Atmospheric Dynamics: lecture 3
Moist convection
Dew point temperature/lapse rate/LCL
Equivalent potential temperature
Conditional and potential instability
Thermodynamic diagram
CAPE
Introduction to Python (Problem 1.11)
problem 1.10 (sec. 1.18); problems in Boxes 1.6 & 1.7 + extra problems
([email protected])
(http://www.phys.uu.nl/~nvdelden/dynmeteorology.htm)
Cumulus clouds and
convective precipitation
Parcel of air rises spontaneously due to hydrostatic instability
Figure 1.27: (Espy, 1841)
Section 1.15 & last week
Vertical acceleration of an air parcel
2
d
δz
g dθ 0
2
Governed by:
=
−
δ
z
≡
−N
δz
2
θ 0 dz
dt
Brunt Väisälä-frequency, N:
€
g dθ 0
N ≡
θ 0 dz
2
N is about 0.01-0.02 s-1
The solution:
€
If
If
€
g dθ 0
<0
θ 0 dz
g dθ 0 €
2
N =
>0
θ 0 dz
N2 =
δz = exp(±iNt )
Exponential growth
instability
oscillation
stability
Radiosonde measurements
De Bilt, 26 Sep. 2013, 12 UTC
Launching a radiosonde at KNMI on 30 November 2012
Radiosonde measurements
De Bilt, 26 Sep. 2013, 12 UTC
θ
Will an air parcel at the surface accelerate upwards spontaneously?
Lifted condensation level (LCL)
Figure 1.27: (Espy, 1841)
LCL
The air parcel will cool as it ascends. At some level it will become
saturated. This level is the Lifted Condensation Level (LCL)
Radiosonde measurements
De Bilt, 26 Sep. 2013, 12 UTC
θ
Will it reach the LCL below this height, i.e. will clouds form?
Lifted condensation level (LCL)
Figure 1.27: (Espy, 1841)
LCL
The LCL can be determined with information of the dewpoint
and the temperature at the surface…
Dew point temperature
=temperature of air if it were cooled to saturation at constant pressure Teten’s formula (based on Clausius Clapeyron equation):
⎧
⎞⎫
⎛
T [°C]
(box 1.5)
es = 611.2 exp⎨17.67⎜
⎟⎬
⎝T [°C]+ 243.5 ⎠⎭
⎩
Substitute es=e and T=Td and “invert”:
€
⎛ e ⎞
243.5 ln⎜
⎟
⎝ 611.2 ⎠
Td [°C] =
⎛ e ⎞
17.67 − ln⎜
⎟
⎝ 611.2 ⎠
Equations of state
Dry air:
pd = ρ d Rd T
(ideal gases)
Water vapour:
e = ρv RvT
ρv
r=
ρd
€
Water vapour mixing ratio:
€
ρv e Rd εe εe
r=
=
=
≈
ρ d Rv pd pd p
€
€
€
Rd
with ε ≡
Rv
Finding the lifted condensation level
Dew point lapse rate
Substitute es=e and T=Td in the Clausius Clapeyron equation:
de
Le
=
dTd RvTd2
€
rv p
Previous slide:
e ≈
ε €
rv is constant (see problem 1.3)!
de
Lrv p
≈
dTd RvεTd2
de rv dp
≈
dTd ε dTd
dTd RvTd2
Therefore:
=
dp
Lp
€
€
dTd
ρgRvTd2
gRvTd2
gT
With hydrostatic equation:
=−
≈−
≈− d
dz
Lp
LRd T
Lε
€
€
Finding the lifted condensation level
Dew point lapse rate
dTd
gTd
10 × 285
-1
≈−
≈−
≈
−0.0018
K
m
dz
Lε
2.5 ×10 6 × 0.62
€
g
Γd ≡
cp
dTd
Γdew ≡
dz
€
Figure 1, Box 1.5
Derive this
expression for the
dry-adiabatic lapse
rate
Dew point lapse rate
dTd
gT
≡ Γdew ≈ − d
dz
Lε
Extra problem
(see box 1.6)
€
Figure 1, Box 1.5
When is
dTd
−
≡ Γdew > Γd ?
dz
Investigate the
consequences for
conditional instability,
cloud formation,
precipitation and
global water cycle if
this is indeed the
case.
