Overview of thermodynamics

Overview of thermodynamics
Chap. 1 Overview of thermodynamics

What is a simple definition of the laws of thermodynamics?
Answer Thermodynamics is the study of the inter-relation between heat, work and internal energy of a system.

Thermodynamics defined
is an empirical science based on a small number of principles is concerned only with macroscopic or large-scale properties of matter provides general relations between quantities such as coefficients of expansion, compressibilities, specific heat capacities, heats of transformation, and magnetic and dielectric coefficients, and their dependence on temperature. indicates which relations must be determined experimentally in order to completely specify the system properties is complementary to kinetic theory and statistical thermodynamics. provides relationships between physical properties of a system

Areas within physics: Where does thermodynamics fit in?
Mechanics Thermodynamics Vibrations and wave phenomena Optics Electromagnetism Relativity Quantum mechanics Source: High School physics text

Areas within Atmospheric Science
Atmospheric dynamics Atmospheric thermodynamics Severe storms Radiation Cloud physics and dynamics Physical climatology General circulation Synoptic meteorology Tropical meteorology Mesoscale meteorology Boundary layer meteor. Remote sensing Hydrology Climate dynamics But there is more:

Thermodynamics deals with energy and the transformations of the nature of energy three fundamental laws (principles): the equation of state (ideal gas law), the first law of thermodynamics (conservation of energy), the second law of thermodynamics (degradation of energy, i.e., entropy increases). “Well then, thermodynamics is simple!”

statistical thermodynamics
Strives to deduce the average properties of a large number of molecules

Thermodynamics and dynamics: Some end results
Dynamics/kinematics (momentum equations for u, v; continuity equation) Thermodynamics (equation of state, hydrostatic equation, thermodynamic equation or First Law): Equations are complete sentences. Equations convey important information, like any other language

Cloud physics An equation set used to simulate ice fog, thin stratus, etc. Aerosols Ice crystals Water droplets Specific humidity Temperature Saturation ratio

Some definitions Energy: This is almost impossible to define precisely. It is not the "ability to do work" unless some conditions are carefully defined. Energy is a property and is defined as such by the First Law .of Thermodynamics

System System: A system is some portion of the universe selected for study. Strictly speaking a system may, or may not, contain matter or energy. For atmospheric applications, systems will always contain both matter and energy. The remainder of the universe is defined as the "surroundings" of the system. Systems may be classified as isolated, closed or open. Isolated system: A system which cannot exchange either matter or energy with its surroundings. The universe is defined by the First Law as an isolated system. Closed system: A system which can exchange energy, but not matter, with its surroundings. Open system: A system which can exchange both matter and energy with its surroundings. The systems in Atmospheric Thermodynamics will be “parcels” of air undergoing transformations in the atmosphere. Thus, these are really open systems, but for the sake of simplicity, they will often be treated as closed systems.

Parcel We will often use the concept of a parcel (which is a closed system) in our discussions of atmospheric thermodynamic processes. A parcel is often defined as a volume of air which does not interact with its environment, i.e., (a) no compensating motions occur in the environment as a consequence of the parcel motions (b) the parcel does not exchange mass with the environment. A parcel can do work on the environment. A parcel is therefore an example of a closed system.

Property Property: A property is an observable. It results from a physical or chemical measurement. A system is characterized by a set of properties. Typical examples are mass, volume, temperature, pressure, moisture content, chemical composition, energy, etc. Properties may be classified as intensive or extensive. Intensive property: A property that characterizes the system as a whole, and is not given as the sum of the property for portions of the system. Pressure, temperature, density, water vapor mixing ratio, and specific heat are examples of intensive properties. Extensive property: A property which is dependent on the extent or physical size of the system, and which is given by the sum of the extensive properties of portions of the system. Examples are mass, volume and energy. [We will avoid this to a large extent by defining specific volume and specific energy, which have units per unit volume or mass, respectively.]

State of a system The state of a system is defined by a set of system properties. A certain minimum number of such properties are required to specify the state. Thus in an ideal gas, any three of the four properties -- pressure, volume, number of moles, and temperature -- define the state of the system. The equation of state defines the relation among these properties.

Process Process: A description in the manner in which a change in state occurs. Four examples are: isothermal process: a change in state occurring at constant temperature; adiabatic process: a change in state involving no transfer of energy between the system and its surroundings; isochoric process: a change in state occurring at constant volume; cyclic process: a change occurring when the system (but not necessarily its surroundings) is returned to its initial state after one or more processes.

Equilibrium A necessary, but insufficient, condition for a system to be in equilibrium is that the system state must be constant with time. In addition, if the system is perturbed by a small amount, then in an equilibrium state the system would retain its initial properties. We will later further distinguish among, stable, unstable and metastable equilibria. Types of equilibria: Physical, dynamical, chemical

Scale and units We will generally adhere to the SI (System Internationale) or MKS (meter/kilogram/second) units, but for convenience we will depart from MKS at times. (After all, we can’t avoid using calories in a course like this.) The appropriate temperature scale will be either Celsius or Kelvin Pressure will be expressed in both mb and Pa (or hPa which is numerically identical to mb).

Aside: What quantities are used to describe physical systems?
Base quantity Name Symbol Length meter m Mass kilogram kg Time second s Electric current ampere A Temperature kelvin K Amount of substance mole mol Luminous intensity candela cd Source:

Force newton (N)  N = kg m s-2
Quantity Unit Length m (or cm, mm, mm) Mass kg (g) Acceleration m s-2 Density kg m-3 Force newton (N)  N = kg m s-2 Temperature Kelvin (K) or Celsius (C) K = C F = 1.8 C + 32 Pressure pascal (Pa) [1 mb = 103 dyne cm-2 = 102 Pa] Pressure is force per unit area, so the unit of Pa is Pa = N m-2 = kg m-1 s-2 (1 mb = 102 Pa = 1 hPa) Also:1 atm = mb = kPa = PSI (lb in-2) = 760 mm Hg (29.92 in) Energy joule (J)  J = N m = kg m2 s-2 [1 cal = J]

calorie calorie or the gram-calorie (cal). This is defined as the heat necessary to raise by 1 K the temperature of one gram of water at a temperature of 15 C. In terms of the MKS units, 1 cal = J. The international calorie was introduced in engineering, in relation to characteristics of water. The thermochemical calorie was formulated by physical chemists and is defined as above. We will at times use the IT cal in this course.

kilogram molecular weight
kilogram molecular weight, or kilomole (kmol) is defined as the molecular weight (M) of a material in kg. Thus, one kmol of water is kg. The number of kmol n in a mass m is given by n = m/M. The number of molecules in one kmol of any material is constant and equal to 103 x Avogadro's number (6.022x1023).

Recommendation Refer to this material frequently throughout the course.

Have any of you been following the developments in redefining planets in our solar system?
Size criterion: 500 mi diameter (this would have sufficient mass to force the planet into a quasi-spherical shape) What about an atmosphere criterion? This implies a mass large enough to hold gas molecules (relation to escape velocity)

Atmospheric Composition
We will consider the atmosphere to be composed of: a mixture of gases as defined in Tables 1 and 2; water substance in the gaseous, liquid, or ice phase; and solid/liquid (aerosol) particles of very small size (10-7 to 10-4 m).

Table 1. 1 Main gas components of the dry atmosphere. (U. S
Table 1.1 Main gas components of the dry atmosphere (U.S. Standard Atmosphere, 1976) Gas Molecular Weight Molar (or volume) fraction Mass (mi) Specific gas constant (Ri, J kg-1 K-1) miRi/m N2 28.013 0.7552 296.80 224.15 O2 31.999 0.2315 259.83 60.15 Ar 39.948 0.0128 208.13 2.66 CO2 44.010 (variable) 0.0005 188.92 0.09 H2O 18.016 0 – 0.07 (highly variable) Sum Avg = Sum = Sum = 287.05

Table 1. 2 Minor gas components of the dry atmosphere (U. S
Table 1.2 Minor gas components of the dry atmosphere (U.S. Standard Atmosphere, 1976) Gas Abbreviation Molar (volume) fraction Neon Ne 18.18 x 10-6 Helium He 5.24 x 10-6 (?) Methane CH4 2 x 10-6 Krypton Kr 1.14 x 10-6 Nitrous Oxide N2O 2.5 x 10-7 Hydrogen H2 0.5 x 10-6 Xenon Xe 0.087 x 10-6 Ozone O3 0 – 10-4 Sulfur dioxide SO2 variable Nitrogen dioxide NO2 0-2 x 10-6 Carbon monoxide CO

Vertical temperature profile for the U. S. Standard Atmosphere
Vertical temperature profile for the U.S. Standard Atmosphere. From Wallace and Hobbs (1977).

Vertical profile of pressure in mb (dashed), density in g m-3 (solid) and mean free path in m (dot-dashed) for the U.S. Extension to International Civil Aviation Organization (ICAO) Standard Atmospere. p(z) = p0 exp(-z/H) r(z) = r0 exp(-z/H) H is the atmospheric scale height (H ~7-8 km) Mean free path r(z) p(z)

Global mean pressure (bold), temperature (shaded), mean molar weight (solid) and number densities of atmospheric constituents, as functions of altitude. Source: U.S. Standard Atmosphere. Taken from Salby (1996).

A standard atmosphere calculator
relationships among height, pressure, density, temperature, and the speed of sound for a dry atmosphere

Pictures that tell (or imply) a “story” (about thermodynamics)
Example pictures Atmospheric thermodynamics in everyday living

Thermodynamics?

Hurricane Dean, 0245 UTC 21 Aug

Hurricane Dean, 1115 UTC 21 Aug

000 WTNT44 KNHC TCDAT4 HURRICANE DEAN DISCUSSION NUMBER 33 NWS TPC/NATIONAL HURRICANE CENTER MIAMI FL AL AM EDT TUE AUG DEAN MADE LANDFALL ON THE EAST COAST OF THE YUCATAN PENINSULA NEAR THE CRUISE SHIP PORT OF COSTA MAYA AROUND 0830 UTC...AND THE EYE IS NOW JUST INLAND. OBSERVATIONS FROM AN AIR FORCE HURRICANE HUNTER PLANE INDICATE THAT THE HURRICANE WAS INTENSIFYING RIGHT UP TO LANDFALL. A PEAK FLIGHT-LEVEL WIND OF 165 KT WAS MEASURED JUST NORTH OF THE EYE. MAXIMUM SURFACE WINDS FROM THE SFMR WERE 124 KT...BUT IT IS HIGHLY LIKELY THAT THE MAXIMUM SURFACE WIND SPEED WAS NOT REPORTED BY THE SFMR INSTRUMENT. A GPS DROPSONDE IN THE NORTHERN EYEWALL MEASURED A WIND SPEED OF 178 KT AVERAGED OVER THE LOWEST 150 METERS OF THE SOUNDING. BASED ON THE DROPSONDE AND THE FLIGHT-LEVEL WINDS...THE INTENSITY IS SET AT 145 KT. A DROPSONDE IN THE EYE MEASURED A CENTRAL PRESSURE OF 906 MB JUST PRIOR TO LANDFALL.

SOME HISTORIC NOTES ARE IN ORDER HERE
SOME HISTORIC NOTES ARE IN ORDER HERE. THE 906 MB CENTRAL PRESSURE IS THE NINTH LOWEST ON RECORD FOR AN ATLANTIC BASIN HURRICANE...AND THE THIRD LOWEST AT LANDFALL BEHIND THE 1935 LABOR DAY HURRICANE IN THE FLORIDA KEYS AND HURRICANE GILBERT OF 1988 IN CANCUN MEXICO. DEAN IS ALSO THE FIRST CATEGORY FIVE HURRICANE TO MAKE LANDFALL IN THE ATLANTIC BASIN SINCE ANDREW OF DEAN WILL WEAKEN AS IT TRAVERSES THE YUCATAN PENINSULA AND THE AMOUNT OF WEAKENING WILL DEPEND ON HOW LONG THE CENTER REMAINS OVER LAND. OUR CURRENT THINKING IS THAT THE CYCLONE WILL STILL BE A BORDERLINE CAT 1/2 HURRICANE WHEN IT EMERGES OVER THE BAY OF CAMPECHE...BUT THERE IS CONSIDERABLE UNCERTAINTY IN THIS FORECAST. ASSUMING THAT THE INNER CORE IS NOT TOO DISRUPTED BY ITS INTERACTION WITH LAND...DEAN SHOULD REGAIN MAJOR HURRICANE STATUS BEFORE ITS FINAL LANDFALL IN MAINLAND MEXICO.

FORECAST POSITIONS AND MAX WINDS
INITIAL 21/0900Z 18.7N 87.8W 145 KT (167 MPH) 12HR VT 21/1800Z 19.1N 90.4W 85 KT...INLAND 24HR VT 22/0600Z 19.6N 93.9W 95 KT...OVER BAY OF CAMPECHE 36HR VT 22/1800Z 20.1N 96.8W 105 KT...INLAND 48HR VT 23/0600Z 20.5N 100.0W 25 KT...INLAND...DISSIPATING 72HR VT 24/0600Z...DISSIPATED

Thermodyamics?

Thermodynamics?

Assignment Solve problems 1-3 (notes, p. 8) Due date: one week, 8/28

Equation of State (Ideal Gas Law)
Chap. 2 Equation of State (Ideal Gas Law)

Quiz: True or False? (Some have caveats)
Atmospheric pressure is just the weight of the atmosphere above us. Absolute zero is the temperature at which all motion ceases. As temperature increases, so does pressure, and vice versa. Cold air is denser that hot air.

Quiz: True or False? (Some have caveats)
Atmospheric pressure is just the weight of the atmosphere above us. True, if dw/dt = 0 Absolute zero is the temperature at which all motion ceases. As temperature increases, so does pressure, and vice versa. Must assume that V = const. Cold air is denser that hot air. Depends on pressure.

From elementary kinetic theory . . .
Assume an ideal gas has the following properties: The molecules are in random motion and obey Newton's laws of motion. The total number of molecules is large. The volume of molecules is negligible relative to the volume occupied by the gas. No appreciable (molecular) forces act on the molecule during a collision. Collisions among molecules are elastic and of negligible duration. In addition, the following assumptions are made regarding the interaction between molecules and a surface membrane or wall, which contains the molecules. The collision with the wall is elastic. There is no loss in momentum, in the direction parallel to the wall, during the collision with the wall (i.e., no friction).

