Post on 20-Dec-2015
Atms 4310 / 7310
Anthony R. LupoTest 3 material
Day 1
Then this vapor equation is: es v=RvT or es = vRvT
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
0 C 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 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 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 or e actual = RH * es
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
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 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
Water…
Day 2
Carbon
Day 2
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
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
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 ).
Day 2
1) To the right of B (in A) water mass is all vapor.
2) B to C vapor and water co-exhist.
3) To the left of C water mass is all liquid.
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
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:
1) Super cooled water
2) Another special case Follow the point D isotherm. This is a critical point.
Day 3
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.
Their critical points are WAY below
terrestrial T and P’s!
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
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
(recall ds = dq/T) and p = es
Day 3
Thus;
And
The definition of latent heat for the phase transformation L 1 to 2 (where 1 = liquid (l) and 2 = gas (v))
deCvdTpddudq s
2
1
2
1
2
1
2
1
deduTdsdq s
Day 2
Latent heat
Recall that during a change of phase, T and es are constant!, so;
2
1
)( dqvlL
1212)12(21 seUUSSTL
Day 3
L(1 – 2) is defined as the latent heat of vaporization.
Thus our equation can be rewritten as:
Density of air is = 1 kg /m3
lveUlUvvlL s
Day 3
Then, if the mixing ratio is = 10 g/kg v must be: 0.01 kg/m3
Density of liquid water is: 1 gm cm-3 or 1000 kg / m3
Flip each to get specific volume, but look….. there’s 5 orders of magnitude between the two.
Day 3
Thus v is 100,000 times larger than l, so the equation becomes
Use the equation of state: es v = Rv T
L(l-v) = RvT + uv – ul
veUlUvvlL s
Day 3/4 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.
How?
dL/dt = RvdT/dt + duv/dt – dul/dt
Day 4 Recall: Defn of specific heat cl = dul / dT
and Cvv = duv / DT
So now we rewrite as:
dL/dt = RvdT/dt + CvvdT/dt – CldT/dt
-or- (by chain rule) dL/dT = Rv + Cvv – Cl
Day 4
and, of course, Rv + Cvv = Cpv
So DL/DT = Cp – Cl = -2369 J/kg K
(aha! – slope of L versus Temperature, so Latent heat is temp dependent!)
Day 4 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:
After applying the snake:
L (l – v) = Lo + (Cp – Cl) (T – To)
Day 4 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 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 d
Day 4
Since during a phase change T and es are constant then;
where;
2
1
)21(LDq
2
1
2
1
2
1
2
1
desdudsTDq
Day 4 then,
L(1-2) = T(S2 – S1) = (u2 – u1) + es(2 – 1)
or rearrange the above to isolate each state, state 2 and state 1
TS2 – u2 – es 2 = TS1 – u1 – es 1
Day 4 During the phase change:
T S – u – es = Constant
- or –
u + es - T S = Constant = G
(J. Willard Gibbs potential)
Day 4 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 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
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 d/dt + des/dt – T dS/dt – S dT/dt
Day 5 Rearrange: dG/dt = du/dt + es d/dt – T dS/dt + des/dt – S dT/dt
On the RHS, the first three terms
are the 1st Law of Thermodynamics!
Day 5 If T dS/dt = du/dt + es d/dt
Then dG = des/dt – S dT/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 Then dG must also be a constant!
If dG is constant;
Then;
2 des/dt – S2 dT/dt = 1 des/dt – S1 dT/dt
Day 5
so;
des/dt (2 – 1) = dT(S2 – S1)
Or
des/dT = (S2 – S1) / (2 – 1)
Day 5 But,
So:
This is it! (Make no mistake)!
Generalized Clausius Clapeyron equation for any phase change from phase 1 to 2.
2
1
2
1
2
1
)21(112
TL
dqTT
dqdsSS
21)21(
TL
dT
des
Day 5 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
In the case of the CC equation:
L(1-2) is L (l-v) 2 = v and 1 = l
es = esw
Day 5
We’ll invoke the same argument we used before, mainly;
v >>> l
So the Clausius Clapeyron Equation becomes:
vTvlL
dTdesw
)(
Day 5
From our equation of state;
esw v = Rv T or v = Rv T / esw
Substitute into our CC:
RvTvlL
dTde
esw
sw2
)(1
Day 5
Or
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).
