Transcript Atms 4310 / 7310

Atms 4310 / 7310
Anthony R. Lupo
Test 3 material
Day 1

Then this vapor equation is:

es av=RvT or es = rvRvT


Saturation or Equilibrium Vapor
Pressure (es)
 “es” is a function of temperature only and
not dependent on the pressure of the other
gasses present
Day 1

The concept of equilibrium vapor
pressure over a plane of pure water
(does the atmosphere “hold” water
vapor?):
Day 1

The Variation of es (es over water and es over ice
– or “on the rocks”) with temperature:
Temperature
esw(hPA)
esi (hPa)
esw - esi
-20 C
1.25
1.03
0.22
-10 C
2.86
2.60
0.26
0C
6.11
6.11
0
10 C
12.27
n/a
20 C
23.37
n/a
30 C
42.45
n/a
40 C
73.77
n/a
Day 1

Graph here:
Day 1
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So, some summary points:

1) es is the maximum possible vapor pressure
for a particular temp. (e most often less)

2) es is dependent on temperature only
(highly non-linear – exponential)

3) Note that a 10o C increase (decrease) in
temperature yields a doubling (halving) of es
Day 1
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

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4) The vapor pressure on a water surface equals es
which depends only on water temperature
5) The actual vapor pressure in the air may range
from 0 mb to es hPa  0 < e < es
6) Thus by definition RH = e/es
* es
or e actual = RH
7) Water vapor will diffuse from regions of higher e
values toward lower (evaporation).
Day 1


Now, you can convert mixing ratio
(g/kg) to e (vapor pressure) on a
thermodynamic diagram.
 Follow Temperature line (go straight
up) to roughly 620 – 630 hPa and the
mixing ratio here will roughly equal e in
hPa!!!
Day 1



Changes of phase of water mass and
associated latent heats
** Recall that as we add heat (specific heat),
we raise the temperature until we reach the
melting point!
 then all heat at 0 C (273.15K) goes into
changing the phase
Day 1



 once all the ice changes to water, then we
raise temperature again,
 until vaporization, then there is a phase
change first before we can raise temperature
again (process also works in reverse)
** Recall 1 cal = 4.186 J
Day 1



** Recall 1 cal = 4.186 J
Raise the temperature of ice or steam:
0.5 Cal / kg (2.09 J / kg)
Raise the temperature of water: 1.0 Cal
/ kg
Day 1/2

Liquid to gas phase transformation:

Latent heat of vaporization  condensation
L = 2.500 x 106 J/kg = 5.972 x 105 Cal/kg

Liquid to solid phase transformation:



Latent heat of fusion   melting
Lf = 3.34 x 105 J/kg = 7.98 x 104 Cal /kg
Day 2
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Solid to gas phase transformation:
Latent heat of ablation  
sublimation
La = Lf + L = 2.834 x 106 J/kg = 6.770
x 105 cal/kg
Day 2


The Phase Change Diagram for a
Water Substance
First thing to note, that at Terrestrial
pressures and temperatures water
exists in all three phases (liquid, solid,
and gas) at the same time.
Day 2
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This has tremendous implications for earth’s weather
(clouds, etc) and climate (recall other parts of the
climate system, the oceans, ice sphere (cryosphere))
and the interactions between various components of
the climate system.
(Hand out phase diagram, in 2-D and 3-D).
Phase diagram  Describes the state of a system in
physical space. In our case (specific volume, vapor
pressure, and Temperature)
Day 2
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Water…
Day 2
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Carbon
Day 2
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 Water behaves as an ideal gas so the
isotherms are hyperbolas in the phase plane
(provided we’re far from phase changes).
Consider these important points:
At point A: all water is in the form of vapor, if
we increase the pressure (with temperature
constant), then volume shrinks roughly in
accord w/ ideal gas law (Boyle’s Law).
Day 2
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
Eventually you can reach a point B: where
increases in the pressure will force some
water vapor to liquefy.
If we reach this point, the a small increase in
pressure forces all vapor to condense out,
then as we move from B to C with little
change in pressure and constant temperature
(we change from a gas to liquid with a huge
decrease in volume)
Day 2
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
This constant P is called saturation
vapor pressure for that particular
temperature (That’s the straight lines
across the parabola).
At point C: all sample is liquid which is
for all practical purposes incompressible
(constant a).
Day 2
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1) To the right of B (in A) water mass is all vapor.
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2) B to C vapor and water co-exhist.
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3) To the left of C water mass is all liquid.
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If we follow triple state isotherm  at the triple point
moisture (again fixed temperature and pressure)
condenses out as liquid and solid
Day 2/3
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Equilibrium triple state (6.11 hPa and 0.0o C).
If pressure and temperature fall below triple
state, then ice and gas will equilibrate

Two Exceptions:
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1) Super cooled water
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2) Another special case Follow the point D
isotherm. This is a critical point.
Day 3
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

At the critical point, the distiction between
vapor and liquid disappears (surface tension
is 0) and there is no more interface.
This occurrs at 211 atmospheres (bars) or
21100 Kpa 211,000 hPa, and 374o C.
Defn: Critical point – above this value it is
impossible to liquefy a gas by compression
(above pc), or cooling (Tc).
Day 3

Aha! This explains why N2 O2 CO2 and Ar do
not condense in our atmosphere.
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Their critical points are WAY below terrestrial
T and P’s!
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The CO2 critical point is 31o C and 74000 hPA,
thus condensation of CO2 could take place
(under special conditions) if it were in
sufficient quantities.
Day 3
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CO2 behaves in Venus mid-upper atmosphere and
below like water vapor does here.
The variation of the latent heat of vaporization
(Ll –v) or condensation (L(v-l)) with
temperature.

Let us integrate the First Law of Thermodynamics
during a change of phase from 1 to 2
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(recall ds = dq/T) and p = es
Day 3
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Thus;
dq  du  pda  CvdT  es da

And
2
2
2
2
 dq   Tds   du   e da
s
1

1
1
1
The definition of latent heat for the phase
transformation L 1 to 2 (where 1 = liquid (l)
and 2 = gas (v))
Day 2
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Latent heat
2
L(l  v)   dq
1

Recall that during a change of phase, T
and es are constant!, so;
L1  2   T ( S 2  S1)  U 2  U 1  es a 2  a1
Day 3
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L(1 – 2) is defined as the latent heat of
vaporization.
Thus our equation can be rewritten as:
Ll  v   Uv  Ul  es av  al 
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Density of air is = 1 kg /m3
Day 3
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Then, if the mixing ratio is = 10 g/kg
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rv must be: 0.01 kg/m3
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Density of liquid water is: 1 gm cm-3 or 1000
kg / m3
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Flip each to get specific volume, but look…..
there’s 5 orders of magnitude between the
two.
Day 3
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Thus av is 100,000 times larger than al,
so the equation becomes
Ll  v   Uv  Ul  es av 
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Use the equation of state: es av = Rv T

L(l-v) = RvT + uv – ul
Day 3/4
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This equation is good for any change from liquid to
vapor regardless of initial and final values of internal
energy (exact differential!)
Although changes occur at constant temperature, we
can look at how L will vary with changes in
temperature.
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How?
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dL/dt = RvdT/dt + duv/dt – dul/dt
Day 4
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Recall: Defn of specific heat cl = dul / dT
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and Cvv = duv / DT
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So now we rewrite as:

dL/dt = RvdT/dt + CvvdT/dt – CldT/dt
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-or-
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(by chain rule)
dL/dT = Rv + Cvv – Cl
Day 4
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and, of course, Rv + Cvv = Cpv
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So DL/DT = Cp – Cl
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= -2369 J/kg K
(aha! – slope of L versus Temperature,
so Latent heat is temp dependent!)
Day 4
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We can then integrate above expression
from T = To = 0 degrees C, and L(l – v)
= Lo (at T=0) to an arbitrary
temperature T:
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After applying the snake:
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L (l – v) = Lo + (Cp – Cl) (T – To)
Day 4
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The product term on the RHS is small for T – To less than 40 C,
thus L(l – v) is approximately Lo for typical weather situations,
and is taken to be 2.5 x 106 J/kg
Thus, L is not strictly a constant, and in Latent heat release
(cloud and precipitation) schemes (advanced ones) this fact is
taken into account.
For water:
Cpv = 1811 J/kg
Cv = 1350 J/kg
Rv = 451 J/kg
Cl = 4186 J/kg
Day 4

Over typical ranges of T here value of L
varies 6%
Temperature (C)
L(l-v) x 106 J / kg
30 C
2.425
20 C
2.45
10 C
2.475
0
2.5 = Lo
-10 C
2.525
-20 C
2.55
-30 C
2.575
Day 4
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So, in the range of 20 to –20 C there is only 2% error
in using Lo, thus to within 98% accuracy L = Lo. This
is good enough to win $50.00 at the bar this
weekend.
The variation of es with temperature (the
Clausius Clapeyron equation)
If as before L(l – v) is the latent heat associated with
a change in phase:
Dq = Tds = du + es da
Day 4
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Since during a phase change T and es
are constant then;
2
2
2
2
1
1
1
1
 Dq  T  ds   du  es  da

where;
2
 Dq  L(1  2)
1
Day 4
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then,

