lecture 1 general concepts

8/2/2019 Lecture 1 General Concepts

http://slidepdf.com/reader/full/lecture-1-general-concepts 1/9

1

A SHORT COURSE ON MICROMETEOROLOGY (F. Castellví, Bahía Blanca, Oct. 2010)

LECTURE 1. GENERAL CONCEPTS

1. The air

The atmosphere has a mass of about 5 × 1018

kg, three quarters of which is within about 11 km of

the surface.

Dry air contains (by volume) 78.09% nitrogen,

20.95% oxygen, 0.93% argon, 0.039% carbon

dioxide, and small amounts of other gases (ozone,

methane,..,water vapour).

Air also contains a variable amount of water vapor,

on average around 1%. Recall that when 1 g of

water vapour condense, the amount of energy

delivered is capable to increase 2.5 C the

temperature of about 1kg of air.

2. General properties

A summary of concepts to recall are the following:

1.- Most of gas in an air parcel is N2 and O2. Therefore, in practice, dry air is considered a

diatomic perfect gas. For convenience, also air (dry air + water vapor + …) is assumed a diatomic

ideal gas. Therefore,

(never seen in textbooks of meteorology or fluid dynamics!)

We do not measure V of air parcels or n for each gas

Where Mw is the molecular weight.

Standard conditions

(T=0 C, P=1 atm).

Molecular Weight

(g/mol)

Density

(kg m-3)

Nitrogen 28.01 1.250

Oxygen 32.00 1.428

Argon 38.98 1.782

Carbon dioxide 44.01 1.977

Other 2.974

Dry air 29.00 1.292

nRT =PV

T r m=T Mw

Rm=PV



2

Therefore,

For dry air, typical notation is; Pd = ρ d r d T ,r d = 287 J kg-1

k -1

.

For water vapour, e = ρ v r vT , r v=461.5 J kg-1 k -1.

2.- For air (dry air +water vapour); P = Pd +e (Dalton’s law) and ρ = ρ d + ρ v . The latter allows to

write:

)1(

)608.11(

m

ma

r

r r r

+

+= where r m = ρ v/ρ d = mv /md [ mass of water vapour over mass of dry air] is the

mixing ratio.

3.- Typically, e / P is about 10-2. The specific humidity, q = mv /m [typical units are gr/kg], in

general, ranges in the following interval: 0.28 (cold climate) – 18 gr/kg (tropics).

When a parcel of air is saturated of water vapour, e is typically denoted by, es(T), and can be

determined as

es(T)=0.611 exp[ (17.27 T ) / (237.3+T )]

For this expression, the units for es(T) are [kPa] and the temperature, T , must be expressed in

Celsius. The saturation vapour pressure is the vapour pressure at which a change in phase can

occur at constant temperature.

To calculate the latent heat of condensation (vaporation), L, in water in the temperature range

from −40 °C to 40 °C the following empirical cubic function (determination coefficient, R2 =

0.999988) can be used

L = − 0.0000614342T 3

+ 0.00158927T 2

− 2.36418T + 2500.79

where T is in °C and L is expressed in kJ/Kg.

4.- Specific heat at constant pressure, Cp. For dry air, Cpd =1005 J Kg-1

K-1

. For water vapour

Cpv=1846 J Kg-1

K-1

. For air, Cp= q Cpv +(1-q) Cpd = (1+0.84q) Cpd .

The psychometric constant, γ , is )()(

ε γ LCpP= , where ζ=rd /rv=(Mw)v /(Mw)a=0.622.

T r =P ρ



3

5.- For same conditions (pressure and temperature), dry air is denser than (humid) air.

ρ= ρa+ ρv , which leads to; 1)]-e(+[Pr T

1=]

r

e+

r

e-P[

T

1=

ava

ζ ρ

Therefore, it is easier to raise a parcel of air rich of water vapour. The virtual temperature, T v, is

defined as the temperature of dry air, for same P, should have in order to have the same density

as moist air with given q and T . Accordingly, ρ =T r

P

va

, and T v can also be expressed as

]P

e)-(1+T[1

]P

)e-(1-[1

T =T v ζ

ζ ≈ =(1+0.61q)T

Pressure mainly changes with height. Two air parcels at same level having different T v informs

that it is easier to raise the parcel which T v is higher. If the two air parcels are not levelled (P is

different), therefore, we have to level them before to make this comparison. The latter also

requires correction for a change in temperature with height.

