# Excercise 14 Problems

Pages: 5 (978 words) Published: July 27, 2013
EXERCISE 14 PROBLEMS—PART I

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?

38 °C

(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)?

Latent heat

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?

Increasing

(b) Why?

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?

Decreasing

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? 7.6 g/kg
(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 vapor is added as the parcel descends down the lee side of the mountain to sea level, is the water vapor content (the mixing ratio) of the parcel higher or lower than before it began to rise over the mountain?

Lower
(b) Why?
The air temperature increases on the lee side of the mountain, the capacity of the air changes. This change allows the air to have a greater capacity for water vapor.
(c) What is the lifting condensation level of this parcel now? The lifting condensation level of this parcel is now the peak of the mountain at it begins its decent, 4000m. EXERCISE 14 PROBLEMS – PART III

1. Calculate the temperature of the parcel at the following elevations as it rises up the wind-ward side of the mountain: (a) 1000’ 71 F
(b) 3000’ 60 F
(c) 6000’ 50.1 F

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?

83.1 F
(b) Why is the parcel now warmer than it was at sea level on the windward side (what is the source of the heat). The parcel is now warmer because the parcel had a release of latent heat during condensation on the wind-ward side of the mountain. This allows the air to heat at a faster rate on the decent of the mountain with adiabatic warming. 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 3000 meters?

Increasing

(b) Why?

As temperature decreases, relative humidity increases until condensation begins.

4. (a) On the lee side of the mountain, is the relative humidity of the parcel increasing or decreasing as it descends from 6000 feet to sea level?

As the air warms on the descent on the lee side of the mountain, the air is warming up. This warm air has a greater capacity than cool air to hold water vapor. The relative humidity goes declines as the air gets warmer.

EXERCISE 14 PROBLEMS – PART IV

5. (a) On the windward side of the...