L(1-2) = T(S2 – S1) = (u2 – u1) + es(a2 – a1)
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or rearrange the above to isolate each state,
state 2 and state 1
TS2 – u2 – es a2 = TS1 – u1 – es a1
Day 4
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During the phase change:
T S – u – es a = Constant
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-
or –
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u + es a - T S = Constant = G
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(J. Willard Gibbs potential)
Day 4
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Gibbs function this is a “fundamental”
Thermodynamic (Gibbs) function for simple
compressible systems (such as an air parcel) of fixed
chemical composition, and using the concept of an
exact differential.
A thermodynamic function provides a complete
description of the thermodynamic state of a system.
In principle, all properties of interest (v,T,P) can be
determined from the function given a suitable set of
boundary or initial conditions.
Day 4
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In plain English: the phase diagram
represents all possible states of system.
G is constant during a phase change (T
and es constant) it has values in accord
with T and es depending on the T and
es at which the phase change takes
place. G(T,es) (recall diagrams?)
Day 4/5
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So, let us look at the variation of G;
Take the derivative with respect to time
(remember to use product rule!)
DG/Dt = du/dt + es da/dt + a des/dt –
T dS/dt – S dT/dt
Day 5
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Rearrange:
dG/dt = du/dt + es da/dt – T dS/dt + a
des/dt – S dT/dt
On the RHS, the first three terms are
the 1st Law of Thermodynamics!
Day 5
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If

Then dG = a des/dt – S dT/dt
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T dS/dt = du/dt + es da/dt
And if G = a constant during phase change at
T and es;
And if G + dg is constant at a phase change
for T + dT, es + des
Day 5
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Then dG must also be a constant!
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If dG is constant;
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Then;
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a2 des/dt – S2 dT/dt = a1 des/dt – S1 dT/dt
Day 5
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so;
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des/dt (a2 – a1) = dT(S2 – S1)
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Or
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des/dT = (S2 – S1) / (a2 – a1)
Day 5
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But,
2
dq 1
L(1  2)
  dq 
T T1
T
1
2
S 2  S1   ds  
1
2

So:
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This is it! (Make no mistake)!
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Generalized Clausius Clapeyron equation for any
phase change from phase 1 to 2.
des
L(1  2)

dT T a1  a 2
Day 5
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Saturation Vapor Pressure over Water
Conditions governing saturation or
equilibrium vapor pressure over a plane of
pure water surface (es), which involves the
phase changes from liquid to vapor
(evaporation) = phase changes from vapor to
liquid (condensation)
Evaporation = condensation
Day 5
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In the case of the CC equation:
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L(1-2) is L (l-v)
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a2 = av
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es = esw
and a1 = al
Day 5
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We’ll invoke the same argument we
used before, mainly;
av >>> al
So the Clausius Clapeyron Equation
becomes:
desw L(l  v)

dT
T av 
Day 5
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From our equation of state;

esw av = Rv T

Substitute into our CC:
or
av = Rv T / esw
1 desw L(l  v)
 2
esw dT T Rv 
Day 5


Or
desw L(l  v) dT

2
esw
Rv T
Invoke the snake yet again! Integrate
from initial reference values, T = To,
where esw = eso to arbitary final values
of T and esw (holding Ll – v constant).
Day 5
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Integrate and put in the limits:
then
 esw 
L(l  v)  1 1 
ln   
  
Rv  T To 
 eso 
 L  1 1 
 Lmv  1 1  
esw  eso exp       eso exp 
  
 Rv  To T  
 R *  To T  
Day 5
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This is the saturation-equilibrium water
vapor curve over water or the Clausius
Clapeyron equation as used in
Atmospheric Science!
We can write this such that; eso = 6.11
hPa and To = 273.15
Day 5
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Saturation Vapor Pressure over Ice
(or “on the rocks”?)

In this case;

L(1 – 2) becomes L(v – i), thus a2 = av and a1 = ai

and av >>> ai
Day 5
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Thus with the same derivation:
 La  1 1  
 Lamv  1 1  
esi  eso exp       eso exp 
  
 Rv  To T  
 R *  To T  

This is the sublimation curve –
saturation or equilibrium vapor
pressure over ICE (on the rocks)!
Day 5 / 6
The phase diagram (my own), see the tail.
10
10
8
6
e
i
ei
i
4
2
0
0
200
243.15
250
300
T
i
350
400
393.15
Day 5/6
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Notes:
1. This Equation is essentially identical to
the previous version of the ClausiusClapeyron equation except that La > L,
(correct?) and both can be integrated
numerically.
2. Recall from Ideal Gas Law that: Rg =
R*/mg or Rv = R*/mv
Day 6

Thus in both versions of CC we can
replace Rv = R*/mv (Hess, and most
other texts show this way) (Also, any
sort of cloud modelling scheme
(Convective schemes, ie., Kuo, Arakawa,
Grell, Cain-Fritch) uses CC relationship
in the form above:
Day 6
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The Equilibrium curve for ice and water
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Here,
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al = aice
a2 = al = 1.00 x 10-3 m3/kg
a1 = ai = 1.09 x 10-3 m3/kg
Day 6
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so (a2 – a1) = (1.0 – 1.09) = -9 x10-5 m3/kg
and this value is roughly constant! (Water
and ice for all intents and purposes are
incompressible substances)
Then the equation becomes:
des
L(i  l )

dT
T al  ai 
Day 6

and that becomes:
des
L(i  l ) dT

5
dt
T 9 x10  dt

and let’s get the snake involved
(integrating from es = eso to es, as T
goes from To to T):
L(i  l )  T 
es  eso 
ln 
5
9 x10
 To 
Day 6
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This is the melting curve!
L(i  l )  T 
es  eso 
ln 
5
9 x10
 To 
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Draw e vs. T (2-D) phase curve (we’re
only looking at a portion of the 3-D
phase diagram).
Day 6
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(we’ll add the melting curve -----)
Note: the melting curve is nearly vertical or
ice melts at T = To for all e!! (Ice and water
exhist in equilibrium, or ice freezes and melts
at constant and equal rates)
 Ice can be melted by increasing the
pressure at constant T, this is how an
iceskater glides over the ice.
Day 6
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Let’s examine the behavior of the curve;

Scenario 1: If T = To,

 then ln (1) = 0 and term drops out.
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Scenario 2: If T < To,

 e goes rapidly to “high” values, since ln (T/To) <
0
Day 6
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Scenario 3: T > To (mathematically Ok,
but physically?)
Q: What’s wrong with this picture?
A: 1) no ice (lnT/To) > 0; and e < 0
cannot happen (Thus, phyically
unrealistic solution!)
Day 6
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A few more points:
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1) We can identify the Triple point.

2) melting curve: to the right of melting
curve only water, to the left only ice.
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3) evaporation curve ends at Critical point
(again distinction between liquid and vapor
disappears)
Day 6
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4) Water boils at 1000 hPa or sfc pressure at T where
vapor pressure = total atmospheric pressure (T =
212 F or 100 C)
5) Thus, as we go up in height (p decreases), thus
vapor pressure = p atmosphere at a lower T.
 if we could evacuate a container, we can boil a
glass of water at room temperature.
 your blood can boil at sufficently low Pressure.
Day 6
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Supercooled water (the exception):
It is a well-known fact that when water is
cooled below 273 K it often does not freeze!
(In fact this is the rule not the exception).
Liquid water below 0 C is called supercooled
water (has the same saturation vapor
pressure as liquid water on es curve!!!!
(Higher than ice since L sub. > Lvapor.).
Day 6
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at –14C: 20% of clouds are water droplets only
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at –8C: 50% of clouds are water droplet only.
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at –40C: pure water will freeze instantaneously 0%
of clouds contain water vapor at all. This is called
spontaneous nucleation.
So, above –40 C water droplets may freeze if they
contact foreign particle (like airplane  icing a major
hazard).
Day 6
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Thus, if you introduce an ice crystal into a cloud of
water vapor, water vapor condenses on the ice (thus
snow!). This is the Bergeron – Findeisen mechanism.
Super cooling process occurs because of the water
droplet may take on a crystal lattice structure similar
to that of ice. These things can grow, but if agitated,
the particle or droplet quickly freezes.
Also if the droplet reaches a critical size, collisions
with foreign bodies become more likely leading to
freezing.
Day 6
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Interpretation and implications of the Es and Esi Curves:
Since the vapor pressure on the sfc of a plane pure water sfc (or any pure droplet larger than 20 microns (vapor is 10 and rain is
100 typically)) equals es, and by definition RH = e / es, and since the flux or diffusion of vapor will be in the direction from
higher to lower vapor pressure, and proportional to the vapor pressure gradient, we can say the following:
Sat. w/r/t a surface of water at T represents supersaturation for cooler T’s and undersaturation w/r/t higher T’s
Evaporation from a warmer water sfc will be significantly greater than from a cooler water surface when both are exposed to the
same atms. Environment! (Since es is higher)
If air is saturated (eair = es(Tair)) and the temp of the water is warmer that T air, evaporation will occur from the water surface
followed by condensation in the air (Steam fog). If on the other hand the saturated air is in contact with water sfc. with a temp.
less that T air, condensation will occur on the cool water sfc.
If air has a vapor pressure = e(air) and within that air there exists vapor at different temperatures, e.g. T1 and T2 where T1 <
T2 and
es(T1) < e air < es(T2)
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the warmer drops will evaporate while the cooler drops will grow due to condensation on
their surfaces
Air with an RH < 100 % [ e.g. e air < es air] is undersaturated w/r/t droplets at temp. = Tair but may be supersaturated w/r/t
drops at temp < T air. Thus equilibrium can exist between cool drops and unsaturated air. (This explains why RH can be less
than 100% during a continuous rain) (Also recall, esw is equilibrium saturation w/r/t water!!).
Thus cold drop can fall though warm unsaturated air and experience negligible evaporation.
Air which is initially unsaturated (e air < es(Tair)) may be brought to a state of saturation by simply lowering it’s temp. untill it’s
actual vapor pressure equals the saturation vapor pressure corresponding to it’s lowered temp.
Saturated air at an initial temp Ti (e air = es(Ti)) will become unsaturated if the air temp is increased.
For two water surfaces at different temperatures but exposed to the same unsaturated air, the evaporation from the warmer
water surface will be much more rapid than from the coll water surface. T2 > T1 and thus es(T2) >> Es(T1) so [es(T2) – e air]
>> [es(T1) – esair]
Day 6
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Saturation W/r/t to water at any temp T represents supersaturation w/r/t and ice surface at the same
temp. (for supercooled water).
If we have the co-existence of supercooled water droplets and ice crystals at the same temperature
the vapor pressure on the water drop will be greater than on the ice crystal surface (esw(T) > esi(T)).
If these two surfaces are simultaneously immersed in air that is slightly undersaturated such that esw
> eair > esi evaporation will occur from the water droplets and deposition on the ice crystals.
(Bergeron-Findeisen-Wegner precip formation process).
Not only does the maximum water vapor capacity of air essentially double with each 10 C increase in
temp, but the actual water vapor in the air (e air) doubles with each 10 C increase in dew point.
An increase of dew point at high temperatures is associated with a much larger increase of actual
water vapor in the air than that associated with the same increase in dewpoint at at low temps. E.g.
inc. dewpoint 5C
If Td inc. from 15 to 20 C, this corresponds to e air inc. from 17.0 to 23.4 hPa
If Td inc. from 0 C to 5 C this corresponds to e air inc from 6.11 to 8.71 hPa
A melting snow and ice surface will be at 0 C and the vapor pressure on that surface will
Be eso = 6.11 hPa. Warmer air passing over those surfaces with a dewpoint greater than
0 C will have an actual vapor pressure greater than 6.11 hPa. As a consequence vapor will
move from the unsaturated air and condense on the cooler – melting ice surfaces with an
associated release of the latent heat of condensation. There will be no evaporation and both
the latent heat flux and the sensible heat flux will be directed to the melting surface, thus
contributing to enhanced melting.
Day 6




