Midterm 1 Review

ATS 620, Fall 2011 October 3, 2011

EQUATION OF STATE (GAS LAW)

Relationships that express intensive parameters (e.g., p1, T) in terms of extensive parameters are called “equations of state.”

Early lab studies determined that p, V, and T could be related by an “equation of state.”

IDEAL GAS LAW- Equation of state for ideal gases, gases where molecules have no interactions, no molecular attractive forces. Molecules have zero volume assumed too.

pV= nR*T= mR*T/M= mRT

n= number of moles= m/M

R* = universal gas constant 8.3143 J mol-1 K-1

M= molecular weight g/mole

First Law is concerned with conservation of microscopic energy input heat .

body does ‘work’ as a result of this heat added; is amount of work done.

First Law - is change in internal energy associated with KE and PE of molecules

In differential form

If , then

Internal energy is a STATE VARIABLE. Change in internal energy is difference between initial and final states. is independent of path between initial and final states.

JOULE’S LAW: Consider the internal energy of an Ideal Gas in the form . Joule’s experiment demonstrated that is a function of only, for an Ideal Gas.

Consider a gas expanding into an evacuated cylinder (vacuum). Hence no work is done and . Let the process also be adiabatic, .

Since , . Therefore the temperature of the gas remains constant. So Joule’s Law states: “if a gas expands without doing any work, under adiabatic conditions the temperature of the gas remains constant.”

Under these conditions, its clear that the volume of the gas changed, as did the potential energy of the molecular configuration. Therefore the internal energy of the gas must be independent of volume.

So we conclude that the internal energy is independent of volume if the temperature of the gas is constant.

Corollary: Molecules of an Ideas Gas do not exert any attractive or repulsive forces on one another.

First Law can then be written as

Another form of the First Law,

ENTROPY

There exists a function called entropy S, of the extensive variables of a system, defined for all equilibrium states and having the following properties

The values assumed by the extensive variables are those that maximize S (at equilibrium)

From the viewpoint of classical thermodynamics, entropy is defined as

where in this case heat is added to a substance undergoing a reversible transformation.

Entropy is a STATE VARIABLE.

Since dq= Tds, the First Law can be written as,

For a substance transforming form state 1-2

Reversible & Irreversible Processes

•  Reversible Transformation- In a reversible transformation, each state (point, configuration, etc.) is an equilibrium state, such that a reversal in the direction of a small change returns the system to its original state (point, configuration etc.)

•  Also, the small departures from equilibrium are associated with vanishingly small changes in ENTROPY. Entropy is a constant (and a maximum) for reversible transformations.

–  Reversible transformations are not associated with any dissipative processes such as friction, electrical resistance, mixing, etc.

–  NO ENERGY LOSS- ALL ENERGY IS TRANSFERRED IN A CONSERVATIVE MANNER BETWEEN THE SYSTEM & SURROUNDINGS.

•  Irreversible Transformation- Irreversible transformations are all processes that are not reversible.

•  Turbulent stirring of liquid •  Free expansion of a gas into a vacuum •  Freezing of super-cooled water

–  All irreversible transformations are accompanied by an increase in entropy.

Clausius-Clapeyron Equation

Aerosol particles Aerosol particle sizes: 10-4 µm to 100 µm (0.01–10 µm focus for this class) Aerosols typically divided into 3 basic categories based on size

r < 0.1 µm Aitken particles (Nuclei mode) 0.1 ≤ r ≤ 1.0 µm Large particles (Accumulation mode) r > 1.0 µm Giant particles (Coarse mode)

-Note values are approximate -Most accumulation mode particles produced by coagulation of Aitken particles

http://www.windows.ucar.edu/tour/link=/earth/Atmosphere/aerosol_cloud_nucleation_dimming.html

Dominant source: Earth’s surface Secondary sources: Volcanoes, outer space N. Hemisphere: produces 61% of aerosol mass S. Hemisphere: produces 39% of aerosol mass

Natural Sources - Mineral dust -Sea salt -Volcanic sulfur (S) -SOA (organic)

Anthropogenic Sources -Wild fire (black carbons, BC) -Bio fuels (BC) -Fossil fuels (BC) -Industry (S)

Cloud Condensation Nuclei (CCN)

•  From previous discussion we know that rather special requirements must be met by an aerosol particle for it to serve as a center for water condensation –  Most “efficient” is a water soluble particle of sufficient mass (Aitken) –  Less “efficient”, wettable larger aerosols (Large/Giant)

•  For 1% supersaturation, a perfectly wettable, but insoluble aerosol must have a radius > 0.1 micron (reduction in curvature term)

•  For a 1% supersaturation, a soluble particle must have a radius of at least 0.01 microns (specifies sufficient salt mass for solute effect)

•  Hence many aerosols do not serve as CCN •  In general, 1% of the aerosols in a continental air mass serve as CCN at 1%

S.S. For maritime air masses, about 10% of the aerosols act as CCN at 1% S.S.