Radiosonde measurements
De Bilt, 26 Sep. 2013, 12 UTC
θ
Determine the lifted condensation level (LCL).
Will clouds form?
Radiosonde measurements
De Bilt, 26 Sep. 2013, 12 UTC
θ
The lifting condensation level is found by solving:
⎞
⎛ g
⎞
⎞
⎛ dT
⎛ dTd
⎞
⎛ 1.8
Ts + ⎜ zLCL ⎟ = Tds + ⎜
zLCL ⎟
16.0 − ⎜ zLCL ⎟ = 8.0 − ⎜ 3 zLCL ⎟
⎝ dz
⎠
⎝ dz
⎠
⎝10
⎠
⎝ c p
⎠
zLCL = 1 km
Convective clouds organized into “streets”
Thursday 26 Sep. 2013
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…..
Section 1.16
Latent heat release in updraught
The rate of heating due to condensation is
dmv
drs
?
mJ = −L
J ≈ −L
dt
dt
rs is saturation mixing ratio
L (=2.5×106 J kg-1) is the latent heat of condensation
€
€
Section 1.16
Latent heat release in updraught
The rate of heating due to condensation is mJ (m is mass of air parcel):
dmv
drs
?
mJ = −L
J = −L
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
drs
≅w
for w > 0;
dt
dz
drs
≅ 0 for w ≤ 0.
dt
€
Section 1.16
Conditional instability
Assume that θ=θ0(z)+θ’, with θ’<<θ0. We have (eq 1.55),
dθ dθ '
dθ
J
≈
+w 0 = .
dt dt
dz Π
{
dθ ' −θ 0 2
=
N w if w ≤ 0;
dt
g
dθ ' −θ 0
≈
N m2 w if w > 0,
dt
g
€
J=0 if w≤0 and J=-Lwdrs/dz if w>0
€ only in the updraught!
Latent heat release
Section 1.16
Conditional instability
Assume that θ=θ0(z)+θ’, with θ’<<θ0. We have (eq 1.55),
{
dθ dθ '
dθ
J
≈
+w 0 = .
dt dt
dz Π
dθ ' −θ 0 2
=
N w if w ≤ 0;
dt
g
dθ ' −θ 0
≈
N m2 w if w > 0,
dt
g
€
J=0 if w≤0 and J=-Lwdrs/dz if w>0
€
N m2
gL drs
≡N +
,
θ 0Π 0 dz
2
Nm is the "moist" Brunt Väisälä frequency
€
dθ '
> 0: positive buoyancy and upward acceleration
If Nm<0 and w>0 then
dt
Section 1.16
Conditional instability
Assume that θ=θ0(z)+θ’, with θ’<<θ0. We have (eq 1.55),
dθ ' −θ 0 2
=
N w if w ≤ 0;
dθ dθ '
dθ 0 J
dt
g
≈
+w
= .
dt dt
dz Π
dθ ' −θ 0
≈
N m2 w if w > 0,
dt
g
{
J=0 if w≤0 and J=-Lwdr€
s/dz if w>0
€
N m2
gL drs
≡N +
,
θ 0Π 0 dz
2
Nm is the "moist" Brunt Väisälä frequency
€ In these circumstances the atmosphere is
Frequently: Nm2 <0 and N2>0.
statically or buoyantly unstable only with respect to saturated upward
motion. This is called conditional instability.
