1. Cumulus, stratus, and stratocumulus are types of low-level, or boundary layer, clouds.

a) Describe the visual appearance of each of these three cloud types for a ground-based observer. Include characteristics that distinguish the three cloud types (you may want to consult another text or the Web).
 

Stratus - Looks fairly uniform and grayish. Stratus may appear in the form of patches, but otherwise does not exhibit individual cloud
elements as do cumulus and stratocumulus clouds.

Stratocumulus - Lumpy cloud elements existing in a relatively flat layer. Elements often are arranged in rows or bands. Overcast stratocumulus has lighter and darker spots that show the cloud elements.

Cumulus  - Detached lumpy clouds, generally dense and with sharp outlines. Often show vertical development in the form of domes, mounds, or towers. Tops normally are rounded while bases are more horizontal.
 

b) Explain the physical basis behind the appearance of these cloud types. Consider the roles of surface heating, vertical velocity, stability, and the widespread humidity profile.

Stratus- These clouds exist in layers in which the widespread relative humidity is about 100% (dewpoint equals temperature). The air is stably stratified, so the vertical motion is widespread gentle uplift. Because the boundary layer is stable these clouds are not much  influenced by the surface.

Stratocumulus - These clouds exist in layers in which the widespread relative humidity is near 100%. However, the layer is
unstable (or neutral) so that it is mixing. Often the whole boundary layer (both below cloud and in cloud) is convecting.  The darker patches
or cloud elements are updrafts with higher liquid water content, while the lighter patches or clear sky portions are downdrafts.

Cumulus -  These clouds exist in boundary layers in which the widespread relative humidity is below 100%.  Surface heating by the sun
is usually occuring, giving rise to an unstable or neutral boundary layer, and vigorous mixing. Thermals (plumes of warmer air) rise from
the surface to the lifting condensation level, where the individual cumulus clouds appear.  Cumulus clouds have much stronger updrafts
(~1 m/s) than stratocumulus clouds (~0.2 m/s). If there is conditional instability in the sounding then the cumulus clouds
are likely to have greater vertical extent, such as cumulus congestus or even cumulonimbus.
 

2. Suppose a sample of marine air has an aerosol size distribution described by

dN/d(lnr) = rdN/dr = c r^-b

from radii r₁ = 0.01 m m to r₂ = 10 mm (with no aerosol outside this range), with constants b = 2.5 and c = 0.025 cm^-3mm^2.5.

(a) Calculate the total number concentration of aerosols (in cm^-3).

Number concentration (N) is the integral of dN = 1/r x c r^-b = c r^-b-1 dr

or N = c/(-b) r^-b = (0.025 cm^-3 mm^2.5 / 2.5) x  r^-2.5 integrated from 0.01 to 10 mm = (1000 - 3x10^-5)  cm^-3 ~ 1000  cm^-3
 

(b) Calculate the mass of aerosols per cubic meter of air (g m^-3). Use a density of 2.0 g cm^-3 for this aerosol material. Hint: find the mass of each aerosol radius and then integrate to find the total mass of aerosol per volume of air.

In this case, we need to multiply the number distribution by the mass of a spherical particle, 4/3 pr³ r

or dM = dN x 4/3 pr³ r  = 4/3 pr³ rc r^-b-1 dr

simplifying, dM = 4/3 prc r^-b+2 dr

integrating gives M = 4 prc / [3 x (-b+3)]  r^-b+3 =  (4 x 3.14 x 2 g cm^-3 x 0.025 cm^-3 mm^2.5)  r^0.5 / 1.5 evaluated over 0.01 to 10 mm

First, it's best to convert 1 cm^-3 to 10^-12 mm^-3 so these units will cancel later....so,

M ~ (4 x 3.14 x 2 x 10^-12 g mm^-3  x 0.025 cm^-3 mm^2.5)  (10 mm)^0.5/ 1.5 =  1.3 x 10^-12 g cm^-3 = 1.3 x 10^-6 g m^-3

(c) Suppose all aerosol with dry radii above r_c=0.03 mm are cloud condensation nuclei (CCN) for typical cloud supersaturations. Calculate the number concentration of CCN for this marine air.

This is just part (a), repeated using a starting point of 0.03 mm

N = (0.025 cm^-3 mm^2.5 / 2.5) x r^-2.5 integrated from 0.03 to 10 mm = (64.1 - 3x10^-5)  cm^-3 ~ 64.1  cm^-3

Note that a very small fraction of the total particles activated. This is good, because it means that the particles will grow pretty large compared to the situation if the same mass of water condensed on a larger number of particles.

3. (a) What is meant by the activation of a cloud condensation nuclei?

Getting past the Kohler curve maximum in saturation vapor pressure, so that the time evolution from this point on is all growth, unless the relative humidity falls below 100%.