Derivation of the equation of state from kinetic theory – in abbreviated form
mvcosq - (-mvcosq) = 2mvcosq change in momentum (wall collision) mv2dnvsinqcos2qdq incremental change from all collisions (mv2/3)dnv integration over all angles dF = (m/3)( v2dnv)dA differential force from molecules p  dF/dA = (m/3) v2dnv definition of pressure + algebra p = mnv2/ simplification using vbar pV = (1/3)Nmv2 assume dN = ndV (uniform distribution) pam = (1/3)N0v2 definition of molar specific volume (3/2)kT = (1/2) mv2 definition of temperature pam = kN0T = R*T substitution a = am / M, definition pa = (R*/M)T = RT, substitution This is the final result

Equation of State from experimental results
Boyles Law: V  p-1 for an isothermal (T = const) process. First Law of Gay-Lussac: V  T for an isobaric (p = const) process Second Law of Gay-Lussac: p  T for an isochoric (volume = const) process

Boyles Law: V  p-1 for an isothermal process
p1V1 = p2V2 (T = const) (from google on “Boyle’s Law”)

First Law of Gay-Lussac (Charles’ Law): V  T for an isobaric process
dV = aV0dT “a” is the coefficient of thermal expansion at constant pressure, a = 1 / 273 deg-1, V0 is the volume at 0 C Integrated form: V – V0 = aV0T V p = const Extrapolation to T = -273 C suggests that V = 0. This temperature is absolute zero Discussion? (theoretical) V0 -273 C 0 C T

Second Law of Gay-Lussac: p  T for an isochoric process
dp = bp0dT “b” is the pressure coefficient of thermal expansion at constant volume (= 1 / 273 deg-1) Integrated form: p = p0(1 + bT) p V = const See Application at the bottom of page 12 (Tsonis). Does such a pressure difference really exist? p0 -273 C 0 C T

Equation of state by inference
Combination of the laws of Boyle and Gay-Lussac: pV/T = p’V’/T’ = A pV = AT, where A is a constant, which can be equated to nR* (or mR) as follows. pV = nR*T = m(R*/M)T = mRT p = rRT pa = RT (a = r-1)

In summary The equation of state (p = RT) is general
Boyle’s Law is a special case of the equation of state: V  p-1 (T=const) First Law of Gay-Lussac (Charles’ Law) is a special case of the equation of state: V  T (p = const) Second Law of Gay-Lussac is a special case of the equation of state: p  T (V = const or  = const)

Dalton’s Law of partial pressures
The equation of state is valid for individual gases, as well as for a mixture of gases that comprise the atmosphere. For the ith gas, the equation of state is: pii = RiT. Dalton’s Law of partial pressures p = ∑pi(T,V) (pi is the partial pressure of gas i), The pressure of a gas mixture is equal to the sum of the partial pressures of each component gas See bottom of p. 17 (Tsonis)

Eq. of State for the atmosphere
The value of R for the dry atmosphere in Eqs. (2.12a,b of the notes) is Rd = J kg-1 K-1 For the dry atmosphere, pa = RdT

Usage of the equation of state
Used to derive the individual ideal gas laws (working backwards from our derivation) Use of Ideal Gas Law Equation to determine the density of a gas; r is difficult to measure directly. [The only instrument that can do this is the “direct detection” lidar which measures backscatter from molecules.] Solve for Partial Pressure of a known amount of gas in a gas mixture (p1 = n1RT/V and p2 = n2RT/V)

Applications of the eq. of state
A 1 liter (L) sample of air at room temperature (25 °C) and pressure (1 atm) is compressed to a volume of 3.3 mL at a pressure of 1000 atm. What is the temperature of the air sample? Use pV/T = const

Eq. of state in graphical form
ischores isotherms isobars

Equation of state for moist air
commonly used in atmospheric thermodynamics: water vapor pressure (e): the partial pressure due to water vapor molecules. (How could this be measured? We will see later that it can be determined theoretically/analytically with the Clausius-Clapeyron equation.) mixing ratio: rv = mv/md specific humidity: qv = (mv/(mv+md)) = rv/(1+rv)

As an aside, we note that the eq
As an aside, we note that the eq. of state applies to water vapor: e = rvRvT p = rmT[(mdRd+mvRv)/(md+mv)] (mass weighted), Rv = R*/MH2O rm = (md+mv)/V = rd + rv Eventually, we derive the eq. of state for moist air, using the new variable Tv: p = rmRdTv

Forms of the equation of state for dry air
pV = NR*T (R* = J K-1 kmol-1) pV = mRdT (Rd = J K-1 kg-1) p = RT Notes: Rd = R*/md

Empirical eq. of state with corrections to account for non-ideal gas
Vander Waals’ equation (p + aV-2)(V - b) = R*T Kammerlingh-Omnes (HW problem on this one) pV = A(1 + B'p + C'p ) A=R*T; B’ from Table 2.1 (p2 term can be ignored to good approximation T (C) B' (10-8 m2N-1) pV/R*T P = 500 mb P = 1000 mb -100 -4.0 0.9980 0.9996 -50 -1.56 0.9992 0.9984 -0.59 0.9997 0.9994 50 -0.13 0.9999

A linearized equation of state
Linearize the equation about a dry reference state The reference state obeys the gas law p0a0=RdT0 Substitue the following into the eq. of state a=ao+a', p=p'+po, T=T'+To, and rv=rv' Then: po(1+p'/po) ao(1+a'/ao) = Rd(1+0.61rv')To(1+T'/To) Take natural log of both sides, expand the log in a Taylor’s series, and ignore the higher order terms. The result is a'/ao = T'/To rv' - p'/po

Example Typical perturbations within a cloud are:
T' ~ 1 K (up to 15 K) rv' ~ 2 g kg-1 (up to 8 g kg-1) p' ~ 0.2 mb (up to 1-2 mb) Thus, T'/To = 1/273 = , rv' = 0.002, and p'/po = 0.2/800 = Discussion Temperature and moisture perturbations are comparable and thus provide the most important contributions to density fluctuations in the cloud (or cloud-free) environment. Only in limited regions of cloud systems does p' exceed mb. [It is the density fluctuations that control cloud dynamical processes.]

2.5 Measurements of temperature, pressure, and water vapor
Temperature: thermometer, thermister, thermocouples, IR emission, microwave emission (O2 band) Density: lidar Sonic temperature time series Pressure: barometer (mercury, aneroid), transducer Water vapor: wet bulb temperature, RH directly, lidar differential absorption, microwave emission Virtual temperature: radio acoustic sounding system (RASS – speed of sound  Tv)

Review of variables Variable symbol Variable Measurable? p Pressure
Barometer, pressure transducer T Temperature Thermometer, thermister, etc. V Volume Special cases only a Specific volume a = r-1 r Density Lidar; eq. of state calculation Tv Virtual temp. Need T and rv to calculate; Radio Acoustic Sounding System -- RASS rv Water vapor density Radiometer (indirectly); eq. of state calculation R Gas constant Given (but it can be estimated)

Useful web links: Wikipedia discussion of the ideal gas law: Hyperphysics, Georgia State Univ.: Gas Law animation

Review, from http://www.grc.nasa.gov/WWW/K-12/airplane/eqstat.html

Back to the quiz Atmospheric pressure is just the weight of the atmosphere above us. Yes, but one needs to be careful with this (dw/dt=0) Absolute zero is the temperature at which all motion ceases. One cannot assume that the ideal gas law is valid at T = 0 K. As temperature increases, so does pressure, and vice versa. This assumes that V = const. (Does not generally apply in the atmosphere) Cold air is denser that hot air. This assumes that p = const. (Generally true in the atmosphere, but be careful!)

Relation between T and p?

Example: p = pd + e = 999.7 + 6.014 = 1005.7 hPa.
If at 0 °C the density of dry air alone is kg m-3 and the density of water vapor alone is x 103 kg m-3, what is the total pressure exerted by a mixture of the dry air and water vapor at 0 °C? Solution: From Dalton’s law of partial pressures, the total pressure exerted by the mixture of dry air and water vapor is equal to the sum of their partial pressures. The partial pressure exerted by the dry air is pd = rdRdT where rd is the density of the dry air (1.275 kg m-3 at 273 K), Rd is the gas constant for 1 kg of dry air (287.0 J K-1 kg-1), and T is K. Therefore, pd = x 104 Pa = hPa Similarly, the partial pressure exerted by the water vapor is e = rvRvT where rv is the density of the water vapor (4.770 x 103 kg m-3 at 273 K), Rv is the gas constant for 1 kg of water vapor (461.5 J K-1 kg-1), and T is K. Therefore, e = Pa = hPa Hence, the total pressure exerted by the mixture of dry air and water vapor is p = pd + e = = hPa.

HW problems Petty 3.1 Petty 3.5 Petty 3.10
Now, show that the density of moist air is less than that for dry air at the same temperature and pressure. Interpret your results. Does this difference have any relevant atmospheric applications? (Hint: Refer to Petty and the previous problem) Determine the number of molecules in a 1 cm3 volume of air having a pressure of 1 atm. Make any other reasonable assumption if required. [Ans: about 3x1019 cm-3 – your answer will be more precise]. (Note, this is similar to problem 3.5 in Tsonis.) (b) What is the mean free path for the average molecule in this volume? Mean free path is determined from Dxmfp = (ns)-1, where n is the number of molecules per unit volume, s = pdo2 is the collision cross section (s is about 3 x cm2 for an air molecule), and do is the diameter of an average molecule. You can check your answer with Fig. 1.1b. At what pressure is the ideal gas law in error by 1%, for air with T = 0 C? [Ans: 17 atm; Hint: Use Table 1.3] (a) Calculate some extremes in air density at the surface for different scenarios. For example, consider (a) International Falls in the winter under high pressure (anticyclone) conditions: T = -40 F, p=1050 mb, rv=0.1 g kg-1; (b) Denver in the summer with T = 95  F, p=850 mb (actual station pressure) and rv=10 g kg-1. (c) What are some practical implications (e.g., aircraft lift, wind drag on a vehicle)? [Fleagle and Businger Prob. 1, ch. 2.] If 106 molecules are required in order to ensure a statistically uniform distribution of velocities in all directions, what is the minimum volume in which the state can be defined at standard atmospheric conditions (p=1013 mb, T=0 C)? [Ans x10-21 m3, which corresponds to a linear distance of 3.34x10-7 m for a cube. Hint: use the definition dN = ndV]. ATS/ES 441 students: You may eliminate two problems (choose from 4-8) of your choice, or if you turn in all problems, I will ignore the lowest scores on two problems.

First Law of Thermodynamics
Chap. 3 First Law of Thermodynamics General Form Atmospheric Science Applications

Some general statements
The energy of the universe is constant The First Law: defines internal energy states that heat is a form of energy. states that energy is conserved The First Law is the second fundamental principle in (atmospheric) thermodynamics, and is used extensively.

Work of expansion Work is part of the First Law, but it can be considered independently What is work? If a system (parcel) is not in mechanical (pressure) equilibrium with its surroundings, it will expand or contract. This involves “work”. Work is defined by the differential dw  fds

Illustration: dW = pAdx = pdV (shaded region of the graph)
Specific work (work per unit mass): dw = pda work is found by integrating this differential over the initial and final volumes V1 and V2: W  area under the curve AB This thermodynamic diagram (p-V) represents the state of the system at every point along the line. (Recall that the p-V diagram is a simple thermodynamic diagram.)

Example 3.1 Calculate the work done in compressing (isothermally) 2 kg of dry air to one-tenth its initial volume at 15 ºC. From the definition of work, W =  pdV. From the equation of state, p = rdRdT = (m/V)RdT. Then W = mRdT  dlnV = mRdTln(V2/V1) (remember the process is isothermal) = (287 J K-1 kg-1)( K)(2 kg)(ln 0.1) = x 105 J. The negative sign signifies that work is done on the volume (parcel) by the surroundings.

Work, cont. The quantity of work done depends on the path taken; work is not an exact differential. If it was, work would depend only on the beginning and end points (or initial and final conditions. Reconsider Eq. (3.a) above, rewriting it as follows (noting that the displacement dx = vdt, where v is the magnitude of the velocity vector): dW = pA(dx) = pA(vdt) Since p = F/A (or F = pA), the above equation becomes dW = Fvdt or dW/dt = Fv Now from Newton’s Law, F = ma = mdv/dt. Substituting this in the above yields or dW/dt = dK/dt (K = ½ mv2) Hmmmm, we just can’t escape dynamics!

Example 3.2 (from problem 3.7 in Tsonis): An ideal gas of p,V,T undergoes the following successive changes: (a) It is warmed under constant pressure until its volume doubles. (b) It is warmed under constant volume until its pressure doubles. (c) It expands isothermally until its pressure returns to its original pressure p. Calculate in each case the values of p, V, T, and plot the three processes on a (p,V) diagram. The three processes are shown in the graph on the right. In the first process, the work is In the second process, the work is zero since volume does not change. In the third process, the value of work is similar to that done in Example As this process proceeds through steps a-c, the temperature increases such that Tb>Ta>T.

More on thermodynamic work

Internal energy and a mathematical statement of the First Law
Consider a system which undergoes a change from some heat input q per unit mass (q=Q/m). The system responds through the work of expansion, w (work per unit mass). The excess energy beyond the work done by the system is q-w. If there is no change in macroscopic or bulk kinetic and potential energy of the system, then it follows from conservation of energy that the internal energy (Du - per unit mass) of the system must increase according to: q - w = Du (simple expression of the First Law) (3.2)

General form of the First Law in terms of various energy terms:
We will sum the various forms of energy which have passed through the system-surroundings boundary and set this new sum equal to the change in the system internal energy, similar to what we did in the previous equation. This expression differentiates thermal (LHS) and non-thermal (RHS) forms of energy: Du = q + Sei, (a more general form of the First Law) where q is again the net thermal energy (per unit mass) passing into the system from the surroundings. [Thermal energy can be defined as the potential and kinetic energies that change as a consequence of a temperature change.] Note that this last expression deals with the classification of energy passing through the system boundary -- it does not deal with a classification of energy within the system.

What is thermal energy (q)?
At this point, it is instructive to define thermal energy. In the atmosphere, thermal energy can include heating/cooling by Radiation (heating by absorption of SW or LW; cooling by emission of LW) latent heating (associated with water phase changes).

u = f(T) (Is this a proof?)
Joule's Law: u depends only on T OK, prove it! From statistical mechanics, for an ideal monatomic gas, the kinetic energy of translations is given by (refer to Chap 2 of Knupp’s notes) pV = (1/3)Nomu2 = (2/3)Ekin = RT, where N0 is Avogadro's number (6.023x1023). Thus, Ekin = (3/2)RT. Since at constant temperature there are no energy changes in electronic energy, rotational energy, etc., the internal energy of an ideal gas is only a function of T. This is also true for polyatomic ideal gases such as CO2 (and more generally for air). u = f(T) (Is this a proof?)

A general form of the First Law (in the differential form)
Dq = du + Dw = du + pDa. (3.4) [What has happened to the Sei term?] The operator "d" refers to an exact differential and "D" to inexact. One property of the inexact differential (e.g., Dw) is that the closed integral is in general nonzero, i.e., [see The first law requires that du be an exact differential -- one whose value depends only on the initial and final states, and not on the specific path. However, from here on, we will ignore (but not forget) this formal distinction between exact and inexact differentials. Aside: An exact differential can also be expressed as, for a function U = U(x,y) (Tsonis, Section 2.1) Illustrate with p-V diagram

Review and applications
Applications of the equation of state, and connection with the First Law. From

Review and example problems
Review the information and example problems at the following web site:

3.4 Specific heats Consider the case where an incremental amount of heat dq is added to a system. The temperature of the system increases by an incremental amount dT (assuming that a change of phase does not occur). The ratio dq/dT (or Dq/DT) is defined as the specific heat, whose value is dependent on how the system changes as heat is input. c = dq/dT [note that C = mc = m(dq/dT)]

Two atmospheric scenarios: constant volume and constant pressure.
For constant volume: cv  (dq/dT)a=const Since specific volume is constant, no work is done. According to the First Law, Eq. (3.1), dq = du (since da = 0) and cv = (du/dT)a=const (3.5) Then du=cvdT and dq = cvdT + pda (3.6)*

For the isobaric process, the specific heat is
cp  (dq/dT)p=const (3.7) In this case, some of the heat added is used in the work of expansion as the system expands against the constant external pressure of the environment. The value of cp must therefore be greater than that of cv. To show this, we can write (3.6) as dq = cvdT + d(pa) - adp = d(u+pa) - adp = dh - adp, where h, the enthapy is defined as h  u + pa. (Enthalpy is discussed further in the following section.)

Since pa = RT from the equation of state (for dry air), the previous equation can be rewritten as
dq = (cv+R)dT - adp = cpdT - adp. (3.8)* If pressure is constant then dp=0 and, using (3.6), we note that cp exceeds cv by the amout R, the gas constant: cp = cv + R (3.9)*

For dry air, the values are:
cv = 717 J K-1 kg-1 cp = J K-1 kg-1 [ = f(T,p); Bolton, 1980] For ideal monatomic and diatomic (air) gases, it can be shown from statistical mechanics theory that the ratios cp:cv:R are 5:3:2 and 7:5:2, respectively. (See Tsonis, p. 32.) The variation of cp with T and p is presented in Table 3.1. Table Dependence of cpd (J K-1 kg-1) on T and p. From Iribarne and Godson (1973). p (mb) T (C) -80 -40 40 1003.3 1003.7 1004.0 1005.7 300 1004.4 1006.1 700 1006.5 1005.3 1000 1009.0 1007.4 Relative variation: ( – ) / = 3.4/ = Why is cp not constant?

3.5 Enthalpy Many idealized and natural processes of interest in atmospheric science occur at constant pressure. An example is evaporation of rain. If heat is added isobarically to a system such that both the internal energy u and specific volume a change, then the First Law (dq = du + pda) can be integrated as Dq = (u2-u1) + p(a2 - a1) = (u2+pa2) - (u1+pa1) = h h1, where enthalpy h is defined as h = u + pa (3.10)

 dq = dh - adp. Upon differentiation, we obtain
dh = du + pda + adp [= dq + adp]  dq = dh - adp. Comparing this with (3.8) we can redefine dh as dh = cpdT (3.11) This can be integrated to give (assuming h=0 when T=0 K) h = cpT. Yet another form of the First Law is thus dq = dh - adp = cpdT - adp (3.12)* Three useful forms of the first law: (3.6), (3.8) and (3.12).

