Atmospheric Science 4310 / 7310

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Atmospheric Science 4310 / 7310. Atmospheric Thermodynamics - II By Anthony R. Lupo. Day 1. The work done by an expanding gas Let’s draw a piston:. Day 1.  Consider a mass of gas at Pressure P in a cylinder of Cross section A Now, Recall from Calc III or Physics:

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### Atmospheric Science 4310 / 7310

Atmospheric Thermodynamics - II

By

Anthony R. Lupo

Day 1
• The work done by an expanding gas
• Let’s draw a piston:
Day 1
•  Consider a mass of gas at Pressure P in a cylinder of Cross section A
• Now, Recall from Calc III or Physics:
• Work = force x distance or Work = Force dot distance
• So only forces parallel to the distance travelled do work!
Day 1
• Then,
Day 1
• But, we know that:
• Pressure = Force / Unit Area
• So then,
• Force = P x Area
Day 1
• Total work increment now:
• Well,
• Area x length = Volume
• soo………………… A * ds = dVol
Day 1
• Then we get the result:
•  Work :
• Let’s “work” with Work per unit mass:
•  Thus, we can start out with volume of only one 1 kg of gas!!!
Day 1
•  And we know that:
• a = Vol/m
• then……..
• Note the “heavy D”
Day 1
• So….
• Very Important!!! This is not path independent, it’s not an exact differential!
•  Thus is is easy to see that if a parcel undergoes expansion, or a2 > a1
Day 1
• Parcel does work expanding against environment!
•  But if the parcel undergoes contraction (a2 < a1)
• Environment is doing work on the parcel!
Day 1/2
• A derivation more relevant to an air parcel
•  Given a spherical parcel of air with radius R, Surface Area = 4pr2 and the volume is 4/3pr3.
Day 2
•  We calculate dW in expanding the parcel from r to r + dr in an environment where the pressure is p!
• Then,
• Work = F x dist = Pressure x Area x distance
Day 2
• Well,
• Dist = dr
• And;
• Area = A + dA
Day 2
• As with the “expanding piston”
• DW = p*A*dr = p * dV
•  Recall from Calc II (or is it physics?)
• dV = 4p r2 dr
Day 2
• Then we’ll deevide by the unit mass again to get:
• dw = p da
• Now the short comings in this problem are obvious:
• 1. (Surf)Area = A + dA = 4 p r2 + 8p r dr
Day 2
• 2. The parcel expands and p decreases;
• so: p = p + dp
• So….. the REAL problem is:
• dW = (p+dp)(A+dA) dr
Day 2
•  Which “foils” out to,
• then assume;
• Dw = p da + dp da + p dA dr + dp dA dr
Day 2
• Which then reduces to (by scale analysis):
• Dw = p da
Day 2
• The First Law of Thermodynamics: A derivation.
•  The First Law of Thermodynamics, is really just a statement of the principle of the conservation of energy.
•  This law can be used as a predictive tool: ie, we can calculate changes in T, a (r), or pressure for a parcel.
Day 2
•  The first law of thermodynamics is associated with changes in these quantities whereas the equation of state relates the quantities themselves, at a given place at a given time (or at a given place for a steady state).
• Forms of energy relevant to our treatment of the 1st Law
• “Energy” is (simply) the capacity to do work!
• for our purposes, a system is an air parcel.
Day 2
• KE (kinetic energy)  associated with motion, energy of motion
• (Definition: “Kinetic” means the study of motion w/o regard to the forces that cause it)
Day 2
• PE (potential energy)  is the energy of position relevant to some reference level, or within some potential (e.g., gravitational) field.
Day 2
• **The location of Z = 0 is arbitrary, but it makes good sense to use sea level.
• Internal energy (IE)  Is the energy (KE and PE) associated with the individual molecules in the parcel.
• **So IE is temperature dependent (use Absolute T)!!!
Day 2
• Deriving the First Law: (Hess pp 25 – 26)
• Statement of Conservation of Energy: “Energy of all sorts can neither be created nor destroyed” (Of course we ignore relativistic theory E = mc2)
• Written in incremental form (here it is!):
Day 2/3
• The “book definition”:
• “Any increment of energy added to a system is equal to the algebraic sum of the increments in organized KE, PE, IE (thermal), work done by the system on it’s surroundings, and whatever forms of electrical and magnetic energy may appear”
• Dheat = DKE + DPE + DIE + DWork + DE&M
Day 2/3
•  We can neglect electrical and magnetic for atmospheric motions we consider (scale analysis).