2
)(TdT
RvvlL
e
de
sw
sw
Day 5
Integrate and put in the limits:
then
ToTRv
vlLe
e
so
sw 11)(ln
TToRLmv
eTToRv
Lee sososw
11*
exp11
exp
Day 5
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 Saturation Vapor Pressure over Ice
(or “on the rocks”?)
In this case;
L(1 – 2) becomes L(v – i), thus 2 = v and 1 = i
and v >>> i
Day 5
Thus with the same derivation:
This is the sublimation curve – saturation or equilibrium vapor pressure over ICE (on the rocks)!
TToRLamv
eTToRv
Laee sososi
11*
exp11
exp
Day 5 / 6
10
0
ei
eii
393.15243.15 Ti
200 250 300 350 4000
2
4
6
8
10
The phase diagram (my own), see the tail.
Day 5/6 Notes:
1. This Equation is essentially identical to the previous version of the Clausius-Clapeyron 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 The Equilibrium curve for ice and
water
Here,
l = ice
2 = l = 1.00 x 10-3 m3/kg 1 = i = 1.09 x 10-3 m3/kg
Day 6
so (2 – 1) = (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:
ilTliL
dT
des
)(
Day 6
and that becomes:
and let’s get the snake involved (integrating from es = eso to es, as T goes from To to T):
dtdT
xTliL
dtdes
5109)(
ToT
xliL
ee sos ln109
)(5
Day 6 This is the melting curve!
Draw e vs. T (2-D) phase curve (we’re only looking at a portion of the 3-D phase diagram).
ToT
xliL
ee sos ln109
)(5
Day 6 (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 Let’s examine the behavior of the curve;
Scenario 1: If T = To,
then ln (1) = 0 and term drops out.
Scenario 2: If T < To,
e goes rapidly to “high” values, since ln (T/To) < 0
Day 6 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 A few more points:
1) We can identify the Triple point.
2) melting curve: to the right of melting curve only water, to the left only ice.
3) evaporation curve ends at Critical point (again distinction between liquid and vapor disappears)
Day 6 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 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 at –14C: 20% of clouds are water droplets
only
at –8C: 50% of clouds are water droplet only.
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 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 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)
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 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.
v = (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 = 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 (e) = 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 = v (R*/mov) Tv where v = Massv /Volv
Equation of state for dry air:
Pd = d (R* / mod) Td where d = Massd / Vd
Day 6
And;
Volv = Vold Tv = Td
Forming the ratio:
TR
TmovR
Pe
d
v
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)
Divide by Volume
mdmvmv
q
dv
v
Volmd
Volmv
Volmv
mdmvmv
q
Day 6 From eqn of state for vapor: e = v (R*/mov) T or v = (mov/R*)(e/T)
For dry air:
P = d (R*/mod) T or d = (mod/R*)(e/T)
Day 6
So,
epde
emov
pd
emov
Te
Rmov
Tp
R
Te
Rmov
q622.0
622.0
mod
modmod
**
*
Day 6
But Pd = P – e
Since e and es << Pt
epte
qsept
eq
378.0622.0
378.0622.0
esRHeqsq
msm
ese
RH
or
omixingratip
eq
%%
622.0
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 = m Rm T
Day 6
where m = (Massm/V)
Recall Dalton’s law:
pm = ptot = Pi
where i = gas 1,2,3, ….
Day 6 Rewrite ideal gas law above:
pm = (Massm/V) Rm T = (Massi/V) Ri T
where i = gas 1,2,3,….
Solving for Rm:
Rm = (Mi/Mm) Ri
Day 6 Separate gasses into dry gasses +
moisture:
Rm = (Md/Mm) Rd + (Mv/Mm) Rv
But, Md = Mm – Mv
So,
Rm = ((Mm – Mv)/ Mm) Rd + (Mv/Mm) Rv
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 = m (1 + 0.61q) Rd T
Day 6 Rearrange (and this is a short cut to getting Tv)
Pm = m Rd (1 + 0.61q) T
or
Pm = m 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 Cv = 717.59 J kg-1K-1
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 Cpv / Cpd = 1.81
So….