6.- First Law of Thermodynamics:

Where q, u and w are heat, internal energy and work.

du = n C v dT . For a perfect gas, C v does not depends on P or T . Therefore, u exclusively depends

on mass of gas and temperature.

dw is positive when the air parcel expands and vice versa, dw = P dV .

dq is positive when the parcel takes heat from its surroundings.

In thermodynamics of the atmosphere are not observed processes at constant volume. However,

during the night some processes can be assumed at constant pressure, such as dew formation.

Adiabatic processes are mainly assumed when air parcels with e / es(T) < 1 have small and

relatively fast vertical displacements. The latter conditions make feasible to assume that the

parcel is not mixed with air of the surroundings and that the pressure of the parcel is adjusted to

w+u=q ∆∆∆



4

the new surroundings as it moves.

When an air parcel rises adiabatically it expands because the

pressure of his surroundings decreases. Consequently, itstemperature also decreases (internal energy per unit mass only

depends on temperature). Therefore, as the parcel moves up, it

becomes closer to saturation and a cloud may be formed. If the

parcel is saturated and still moves up, the process is not

adiabatic [parcel is enriched of heat delivered in the change of

phase vapour-liquid and may loss some mass (droplets)]. It is called pseudo-adiabatic and it is not

a process at constant entropy.

The First Law of Thermodynamics in micrometeorology is often called the surface energy

balance and it is written as:

( Rn-G)= H+LE+other energy flux terms

Rn is the net radiation, G is the soil heat flux, H is the sensible heat flux, LE is the latent heat flux, and other energy terms refers to dissipation, to photosynthesis, storage ….Typical units;

Wm-2.

Note that the energy balance refers to a volume of air, so those terms are flux densities. However,

this terminology is not often used.

Exercices:

a.- During a calm night, the air temperature is T=0 C. A moist air parcel, which mass is 1kg, has1g of water that condenses and the parcel becomes dry. Determine the temperature rise of the

parcel.

b.- Consider a mass, m=1kg, for a given perfect gas. Derive the following expression valid for an

adiabatic process:

z

P

z

PV

z

T C p

∂

∂=

∂

∂=

∂

∂

ρ

1

What is the assumption made?.



5

c.-The potential temperature, θ, is the temperature which would result if a parcel located at a

given height, z, T and P were brought adiabatically to a standard pressure level P0 (=1000 mb).

Derive the following expression

k

PPT

= 0θ where ( )q

Cpr

Cpr k

d

d 23.01−

=

=

7.- Similar to θ, it is defined the potential virtual temperature, θv

k

vvP

PT

= 0θ where

=

d

d

Cp

r k

If two air parcels are not at same level. Determination of θv (they are now at a reference level, P0)

indicates to us which of them is easier to rise (the higher θv).

8.- Considering a dry parcel having a vertical displacement (adiabatic) in a hydrostatic body of air

( zg-=P ∂∂ ρ ), the rate at which its virtual temperature changes with height, γ [=− z

T v p

∂

∂; where

index p denotes air parcel ], is γ = 1 K/hm. Therefore, if one considers that the parcel does not

exchange moisture with the surrounding body air as it displaces ( z

q

∂

∂=0), arrives to,

z

T v p

∂

∂=

(1+061q) z

T p

∂

∂. The latter expression indicates that for parcels far from saturation, in practice, it

can be assumed, γ ≈ z

T -

p

∂

∂.

Exercice: Derive the following expression for γ;

hmK 1

C g=

zT -

p

p ≈∂

∂≈γ

Note that, considering a huge layer of the atmosphere (ex. the troposphere), we do not observe

fast vertical displacements. Therefore, as a first approximation, the atmosphere can be assumed a

hydrostatic system.

9.- As a first approximation, the atmosphere may be considered vertically thermal stratified. The

mean temperature of thin atmospheric layers changes with height and the gradient is denoted as,



6

z

T -=

∂

∂α .The next table shows few characteristics for the mean actual lapse rate, α , for different

layers (values for mid latitudes).

As a rule of thumb, in the troposphere α=0.65 K/hm. But thisvalue is far to be close for layers close to the surface. The latter

due to continuous energy and scalar exchanges between the

ground-vegetation-atmosphere. For example, during the night,

typically α is positive (air temperature increases as we move

up, this is called inversion).