3.6 Example: an isothermal process and reversibility
Consider a fixed mass (m=const) of ideal gas confined in a cylinder with a movable piston of variable weight. The piston weight and its cross-sectional area determine the internal pressure. Assume that the entire assembly is maintained at T=const (an isothermal process). Let the initial pressure be 10 atm and the initial volume Vi be 1 liter (L). Consider three different processes in going from A to B (below right)

Process 1: The weight of the piston is reduced to change the cylinder pressure to 1 atm. The gas will expand until its pressure is 1 atm, and since pV=mRT=const, the final volume Vf will be 10 L (see Fig. 3.1). The work of expansion is W = pdV = psurr(Vf-Vi) = 1 atm * (10-1) L = 9 L-atm. This is the work done on the surroundings. Process 2: This will be a two-stage process: (i) Decrease (instantaneously) the cylinder pressure to 2.5 atm; then the volume will be 4 L, since this is similar to Process 1. (ii) Then further decrease the pressure (instantaneously) to 1 atm with a volume of 10 L. The work is the sum of these two processes: W = p1DV1 + p2DV2 = 2.5 atm * 3 L + 1 atm * 6 L = 13.5 L–atm Process 3: The pressure is continuously reduced such that the pressure of the gas is infinitesimally greater than that exerted by the piston at every instant during the process (otherwise no expansion would occur). Then we must apply the integral form of work to get W = pdV = mRTln(V2/V1) = L-atm. Note that pV=mRT=10 L-atm = const in this example.

This last process is reversible and represents the maximum work.
Definition: A reversible process is one in which the initial conditions can be reproduced after a system goes through at least one change in state. In process 3 above, the initial state of the system can be realized by increasing the pressure continuously until the initial volume is attained. Note that the value of work for the reversible process depends only on the initial and final states, not on the path. For the most part, idealized atmospheric processes are reversible since parcel pressure is assumed to be equal to (i.e., differs infinitesimally from) the ambient pressure.

Wikipedia definition of reversibility
In thermodynamics, a reversible process, or reversible cycle if the process is cyclic, is a process that can be "reversed" by means of infinitesimal changes in some property of the system without loss or dissipation of energy. Due to these infinitesimal changes, the system is at rest during the whole process. Since it would take an infinite amount of time for the process to finish, perfectly reversible processes are impossible. However, if the system undergoing the changes responds much faster than the applied change, the deviation from reversibility may be negligible. In a reversible cycle, the system and its surroundings will be exactly the same after each cycle.[2] An alternative definition of a reversible process is a process that, after it has taken place, can be reversed and causes no change in either the system or its surroundings. In thermodynamic terms, a process "taking place" would refer to its transition from its initial state to its final state.

3.7 Poisson’s Equations An adiabatic process is defined as one in which dq=0. The two advanced forms of the 1st Law (which are related by the equation of state) become 0 = dq = cvdT + pda 0 = dq = cpdT - adp Using the equation of state in the form pa=RdT (dry atmosphere), the above relations can be manipulated to get the following differential equations: 0 = cvlnT + Rddlna, [T, a] 0 = cpdlnT + Rddlnp, [T, p] (let’s look at this one) 0 = cvdlnp + cpdlna, [p, a] where the third expression was obtained using the equation of state. Integration yields three forms of the so-called Poisson’s Equations: Tah-1 = const (TcvaRd = const) Tp-k = const (Tcpp-Rd = const pah = const (pcvacp = const) k = Rd/cp = and h = cp/cv = 1.403

which are known as polytropic relations.
The last of the three above equations (pah = const ) has a form similar to that of the equation of state for an isothermal atmosphere (in which case the exponent is 1). These relationships can be expressed in the more general form pan= const, which are known as polytropic relations. The exponent n can assume one of four values: For n= 0, p = const isobaric process For n = 1, pa = const isothermal process For n = h, pah = const adiabatic process For n =  isochoric process Refer to Tsonis, pp

3.8 Potential temperature and the adiabatic lapse rate
Potential temperature q is defined as the temperature which an air parcel attains upon rising (expansion) or sinking (compression) adiabatically to a standard reference level of p0 = 100 kPa (1000 mb). We use the (T,p) form of the First Law (3.5) , assuming an adiabatic process (dq=0) dq = 0 = cpdT - adp. Incorporate the equation of state, pa=RdT, to eliminate a, and rearrange to get Now integrate over the limits, in which a parcel has a temperature T at pressure p, and then end with a (potential) temperature q at the reference pressure p0. Although not strictly correct, we assume the cp and Rd are constant.

or . Take the antilog of both sides and rearrange to isolate potential temperature (q): (3.13a) [This is also called Poisson's equation, since it a form of Poisson’s equations, e.g., Tp-k = const] For dry air, k = Rd/cp = 287/ = [= 2/7 for a diatomic gas – from kinetic theory]. This value changes somewhat for moist air because both cp and R (Rd) are affected by water vapor (more so than by T,p), as we shall see in the Bolton (1980) paper.

Potential temperature has the property of being conserved for unsaturated conditions (i.e., no condensation or evaporation), assuming that the process is adiabatic (i.e., no mixing or radiational heating/cooling of the parcel). For a moist atmosphere, the exponent k in Eq. (3.13a) is multiplied by a correction factor involving the water vapor mixing ratio rv, and q is expressed as (see Bolton 1980, eq. 7) (3.13b)* where rv is the water vapor mixing ratio expressed in kg kg-1.

Dry Adiabatic Lapse Rate
An associated quantity, the dry adiabatic lapse rate, which is used to evaluate static stability. The term "lapse rate" refers to a rate of temperature change with height (or vertical temperature gradient), i.e., T/z. [Aside: It is important to differentiate the static stability of the atmosphere, as given the the vertical gradient of temperature, T/z, from the Lagrangian temperature change that results when a parcel moves adiabically in the vertical direction. The parcel change of temperature would be dT/dt = (dT/dz)(dz/dt) = w(dT/dz).]

Our starting point is once again the First Law (3
Our starting point is once again the First Law (3.5) with dq = 0 (adiabatic process). dq = cpdT - adp = 0. For a hydrostatic atmosphere (hydrostatic implies no vertical acceleration, and will be defined more fully later) the vertical pressure gradient is dp/dz = p/z = -rg = -g/a (hydrostatic equation) Solving the above for a and substituting into the First Law, we obtain cpdT + gdz = 0.

(dT/dz)d = -g/cp = Gd = -9.81 m s-2 /1005.7 J K-1 kg-1 [J = kg m2 s-2]
Thus, the value of the dry adiabatic lapse rate (Gd) is (dT/dz)d = -g/cp = Gd = m s-2 / J K-1 kg-1 [J = kg m2 s-2] = K km (3.14) Again, one should be aware that this value changes slightly for a moist (subscript m) atmosphere (one with water vapor), since the addition of water vapor effectively yields a modified value of the specific heat at const pressure, which has the following dependence on water vapor: cpm = cpd( rv), where rv is in units of kg kg-1 (Bolton, 1980). (We will see this difference in graphical form later.) Specifically, for moist air, Gm = Gd / ( rv)  Gd ( rv).

Some uses of q Atmospheric structure; conserved for subsaturated motion; atmospheric bores Vertical cross sections Analysis of static stability

3.9 Heat capacities of moist air; effects on constants
The exponent of Poisson's equation (k = Rd/cp) requires adjustment when water vapor is present. Why? The water vapor molecule (H2O) is a triatomic and nonlinear molecule, whose position can be described by 3 translational and 3 rotational coordinates. Dry air is very closely approximated as a diatomic molecule (N2, O2) (See web site The specific heats for water vapor are therefore quite different from (much larger than) that of dry air: cwv = 1463 J K-1 kg-1 (w subscript designates the water phase) cwp = 1952 J K-1 kg-1, For Poisson's eq. (3.10a) the exponent Rd/cp is adjusted using the correction term (Rd/cp)(1-0.28rv) (Bolton 1980), where the water vapor mixing ratio rv is expressed in kg kg-1. Also the "constants" Rd and cp can be corrected for moist air as follows: cpm = cpd( rv), cvm = cvd(1+0.97rv), Rm = Rd( rv).

3.10 Diabatic processes, Latent Heats and Kirchoff's equation
diabatic process  dq  0. Two examples of diabatic heating/cooling absorption/emission of radiation; heating/cooling associated with water phase changes In the moist atmosphere, there are cases where heat supplied to a parcel without a corresponding change in temperature. Under such conditions, the water substance is changing phase, and the change in internal energy is associated with a change in the molecular configuration of the water molecule, i.e., a change of phase. (latent heating)

Lil = 0.334 x 106 J kg-1 latent heat of melting
Notation on the latent heating terms: the two subscripts define the change in phase of the water substance. Lvl is the latent of condensation with the subscript vl denoting a change in phase from vapor (v) to liquid (l). Thus, condensation is implied from the order of subscripts: vapor to liquid. For the sake of simplicity we can define the latent heats as follows: Lvl = 2.50 x 106 J kg-1 (0 C) latent heat of condensation (function of T) Lvl = 2.25 x 106 J kg-1 (100 C) (yes, Lvl does vary by 10% between C) Lil = x 106 J kg-1 latent heat of melting Lvi = 2.83 x 106 J kg-1 (0 C) latent heat of deposition Note: Lvi = Lvl + Lil In general, these terms are defined for p=const . . . but Lvl does vary with temperature!

Why is L a function of temperature?
From the First Law: L is related to an enthalpy change, i.e., L = Dh. [Proof: Since dp=0, the First Law can be written as dq = cpdT = dh. Also recall that enthalpy can be derived from the First Law.] To examine this temperature dependence, expand the differential dh, based on the definition of the total derivative (note that h = f(T,p): dh = (h/T)p dT + (h/p)T dp (definition of the exact differential), and apply this to two states a and b (Dh=L=hb-ha): d(Dh) = (Dh/T)p dT + (Dh/p)T dp. For an isobaric process, only the second term vanishes and we have d(Dh)p ≡ dL = (Dh/T)p dT = (hb/T)pdT - (ha/T)pdT = (cpb - cpa)dT. This latter equivalence is based on the definition of specific heat (see Section 3.4), cp  dq/dT = dh/dT for an isobaric process.

From the previous equation, we can write Kirchoff's equation
(L/T)p = Dcp, (3.15)* Thus, the temperature dependence of L is related to the temperature dependence of cp. Bolton (1980) provides an empirical equation that has a linear form for the temperature correction of Lvl: Lvl = ( aTc) x 106 J kg-1, (3.16) where a = C-1 and Tc is the dry bulb (actual air) temperature in C. T (C) Liv (106 J kg-1) Lil (106 J kg-1) Llv (106 J kg-1) -100 2.8240 -80 2.8320 -60 2.8370 -40 2.8387 0.2357 2.6030 -20 2.8383 0.2889 2.5494 2.8345 0.3337 2.5008 20 2.4535 40 2.4062

3.11 Equivalent potential temperature and the saturated adiabatic lapse rate
Equivalent potential temperature (approximate form) Saturation  condensation  latent heating by condensation (Lvl, multiplied by the mass of water vapor condensed) Expressed by a change in the saturation mixing ratio, rvs [Question: Is this an adiabatic process since dq ≠ 0?] Starting point – First Law (dq is now nonzero due to latent heating) dq = -Lvldrvs = cpdT - adp (3.17)

Substitute the equation of state pa=RT for a, and rearrange terms:
Take the log differential of Poisson's eq. (3.10a): or

Combine the preceding with eq (3.14):
Physical interpretation: the latent heating changes the potential temperature of the parcel, such that a reduction in rvs (drvs < 0) corresponds to a positive dq. The LHS of the preceding equation is cumbersome to integrate, as it currently stands, because Lvl = Lvl (T). With the use of a graphical (thermodynamic) diagram, it will later be shown that , (this is an approximation, but it provides an exact differential) Then we can integrate the following

Note the approximations used here:
assuming that  -> e as rvs/T -> 0, to get or (approximate form) (rvs in kg kg-1) (3.18) which is an approximation for equivalent potential temperature, e. Note the approximations used here: assumed that cp and Lvl are independent of rv and/or T and p; made the approximation in the differential (Lvl/cpT)drvs  d(Lrvs/cpT). In essence, we have assumed that Lvl is independent of temperature, which sacrifices precision in the integrated form. Tsp in Eq (3.18) is the temperature of the parcel's saturation point (SP), traditionally called the lifting condensation level or LCL, and rvs is the mixing ratio at the LCL (or alternatively, the actual water vapor mixing ratio of the parcel).

A semi-empirical formula for qe, superior to Eq. (3.15) and accurate
to within ~0.5 K, is (within 0.5 K) (rvs in kg kg-1) (3.19)* Better! (I don’t recall the source of this, but it is given as Eq. (2.36) in Rogers and Yau Note that the numerical value 2675 replaces the ratio Lvl/cp, so this implies some constant values for cp, and especially Lvl. This form is good for quick, relatively accurate calculation of qe. An accurate calculation of qe requires an accurate determination of Tsp, q, cp, and an integrated form that does not assume that Lvl is consant. These steps are detailed in the paper by Bolton (1980). Note: L/cp = 2.5 x106 / = 2488 L/cp = 2675; L = 2675 cp = 2675 x = 2.69 x 106

3.11.2 Equivalent potential temperature (accurate form)
Because e is conserved for moist adiabatic processes, it is widely used, and its accurate calculation has received much effort. An analytic solution is not possible. The approximate form derived in the previous section may produce errors of 3-4 K under very humid conditions (i.e., large rv). Refer to the paper of Bolton (1980) for a presentation of the accurate calculation of e. We will consider this in some detail. [Assignment: Read the paper by Bolton (1980).] Bolton’s curve-fitted form of qe is (more clumsy with the calculator, but is easily coded) (3.20) (within 0.04 K) (rv in g kg-1)

3.11.3 Saturated (psuedo) adiabatic lapse rate – preliminary form
The derivation of the saturated adiabatic lapse rate is complicated and requires advanced concepts/tools that are developed in later chapters. The derivation below is first order only. A complete derivation will be presented in Chapter 6 (my notes). When water droplets condense within an ascending parcel, two limiting situations can be assumed: the condensed water immediately leaves the parcel (the irreversible pseudo-adiabatic process), or all condensed water remains within the parcel (the reversible saturated-adiabatic process). In reality, the processes acting within clouds lie somewhere in between. Here we consider the pseudo-adiabatic process. With the addition of latent heating, one may anticipate that a rising parcel undergoing condensation will cool less rapidly that an unsaturated parcel. Our starting point is (3.14), but with the hydrostatic equation term gdz substituted for adp in (3.14). We can then write -Lvldrvs = cpdT + gdz. (remember dp/dz = -rg)

Ignore the effects of water vapor being heated along with the dry air and
write the above as Applying the chain rule to drvs/dz The second term g/cp defines the dry adiabatic lapse rate. The first term is new, and is the somewhat messy water vapor term (first term on the RHS). Solving the above for dT/dz, the approximate saturated adiabatic lapse rate is given as A functional relationship for rvs is obtained from the Clausius-Clapeyron equation, to be considered Chapter 5 (my notes). We also note that the magnitude of Gs is not constant, but decreases (nonlinearly) as T increases. This is not obvious from Eq. (3.21), but will become more apparent when we examine and analyze the Clausius-Clapeyron equation. (3.21)

More on the saturated adiabatic lapse rate
Pseudo adiabatic process – all condensed water leaves the parcel; irreversible Saturated adiabatic process – all condensed water is carried with the parcel; reversible More on this in Chap. 6

More on qe Recall the approximate form (for illustrative puroses only): rvs is the initial parcel water vapor mixing ratio (rv) Tsp is the so-called “saturation point” temperature. Bolton defines this is TL.

How is Tsp determined? Tsp will be derived in Chap. 6
For now, use Bolton’s formula: e is water vapor pressure; T is temperature in K f is relative humidity

3.12 Atmospheric Thermodynamic Diagrams
Refer to Chapter 9 in Tsonis. thermodynamic diagram - a valuable tool to illustrate the relationship between dry adiabats, saturated adiabats and other thermodynamic variables A number of thermodynamic diagrams used for atmospheric applications have been constructed. We have considered only the simple p-V diagram so far (e.g., Fig. 3.1). Any diagram is based on two thermodynamic coordinates. With two thermodynamic variables being defined, other variables and processes can be determined based on the Equation of State, First Law, and ancilliary relations that we have considered, or will consider. For example, the dry and saturated adiabatic lapse rates can (and must) be drawn on an atmospheric thermodynamic diagram to illustrate static stability. Once the coordinates are defined, then other isolines (e.g., saturation mixing ratio, dry/saturated adiabats, isotherms, isobars) can be defined and drawn on the diagram to graphically represent atmospheric processes and evaluate static stability.

The most commonly used diagrams are the skew-T and the tephigram
The most commonly used diagrams are the skew-T and the tephigram. Interestingly, it seems as though the "tropical" meteorologists favor the tephigram (with the exception of Rogers and Yau, who are Canadians). The skew-T is most widely used in the research and operational sectors in the U.S. An ideal atmospheric thermodynamic diagram has the following features: area equivalence: the area traced out by some process, e.g., the Carnot cycle, is proportional to energy; a maximum number of straight lines; coordinate variables that are easily mearured in the atmosphere; a large angle between adiabats and isotherms; a vertical coordinate that is approximately linear with height. The tephigram and skew-T closely satisfy nearly all these criteria. The table below summarizes the important aspects for some diagrams. Note that the skew-T, which we will use in this class, exhibits most of the ideal properties. See Irabarne and Godson (1973, pp ; handout) and Tsonis (Chapter 9) for a more complete discussion.