Cpm = Cpd(1 –q + 1.81 q)
and combining “q” terms;
Cpm = Cpd(1 + 0.81q)
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,
Ok, Let’s move on!
CvdCvmRdRmCpdCpm ;;
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:
Recall: LHR is a process by which
we can add internal energy and pressure work
dtdp
mdtdT
Cpmdtdq
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:
dtdp
mdtdT
CpmmsdtLmsd 1
Day 7
use our approximations:
1) L = Lo 2) Cpm = Cpd 3) 1+ms = 1
and the equation becomes:
Day 7
dtdp
mdtdT
Cpdtmsd
Lo
•and the equation becomes:
•Now, pressure and height again related via hydrostatic balance:
dtdz
gdtdp
m
Day 7
Then, the 1st Law becomes:
but by “chain” rule:
gdzdT
Cpdz
dmLo
ordtdz
gdtdT
Cpdtmsd
Lo
s
Day 7
Ok seriously,
substitute
dzdT
dTdm
dzdm ss
gdzdT
CpdzdT
dT
dmLo s
Day 7 solve algebraically for dT/dz
**Now we’ve derived a “moist
adiabatic lapse rate” which answers the question “how is moisture modifying the dry adiabatic lapse rate?”.
Cpg
dTdm
CpLo
dzdT s
1
Day 7 Viola!
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.
dT
dm
Cp
Lodz
dT
s
d
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, (m) will approach d 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 (m) is significantly less than (d).
Day 7
Recall, a Stuve:
At p = 1000 hPa and T = 20 C (m) ~ 4.3 C/km
at p = 500 hPa and T = -20 C (m) ~ 7.8 C/km
Day 7
at p = 100 hPa and T = -60 C (m) ~ (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!
d/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 average or the original layer. (again, d/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 .
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 (top and bottom). 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 = avg. 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
Fz = 0
Consider a parcel of volume (V = xyz) 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 = Fz = Force due to gravity + Force due to pressure gradient (across a Vol.)
0 = Fz = -Mg + pl(xy) – pu(xy)
- but -
pu = pl + (p/z) z
Day 8 0 = Fz = -Mg + pl(xy) – (pl(xy) + (p/z)
xyz)
deevide by Mass….
0 = - g –[(p/z) (V / Mass)]
Q: What is Volume / Mass, especially a unit Mass?
A: Specific Volume ()
Day 8
Thus, the limit of the above equation as delta 0 is:
!!
Question: Does;
gzp
zp
gzp
g
1
0
dzdp
zp
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,
zp
yp
wv
xp
wu
dzdt
tp
dzdp
becomes
pVtp
dtdp
33
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……
then:
zp
yp
wv
xp
wu
dzdt
tp
dzdp
dzdp
zp
Day 8 The local change in pressure with height
equals the total change in pressure with height.
Thus,
Note the change in notation! This is hydrostatic balance in an atmosphere at rest! In such an atmosphere, p = p(z) only!
gdzdp
zp
Day 8 The atmosphere in motion and
hydrostatic balance.
This time we’ll use Newton’s 2nd law!
Which is;
F = ma,
Day 8
where,
a = dV/dt or the (3-D) velocity vector.
Recall;k
dtdw
jdtdv
idtdu
adtVd
and
kwjviuV
ˆˆˆ
ˆˆˆ
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) gravitational force – directed downward 2) vertical component of PGF – upward 3) vertical component of coriolis force 4) 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 (k-component):
(3)ViscFricuzp
gdtdw
)cos(2
1
Day 9
Let’s use scale analysis to justify our final result!
Acceleration term:
dw/dtzww
dzd
dzdw
wdtdz
dzdw
22
21
2
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 w5
2 m s-2
Gravity term
C’mon folks! 10 m s-2
Day 9 Pressure gradient term:
= 1 kg m-3
p/z 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
2cos = 2sin and is approximately 10-4 s-1
10-3 m s-1
Day 9 Terms versus scale for each scale
of motion:Horizontal
ScaleW
(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:
so with a very small error: (0.01% for synoptic scale features).
zp
g
1
0
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.
gzp
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;
So;
P = Po + p’(x,y,z,t) = o + ’(x,y,z,t)
t
to
dtQ 0
Day 9 It has been shown (Dr. N. Phillips, 196?) that the
variable parts of P and are very small compared to the mean parts (This should make sense to us).