Review and tidy up (Expressions of
Water Vapor content of air)
Vapor Pressure  partial pressure due to H2O
vapor
e (hPa)
es, esi
Mixing ratio  mass of water vapor to the
unit mass of dry air
Day 6





Mixing ratio  mass of water vapor to the
unit mass of dry air
w = r = m = (mv / md) (kg / kg -or - g /
kg)
also ws, rs, ms
Specific humidity mass of water vapor to
the mass of dry air + vapour
q = ( mv/ md + mv)
also qs
Day 6




Relative humidity  The potential of the
atmosphere to “hold” water vapor.
RH% = (e/es) x 100% or m/ms = q/qs
Absolute humidity  the amount of water
mass in the air.
rv = (density of vapor  mass / Volume) (kg
/ m3)
Day 6



Defn: Dew point temperature (Td) = the temperature
to which air must be cooled at constant pressure and
constant water vapor content to bring it to a
saturated state. (e air = es(Td))
Another Defn: Wet Bulb Temperature (Tw) = the
temperature to which air may be cooled isobariacally
by evaporating water into it until it is saturated. The
air provides the L (evap) as the vapor pressure
increases and T decreases.
Td <= Tw <= T
Day 6

Defn: The LCL = the pressure to which an
unsaturated air parcel must be lifted in an
unsaturated or dry adiabatic process in order to bring
it to saturation.
Dry adiabatic  G = dT/dt = -g / Cp



Defn: Virtual Temperature (Tv) = the temperature of
dry air at the same pressure and density as that of
moist air.
Tv > T.
Day 6

Defn:
Eqivalent potential temperature (qe) = The
temperature a sample of air would have if all off its vapor
content were to be condensed to liquid during and isobaric
process and the release of latent heat were added to the air
temperature, or LHR  Internal energy

Relationships between moisture variables (water vapor)

Mixing ratio versus vapor pressure:

Definition – Mixing ratio

m = Mv/Md
Day 6

Equation of state for water vapor:

e = rv (R*/mov) Tv
where

Equation of state for dry air:

Pd = rd (R* / mod) Td

where

rv = Massv /Volv
rd = Massd / Vd
Day 6

And;

Volv = Vold  Tv = Td

Forming the ratio:
R*
rv
T
e
mov

R*
P
rd
T
mod
Day 6

Density is mass per unit volume, so the
densities become mixing ratio (cancel out the
Volume).

R* and T cancel and mov (18.016) / mod
(28.97) = 0.622

So,

e/Pd = m / 0.622
Day 6


But,
Pd = P – e

So,

m = 0.622e / (P-e)

ms = 0.622es/ (P – es)
Day 6


But,
e <<< P

10 hPa <<< 1000 hPa

So near surface:

m = 0.622e / P or ms = 0.622 es / P
Day 6

Specific humidity (q) versus vapor
pressure (e)
mv
q
mv  md

Divide by Volume
mv
rv
mv
q
 Vol

mv  md mv  md r v  r d
Vol Vol
Day 6

From eqn of state for vapor:

e = rv (R*/mov) T or rv = (mov/R*)(e/T)

For dry air:

P = rd (R*/mod) T or rd = (mod/R*)(e/T)
Day 6

So,
mov
mov e
e
*
0.622e
mod
R
T
q


mod p mov e
mov
pd  0.622e

pd

e
R* T R* T
mod
Day 6

But Pd = P – e
0.622e
0.622e
q
 qs 
pt  0.378e
pt  0.378e

Since e and es << Pt
0.622e
q
 mixingratio
p
 or 
e
m q
RH %  
  e  RH %  es
es ms qs
Day 6


How does R vary with moisture content?
Again, moisture throws a wrench in things, like the
ideal gas law (Rd), and the latent heats vary w/temp.
See how complicated moisture makes wx and
modelling.

Consider ideal gas law for moist air:

pm = rm Rm T
Day 6

where rm = (Massm/V)

Recall Dalton’s law:

pm = ptot = S Pi

where i = gas 1,2,3, ….
Day 6

Rewrite ideal gas law above:

pm = (Massm/V) Rm T = S(Massi/V) Ri T

where i = gas 1,2,3,….

Solving for Rm:

Rm = S(Mi/Mm) Ri
Day 6

Separate gasses into dry gasses + moisture:

Rm = (Md/Mm) Rd + (Mv/Mm) Rv

But,

So,

Rm = ((Mm – Mv)/ Mm) Rd + (Mv/Mm) Rv
M d = Mm – Mv
Day 6

Then in first RHS term:

Rm = (1 – (Mv/Mm)) Rd + (Mv/Mm) Rv

But by definition (Mv/Mm) = q (specific humidity)

So,

Rm = (1 – q) Rd + q Rv
Day 6

Express Rv using “strategic 1” or Rd / Rd:

Rv = Rd (Rv/Rd)

and (Rv/Rd) = 461.5/ 287.04 = 1.61

So,

Rm = (1 – q) Rd + 1.61 q Rd
Day 6

Finally,

Rm = (1 + 0.61 q) Rd


Then we can insert this into the Ideal Gas
law:
Pm = rm (1 + 0.61q) Rd T
Day 6





Rearrange (and this is a short cut to getting Tv)
Pm = rm Rd (1 + 0.61q) T
or
Pm = rm Rd Tv (since q approx. m)
The “modeler’s special” – avoids variations in R.
Day 6

The Effects of water Vapor on the
Specific Heats of moist air!!!

Consider:


 Cp and Cv are the specific heats for dry air
at constant pressure and constant volume:
Cp = 1004.63 J kg-1K-1 and
kg-1K-1
Cv = 717.59 J
Day 6




Q: What is Cpm or Cvm?
Consider 1 kg of moist air which contains X
kg of water vapor, so lets consider during an
isobaric process:
Dhm/Dt = (1-q) Dhd/Dt + q Dhv/Dt (1)
well dh/dT= Constant
Day 6

Q: What is this (dh/dT)?

A: The definition of specific heat!

So, let’s deeevide equation (1) by dT.

dhm/dT = (1-q) dhd/dT + q dhv/dT
Day 6




then, using our definition of Cp:
Cpm = (1- q) Cpd + q Cpv
Well let’s yank out Cpd:
Cpm = Cpd (1 – q + (Cpv/Cpd) q)
Day 6



where,
Cpv = 1820 J / K kg
Cpd = 1004.63 J/ K/ kg

So….