(b) Using the shape of the Kohler curves as given by

e'_r/e_s = 1 + A(rT)^-1 - B m_s/r³

derive the radius at which a salt aerosol particle becomes activated. Also derive a simple expression for the supersaturation at which this occurs. Hint: Think about the point at which activation occurs on a Kohler curve and then use calculus to find a mathematical expression for this point. For a droplet in equilibrium the relative humidity is RH/100 = e'_r/e_s.

d(e'_r/e_s)/dr = - AT^-1r^-2 + 3B m_sr^-4 = 0

this simplifies to r^-2(3 B m_sr^-2 - AT^-1) = 0.

We know the first two solutions for +/- infinity aren't the ones we care about, so it is the parenthetical expression that we grab...so that

3 B m_sr^-2 = AT^-1

or, r^-2 = A/(3BT m_s)

or, r = (3BT m_s/A)^1/2

Plug this value back into the original expression to derive the supersaturation, and recognize that:

S = (e' - e_s)/e_s = e'/e_s - 1 = A(rT)^-1 - B m_s/r³ = AT^-1 A^1/2 (3BT m_s)^-1/2   - B m_s A^3/2 (3BT m_s)^-3/2

This simplifies to A^3/2 [3 (27 B T³ m_s)^-1/2 - (27 B T³ m_s)^-1/2] = 2A^3/2 (27 B T³ m_s)^-1/2

(c) Calculate the activation radius in (m) and supersaturation (in percent) of a sodium chloride particle of mass m_s=10^-15 g. Use A = 3.25 x 10^-7 m K for water at 0 ^oC and B = 1.47x10^-7 m³ g^-1 for NaCl. You should check your results with the Kohler curve figure from the book.

Plug in -

r = 0.61 x 10^-6 m

S = 0.0013 (or 0.13% supersaturation)

4. A concentration of N=100 cm^-3 haze droplets all activate at a radius of 0.3 m m and grow in an updraft having an average supersaturation of 0.1% (S=0.001). The temperature is -15 ^oC and pressure is p = 800 mb, so the diffusion constant is D= 2.54x10^-5 m²/s.

(a) What is the radius of the cloud droplets after 10 minutes?

integrating the growth equation, dr/dt = 1/r (G_lS), assuming that S and G_l are constant, yields:

r = (r_o² + 2G_lSt)^1/2

where r_o is the radius at t=0.

G_l = De_s/(r_lR_vT) = (2.54x10^-5 m² s^-1 * 191 Pa)/(1000 kg m^-3 * 461 J K^-1 kg^-1*258 K)  =  40.7 mm² s^-1

thus, r = (0.3 mm² + 2*0.001*40.7 mm² s^-1)^1/2 = 7.0 mm

(b) What is the liquid water content (g m^-3) of the cloudy parcel at this time?

Liquid water content is mass of cloud droplets per unit volume of air, or the mass of one droplet multipled by the number concentration of droplets

W = 4/3 p r³ p_l N = 4/3 p (6.7 x 10^-4 cm³)(10⁶ g m^-3)(100 cm^-3) = 0.14 g m^-3

(c) How would the growth rate and liquid water content compare for a warmer cloud with the same average supersaturation? Explain.

The droplet drowth rate (hence, liquid water content) would be larger for a warmer cloud with the same level of supersaturation because there is more water in the vapor phase, and the growth rate is proportional to the vapor pressure of water which increases strongly with temperature.

5. (5710 question) In a real updraft the supersaturation is not constant because there is a competition between the supersaturation increasing as the parcel rises (and cools) and the supersaturation decreasing as the water vapor condenses onto droplets.

(a) Derive the following expression for the droplet growth rate assuming all of the cloud droplets are the same size:

 dr/dt = D(r _a/r_l) a (t/r) - 4p /3 (DNr²)

where N is the number concentration of droplets, r is the droplet radius, t is time, r _a is the density of air, r _l is the density of liquid water, D is the diffusion coefficient, and a is the rate of change of the saturation mixing ratio, w_s=w_s(t=0)-at. Hint: start with the condensational growth equation in terms of vapor density, rather than supersaturation.
 

The way to solve this problem is to note that the reason the supersaturation does not stay high is that droplets take up water, thereby removing it from the vapor phase. The sum of water mass in the vapor and solid phases is constant (unless you can think of some mechanism to increast the water...like drops falling from above, but let's assume that this is a new cloud and these are the only droplets in the vicinity).

so, w_tot = w_vap + w_liq = w_v + w_l

starting with the equation for growth by condensation

r dr/dt = D/p_l [r(infinity) - r(r)]

recall that the density of water vapor at infinity is the 'ambient water vapor density) and that at the radius r of the droplet is the saturation density (which we can get from the Clausius Clapeyron equation)

r dr/dt = D/p_l [r_v - r_s]