Table 3.3. Summary of thermodynamic diagram properties.
Abscissa Ordinate Straight line characteristics Angle between adiabats and isotherms isobars adiabats isotherms Skew-T, ln p T ln p Yes no nearly 90 (variable) Tephigram ln q 90 Stuve pk Psueo-adibatic -pkd ~45 Clapeyron a -p No small Emagram -ln p yes Refsdal ln T -Tlnp

Various isolines are drawn on thermodynamic diagrams, including isobars, isotherms, lines of constant saturation mixing ratio, dry adiabats and saturated adiabats. On-line references: Skew-T diagram: A good tutorial with bad graphics: Another one: An excellent resource to access recent and archived soundings (U. Wyoming): Available commercially:

Applications representation of vertical profiles of temperature and moisture (i.e., soundings) evaluation of static stability and potential thunderstorm intensity quick estimation of derived thermodynamic quantities such as: relative humidity, given the temperature (T) and dewpoint temperature (Td) mixing ratio, given Td and pressure p, potential temperature and equivalent potential temperature, given p, T and Td others determination of cloud/environment mixing processes determination of thickness ( mb thickness) mixing processes (advanced application)

Sources of skew-T diagrams (real-time and historical)
1) NCAR/RAP – the best Skew-T on the web: 2) University of Wyoming – flexible site, data, skew-T or Stuve diagram; historical data 3) Unisys Other valuable information: GOES satellite sounding page – good information on skew-T’s and their applications. We will examine many of these during this course. RAOB program:

Soundings plotted on a skew-T, ln p diagram

Variations in Tv vertical profiles around an atmospheric bore

So what is a bore? x

Schematic of a bore

A more detailed analysis on the skew-T

Animation of parcel ascent
Buoyancy and CAPE

Chap. 3 HW Problems 1-7 in notes, plus Petty 5.4, 5.7, 5.10
ATS 441 students may waive number 1

Examples 3.39 (W&H) A person perspires. How much liquid water (as a percentage of the mass of the person) must evaporate to lower the temperature of the person by 5 °C? (Assume that the latent heat of evaporation of water is 2.5 x 106 J kg-1, and the specific heat of the human body is 4.2 x 103 J K-1 kg-1.) 3.38 (W&H) Consider a parcel of dry air moving with the speed of sound, cs = (gRdT)½, where g = cpcv = 1.40, Rd is the gas constant for a unit mass of dry air, and T is the temperature of the air in degrees kelvin. (a) Derive a relationship between the macroscopic kinetic energy of the air parcel Km and its enthalpy H. (b) Derive an expression for the fractional change in the speed of sound per degree Kelvin change in temperature in terms of cv, Rd, and T.

2nd Law of Thermodynamics
Chapter 4 2nd Law of Thermodynamics

4.1 General statement of the law
The First Law is an empirical statement regarding the conservation of energy. The Second Law is concerned with the maximum fraction of heat that can be converted into useful work. The second law may be stated in several different ways, such as : Thermal energy will not spontaneously flow from a colder to a warmer object. (What is thermal energy?) The entropy (defined below) of the universe is constantly increasing. Thus, the second law is not a conservation principle (as in the 1st Law), but rather is a law defining the direction of flow of energy. In the following we will see that entropy and energy are closely related

4.2 Entropy Entropy is a state function defined by (per unit mass)
(defined for a reversible process) Note: entropy is also represented by the symbol f The second law defines entropy as a state function (see Petty, Section 6.1) and permits the following statements: For a reversible process the entropy of the universe remains constant. For an irreversible process the entropy of the universe will increase. Thus, a more general definition of entropy is Note that the Second Law does not address anything specifically about the entropy of the system, but only that of the universe (system + surroundings).

Chap. 6 in Petty is short and sweet
Chap. 6 in Petty is short and sweet. He has a good explanation of the second law and its relation to thermodynamic equilibrium in Section Read this carefully. Key Fact: Within any isolated system that is not at equilibrium, the net effect of any active process is always to increase the total entropy of the system. A state of equilibrium in therefore reached when the total entropy of the system has achieved it maximum possible value. At this point, no further evolution of the system state variables is possible.

Definition of reversibility (revisited)
A system process is defined as reversible if a system, after having experienced several transformations, can be returned to its original state without alteration of the system itself or the system's surroundings. A reversible transformation will take place when a system moves by infinitesimal amounts, and infinitesimally slowly, between equilibrium states such that the direction of the process can be reversed at any time. Remember that in a reversible process the deviation from equilibrium is infinitesimal. [Refer to the work of expansion problem considered previously in Section 3.6.] In a reversible process, the entropy of the universe (i.e., the system plus surroundings) remains constant. We can examine reversible processes theoretically, but do reversible processes actually take place in the atmosphere?

Thermodynamic process can be classified into one of three categories:
natural reversible impossible. Natural processes are more or less irreversible. Examples include: friction - associated heating warms the surroundings (frictional heating in a hurricane is real) unrestrained expansion (expansion of a gas into a vacuum) - again the surroundings are modified heat conduction in the presence of a temperature gradient (surface heating/cooling) chemical reactions (e.g., the combination of two atoms of H and and one atom of O in the production of H20) turbulent mixing and molecular diffusion of pollutants and aerosols freezing of supercooled water precipitation formation - removes water and heat from an air parcel mixing between a cloud and the subsaturated atmosphere deliquescence behavior of NaCL

A closer look at entropy:
In the p-V diagram below , isotherms are distinguished by differences in temperature and the adiabats by differences in potential temperature q. There is another way of distinguishing differences between adiabats. In passing from one of the adiabats (q1 or q2) to another along an isotherm (this is actually one leg of the Carnot cycle, see also the appendix), heat is absorbed or rejected, where the amount of heat Dqrev depends on the temperature of the isotherm. It can be shown that the ratio Dqrev /T is the same no matter what isotherm is chosen in passing from one adiabat to another. Therefore, the ratio Dqrev/T is a measure of the difference between the adiabats – and this is also the difference in entropy s.

Using the definition of entropy from Eq. (4
Using the definition of entropy from Eq. (4.1), the first law can be expressed as dq = Tds = du + pda. When a substance passes from state 1 to state 2, the change in entropy is found by integrating (4.1):

The relation between s and q -- Another cool derivation
How is entropy related to the more commonly-used atmospheric variables? Combine the equation of state, pa = RT, with the first law in the form dq = cpdT - adp. We can then write Taking the log differential of Poison's Eq. (potential temperature) we can write Since (4.2) and (4.3) have identical right-hand sides (RHS), they can be equated:

Now we have a more intuitive definition of entropy!
Thus the entropy function can be expressed in terms of potential temperature as (differential entropy) or s = cp lnq + const. From this we can see that transformations in which entropy is constant are also processes in which the potential temperature of an air parcel is constant. Such processes are called isentropic (adiabatic) processes. Analyses using the variable q are similarly called isentropic analyses, and lines of constant q are termed isentropes. An example of an isentropic analysis, and a corresponding temperature analysis, is shown in Fig. 4.2.

Conversion between T and q, as shown in a vertical cross section of each
tropopause Solid contour lines Upper panel: isotherms Lower panel: isentropes The difference: T  q transformation Note how the tropopause is better defined in the insentropic analysis. cold front Fig Analysis of (a) temperature and (b) potential temperature along a vertical section between Omaha, NE and Charleston, SC, through the core of a jet stream. In each panel, wind speed in m s-1 is indicated by the dashed contours. Taken from Wallace and Hobbs (1977).

4.3 A generalized statement of the second law (review)
Calculation of entropy requires an equivalent reversible process. [But all natural processes are irreversible since they move a system from a nonequilibrium state toward a condition of equilibrium.] The second law can be stated more generally in terms of the following postulates: There exists a function of state for a system called entropy s. s may change as the system: (a) comes into thermal equilibrium with its environment or (b) undergoes internal changes within the body. The total entropy change ds can be written as the sum of external (e) and internal (i) changes ds = (ds)e + (ds)I 3) The external change (ds)e is given by (ds)e = dq/T. 4) For reversible changes, (ds)i = 0, and for irreversible changes, (ds)i > 0. .

Thus, ds = dq/T for reversible changes ds > dq/T for irreversible changes Combining these two gives the generalized form of the first law as Tds  du + pda, (4.4)* where the equality refers to reversible (equilibrium) processes and the inequality to irreversible (spontaneous) transformations.

Examples 4.4.1 Some idealized entropy change processes
[Note: in the examples below we are beginning with the first law dq = du + pda or dq = cpdT - adp] isothermal expansion of an ideal gas For an isothermal process du = 0 and the work of expansion (determined previously) is pda = nRTln(a2/a1). Ds = Dq/T = nRln(a2/a1). [Proof: Since a = RT/p, da = (R/p)dT - (RT/p2)dp. Then -pda (=dw) = -RdT + RTdlnp] If the final specific volume a2 is greater than the initial a1 then the entropy change is positive, while for a compression it is negative.

b) adiabatic expansion of an ideal gas
For a reversible adiabatic expansion dq=0 and the entropy change is ds=0. This is the isentropic process defined previously.

c) heating of an ideal gas at constant volume
By defintion, da=0. Then ds = dqrev/T = cvdT/T = cvdlnT.

d) heating of an ideal gas at constant pressure
For a reversible process ds = dqrev/T = cpdT/T = cpdlnT.

e) entropy changes during phase transitions
For a phase transition carried out reversibly, Ds = Dhtransition/Ttransition. [Recall that Dh = L = cpDT for a phase change which occurs at constant pressure.]

4.4.2 A more comprehensive example: The entropy change in an irreversible process
Consider the isothermal expansion of an ideal gas: p = 1 atm, T = K, and V = liters per mole. Let this system expand isothermally against a constant external pressure of 0.5 atm. The final volume is liters and the work done is pext(V2-V1) = 0.5(22.412) = L atm = cal = 1135 J (1 cal = J). This is the heat that must be supplied from an external reservoir to maintain isothermal conditions. Since this process is irreversible, the entropy change of the system is not dq/T. Rather, we must find a reversible process from the initial to final state. In this case we refer to Example (a) above (isothermal expansion) in which dqrev = RTln 2 = 1573 J. The change in entropy of the reversible process is thus Dq/T = 1573 J / K = 5.76 J K-1.

4.4.3 The phase change entropy
At K (0 C) the entropy of melting of water is Lil/Tf = 3.34 x 105 J kg-1 / K = 1223 J K-1 kg-1, while at K the entropy of vaporization is Llv/T = 2.25 x 106 J kg-1 / K = 6031 J K-1 kg-1. Note the large difference between these two entropies. Why? This entropy change is due primarily to two effects: (1) the entropy associated with the intermolecular energy and (2) configurational entropy. Further explanation: For the conversion of ice into water there is little change in the intermolecular entropy term and an increase in configuration entropy in transforming to a slightly less ordered system. However, in evaporation there is a large change in intermolecular entropy (the molecules in the gas are spaced far apart and are subject to little interaction compared to molecules in the liquid phase) as well as a large change in configurational entropy in going from a somewhat ordered liquid to a nearly completely disordered gas.

Example: Calculate the change in entropy when 5 g of water at 0 C are raised to 100 C and then converted to steam at that temperature. We will assume the latent heat of vaporization is 2.253x106 J kg-1 at 100 C. (Note that we will use the extensive forms – capital letters – since mass is involved.) Step 1: Compute the increase in entropy resulting from increasing the water temperature from 0 to 100 C: Here, dQrev = m(dqrev) = mcwdT where m is mass and cw is the specific heat of water. It we assume cw to be constant at 4.18x103 J kg-1 K-1 then = 20.9 ln(373/273) = 6.58 J K-1

Step 2: Compute the change in entropy from conversion of 5 g of water to steam, which involves a latent heat term. This is DS2 = mLvl/T = (.005 kg)(2.253×106 J kg-1)/373 K = 30.2 J K-1. The sum of these components gives the total increase DS DS = DS1 + DS2 = = J K-1.

4.5 The free energy functions
The first law is a conservation statement . . . The second law governs the directions of thermal energy transfer and also permits the determination of the reversibility of a process. It is desirable to have a function or set of functions which will describe for a system the likelihood of a given process and the conditions necessary for equilibrium. Since there are really only two basic thermodynamic functions (u and s), we can on the basis of convenience define additional functions that may be based on u or s [Wait a minute -- It may not be clear why u and s are so basic. Think about this.] Is this true??? These functions can then be used to define equilibrium conditions for processes to be considered later.

4.5.1 Helmholtz free energy The Helmholtz free energy is defined as
f  u - Ts. In differential form, we have df = du - Tds - sdT (4.5) Combining this with Eq (4.4) (Tds=du+pda -- recall that the equality implies the reversible condition here) gives df = -sdT – pda If a system is in equilibrium and both T and a are constant, then df = 0. For a system which undergoes a spontaneous (irreversible) process, we have df < -sdT - pda and df < 0. Thus, a system at constant T and volume (a) is in a stable equilibrium when f attains a minimum value. For this reason, the Helmholtz free energy is sometimes called the thermodynamic potential at constant volume.

4.5.2 Gibbs free energy In this case we will derive the Gibbs free energy from the First Law using the form (after all, let’s see how fundamental Gibbs free energy is): dq = du + pda. Integration between the limits associated with a phase change, we get L   dq =  du +  pda Assuming p=const and through some simple rearrangement we can obtain where the subscripts 1 and 2 denote the two phases. Rearranging to combine like subscripts yields the following equality regarding the energy between the two phases: u1 + pa1 - Ts1 = u2 + pa2 - Ts2

Based on the above, the Gibbs free energy is defined as (per unit mass)
g = u - Ts + pa [= f + pa] In differential form, dg = du - Tds - sdT + pda + adp (4.6) Again, we can use (4.4) (Tds = du + pda, in reversible form) to obtain dg = -sdT + adp. In this case, if T and p are constant, for a body in equilibrium we have dg = 0. For an irreversible process, dg < -sdT + adp. Thus, dg < 0 in an irreversible, isobaric, isothermal transformation. Gibbs free energy is also called the thermodynamic potential at constant pressure. We will find that g is very useful for phase changes which occur at constant T (isothermal) and p (isobaric).

4.5.3 The free energy functions and total work
At this point it is instructive to relate g and f to the external work that a system can perform under various conditions. So far we have assumed that the only work term is that of expansion, pda. There are other forms of work that we will consider, however. [Recall the strange ∑ei term in Eq. (3.3) at the top of page 3, Chapter 3: Du = q + ∑ ei] For example, the creation of a surface in the nucleation (formation) of water droplets and ice crystals will be of interest to us. In this more general form, the First Law can be written as dq = du + dwtot, and for a reversible transformation Tds = du + dwtot (4.7) where the total work is dwtot. If we combine the above with (4.5) and assume an isobaric condition, we find dwtot = -df - sdT. Furthermore, for an isothermal process, dwtot = -df. Thus, the total external work done by a body in a reversible, isothermal, isobaric process is equal to the decrease in Helmholtz free energy of the body.

If da (this variable is a and not a) is the external work done by a unit mass of a body over and above any work of expansion (pda), i.e., da  dwtot - pda , then we can use (4.6), (4.7) and the above to write da = -dg -sdT + adp. For an isothermal, isobaric process, da = -dg. The thermodynamic functions f and g have important applications in problems involving phase changes in the atmosphere. In particular, these functions will be utilized later in this course when we consider the formation (nucleation) of water droplets from the vapor phase. [In other words, don’t forget about g!]

Appendix: The Carnot Cycle: Highlights
The Carnot cycle may be one the the most popular examples used in the study of (general) Thermodynamics. ( See Petty, pp ). The Carnot cycle illustrates several aspects of the Second Law, and also defines thermodynamic efficiency. The Carnot cycle is a sequence of 4 component processes, two isothermal and two adiabatic. These component cycles are interlaced as follows: reversible isothermal expansion at T = T1 reversible adiabatic compression at q = q1 reversible isothermal compression T = T2 reversible adiabatic expansion q = q2 increasing p p2 p1 Fig. A.1 illustrates these paths as they would appear on a skew-T, ln p diagram. (note, Tsnonis uses a p-V diagram.)

The quantitative measures of work, internal energy change, and heat input along each leg are detailed in Petty, pp Take some time to examine these. From the Carnot cycle, the thermodynamic efficiency can be defined as Efficiency is zero when T1 = T2, and is maximized when T2 << T1.