Section 1.16
Equivalent potential temperature
Previous slide:
N m2
gL drs
≡N +
,
θ 0Π 0 dz
2
Define a pseudo- or moist adiabatic process in which a
“equivalent potential temperature”, θe, is constant. That is, θe €
is constant following saturated ascent. Extra problem:
Simply define:
2
Nm
then
€
€
g d (θ e )0
=
(θ e )0 dz
⎛ Lrs ⎞
θ e ≈ θ exp⎜ ⎟
⎝θΠ ⎠
Show that equivalent
potential temperature is
approximately
conserved in saturated
upward motion.
cold and dry air
Warm conveyor belt
cold and dry air
Warm conveyor belt
Tropical cyclone “Nadine”: warm/moist core
cold and dry air
Cyclone with cold/dry core
Section 1.16
Equivalent potential temperature
⎛ Lrs ⎞
θ e ≈ θ exp⎜ ⎟
⎝θΠ ⎠
Equivalent potential
temperature unsaturated air parcel:
€
actual mixing ratio
⎛ Lr ⎞
⎛ Lr ⎞
θ e = θexp⎜
⎟
⎟ = θexp⎜
⎝θΠ LCL ⎠
⎝ c pTLCL ⎠
(LCL: lifting condensation level)
€
Potential Instability:
θ
De Bilt, 26 Sep. 2013, 12 UTC
∂θ e
<0
∂z
θe
€
∂θ
<0
∂z
€
Potential Instability (PI) up to 650 hPa (3600 m)
€
∂θ e
<0
∂z
Three-dimensional convection:
open convection cells
Special type of threedimensional flow!
Narrow rings of clouds
(upward motion)
surrounding broad clear
areas (downward
motion)
satellite image, 18 April 2001, 1245 UTC
Precipitation pattern associated
with open convection cells
radar image, 18042001, 1245 UTC
De Bilt,
12Z 18 Apr ‘01
Absolute instability:
∂θ
<0
∂z
-----------------------------------------------------------------------------!
PRES
HGHT
TEMP
DWPT
RELH
MIXR
DRCT
SKNT
THTA
THTE
THTV!
hPa
m
C
C
%
g/kg
deg
knot
K
K
K !
-----------------------------------------------------------------------------!
1004.0
4
7.8
-0.2
57
3.77
350
16 280.6 291.3 281.3!
1000.0
36
7.2
-0.8
57
3.62
350
16 280.4 290.6 281.0!
987.0
143
5.2
-2.8
56
3.17
349
17 279.4 288.4 279.9!
925.0
668
0.0
-6.0
64
2.65
345
21 279.3 286.9 279.8!
903.0
860
-2.1
-5.4
78
2.84
346
22 279.1 287.2 279.6!
850.0
1338
-5.7 -10.7
68
2.00
350
23 280.2 286.0 280.5!
807.0
1742
-9.4 -13.3
73
1.70
355
25 280.4 285.5 280.7!
801.0
1800
-9.9 -13.7
74
1.67
354
25 280.5 285.4 280.8!
777.0
2034 -10.9 -25.9
28
0.60
348
25 281.9 283.8 282.0!
759.0
2214 -11.5 -31.5
17
0.36
344
25 283.1 284.3 283.2!
703.0
2796 -16.1 -36.1
16
0.25
331
25 284.3 285.1 284.3!
700.0
2828 -16.3 -32.3
24
0.37
330
25 284.4 285.6 284.5!
688.0
2958 -17.1 -22.1
65
0.95
327
25 284.9 287.9 285.1!
682.0
3024 -17.5 -25.5
50
0.71
325
24 285.2 287.4 285.3!
672.0
3135 -18.5 -26.5
49
0.66
323
24 285.3 287.4 285.4!
662.0
3246 -19.5 -26.1
56
0.69
320
23 285.3 287.6 285.5!
647.0
3417 -21.1 -25.6
67
0.74
320
23 285.4 287.8 285.6!
638.0
3520 -21.9 -24.7
78
0.81
321
24 285.7 288.3 285.8!
616.0
3778 -23.3 -33.3
39
0.38
321
24 286.9 288.2 287.0!
598.0
3995 -24.7 -32.7
47
0.41
322
24 287.8 289.1 287.8!
577.0
4255 -26.7 -37.7
35
0.26
322
24 288.4 289.3 288.4!
568.0
4368 -26.3 -50.3
9
0.07
323
24 290.1 290.4 290.2!
500.0
5280 -32.7 -51.7
13
0.07
325
25 293.1 293.4 293.1!
€
Problem
Compute the height of the LCL.
Is this level reached spontaneously?
De Bilt,
Absolute and Conditional Instability:
18 -----------------------------------------------------------------------------!
