~

,

~ NCArt)e,..: ~ ~

it

SATURATION AND ATMOSPHERIC STABILITY

Materials Needed • calculator

L

Use Table 6-J tojind the saturation mixing ratio (m,).of air samples Ii and B.

• ruler A:

m. =

g kg-I

Introduction How do douds form? The concept of saturation, introduced in the previous lab. is important here. The two baste means by which air achieves saturation (relative humidity

= 100%) are cool-

2.

ing and increasing water vapor content. In this lab we will examine several common atmos-

What are the

mixing ratio

(m) values

of samples

A

andB?

pheric processes that will lead to either or both of these phenomena. We will focus particularly on how air changes as it rises from the earth's

B:

m=

gkg-'

surface and how these changes relate to atmospheric stability. J.

Use the temperature ana mixing ratio data to plot

the two air samples in Figure 6-1.

Mixing Saturation occasionally occurs when two air masses rnix, Mixing usually involves a change in both water vapor content and temperature. Consider, for example, two unsaturated airsamples, A and B: Air Sample

A

B

Temperature Relative humidity

lO·C 75%

38"C 75%

--

Assume that equal parts. of these two air

masses mix together. The resulting temperature is halfway between 1O"Cand 38°C, and the mixing ratio is halfway between the values that you fouad.for the two samples.

Table 6-1 Saturation Mixing Ratio (9 kg·1)

at Sea-Level Pressure as a Function of Dry-Bulb Temperature ("C)

eel

60 ...------

(g/kg)

-40 -35

0.118 0.195

-30

0.318

-25 -20

0.510 0.784

-18 -16 -14

0.931 1.102

50

1.300 1529

-12

40

.j>

-10

1.794

-8 -6

2.009 2.450

~

"4

2.852

-2

3.313

a:

o

3.819

2

4.439

4 6 8 10

5.120 5.894 6.771 7.762

12 14

8.882 10.140

16 18 20 22 24

11.560 13.162 14.956 16.963 19.210

26

21.734

28

24.557

30

27.694

.2

10

30

)

0>

c

'x ~

20

10

ou.~~~all~~~~~~~~ -40

-30

-20

-io

to

20

30

40

Figure';.1

8.

4.

Plot the mixed air sample in Figure 6-1.

5.

What is the new temperature?

6.

Saturation mixing ratio?

_ ~

Relative humidity? _--"

.~ample.Jtt contraiisare one example of this as warm exhaust CQrttainingmoisture mixes with cold air. Some contrails disappear quicklY, others linger; What doyou think determines the life span· of a contrail?

_

__

9: The nonlinear relationship between temperature and saturation makes it possible for two unsaturated air samples· tontix and form Ii new saturated air

49.815.

7. Mixing ratio?

0

Temperature (°0)

32 31.213 34 35.134 36 39.502 3844.381 40

.••.••• - •.•••••••••• --------,

Adiabatic Cooling Saturation occurs more commonly when air is

cooled to its dew-point temperature than when air masses mix. In nature, this cooling often occurs when air rises. Consider a parcel of air

;3000m--

that is forced to rise and does not mix with the

surrounding environment. As the parcel rises atmospheric pressure around it decreases, allowing the parcel to expand (Figttre 6-2}. Since the parcel has the same number of molecules

2000m--· ._.

but occupies more volume, its average internal

energy (i.e.•temperature) decreases.

I

1000m

..

h. The "collision» or convergence of surface air masses. c. The contact of air masses ofunlike temperature along warm and coldfronts,

e.

@

Surface Height .. 1-

s

..c•.... decreasing at the cpsrce.ll. m.peratUre dry adiabatic lapse rate: .• 1°C/1 00 rn

Air can be forced to rise because of: a. Intense surface heating.

,--_....,..._

••••••• --------

Figure 6-2. Adiabatic cooling

of anunsaturat4ld air parcel.

d. A topographic barrier (such as a mountain range). e. Upper-air divergence (a concept we will

discuss.in a later lab). As an unsaturated air parcel rises its temperature will decrease at the d1)'adiabatic lapse rate (DALR) of approximately lOoe per kilometer. 10. Calculate the temperature of an un.~aturaud air

parcel at IOO-m increments as it L~forcedto rise from the earth's sutjaa, where its temperature is 35°C, Height (tn)

Temperature

Itis possible that rising air parcels will cool

to the dew-point temperature. When this happens, a parcel becomes saturated-and condensationor clouds form. The height at which this occurs is called the lifting ccmdcnsatWn level (LCL). If the air parcel rises above the LeL, it.cools at a slower rate referred.to as the saturated,moistor wet adiabatiC lapse rate (WALR). The wet adiabatic lapse.tate.ranges between 5·Cartd9°C per kilometer, Itis slower than the dry adiabatic lapse

1000(1 km)

rate because Iatent.heat is released Within the

900

parcel.as water vapor condenses, a warming that partially offsets adiabatic cooling. The WALR varies because the amount of condensation

800 700

depends both on the amount of water vapor in the parcel and on atmospheric pressure.

600 500

400 300 200 100

Earth's surface

__

I

d!

A ..

l!SC

•••• ..;;.. ..

------

....

Table 6-2 PartlEdA Temperature

Parcel 8 Temperature ("C)

Height (km)

(0C)

5.0 4.5

II--------_--+---~--.,-.-...~+----.·-. ---.-----.~"--~4.0

3.5 ______

3_.~O__ ~ __

+,~.~---------~ .. -----

2.5 2.0.