Then;
P’ << Po
’ << o
and for a resting atmosphere: p’ = ’ =0
Day 9/10 Thus, o 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:
Then, we have defined Po and o such that the hydrostatic relation is satisfied, given Po(z) gives o(z), and vice-versa.
gdz
dpo
o
Day 10
So, if we are going to replace the vertical component of Newton’s 2nd Law by the hydrostatic approximation,
gzp
1
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:
zp
dtdw
1
Day 10
Well we know;
so this is where we need to go;
zp
z
p
zp
g o
111
zp
g
1
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.
ViscFricuzp
gdtdw
cos2
1
Day 10
Let’s rewrite each variable in perturbation form, starting with density and then assume that ’/o << 1
Which gives for density:
ooooo
ooo
1...1
1
1111
32
1
Day 10
However, we’ll only neglect second order and higher terms in order to get:
Now hydrostatic balance:
ooo
11
1
gpp
zg
zp
oo
11
Day 10
Becomes:
By “Foil”
Term 1 Term 2 Term 3 Term 4
gzp
z
po
oo
1
1
011
22
g
zp
z
p
zp
z
p
o
o
oo
o
o
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 + ’/o)-1. Thus we have left terms 2 and 3:
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!):
01
2
z
p
zp o
oo
Day 10 Viola!
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.
01
gzp
oo
Day 10 So we get :
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!
ViscFricuzp
gdtdw
oo
cos21
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.
01 gzp
Day 10
But, can we justify using:
?
as we did for a resting atmosphere, as also being valid for an atmosphere in motion?
dzdp
zp
Day 10
In other words, how much error is involved replacing :
With?zp
dzdp
Day 10
Start with p = p(x,y,z,t) again!
thus:
zp
yp
wv
xp
wu
dzdt
tp
dzdp
becomes
pVtp
dtdp
33
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 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:
zp
yp
wv
xp
wu
dzdt
tp
dzdp
Day 10
on the mesoscale, it’s about 5%.
So, to 1% for the atmosphere in motion:
dzdp
zp
gdzdp
zp
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 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 = Rd T Density is constant: p = o Rd T Differential form of this equation:
dp/dt = o Rd dT/dt
Day 11 Then, from the hydrostatic equation (apply
chain rule):
dp/dt = -go dz/dt
Aha!! dp/dt? Can we set these equal?!
Thus:
-go dz/dt = o Rd dT/dt
Day 11 -or-
dT/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 d/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
L cold 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
L warm 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
H cold 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
H жарко 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) = o + ’(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:
gdz
dpo
o
Day 12
and we saw that in the vertical equation of motion that:
which can be replaced by:
01 gzp
01
gzp
oo
Day 12
and we further stated that the vertical equation of motion could be written as:
10-6 10-1 10-1 10-3 10-6
ViscFricuzp
gdtdw
oo
cos21
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:
10-2 10-1 10-1 10-3 10-6
Coriolis will still be rejected! But, can we neglect the acceleration term now? (NO!)
ViscFricuzp
gdtdw
oo
cos21
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: e, Pe, and we
Where Pe and re are environmental pressure and density and they are related by:
gdzdp
ee
Day 12 The environment, as we saw before and stated
above, are virtually functions of z only, which nearly correspond to po and o.
Also let’s assume that the vertical motions are small or zero (w = 0).
Inside the parcel, we will let the total p and , which can vary in x,y,z,t be written as:
P*(x,y,z,t) and *(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) * = e(x,y,z,t) + ’(x,y,z,t)
Thus:
*
*
ppp
and
e
e
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:g
zp
dtdw
ee
1*
Day 12 Since, e is approximately * then we
can rewrite the equation as:
w* is the vertical motion of the parcel and dw/dt is the vertical acceleration, which in this case is significant!
gzp
dtdw
**
* 1
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 * = e + ’
Day 12 The equation;
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!
g
z
pp
dtdw
gzp
dtdw
ee*
**
*
**
1*
1*
Day 12
The non-hydrostatic pressure effect is neglected so, the equation becomes:
and we’ll call this, the bouyancy equation! (does this look familiar?)
g
dtdw e
*
**
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 * < e, then bouyancy term is positive and the acceleration term is > 0, and the parcel accelerates upward in time increasing the w* over time.
