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<H2 align=3Dcenter>SUPPLEMENTAL INFORMATION...IN GREATER DEPTH</H2>
<H4 align=3Dcenter>To complement the Daily Summary for Thursday, 15 =
October=20
2009</H4>
<H3 align=3Dcenter>THE ADIABATS ON A ST=DCVE DIAGRAM </H3>
<HR>

<P>The St=FCve diagram (a blank St=FCve diagram can be obtained from the =
<I>AMS=20
Weather Studies website</I>) is designed to permit evaluation of various =

atmospheric process problems using simple graphical techniques. The =
fundamental=20
coordinates on this diagram include temperature (horizontal axis) and =
pressure=20
variations in the atmosphere (vertical axis). These coordinates appear =
also on a=20
simplified diagram. </P>
<P>The primary vertical coordinate on this diagram is plotted in =
pressure units,=20
decreasing from 1000 mb to 100 mb, instead of height because the desired =

temperature calculations actually involve pressure changes. This scale=20
conversion between height and pressure has been performed for you. If =
you refer=20
to the simplified diagram provided on page 2B-3 (<I>Weather Studies=20
Investigation 2B</I>), you will note the additional scale along the =
right margin=20
that contains the approximate geometric altitude of various pressure =
levels,=20
using a typical, reference atmosphere. </P>
<H4>DRY ADIABATS</H4>
<P>The dry adiabatic lapse rates are constructed on the St=FCve diagram =
as=20
straight lines that slant from lower right toward upper left. These =
lines are=20
often simply called "dry adiabats" to identify this process lapse rate. =
If you=20
would lift a dry (unsaturated) air parcel from a known initial point =
defined by=20
its temperature and pressure, to a final point, you could trace the =
amount of=20
cooling on the nearest dry adiabat. </P>
<P>On a blank St=FCve diagram, locate an air parcel as a point at sea =
level that=20
has a temperature of 17=BAC and air pressure of 1000 mb. Suppose that =
you lifted=20
this air parcel to an altitude of 1000 meters; using the approximation =
that air=20
pressure decreases by 1 mb for every 10 meters ascent, the air pressure =
at 1000=20
meters would be approximately 900 mb. Now trace along the dry adiabat =
from the=20
horizontal line marked 1000 mb to that marked 900 mb to determine how =
the parcel=20
cools dry adiabatically. The final temperature is 8=BAC; because of the=20
approximations used, this graphically determined value is not =
significantly=20
different from the value of 7=BAC that you would calculate from a 10 =
Celsius=20
degree cooling over 1 kilometer. Continue along this adiabat and prove =
to=20
yourself that at a pressure of 700 mb (an altitude of approximately 3000 =
meters)=20
the temperature of your air parcel would have cooled to roughly -11=BAC. =
Finally,=20
move back along the same adiabat to simulate sinking, compression and =
heating.=20
Verify that the parcel would return to 17 degrees C by the time you =
reached the=20
surface pressure of 1000 mb. </P>
<P>Note that you can also determine the response of parcels whose =
initial=20
conditions do not lie on a dry adiabat by using interpolation. For =
example, try=20
lifting a parcel with a temperature of 20=BAC and air pressure of 1000 =
mb first to=20
900 mb and then to 700 mb. By tracing along an imaginary line parallel =
to the=20
dry adiabat you should be able to find that at 900 mb (approximately =
1000=20
meters) the temperature would become 10.5=BAC and at a pressure of 700 =
mb (an=20
altitude of approximately 3000 meters) the temperature of your air =
parcel would=20
have cooled to roughly -10=BAC. </P>
<H4>SATURATION ADIABATS</H4>
<P>The saturation adiabatic lapse rates appear on a St=FCve diagram as a =
set of=20
dashed curves (also called "moist adiabats") with slopes that only =
become=20
tangent to the dry adiabats at low pressures and cold temperatures when =
most of=20
the vapor has been removed from the parcel and little latent heat =
release=20
occurs. This second process lapse rate varies between 2 and 9 Celsius =
degrees=20
per kilometer; the value depends upon the amount of vapor, as well as =
the=20
temperature and pressure of the air parcel. Warm moist air parcels can =
contain=20
large quantities of vapor that liberate large amounts of latent heat =
when=20
lifted; cool parcels have less vapor and less latent heat. A =
representative=20
estimate of the saturation adiabatic lapse rate is 6 Celsius degrees per =

kilometer. </P>
<P>Suppose that you assume that your air parcel had been saturated =
before it was=20
lifted from the surface (1000 mb at 17=BAC). Now follow the curve =
passing through=20
that point that portrays the temperature changes that occur upon a =
saturated air=20
parcel when lifted. If you ascend 1000 meters to 900 mb, notice that the =

saturated parcel temperature was 12=BAC, or 4 degrees war mer than when =
the parcel=20
was dry. The temperature difference results from warming produced by the =
release=20
of latent heat into the parcel as vapor is condensed. Continuing to 3000 =
meters=20
or 700 mb results in a final temperature of 4=BAC. </P>
<P>In situations where a saturated air parcel is forced to sink and no =
cloud=20
droplets evaporate into the parcel, the parcel will warm at the dry =
adiabatic=20
lapse rate because no phase change is involved. In other words, the air =
parcel=20
is "drying out". Sinking causes compression and adiabatic heating, which =
means=20
that the saturation vapor pressure increases, and if the actual moisture =
content=20
(i.e., the vapor pressure) were held fixed, the relative humidity =
decreases.=20
Cases do occur where evaporation of cloud droplets or precipitation =
causes the=20
sinking air parcel to warm at the saturation adiabatic lapse rate =
because the=20
parcel remains saturated until all droplets have evaporated. </P>
<H4>PUTTING IT ALL TOGETHER</H4>
<P>By using radiosonde observations of initial atmospheric conditions of =
air=20
temperature, dewpoint and wind data at the various levels, =
meteorologists can=20
employ the St=FCve diagram (or one of its counterparts) to predict =
conditions in=20
the troposphere, and inside clouds, as weather systems move and =
evolve.</P>
<HR>

<P><I>Return to the <A=20
href=3D"http://www.ametsoc.org/amsedu/online/archive/course/09_fall/f09w0=
6r_sum.html">Thursday=20
Daily Weather Summary </A></I></P>
<P><I>Prepared by Edward J. Hopkins, Ph.D., email <A=20
href=3D"mailto:hopkins@meteor.wisc.edu">hopkins@meteor.wisc.edu</A> =
<BR>=A9=20
Copyright, 2009, The American Meteorological Society. =
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