Cpm = Cpd(1 –q + 1.81 q)

and combining “q” terms;

Cpm = Cpd(1 + 0.81q)
Cpv / Cpd = 1.81
Day 6

Well, we now have;

Rm = Rd(1+0.61q)

and we know that:

Cpm = Cvm + Rm
Day 6

-or-

Cvm = Cpm – Rm

Substituting our relationships for Cpm and Rm:

Cvm = Cpd (1+0.81q) – Rd(1+0.61q)
Day 6

A bit ‘o algebra:

Cvm = (Cpd – Rd) + q(0.81 Cpd – 0.61Rd)

And a bit more…..



Cvm = Cvd + qCvd(0.81(Cpd/Cvd) – 0.61(Rd/Cvd))
At last!!!!
Cvm = Cvd ( 1 + 0.89q)
Day 6




There is always more than one way to skin a
cat!
Another way (start back with heat added):
Dhm = (1-q) dhd + qdhv
Assume constant volume process C = dh/dT
Day 6

Cvm = (1 – q) Cvd + q Cvv

Then a bit o’ algebra:

Cvm = Cvd (1 – q + (Cvv/Cvd)q)

And recall these constants:

Cvv / Cvd = 1350 (J kg-1 K-1) / 717 (J kg-1 K-1) = 1.89
Day 6

“and it can easily be shown that….”

Cvm = Cvd(1 + 0.89q)

as before!!
Buuut, this is also an approximate form
since unlike Cpv, Cvv varies with
temperature!
Day 6

In Summary:

Cpm = Cpd(1 + 0.81q)
Rm = Rd(1 + 0.61q)
Cvm = Cvd(1+ 0.89 q)

Within 98% accuracy,


Cpm  Cpd ; Rm  Rd ; Cvm  Cvd

Ok, Let’s move on!
Day 7



The saturated Adiabatic or
Pseudoadiabatic Process
(What we’re really after is the moist adiabatic
lapse rate)
Consider a sample of saturated air containing
ONE KILOGRAM of dry air and “ms” grams of
moisture.
Day 7

 If the temp of air fall slightly (dT/dt < 0),

then “ms” must also decrease dms/dt < 0

 The decrease in saturated mixing ratio will
produce condensation, and be accompanied
by the release of latent heat of condensation
to the amount of d(L ms) < 0 where L is the
latent heat (our parameterization).
Day 7


The 1st law of thermodynamics for
saturated air:
dq
dT
dp
 Cpm
 am
dt
dt
dt
Recall: LHR is a process by which we
can add internal energy and pressure
work
Day 7

also we must consider:

we had 1 kg + ms of air or (1 + ms),

but we want to work with unit mass, so
the 1st Law becomes:
d  Lms 
dT
dp

 1  ms Cpm
 am
dt
dt
dt
Day 7

use our approximations:

1) L = Lo
2) Cpm = Cpd
3) 1+ms = 1

and the equation becomes:


Day 7
•and the equation becomes:
d ms 
dT
dp
 Lo
 Cp
 am
dt
dt
dt
•Now, pressure and height again related via hydrosta
dp
dz
  rm g
dt
dt
Day 7

Then, the 1st Law becomes:
d ms 
dT
dz
 Lo
dt
 Cp
dt
g
 or 
dms
dT
 Lo
 Cp
g
dz
dz

but by “chain” rule:
dt
Day 7

Ok seriously,
dms dms dT

dz
dT dz
substitute
dms dT
dT
 Lo
 Cp
g
dT dz
dz
Day 7

solve algebraically for dT/dz
dT  Lo dms 
g
1 

dz  Cp dT 
Cp

**Now we’ve derived a “moist adiabatic
lapse rate” which answers the question
“how is moisture modifying the dry
adiabatic lapse rate?”.
Day 7



Viola!
Gd
dT

dz
 Lo dm s 
1 

 Cp dT 
Then, as you showed in Lab #3 back in
September, the second term in the
denominator is a pure number > 0!!
 Then, the denominator is > 1.
Day 7



So the saturated adiabatic lapse rate is
always less than the dry adiabatic lapse rate
(-g/Cp) or –9.8 C / km.
Thus, G(m) will approach Gd when dms/DT is
small!!!
 Let math talk to you, and remember (T,e)
phase graph.
Day 7



Q: Where does this occur in our
Atmosphere???
A: At low temperatures, typically in the
upper troposphere.
Then, at high temperatures, dms/dT is large,
then G(m) is significantly less than G(d).
Day 7





Recall, a Stuve:
 At p = 1000 hPa and T = 20 C
 G(m) ~ 4.3 C/km
 at p = 500 hPa and T = -20 C
 G(m) ~ 7.8 C/km
Day 7



 at p = 100 hPa and T = -60 C
 G(m) ~ G(d)
** Recall: when a parcel is dry, we
move dry adiabatically, but when we
reach the LCL and above; parcels will
ascend moist adiabatically!
Day 7



Two extreme scenarios (and four points):
1) A reversible saturated process  cloud droplets
(parcel content changes, but liquid water stays
available for re-evaporation during decent and
warming)
2) An irreversible saturated process  during ascent
saturation reached and vapor precipitates out (earlier
convective parameterizations – e.g., Kuo scheme,
assumed this), thus the air parcel changes its content
permanently.
Day 7


3) Reality “bites”, it’s somewhere in
between at any one time.
4) We model m < = ms or e < = es, but in
reality e(m) can exceed es(ms). This is super
saturation. We would need to look at cloud
microphysics to form relastic models of this,
but this is beyond the course scope right
now and we won’t discuss it (take 4510 or
4550).
Day 7/8




Mixing of Air layers by Turbulent - Convective
Mixing
 Let’s look at the development of a layer with
uniform conservative properties after a thorough
mixing!
dq/dp = 0
This will happen in an afternoon PBL for example.
Day 8




The following factors and processes can
influence the vertical distribution of
temperature in a layer (stability)
Temperature
1) The type of air mass and it’s temperatures, which
are then dependent on the synoptic and large scale
patterns
2) Warm or cold air advection as a function of height
Day 8

3) The PBL and/or the thermal
characteristics of the surface

4) Variation of net radiation with height.

5) PBL sensible heating

6) Latent heating (PBL)
Day 8



Specific humidity (moisture):
1) Specific humidity of the air mass
2) Horizontal advection of specific
humidity
Day 8

3) Type of surface and it’s temp.

4) Rate of evaporation

CASE I: Unsaturated air remaining
unsaturated (and No heating)

1) mixing produces a uniform theta,
through layer equalling q average or the
original layer. (again, dq/dp = 0)
Day 8



2) thorough mixing also produces a uniform
mixing ratio value equal to the m acts to
average of the original layer
(analogous to parcel theory) except for a
layer
Step 1: Air displaced upward and downward
at constant q.
Day 8



Step 2: Now lift air and mix thoroughly
in horizontal.
Q: Why does air in a well-mixed layer
have a constant value of theta?
A: Consider the mixing to take place in
an idealized way
Day 8

1) parcels from all levels within the
layer are randomly rearranged in the
vertical (through bulk or convective
mixing, turbulent transport), i.e., at
any level within the mixed layer we
find parcels from all other levels within
the layer. This defines the temperature
range at each level within our layer.
Day 8

2) next at each level, the air is
homogenized (mixed) such that the original
parcels are no longer identifiable and air at
each level has a new temperature and
mixing ratio. This results in a temperature
sounding somewhere between the limits
(qtop and qbottom). Note cooling at the top
of the layer and warming at the bottom.
Day 8


3) Repeat the above process after defining
“new” parcels throughout the layer. We get
theta that are closer together now than
before. This cycle will continue until all the air
in the layer has one value of theta and mixing
ratio
The actual mixing process occurs in such a
fashion except parcels are mixed as they are
rearranged.
Day 8


(Initial mixing ratio and final mixing ratio)
Case II: Initially an unsaturated layer
becomes saturated in the in the upper
portions after mixing: the mixing
condensation level.
Day 8

If thorough mixing to q = avg. q and m
= avg. m produces a condition aloft
where Td > T, condensation and cloud
formation will occur in the
supersaturated layer above the level
where T = Td (which is called the
mixing condensation level) 
Stratiform cloud formation.
Day 8



Hydrostatics and the hydrostatic
approximation
Hydrostatics: A physical derivation (no equations).
 The hydrostatic equation in a resting atmosphere
(consider an atmospheric or any fluid parcel of unit
volume in static equilibrium with respect to the
vertical forces acting on it (i.e., at rest  no
horizontal forces!))
Day 8



S Fz = 0
Consider a parcel of volume (DV =
DxDyDz) and of total mass = M.
Our convention: Let’s consider forces +
if they act in the + Z direction
Day 8

Force diagram
PGF

Gravity

Day 8

0 = S Fz = Force due to gravity + Force due
to pressure gradient (across a Vol.)

0 = S Fz = -Mg + pl(DxDy) – pu(DxDy)

- but -

pu = pl + (Dp/Dz) Dz
Day 8

0 = SFz = -Mg + pl(DxDy) – (pl(DxDy) + (Dp/Dz)
DxDyDz)

deevide by Mass….

0 = - g –[(Dp/Dz) (DV / Mass)]

Q: What is Volume / Mass, especially a unit Mass?

A: Specific Volume (a)
Day 8

Thus, the limit of the above equation as
delta  0 is:
p
1 p
p
0  g a
 g 
   rg
z
r z
z

Question: Does;
p dp

z dz
!!
Day 8

Many in our field write the hydrostatic
balance relationship making this assumption.