Tsnonis mentions two postulates that originate from this, and these are alternate statements of the Second Law: Kelvin’s postulate: It is impossible for a thermal engine to accomplish work at only one temperature (p. 52, Tsnois). Clausius’s postulate: A transformation that permits heat transfer from a cold body to a hot body is impossible (p. 53, Tsonis). (Is this not a restatement of the 2nd Law?) Recall that the First Law does not address the possibility of transformations; it only quantifies them, even if they are impossible. (Think of the First Law as the smart person who has no common sense, and the Second Law as the wise person who has abundant common sense.) Question for discussion: How does the Carnot Cycle illustrate the way in which a heat pump (or refrigerator) works.

Another example problem:
Calculate the change in entropy when 1 mol of an ideal diatomic gas initially at 13 °C and 1 atm changes to a temperature of 100 °C and a pressure of 2 atm.

Find the change in air pressure if the specific entropy decreases by 50 J kg-1 K-1 and the air temperature decreases by 5%. ds = cpdT/T - Rdp/p Rearrange: dp/p = (cp/R)dT/T - ds/R Integrate: ln(pf/pi) = (cp/R)ln(0.95Ti/Ti) - s/R Insert values: ln(pf/pi) = (1005.7/287.05)ln(0.95Ti/Ti) - (-50)/287.05 = ( ) = pf/pi = 0.994 pf = 0.994pi

HW problems Notes: Problems 1-4
Hint for No. 1a: Internal energy has 3 components

The water-air heterogeneous system
Chapter 5 The water-air heterogeneous system

On-line resource on-line analytical system that portrays the thermodynamic properties of water vapor and many other gases Note: if you are motivated, search the web for resources that relate to this course, and bring them to my attention.

Review on equilibrium thermodynamic equilibrium -- a system is in thermal equilibrium (no temperature difference) mechanical equilibrium -- a system in which no pressure difference exists between it and its environment chemical equilibrium -- condition in which two phases coexist (water in this case) without any mass exchange between them Example of chemical equilibrium A closed container initially filled partly with water in a vacuum will be in chemical equilibrium when p = es (the saturation vapor pressure). At this point, the number of molecules passing from liquid to vapor equals the number passing from vapor to liquid. Vapor in equilibrium (vacuum initially) pequil  es water

A picture of what we are considering
A picture of what we are considering. Water in an initial vacuum will eventually reach an equilibrium. Equilibrium is achieved in the saturated state. We will examine this more closely. Taken from Petty, Fig. 7.1

Motivation We require an expression for the saturation vapor pressure (or saturation mixing ratio as a function of temperature in order to develop a derivation for the saturated adiabatic lapse rate. It turns out that saturation vapor pressure is a function only of temperture, i.e., es = es(T). Fig. 7.2 from Petty. Plot of the functional form of es(T). Note the difference between ice and water for T < 0 C. This is important in the growth of ice in clouds. Fig Same as fig. 7.2, except es on the vertical axis is linear.

5.2 Chemical potential If a single molecule is removed from a material in a certain phase, with the temperature and pressure remaining constant, the resulting change in the Gibbs free energy* is called the chemical potential of that phase. In other words, the chemical potential can be defined as the Gibbs free energy per molecule. The chemical potential of an ideal gas is defined by m = m0 + RTlnp, [dm = RTdlnp = RTdp/p = adp] where m0 is the chemical potential for p = 1 atm. For an ideal gas, Dg = m2 - m1 = RTln(p2/p1) [= -wmax] * Recall, dg = -sdT + dp for a reversible transformation (g is Gibbs free energy)

Example (Taken from Wallace and Hobbs, 1977, pp. 101-102):
Derive an expression for the difference in chemical potential between water vapor and liquid water, mv - ml, in terms of the bulk thermodynamic properties: e (the partial pressure of water vapor) and T. For a (vapor) pressure change de, we have (from dg = -sdT + adp), for vapor and liquid, respectively, dmv = avde dml = alde, where av is the specific volume of a water vapor molecule and al is the specific volume of a liquid water molecule. Then it follows, by combining the equations above, that d(mv - ml) = (av - al)de  avde (5.1) since av >> al. [Values for av and al are given below.] We will apply this approximation (av >> al) many times in subsequent derivations.

The equation of state for water vapor (applied to just one water vapor molecule) is
eav = kT, (5.2) where k, Boltzmann's constant (1.381 x J K-1 molecule-1), is used in place of the gas constant to represent one molecule. (Recall that Rd = 287 J K-1 kg-1) Now solve (5.2) for av, and insert the result into (5.1) to get d(mv - ml) = kTde/e. (5.3) Since mv = ml at e = es (saturation, which is equilibrium) we can integrate (5.3) as follows: or (5.4)* This result will be used later in development of the theory of nucleation of water droplets from the vapor phase. [Nucleation is therefore closely related to thermodynamics.]

5.3 Changes in state We will now consider various equilibria for water in the atmosphere. 5.3.1 Gibbs phase rule For a system of C independent species and P phases, the Gibbs phase rule states that the number of degrees of freedom F (i.e., the number of state properties that may be arbitrarily selected) is F = C - P (5.5) If we deal with a single component system (C=1) such as water then we can form the table below. P (number of phases) F (degrees of freedom) 1 2 3 Water vapor eq. of state Equilibrium: gas-liquid Triple point

For a single existing phase (such as water vapor) we may arbitrarily choose p (e) and T (i.e., the equation of state). Then F = 2. For two phases, the choice of either p or T determines the other (F = 1), and for three phases (i.e., the triple point) there is no choice, i.e., this point has specific values for p and T (F = 0). The phase diagram for water is depicted in Figs. 5.1 and 5.2. Along the two coexisting phase lines, one degree of freedom is possible, within each phase two degrees of freedom exists, and with three coexisting phases (the so-called triple point) there are none. We will now consider the phase lines. Water vapor pressure

A closer look: P (number of phases) F (degrees of freedom) 1 2 3
Equation of state: ea = RvT Clapeyron Equation Clausius-Clapeyron Equation: es = f(T) No equation (a fixed point)

3-D surface for water Key points on a p-T diagram for water (Figs. 5.1, 5.2 and 5.3): triple point: pt = Pa = mb; Tt = K = 0.01 C ait = x 10-3 m3 kg-1 awt=1.000 x 10-3 m3 kg-1 avt = 206 m3 kg-1 critical point: Tc = 647 K = 366 C; pc = atm ac = 3.07x10-3 m3 kg-1 Fig Thermodynamic surface for water. Taken from Iribarne and Godson (1973).

Clausius-Clapeyron Equation
Fig P-T phase diagram for water. Stable phase boundaries are shown by solid lines; metastable phase boundaries are indicated by dashed lines. Note that the ordinate in exponential in pressure, in contrast to the linear p dependence in Fig From Young (1993), p. 401.

5.3.3 Equilibrium between solid (ice) and liquid water - The Clapeyron Eq.
The lines drawn in Figs represent the equilibrium condition between two phases of water. These are important in our study of thermodynamics, and the equilibrium line between the liquid and vapor phase is particularly valuable. These equilibrium conditions can be expressed analytically, and the basis for the functions derived in the following is the First Law, from the the Gibbs free energy is derived. At equilibrium dg=0, so therefore g(solid) = g(liquid). [dq = -sdT + adp] -sliquiddT + aliquiddp = -ssoliddT + asoliddp Define the following: Ds = ssolid - sliquid Da = asolid - aliquid Now combine the terms the multiply dT and dp.

Then Ds dT = Dadp. We solve this as follows, and utilize the definition of entropy change for a phase transition (Ch. 4): dp/dT = Ds/Da = -DHfusion/(TDa) Clapeyron Equation (5.6) where the latter equality is based on the entropy change associated with a phase change (Chap. 4): Ds = -DHfusion/T Since DH > 0 (heat is liberated upon freezing) and Da > 0 (freezing water expands), the slope dp/dT < 0, as shown in Figs. 5.1 and 5.2. One implication: If pressure is increased, then melting can occur. An example is the ice skate, which works so well since a water film forms between the skate blade and the ice surface as the pressure (F/A) rapidly increases.

5.3.4 Liquid/vapor equilibrium - the Clausius-Clapeyron Eq.
In this case Da = aliquid - avapor  -a vapor. (avapor = 206 m3 kg-1 ; aliquid = 1.00 m3 kg-1) Then, from the equation of state, we can write -av = -RvT/es (es is used in place of p since we are concerned with equilibrium at saturation) Then des/dT = -DHvap/(Tav) = DHvapes/(RvT2) or dlnes/dT = Lvl/(RvT2). (5.7) Clausius-Clapeyron Eq. (see also Figs. 7.2, 7.3, 5.1, 5.2)

Recall that Lvl is not a constant, but varies by ~10% over the temperature interval (0, 100 C), and by about 6% over the interval (-30, 30 C). (See Table 5.1) However, to a first approximation, we can assume that Lvl is constant and integrate Eq. (5.7): which is solved as where T0 = K and es0 = 6.11 mb (from lab measurements).

Rewriting this expression and grouping the constants into new constants A and B yields the simple, approximate expression: es(T) = Ae-B/T, (5.8) where A = 2.53 x 108 kPa and B = 5.42 x 103 K. But beware: This form is acceptable to use for quick calculations, but precise calculations require a derivation that takes into account Lvl = Lvl(T). A more elegant derivation of the Clausius-Clapeyron equation given in Rogers and Yau, pp We will also consider this.

In order to get a more accurate representation of es(T), the temperature dependence can be incorporated using a linear correction term involving T in an expression for Lvl, and then substitute this in Eq (4.10) and integrate. A more accurate form is (see HW problem) log10 es = /T log10T (es in mb and T in K) Bolton (1980) determined the following empirical relation by fitting a function to tabular values of es(T). This form is accurate to within 0.1% over the temperature interval (-35 C, +35 C): (5.9)

Liquid-vapor equilibrium line
The Clausius Clapeyron Eq. is represented by an exponential function: Liquid-vapor equilibrium line where B = and C = (from the Bolton empirical relation on the previous page. Know this derivation; also look over Petty Section 7.3, pp

Table 5.1. Saturation vapor pressures over water and ice
T (C) es (Pa) ei (Pa) Llv (103 J kg-1) Liv (103 J kg-1) -40 19.05 12.85 2603 2839 -35 31.54 22.36 -30 51.06 38.02 2575 -25 80.90 63.30 -20 125.63 103.28 2549 2838 -15 191.44 165.32 -10 286.57 259.92 2525 2837 -5 421.84 401.78 611.21 611.15 2501 2834 5 872.47 2489 10 2477 15 2466 20 2453 25 2442 30 2430 35 2418 40 2406

Accuracy of es(T) = Ae-B/T vs.
Relatively accurate (with ~1%) for the T interval [-20, 20 C] 1% error

Expression for saturation vapor pressure over ice is similar to that over water:
Why do the constants differ? The difference (es - ei) is important in the growth of ice crystals Fig. 7.6

To what temperature can water in a pan be heated?
It depends on _____________ Consider Problem 7.1

To what temperature can water in a pan be heated?
It depends on pressure

Problem 7. 1 By consulting Fig. 7. 2 and Fig. 1
Problem 7.1 By consulting Fig. 7.2 and Fig. 1.7, estimate the altitude at which the boiling point of water drops to room temperature (25 C). Find es from Fig. 7.2, or the preceding table: es (25 C) = 32 mb Determine z from the relation: p(z) = p0 exp(-z/H) Rewrite to solve for z: ln (p/p0) = -z/H z = -H ln(p/p0) (let H = 8 km) z = -H ln (32/1013) = m (-3.45) ≈ 27 km

5.4 Water vapor measurement parameters (some previously defined)
Common variables used for measurement of moisture: mixing ratio (rv) rv= mv/md = ee / [p-e]  ee/p (5.10) specific humidity (qv) qv = mv/(mv+md) = rv/r = rv / (rd+rv) = ee / [p-(1-e)e] (using the equation of state for rv and rd)  ee/p Thus, qv  rv. vapor pressure (e) (connection to Clausius-Clapeyron Eq.) -- Don’t confuse e with es relative humidity (f) f = rv / rvs(T,p)  e/es(T) = rvs(Td)/rvs(T) (5.11) Be careful - f can be expressed in % or as a fraction (e.g., 66% = 0.66) virtual temperature (Tv) (defined previously; not really used to measure water vapor) Tv  T( rv) dewpoint temperature (Td): temperature at which saturation occurs; defined as a process in Chap. 6; Td = f(rv, p)

Some problems to work in class:

HW Problems Notes: 2, 3, 4 Petty 7.2, 7.6
Note: Section 7.5 and following will be addressed next in my Ch. 6-7 notes.

Chap. 6 ATMOSPHERIC THERMODYNAMIC PROCESSES
[see also Petty, Section , pp ]

Objectives: Develop other important thermodynamic variables and applications of the fundamental relations that we have considered to this point. Use and applications the skew-T diagram to examine atmospheric processes. Examine some important atmospheric thermodynamic processes. In particular, we will explore the behavior of water vapor and its effects on atmospheric processes.

Saturation is common in the atmosphere
Near the surface on clear, calm nights

BNA LCH Formation, evolution, and movement of clouds

Depiction of saturation in a skew-T, ln p sounding (1200 UTC, 13 Sept 2007, Lake Charles, LA)

Thick Ci clouds

6.1.1 Some processes that define additional thermodynamic variables
There are four natural processes by which saturation can be attained in the atmosphere. These are: isobaric cooling (dq0, rv=const), e.g., by radiative cooling (diabatic cooling, dq < 0), in which the temperature T approaches the dew point temperature Td; evaporational cooling (dq0, rvconst) in which a decrease in T and an increase in Td result in the wet-bulb temperature Tw, (at which point the air is saturated); adiabatic cooling (dq=0, rv=const) in which saturation is produced at the saturation point temperature[1] Tsp by adiabatic expansion; mixing of two air masses – in this case saturation can be analyzed from a “saturation point” mixing analysis. [1] The saturation point temperature is also known as the temperature of the lifting condensation level (Tlcl) or the isentropic condensation temperature (Tc).

Saturation adiabat Dry adiabat Figure Illustration of processes by which saturation may be achieved in the atmosphere. This skew-T diagram also illustrates the graphical method to determine Td, Tw and Tsp. Illustration of Normand's rule. This is an important figure!!!

Upper air observations

isobaric process Isobaric cooling and the dew point temperature, Td.
(radiational) cooling occurs in the presence of constant water vapor (e=const or rv=const). Under clear sky conditions the radiational cooling frequently reduces the surface temperature to the dewpoint temperature. Cooling is greatest at the surface, as illustrated in Fig. 6.1, a 1200 UTC sounding from Salem, Oregon. T and Td are nearly superimposed (i.e., the air is saturated) and fog was reported in the region. This sounding was obtained from the web site

An even better example of a surface based T inversion and saturation

T  Td (RH  100%) Low winds in the NBL + clear skies + dry air above 1 km = maximum surface cooling Fluctuating wind (turbulent bursts) Relation between Dqrad and DT

Further illustration: For the next two weeks, refer to T, Td time series over night from Derivation of Td: dp=0, dq0 and dh=dq. Physical process: As isobaric cooling proceeds with no change in the absolute moisture content, a temperature is reached in which the air just becomes saturated (T=Td, or e = es). Consider the relationship between Td and the relative humidity f. We can write rv = rvs (p,Td) (definition) and use an expression for es(T) and the approximate relation rvs=ees/p. We begin with the integrated approximation of the C-C equation: es = Ae-B/T (6.1) Take the natural log of each side (i.e., ln es = ln A - B/T), utilize the approximate formula es=prvs/e, and then solve for T (which is Td in this case).

The approximate analytical expression for Td in terms of rv and p can then be expressed as
(6.2) This relation explicitly shows that Td is a function of rv and p. Given values of rv and p, one can graphically determine Td on a skew-T as shown in Fig. 6.2 below. As an extension of this problem, we will consider fog formation in Section The accuracy of this relation is dependent on the accuracy of es = Ae-B/T [accurate to about 1% in the -20 to +20 C interval]

From Petty.

Relationship between Td and relative humidity f.
Determine how relative changes in Td are related to relative changes in f. Again, we utilize the formula and take the log differential to get dlne = dlnp + dlnrv. (6.3) Combine (6.3) with the Clausius-Clapeyron equation, written in differential form (here e = es since T=Td) (6.4) to obtain, after some rearranging, the following: , Depends on relative variations in p and rv

Dividing both sides by Td yields
where the latter approximate equality is obtained from the term [RvTd/Lvl] by assuming Td=270 K, Lvl=2.5x106 J kg-1, and Rv= 461 J kg-1 K-1 . This result indicates that the relative increase in Td (here, relative refers to the ratio dTd/Td, or an incremental change relative to the total value) is about 5% the sum of the relative increases in p and rv. We now integrate the C-C eq. (6.4) above to get

We then solve for so-called dewpoint depression (T-Td), use decimal logarithms [using the definition that log10 x = ln x / ln 10 = ln x], and substitute for constants to get (T-Td) = 4.25x10-4 TTd(-log10f) For TTd =2902 (i.e., assuming T = 290 K and Td = 290 K) we have T-Td  35(-log10f) (6.5) Then for f=0.8, (T-Td)  3.5 C = 6.3 F. Thus, a change of (T-Td) every 1 F translates to a change in f of about 3.2%. This verifies my general rule of thumb that [at least] for f > 0.8, the dewpoint depression, T-Td, is 1 F for every 3% change in f, for f  100%. For example, if T=75 F and f=0.88, then Td71 F. Observational question: What is the range of Td in the atmosphere? What is the upper limit of Td, and where would this most likely occur?