Consider an atmospheric layer at rest and thermal stratified. The diagram of the forces ( positiveupwards) involved in a parcel is the following

Fz - Fz+dz - W = Fnet

Therefore, the vertical acceleration of the parcel, p z&& , may be determined by the following

expressions:

−−≈

−

∂

∂=

∂

∂−

∂

∂=

∂

∂−

p

ps

ps

ps

s ps

p g z

p

z

p

z

p

z

pg-= z

ρ

ρ ρ

ρ ρ

ρ ρ

ρ ρ ρ

111&&

where ρ s and ρ p are the mean air density of the layer (surroundings) and the parcel, respectively.

The latter expression can be re-written as,

) z-(z)-(T

g

-=gT

T -T

= z 0vvv

vvp

p α γ &&

Aprox*. Layerdepth (km)

Sign for α Name of theLayer

[0,11]

[11,51]

[51,80]

[80,..]

Positive

Null

Negative

Null

Positive

Nul

Negative

Troposphere

Tropopause

Stratosphere

Stratopause

Mesosphere

Mesopause

Termosphere

dz

Fz

Fz+dz

Weight



7

Where z0 is a reference height. Note that for a given height, z, the pressure is constant. Thus, by

virtue of the equation of state and definition of virtual temperature, the quantity, ( ρ T v), remains

constant at z. The right expression is obtained by assuming the initial condition that at z0 the

virtual temperature of the air parcel and its surrounding is the same. That is, in terms of T v it isnot possible to delimitate the boundary of the air parcel at z0. Therefore,

Tv = Tv(z) = Tv(zo) – αv( z- z0) where

∂

∂−=

z

T v

vα

Tvp = Tvp (z) = Tvp(zo) - γ( z- z0)

Exercise: Derive the following two expressions;

+

∂

∂=

∂

∂γ

θ θ

z

T

T z

v

v

vv

z

) z-(z g= z v

v

0

p∂

∂−

θ

θ &&

The latter expression derived in the exercise, which resembles the Hook’s law, indicates that

when

∂

∂

z

vθ = 0, the atmosphere is statically neutral. When it is negative, therefore, the

atmosphere is unstable and vice versa.

10.- Often the vertical profile of the humidity is not available. However, for many practical

purposes, within the first 10 m above the ground, it is not necessary to distinguish between T and

θ , and between θ and θ v. Therefore, for thin layers close to the surface the stability is mainly

explained by the sign of (γ−α), as shown in the next Figures. Such simplifications often are used

to remind that the troposphere is a thick stable layer (i.e., the mean value for α, is α=0.65 K/hm.)

which has a capping inversion (i.e., α<0 in the stratosphere).



8

11.- The next photographs and artistic graphs show consequences of the thermal stability in a

stratified flow at rest (From Ogawa et al., 1992. Atmos. Env., 16, 1419-1433, and

http://ocw.usu.edu/Forest__Range__and_Wildlife_Sciences/Wildland_Fire_Management_and_Planning/Unit_7_Atmospheric_Stability_and_Instabi

lity_1.html, respectively)

Spatial evolution of traces in a lab experiment. Unstable case (left). Stable case (right)

If the atmosphere is unstable, any parcel of

air that is lifted will tend to rise like a hot air

balloon. As this parcel rises, it decreases in

temperature at a rate of γ, and will continue

to rise until it reaches a level in which its

new temperature equals that of the

surrounding air.

Stable air tends to resist vertical air

movement. If a horizontally moving parcel

of air is lifted or forced to rise, as over a

mountain, that parcel will tend to settle back

to its original level. It is heavier than the air

around it; therefore, it will sink back, if

possible, to the level from which it

originated.

If the atmosphere is neutral; that is, the

actual temperature lapse rate equals the dry

adiabatic lapse rate, a parcel of air that is

lifted will be neither heavier nor lighter at a



9

different altitude. As this parcel is forced up, it decreases in temperature at a rate of 1K per 100

m. The surrounding air at the new altitude will have the same temperature, and it will remain

neutral.

As in the lab experiments, the smoke

reveals the stability of the surface layer.

How the plume is dispersed is useful to

quantify the effect of contamination.

Exercises.

1. Which picture was taken under stable conditions?

2. At night, imagine a bubble of air initially traveling in the mean flow that suddenly moves down

towards the surface and it encounters an inversion. ¿How is the expected movement’s bubble?,

and may it have some consequences at the surface?.

lecture 1 general concepts

Documents