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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

(a.j.vandelden@uu.nl) (http://www.phys.uu.nl/~nvdelden/dynmeteorology.htm)

Problem 1.12

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;

€

drsdt

≅ 0 for w ≤ 0.

€

J ≈ −L drsdt

Section 1.14

€

m ≈ md = constant

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Conditional instability

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

Assume that θ=θ0(z)+θ’, with θ’<<θ0. We have (eq 1.23/1.33),

€

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

€

Nm2 ≡ N 2 +

gLθ0Π0

drsdz,

(latent heat release only in the updraught!!!)

If Nm<0 and w>0 then

€

dθ 'dt

> 0: positive buoyancy and upward acceleration

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.

Simply define:

€

Nm2 =

gθe( )0

d θe( )0dz

€

θe ≈θ expLrsθΠ⎛ ⎝ ⎜

⎞ ⎠ ⎟ then

Extra problem for exercise session: Show that equivalent potential temperature is approximately conserved in saturated upward motion.

€

Nm2 ≡ N 2 +

gLθ0Π0

drsdz,

Section 1.14

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Warm conveyor belt

cold and dry air

Warm conveyor belt

cold and dry air

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Tropical cyclone “Nadine”: warm/moist core

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Cyclone with cold/dry core

cold and dry air

Equivalent potential temperature

€

θe ≈θ expLrsθΠ⎛ ⎝ ⎜

⎞ ⎠ ⎟

Equivalent potential temperature unsaturated air parcel:

€

θe =θexp LrθΠLCL

⎛

⎝ ⎜

⎞

⎠ ⎟ =θexp

LrcpTLCL

⎛

⎝ ⎜

⎞

⎠ ⎟

(LCL: lifting condensation level)

actual mixing ratio

Section 1.14

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Explanation: constant θ

constant θe

constant satura-tion mixing ratio

isotherm

constant pressure €

θ =Tprefp

⎛

⎝ ⎜

⎞

⎠ ⎟

R /cp

€

θe ≈θ expLrsθΠ⎛ ⎝ ⎜

⎞ ⎠ ⎟

€

rs ≈RdesRv p

?

€

T

“Tephigram”:

dew point temperature

temperature

undiluted parcel

diluted parcel

Tephigram

LCL

LNB

Parcel of air is lifted from the ground. What “temperature-profile” does it follow?

LNB=level of no buoyancy

LCL=lifted condensation level

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Tephigram Close-up

Dew point temperature (environment)

Temperature (environment)

20°C 10°C 30°C

0°C

diluted parcel

undiluted parcel

800 hPa

1000 hPa

700 hPa

LCL

900 hPa dTd/dz=1.8 K/km

dT/dz=9.8 K/km

A case study of convection on 10 July 2013

x

x

Paris

Barcelona

where will cumulus convection occur?

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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

Paris

Compute the height of the LCL

Will cumulus clouds (and thunderstorms) form over Paris?

Dewpoint temperature

Temperature

(Paris) 00 UTC 10 July 2013

LCL at 550 m

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Dewpoint temperature

Temperature

(Paris)

€

∂θe∂z

< 0

€

Potential instability if ∂θe∂z

< 0

€

∂θe∂z

< 0

00 UTC 10 July 2013

Dewpoint temperature

Temperature

(Paris) 12 UTC 10 July 2013

Absolutely unstable surface layer, but drier LCL

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Dewpoint temperature

Temperature

00 UTC 10 July 2013

LCL

CAPE= 149 J/kg

Dewpoint temperature

Temperature

12 UTC 10 July 2013

LCL

CAPE= 536 J/kg

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Convective Available Potential Energy (CAPE)

€

dwdt

≈ w dwdz

= Bg.

Assuming a stationary state and horizontal homogeneity we can write:€

B ≡ θ 'θ0

€

F = md2zdt2

≈ mg θ 'θ0

≡ mgB

Force on air parcel:

B = buoyancy

or

€

wdw = Bgdz.

Section 1.15

Convective Available Potential Energy (CAPE)

A 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.

Section 1.15

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A case study of convection on 10 July 2013

x Paris

x Barcelona

A case study of convection on 10 July 2013

x Paris

x Barcelona

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A case study of convection on 10 July 2013

x Paris

x Barcelona

A case study of convection on 10 July 2013

x Paris

x Barcelona

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A case study of convection on 10 July 2013

x Paris

x Barcelona

A case study of convection on 10 July 2013

x Paris

x Barcelona

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A case study of convection on 10 July 2013

x Paris

x Barcelona

A case study of convection on 10 July 2013

x Paris

x Barcelona

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A case study of convection on 10 July 2013

x Paris

x Barcelona

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.

Homework:

*Download the tephigram from: http://www.staff.science.uu.nl/~delde102/tephigram.pdf

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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)