April 2001 12 UTC
PRES
HGHT
TEMP
DWPT
RELH
MIXR
DRCT
SKNT
THTA
THTE
THTV!
hPa
m
C
C
%
g/kg
deg
knot
K
K
K !
-----------------------------------------------------------------------------!
1004.0
4
7.8
-0.2
57
3.77
350
16 280.6 291.3 281.3!
1000.0
36
7.2
-0.8
57
3.62
350
16 280.4 290.6 281.0!
987.0
143
5.2
-2.8
56
3.17
349
17 279.4 288.4 279.9!
925.0
668
0.0
-6.0
64
2.65
345
21 279.3 286.9 279.8!
903.0
860
-2.1
-5.4
78
2.84
346
22 279.1 287.2 279.6!
850.0
1338
-5.7 -10.7
68
2.00
350
23 280.2 286.0 280.5!
807.0
1742
-9.4 -13.3
73
1.70
355
25 280.4 285.5 280.7!
801.0
1800
-9.9 -13.7
74
1.67
354
25 280.5 285.4 280.8!
777.0
2034 -10.9 -25.9
28
0.60
348
25 281.9 283.8 282.0!
759.0
2214 -11.5 -31.5
17
0.36
344
25 283.1 284.3 283.2!
703.0
2796 -16.1 -36.1
16
0.25
331
25 284.3 285.1 284.3!
700.0
2828 -16.3 -32.3
24
0.37
330
25 284.4 285.6 284.5!
688.0
2958 -17.1 -22.1
65
0.95
327
25 284.9 287.9 285.1!
682.0
3024 -17.5 -25.5
50
0.71
325
24 285.2 287.4 285.3!
672.0
3135 -18.5 -26.5
49
0.66
323
24 285.3 287.4 285.4!
662.0
3246 -19.5 -26.1
56
0.69
320
23 285.3 287.6 285.5!
e
647.0
3417 -21.1 -25.6
67
0.74
320
23 285.4 287.8 285.6!
638.0
3520 -21.9 -24.7
78
0.81
321
24 285.7 288.3 285.8!
616.0
3778 -23.3 -33.3
39
0.38
321
24 286.9 288.2 287.0!
598.0
3995 -24.7 -32.7
47
0.41
322
24 287.8 289.1 287.8!
577.0
4255 -26.7 -37.7
35
0.26
322
24 288.4 289.3 288.4!
568.0
4368 -26.3 -50.3
9
0.07
323
24 290.1 290.4 290.2!
500.0
5280 -32.7 -51.7
13
0.07
325
25 293.1 293.4 293.1!
∂θ
<0
∂z
Absolute Instability (AI) up to 800 hPa
€
∂θ
<0
∂z
Conditional (potential) Instability up to 568 hPa
€
South America
Manaus
Station latitude: -3.15
Station longitude: -59.98
Station elevation: 84.0!
82332 SBMN Manaus (Aeroporto) Observations at 00Z 08 Oct 2004
------------------------------------------------------!
PRES
HGHT
TEMP
DWPT
RELH
MIXR
DRCT
SKNT
THTA
THTE
THTV
hPa
m
C
C
%
g/kg
deg
knot
K
K
K !