•  We anticipate that CCN concentrations will increase with supersaturation.

Importance of Giant CCN (GCCN) •  For warm rain, given the same liquid water content, greater

concentrations of GCCN: –  Initial broadening of the cloud droplet spectra by creating larger cloud droplets –  Enhances collision and coalescence processes –  This tends to enhance warm rain precipitation processes

•  (Hobbs et al. 1970; Eagan et al. 1974; Braham et al. 1981; Rosenfeld et al. 2002)

•  Therefore, warm rain precipitation formation is dependent upon aerosol particle size distributions as well as number concentrations

The next step is to combine equations (4) & (7) with the Clausius Clapeyron equation and the Ideal Gas Law to write in terms of the supersaturation of

the environment. After much algebra,

Is the saturation ratio of the environment.

This equation is written in the simpler form as,

molecular weight of water

All other variables have been previously defined

Houghton  (1985)  

Kelvin  Curve  

Curves  are  for    aerosol;  ammonium  sulfate.    Do=ed  curve  is  for    

Köhler  Curves  

Important points to glean from the Kohler curves:

1.  As the dry particle size increases, the critical supersaturation decreases (for a chosen composition, it is easier to “activate” larger dry particles).

2.  For similar-sized dry particles, but with different composition, those creating the largest suppression of water vapor pressure over the solution are the easiest to activate. (This means solutes that dissociate are easier to activate than those that do not, and those that dissociate into more than two ions are easier to activate than those that dissociate into only 2) CAVEAT: The statement ignores the influence of the dry particle density. In other words, particles of the same dry size but different densities contribute different amounts of mass to the solution, which has to be considered.

3.  The size of the droplet at activation decreases with decreasing dry particle size (since the wetted size is so small, the Kelvin effect is large and this leads to high required critical supersaturations)

4.  For large dry particles, the size at activation is very large and the Kelvin (curvature) effect very small – it is very easy to activate these particles: they could even be (nearly) insoluble and still activate (“giant” CCN)

This  results  suggests  that  a  plot  of  ln  sc  vs.  ln  rd,  for  a  chosen  composiCon  of  dry  parCcle,  should  yield  straight  lines,  with  slopes  of  -­‐3/2  

This  line  is  for  pure  water  

(insoluble  parCcle)  

This  line  is  for  a  “salt”  similar  to  

NaCl  

dry  diameter,  cm  

Utility of κ When  κ =  0,  aw  =  1    pure  water  

Can  see  that  κ  “scales”  the  water  content  at  a  given  aw  (for  a  bulk  soluCon,  aw  =  RH)  

κ  values  for  “real”  atmospheric  par9cles  range  from  0  to  1.2  

Shaded  area  =  reported  range  for  ammonium  sulfate  

Consider the following situation:

r

R

Droplet of mass m and radius r is embedded within a steady-state field of water vapor

Droplet density

Drop mass

To calculate the droplet growth rate by condensation, surround this droplet with an imaginary sphere of radius R, write a mass balance, and use the Fickian Diffusion Law to compute the flux of water vapor through the imaginary spherical surface, towards the droplet’s surface.

For steady-state conditions and no “storage” (accumulation) of water vapor in the region around the droplet, this vapor flux must be equivalent to the growth rate of the droplet.

The unsteady-state diffusion of species A (here, water) to the surface of a stationary particle of radius Rp is described by

where c(r,t) is the molar concentration of A (moles vol-1), and JA,r is the flux of A (units: moles area-1 time-1) at any radial position r. (The equation arises from a mass balance in a spherical shell around the particle.) Fick’s Law for our problem can be well approximated by

€

∂c∂t

= −1r2

∂∂r

r2JA ,r( )

€

JA ,r = −DVdcdr

DV is the diffusivity of vapor in air

€

∂c∂t

= −∂∂x

JA ,x( ) =∂∂x

DV∂c∂x

⎛

⎝ ⎜

⎞

⎠ ⎟ (Cartesian)

Combining, we get the governing equation,

We need one initial condition, and two boundary conditions, to solve this. We use

We can find a solution for this full problem (see Chapter 11 of the text by Seinfeld and Pandis), and putting in typical values for the kinds of problems we’ll be trying to solve, we find out that the transient is very short. So it is valid to assume we get to a steady-state situation very quickly, and that we can set the time derivative to zero. The steady-state solution is,

(*)

The total flow of A (moles time-1) toward the particle is

So, take the derivative of the solution (*), plug into the equation for JA,r, and find

€

∂c∂t

= DV∂ 2c∂r2

+2r∂c∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ = DV∇

2c

€

c(r,0) = c∞c(∞,t) = c∞c(Rp,t) = cs

€

c(r) − c∞cs − c∞

=Rp

r

€

J = 4πRp2 (JA )r=Rp

€

J = 4πRpDV (c∞ − cs)

What happens as r = Rp and as r gets very large?

