Excercise 14 Problems
1. Calculate the temperature of the parcel at the following elevations as it rises up the wind-ward side of the mountain:
(a) 1000m 20 °C
(b) 2000m 10 °C
(c) 4000m -2 °C
2. (a) After the parcel of air has descended down the lee side of the mountain to sea level, what is the temperature of the parcel?
(b) Why is the parcel now warmer than it was at sea level on the windward side (what is the source of the heat energy)?
3. (a) On the windward side of the mountain, is the relative humidity of the parcel increasing or decreasing as it rises from sea level to 2000 meters?
When air rises, its pressure decreases, so it expands and cools adiabatically.
4. (a) On the lee side of the mountain, is the relative humidity of the parcel increasing or decreasing as it descends from 4000 meters to sea level?
b) Why? When the temperature of a parcel of air increases, its relative humidity
As the air goes down the lee side of the mountain, the air actually rises in temperature. This warmer air increases the water vapor capacity or the air.
EXERCISE 14 PROBLEMS – PART II
5. (a) On the windward side of the mountain, should the relative humidity of the parcel change as it rises from 2000m to 4000m?
(a) No, the relative humidity should not change.
(b) The elevation at which the parcel of air reaches its dep point temperature is called the lifting condensation level. The LCL in this example was 2000m, so the relative humidity is at 100% already.
6. As the air rises up the windward side of the mountain:
(a) What is the capacity (saturation mixing ratio) of the air at 2000 meters?
(b) What is the capacity of the air at 4000 meters?
Approx 3.2 g/kg
7. What is the capacity of the air after it had descended back down to sea level on the lee side of the mountain?
Approx 4.2 g/kg
8. (a) Assuming that no water