But,

Is it true?

Given: p = p(x,y,z,t), show the above
condition!
Day 8

Well,
dp p 
  V3   3 p
dt t
 becomes 
dp p dt u p v p p




dz t dz w x w y z
Day 8



We’ve assumed a resting atmosphere
so….
u = v = w = 0 and,
pressure does not change in time,
and
Day 8

there are no pressure gradients, so……

dp p dt u p v p p




dz t dz w x w y z
then:
p dp

z dz
Day 8



The local change in pressure with height
equals the total change in pressure with
height.
Thus,
p dp

  rg
z dz
Note the change in notation! This is
hydrostatic balance in an atmosphere at rest!
In such an atmosphere, p = p(z) only!
Day 8

The atmosphere in motion and
hydrostatic balance.

 This time we’ll use Newton’s 2nd law!

Which is;

SF = ma,
Day 8

where,

a = dV/dt or the (3-D) velocity vector.

Recall;

V  uiˆ  vˆj  wkˆ
 and 

dV  du ˆ dv ˆ dw ˆ
a i
j
k
dt
dt
dt
dt
Day 8/9

 Then, the acceleration equals the sum of the forces per unit
mass:

Q: And these forces are?

A: pressure gradient force, coriolis force, gravity, friction, etc.




1)
2)
3)
4)
gravitational force – directed downward
vertical component of PGF – upward
vertical component of coriolis force
vertical component of viscous and Frictional forces
Day 9



**Well, 1) and 2) we know from our
previous look at hydrostatics for the
atmosphere at rest.
So write the equation of motion (kcomponent):
dw
1 p
 g 
 2 cos( )u  Fric  Visc
dt
r z
(3)
Day 9



Let’s use scale analysis to justify our
final result!
Acceleration term:
dw/dt
dw dz
dw d  w2  1 Dw2
w
  
dz dt
dz dz  2  2 Dz
Day 9

if z = 0 and w = 0 at the surface,

and z is roughly 5 km up (5 x 103 m)

(w52 – 0) / (2 x 5 x 103)  10-4 w52 m s-2

Gravity term

C’mon folks!  10 m s-2
Day 9

Pressure gradient term:

r = 1 kg m-3



Dp/Dz  100 hPa/1 km = 104 Pa/103 m = 10 kg m-2
s-2
Then the term is on order of:
10 m s-2
Day 9

Coriolis Term:

u = 10 m/s



at 45o N or S
2cos = 2sin and is approximately 10-4 s-1
10-3 m s-1
Day 9

Terms versus scale for each scale of
motion:
Horizontal
Scale
W
(m/s)
dw/dt
g
PGF
Coriolis
Error(%)
planetary
0.01
10-8
10
10
10-3
0.01
synoptic
0.1
10-6
10
10
10-3
0.01
mesoscale
1
10-4
10
10
10-3
0.01
microscale
10
10-2
10
10
10-3
0.1
Day 9

Where % error is:
neglected terms
-------------------- x 100%
retained terms
Day 9

Thus we can show that Equation (3)
can be approximated by:
1 p
0  g 
r z

so with a very small error: (0.01% for
synoptic scale features).
Day 9

So, hydrostatic approximation (Newton’s
2nd law – vertical component) for an
atmosphere in motion is excellent
approximation similar to atmosphere
not in motion.
p
  rg
z
Day 9




A closer look at scale analysis
(Prove that for an atmosphere in motion,
hydrostatic balance is a relationship).
We can split any variable into it’s mean field
plus a perturbation:
Q = Qo + Q’(x,y,z,t)
Day 9

Where;
t
 Qdt  0
to



So;
P = Po + p’(x,y,z,t)
r = ro + r’(x,y,z,t)
Day 9


It has been shown (Dr. N. Phillips, 196?) that the
variable parts of P and r are very small compared to
the mean parts (This should make sense to us).
Then;

P’ << Po
r’ << ro

and for a resting atmosphere: p’ = r’ =0

Day 9/10

 Thus, ro and Po would be in exact
hydrostatic balance since all other terms in 3rd
equation of motion go to zero in a resting
atmosphere, thus the 3rd equation reduces to:
dpo
  ro g
dz

 Then, we have defined Po and ro such that
the hydrostatic relation is satisfied, given
Po(z) gives ro(z), and vice-versa.
Day 10

 So, if we are going to replace the
vertical component of Newton’s 2nd
Law by the hydrostatic approximation,
1 p

g
r z
Day 10

then we must really show that dw/dt,
as well as the other neglected terms are
much smaller than the perturbation or
variable part of the pressure gradient
force term:, ie we must show that:
dw
1 p
 
dt
r z
Day 10

Well we know;
1 p
1 po 1 p
g 


r z
r z r z

so this is where we need to go;
1 p
g
r z
Day 10


 Thus the straightforward scale
analysis we did was perhaps a bit
misleading.
 Let us rewrite the vertical component
of the equation of motion.
dw
1 p
 g 
 2 cos u  Fric  Visc
dt
r z
Day 10


Let’s rewrite each variable in
perturbation form, starting with density
and then assume that r’/ro << 1
 Which gives for density:
1
1
1  r 

 1   
r ro  r  ro  ro 
 1
1  r  r  2 r 3
1  

 ...  
ro  ro ro ro
 ro
1
Day 10

However, we’ll only neglect second
order and higher terms in order to get:
1  r 
1
1   
r o  r o   r o  r 

Now hydrostatic balance:
1 p
1


 g  
 po  p  g
r z
 ro  r  z
Day 10

Becomes:
1  r   po p 
 1  

 g
r o  r o  z z 
By “Foil”
1 po 1 p r  po r  p


 2
 2
g 0
r o z r o z r o z r o z
Term 1
Term 2 Term 3 Term 4
Day 10


 Term 1 and Gravity cancel (recall
hydrostatics in a resting atmosphere!?)
 But Term 4 has two “primed” terms
(thus smaller than terms with one
“prime)” is neglected since it is small
compared to the two first order terms
(single primed).
Day 10


 This is consistent with truncating the
expansion for: (1 + r’/ro)-1. Thus we have
left terms 2 and 3:
1 p r  po

 2
0
r o z r o z
So we use this expression and replace into
the second RHS term (formerly known as
Term 3) – g from hydrostatic balance in a
resting atmosphere (again!):
Day 10



Viola!
1 p r 

 g 0
ro z ro
 This is it. Hydrostatic balance! We have
perturbation form (perturbation pressure gradient
and reduced gravity [buoyancy] term)!
So, in the equation of motion we can redo the scale
analysis. For synoptic-scale motions, these terms
have magnitudes of only 0.1 m s-2 not 10 m s-2 as
we did in our table a bit ago.
Day 10

So we get :
dw
r
1 p
 g
 2 cos u  Fric  Visc
dt
ro
ro z


10-6 10-1 10-1
10-3
10-6
 So even for the perturbation or variable
parts of the pressure and density fields we
still have, to an error of 1% for synoptic–
scale motions!
Day 10

However, for storm scale (cb’s, strong storms,
and cu) we have an error of roughly 10%.
For the synoptic –scale we have shown the
hydrostatic approximation to be valid even for
a more realistic analysis.

So, we have justified, that:

Even for an atmosphere in motion.
1 p

g 0
r z
Day 10



But, can we justify using:
p dp

z dz
?
as we did for a resting atmosphere, as
also being valid for an atmosphere in
motion?
Day 10


In other words, how much error is
involved replacing :
p
z
With?
dp
dz
Day 10


Start with p = p(x,y,z,t) again!
thus:
dp p 
  V3   3 p
dt t
 becomes 
dp p dt u p v p p




dz t dz w x w y z
Day 10





 so, maybe the total derivative and partial
derivative forms are not equal in this case!
 buuut, can we show that they are approximately
equal? Well, lessee…..(scale analysis)
on the synoptic scale:
u = v = 10 m/s
w = 0.1 m/s
and u / w = v / w = 100
Day 10


horizontal pressure gradients:
1 hPa / 100 km = 10-5 hPa / m

vertical pressure gradients:

100 hPa / 1 km = 0.1 hPa / m

and pressure changes with time:

1 hPa / 3 hr ~ 10-4 hPa / s
Day 10

So…

10-1

dp p dt u p v p p




dz t dz w x w y z
10-3
10-3
10-3
10-1
 Then, the smaller terms (neglected) are
only 1% that of the larger terms (retained).
So, to 1% error on the synoptic scale:
Day 10
p dp

z dz


on the mesoscale, it’s about 5%.
So, to 1% for the atmosphere in
motion:
p dp

  rg
z dz
Day 10

This is the hydrostatic relationship. In
the real atmosphere, which is in motion,
this is an approximation. A darn good
one, but an approximation
nonetheless!
Day 10/11





Implications and conclusions of using scale analysis for synoptic-scale motions
Meteorologically significant vertical accelerations and vertical motions result from a very
small (almost infinitesimal) difference between two vertically opposing forces (PGF z and
gravity)
To a high degree of accuracy, the upward directed pressure gradient force equals the
downward directed gravitational force – the hydrostatic approximation.
To a high degree of accuracy, the actual pressure variations with height in the real
atmosphere equal those required for hydrostatic equilibrium. Thus the actual pressure at
any level is essentially the hydrostatic pressure at that same level! Even in the presence of
meteorologically significant variations or motions.
Since the vertical acceleration dw/dt and the associated vertical motions w = dz/dt are the
result of such small imbalances between the vertical PGF and gravity, it is not possible to
calculate dw/dt or w from measurements of PGF or g unless they are known to an
accuracy much greater than is possible with routine observations of these quantities. Thus,
other methods of calculating w or w must be found.
Day 11



Clean up: The “autoconvective” lapse
rate
 We derived a hypsometric equation from
hydrostatic balance and then examined a
constant temperature and constant lapse rate
atmosphere.
 How about a constant density atmosphere
(homogeneous atmosphere)?
Day 11

 The equation of state:

p = r Rd T

Density is constant:

p = ro Rd T

 Differential form of this equation:

dp/dt = ro Rd dT/dt
Day 11


Then, from the hydrostatic equation (apply chain
rule):
dp/dt = -gro dz/dt

Aha!! dp/dt? Can we set these equal?!