A reading assignment:

b) Isobaric wet-bulb temperature (Tiw)
We will consider the isobaric wet-bulb temperature Tiw here – there is also an adiabatic wet-bulb temperature, Taw. The wet-bulb temperature is achieved via the process of evaporation. Practical examples of Tiw are evaporation of rain and the evaporation of the wet bulb wick on the sling psychrometer, a device which measures the dry and wet-bulb temperatures. While the process is isobaric (ideally), the parcel gains rv at the expense of a decrease in T. Assuming that a parcel of unit mass (1 kg) contains rv kg kg-1 of water vapor, we can write from the First Law (p=const) dq = cpd( rv)dT = cpmdT [cp=cpd] The heat loss from evaporation (including a mass rv of water vapor) is (1+rv)dq = -Llvdrv Equating the two expressions above gives cpddT = -Llvdrv[1/(1+rv)][1/(1+0.9rv)]  -Llvdrv(1-1.9rv) cpdT  -Llvdrv (within ~2%, since rv ~ 0.01) (6.7)

6.1.1 Some processes that define additional thermodynamic variables
There are four natural processes by which saturation can be attained in the atmosphere. These are: isobaric cooling (dq0, rv=const), e.g., by radiative cooling (diabatic cooling, dq < 0), in which the temperature T approaches the dew point temperature Td; evaporational cooling (dq0, rvconst) in which a decrease in T and an increase in Td result in the wet-bulb temperature Tw, (at which point the air is saturated); adiabatic cooling (dq=0, rv=const) in which saturation is produced at the saturation point temperature[1] Tsp by adiabatic expansion; mixing of two air masses – in this case saturation can be analyzed from a “saturation point” mixing analysis. [1] The saturation point temperature is also known as the temperature of the lifting condensation level (Tlcl) or the isentropic condensation temperature (Tc).

Assume that Llv and cp are constant (which is a good assumption since the temperature reduction DT=T-Tiw associated with evaporation is typically <10 K) Integrate the above to get the wet-bulb depression (T – Tiw) T-Tiw = (Lvl/cp)(rvs(Tiw,p) - rv). Use the Clausius-Clapeyron formula (5.5) [es(T) = e-B/T] to get an iterative formula for Tiw: Tiw = T - (Llv/cp)[(e/p)Ae-B/Tiw - rv], (6.8) where T and rv are the initial parcel values.

On the psychometric equation (Bohren and Albrecht, pp 282-284):
This equation provides a relation between vapor pressure (e) and the wet bulb depression, (T-Tiw). T is termed the dry-bulb temperature, and Tiw the wet bulb temperature. Both can be measured with a sling psychrometer (or an Assman aspirated psychrometer, see Fig. 7.19); these measurements are used to determined the vapor pressure e. (In the “old days” tables were used.) Derivation? Same starting point: cpdT  -Llvdrv

Sling psychrometer (hand-held)
Handheld Relative Humidity and Temperature Meter

How good is it? Possible to achieve an inaccuracy of <1% RH
Sensitivity increases markedly as T increases, and slightly as the RH decreases Assmann psychrometer Can be used to check other instruments

Psychrometric chart Tables are also available

Measurement of relative humidity
Ex: Vaisala HMP-45C

Problem 7.11 using the skew-T for an approximate solution
8.6 g/kg 16.5 g/kg Given: T = 20 C Td = 10 C p = 900 hPa Find f f = rv/rvs = rvs(Td)/rvs(T) = 8.6 / 16.5 = 0.52

Equivalent temperature (Tie)
This is the temperature achieved via isobaric (p=const) condensation (latent heating) of all water vapor. Tie is a fictitious temperature – there is no atmospheric process that is associated with it. (In fact Tsonis notes that Tie is the reverse of an irreversible process associated with Tiw.) Thus, this is also referred to as the isobaric equivalent temperature (Tie). This process is similar (but opposite) to that of the isobaric wet-bulb temperature, Tiw, so the same equation applies. In this case, integration of (6.7) gives or Tie = T + Llvrv/cp. (6.9) How does the isobaric equivalent temperature differ from the adiabatic equivalent temperature? What is the relation between adiabatic equivalent temperature and adiabatic equivalent potential temperature? [Brief discussion here.] Tie and Tiw are related by Eq. (6.7) and represent the respective maximum and minimum temperatures that an air parcel may attain via the isenthalpic (adiabatic and isobaric) process.

d) Saturation point temperature (Tsp)
This is also called the "isentropic condensation temperature" (Tc) as defined by Bolton (1980), or the more classical temperature of the lifting condensation level (Tlcl). Tsp is achieved via adiabatic lifting (cooling by expansion). The value of Tsp is easily found graphically on a skew-T diagram (see Fig. 6.2). Recall that the adiabatic equation can be derived from the First Law and equation of state to get cpdT = RdT(dp/p). Also recall that the integrated form is Poisson's equation (T/T0) = (p/p0)k (6.10) We now note that Tsp = Td(rv,psp). Substitution of (6.4), the expression for Td, into (6.10) gives an iterative formula (derivation given in Rogers and Yau 1989):

More accurate (and explicit) empirical expressions are given by Eq
More accurate (and explicit) empirical expressions are given by Eq. (21) in Bolton (1980): (using vapor pressure e) What about a form that has mixing ratio as an input? Use rv = e/p  e = rvp/ (using relative humidity f) For these two formulations, Tsp is in C, T in deg K, f in %, and e in mb. The graphical method of determining Tsp is known as Normand’s Rule (p. 207 in Petty), illustrated in Fig. 6.2 (and Petty Fig. 7.16) (To be clarified in class.)

Back to Problem 7.9e:

We have been considering saturation by adiabatic expansion:
adiabatic expansion  cooling  tendency towards saturation Taking the log differential of relative humidity, f=e/es we get dln f = dlne – dlnes Note that the ratio e/p = Nv is constant during ascent, which is equivalent to saying that the mixing ratio rv = ee/p is conserved. Furthermore, from Poisson’s equation, Tp-k is constant (i.e., q is conserved). Since e = Nvp, then Te-k = Nv-k x const = new const. Taking the log differential of the above, we obtain dlnT = kdlne (or dlne = k-1dlnT).

Use the differential form of the C-C eq.
to write The first term on the RHS is the change due to a decrease in p (and e) The second term represents the change in f from a decrease in T and es(T). These terms have opposite signs; therefore, adiabatic expansion could increase or decrease f.

To clarify this point, we can write the above to represent the slope, df/dT, the sign of which we want to determine: In other words, we are asking, “How does f vary with T?” This equation shows that df/dT < 0 (i.e., f increases when T decreases) when the condition cpT < eLvl or T < eLvl/cp  1500 K.

Adiabatic expansion and condensation
Does adiabatic expansion necessarily produce condensation? Consider the following example hypothetical problem: Condensation of water can occur in updrafts because the saturation mixing ratio decreases in adiabatic ascent. This property of water can be attributed to the high value of latent heat of condensation. It has long been speculated that there may be trace gases which, because of low values of L, would condense in downdrafts (Bohrens 1986). Show that the criterion that must be satisfied if vapor is to condense in downdrafts (adiabatic compression) is L < cpT/e. Solution: From the definition of f=e/es, we can write dlnf/dz = dlne/dz – dlnes/dz. Since e = rvp/e (and mixing ratio rv is constant), dlne = dlnp. Then dlnf/dz = dlnp/dz – (dlnes/dT)(dT/dz). We now use the C-C equation dlnes/dT = L/(RvT2) and insert into the previous equation: df/dz = p-1dp/dz – (L/RvT2)dT/dz. Recall that the dry adiabatic lapse rate (dT/dz here) is dT/dz = -g/cp. Also, p-1dp/dz = p-1g/a = g/RdT. Substitution of these into the previous yield dlnf/dz = g/RdT – gL/(cpRvT2) = g/(RdT)[1 -LRd/(cpRvT)] = g/(RdT)[1-(Le/cpT)]. Thus, if f increases with decreasing height, the term in brackets should be > zero, i.e., 1 – Le/CpT > 0. Rewriting, L < cpT/e is the criterion for saturation upon descent. For the atmosphere, cp = 1005, T = 290, and e = 0.622, we have L < 4.7x105 J kg-1. This is clearly not satisfied for water, but is possible for some volatile substances.

Determination of cloud base from the dew point depression, (T-Td)
There is a practical application that is closely assocated with Tsp. In this application we will derive a relationship between the height of Tsp and the surface dewpoint depression, T-Td. Assume that a surface parcel rises (adiabatically) until condensation occurs (this defines a Cu cloud cloud base). In reality, rv typically exhibits a negative vertical gradient, because the source of rv is surface evaporation, and the sink is mixing from above. The relation that we derive will provide a useful formula for estimation of the base of cumulus clouds, given a measurement of (T-Td) at the surface. Recall that the lapse rate for a subsaturated parcel is given by the dry adiabatic lapse rate (g/cp), approximately 10 K km-1. To estimate the height at which condensation occurs, we need to examine the variation of Td along a dry adiabat. This is given by the C-C eq.

Using the relation dlnT = kdlne (k = Rd/cpd) we can rewrite the above as
For T  Td  273 K, and using finite differences, we obtain the approximate relation DTd  (1/6)DT, i.e., the magnitude of the Td decrease is about one sixth that of the adiabatic lapse rate for a parcel undergoing adiabatic ascent (see Fig. 6.3). Fig Illustration of the relation between decreases in T and Td during adiabatic lifting of a subsaturated parcel

This provides a basis for the following equations 7.29 and 7.30.
More on the saturation point (SP) or lifting condensation level (LCL) Reference to Section 7.6 in Petty (p. 191) We have developed a precise relationship for Tsp Some approximate relationships for psp and zsp: Psp  pexp(-0.044Td) (7.29) where Td is the dewpoint depression (T - Td) in deg C. zsp  (T-Td) / (7.30) Problem 7.13 T = 30 C, f = 0.70 Find Td: First, find rv using rv=rvs(0.70) = es(30 C)(0.70)/p Then use Eq. (6.2) from my notes: Find zsp = (T-Td)/8 This can be done with the skew-T (next slide)

Problem 7.13 using the skew-T for an approximate solution
19.6 g/kg T = 30 C f = 0.70 rvs = 28 g/kg (skew-T) rv = 28(0.70) = 19.6 Then Td = 24.5 C (skew-T) Zsp = (T-Td) / 8 = 5.5 / 8 = 0.7 km *Assumed p = 1000 hPa

6.2.1 Derivation of the reversible saturated adiabatic lapse rate*
We have considered a related topic in the derivation of the pseudo adiabatic lapse rate and qe. In Chap. 3 (notes), we considered a preliminary form of the pseudo-adiabatic lapse rate, Eq. (3.21), reproduced here: (3.21) The term in the denominator required the Clausius-Clapeyron Equation to provide an expression for drvs/dT. This term is related to the magnitude of latent heating within the saturated parcel. As shown in Fig. 6.4, the local lapse rate along the saturated adiabat in the lower right side (warm, high water vapor content) is relatively low, while at low pressure and cold temperature (upper part) the local value of the saturated adiabat approaches that of the dry adiabat. * Corresponding material is in Petty, Section 7.7

Figure Variation in the local value of T/z along the saturated adiabat (bold solid line) on the Skew-T, ln p diagram.

One can make two limiting assumptions regarding the condensed water:
i) It is carried along with the parcel. ii) It immediately leaves the parcel (by removal – called sedimentation – from the precipitation process). In reality, the clouds in the atmosphere are somewhere in between these two extremes. The latter process is defined as the pseudoadiabtic process and simplifies matters since one need not consider the heat content of the condensed water (condensate) that is carried with the parcel. This process (and lapse rate) was considered in Section , Eq (above). We also note that the pseudoadiabtic process is irreversible, whereas the saturated adiabatic process (i) is reversible. It turns out that the lapse rate defined by each is nearly the same, but the differences may be important in some situations.

The starting point differential equation (based on the First Law) for the reversible process has an extra term that expresses that amount of heat contained by the condensate: (6.11) where ra is the weight of condensed water per gram of air (this will be referred as the adiabatic liquid water content). Our starting point for the pseudo-adiabatic lapse rate was a simplified form of (6.11), already having the approximation Lvl=const. We also ignored the contribution of enthalpy from moist air, the second term within the brackets on the RHS of (6.11). The key term here is the first term on the LHS, cwradT, the heat stored by the condensate.

Introducing the equation of state (p=rRdT) and then expanding terms in (6.11) yields
The final form (taken from Iribarne and Godson, 1973) is A1 where Rm = Rd(1+0.61rv), the total water mixing ratio rtw = rvs + ra (ra is the adiabatic mixing ratio of condensed water), and Gd = g/cpd. Refer to the handout, copied from Chapter 6 of the book Atmospheric Thermodynamics (Bohren and Albrecht (1998) for details of the derivation.

The formula for the pseudoadiabatic lapse rate is (assuming that no liquid water remains with the parcel, i.e., rtw - rvs = ra = 0 in the above equation). A2

One can also make the following approximations in Eq. A2:
to get the approximate form for the pseudoadiabatic lapse rate (also letting Rd ≈ Rm): A3 This is an approximation for the local slope and should never be used be used for calculations requiring an accurate parcel T, since the errors will accumulate to unacceptably high values in the integration. Compare with Eq from Bohren & Albrecht

Example (from Iribarne and Godson, p. 158):
This will illustrate the errors for a saturated parcel with the following values: T = 17 C, p = 1000 mb, es = 19.4 mb, rvs = kg kg-1. Also assume that ra = kg kg-1. Equation A1, A2 and A3 yield the following values: Gs-rev = 4.40 K km-1 Gs-pseudo = 4.42 K km-1 Gs-approx = 4.50 K km-1 These differences are appreciable. Another example take from Emanuel (1994, p. 133) illustrates how the differences can accumulate, as shown below in Table A1. Also refer to Bohren and Albrecht, Fig. 6.3, reproduced below.

Table A1. Parcel temperature obtained from Gs-rev (Eq
Table A1. Parcel temperature obtained from Gs-rev (Eq. A1) and Gs-pseudo (Eq. A2), for a saturated parcel ascending with initial conditions p = 950 mb, T = 25 C (very wet) (From Emanual 1994) p (mb) Ts-rev (K) Ts-pseudo (K) Ts-rev-Ts-pseudo (K) Tr-rev – Tr-pseuo 950 298.15 800 292.36 292.35 0.01 -1.04 700 287.77 287.73 0.04 -1.75 600 282.32 282.22 0.10 -2.44 500 275.59 275.36 0.23 -3.07 400 266.78 266.27 0.51 -3.51 300 254.10 252.90 1.21 -3.39 200 233.30 230.32 2.98 -1.76 100 195.77 189.96 5.81 1.70 The right-most column in Table A1 is the difference between density temperatures (Tr) for the two processes. Density temperature is the temperature dry air would have to yield the same density as moist, cloudy air (i.e., virtual temperature including water vapor and cloud condensate – more on this later).

Notes: Perhaps the most common application involves the calculation of convective available potential energy (CAPE), which requires an accurate calculation of the parcel Ts during saturated ascent. CAPE is discussed in Chap. 8 of the Notes and pp of Petty. If we recognize that qe is conserved along a saturated adiabat, then a numerical calculation of Ts should conserve qe. This can be used as a check for the accuracy of the Ts calculation. After having considered the differences between the reversible and pseudo saturated adiabatic processes, we can ask the question “Which is the best to use?” We cannot fully answer this question until we have a greater understanding of cloud physics and microphysical structure of clouds. Effects of freezing. If we use the latent heat of deposition, then a difference will exist between Gs in Eq. A3 above for condenstation and deposition. This difference is shown in the figure below.

Relative orientation or slope (magnitude) of dry and saturated adiabats on the skew-T diagram. (Fig. 6.7) The magnitude of the saturated adiabatic lapse rate is slightly less than that of the pseudoadiabatic lapse rate. The magnitude of the dry adiabatic lapse rate (for dry air) is slightly greater than its counterpart for moist (rv nonzero) air, since the effective cp is larger [cp( rv)]. The ratio of the dry to moist adiabatic lapse rates is For a mixing ratio of 20 g kg-1, this ratio is Thus, a difference up to ~2%, or 0.2 K km-1, is possible. This is potentially significant!