----------------------------------------------------------------------------1002.0
84
27.0
25.4
91 20.90
0
0 300.0 361.6 303.7
1000.0
101
27.0
25.8
93 21.47
0
0 300.1 363.5 304.0
935.0
693
22.2
21.8
98 17.94
65
3 301.1 354.1 304.3
925.0
787
22.2
21.4
95 17.69
75
3 302.0 354.4 305.2
911.6
914
21.6
20.3
92 16.74
75
6 302.6 352.4 305.7
850.0
1521
18.6
15.0
80 12.77
70
12 305.6 344.1 308.0
791.2
2134
15.9
11.3
74 10.77
100
12 309.1 342.1 311.1
789.0
2158
15.8
11.2
74 10.70
100
12 309.2 342.0 311.2
763.3
2438
13.6
11.0
84 10.89
100
12 309.8 343.2 311.8
736.0
2745
11.2
10.7
97 11.10
102
11 310.4 344.6 312.4
700.0
3164
9.4
7.7
89
9.51
105
9 312.9 342.6 314.6
657.0
3688
7.2
3.6
78
7.60
103
9 316.1 340.3 317.6
611.6
4267
3.5
0.9
83
6.75
100
10 318.4 340.2 319.7
541.0
5258
-2.7
-3.6
94
5.45
54
8 322.3 340.4 323.4
500.0
5880
-5.7
-6.4
95
4.77
25
6 326.0 342.1 327.0
486.5
6096
-6.8
-7.5
95
4.51
25
6 327.2 342.5 328.1
481.0
6185
-7.3
-7.9
95
4.41
26
6 327.7 342.7 328.6
480.0
6201
-5.1
-5.6
96
5.28
26
6 330.6 348.6 331.6
476.0
6267
-7.3
-7.9
95
4.46
26
6 328.7 343.9 329.6
400.0
7610 -15.3 -19.5
70
2.05
35
3 335.0 342.5 335.4
367.6
8230 -19.5 -24.6
64
1.43
335
5 337.6 342.9 337.9
300.0
9720 -29.7 -36.7
51
0.55
215
4 343.4 345.6 343.5
281.0 10182 -32.5 -38.5
55
0.49
219
5 345.9 347.9 346.0
250.0 10990 -39.9 -47.9
42
0.20
225
6 346.6 347.5 346.6
200.0 12470 -53.3 -60.3
42
0.06
220
12 348.2 348.5 348.2
150.0 14260 -68.9 -75.9
36
0.01
265
12 351.2 351.2 351.2
126.0 15281 -77.3 -83.3
38
0.00
354.0 354.0 354.0
Manaus (Brazil)
No absolute instability;
Only potential instability!!
How far up will a parcel at the ground have
to rise to reach its lifting condensation level?
Oct. 7 2100 UTC
tropical Cu convection
Layered clouds
Forced lifting in the ITCZ
M
“Tephigram”:
⎛ pref ⎞R /c p
θ = T ⎜
⎟
p
⎝
⎠
Explanation
constant θ
constant pressure
constant θe ⎛ Lrs ⎞
θ e ≈ θ exp⎜ ⎟
⎝θΠ ⎠
constant saturation mixing ratio
R e
rs = d s ?
Rv p
€
isotherm
€
T
€
€
Typical profiles in the tropics
⎛ pref ⎞R /c p
θ = T ⎜
⎟
⎝ p ⎠
Dew point temperature
temperature
€
€
⎛ Lrs ⎞
θ e ≈ θ exp⎜ ⎟
⎝θΠ ⎠
Typical profiles in the midlatitude “summer”
18 April 2001 12 UTC
⎛ pref ⎞R /c p
θ = T ⎜
⎟
⎝ p ⎠
Dew point temperature
temperature
€
tropopause
€
⎛ Lrs ⎞
θ e ≈ θ exp⎜ ⎟
⎝θΠ ⎠
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
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
1000 hPa
dTd/dz=1.8 K/km
10°C
20°C
30°C
Convective Available Potential
Energy (CAPE)
Force on air parcel:
d 2z
θ'
F = m 2 ≈ mg ≡ mgB
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.
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
€
€
De Bilt,
12Z 18 Apr ‘01
Potential Instability:
∂θ e
<0
∂z
-----------------------------------------------------------------------------!
PRES
HGHT
TEMP
DWPT
RELH
MIXR
DRCT
SKNT
THTA
THTE
THTV!
hPa
m
C
C
%
g/kg
deg
knot
K
K
K !
-----------------------------------------------------------------------------!
1004.0
4
7.8
-0.2
57
3.77
350
16 280.6 291.3 281.3!
1000.0
36
7.2
-0.8
57
3.62
350
16 280.4 290.6 281.0!
987.0
143
5.2
-2.8
56
3.17
349
17 279.4 288.4 279.9!
925.0
668
0.0
-6.0
64
2.65
345
21 279.3 286.9 279.8!