• ** Recall: our system is the air parcel!
•  Consider the LHS; “increment of energy added to a system” can be broken down into “heat added to the system” + work done on the system by the surroundings”.
• We want to simplify, if possible to quantify our physical principle (the conservation of energy).
Day 3
•  Kinetic Energy: We first note that increments of Kinetic energy are small compared to changes in internal energy and increments of work done. For example, consider a parcel of air changing speeds from 10 m/s to 50 m/s
• DKE = ½ (50 m/s)2 – ½ (10 m/s)2 or 1.2 x 103 m2 sec -2
Day 3
• Next, let us consider a change of 20 C for a parcel of air at a constant volume, a change of internal energy:
• DIE = Cv DT = 717 J K-1 kg-1 x 20 K
• or 1.4 x 104 m2 sec-2
Day 3
• Then, let’s consider the change in PE at, oh, say 500 hPa,
• So,
• DPE = g*Dz = 10 m s-2 * 60 m = 600 m2 s-2
Day 3
• Pressure work;
• We have dw = p da and pa = RT
• so dW = (RT/a) da = RT d[ln(a)]
• Since T = Const, we can write:
• = ln 0.1 * 287.04 J/kg K * 280 K
• W = 8.47 x 103 m2 s-2
Day 3
• So, we’ve shown that:
• Dh/dt = IE + PE + KE - Work done on the system by the surroundings + work done by the system on the surroundings+ EE + ME
•  We neglected EE and ME by scale analysis, and we can neglect KE and PE by scale analysis (or do we)?
Day 3
• Well,
• DKE = DKE(horiz) + DKE (vert)
• ***KE vertical is very tiny, by several orders of magnitude (at least 4)!
• Work done on the system by surroundings (W):
• -(Work horiz + Wvert)
Day 3
•  The Horizontal pressure gradient is responsible for changes in wind speed.
• The pressure gradient is in DKE and Wh (which is horizontal wind times distance covered)
• Then, DKE must cancel with Wh!
Day 3
•  Well, DPE = gdz
• and,
• Wv = vert. PGF x distance = - (1/r) (dp/dz) (dz)
• If hydrostatic balance is assumed to hold then g = -(1/r) (dp/dz).
Day 3
• Thus, DPE + Wv = gdz – g dz = 0 (and these cancel)!
•  So in the incremental form of the first law we are left with:
• Dh/Dt = IE + work done by the system on the surroundings (1)
• **(note, the “heavy D”)
Day 3/4/5
• Then if we
• define heat added as dh/dt (dq/dt) or just dh or dq
• Internal energy as “du”,
• and work done by the system as “dw”, then:
• We can quantify first law (this is it)!  ?
• Dh = du + Dw or du = dh - dw
Day 5
• The First Law of Thermodynamics and Entropy.
• What happens when heat is added to parcel?
• 1st Law: dh = du + dw
•  Heat can be added (or subtracted, of course)! How?
• Conduction, convection, radiation, or phase change (All “Diabatic” heating)
Day 5
• Diabatic processes:
• Sensible heating  warm to cold by contact
• Radiative heating  “Short wave in” or “long wave out”
• Latent Heating  phase changes of water mass
Day 5
•  So we can say formally “if a small quantity of heat* is added to the system (parcel) some goes into increasing the int. energy of the parcel, and some gets expended as the parcel does work (dw) on the surroundings”.
• *(Note: heat is NOT temperature!)
•  We know the 1st law, however, how do we express in terms of state thermodynamic variables (T, a, or P?) so we can obtain useful expressions for atmospheric calculations?
Day 5
•  Unfortunately, there are no formal, or analytical, mathematical expressions for dh (especially for atmospheric processes – these typically “parameterized” (which is an empirical “fudge factor”)).
•  Thus, we are reduced to solving as a residual, and in the 1st Law leave as dh/dt or dq/dt (Q-dot).
• Well, we showed that the work done by an expanding gas could be derived by considering a piston (closed controlled system), or an air parcel.,
Day 5

Thus, dw = p da.