6.2.2 Adiabatic liquid water content (see also p. 222 in Petty)
A measure of water condensed along the saturated adiabat. Expressed in g m-3 by multiplying rca by the density, r, of moist air. Typical values of rc in clouds are in the range 0.5 to 3 g m-3. During saturated adiabatic ascent, a reduction in water vapor (-drvs) is a gain in cloud condensate (rc), such that the total water is conserved. Thus, dc = rdrca = -rdrvs. (6.12) c can be derived from the adiabatic equation. It is related to the difference between the dry and saturated adiabatic lapse rates (this is a homework problem). c is easily found graphically as shown in the figure below. Its value requires knowledge of cloud base (saturation point) conditions (rvs,p) and is computed as  = rm [rvs(Tsat,p)qe=const - rvs(Tsp,psp)] Tsat is evaluated along a saturated adiabat (line of constant qe).

Figure 6.4. Graphical determination of adiabatic liquid water content.

Adiabatic liquid water content illustration from Petty

6.2.3 Relation between qe and qw
qe was derived in increasing levels of accuracy in this chapter, and in Chap. 3. One can determine w if Tw is known, and then calculate the temperature of the parcel after it descends along a saturated adiabat to 1000 hPa (mb). In a sense, this process is the opposite of that by which qe is achieved, where the parcel ascends along a saturated adiabat until all water vapor is removed by condensation. This numerical computation of either of these processes would require use of Eq. A1 or Eq. A2. The derivation of qw begins with the assumption that the parcel is at its saturation point. We start with the differential equation and integrate

The result is (A5) This is another transcendental equation that requires iteration. Since the exponent term is less than ~0.2, (A5) can be approximated as (A6) If we can further assume that the ratios q/Tsp and q/qw are near unity, then further approximation provides the following: (A7) This reveals the relationship between qw and qe, both of which are conserved for saturated adiabatic processes

Heat Index (HI) and thermodynamics
Good descriptions: Is there any correlation between HI and Tie? Extra credit (25 pts): Create a scatter plot of HI vs. Tie (or some other appropriate thermodynamic variable) for a range of temperatures and RH (e.g., T = 80, 85, 90, 95 and RH = 30, 40, 50, 60, 70, 80, 90% for each T)

6.2.4 The wet equivalent potential temperture (qq)
See RY, pp for derivation. qq is conserved for saturated processes, and has been used to analyze mixing processes in clouds. More on applications of this will be considered later.

Mixed layers (as in a convective boundary layer)
a) Unsaturated mixed layer (Fig. 7.24b,c) b) Cloud-topped mixed layer (Fig. 7.24d)

6.4 Example computations using the skew-T diagram
Given: p=900 hPa, T = 25 C, Td = 15 C (variables commonly reported) Find: rv, f, q, Tv, Tsp, Te, Tae, qe, Tw, qw, c(at 50 kPa)

Tsp < Td < Tw < T < Tv < Tie < Tae
The following figure and table (i) summarize the relative magnitudes of the basic and derived thermodynamic variables that we have considered to this point and (ii) illustrate their graphical determination. The non-potential temperatures exhibit the inequality: Tsp < Td < Tw < T < Tv < Tie < Tae Variable Process Isbaric warming or cooling (without condensation or evaporation) such as radiation Isobaric evaporation or condensation Non-saturated adiabatic expansion or compression Saturated adiabatic expansion or compression (reversible) f NC C e or Td rv Tw or Tae q qe or qw Tsp T NC – not conserved ; C – conserved

qw Figure Summary of temperature/humidity parameters, plotted on a tephigram. Line orientations are similar for a skew-T, ln p diagram. (After Iribarne and Godson, 1973)

Example problems: From Petty: 7.13 7.14 7.16 7.17 7.22

Problem 7. 14: The definition (7
Problem 7.14: The definition (7.55) of equivalent potential temperature assumes that all condensation results in the appearance of liquid water, regardless of the temperature. If freezing occurs, additional latent heat is released, which further raises the temperature of the parcel. Assume that a parcel is at saturation at a temperature of freezing and a pressure of 900 hPa. a) What is the equivalent potential temperature e according to the traditional definition? b) By how much would e increase if all condensate were assumed to freeze? Answer: 1.5 K See Eq (p. 203) and make the correction Assume that the condensate freezes instantaneously at a fixed pressure Then dq = -Llidrvs = cpdT Note: The value of rvs at 0 C and 900 mb is about 4.3 g kg-1 = kg kg-1 Integrate and solve for T: T = -Lrvs / cp = x 105 J kg-1 ( ) / J K-1 kg-1 = 1.42 K But note: cpm = cpd ( rv) = ( * ) =

Problem 7.17: The equivalent potential temperature of a parcel is
310 K. (a) Graphically, estimate its wet-bulb potential temperature w. (b) If its potential temperature q is 300 K, graphically find its mixing ratio w. Do we need to assume a pressure??? No. The pressure is implicitly given. Use a skew-T and find the saturated adiabat corresponding to e = 310 K. Then find the value of T along this adiabat at p = 1000 mb. Find the dry adiabat corresponding to  = 300 K. The intersection of this dry adiabat with the saturated adiabat found in part (a) determines the SP of this parcel, and hence the value of rv (w).

Homework: Problems 1-5 (my notes)

Common thermodynamic processes in the atmosphere
Chap. 7 Common thermodynamic processes in the atmosphere

We will consider: Real thermodynamic processes of importance in the atmosphere. Underlying theme: latent heating or cooling Link to cloud microphysical processes, (examined later in this course).

Important temperatures
Example: Latent heating components in deep convection Each latent heating component is associated with a phase change involving various particles and microphysical processes warming Important temperatures cooling

7.2 Fog formation Several thermodynamic processes can produce fog:
a) local diabatic cooling via radiational cooling (e.g., radiation fog), b) addition of water vapor (e.g., evaporation fog), c) mixing of air parcels having different temperatures (e.g., advection fog). For radiation fog, after saturation is attained, a negative heat flux (i.e., the upward conduction of cold air from the ground surface, from emission of long-wave radiation) results in condensation near the surface, i.e., production of fog layers.

Ground fog in the early morning

Consider three of six equations that were used by Girard and Jean-Pierre (2001) to form a simple cloud/fog model: Local change diffusion Sources/sinks Water vapor mixing ratio Temperature Physical basis? What familiar equation do you see here? Saturation ratio Conservation equations for water vapor mixing ratio (q), temperature (T) and saturation (d, think of this in terms of relative humidity)

Now, back to the radiation fog problem:
From the isobaric form of the First Law (radiational cooling – net emission of long-wave radiation – is related to dq) dqrad = dh = cpdT + Lvldrvs (dp=0) (7.1) Using the relation between rvs and es rvs = ees/p -> drvs = edes/p (dp=0) together with the Clausius-Clapeyron eq. des/es = LvldT/(RvT2) we can write drvs = [e/p]des = [eLvles / pRvT2]dT (7.2)

Substitution of Eq. (7.2) into Eq (7.1) yields
(7.3) Or, using the alternate version of (7.2) we can write (7.4) Eq (7.3) can be used to compute the DT if a corresponding Dq (IR radiational cooling) can be measured. The decrease in saturation vapor pressure (es ) can be obtained from (7.4).

The mass of water vapor per unit volume can be obtained from the equation of state for water vapor:
rv = es/RvT Differentiation of this equation yields (7.5) (This approximation is accurate to within ~5%; which can be shown with the C-C eq. or even with a more simplistic scale analysis.) As cooling produces condensation in the fog, the differential amount of condensate (mass per unit volume) can be found using (7.5) with the C-C eq. as (7.6)

Example: Using (7.6), find the cooling required to form a fog liquid water content of 1 g m-3 (a vary large value) if the air is saturated at 10 C. From (7.6) we have DT = -DM[Rv2T3/Lvles(T)] = kg m-3 [(461 J K-1 kg-1)2(283 K)3/(2.5x106 J kg-1)(1227 Pa) = K

Using (7.5) the DM relates to Des by -Des = RvTDrc.
For DM=1 g m-3, Des is 1.3 to 1.4 mb over a broad range of T. Examination of a plot of es vs. T (Fig. 7.3), shows that the DT required for a given production of condensate (DM) increases as T decreases. This suggests that thick fogs are less likely at low T. Some considerations of the previous statement: Is this consistent with your observation? Fog tends to be more frequent during the cold months. Why? (What “forcing” is required for fog formation? A good answer requires some knowledge on radiative transfer.) Why are ground fogs more prevalent in September than in June (at least for the Huntsville region)?

7.3 Effects of freezing in a cloud system (pp. 124-126, I&G)
From the First Law, freezing of supercooled water within a cloud will produce a temperature increase (assuming all supercooled liquid water is instantaneously converted to ice – and this is a good assumption) according to dq = -Lildrc = cpdT (isobaric process) (7.8) This is only the first order approximation and does not account for all the physics. Rather, three processes (enthalpy components) should be considered: 1) Latent heat of freezing: DH1 = - Llidrc = Llidrice (drc = -drice) (7.9a) 2) Depostion of water vapor on the newly formed ice particle (this occurs because of the difference in esv(T) and esi(T) – see Table 5.1): DH2 = -Lvi[(rvs(T)-rvi(T)] (7.9b) 3) Absorption of the latent heat by dry air, water vapor and the newly-formed ice condensate: dry air water vapor ice condensate DH3 = [cpd + rsi(T)cpv + riceci]  (T - T) (7.9c)

The saturation vapor mixing ratios in (7
The saturation vapor mixing ratios in (7.9b) be related to vapor pressure, utilizing the often-used relation (7.10) along with the Clausius-Clapeyron equation (dlnes = (Lvl/RvT2)dT), to express esi(T) in terms of the initial temperature (T) and the temperature difference (T-T), assuming that (T-T) is small enough to be treated as a differential, e.g., (7.11) Combining 7.10 and 7.11 into 7.9b: (7.12)

From heat balance considerations (latent heating is balanced by an increase in H) we can write
DH1 + DH2 + DH3 = 0 and then solve for DT = (T-T) to get (7.13) The cp term includes rsi(T) which is not known, but can be obtained if desired by numerical solution (successive approximations). The contributions from this term are small since rsi(T)cpv << cpd. When rvs and rsi are small, (7.13) can be well approximated by DT = riceLvi/cp (7.14)

This latent can be an important source of local heating in convective cloud systems since the heating proceeds relatively rapidly over a relatively shallow depth. Such rapid conversion of supercooled water to ice is termed glaciation. For example, if 5 g kg-1 of supercooled water is converted to ice, then from (7.14), DT = 1.7 K, which in general is a significant increase in bouyancy. This concept was vigorously pursued during the 1970's over south Florida, where cloud systems were studied and seeded in order to evaluate the dynamic response* (and subsequent upscale cloud system growth) of rapid glaciation in the mixed phase region of clouds. (The mixed phase region is defined as the region, typically between temperatures of -20 and 0 C, where ice and supercooled water coexist, within updraft regions of cloud systems). * In this dynamic response, it was hypothesized that the net latent heating would accelerate updrafts, thereby reducing pressure near the surface, which in turn would increase mass convergence in the boundary layer.

7.4.1 Melting in stratiform precipitation (see also Rogers and Yau, pp. 197-203)
Stratiform precipitation is a common. One important characteristic of melting is that it proceeds relatively rapidly over a relatively shallow depth, typically in the range m. Local cooling rates within melting regions can be appreciable. Before proceeding with cooling by melting, we will first examine the factors that govern the local rate of cooling within mesoscale precipitation systems. Starting with the First Law, we have dq = cpdT - adp. Dividing both sides by the time differential dt and solving for dT/dt (we are after rates of cooling here) yields (7.15) We now use the equation of state to substitute for a [a=RT/p], and use a definition from atmospheric dynamics, wdp/dt. Eq. (7.15) then becomes (7.16) rate of diabatic heating (cooling by melting)

To find the local change we decompose the total derivative into the local and advective changes:
where V is the horizontal wind and s is in the (horizontal) direction of the flow. Substituting this into (7.16) and solving for local term T/t gives i ii iii (7.17) Local changes in T are accomplished by horizontal temperature advection, vertical motion and diabatic heating. In actual precipitation systems, there are instances where effects from term (iii) dominated by terms (i) and/or (ii).

For melting (term iii), the temperature change can be found from a form similar to that of Eq (7.14), ri dq = riLil = ricp dT, where ri is the mixing ratio of ice precipitation. For melting, (dq/dt)dia = Lil * (rate of precipitation), and we can write DmiLil = rdcpHADT where mi is the mass of ice, rd the density of air, H the melting depth and A the unit area. [Also used rd = md/V = md/HA). Converting mi to a precipitation rate R (mm hr-1), and dividing through by Dt yields (DT/Dt) = RLil / cprdH (average over depth H and Dt) (7.18)

For a precipitation rate of 2 mm hr-1 (a modest rate in stratiform precipitation) and H=400 m, a cooling rate of 1.6 K/hr is produced. The depth H over which the melting occurs is a function of ice particle size, precipitation rate R and relative humidity. One consequence of this prolonged diabatic heating is the formation of an isothermal layer, increased static stability within this layer, and often a decoupling of the atmosphere above and below the melting layer. [Insert a conceptual picture here] [Also think about the initial value problem.]

7.4.3 Thermodynamics within thunderstorm downdrafts
Consider an air parcel moving downward within a precipitation environment in the lower levels of a precipitating cumulonimbus cloud system. Diabatic cooling sources include evaporation and melting. Changes in q (following a parcel, so we consider the total derivative) are accomplished by evaporation of water, melting, and sublimation of ice: total evaporation of: sublimation of melting of change rain cloud ice particles ice particles (7.19) qo, To are initial parcel values of q and T, VDrv is the rate of evaporation of rain, VDcv is rate of evaporation of cloud, VDgv is rate of sublimation of ice, MLgr is rate melting of ice

Chap. 9 Surface Thermodynamics and Nucleation of Water Droplets and Ice Crystals

Nucleation: the formation of cloud droplets and ice crystals, involving a change of phase (vapor to liquid, vapor to solid, liquid to solid) Question: What physical processes promote the nucleation within warm and cold clouds? Warm cloud – T > 0 C everywhere; or, a cloud that is entirely liquid. Cold cloud – T < 0 C somewhere; or, a cloud that has some ice.

The nucleation of a water droplet and growth to a raindrop spans four orders of magnitude in size, ranging from O(0.1 mm) for a CCN to O(1 mm) for a rain drop 103

9.1.1 Review of Gibbs and Helmholtz free energy
From Chap. 4: Gibbs free energy g = u - Ts + pa dg = du - Tds - sdT + pda + adp = -sdT + adp. For a phase change we need to consider two additional terms in the Gibbs free energy: 1) changes in chemical potential (associated with phase change) 2) work connected to the formation of a physical surface (i.e., a discontinuity).

We have considered thermodynamics in terms of bulk properties (e. g
We have considered thermodynamics in terms of bulk properties (e.g., p, T, r, rv, etc). When cloud droplets or ice crystals form, a surface (discontinuity) is formed. Thus, we need to examine surface effects. When the area of an interface such as the surface of a liquid and its vapor is increased, molecules from the interior have to be brought to the surface. In this process work must be done against the cohesive forces in the liquid. The surface tension is defined as the work per unit area (A) done in extending the surface of the liquid (units of energy per unit area). The surface energy is increased by an amount equivalent to the work done (it takes work to expand a water droplet), so the surface tension may be written as s = g/A Some careful thought on the previous statements would suggest that the First Law applies here, since work and energy are involved. This is, in fact, the reason why we introduce the Gibbs free energy as a “tool” to examine this problem.

The Gibbs function takes on two additional terms related to the additional energy and work alluded to in the previous section: chemical surface potential tension dg = -sdT adp S midni (g/A)dA, (9.1) where s is entropy, ni is the number of moles of the ith component, and m represents the chemical potential (at the surface), which is the Gibbs free energy per molecule. Recall from Chapter 3 (Notes, p. 3) that a preliminary form of the First Law, q = Du + ∑ei The second term on the RHS, ∑ei, is related to the work of expansion plus other forms of work, such as last two terms on the RHS of Eq. (9.1).

The quantity (g/A) at constant temperature, pressure and composition is termed the surface tension s. It is a quantity easily measured for liquids. To compute the surface energy one makes use of the Gibbs-Helmholtz equation dg = dh - Tds = dh + T(g/T). We will also define hs, the total surface energy, as hs = s - T(s/T). The surface tension of water depends on temperature according to the empirical relation s = T (units: erg cm-2; T in C; recall that 1 J = 107 erg). A plot of this function is provided on an attached figure (Fig. 5-1 from Pruppacher and Klett, 1978). At 20 C s is erg cm-2 and the total surface energy at this temperature is 112 erg cm-2.