Dividing both sides by density of air,

r/r_a (dr/dt) = D/r_l [w_v - w_s] = D/r_l [w_tot - w_l - w_s]
 

Substituting in the linear expression for the time dependence of water saturation, w_s = w_s(t=0) - at, and noting that at t=0, there is no liquid water, so w_tot = w_s(t=0), we get:

r/r_a (dr/dt) = D/r_l [w_s(t=0) - w_l - w_s(t=0) + at] = D/r_l [at - w_l]

w_l is just 4/3 p r³ N (r_l /r_a) where N is the number density of droplets

Substituting and rearranging gives us the desired equation.

dr/dt = D(r _a/r_l) a (t/r) - 4p /3 (DNr²)
 

(b) Numerically integrate the above growth equation. Use a cloud base pressure of p=800 mb and temperature of T=10 ^oC. Obtain the constant a from assuming a linear change in saturation mixing ratio (w_s) for a small change in height ((z)

a = [w_s(z_b) - w_s(z_b + D z)](v_u/D z)

 where z_b is the cloud base height and vu is the updraft speed, and (z=500 m. The saturated adiabatic lapse rate for these conditions is about dT/dz= -5.0 ^oC/km. Assume the diffusion coefficient is constant at D=3.0x10^-5 m²/s and the air density is constant at r _a= 0.98 kg m^-3. Assume the CCN are relatively large, so the droplets activate at rather low supersaturations, and the starting droplet radius is r₀=1.0 m m.
Plot the droplet radius and supersaturation as a function of time for three cases: 1) N=100cm^-3 and v_u=1.0 m/s, 2) N=100 cm^-3 and v_u=0.4m/s, and 3) N=200 cm^-3 and v_u=1.0m/s. The maximum time should be t_max = D Z/v_u.

First, we need to determine the constant "a". The best way to get ws is to use the Clausius Clapeyron equation, in this case using 10 ^oC at 800 hPa (pressure at z_b, the base of the cloud) and the 7.5 ^oC at z_b + 500 m (10 ^oC - 5 ^oC km^-1 x 0.5 km).
 

e_s(10^oC) = 6.11 hPa x exp [2.5x10⁶/461(1/273.15 - 1/283.15)] = 12.3 hPa

es(7.5 ^oC) = 6.11 hPa x exp [2.5x10⁶/461(1/273.15 - 1/280.65)] = 10.4 hPa

Use the hydrostatic equation (dlnP = -g/RT dz), with T = ~10^oC, g=9.8 m s^-2, and R = 287 J kg^-1 K^-1, to get P₂ ~ 753 hPa, the pressure at z_b + 500 m.

Calculate the volume mixing ratios:

800 hPa:   vmr_H2O = 12.3/(800-12.3) = 0.0156
753 hPa:  vmr_H2O = 10.4/(753-10.4) = 0.0140

Convert to mass mixing ratios by multiplying by 0.622

w_s(800 hPa) = 0.00970 kg kg^-1
w_s(753 hPa) = 0.00871 kg kg^-1

so a = (0.00970 - 0.00871) x 1 m s-1 / 500 m = 1.98 x 10^-6 s^-1 for the case of an updraft velocity of 1 m s^-1, and 0.8 x 10^-6 s^-1 for the case of an updraft of 0.4 m s^-1.

Now, to the iteration! This isn't so hard, if you remember how to do it.

r_n+1 = r_n + (dr/dt)_n Dt

Example, for the first iteration,

r₁ = r_o + (dr/dt)_o Dt

(dr/dt)_o in units of m s^-1 = D(r_a/r_l) a (t/r_o) - 4p /3 (DNr_o²) = 3.0x10^-5 m²  s^-1 [(0.99 kg m^-3/1000 kg m^-3) 1.98 x 10^-6 s^-1 (t_o/r_o) - 4.20 (10⁸m^-3)(1x10^-6 m)²] = 5.9 x 10^-14 m (t_o/r_o) - 1.26 x 10⁴ r_o²

You can take 1 second time steps without incurring a large accumulation error, but if you try this with 10 second steps, you'll go negative (the first term starts out bad and gets worse)

For the first question (v = 1 m s^-1 and N = 100 cm^-3) I get 13.2 mm after 500 seconds.
For the second question (v = 0.4 m s^-1and N = 100 cm^-3) I get 9.77 mm after 500 seconds
For the third question (v = 1 m s^-1 and N = 200 cm^-3) I get 10.5 mm after 500 seconds

Supersaturation can be determined from the equation
 
S = [(w_t - w_l)/w_s - 1] where w_l is determined from 4/3 p r³ N r_l /r_a and w_t - w_s = at

Also, note that w_t = 0.00969

so, S = (0.00969 - w_l)/(0.00969 - at) - 1

This gives 0.166% supersaturation at the peak at ~15.1 seconds for the first question, 0.094 at 18 seconds for the second, and 0.117% at 10.5 for the third.