Thus:

-gro dz/dt = ro Rd dT/dt
Day 11




-ordT/dz = g/Rd = 34.2 C /km
This is what we call the “autoconvective”
lapse rate!
 at this lapse rate, overturning breaks out
sponaneously, or we get spontaneous
convection. No forcing is necessary!!!
Day 11



 Now, typically density falls off w/height.
 But Above strongly heated surfaces
(especially very close to the ground) density
can increase with height leading to dr/dz > 0
in a shallow layer.
 This is responsible for mirages on the road
or in the desert.
Day 11

Hydrostatics and Pressure systems

Warm Core Lows:

Central part of system warmer than the
outside part. Example: hurricane,
developing cyclones (baroclinic)
Day 11


Warm Core Low (x, z cross-section):
H
cold
L
warm
cold
Day 11


Cold Core Lows:
Central part of system colder than the
outside part. Example: occluded
cyclones, tropical depression
(equivalent barotropic)
Day 11


Cold Core Low (x, z cross-section):
L
warm
L
cold
warm
Day 11


Warm Core Highs (:
Central part of system warmer than the
outside part. Example: subtropical
highs, blocking (maintenance phase)
(equivalent barotropic)
Day 11


Warm Core High (x, z cross-section):
H
cold
H
warm
cold
Day 11


Cold Core Highs:
Central part of system colder than the
outside part. Example: arctic highs,
synoptic scale highs (baroclinic)
Day 11


Cold Core High (x, z cross-section):
L
жарко
H
холодно
жарко
Day 11

Vertical Stability and instability and
convection

The parcel method of layer stability analysis
(assume):


parcels of air are vertically displaced within an
environmental layer which is in hydrostatic
equilibruim (or hydrostatic is valid)
the displaced parcels do not mix with their
environmental air (No entrainment) no mixing of
mass.
Day 11




Defn: Hydrostatic or static stability of an
atmospheric layer
 an analysis of the consequences of vertical
displacement of parcels within the layer.
Classification:
1. Stable Stratification  displaced parcels
return to original position.
Day 11/12

2. Unstable stratification  displace parcel acceleration away
from the original position

3. Neutral Equilibration.  No acceleration after displacement

Now, the Non-hydrostatic vertical equation of motion:



Previously we stated that:
P = Po + p’(x,y,z,t)
r = ro + r’(x,y,z,t)
Day 12


 Where p’ << Po, thus P and Po are on
nearly the same order of magnitude, but the
primed quantities are variable in space and
time.
Thus, we saw for a resting atmosphere that:
dpo
  ro g
dz
Day 12


and we saw that in the vertical equation
of motion that:
1 p

g 0
r z
which can be replaced by:
1 p r 

 g 0
ro z ro
Day 12

and we further stated that the vertical
equation of motion could be written as:
dw
r
1 p
 g
 2 cos u  Fric  Visc
dt
ro
ro z

10-6
10-1
10-1 10-3
10-6
Day 12


 and after our TRUE Scale analysis, we
derived the perturbation form of the
hydrostatic relationship.
 Now we place the orders of magnitude of
each term for the STORM SCALE
(MICROSCALE), which could represent a
storm or Convective motions.
Day 12

Microscale:
dw
r
1 p
 g
 2 cos u  Fric  Visc
dt
ro
ro z


10-2
10-1
10-1 10-3
10-6
Coriolis will still be rejected! But, can
we neglect the acceleration term
now? (NO!)
Day 12

A thought experiment:

Let us consider a parcel of air embedded in an
environment of differing density.

In this context, the parcel could represent a cloud or
rising thermal.

We will assume the environment is either at rest or a
slowly moving region in which the hydrostatic
approximation is valid.
Day 12

So let’s define the situation:

Environment: re, Pe, and we

Where Pe and re are environmental
pressure and density and they are
related by:
dp
e
dz
  re g
Day 12


The environment, as we saw before and stated
above, are virtually functions of z only, which nearly
correspond to po and ro.
 Also let’s assume that the vertical motions are
small or zero (w = 0).

Inside the parcel, we will let the total p and r, which
can vary in x,y,z,t be written as:

P*(x,y,z,t) and r*(x,y,z,t).
Day 12

 These correspond to the total p and r, thus
the small difference can be expressed as:
P* = pe(x,y,z,t) + p’(x,y,z,t)
r* = re(x,y,z,t) + r’(x,y,z,t)



Thus:
r   r e  r
*
 and 
*

p  pe  p
Day 12


Now, since we do not assume w* is
small for the parcel, we can write the
vertical equation of motion using the
“partitioned” form as before: (this is
frequently done in modelling problems)
The equation:
dw*
1 p r 

 g
dt
r e z r e
Day 12

Since, re is approximately r* then we
can rewrite the equation as:
dw
1 p r 
 *
 *g
dt
r z r
*

w* is the vertical motion of the parcel
and dw/dt is the vertical acceleration,
which in this case is significant!
Day 12

 This is the equation that would appear in a numerical model!

 Thus, we are assuming that the rising parcel is not
hydrostatically balanced from the get-go!!!

The RHS terms are the non-hydrostatic PGF. The second term
is reduced gravity (bouyancy term).

 Since the parcel could represent a rising thermal, we can see
that convective and cloud-scale vertical motions and
accelerations are driven by non-hydrostatic PGF and bouyancy
differences!!!
Day 12



The Bouyancy equation
 In deriving the “parcel” form of the 3rd equation
of motion, we have neglected friction, viscosity, and
coriolos forces! (Meso and Micro-scale motions!)
 The physical picture we’ve conjured up is also
somewhat related to ‘parcel theory’ in that we have
assumed no mixing of the environment with the
parcels of the ascending and descending air (no
entrainment).
Day 12



Then, it is consistent with assuming:
We will now add another assumption for the
parcel method analysis consistent with ‘parcel
theory’.
* The parcel pressure (not to be confused
with Dalton and “partial’ pressure) equals the
environmental pressure (pe) at all levels as
the parcel moves vertically.
Day 12




Of course;
assumption 1) that environment hydrostatically
balanced introduces LITTLE error (good
assumption!).
and assumption 2) that no mixing of the parcels
occurs, is a fair to poor assumption depending on the
strength of the vertical motions, (mixing and
entrainment do occur).
 However, we do get useful results and sufficient
understanding of vertical atmospheric motion is
gained.
Day 12



Thus, assumption 3) is a good assumption for
most situations (convection), but for deep
and intense thunderstorms, this assumption
can also fail (some error here).
So, let’s rewrite our vertical equation of
motion:
where; p* = pe + p’ and r* = re + r’
Day 12

The equation;
dw *
1 p r 
 *
 *g
dt
r z r
dw *
1   p *  pe  r *  r e 
 *

g
*
dt
r
z
r

 Now we introduce assumption (3) into this
equation, this means p* = pe at ALL levels,
thus the pressure gradient term will have to
go to ZERO!
Day 12

The non-hydrostatic pressure effect is
neglected so, the equation becomes:

r
dw *

dt

*
 re 
r
*
g
and we’ll call this, the bouyancy
equation! (does this look familiar?)
Day 12



 This equation states quite clearly that the vertical
accelerations on this scale are proportional to the
desity difference between the parcells and the
environment.
 If r* < re, then bouyancy term is positive and
the acceleration term is > 0, and the parcel
accelerates upward in time increasing the w* over
time.
 If r* > re, then bouyancy term is negative and
the acceleration term is < 0, and the parcel
accelerates downward in time increasing the w* over
time.
Day 12



Archimedes Principle:
 Using the relationship above, and
stating that re = environmental mass /
volume, and r* = parcel mass /
volume.
Then, we have;
dw*
Menv  Mparcel
Me  M *
g
g
dt
Mparcel
M*
Day 12



which is also;
*
dw
M*
 gMe  gM *
dt
The LHS is now the Bouyancy force!
Thus, the force is just the difference between
the weight of the parcel (gm*) and the
weight of the fluid it displaces (gMe).
Day 12



The use of the parcel method for
stability analysis
We can rewrite the bouyant force in terms of
Virtual temperature, thus giving us a familiar
relationship. (Bouynacy term in
determination of CAPE)
Pe
re 
RdTve
From equation of state;
P*
r* 
RdTv *
Day 12

and using assumption (3)
(environmental P is approximately
parcel P);
*
dw * Tv  Tve 

g
*
dt
Tv

the vertical acceleration of the parcel is
directly proportional to Tv excess or
deficit. This is the definition of CAPE.
Day 12


