Physics 137 - Introduction to the Atmosphere and Weather                                                                            Homework Key - Assignment #7
Revised - 11/02/09

7-QT3. Explain why very small cloud droplets of pure water evaporate even when the relative humidity is 100 percent.

    Because of the curvature of a droplet's surface, the molecules on that surface are more exposed to their environment and are less tightly bound by surface tension forces and therefore evaporate more readily than molecules on a plane surface. For that reason the saturation vapor pressure of a surface increases slightly with the curvature of that surface and the relative humidity (defined with respect to a standard flat surface) at which that surface is in equilibrium increases to slightly over 100 percent.

7-QT8.  Raindrops rarely grow larger than 5 mm. Two reasons were given on p.167. Can you think of a third? (Hint: See the Focus section on p. 175, and look at the shape of a large drop.)

    Large droplets tend to be flattened in shape, with the amount of flattening increasing with size. Because flattening increases the surface area of a drop it also decreases the surface tension forces by which the droplet is bound together thereby weakening the drop, enabling it to split into fragments.

7-QT12.  When falling snowflakes become mixed with sleet, why is this condition often followed by the snowflakes changing into rain?

    A transition from snowflakes to sleet indicates that warmer air aloft has moved into a region even though the surface air remains below freezing. This may be a precursor to further influx of warm air into the region which may ultimately increase the surface temperature, converting the sleet to rain.

7-PE1.  In the daily newspaper, a city is reported as receiving 1.32cm (0.52in.) of precipitation over a 24-hour period. If all the precipitation fell as snow, and if we assume a normal water equivalent ration of 10:1, how much snow did this city receive?

    1.32 cm (0.52 in.) × 10 = 13.2 cm (5.2 in.) of snow.

7-PE3.  (a) How many minutes would it take drizzle with a diameter of 200 µm to reach the surface if it falls at its terminal velocity from the base of a cloud 1000 m (3300 ft) above the ground? (Assume the air is saturated beneath the cloud, the drizzle does not evaporate, and the air is still.) (b) Suppose the drizzle in problem 3a evaporates on its way to the ground. If the drop size is 200 µm for the first 450 meters of descent, 100 µm for the next 450 meters, and 20 µm for the final 100 meters, how long will it take the drizzle to reach the ground if it falls in still air?

    (a) From Table 7.1, p.166, 200-µm drizzle has a terminal velocity of 0.70 m/s. Hence the elapsed time is Dt = 1000 m/0.70 m/s = 1429 s = 23.8 m.

    (b) From Table 7.1, p.166, 200-µm drizzle has a terminal velocity of 0.70 m/s; 100-µm drizzle has a terminal velocity of 0.27 m/s; and 20-µm drizzle has a terminal velocity of 0.01 m/s. The elapsed time is therefore Dt = 450 m/0.70 m/s + 450 m/0.27 m/s + 100 m/0.01 m/s = 643 s + 1667 s + 10000 s = 12310 s = 205.2 m = 3.42 hr = 3^hr 25^m 10^s.

7-PE4.  Suppose a large raindrop (diameter 5000 µm) falls at its terminal velocity from the base of a cloud 1500 m (about 5000 ft) above the ground. (a) If we assume the raindrop does not evaporate, how long would it take the drop to reach the surface? (b) What would be the shape of the falling raindrop just before it reaches the ground? (c) What type of cloud would you expect this raindrop to fall from? Explain?

    (a) From Table 7.1, p.166, a 5000-µm raindrop has a terminal velocity of 9.0 m/s. Hence the elapsed time is Dt = 1500 m/9.0 m/s = 167 s = 2.8 m = 2^m 47^s.

    (b) An oblate spheroid (slightly flattened sphere).

    (c) The very large size implies that the cloud of origin was a cumulonimbus cloud or possibly a cumulus congestus.