So the 1st Law is now:

Day 5
• Internal Energy
•  With the pressure work term expressed in terms of state variables p and a, we must now consider that internal energy may be dependent on (T and a) or (T and p) (since p and a can be related)
• So: Consider that u = u(T,a) or u = u(T,p)
Day 5
• The expression
Day 5
• Joule’s Law; and his experiment conceptualize this term
• Consider an insulated container with two chambers, one filled with Gas at pressure P = Po, and the other a vaccuum at P = 0. (connected by valve)
• Insulated chamber, thus Q dot = 0. (No surroundings also, so dw = 0).
Day 5
• Draw:
Day 5
• Then the valve is opened and gas expands into vacuum chamber (let ‘er rip!) P = 0. (No temp change occurs), then from first law:
• Hmm, Dq = du + dw
• Well:
• 1. Dq = 0 (no heat added)
• 2. dw = 0 (since chamber cannot expand)
Day 5
• Thus, du must be 0!! 
• So,
Day 5
• Thus:
• Buuut, Joules experiment also showed that dT = 0 soooo,
Day 5
•  This (celebrated) result is known as Joule’s law, which is strictly true for an Ideal gas.
• Thus,
Day 5/6
•  Thus, u = u(T) internal energy is a function of T only!!!
• Q: So what did Joule’s experiment show?
• A: Joule’s experiment, separate from Joules’ law showed that heat and mechanical energy are two forms of the same thing.
Day 6
• Next, let’s discuss the concept of specific heat or specific heat capacity:
• Suppose we add small amount of heat (dq)
• Temp changes from T to T + DT
• Then, if no phase change occurs, the ratio of:
• dq/dT = Constant
Day 6
• This is the definition of specific heat (which is unique to each substance or gas).
• Add lots o’ heat and small dT
• then dq/dT = some big number
• Example: Metal – heat is dispersed/conducted readily throughout the substance. (add little heat get larger dT) small heat capacity, small Cp
Day 6
• Wood – large heat small dT inside (specific heat large)
• Then a:
• Conductor  small specific heat
• Insulator  large specific heat
Day 6
• We are interested in the behavior of gasses.
• The specific heat of a gas:
• dq/dT = Constant
Day 6
• will have different values depending on what happens to p or a as gas receives heat, thus:
• Cv = (specific heat at constant Volume) =
• 717.59 J kg-1 K-1
• Cp = (specific heat at constant pressure) =
• 1004.63 J kg-1 K-1
• (learn ‘em, live ‘em, love ‘em)
Day 6
• Important Points:
•  Thus, for Cv gas cannot expand as heat is added (constant Volume), but Pressure can increase
•  Thus, for Cp pressure is kept constant as heat is added, but gas is allowed to expand (a increase)
Day 6
• But, as we saw…..
• Cp > Cv for any given gas (why?)
• A: since as heat is added and temperature rises, but pressure is kept constant, then some heat added will have to be expended to do the work, as material expands against the constant pressure environment!
Day 6
• In plain English: A larger quantity of heat must be added to the same amount of gas at constant P (press work term gets some), than if the volume kept constant (no pressure work term).
• ***Specific heats are important since for ideal gas they are constants.
Day 6
• If constant volume (press work term =0)
• Dq = du
• if u = u(T), then,
• Cv = Dq/dT = CvdT = Dq (substitute upstairs).
Day 6
• Or….
• Du = CvdT
• u = CvT + Const. (having applied the snake)
• Now the celebrated result! ……
Day 6
• The 1st Law……
• (1st law [pure partitioned form] is in terms of P, a, and T!!!)
• What about const pressure process?
Day 6
• Ugh! Let’s see, there’s “p” above, so maybe:
• (Q: where’s the other RHS term?)
Day 6
• Insert into 1st law and:
•  If constant P process,
• then,…….. dp = 0
• and,
• dq = (Cv + R) dT
Day 6/7
• or,
• dq/DT = Cv + R (which is definition of Cp!)
• So,
• Cp > Cv as expected (as reasoned out before!).
• 717.59 + 287.04 = 1004.63.
Day 7
• Adiabatic process (Dq/Dt = 0)
• Recall “Adiabatic” means that no heat is added or removed to a system (parcel) (from the surrounding air).
• Thus, “Diabatic” processes describe heat input or extraction. This heat must be distributed (apportioned) between internal energy and pressure work!
Day 7
• If a “parcel” is in motion, we call this adiabatic motion. This says absolutely nothing about how the temperature of the parcel changes recall these are different concepts.