1.5

._---""-,,_._--,, .... __ ._----+---------+--._--_._-_.--_.

__ ... -... --.-" .."..

1.0

11--------------------------------+-----_·_-------0.5 28"C

surface

11. Consider air parcel A, which is 28~Cat the sur.face. It isforced to rise to 5 km. The liJtingcondensation level is 1.5 km, above which the parcel cools at an

10"C

In the previous example,

sation

level

was

provided for

the lifting conden-

you. We Can calcu-

late the LeL from surface temperatureand

average wet adiabatic lapse rate of 5°C per kilome-

dew-point temperature. TheLCL generally

ter. Fill in the left column oITable 6-2 indicating the parcel's changing temperature.

occurs at the height where air has cooled to the dew-point temperature. dew-point

12. Now coflsic1er ai-rparcel B with a swfacetemperature of 1O·C that is Joreed to rise. Jt too reac1tes the LeL at 1.5 km, but the average wet adiabatic lapse rate in this case is 7'C per kilometer. Fill in the right column oj the table indicating temperature change in the lower 5 kilometers.

like temperature,

temperaturealsodecreases

heigh t because of decreasing pressure.

with

Asa rule

of thumb. dew point drops approximately 2 C pet kilometer. Consider the example shown in Figure 6-3. An air parcel at sea level has the following charu

acteristics; Temperature = 30·C. Dew-polar temperature 14·C. Pressure= 101{) mb, If

=

13. \4'hy would the warmer pareelcool at a slower rate' between 1.5 anll 5 km?

the parcel is forced to rise, its temperature

drops

at the dry adiabatic lapse rate {IO·Cper

kilometer) until it becomes saturated. The drop in dew-point

temperature

{2°C per kilometer)

can

be found by following a mixing ratio line associated with dew-point temperature at the surface, The point-where the lines 'intersect is the lifting condensation level.

.. " .....

_.---' -.-

....

~-.~~-

Mlxln9 Ratio Lines

Height

2} g kg-1

3000m

Lifting condensation

level ......~ Sea level

;-,;~ >

-'

-----c-'-~~~~~t--.---~.~-,~.------30"0

14"C

Temperature

Dew point

•• ~

Figure 6-3. Lifting condensation .Ievel.

"'~ ,~

,~

... ~

14. Fill in Table 6-3 below to show thi temperature and dew-point temperature of the forced air parcel .

lOx -lx = T - TJ 8x=30~14 8x:; 16 Now,

recalling the units in our equation: i6°C

If we know the surface temperature and dew point of an air parcel, we can use the following -

.,'y!

x=8°,C krn"

x=2km

equation to calculate the point of intersection: Ti.- 2x"" T -lOx The left side of the equation represents the

Nowthat you understand the concept, here's a

decrease of dew-point temperature with height; the right side represents the temperature

shortcut:

decrease with height. At some height x these expressions are equal. Solving algebraically, we

or

find: LCL (meters)= 125 [T("C) -T.("C)} Table 6-3 Height(m)

Temperature (toC)

Dew-point Tempel'ature(T~"C)

2000 1500 1000 500

earth's surface

30·C

14·C

t.;,;; F.··i' \'.; r.. ·. .,

15. calculate

the lifting condensation level (LeL) fOT I

the two examples below-.

Example A ExampleB

Surface

Surfa.ce Dew

Temp. (T)

Point (TJ)

32°C .32"C

20"(

LeI

Wee

16. What do the above calculationstellyou about the )·elationshipbetween surface del1.1 point and cloud height?

Optional

A Mountain

•.••"¥

Example

If air continues to rise beyond the lifting conden- . sation level, it cools at the wet adiabatic lapse rate

Ci.' :;; ~;;-:

{WALR).Because water vapor condenses out of

I'".~,," 1J.V.i

the rising air, temperature and dew-poinr temperature both decrease at the wet adiabatic lapse rate to the top of the mountain, and relative humidity remains at 100%. As air sinks on the leeward side of the mountain assume that its tei11peratureinq-easesat the dry adiabatic lapse rateand its dew-point ternperatureincreases at . 2~Cper kilometer. Consider an ,lit parcelforcedovera mountain range (figure 6-4). At the base of the wlnd-

", ELR> WALR). Again, notice that the

resulting cloud patterns reflect the tendency for of

vertical development only in unstablelayers

km

23. Enter the ,;"rect parcel temperatures and parcel

Jew poinis to complete Table 6-4. (Assume that the WALR ~ S"C/km);

the atmosphere (above 3300 meters in this example).

Table 64

____ tt~!gh! _

~~y.lr~!l~~!l!ill.'-T~m'p·._, .._..__

__ .__ ~,.9.!s.rn_,_.....

-28°e

__

-25°(

.__ ],?JrrL...

.sm

--=1.f~(::.__ ._ "" ~J9°C:_._.

.___._. ] .0

___ _____Q,?J m... . __.... ..Q.OI_........ _ ,__.__.__ ._. _ _.._. ..,_ ' 2?~c:._-__---------- __-._..-.--.-

0.5km

_ .__

_.

.._ .. , ..__.§~c:.......__.......~.o>,

surface

.

-l1SC'

o> .

._=_1l0~ __-c-~:1_Q:c:

____ .?;?_~r.rl__..~_.

..::t4~~ ..............• _

-HoC.

:J!?~