EPS5 - Problem Set 4 Solutions (50 points) Due March 3, 2010

1. Conceptual Questions (12 Points)(a) Explain in simple terms what the dry adiabatic lapse rate represents. (3 points)

It represents the rate at which the temperature of an air parcel changes with altitude whereno condensation of water takes place and the parcel does not exchange heat with the sur-rounding air.

(b) In order for a parcel to move up adiabatically, it has to do work on the atmosphere. Wheredoes the energy to do this work come from? (3 points)

When the air parcel rises adiabatically, the parcel pushes against the force due to the pressureof the surrounding atmosphere (it expands). In order to do this work, the air parcel requiresenergy. Since there is no heat exchange, this energy must come from the kinetic energy of themolecules within the air parcel. This kinetic energy is directly proportional to temperature(E=3/2 kT), so as we take some of this energy to do work (i.e., expand), the temperature ofthe air parcel has to decrease.

(c) If we add heat to the air near the ground and the atmosphere is initially neutrally stable, whatwill happen? Will this cause the atmosphere to remain neutrally stable or become unstable? (3points)

The atmosphere will become unstable. In a neutrally stable atmosphere, the lapse rate fol-lowed by the parcel is the same as the environmental lapse rate. If we add heat, the air nearthe ground will become lighter (warmer temperature, smaller density), so the net buoyancyforce will be positive. The air will start to rise and it will continue to rise as long as it stayswarmer than its surroundings and its net buoyancy force stays positive.

(d) The moist lapse rate is less negative than the dry adiabatic lapse rate. This means that anair parcel that follows the moist lapse rate will be warmer than an air parcel that follows the dryadiabatic lapse rate when both air parcels start at the same temperature. Where does this extra heatcome from? Explain. (3 points)

It comes from the latent heat of condensation. Water vapor condenses when it reaches thedew point. The phase change from gas to liquid (or solid) that occurs during condensationrequires release of energy, and it is this energy release (i.e., latent heat of condensation) thatwarms up the surroundings.

2. Implications of atmospheric stability (15 Points)(a) For each segment (i.e. A, B, C) on the graph shown below, calculate the lapse rate of theatmosphere and indicate if the atmosphere is stable or unstable for dry air parcels. (3 points each)For each segment, we calculate the lapse rate as Γ = −∆T/∆z:A: -∆T/∆z = -(300-298 K)/(1-0 km) = -2 K/km. The atmosphere is stable.B: -∆T/∆z = -(288-300 K)/(3-1 km) = 6 K/km. The atmosphere is conditionally stable sincethe lapse rate is less than the dry adiabatic lapse rate, but greater than the smallest moistadiabatic lapse rate (3K/km).C: -∆T/∆z = -(252-288 K)/(6-3 km) = 12 K/km. The atmosphere is unstable.

1

EPS5 - Problem Set 4 Solutions (50 points) Due March 3, 2010

(b) If the meteorological conditions shown in the figure above persist over an area with surface airpollution sources, is the region at risk for a severe pollution episode? Explain whether you thinkthe surface air will stagnate or ventilate, and whether this will cause pollution to accumulate. (3points)

Yes. There could be a severe pollution episode because the the inversion leads to stable air atthe surface (region A), which will inhibit the dilution of surface emissions with clean air fromabove. The emissions will remain near the surface where they could harm people, plants, andanimals.

(c) Large, buoyancy-driven, convective storms - which can produce tornadoes, hail, and lightningand sometimes exceed 100 km in diameter - are relatively common over the central US duringsummertime. Which of the two scenarios below is more favorable for the formation of severeconvective storms? Explain why. (3 points)

Figure A shows an unstable atmosphere. Those conditions make convective activity likely.While severe convective storms will not always materialize under unstable conditions they

2

EPS5 - Problem Set 4 Solutions (50 points) Due March 3, 2010

also require moisture and certain configurations of fronts the stable conditions shown infigure B make such storms unlikely.

3. The Two Faces of Mt. Rainier (23 Points)With a summit elevation of 4394 m, Mt. Rainier is the second tallest peak in the contiguous UnitedStates (and only 27 m smaller than the tallest). Its dramatic topography (along with that of thesurrounding Cascade Range) plays a large influence on local weather patterns. In this problem,we’ll see how Mt. Rainier influences the climate of two nearby cities: Olympia, located to thewest of the peak, and Yakima, located to the east. Despite their proximity to one another andsimilar average surface temperatures, these cities experience very different weather, as shown inthe following table (data from www.weatherbase.com). Throughout the problem, you can assumethe adiabatic lapse rates are Γdry = 9.8 K/km and Γmoist = 6.4 K/km.

Yakima OlympiaLatitude (◦N) 46.6 47.0Elevation (m) 325 29Distance from Mt. Rainier (km) 126 (to east) 120 (to west)Average Annual Temperature (◦C) 10 10Average Relative Humidity (%) 71 89Average Yearly Precipitation (cm) 21 128Average Number of Cloudy Days 164 229

(a) Assume air blows from west to east, starting at Olympia, and rises over Mt. Rainier. If therewere no moisture in the air, what would the temperature be at the summit? You can assume the airoriginally is at the average surface temperature of Olympia. (2 points)

We use the dry adiabatic lapse rate, starting from the temperature at Olympia.

TRainier = TOlympia − Γdry ∗ (zRainier − zOlympia)

= 10◦C − (9.8◦C/km)(4.394km− 0.029)

= −33◦C

(b) If this dry air then descended over Yakima, what would its temperature be when it arrived atYakima? (2 points)

We again use the dry adiabatic lapse rate. This time, the air starts at the temperature of thesummit and warms as it descends over Yakima.