• Important!!! Dq/Dt = 0 does not mean dT/dt = 0 (adiabatic and isothermal processes are not the same!)
• Many atmospheric motions can approximate adiabatic motion for a few minutes, up to one day, as long as we’re away from precipitation areas.
Day 7
• Parcel theory
• In atmosphere: molecular heat transfer important only for first few centimeters of air closest to the ground. Vertical mixing is mainly the result of bulk tranfer by “air parcels” of varying sizes (depending on the scale we’re talking about).
• So, let’s consider a parcel that’s of infinitesimal size! (Compared to whole atms)
Day 7
• 1) parcel is thermally insulated from it’s environment, so that temp changes adiabatically as it rises or sinks.
• 2) The parcel is at exactly the same pressure as it’s environment (at same level), Pparcel = Penv
• 3) Environment is in hydrostatic balance
• 4) Moving slowly enough such that the KE is small.
Day 7
• **For real air parcels one or more of these assumptions are likely to be in violation, however with this simple conceptual model, it is easier to understand some of the physical processes that influence vertical motions and mixing in the atmosphere.
Day 7
• Relate 1st law to air movement in vertical
•  If we know that p = p(x,y,z,t) and dp/dt = w
• Also, for synoptic scale features V >>> w
• such that dp/dt = -rg dz/dt
Day 7
• So for synoptic features in hydrostatic balance:
Day 7/8
• Assuming this we can write first law in the following form:
•  This is another form of the 1st law valid for synoptic – scale motions where
• M = St = CpT + gz = CpT + F
Day 7/8
• St is called the “dry static energy” (Internal + Potential energy).
• F Is geopotential (m2 s-2)
• CpT  is enthalpy or sensible heating
• This is a form of the first law that relates temperature change to height change!
Day 8
• In q coordinates this is called the Montgomery Streamfunction (M) (and we shall talk about this more in 4320/7320)
• Stream function defines air motions assuming a balance condition, thus these are streamfunctions for adiabatic atmosphere. “M” is used in PV framework!
Day 8
• The dry adiabatic process change rate and the dry adiabatic lapse rate!
• Consider the 1st law written in the following form:
Day 8
• Thus for an air parcel dry static energy is conserved!
• And, (if we invoke “il serpente”)
• CpT + F = Constant
• Now, Let us consider a parcel of air undergoing adiabatic motion, how do we arrive at the dry adiabatic lapse rate?
Day 8
• Since:
• F =g dz/dt,
• And,
• CpdT/dt + gdz/dt = 0,
Day 8
• Which means that,
• CpdT/dt = -gdz/dt
• Thus we get
• dT/dz = -g/Cp = Gd
• (the dry adiabatic lapse rate! – which is valid in any planetary atmosphere!)
Day 8
• What assumptions did I make in deriving Gd?
• hydrostatic balance, and
• air is undergoing adiabatic vertical motion.
• Thus as a parcel rises, temp changes by a constant rate g/Cp which is approximately 10 K/km
Day 8
• Or: 9.81 / 1004.63 which is 9.76 C(K) / km. (T decreases/increases as parcel rises/sinks)
• 5.4 F / 1000 ft (good question for test or HW!)
• Recall, for adiabatic process, no heat is gained or lost, but if we move up 1 km, temp certainly does change!
Day 8
• The notation dT/dZ refers to vertical motion following a particular parcel of Air!!
• ***Recall that in Atmospheric Science dT/Dz is minus, but it’s customary to refer to a drop in T with height as positive, so 10 C/ km represents a 10 degree drop with 1 km height.
Day 8
• This compares with the observed Lapse rate of 6 to 7 degrees C / km. Thus parcels moving upward at the dry adiabatic lapse rate quickly become colder than their surroundings.
• Q: What does this say about the Atmosphere?
• A: So if the atmosphere is dry much of the time, then much of the time, the troposphere is inherently stable!
Day 8
• Now suppose the air parcel contains water vapor:
• a. Water vapor will condense as parcel rises (main cloud producing mechanism).
• b. Water vapor condenses releasing latent heat to the parcel. (Adding heat not same as changing temperature)
Day 8
• 1st Law:
• So when heat is added to the parcel, some goes into increasing the temperature, and some into changing the pressure.
• Q: So what happens if water vapor begins to condense within the rising air parcel above?