The transferred moles time-1 increase as Rp increases, and are proportional to the difference between the surface concentration and the far-field concentration “Maxwellian flux” (1877)

Now write a mass (or here, mole) balance on the growing (or evaporating) droplet:

This yields an equation for the time rate of change of the drop radius,

Which we could also express in terms of the vapor density (mass vol-1)

Or as a rate of change of the droplet mass,

The diffusivity of water vapor can be expressed as

€

ρlMw

ddt

43πRp

3⎛

⎝ ⎜

⎞

⎠ ⎟ = J = 4πRpDV (c∞ − cs)

€

dRp

dt=1Rp

DVMw

ρl(c∞ − cs)

€

dRp

dt=1Rp

DV

ρl(ρ∞ − ρs)

€

dmdt

= 4πRpDV (ρw,∞ − ρweq )

ρ∞ and ρw,∞ are the same thing; w used to indicate water vapor here; eq means the equilibrium value over the drop

€

DV =0.211p

T273⎛

⎝ ⎜

⎞

⎠ ⎟ 1.94

DV in units of cm2 s-1, T in K, p in atm

Geometry (from Lamb & Verlinde)

Imagine we have a bimodal spectrum, produced by some means: So drops are falling with DIFFERENT fallspeeds

rL rL=radius of collector (drop)

rs=radius of collected droplets

Area of cylinder that is swept out

Volume of cylinder that is swept out, per unit time:

Continuous Collection Equation

Derive an expression for the growth rate of a droplet by collection of small droplets -

1.  Assume small collected droplets are uniformly distributed in the cloud. 2.  Collector drops of the same size collect droplets at the same rate, that is

collector drops of a given size all grow at the same rate.

Drops of size R have the same probability for undergoing collisions with small droplets in given time

We need to consider the portion of the cloud liquid water content that is collected by the drop. The mass to be collected resides within an imaginary cylinder swept out by the collector drop.

terminal fall speed

Liquid water content

Volume swept out per unit time

Assuming cloud drafts are small

collection efficiency

This is the Continuous Collection Equation

Mass of water in volume collected

Since

Equation (1) becomes

For spherical water droplets,

Therefore,

Since increases with , and increases with ,

Stochastic Model of Drop Growth •  In this section we introduce the stochastic model for drop growth. This stochastic model differs

from the continuous growth model in that it relaxes the assumption that all drops of a given size—R—grow at the same rate. The stochastic model better predicts the evolution of the drop size spectra observed in clouds, and in particular, the production of large precipitation-sized drops. The stochastic model allows large drops to form much faster compared to the continuous model.

•  Continuous Model: Each and every drop of a given size R collects the same number of droplets in given time Δt. In this model no information on cloud droplet size is carried, other than a fixed value.

•  Quasi-stochastic Model: Allows a specific fraction of the drops of size R (say 50%) to undergo collisions with droplets in a given time period Δt. This model assumes that a finite fraction of the drops of size R will collect one or more droplets and the remainder of the drops of size R will collect no cloud droplets in this same time period. In this model the cloud droplets may be randomly distributed in space.

•  Stochastic Model: Assigns a probability (not all drops have the same probability for collisions in time Δt) that a given collector drop will experience a collection event in time Δt. The positions of the cloud droplets are assumed to change randomly with time, leading to the probabilistic nature of the collection process.

Each of these models are illustrated conceptually in the following figure.

Drop breakup

Explosive deepening of notch marks onset of “break-up” phase

Notch deepens into annular ring with “bags” attached to it.

Break up can also be modelled as a physical process which occurs when the drag stress (between air and falling drop) exceeds the surface tension stress

Models for drop break up due to collisions with other drops

Energy for breakup supplied by momentum associated with the colliding drops

Neck, sheet, disk, bag. Sheet breakup most common.

Collision-Coalescence Growth Since we know that condensation growth alone cannot produce precipitation-sized drops (> 100 µm) in reasonable cloud lifetimes, we must invoke other mechanisms to explain the production of precipitation in ‘warm’ clouds. Warm clouds contain no ice and their summit temperatures are greater than 0°C.

Invoke the collision-coalescence mechanism

Collisions between cloud droplets can occur to produce a larger droplet by coalescence with neighboring droplets

These initial collisions are motivated by the presence of different-sized cloud droplets -- this size distribution can be caused by nucleation on both small (hygroscopic) and large (wettable) CCN. (Broadening of the cloud droplet size distribution is discussed later)

Once a larger cloud droplet is formed, it will collide with other droplets and thus grow rapidly.

- CCN spectra (giant nuclei)

- Broadening via turbulence

- Inhomogeneous mixing Three mechanisms that act to broaden the droplet size distribution.