If the virtual temperature and ambient
temperature are nearly the same, then;
dw * Tv*  Tve 

g
dt
Te
and a fair to poor approximation;
dw * T  Te 

g
dt
Te
*
Day 12


What magnitude of density, and/or virtual
temperauture differences between the parcel
and the environment will produce
meteorologically significant vertical
accelerations?
dw *
Dr
DTv
g
g
r*
Tv *
Let’s play with some numbers! dt
 so 
1 dw * Dr DTv


g dt
r * Tv *
Day 12



Ok:
dw *
Dr
DTv
g
g
dt
r*
Tv *
 so 
1 dw * Dr DTv


g dt
r * Tv *
 Consider the case of an air parcel that is
originally at the surface,
where;
z = 0 and w = 0;
Day 12



accelerating upward to reach the 1 km
level with
w = 4.5 m/sec (10 mph) ~ 9 kts.
 This is a typical vertical velocity for
air ascending into the base of a
thunderstorm cloud!
Day 12

and
 w *2 
d

2 
dw * dw * dz
dw *


 w*

dt
dz dt
dz
dz
 so 
w *2 20 m 2
m2
D

 10 2
2
2
2 s
s
m2
10 2
Dw
2 m
s
 3  10 2
Dz 10 m
s
Day 12

then
Dr DTv 1 dw * 10
1




r * Tv * g dt
10 1000
 so 
2
DTv
1

Tv * 1000
Day 12



 So, if you choose a reasonable surface virtual
temperature of 300 K for a parcel at the surface, it
needs only to be 0.3 K warmer than it’s environment
to generate significant vertical motions!
 Thus, the virtual temperature excess of only 0.3 C
will yield a w of 4.5 sec in just 1 km.
 This could result even if the parcel and the
environment are of the same temperature, but the
parcel contains 1.8 g/kg more water vapor (Note, the
previous result uses: Tv = T + w/6 off the
thermodynamic plot)
Day 12



 We can examine the accelerations of
parcels of air displaced from their original
positions in terms of a comparison between
the environmental lapse rate and the parcel
lapse rate as well.
Expressing at z = zo,
Tv* = Tve = Tvo
Day 12


After displacement:
z = zo + Dz
Tve  Tvo  Ge Dz
 where 
Tve
Ge  
z
Tv*  Tvo  G * Dz
 where 
Tv *
G*  
z
Day 12

Since;
dw * Tv

dt

 Tve 
g
*
Tv
*
 substitute relationship on the right
into this dw * Tv  G*Dz  Tv  Gv Dz 
dt
g
o
v
o
Tve

Gv*  Gve 
dw *
  gDz
dt
Tve
e
Day 12

 Thus, showing (gDz is positive) then,

 if environmental lapse rate > parcel lapse rate:


+ acceleration (environment gains energy from parcels)
 if environmental lapse rate < parcel lapse rate:

-- acceleration (environment loses energy to parcels)
Day 12


Recall: ”S” term in the First Law of
Thermodynamics is also a G(e) – G(d)
problem!
Let’s show that we can approximate Tv
lapse rate as the environmental lapse
rate.
Day 12



And
Tv  T (1  0.61m)
Tv  T  0.61mT
Tv T
T
m

 0.61m
 0.61T
z
z
z
z
m
Gv  G  0.61mG  0.61T
z
Tiny
Term on order of
Term
10-3
Day 12

If T = 280 K, and a typical Dm/Dz = 2 x 10-3

Then:

C
G  Gv  G  0.34
km
 Since in practice we use the buoyancy equation in
this form mainly to determine the sign of dw*/dt as
opposed to calculating it’s magnitude, this
approximation is not dangerous!
Day 12

Quickly review layer stability:

1. Ge < Gm < Gd  absolutely stable

2. Ge = Gm < Gd  neutrally stable (moist atmosphere)
stable (dry air)

3. Gm < Ge < Gd  conditionally unstable


4. Gm< Ge = Gd absolutely unstable (moist atmosphere)
neutrally stable (dry air)
5. Gm < Gd < Ge  absolutely unstable
Day 12


Hydrostatic stability criteria in
terms of vertical variations in
potential temperature.
Recall stability term in First Law of
Thermodynamics:
T q
S  Gd  Ge  
q p
Day 12


Case I: Dry Neutral (unsaturated air)
Recall that an unsaturated adiabatic
lapse rate G(d) is also one of constant
potential temperature. So for G(d):
T q
S  Gd  Ge  Gd  Gd  
0
q p
 where 
Ge  Gd
Day 12

Then:

and since;

then,
q
0
p
p   rgz
q
0
z
Day 12

Case II: Absolutely unstable (unsaturated air)

Ge > G d
T q
S  Gd  Ge  
 (negative)
q p
 where 
Ge  Gd

Then, if S is negative:
q
0
p
Day 12

and p decreases going up, then q is also
decreasing going upward (!), and then,
q
0
z


Case III: Absolutely stable (unsaturated air)
Ge < G d
T q
S  Gd  G e  
 ( positive)
q p
 where 
Ge  Gd
Day 12


Then, if S is positive:
q
0
p
and p decreases going up, then q is
increasing going upward (!), and then,
q
0
z
Day 12



 These same cases apply for moist air
as well! Consider:
Case IV: Moist Neutral (saturated air)
Recall that for saturated air, we note
that either qe or qwb are constant w/r/t
height. So for this case:
Day 12

Then:
T qe
Se  Gm  Ge  Gm  Gm  
0
qe p
 where 
Ge  Gm

and
qe qwb
,
0
p p
Day 12

and since;
p   rgz

then,
qe qwb
,
0
z z
Day 12



Case V: Absolutely unstable (saturated
air)
Ge > Gm
T qe
S  Gm  Ge  
 (negative)
qe p
 where 
Ge  Gm
Then, if S is negative:
qe qwb
p
,
p
0
Day 12

and p decreases going up, then qe is also
decreasing going upward (!), and then,
qe qwb
,
0
z z


Case VI: Absolutely stable (saturated air)
Ge < G m
T qe
S  Gm  Ge  
 ( positive)
qe p
 where 
Ge  Gm
Day 12

Then, if S is positive:
qe qwb
,
0
p p

and p decreases going up, then qe is
increasing going upward (!), and then,
qe qwb
,
0
z z
Day 12




Layer displacement and stability changes
Unsaturated displacement of a layer of constant mass
(DP)
Ascent  decreased stability
Stability is proportional to potential temp lapse (z
coordinates, but could be in p coordinates)
Day 12

 Thus since potential temp conserved
and the change in potential temp
conserved and change in height larger,
then potential temp lapse smaller (more
unstable)
Day 12

Ascent (Dp or Dz increased)
Day 12

Descent  increased stability
Day 12


Layer Lifting to total saturation
(Potential-Convective Instability)
Consider an arbitrarily layer (initially dry
in some or all of it) to be lifted until it
has reached a saturated state at all
levels.
Day 12


If after reaching saturation, its final lapse rate
exceeds the saturated adiabatic process lapse
rate the layer is now unstable (since the
whole layer is saturated, we need only
compare to moist adiabatic lapse rate now,
NOT the dry adiabatic lapse rate!).
With this result the layer is said to possess
potential instability and did so even before
lifting.
Day 12


Let’s recall that in atmosphere, the moisture
concentration (mixing ratios, and RH) are
higher closer to ground, thus lower level will
saturate first, and the upper levels will
saturate later.
Thus while top of the layer is cooling at dry
adiabatic lapse rate, the bottom is cooling at
moist adiabatic laspe rate over a longer
period of time.
Day 12




So:
1) to possess convective instability, Ge must
exceed the Gm after lifting.
2) saturated lapse rate is constant qe or qwb.
3) any layer in which Ge exceeds Gm, will also
display a decrease of qe or qwb with height,
Day 12



4) during lifting, unsaturated or saturated,
potential temperature variables are
conserved.
5) It follows that the layer had a decrease
of qwb or qe.
qe qwb
qe qwb
,
0
,
0
If in any layer z z
or p p
, the layer possesses potential-convective
instability and does so even before lifting!!!
Day 12



Layer stability due to changes in
horizontal Divergence / Convergence:

1 r
in x,y,z,t: 
  V
r z
in x,y,p,t

0   V
Day 12

horizontal divergence (column shrinks
 increasing stability)
Day 12

horizontal convergence (column
stretches  decreasing stability)
Day 12

3. Combined vertical displacements and horizontal convergence
or divergence

 due to the presence of the lower surface boundary -normally in the mid to lower troposphere

Case I

horizontal divergence  subsidence  both tending to
increased stability

e.g., deep high pressure area  boundary layer (1 km)
Day 12




horizontal divergence  subsidence inversions and
increased stability  dissipating or suppresses
cloudiness and precipitation.
Case II:
horizontal convergence  lifting or ascent  both
contribute to decreased stability
e.g., low pressure area  boundary layer
convergence  lifting  decreased stability 
increased cloudiness and or precipitation
Day 12

4. rapid destabilization of an
atmosphere possesing potential convective instability due to low level
(bndry layer) horizontal convergence
and lifting (see drawing)
Day 12

Local temperature and stability
changes at a point in space

Recall:
T = T (x,y,z,t)

dT T dt T dx T dy T dz
and,




dt t dt x dt y dt z dt
Day 12

Recall from 1st law (x,y,p,t):
 a dp
dT
dp
dT
Q
Q  Cp
a



dt
dt
dt Cp Cp dt

and using the hydrostatic
approximation, then:
dp   rgdz
dT Q
g dz


dt Cp Cp dt
Day 12

and, thus expanding dT/dt

T
T
g Q
 V  T  w  w 
t
z
Cp Cp

but, by definition:
T
g
 Ge 
 Gd
z
Cp
Day 12


and substitute into the first law:

T
Q
 V  T  wGe  Gd  
t
Cp
to get “S” term in z - coordinates.
 T g 
S  Ge  Gd   
    Sw
 z Cp 
Day 12

and the local rate of temperature change in zcoordinates is:

T
Q
 V  T  Sw 
t
Cp
Stability change equation.