The change of surface tension with temperature is always negative, and the surface tension becomes zero at the critical temperature (647 K; see Figs in Chap 5 of the my notes). Surface tension for pure water can be modified by the presence of dissolved salt such as NaCl. Fig. 5-2 (from Pruppacher and Klett, attached sheet) presents data for sodium chloride (NaCl), (NH4)2SO4 (ammonium sulfate), NH4Cl (ammonium chloride), and NH4NO3 (ammonium nitrate) solutions.

9.1.4 Hydrostatic pressure difference over a curved surface
Relation between work of expansion and surface tension. For an increase in radius of a spherical bubble of radius r, it is necessary to balance the work of expansion (pdV) by the surface tension, i.e., pdV = 2sdA. If the radius is to be increased, then the pressure within the sphere must be greater than the pressure outside, and the above equation can be written as pi - po = 2s(dA/dV) = 4s/r. (dA/dV = 2/r for a sphere.) For a liquid drop (which has one surface or interface) rather than a bubble (which has two interfaces) the pressure difference is pi - po = 2s/r (9.2) This is a special case of the LaPlace equation pi - po = s(r1-1 + r2-1) where the r's are the principal radii of curvature.

For a sphere the two radii are equal.
For a flat surface, the radii are infinite and no pressure difference exists. The important aspect of this discussion is that a pressure difference is associated with a curved surface, according to Eq. (9.2). This pressure difference obeys the r-1 relationship and is therefore greater for small drops.

9.2. Nucleation of cloud droplets (vapor  water)
Homogeneous vs. Heterogeneous nucleation will be considered. Homogeneous nucleation Definition: spontaneous formation, through the (improbable) collision of a sufficient number of water vapor molecules, to form an embryonic drop of sufficient size to remain intact (or persist). Assume that this droplet has a volume V and surface area A. The change in Gibbs free energy (due to condensation) of the system is nV(mv-ml). The work involved is As, where s is the surface tension, or the work required to create a unit area of the vapor-liquid interface. This is not an equilibrium transformation, that is As  nV(mv-ml).

The change in g, (Dg) will differ from the work term according to:
Dg = As - nV(mv-ml), where Dg is the net increase of energy of the system due to the formation of the drop. We previously determined (Section 5.2) that the change in chemical potential (mv-ml) can be expressed in terms of bulk thermodynamic quantities (T and relative humidity, or saturation ratio S=e/es) as (mv-ml) = kTln(e/es). We can then write Dg = As - nVkTln(e/es). For a spherical water droplet or radius r, this becomes (using relations for the volume and area of a sphere) Dg = 4pr2s - (4/3)pr3nkTln(e/es) (9.3)

Dg = 4pr2s - (4/3)pr3nkTln(e/es). (9.3)
Surface Change in effects chemical potential The equilibrium vapor pressure over a curved surface exceeds that over a flat surface at the same temperature. Eq. (9.3) is a illustrates this. The first term is related to surface effects and the second depends on changes in chemical potential. If e<es then the second term is positive and Dg grows in a nonlinear fashion as r increases. Since a system approaches an equilibrium state by reducing Dg, the formation of droplets is not likely under subsaturated conditions. However, random collisions of water molecules do produce embryonic droplets that continually form and evaporate in the atmosphere, but these are not visible as cloud droplets. When the air is supersaturated (e.g., by adiabatic ascent/cooling) then e > es, and the second term is negative. In this case, Dg attains a maximum value at some critical radius R (see attached Fig 4.10 from W&H). We find R by differentiating (9.3), d(Dg)/dr, and setting the result to zero. Do this yields Kelvin's formula: R = 2s / nkTln(e/es) (9.4) Fig 4.10 (W&H) below plots Eq. (1) for positive and negative values of ln(e/es) and defines R, the critical radius. The free energy becomes zero when r = (3/2)R.

The table below presents some calculations at T=20 C (compare with Table 6.1 of R&Y).
Table The dependence of R and Dgmax on the saturation ratio e/es. ____________________________________________________________ e/es R (mm) Dgmax (erg) x x10-7 x x10-8 x x10-9 x x10-9 x x10-10 x x10-11 x x10-12 x x10-12 x x10-12

9.2.2 Concentration of critical embryos
A critical embryo is defined as a water droplet of radius r = R. To determine the concentration of critical embryos, we utilize the Boltzman relation (9.5) where nR is the number of critical embryos per unit volume while nv is the number of individual water vapor molecules per unit volume. To very good approximation we have assumed that the number of water vapor molecules tied up in the embryos of the various sizes is negligible with respect to the number of water molecules in the vapor phase.

From Table 9.2 below we can see that there are essentially no critical embryos in the free atmosphere. Table Number of critical embryos (nR) of pure water at T=20 C. e/es nR (per liter) x10-61 x10-13 x10-2

9.2.3 Nucleation rate The nucleation rate is defined as the number per unit volume of supercritical droplets (i.e., droplets with r>R). This rate is given by the product of the critical embryo number concentration nR and the collision rate at which vapor molecules strike the embryo surface. The collision rate can be determined from the kinetic theory of gases as the number of gas molecules striking a unit area of substance per unit time: (9.6) where e is in mb. For water this reduces to n = 6.21x1021eT-1/2. Collision rate is directly proportional to the water vapor pressure (e), a result which may be inituitive (or at least consistent with simple thinking). The surface area of the critical embryo is 4pR2 and hence nc = 7.81x1022R2eT-1/2. The initial nucleation rate, J, is given by J = nRc. This is an initial rate before the supply of water vapor is reduced by droplet formation. An attached table gives J values for the condensation of water at 288 and 273 K as a function of S=e/es. What do you conclude?

9.2.4 Homogeneous nucleation of ice (crystals) from water vapor (vapor  ice)
Deposition process vs. condensation considered previously. The latent heating is greater. The same theory described for the case of water applies here. Precise values of the interfacial tension of ice against water do not exist. This value is greater than that of the surface tension of water. If realistic values are used, we find that even at low temperatures of -60 C that the critical concentrations are zero (and less than that for vapor-liquid). Thus, homogenous nucleation of ice does not occur in the atmosphere.

9.2.5 Homogeneous nucleation of ice from liquid water (water  ice)
This represents the second path that is theoretically available for homogeneous ice nucleation. Two unknowns (I think this is true -- I am paraphrasing from unpublished notes from 1977): (1) The interfacial tension between liquid water and ice is not well known, and (2) a simple theory for collision rates does not exist. We can guess that the interfacial energy for ice-water is in the range erg cm-2. (This compares to ~100 erg cm-2 for ice-vapor and ~73 erg cm-2.) Based on laboratory experiments, in situ cloud measurements, and remote sensing of clouds (e.g., lidar with depolarization capability) virtually all (cloud) water is frozen at -40 C. Thus, homogeneous nucleation from cloud water to ice crystals is likely, but probably not dominant, in cloud systems.

9.2.6 Summary of homogeneous nucleation
The only homogeneous nucleation process active in clouds is through the water-ice phase (freezing by -40 C) transition. Measurements of S (=e/es) within clouds rarely exceed 1.01 (near cloud base). In rare instances, S may approach 1.1 in real clouds with strong updrafts. (We will consider a conservation equation for S in Chap. 7.) This implies that homogeneous nucleation does not occur, since the required S for embryonic droplets at radius 0.1 mm exceeds

Review of homogenous nucleation
Dg = 4pr2s - (4/3)pr3nkTln(e/es). R = 2s / nkTln(e/es). Increase E (g) in the energy of a system due to the formation of a water droplet of radius R from water vapor with a pressure e; es is the saturation vapor pressure with respect to a plane surface of water at the temperature of the system. From Wallace and Hobbs. The relative humidity and supersaturation (both with respect to a plane surface of water) at which pure water droplets are in (unstable) equilibrium at 5 C. From Wallace and Hobbs.

9.3 Heterogeneous nucleation of water droplets
The formation (homogeneous nucleation) of droplets in a pristine atmosphere is a difficult process requiring appreciable supersaturation. For heterogeneous nucleation, the presence of foreign particles relaxes the constraints that are present for homogeneous nucleation.

9.3.1 Atmospheric aerosol particles (AP)
Aerosols are comprised of particles which may exist in solid or liquid (excluding water) form. Their concentration and chemical make-up exhibit considerable variability. Aerosol sizes range from ~10-4 to ~100 mm, and concentrations may range from 10-6 to 107 cm-3. It is convenient to divide AP into 3 size categories: Aitken particles (r < 0.2 mm), large particles (0.2 < r < 2.0 mm), giant particles (r > 2.0 mm). As shown in the Fig 6.3 of R&Y, the smallest AP are most numerous. [This distribution was measured in an interior continental setting near Miles City, Montana.] We also note that Aitken nuclei exist in higher concentrations over continents than over oceans, a facet related to the source region. Over continental areas, AP concentrations drop off sharply with increasing height, indicating that the surface acts as a source. (See Fig. 4.1 from Wallace and Hobbs).

Sources of AP include both natural and man-made (anthropogenic) sources (see Table 4.1 and Fig. 4.7 from Wallace and Hobbs, and reproduced below). A large fraction of Aitken nuclei originates from combustion processes. Other sources include gas-to-particle conversion, wind-blown dust and even extra-terrestrial particles (meteoric debris). Aerosol sinks include losses on the small end from coagulation, on the large end from sedimentation (fallout), and from scavenging (capture) from cloud droplets and precipitation. This latter process may account for 80% or more of the total mass of aerosols removed from the atmosphere. (CCN, diffusiophoresis, impaction for r > 2 mm) As mentioned above, a wide variety of aerosols exist within the atmosphere. Some of these serve as nuclei to promote the formation of (a) water droplets from the vapor, (b) ice crystals from the vapor and (c) ice crystals from water droplets.

9.3.2 Cloud condensation nuclei (CCN)
Defined as AP which are capable of initiating drop formation at relatively low supersaturation. CCN represent a very small subset of the total AP. CCN concentrations vary from generally low values over oceans (~100 cm-3), to high values over continental areas (~600 cm-3). The size range encompasses the interval mm, which is the so-called accumulation mode defined by R&Y on p The accumulation mode corresponds to the peak in the surface and volume curves of Fig This peak is also seen in Fig. 4.7 from Wallace and Hobbs. Note that the CCN number is a function of supersaturation (S-1), where S = e/es. This dependence can be expressed as NCCN = C(S-1)k (9.7) where C and k are approximately constant but vary considerably as indicated on p. 95 of R&Y. e.g., maritime air: <C<300 cm-3; 0.3<k<1.0, continental air: 300<C<3000 cm-3; 0.2<k<2.0.

Azores Marine air mass Clean Arctic air Cloud condensation nucleus spectra in the boundary layer from measurements near the Azores in a polluted continental air mass (top line), in Florida in a marine air mass (middle line), and in clean air in the Arctic (bottom line). From Wallace and Hobbs.

CCN concentrations at a particular location can exhibit appreciable variation.
Over oceans, CCN concentrations are nearly uniform with height as shown in Fig. 9-4 from P&K, in contrast to continental regions which CCN decrease with height (Fig. 4.1 from Wallace and Hobbs). It is estimated that only 1-10% (or less) of all aerosols serve as CCN as shown in the table below. Table 9.2: Comparison between total concentration of aerosol particles and concentration of cloud condensation nuclei activated at 1% supersaturation at various locations. Taken from Pruppacher and Klett (1978). ______________________________________________________________________ Location Aitken particles (cm-3) CCN (cm-3) Washington D.C , (Allee, 1970) , 57, 50, Long Island, NY 51, (Twomey and Severynse, 18, 1964) , Yellowstone National Park , (Auer, 1966)

9.3.3 Nucleation onto water-soluble CCN
controlled by the mass and chemistry of the water-soluble component deliquesence onto hygroscopic particles at RH < 100% The most efficient CCN are those that are both hygroscopic and water soluble. Such CCN serve as very efficient centers onto which water vapor can condense, with the nucleated hetergeneous drop then attaining a critical size. Salt particles (NaCl - prominent over oceans) and ammonium sulfate particles [(NH4)2SO4 - prominent over continents] are particularly efficient CCN. (Consider NaCl here. Fig. from Wallace and Hobbs) The aerosols that are soluble in water effectively reduce the saturation required to achieve critical radius. This occurs for the following reason. Since e is proportional to the number of molecules at the surface, we see that if a fraction of water molecules is replaced by the agent being dissolved (an ion), then the fractional change in e is e'/es = f = n0 / (n0 + n')  1 - n'/n0 (n'<<n0) (9.8) where e' is the saturation vapor pressure over a solution droplet containing a kilomole fraction of pure water and es is the vapor pressure over a pure water droplet of the same r and T, n0 and n' are the number of water and solute molecules.

In many solutions, dissolved molecules are dissociated, so (9
In many solutions, dissolved molecules are dissociated, so (9.8) is modified by the van Hoff factor i (~2). Then n = iN0M/ms (N0 - Avogadro's no., M - mass of solute, ms - molec. weight of solute) n0 = N0m/mv (m - mass of water, mv - molec. wt. of water) Since m = V x rL = (4/3)pr3rL, then (9.8) can be written as e'/es = 1 - [3imvM/4prLms] r-3 = 1 - br (9.9) From the Kelvin Eq. (9.4) we can write S = e/es = exp[(2s)/(RvTrLr)] = ea/r. For the dissolved substance the above eq. is modified to S = e'/es = exp[(2Mr)/(RvTrLr)]. or es'(r)/es() = [1 - br-3]ea/r.

For small a/r, ea/r = 1 + a/r + (a/r)2/2 + . . .  1 + a/r, we can write
es'(r)/es() = (1-br-3)(1+ar-1) = 1 - br-3 + ar-1 - abr-4. If we ignore the last term of the above equation then (with justification) es'(r)/es()  ar br-3 (9.10) curvature term solute term A plot of Eq. (9.10) yields the Kohler curve, an example of which is given in Fig of Wallace and Hobbs (attached) and in Fig. 6.2 of R&Y. Note that for small r the (negative) solute term dominates. Eq. (9.10) can be solved to find the critical radius and corresponding superaturation at the critical radius: r* = [3b/a]1/2; S* = 1+[4a3/27b]1/ (9.11), (9.12)

Fig Equilibrium saturation ratio of a solution droplet formed on an ammonium sulfate condensation nucleus of mass g. Taken from Rogers and Yau (1989, their Fig. 6.2).

Variations of the relative humidity and supersaturation adjacent to droplets of (1) pure water (blue) and adjacent to solution droplets containing the following fixed masses of salt: (2) kg of NaCI, (3) kg of NaCI, (4) kg of NaCl, (5) kg of (NH4)2S04, and (6) kg of (NH4)2S04. Note the discontinuity in the ordinate at 100% relative humidity. From Wallace and Hobbs. Kohler curves 2 and 5 from the figure in the upper left. Curve 2 is for a solution droplet containing kg of NaCl, and curve 5 is for a solution droplet containing kg of (NH4)2S04. From Wallace and Hobbs.

Nucleation onto a water-insoluble CCN
Many CCN are a mixture of soluble and insoluble AP. In this case we must consider the effects of the water/solid surface. We can write the Kohler Eq. from the soluble result as ln (e/es) = 1 + a/r - b/(r3-ru3) where ru is the equivalent radius of the water-insoluble sphere. If we plot the supersaturation required for activation, we find that the initial size needs to be >0.1 mm.

Nucleation onto a water-insoluble, partially wettable CCN
We will ignore the mathematical development and only mention some research results here. In this case the critical saturation ratio is highly dependent on contact angle, which is defined in Figure 9-6 (attached) from Pruppacher and Klett. It can be shown that CCN under this category are probably not important unless the contact angle is less than deg. For example, if we consider that the (S-1) value reached in clouds is typically smaller than 3%, we must require that AP of r >.01 mm have contact angles less than 12 deg. Little is known about actual contact angles for AP. However, available measurements suggest that silicate particles are not likely to serve as CCN. An aerosol defined as hydrophilic if it has a contact angle of 180. In the real atmosphere, actual CCN are composed of a combination of the above due to processes of coagulation, etc.

Review of heterogenous nucleation of water droplets
Either of these equations can be used to define the (unstable) equilibrium value of R or e/es.

The presence of dissolved salt reduces the equilibrium vapor pressure to e’/e = f, where e’ is the reduced vapor pressure. It can be shown that This is multiplied with the Kelvin Eq. to get Plots of e’/es vs. r are called Kohler curves

Assignment Problems 6.3 and 6.10 in Rogers and Yau

Overview of thermodynamics

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