903.0
860
-2.1
-5.4
78
2.84
346
22 279.1 287.2 279.6!
850.0
1338
-5.7 -10.7
68
2.00
350
23 280.2 286.0 280.5!
807.0
1742
-9.4 -13.3
73
1.70
355
25 280.4 285.5 280.7!
801.0
1800
-9.9 -13.7
74
1.67
354
25 280.5 285.4 280.8!
777.0
2034 -10.9 -25.9
28
0.60
348
25 281.9 283.8 282.0!
759.0
2214 -11.5 -31.5
17
0.36
344
25 283.1 284.3 283.2!
703.0
2796 -16.1 -36.1
16
0.25
331
25 284.3 285.1 284.3!
700.0
2828 -16.3 -32.3
24
0.37
330
25 284.4 285.6 284.5!
688.0
2958 -17.1 -22.1
65
0.95
327
25 284.9 287.9 285.1!
2
-2
682.0
3024 -17.5 -25.5
50
0.71
325
24 285.2 287.4 285.3!
672.0
3135 -18.5 -26.5
49
0.66
323
24 285.3 287.4 285.4!
-1
662.0
3246 -19.5 -26.1
0.69
320
23 285.3 287.6 285.5!
256
647.0
3417 -21.1 -25.6
67
0.74
320
23 285.4 287.8 285.6!
638.0
3520 -21.9 -24.7
78
0.81
321
24 285.7 288.3 285.8!
616.0
3778 -23.3 -33.3
39
0.38
321
24 286.9 288.2 287.0!
598.0
3995 -24.7 -32.7
47
0.41
322
24 287.8 289.1 287.8!
577.0
4255 -26.7 -37.7
35
0.26
322
24 288.4 289.3 288.4!
568.0
4368 -26.3 -50.3
9
0.07
323
24 290.1 290.4 290.2!
500.0
5280 -32.7 -51.7
13
0.07
325
25 293.1 293.4 293.1!
€
CAPE=15 J/kg
If CAPE=15 m s w =5.5 m s Potential Instability (PI) up to 568 hPa
82332 SBMN Manaus (Aeroporto) Observations at 00Z 08 Oct 2004
------------------------------------------------------!
PRES
HGHT
TEMP
DWPT
RELH
MIXR
DRCT
SKNT
THTA
THTE
THTV
hPa
m
C
C
%
g/kg
deg
knot
K
K
K !
----------------------------------------------------------------------------1002.0
84
27.0
25.4
91 20.90
0
0 300.0 361.6 303.7
1000.0
101
27.0
25.8
93 21.47
0
0 300.1 363.5 304.0
935.0
693
22.2
21.8
98 17.94
65
3 301.1 354.1 304.3
925.0
787
22.2
21.4
95 17.69
75
3 302.0 354.4 305.2
911.6
914
21.6
20.3
92 16.74
75
6 302.6 352.4 305.7
850.0
1521
18.6
15.0
80 12.77
70
12 305.6 344.1 308.0
791.2
2134
15.9
11.3
74 10.77
100
12 309.1 342.1 311.1
789.0
2158
15.8
11.2
74 10.70
100
12 309.2 342.0 311.2
763.3
2438
13.6
11.0
84 10.89
100
12 309.8 343.2 311.8
736.0
2745
11.2
10.7
97 11.10
102
11 310.4 344.6 312.4
700.0
3164
9.4
7.7
89
9.51
105
9 312.9 342.6 314.6
657.0
3688
7.2
3.6
78
7.60
103
9 316.1 340.3 317.6
611.6
4267
3.5
0.9
83
6.75
100
10 318.4 340.2 319.7
541.0
5258
-2.7
-3.6
94
5.45
54
8 322.3 340.4 323.4
500.0
5880
-5.7
-6.4
95
4.77
25
6 326.0 342.1 327.0
486.5
6096
-6.8
-7.5
95
4.51
25
6 327.2 342.5 328.1
481.0
6185
-7.3
-7.9
95
4.41
26
6 327.7 342.7 328.6
2 -2
480.0
6201
-5.1
-5.6
96
5.28
26
6 330.6 348.6 331.6
476.0
6267
-7.3
-7.9
95
4.46 -1 26
6 328.7 343.9 329.6
400.0
7610 -15.3 -19.5
702
2.05
35
3 335.0 342.5 335.4
367.6
8230 -19.5 -24.6
64
1.43
335
5 337.6 342.9 337.9
300.0
9720 -29.7 -36.7
51
0.55
215
4 343.4 345.6 343.5
281.0 10182 -32.5 -38.5
55
0.49
219
5 345.9 347.9 346.0
250.0 10990 -39.9 -47.9
42
0.20
225
6 346.6 347.5 346.6
200.0 12470 -53.3 -60.3
42
0.06
220
12 348.2 348.5 348.2
150.0 14260 -68.9 -75.9
36
0.01
265
12 351.2 351.2 351.2
126.0 15281 -77.3 -83.3
38
0.00
354.0 354.0 354.0
Manaus (Brazil)
CAPE=2839 J/kg
If CAPE=2839 m s w =? m s Sounding Bordeaux
Problem 1.10, p. 71
Spanish plume
(warm and dry)
Air parcels with high relative humidity must be forced to rise through the stable layer
