HOMEWORK 6 SOLUTIONS

1. Sandy is experiencing a phenomenon known as the sea breeze. During the night, land cools down much more rapidly than the ocean. So, (refer to diagram 1 below) as temperature decreases, density increases. From the hydrostatic equation, we know that this will reduce the height of pressure surfaces over land. This means that air aloft will experience a force from the sea to land. But, as soon as mass gets transferred over the column of air over land, the surface pressure over land will increase. So, the isobars will be aligned as in diagram 1. So, in the early morning, Sandy experiences wind blowing in her face as she bicycles inland. During the evening (refer to diagram 2) the reverse situation occurs, as the temperature in the column of air over land increases. So, again Sandy experiences a headwind.

2. a) In the problem, it is given that the mixing ratio is 10 g/kg. From the adiabatic chart, we find the point that corresponds to 800 mb and 20 [[ring]]C, and see that it also corresponds to a saturated mixing ratio of 18 g/kg. So, the relative humidity (mixing ratio divided by saturated mixing ratio) is 10 / 18 = 0.55, or 55%.

b) Because the parcel is not yet saturated, it will rise at the dry adiabatic lapse rate. So, following parallel to one of the dry adiabats, we find the temperature at 700 mb to be about 9 [[ring]]C. This point corresponds to a saturated mixing ratio of 10 g/kg. Because no water has entered or left the parcel, and no water has condensed yet, the actual mixing ratio is still 10 g/kg, so the parcel is now saturated. The relative humidity is 100%.

c) Now the parcel is saturated, so if it cools any more it will be condensing. So, it will cool at the moist adiabatic lapse rate. Following parallel to a wet adiabat up to 550 mb, it looks like the temperature is around -1 [[ring]]C. As the parcel is still saturated, the relative humidity is still 100%. (Note here that mixing ratio only takes into account the amount of water vapor that the parcel contains, so there are still 10 g/kg of water in the parcel, but some of that is condensed).

d) If none of the condensed water rains out, then the parcel will stay saturated as it sinks (following a wet adiabat) until all the water that originally condensed has evaporated. At that point, it will follow a dry adiabat back to the surface. Notice that as nothing inside the parcel has changed, it will follow the exact same path back to the surface as it took to get to 550 mb. So, the temperature and relative humidity are 20 [[ring]]C and 55%, respectively.

e) If all the condensed water rains out of the parcel, then as soon as it sinks a little bit, it will no longer be saturated, so it will follow a dry adiabat back to the surface. So, following a dry adiabat back to the surface, one finds the surface temperature to be 29 [[ring]]C. The relative humidity can be found through first finding the mixing ratio at the surface. Recall that some of the initial 10 g/kg fell out of the parcel as rain. Back at 550 mb, the saturated mixing ratio is about 7 g/kg, and the relative humidity is 100%. So, the actual mixing ratio must be 7 g/kg at 550 mb. As there is no more water to evaporate into the air on the descent, the mixing ratio at the surface must also be 7 g/kg. The saturated mixing ratio is seen to be around 35 g/kg, so the relative humidity at the surface is 7 / 35 = 0.20, or 20%.

f) As the parcel lifts, it cools, and the saturated mixing ratio decreases. At some point, the saturated mixing ratio is equal to the actual mixing ratio (relative humidity = 100%), so the parcel will start to condense if cooled any more. At this point (where the relative humidity first reaches 100%), we will observe the beginning of a cloud. For this parcel, the cloud base (or lifting condensation level) occurs at 700 mb.

g) Winds that form in this way end up much warmer and drier than they started out to be. Notice that after raining out all the condensed water, this air parcel was at 29 [[ring]]C and 20% relative humidity, compared to the initial 20 [[ring]]C and 55% relative humidity. So, they are much more likely to be warm enough to melt snow, and dry enough for sublimation. Thus, they are called "snow eaters".

3. a) In order for air parcels to expand, they must push against the surrounding air. This means that the parcel is using some of its internal energy (which determines its temperature) to do work on the surrounding air. This decreases the internal energy of the parcel, which means that the temperature decreases.

b) If a parcel is saturated, it will still use the same amount of internal energy to do work on its surrounding air as if it were dry. But, as it cools the saturated mixing ratio decreases below the actual mixing ratio, which means that the difference must be condensed. Recall from other homework problems, labs, and lectures, that as water condenses, it releases energy in the form of latent heat, that is realized as an increase in internal energy. So, the first process is a sink of energy for the parcel, and the second parcel is a source of energy for the parcel. The first process wins out, and the parcel cools, but not as much as if it were dry (notice that the dry parcel is the exact same except without the source of internal energy).

4. a) As warm, moist air leaves one's mouth, it mixes with the cold surrounding air, and cools. This mixing decreases the saturated mixing ratio, causing the air to become saturated, and a cloud forms. As the water droplets scatter light, they appear white and you can see your breath.

b) The warm lake warms and moistens the air directly above the lake. As cool air blows over the lake, it mixes with this warm, moist air, bringing the air over the lake to saturation. Recall figure 5.11 from page 118. As warm air mixes with colder surrounding air, it follows a linear path from one point to the next. At some point, this line will cross the saturation vapor pressure line (the saturation mixing ratio decrease with temperature is much more rapid than a linear decrease, in this range of temperatures), and water vapor inside the parcel will condense.

5. a) Humid air is actually lighter than dry air. From the discussion on page 123, we see that typical air has a molecular weight of about 29 (the units are in grams / mole, but that isn't as important here). As water has a molecular weight of 18 g/mole, moist air is actually less dense than dry air.

b) Michela feels less comfortable in New Orleans because the air is much more humid in New Orleans than in Phoenix. Our body cools off through evaporation of sweat. Recall that it takes energy for liquid water to evaporate, so energy that would ordinarily be used to warm our body is now used to evaporate sweat. As Phoenix is much dryer than New Orleans, more evaporation will take place in Pheonix. So, even though it is warmer in Phoenix, Michela feels cooler due to the latent heat of evaporation.