If * > e, 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 e = environmental mass / volume, and * = parcel mass / volume.
Then, we have;
***
MMMe
gMparcel
MparcelMenvg
dtdw
Day 12 which is also;
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).
***
gMgMedtdw
M
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)
From equation of state;e
ee RdTv
P
**
*RdTvP
Day 12 and using assumption (3)
(environmental P is approximately parcel P);
the vertical acceleration of the parcel is directly proportional to Tv excess or deficit. This is the definition of CAPE.
g
TvTvTv
dtdw e
*
**
Day 12
If the virtual temperature and ambient temperature are nearly the same, then;
and a fair to poor approximation;
g
Te
TvTv
dtdw e
**
g
TeTT
dtdw e
**
Day 12 What magnitude of density, and/or
virtual temperauture differences between the parcel and the environment will produce meteorologically significant vertical accelerations?
Let’s play with some numbers!
***1
***
TvTv
dtdw
g
so
TvTv
ggdtdw
Day 12 Ok:
Consider the case of an air parcel that is originally at the surface,
where; z = 0 and w = 0;
***1
***
TvTv
dtdw
g
so
TvTv
ggdtdw
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
22
3
2
2
2
2
2
22
2
1010
10
102
202*
2*
**
**
sm
msm
zw
sm
smw
sodz
wd
dzdw
wdtdz
dzdw
dtdw
Day 12
then
10001
*
10001
1010*1
**
2
TvTv
so
dtdw
gTvTv
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 + z
z
Tv
where
zTvTv
ee
eoe
zTv
where
zTvTv o
**
**
Day 12
Since;
substitute relationship on the
right into this
g
Tv
TvTv
dtdw e
*
**
e
ev
e
eovo
Tv
vzg
dtdw
Tv
zvTvzTvg
dtdw
*
*
*
*
Day 12 Thus, showing (gz 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 (e) – (d) problem!
Let’s show that we can approximate Tv lapse rate as the environmental lapse rate.
Day 12 And
Tiny Term on order of
Term 10-3
zm
Tmv
zm
TzT
mzT
zTv
mTTTv
mTTv
61.061.0
61.061.0
61.0
)61.01(
Day 12 If T = 280 K, and a typical m/z = 2 x 10-3
Then:
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!
kmC
v 34.0
Day 12 Quickly review layer stability:
1. e < m < d absolutely stable
2. e = m < d neutrally stable (moist atmosphere)stable (dry air)
3. m < e < d conditionally unstable
4. m< e = d absolutely unstable (moist atmosphere) neutrally stable (dry air)
5. m < d < e absolutely unstable
Day 12
Hydrostatic stability criteria in terms of vertical variations in potential temperature.
Recall stability term in First Law of Thermodynamics:
pT
edS
Day 12
Case I: Dry Neutral (unsaturated air)
Recall that an unsaturated adiabatic lapse rate (d) is also one of constant potential temperature. So for (d):
de
where
pT
ddedS
0
Day 12
Then:
and since;
then,
0p
zgp
0z
Day 12 Case II: Absolutely unstable (unsaturated
air)
e > d
Then, if S is negative:
de
where
negativep
TedS
)(
0p
Day 12 and p decreases going up, then is also
decreasing going upward (!), and then,
Case III: Absolutely stable (unsaturated air)
e < d
0z
de
where
positivep
TedS
)(
Day 12
Then, if S is positive:
and p decreases going up, then is increasing going upward (!), and then,
0p
0z
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 e or wb are constant w/r/t height. So for this case:
Day 12
Then:
andme
where
pe
eT
mmemSe
0
0,
pwb
pe
Day 12
and since;
then,
zgp
0,
zwb
ze
Day 12
Case V: Absolutely unstable (saturated air)
e > m
Then, if S is negative:
me
where
negativepe
eT
emS
)(
0,
pwb
pe
Day 12
and p decreases going up, then e is also decreasing going upward (!), and then,
Case VI: Absolutely stable (saturated air)
e < m
0,
zwb
ze
me
where
positivepe
eT
emS
)(
Day 12
Then, if S is positive:
and p decreases going up, then e is increasing going upward (!), and then,
0,
pwb
pe
0,
zwb
ze
Day 12 Layer displacement and stability changes
Unsaturated displacement of a layer of constant mass (P)
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 (p or z 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, e must exceed the m after lifting.