TY akima = TRainier + Γdry ∗ (zRainier − zY akima)

= −33◦C + (9.8◦C/km)(4.394km− 0.325)

= 7.1◦C

(c) In reality, we see from the table that assuming the air over Olympia is dry is a bad assumption.Why would you expect the relative humidity at Olympia to be so high? (1 point)

Olympia is located very close to the coast, so air that arrives at Olympia has just passed overthe ocean, picking up moisture.

3

EPS5 - Problem Set 4 Solutions (50 points) Due March 3, 2010

(d) Given the relative humidity and surface temperatures at Olympia and Yakima, what is the vaporpressure at each city? (2 points)

We can get the saturation vapor pressure Ps from the Clausius-Clapeyron curve (or equation)then use that in the definition of relative humidity to get the water vapor pressure of the air:RH = PW

PS∗ 100 or PW = PS ∗ RH

100. Both cities are at the same temperature (10◦C), so they

both have the same saturation vapor pressure of about 12 mb. Because they have differentrelative humidities, they have different water vapor pressures.Yakima: PW = (12mb) ∗ 71

100= 8.5mb.

Olympia: PW = (12mb) ∗ 89100

= 10.7mb.

(e) What are the dew point temperatures at each city? (2 points)

Now that we know the vapor pressures, we can read the dew point temperatures from theClausius-Clapeyron curve (or equation).Yakima: At 8.5 mb, the dew point temperature is about 4.5◦C.Olympia: At 10.7 mb, the dew point temperature is about 7.9◦C.

(f) As the air flows westward, it is forced upwards over Olympia by Mt. Rainier. At what altitudewill clouds form over Olympia? At what temperature? As the air rises below the condensationlevel, which lapse rate should you use? Justify your choice. You can assume the dew point de-creases with altitude at a rate of Γdew = 2 K/km. (4 points)

Use the dry adiabatic lapse rate below the cloud base because no moisture is condensing.Clouds will form when the dew point temperature is equal to the air temperature.We can get the dew point temperature at this unknown altitude using the dew point lapserate and the air temperature at the same altitude using the dry lapse rate:

Tdew = 7.9◦C − (2◦C/km)(z − 0.029km)

Tair = 10◦C − (9.8◦C/km)(z − 0.029km)

We then solve the system of equations for z:

Tdew = Tair

7.9◦C − (2◦C/km)(z − 0.029km) = 10◦C − (9.8◦C/km)(z − 0.029km)

7.9◦C − (2◦C/km)(z) + 0.058◦C = 10◦C − (9.8◦C/km)(z) + 0.28◦C

2.3◦C = (7.8◦C/km)(z)

z = (2.3)/(7.8)km

z = 0.298km

z = 298m

We can now calculate the temperature at this altitude, still using the dry adiabatic lapse rate:

Tair = 10◦C − (9.8◦C/km)(0.298 − 0.029km) = 7.4◦

(g) What is the air temperature at the summit of Mt. Rainier? Which lapse rate should you useabove the level of the cloud base? (3 points)

4

EPS5 - Problem Set 4 Solutions (50 points) Due March 3, 2010

Above the cloud base, we use the moist adiabatic lapse rate because water is condensing,adding heat to the environment. We already know the temperature of the air at the cloudbase, so we just need to find the temperature change from that level to the summit:

TRainier = Tcloud − (6.4◦C/km)(z)

TRainier = 7.4◦C − (6.4◦C/km)(4.394 − 0.298)

TRainier = −19◦C

(h) As the air descends on the eastern side of Mt. Rainier, it warms adiabatically. Calculate thetemperature of the air when it reaches Yakima. Justify your choice of dry or moist lapse rate. (3points)

We need to use the dry adiabatic lapse rate because no precipitation forms while the air de-scends. As soon as the air starts to descend, it warms adiabatically, bringing its temperatureabove the dew point temperature. As it continues to descend, its temperature increases morequickly than the temperature of the dew point, so it continues to follow the dry lapse rate allthe way down.

TY akima = TRainier + (9.8◦C/km)(z)

TY akima = −19◦C + (9.8◦C/km)(4.394 − 0.325)

TY akima = 20◦C

(i) Explain why your answers from parts (b) and (g) are different. What heat source or sinkexplains the difference? Why might your answer to part (g) diverge from the average value givenin the table? (3 points)

NOTE: Obviously there was an error in the question - it should have said (b) and (h). Credit willbe given for answers comparing (a) and (g) - the summit temperatures - or (b) and (h) - the Yakimatemperatures. Next time, please email for clarification if confusion arises!

The air is about 13◦C warmer due to the effect of condensation. The heat comes from thelatent heat of condensation. There are a number of reasons for the divergence from the mea-sured annual average temperature. We’re only looking at one weather pattern and assumingit is the sole explanatory force for the surface temperature. While eastward flow is dominant,Yakima can be influenced be a variety of flow regimes. We’ve also assumed air starts withaverage surface temperatures, but flow over Mt. Rainier may vary seasonally, accompaniedby seasonal swings in the original air parcel temperature. And of course, Yakima clearlydoes have cloudy and rainy days, they just aren’t as prevalent as those in Olympia due to themountain forcing.

(j) Think about the different lapse rates you used for each part of the problem. Can you explainwhy cloudy days are 70% more common over Olympia than over Yakima? (1 point)

We argued that clouds from easily on the western side of Mt. Rainier because air is forcedupwards over the mountain, causing it to cool to the dew point temperatures. This forcingpattern forms clouds to the west of the mountain; i.e., over Olympia. On the eastern side (i.e.,over Yakima), on the other hand, the air descends and warms adiabatically, making cloudformation much less likely. This can partially explain the difference in cloudiness over thetwo cities.

5