• A: The latent heat of condensation added to parcel, (counteracting the adiabatic cooling) and the rate of temp decrease will be something less than 9.76 C/km
Day 9
• Modifying the 1st Law  Entropy and Potential Temperature
• (we need to show that differentials on RHS are exact)
• Suppose an air parcel undergoes some process or change, which we can show as a curve form an initial state A to final state B on a thermodynamic diagram.
Day 9
• **Recall thermodynamic diagram does not show the actual Physical Path in Time or Space (it’s a point location!)
• Thus, the process is reversible! (In higher level classes, reversible will be referred to as a “Newtonian” one)
• Irreverisble process  “Hamiltonian” or sensitively dependent on intial conditions, “chaotic”.
• We want to evaluate the loss or gain of heat (dq – recall no analytic expression) as parcel goes from A to B on the diagram.
Day 9
• We write first law:
• Change in heat during the entire “lifting” or “sinking” process (finite process), is (invoking the snake):
Day 9
• Thus we go from point at (T at this point and P, both known!) to point B, where we know the P, but NOT T! We will predict this.
• When an integral of a quantity along any curve depends only on its value at the endpoints of the curve, like:
• Recall: that this is the definition of an exact differential, hence dT/dt is an exact differential!
Day 9
• However, pda/dt depends on the history and “initial conditions” of the parcel, thus it is path dependent (not an exact differential).
• The parcel could go through a phase change of a gas, etc. which also changes the very makeup of the parcel. Also, T functionally depends on P and a). The path from A to B makes a big difference!
Day 9
• These are irreversible processes!
• Mathematically speaking, this integral is “improper”. We can’t evaluate it. (Bummer! )
• Important! ***Thus, the adiabatic problem is really a problem in prediction! Recall, we said we could use the 1st Law in a predictive capacity)
• So, let’s revisit problem just discussed. It’s not an “esoteric” problem, but a real problem, i.e. for numerical modeling and weather forecasting.
Day 9
• OK, Suppose we know the state of a parcel at start (A):
• So we know:
• At point A  P = Pa and T = Ta for the air parcel.
• We want to predict what T will be in (for example 12-hr), so we want to calculate Tb
Day 9
• Suppose we can deduce where (B) will be approximately in 12-hr by some model forecast of Pb.
• At point B  P = Pb and T = ????
• Thus, our problem is:
• Given: Ta and Pa and Pb
• (Forecast) get: Tb.
Day 9
• We have to ass\u\me:
• Parcel undergoes an adiabatic and reversible process from A to B.
• Use unmodified 1st law: (to get:)
Day 9
• Whoa!
Day 9/10
• **This says there’s a tradeoff between internal energy and pressure work!
• To evaluate Tb we need to know path on diagram from A to B, we know Pa and have a prediction for Pb, but what is the path???? We can’t solve the problem! Double Bummer!! 
• Let’s try alternate form Cp form of 1st law!
Day 9/10
• Q: will this help?
• A: NO! Doesn’t help since (again adiabatic (dq =0))
• So we get:
Day 10
• Whoa again!
Day 10
• Still don’t know the integral’s path!!! Arrrgh!!!!
• Modification of 1st Law into “perfect” differential form
• The 1st Law:
• We rewrite the 1st law in terms of a quantity called “change of specific entropy”
Day 10
• We rewrite the 1st law in terms of a quantity called “change of specific entropy”
• Ds/Dt = dq/dt x (1/T)
• So we need to deevide 1st Law by T:
Day 10
• RHS: P/T ??? Why that’s just R/a!!!
• And:
Day 10
• Now we have perfect differential form!
• Works with other form as well:
Day 10
• Important!!!
• So now you see that dq (change in heat) is not an exact differential, however, change in specific entropy is exact!!!
• All we did was divide by T, (to eliminate T dependence in the pressure work term.)
Day 10
• How does this help? Unmodified form, we could not do integral since we did not know path from A to B.
• Modified form of first law to get Tb
• Change in specific entropy:
Day 10
• Here it is?
• The “improper” integral is gone!! We can assume adiabatic and continue on!!!
Day 10
• Use ideal gas law to get rid of a.
Day 10
• ???
• Hint: What’s this look like?
• So we get to with our prediction problem:
• Tb = Ta(Pb/Pa)k
Day 10/11
• The resolution of our “problem in prediction”
• So the change in specific entropy was calculated from knowing T and a at the start and end of the process, but NOT knowing the history of the process, or the curve itself.