2) saturated lapse rate is constant e or wb.
3) any layer in which e exceeds m, will also display a decrease of e or wb 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 wb or e.
If in any layer or , the layer possesses potential-convective instability and does so even before lifting!!!
0,
zwb
ze 0,
pwb
pe
Day 12
Layer stability due to changes in horizontal Divergence / Convergence:
in x,y,z,t:
in x,y,p,t
Vz
1
V
0
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)
and,
dtdz
zT
dtdy
yT
dtdx
xT
dtdt
tT
dtdT
Day 12
Recall from 1st law (x,y,p,t):
and using the hydrostatic approximation, then:
dtdp
CpCpQ
dtdT
dtdp
dtdT
CpQ
dtdz
Cpg
CpQ
dtdT
gdzdp
Day 12
and, thus expanding dT/dt
but, by definition:
CpQ
Cpg
wzT
wTVtT
dCpg
ezT
Day 12
and substitute into the first law:
to get “S” term in z - coordinates.
CpQ
dewTVtT
SwCpg
zT
deS
Day 12 and the local rate of temperature
change in z-coordinates is:
Stability change equation.
(Doswell: http://www.nssl.noaa.gov/~doswell/)
CpQ
SwTVtT
Day 12 First, take the partial derivative of the
First Law of Thermodynamics w/r/t z;
If an incompressible atmosphere, then partial w / partial z is the vertical divergence, which equals the negative of the horizonal convergence.
CpQ
zS
zw
wzS
TVzt
Tz
Day 12 (Note also, I switch order of
differentiation on LHS)
And this is the Stability change equation!
Q: What’s it all mean? How do we interpret this?
CpQ
zSVw
zTV
ztS
t hee
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 e 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 - dp/dt)-or-
dtdp
dtdT
CpM
dtdms
MLdtdM
memsLdtdM
TeTCp
Day 12
Let us divide equation 1 by M and invoke the “chain rule”:
dzdp
dzdT
Cp
dzdms
LdzdM
MmemsL
dzdM
MTeTCp
ent
11
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
m
dms/dz = dms/dT(dT/dz) ent.
Cpg
dzdT
dTdms
CpL
dzdT
dzdM
Mmems
CpL
dzdM
MTeT
entent
11
Day 12
Now solve for the entrainment lapse rate!
dTdms
CpL
memsCpL
TeTdzdM
M
dTdms
CpLCpg
dzdT
then
dTdms
CpL
dzdM
Mmems
CpL
dzdM
MTeT
Cpg
dzdT
ent
ent
1
1
1
1
11
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 m in the second term as well, can you see it? (It might help to “strategically multiply that term by 1).
Day 12
And then
d
memsCpL
TeTdzdM
Mmm
dzdT
ent
1
Day 12
and reduce:
memsCpL
TeTdzdM
MCent
where
mCdzdT
and
memsLTeTCpgdz
dMM
mdzdT
entent
ent
11
111
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:
The lower bound is true ALL the time, while the upper condition is true MOST of the time.
dm ent
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 Tenviron. = 10 oC and
RH is 0.70 (70%)
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 e or wb 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:
Then divide by Cp, and chain rule as before blah, blah, blah, ……
dtdp
dtdT
CpMdtdM
TeTCp
Cpg
dzdT
dzdM
MTeT
entunsat
.
1
Day 12 Then solve:
There is only a sensible heating term and as such ue will always be greater than d!
dzdM
MTeT
gCp
ddzdT
thendzdM
MTeTd
dzdT
or
dzdM
MTeT
Cpg
dzdT
entunsat
entunsat
entunsat
1)(1
1)(
1)(
.
.
.
Day 13
See you in 4320!