Surface weather map of 20
July 1992, 15 UTC. The
position of a surface station is
indicated by a square. The
number inside the square
indicates the cloudiness (in
octas). Also indicated are the
temperature (°C) (upper left),
the dew point temperature (°C)
(lower left) and the pressure
(hPa-1000) (upper right). Also
shown are sea level isobars
drawn every 1 hPa (thick line
corresponds to 1012 hPa),
according an objective analysis
scheme. The letters 'Za"
indicate the position of
Zaragoza. Arrows indicate the
movement and sources of
moisture for the thunderstorm.
The confluence line is
indicated by a "hooked" solid
line
Infrared Meteosat satellite
image, 20 July 1992,19:30
UTC. Illustration of the
outbreak of summer
thunderstorms due to
forced lifting of potentially
unstable air in advance of
a cold front over southwest Europe.
Height, temperature and temperature
advection at 850 hPa on 20 July 1992, 12
UTC. The height, labeled in units of m, and
the temperature, labeled in units of °C, are
shown in panel (a), while temperature
advection , labeled in units of 10-4 K s-1, is
shown in panel (b).
Homework:
Problem 1.10
Plot the data shown in table 1.2 in a tephigram (figure 1.30).
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 1.5. 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 1.6. Verify the value of θe at
the surface using eq. 1.78. Estimate the value of CAPE.
Repeat this for the data in table 1.3.
Next week:
large scale circulation
geostrophic balance, vorticity
and potential vorticity
11:00-12:45: Room David de Wied Gebouw (DDW) 1.05
13:30-16:00: Room 165 BBG
http://www.staff.science.uu.nl/~delde102/WeatherDiscussions.htm
Next week
•  sections 1.19 to 1.24 and Boxes 1.7 and 1.8.
(geostrophic balance and thermal wind
balance: (in)stability, cyclones)
•  Problems 1.11, 1.12 and Box 1.7
First project: construct a numerical model of an inertial oscillation
Solve this in the programming language Python
Python programming language
Python: www.python.org,
E-book on the use of Python in atmospheric sciences: http://www.johnny-lin.com/pyintro/
Package for scientific computing in python (Numpy): http://www.numpy.org/
Package for plotting in python (Matplotlib): http://matplotlib.org/
This site is useful in demonstrating how to make plots and how to customize your plots:
http://www.loria.fr/~rougier/teaching/matplotlib/
The Canopy Python distribution might now be an easy way to get to know python:
https://www.enthought.com/products/canopy/
If you want to plot some figures that you want to use in latex the best way is to save your
figure as a pdf using the plt.savefig command. That way the plot scales with the resolution
so your figure will appear smooth in latex.
You can use the pandas package (http://pandas.pydata.org/) to load text files, excel sheets
and other databases easily into python and convert them to pandas dataframes. An easy
way to load external data into python.
Example:
Problem 0.1
Python
script of
solution to
problem 0.1

(Part - I ) FOR EXTERNAL STUDENTS (2013

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