• Important!!! In our case: ds = dq = 0, so the process was reversible!!! (Able to be displayed as a curve on the diagram). Reversible No phase changes!!!
Day 11
• Again:
• Reversible (Newtonian processes) path can go both ways, final state not dependent on specification of initial conditions. (adiabatic process) potential temp graphically
• Irreversible (Hamiltonian processes), path dependent, final state highly dependent on the specification of initial conditions. (Diabatic process) equivalent potential temp graphically
Day 11
• This is the whole crux of numerical weather prediction, and the whole concept behind ensemble forecasting. (Take intro. to Chaos theory).
• An asside: If you have an adiabatic and irreversible process, then the entropy can increase OR:
• Ds > dq/T This is the second law of thermodynamics.
• So the power of rewriting the first law in exact form is that knowing the initial state, we can get to the final state by eliminating our dependence on path (Makes the modeler’s job much simpler).
Day 11
• Ok, now suppose we have state A, where we have Ta and Pa and we end up at state B where Tb = Tref, and Pb = 1000 hPa, we have
• T(1000 hPa) = T (1000 hPa/p)k
• **This is our definition of potential temperature! Bring the parcel of air adiabatically (and reversibly) from it’s state down to the reference state.
Day 11
• ***This is the true derivation! Even though you can take a shortcut, via the first law in inexact form, we escape the nasty consequences of improper integrals from the fact that ds = dq = 0.
• Adiabatic process = isentropic process (all defined for a dry atmosphere, however, for an unsaturated atmosphere the errors are small even if there is H2O vapor, T is close enough that we don’t have to be accurate)
Day 11
• Now prove that isentropic process means constant potential temperature:
• (In meteorology there is always more than one way to “skin a cat”)
• We want to show dT/DZ = -g/Cp
• We did this using the dry static energy relationship last time:
• CpT + F = M
Day 11
• Now potential temperature equation:
• q = T(Po/P)k
• Take the natural log:
Day 11
• Then take d/dt of this expression.
• Q: What happens to middle term on RHS?
• A: It disappears! Why?
Day 11
• After a bit ‘o algebra:
• dT / dp = a /Cp
• This is the adiabatic lapse rate in (x,y,p) coordiates.
Day 11
• I can take hydrostatic balance relationship:
• dp / dz = -rg
• and we can apply the chain rule! (Can you see it?) a and r are recipricols of each other.
• dT/dz = - g /Cp
Day 11
• Thus for an isentropic process: theta is conserved! (Any process following dry adiabat)!
• The potential temperture relationship is extremely useful, and it’s a very powerful way to examine atmospheric processes (in isentropic coordinates and assuming adiabatic). Since most processes on synoptic scale are close to isentropic.
Day 11
• The first law and entropy gives us a powerful compliment to eqn. of state, and hydrostatic approx. This is the whole foundation of “PV thinking”
• In fact Dr. Rainer Bleck (U Miami) suggested at a meeting that if meteorologists scrap everything we’ve been taught and just learn isentropic and PV analysis, we’d know all we need to know about the atmosphere.
Day 11
• To show that the first law and eqn of state are complimentary, the eqn. Of state can take on “simplified” forms under the assumption of adiabatic, and hydrostatically balanced flows:
• They are:
• TP-k = const. Pa(Cp/Cv) = constant aT(Cv/R) = constant
Day 11
• The first coming from Tp-k = qPo-k = constant.
• But remember, there’s several ways to get each!
• Bonus: Show mathematically that you can convert mixing ratio to vapor pressure by following a T line to 622 hPa!
Day 12
• Bluestein pp 195 – 200 on dry thermodynamics
• Given the first Law in form:
Day 12
• The equations
• (1) (2) (3) (4)
Day 12
• where a/Cp is the dry adiabatic lapse rate Gd, and the partial of T w/r/t to p is, of course the environmental lapse rate Ge.
• Recall: We’ve discussed using this form to estimate “vertical motion”:
Day 12
•  We can assume adiabatic conditions again, although this expression would also apply to a non-adiabatic atmosphere as well):
• Thus the local rate of change of temperature is due to:
Day 12
• The Equation:
• A B C
• Vertical temperature advection (now in term B)
• Adiabatic temperature changes due to vertical motion and atmospheric stability (term B)
• Diabatic heating term C.