(Doswell:
http://www.nssl.noaa.gov/~doswell/)
Day 12

First, take the partial derivative of the
First Law of Thermodynamics w/r/t z;

 T 
S
w
  Q 
  V  T   w 
S  
z t z
z
z
z  Cp 

If an incompressible atmosphere, then
partial w / partial z is the vertical
divergence, which equals the negative
of the horizonal convergence.
Day 12

(Note also, I switch order of differentiation on
LHS)


Ge S 
Ge
  Q 

  V  T  
w    Vh S   
t
t z
z
z  Cp 


And this is the Stability change equation!
Q: What’s it all mean? How do we interpret
this?
Day 12


The Left-hand Side:
 If the lapse rate increases with
respect to time, (larger negative value),
this implies decreasing S, which implies
decreasing stability.
Day 12

The Right-hand Side:

Term A (The differential temperature advection
term):

 contributes to increased instability if:




a) CAA over WAA
b) stronger CAA aloft over weak CAA
c) or weak CAA or stronger WAA
Basically, we need to warm the bottom of a layer
faster than the top!
Day 12



Term B (the vertical advection of stability
term)
 upward motion (+w) and Ge increasing
with height  stabilization (more stable air
from below advects upward), but be careful,
because upward motion causes adiabatic
cooling!
(opposite for downward motion)
Day 12




Term C (Convergence / Divergence
term)
 S is almost always positive so:
convergence  destabilitzation
divergence  stabilization
Day 12

Term D (Differential diabatic heating term)

Let’s take two examples:



solar heating:
warm the bottom of layer more rapidly
decreasing the general satibility of the lower
atmosphere.
b)
LHR at the top of a layer:
Day 12

LHR is warming the top of the layer
more rapidly thus increasing the
stability in the layer (but decreasing the
stability in the layer above  Potential
Vorticity generation)
Day 12


Entrainment: a mathematical
solution
We’re going to derive an entrainment
equation! (FUN!)
Day 12




To begin:
 assume that a cloud filled parcel and some
entrained environmental air (could be dry) contitute
the thermodynamic system (mixing of air violates 2nd
assumption of parcel theory)
 consider a cloud mass M which rises from level z
where it has properties T,P, and m
 now it’s lifted to z + dz
Day 12

 and in the process entrains environmental air dM/dt which
has a temp Te and me.

 Now the parcel has properties M + dM/dt, T + dT/dt, m +
dm/dt, and z +dz/dt.

Recall in deriving the moist adiabat, we parameterized LHR. We
need to parameterize other processes as well:


 The total sensible heat required to raise the mass of
entrained environmental air dM from its initial temperature Te to
parcel temperature T is:
dQ1/dt = Cp(T - Te) dM/dt
Day 12




 this is supplied by the parcel air (so dQ1 will have
a negative sign)
The total latent heat required to bring the
entrained air dM to a state of saturation at it’s new
mixing ratio ms is:
dQ2/dt = L(ms - me) dM/dt
 where ms is the saturated mixing ratio of the
mixed parcel at temperature T
Day 12



 This latent heat must also be supplied by
the parcel (so dQ will have a negative sign).
 The total latent heat released
simultaneously to the parcel due to it’s
saturated ascent and pressure reduction is:
dQ3/dt = -ML dms/dt
Day 12


Thus the first law applied to the mixed
parcel is:
-dQ1/dt - dQ2/dt + dQ3/dt = M
(CpdT/dt - a dp/dt)
-ordM
dM
dms
 Lms  me 
 ML

dt
dt
dt
dT
dp
M  Cp
 a 
dt
dt 

 Cp T  Te 
Day 12

Let us divide equation 1 by M and
invoke the “chain rule”:
1 dM
1 dM
dms
 CpT  Te 
 Lms  me 
L

M dz
M dz
dz
dT
dp
Cp
a
dz ent
dz
Day 12

then;
1) replace a dp/dz with g from hydrostatic balance
2) deevide by Cp
3) and use chain rule again as we did to derive Gm


dms/dz = dms/dT(dT/dz) ent.
1 dM L
1 dM
 T  Te

ms  me

M dz Cp
M dz
dT
L dms dT
g


dz ent Cp dT
dz ent Cp
Day 12

Now solve for the entrainment lapse
rate!
dT

dz ent

g
1 dM L
 T  Te 

ms  me  1 dM
Cp
M dz Cp
M dz
L dms 

1 

 Cp dT 
 then 
dT
dz ent
1 dM 
L

g
T  Te  ms  me 


M dz 
Cp
Cp



L dms 
L dms 


1 

1 

 Cp dT 
 Cp dT 
Day 12



Recall: Isn’t g/Cp just the dry adiabatic lapse
rate?
 also the first term on the right hand
side is the moist adiabatic lapse rate.
 There is a Gm in the second term as
well, can you see it? (It might help to
“strategically multiply that term by 1).
Day 12

And then
dT
dz ent
 1 dM 
L

  M dz  T  Te   Cp ms  me 


 Gm  Gm 
Gd




Day 12

and reduce:
dT
1 dM 1


 Gm 1 
CpT  Te  Lms  me 
dz ent
 M dz g

 and 
dT
 Cent Gm
dz ent
 where 
Cent  1 
1 dM 
L


T

Te



ms

me



M dz 
Cp

Day 12

This is the lapse rate that is used in convective
parameterization schemes!

Now:


Cent  Is the entrainment factor or correction to
be applied to the moist adiabatic lapse rate
(established rate). This accounts for entrainment.
Cp(T-Te)  is the sensible heat or internal energy
transfer due to parcel mixing.
Day 12

L(ms – me)  is the latent heat transfer due to parcel
mixing.

(1/M) (dM/dz)  is the mass entrainment factor.

Now, Cent is always a positive number, thus the
moist adiabatic lapse rate is modified such that:
Gm  Gent  Gd

The lower bound is true ALL the time, while the
upper condition is true MOST of the time.
Day 12

The temperature decrease with increased height of
an entraining saturated parcel of cloud will always be
greater than the non-entraining parcel rate. The
increase in lapse rate is dependent on three factors!

1) Mass entrainment rate

2) the temperature difference between parcel and
environment

3) environmental dryness.
Day 12




 The dryer and cooler the environmental air the
more stabilizing the impact of the environment
 Let’s take a look at some typical values and
evaluate each term
Let P = 700 hPa Tparcel = 11 oC
and
RH is 0.70 (70%)
Tenviron. = 10 oC
Day 12

 Then the Cp (T - Te) term is:

on order of 1000 J /kg

Next term:


ms at 11 oC = 11.8 g kg-1
me = 0.70 x ms(10 oC) = .7 x 11.1 = 7.8
Day 12



so,
L (ms - me) = 2.5 x 106 ( 0.0118 0.0078) ~ 10,000 J kg-1
 so typically the latent heating term is
10 times the sensible heating term!
Day 12




 the 1/M dM/dz , or fractional change in mass
relative to the whole, may range from 1% for large
convective elements to 100% for smaller elements
(Tp – Te) can range from 0.1 to 5 oC
and (ms - me) may range from 0.1 to 10 g/kg
even if environmental air is saturated, the ratio of the
LHR to SNS term is ~ 2.5
Day 12

applicability of non-entraining  pure parcel method vs. the
entraining marcel method:

100 % for small trade wind CU (Stommel)

1% for large Cb

thus, observed entrainment rates become less important for
large size convective events.

 It has been observed that in the tops of large CB’s the qe or
qwb of the updraft air has the same value as the value of sub
cloud air, thus no - dilution.
Day 12



 So for even for extreme events, the parcel
method can be applicable, reasonably
speaking.
The Unsaturated environment
unsaturated entrainment we can proceed
as before, but we leave out the LHR.
Day 12

Dry entrainment only:
dM
dT
dp 

 CpT  Te
 M  Cp
a 
dt
dt
dt 


Then divide by Cp, and chain rule as
before blah, blah, blah, ……
1 dM dT
g
 T  Te


M dz dz unsat .ent Cp
Day 12

Then solve:
dT
g
1 dM

 (T  Te)
dz unsat .ent
Cp
M dz
 or 
dT
1 dM
 Gd  (T  Te)
dz unsat .ent
M dz
 then 
dT
1 dM 
 Cp
 Gd 1 
(T  Te)
dz unsat .ent
g
M dz 


There is only a sensible heating term
and as such Gue will always be greater
than Gd!
Day 13

See you in 4320!