1. A remarkable new device that follows trajectories of air parcels in the earth's atmosphere was used to measure the following vertical temperature profiles for two air parcels separated horizontally by 10 km.

(20 points) From the temperature profiles alone, use principles presented in this class to estimate the temperatures at points A, B, and C. Describe how you arrived at your answers.
 
Both parcels rise up to 2 km with a constant lapse rate (the figures aren't labeled, so we don't know what this lapse rate is). However, above 2 km, the second parcel cools at a slower rate as it rises than the first parcel, so we can predict that the second parcel becomes "wet" at 2 km, whereas the first remains dry. So, using the dry adiabatic lapse rate of 10 degrees per km, and the wet rate of about 5-6 degrees per km, we estimate that the first parcel will be 30 - 7*10 = -40 C at 7 km (point A), whereas the second parcel will be 30 - 2*10 = 10 C at 2 km (point B) and 10 - 5 (or 6) *5 km = -15 to -20 C at point C.
 
2. (30 points)

(a) Using your answers to Problem 1, determine the potential temperatures of parcel I at point A and parcel II at point C (note: you can assume that the ambient pressure at 7 km is the same for both parcels).

We know that for a dry adiabatic process (first parcel), the potential temperature is constant, so parcel I at point A has a potential temperature of about 30 C (or about 303 K). Note, this isn't exactly right because sea level pressure is 1013 mbar, not 1000 mbar, but the difference is small...only 1 degree (i.e. the "true" potential temperature is 302 K because the reference pressure is taken to be 1000 mbar, not 1013).

From this, we can estimate the pressure at 7 km using Poisson's equation,

303 = T x (1000/P)^0.286

Solving for P where T = -40 C (233 K) we find P = 399 mbar. Let's use 400 mbar to solve the problem just to make things a little bit easier.

So for parcel II at point C (7 km) the pressure is 400 mbar, but because it's temperature is higher than parcel I's, it will have a higher potential temperature.

theta = 258 (1000/400)^0.286 = 335 K (or 329 if you used 6 degrees/km for wet ALR)

Note: you didn't even have to calculate the pressure to get this result...the ratio of the 7 km and surface pressures for the two parcels are the same, namely
303 K/233 K = 1.30, this because the pressures are the same.

In case you didn't know how to get pressure (or eliminate it), you could have used the scale height to calculate the pressure. In this case, a good temperature to use is temperature of parcel I at 3.5 km (half-way point), or 268 K. H = RT/g = 7.8 km. Then, using the hypsometric equation you'll get 413 mbar, so that the potential temperature of parcel I is 300 K and parcel II is 324 - 330 K.

(b) If the air parcels were to be compressed adiabatically to a surface pressure of 1000 mbar, what would their temperatures be?

This is the definition of potential temperature. That is, the answers for (a) will be the answers for (b).

 
(c) Using a scale height of 8.58 km, appropriate for 0-2 km, calculate the potential temperature of air parcel II at Point B. Explain why this value is different from the value that you obtained in 2(a).

P (2 km) = 1013 x exp^(-2/8.58) = 802 mbar

So theta = 283 * (1000/802)^0.286 = 305 K. Note, this is essentially the value that you got in part A for the dry parcel...that is, because parcel II rises at the dry adiabatic lapse rate up to 2 km, it has the same potential temperature as parcel I.

The reason this potential temperature is smaller than the value calculated at point C is that the wet parcel has released latent heat that has warmed the air compared to what it would have been if it rose dry adiabatically.
 

3. (25 points) (a) Using equation 2.76 from the text for equivalent potential temperature, determine the saturation mixing ratio (w_s) for air parcel II. (note: this is equivalent to the mass mixing ratio).

Equivalent potential temperature is defined as the potential temperature that a parcel of air would have if the relative humidity was 0% (that is, if all the latent heat from the water in the parcel was released into the air). Therefore, it must be larger than the potential temperature. Because of the differences in potential temperatures you calculated at points B and C for parcel II, we need some sort of term that is invariant with altitude. This is the equivalent potential temperature. So, if the parcel is mostly dry by 7 km (we'll see later that it is a good assumption), we can assume that the equivalent potential temperature is the potential temperature. From equation 2.76,

330 = 303*exp (Lw/CpT)
where L is the latent heat of evaporation (2.5 x 10^6 J/kg), Cp = 1000 J/deg-kg.

For a temperature of 283 K (T at the dew point), we find w=0.0097 (or 9.7 g/kg).
 

(b) What was the dew point of parcel II at the surface before it began to rise?

This is the mass mixing ratio, which corresponds to a vapor pressure of about 15.5 mbar (.0097 x  29/18 x 1000) at surface pressure. This is a dew point of 14 C (See saturation table).

Note, you could have determined this same value from the fact that the dew point lapse rate is 2 deg/km, and the lifting condensation level (LCL) was at 2 km...so take the temperature there (10 C) and add 2x2 = 4 deg, so the dew point at the surface was 14 C.

4. (25 points) A radiosonde measures the following ambient temperature profile (i.e. not the temperature of an individual air parcel). Also depicted on the graph are the adiabatic temperature profiles for dry air and air with a relative humidity of 50%.
 
(a) Describe what would happen to dry air and that with relative humidity of 50% at point B if they were to start to rise (for example, due to traveling over a surface topographic feature, like a small mountain).

Dry air at point B, if it continues to rise, will be colder, and therefore more dense, than the surroundings. Therefore, it will sink back down to the point where the temperature is the same as the surroundings. Air that is humid will find itself warmer than the surroundings above point B, and will therefore continue to rise without a restoring force (a situation that we call "unstable").
 
(b) Is this ambient temperature profile stable or unstable with respect to the dry parcel? Is the profile stable with respect to a wet parcel? Why?

Because a dry parcel of air will not rise if the surroundings are warmer, the ambient temperature profile is a stable one. That is, dry air cannot mix vertically into this surrounding air because the dry air is more dense, and won't rise. However, the wet parcel, because it is warmer than the ambient air, will continue to rise...and from conservation of mass (where there is air rising there must be air somewhere else that sinks), the ambient temperature profile is not stable...that is, the air will turnover. We call this an unstable situation with respect to the humid air. This situation, where the profile is stable with respect to dry air (i.e. dry air can't turnover) and unstable with respect to the humid air, is called "conditional instability".