temperature
LCL
Absolutely unstable surface layer,
but drier
9 9/25/15 00 UTC 10 July 2013
Temperature
Dewpoint temperature
CAPE= 149 J/kg
LCL
12 UTC 10 July 2013
Temperature
Dewpoint temperature
CAPE= 536 J/kg
LCL
10 9/25/15 Section 1.15
Convective Available Potential
Energy (CAPE)
Force on air parcel:
F=m
€
d 2z
θ'
≈ mg ≡ mgB
2
θ0
dt
B = buoyancy
B≡
θ'
θ0
Assuming a stationary state and horizontal €
homogeneity we can write:
dw
dw
≈w
= Bg.
dt
dz
€
or
wdw = Bgdz.
€
Section 1.15
Convective Available Potential
Energy (CAPE)
wdw = Bgdz.
A parcel starting its ascent at a level z1 with vertical velocity w1, will
have a velocity w€2 at a height z2 given by
w 22 = w12 + 2 × CAPE,
z2
CAPE ≡ g ∫ Bdz.
z1
€
€
11 9/25/15 A case study of
convection on
10 July 2013
x
Paris
Barcelona x
A case study of
convection on
10 July 2013
x
Paris
Barcelona x
12 9/25/15 A case study of
convection on
10 July 2013
x
Paris
Barcelona x
A case study of
convection on
10 July 2013
x
Paris
Barcelona x
13 9/25/15 A case study of
convection on
10 July 2013
x
Paris
Barcelona x
A case study of
convection on
10 July 2013
x
Paris
Barcelona x
14 9/25/15 A case study of
convection on
10 July 2013
x
Paris
Barcelona x
A case study of
convection on
10 July 2013
x
Paris
Barcelona x
15 9/25/15 A case study of
convection on
10 July 2013
x
Paris
Barcelona x
Homework:
Problem 1.12
Plot the data shown in table 1.2 in a tephigram* (figure 1.29).
Determine from the tephigram the lifting condensation level
(LCL) of an air parcel at the ground. Determine the height of
the LCL from the theory described in Box 1.4. Determine the
equivalent potential temperature of the air parcel. Will this air
parcel reach the LCL spontaneously? Once it has reached
the LCL, over how large a vertical distance will it rise?
Estimate this vertical distance from the tephigram and also by
using the theory of Box 1.5. Verify the value of θe at the
surface using eq. 1.99. Estimate the value of CAPE.
*Download the tephigram from: http://www.staff.science.uu.nl/~delde102/tephigram.pdf
16 9/25/15 Next week (40)
•  sections 1.17 to 1.19 (Coriolis force, inertial
oscillations, a second mode of convection
(sea-breeze), geostrophic balance and
thermal wind balance)
•  Exercise session on Wednesday in Minnaert
025 (see problem 1.12 en problem on slide 7
of this lecture)
17