1. (a) Calculate the vapor pressure of water, the relative humidity, and the water vapor mixing ratio (both volume and mass) of an air parcel at sea level that has a temperature of 28 ^oC and a dewpoint temperature of 15 ^oC.

Method 1 - Clausius-Clapeyron
Using the dew point temperature, we can determine the true vapor pressure of water in this air parcel at 1013 hPa total pressure.

e_s(15^oC) = 6.11 hPa x exp [2.5x10⁶/461(1/273.15 - 1/288.15)] = 17.2 hPa

Similarly, the total potential water that this parcel can hold at 28^oC is

e_s(28^oC) = 6.11 hPa x exp [2.5x10⁶/461(1/273.15 - 1/301.15)] = 38.7 hPa

The relative humidity is thus 17.2/38.7 x 100% = 44.4%

The mixing ratios of water are:

The volume mixing ratio is just the ratio of the partial pressure of water (17.2 mbar) to the pressure of dry air (1013 mbar - 17.2 mbar) = 0.0173. Note, if you forget to correct for dry air, the result isn't very different (0.0170).

The mass mixing ratio is the volume mixing ratio multiplied by the ratio of the molecular weight of water (18 g/mole) to the molecular weight of air (29 g/mole), or 0.0173 x 18/29 = 0.0107 kg/kg....or 10.7 g/kg.

Method 2 - Alternatively you can do this with a saturation vapor pressure curve

From this curve we find e(sat)~ 16 mbar at 15 C, and e(sat) ~ 38 mbar at 28 C, so the relative humidity is 16/38 x 100% = 42%, the rest follows. This isn't as accurate as Method 1.
 

(b) If the air in part (a) rises adiabatically to an altitude where the pressure is 800 mb, what are the water vapor mixing ratio, the water vapor pressure, the relative humidity, and the dewpoint?

Potential temperature will be constant for an adiabatic process. The potential temperature at sea level is  theta = 301(1000/1013)^0.286 = 300 K. Therefore, the temperature at 800 mbar can be found from the equation 300 = T(1000/800)^.286, so that T = 281 K. As the parcel expanded, so too did the vapor pressure of water decrease in proportion to the pressure change, so it fell from 17.2 hPa to 17.2(800/1013) = 13.6 hPa.

Note that for a temperature decrease of 20^oC the air must have risen about 2 km (the dry adiabatic lapse rate is 10 ^oC/km). Thus, the dew point decreased about 4 ^oC (the dew point lapse rate is about -2^oC per km), or down to about 11 ^oC. This implies that the air has become saturated (the dew point temperature is larger than the dry adiabatic temperature...so at a somewhat higher pressure, the air condensed).

Using the assumed vapor pressure of water (13.6 hPa), and Clausius-Clapeyron, we determine a T_dew of

1/T = 1/273.15 - (461/2.5x10⁶)ln(13.6/6.11) = 284.6

Note that this is a few degrees above the temperature, so water is saturated (supersaturated, in fact). If there is nothing to condense on, the actual relative humidity can be determined from the vapor pressure (13.6 hPa) and the 100% saturation value determined from C/C equation:

e_s(281 K) = 6.11 hPa x exp [2.5x10⁶/461(1/273.15 - 1/281)] = 10.6 hPa

RH = 13.6/10.6 x 100% = 128%

It is likely that some water vapor has started to condense, so the temperature is a bit higher than the dry adiabatic temperature...let's say it's about 282-283 K. So maybe about 5% of the water has condensed onto particles (see problem 2).

2. In a Chinook wind storm Boulder experiences strong westerly winds, and the air often has descended from the continental divide. There are often orographic clouds caused by flow of air forced upward by the mountains to the west of Boulder. Use a thermodynamic diagram and the information below to investigate the air flowing from Grand Junction (in western Colorado) to Boulder.

(a) Grand Junction, P = 850 mb, T = 11.5^oC, RH = 70%. Use the diagram to find the water vapor mixing ratio, the dewpoint, and the potential temperature of this air at the start of its journey over the mountains.

Plot the point P=850 mbar, T = 11.5 C, and follow the constant water vapor mixing ratio line down to 10 g/kg. This is the saturated amount of water...since the relative humidity is 70%, the actual amount of water is 7g/kg. The dew point temperature is thus about 5-6 degrees at 850 mbar, and the potential temperature is just under 300 K...about 298 K.

(b) As the westerly flow forces the air parcel up over the mountains, find the pressure level and temperature at which the parcel becomes saturated. What is this level called?

Follow the 298 K dry adiabat (just under the 300 line) until it crosses the dew point, which drops about 2 degrees per km...and they intersect at about 785 mbar. This is where the temperature matches the dew point, i.e. water condenses (the "lifting condensation level"). The temperature is about 4-5 degrees.

(c) Find the temperature, the water vapor mixing ratio, the potential temperature, and the equivalent potential temperature when the air reaches the continental divide at 600 mb. How much water vapor has condensed (g/kg)?

Because water has condensed at 785 mbar, the temperature of the rising parcel of air follows the saturated adiabat up to 600 mbar, where the temperature is about -7.5 degrees. Here, the saturated water vapor mass mixing ratio is 4 g/kg, so about 3 g/kg has condensed into particles. The potential temperature is 308 -309 K (you can read it off the chart). The equivalent potential temperature is the value of the saturated adiabat...320 K
 
 

(d) Check the validity of your graphical results by calculating the equivalent potential temperature for the initial parcel at 850 mb and the parcel at 600 mb. Remember, for unsaturated parcels use the temperature and saturation mixing ratio at the LCL. Why are the two q_e values at Grand Junction and at the continental divide equal?

At 600 mbar, you can calculate the equivalent potential temperature from

Theta(e) = 308 x exp(2.5x10^6 x 0.004/1004x265) ~ 320 K

The equivalent potential temperature at 850 mbar requires us to use numbers from the LCL...or

Theta(e) = 298 x exp(2.5x10^6 x 0.007/1004 x 278) ~ 317 K.

These values are essentially equivalent, they differ because we have visually estimated temperatures, humidities, etc.
 

(e) The air parcel continues its trajectory down to Boulder at 850 mb. Assume all the condensed water precipitated out on the west slope of the mountains, so that the parcel is unsaturated as it flows down to Boulder. (Actually some portion of the cloud would evaporate on the way down.) Use the diagram to find the air temperature and the relative humidity in Boulder. How much warmer is the air in Boulder than in Grand Junction? Why is it warmer, even though it started and ended at the same pressure?

See Stuve Diagram

Follow the dry adiabat from 600 mbar to 850 mbar, and the temperature becomes 294 K at 850 mbar. That's about 10 degrees warmer than the starting temperature at Grand Junction. The RH is now about 25%...much more dry! This is all because the water condensed out and released latent heat.

3. (5710 question) The saturation vapor pressure with respect to an ice surface is lower than with respect to a water surface. This fact is very important for cloud microphysics.

(a) Use the Clausius-Clapeyron equation to derive an expression for the difference in saturation vapor pressures e_s-e_si as a function of temperature. You may assume that the latent heat capacities are constant.

e_s(ice) = 6.11 exp [L_i/R_v(1/273.15 - 1/T)]
e_s(liq) = 6.11 exp [L_l/R_v(1/273.15 - 1/T)]

The difference between these two is what we are interested, and we know the vapor pressure over the liquid will be higher than over the solid (because L_i > L_l).

De_s = e_s-e_si = C {exp [L_l/R_v(1/273.15 - 1/T)] - exp [L_i/R_v(1/273.15 - 1/T)] }

(b) Find an expression for the temperature of maximum difference in saturation vapor pressures. Calculate this temperature.

Taking the derivative of the expression above with respect to temperature, and setting equal to zero (for maxima/minima), we get:

d(De_s)/dT  =  - C {(L_l/R_vT²) exp [L_l/R_v(1/273.15 - 1/T)] - (L_i/R_vT²) exp [L_i/R_v(1/273.15 - 1/T)] } = 0

= CL_l/R_v {(1/T²) exp [L_l/R_v(1/273.15 - 1/T)] - (L_i/L_lT²) exp [L_i/R_v(1/273.15 - 1/T)] } = 0

or, (1/T²){exp [L_l/R_v(1/273.15 - 1/T)] - (L_i/L_l) exp [L_i/R_v(1/273.15 - 1/T)]} = 0

The 1/T² term must be the minima at T -> infinity and negative infinity, so we can set the term in {} to zero to see if this is the maximum.
 

exp [L_l/R_v(1/273.15 - 1/T)] = (L_i/L_l) exp [L_i/R_v(1/273.15 - 1/T)]

Solve this by taking logs:

L_l/R_v(1/273.15 - 1/T)  = ln(L_i/L_l) + [L_i/R_v(1/273.15 - 1/T)]

simplify:

(1/273.15 - 1/T)  =   (R_v/L_l)ln(L_i/L_l) + L_i/L_l(1/273.15 - 1/T)

Solve for 1/T:

1/T [1 - L_i/L_l] = 1/273.15 [1 - L_i/L_l] - (R_v/L_l)ln(L_i/L_l)

or 1/T = 1/273.15 - (R_v/L_l)ln(L_i/L_l)/[1 - L_i/L_l]

For L_i = (2.5x10⁶ + 3.34x10⁵)  = 2.834 x10⁶
and L_l = 2.5x10⁶ and R_v = 461 (all in MKS units), we get

1/T = 3.66 x10^-3 + 1.73 x10^-4 = 3.83 x 10^-3 or T = 261 K (or -12.3^oC)

Note that this is a little low (it appears to be about -11 ^oC in Figure 2.7 on page 73 of the text). What is wrong? Probably that we used the
values for latent heats at 0 ^oC and these are a little different at -10^oC (i.e. we didn't account for the T dependence of the latent heats).

FOR THOSE OF YOU WHO CAME TO MY OFFICE FOR HELP, I goofed when I told you Li was 2.25x10⁶. I was reading page xvii too fast. In fact, L_i = L_l + L_f. I apologize for leading you astray.

(c) What is the percent saturation with respect to ice at this temperature (100(e_s/e_si)?
 

e_s(ice) = 6.11 exp [2.83x10⁶/461(1/273.15 - 1/261)]  = 2.15 hPa

e_s(liq) =  6.11 exp [2.5x10⁶/461(1/273.15 - 1/261)]  = 2.42 hPa

This is a 12.6% supersaturation (2.42-2.15)/2.15 x 100%