Day 12
• Now let’s get a closer look at: Static stability (term B):
• and let’s use the relationship for potential temperature:
Day 12
• “logrithmic differentiate” (and a bit o’ algebra):
Day 12
• We get a relationship for static stability, which we name “S”
• This is static stability (e.g, Zwack and Okossi, 1986; Lupo et al., 1992)! The difference between the dry adiabatic and environmental lapse rate! Don’t we calculate this graphically?
Day 12
• ***Also: static stability defined as “sigma” s
• You’ll find this form of the First Law in Bluestein (1992):
Day 12
• If the atmosphere is “dry-neutral” then obviously:
• S = 0  Since Ge = Gd a/Cp = dT/dp
• Let’s take a look at Static stability (S) vs. Vertical motion (w)
Day 12
• To do this, let’s isolate these two variables, so assume that:
• The local rate of change in T is constant (C).
• The temperature advection is zero.
• The diabatic heating is zero.
• Q: Is this unreasonable?
• A: maybe, maybe not.
Day 12
• Then the First Law becomes:
• C = Sww = C/S
• Important Definition: Static Stability “S” can be defined as the “resistivity” of the atmosphere to vertical overturning (motion)!
Day 12
•  How to interpret: for the same amount of heating, a larger (smaller) stability (more[less] stable air) resists vertical motion and produces a smaller (larger) w.
• So, in the case of large static stability, expansion and compression of air overwhelm, vertical temperature advection, or the vertical advection may act in the same sense as T (cool air under warm)!
Day 12
• Q: Where in the atmosphere (Homosphere) is S very large and consequently, vertical motions are very small?
• We have looked at the 1st Law in (x,y,p) coordinates.
Day 12
• In (x,y, q) coords. It’s:
• A B C
• Where term;
• A - is the advection of potential Temp.
• B - is the vertical advection of potential temp (stability term)
• C - Is the diabatic heating term
Day 12
• This version is valid for synoptic scale process only. On smaller scales, on scale of convection, where hydrostatic balance does not hold, we must use (x,y,z) thermo dynamic equation.
Day 13
• Baroclinic atmosphere vs. Barotropic atmosphere.
• An atmosphere in which density is a function of pressure and temperature is called a baroclinic atmosphere. P (mass) and r (thermal) fields cross to form solenoids:
Day 13
• …soleniods (avoid the ‘noid?)
• (Gradient of density and pressure not parallel).
Day 13
• This is where all the work is done in the atmosphere, and this is where vorticity is generated!
•  This is where Available potential energy is converted to Kinetic by cyclones and anticyclones! (Midlatitudes)
Day 13
• Some math…
Day 13
• An atmosphere in which density is a function of pressure only is a barotropic atmosphere.
• A Barotropic atmosphere is isothermal, thus there is NO advection of temperature. There are no vertical wind shears and NO solenoids (avoid the ‘noid!).
Day 13
• (Gradient of density and pressure parallel) (Tropics may nearly mimic at times)
Day 13
• Absolute vorticity conserved! No Available Potential Energy. A barotropic atmosphere is a steady state, basic state atmosphere.
• Equivalent Barotropic  isotherms are parallel to the pressure lines. There are horizontal temperature advections.
Day 13
•  However there is vertical shear. (thus density and pressure gradients nearly, but not quite parallel.
• Some examples of this type of atmosphere:
• The Tropics
• cutoff lows
• Blocking anticyclones
• We’ll talk more about these ideas in dynamics.
Day 13
• The thermodynamics of Moist air
• (Read Bluestein 200 – 223, and Hess ch 4 and 5)
• The equation of state for an atmosphere of water vapor only (A “water world?” – sorry Kevin Costner):
• Pv = R*/mvrvTv
Day 13
• Let’s call vapor pressure Pv  e
• and Rv is 
• R*/mv = 8314.3 J/K kg / 18.016 = 461.5 J/K kg
Day 13
• Thus, the equation of state is: e av=RvT or e = rvRvT
•  This is the equation of state for water vapor itself or as a constituent of moist air.
• Next, consider moist air + dry air, and now the parcel is saturated or e= es
Day 13
• Then this vapor equation is:
• esav=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 13
• The concept of equilibrium vapor pressure over a plane of pure water (does the atmosphere “hold” water vapor?):
Day 13
• The Variation of es (es over water and es over ice – or “on the rocks”) with temperature:
Day 13
• Graph here: