Essay on Excercise 14 Problems

982 Words Aug 12th, 2013 4 Pages
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,
…show more content…
The relative humidity is at 100% and starts forming condensation at 3000’ ft.

6. As the air rises up the windward side of the mountain: (a) What is the capacity (saturation missing ratio) of the rising air at 3000 feet?

11.1 g/kg

(b) What is the capacity of the air at 6000 feet?
7.6 g/kg

7. What is the capacity of the air after it has descended back down to the sea level on the lee side of the mountains?

Approx 24 g/kg

8. (a) Assuming than 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?

Higher

EXERCISE 14 PROBLEMS – PART IV

(b) Why?
The parcel starts descending at 6000’ feet, the temperature at this height is 50.1 F, and is above the original LCL. The new LCL is at 6000’ feet as the parcel is starting its descent down the lee side of the mountain. The mixing ration to start to movement up the mountain was approx 18.9 g/kg and the descent capacity is approx 7.6 g/kg. (c) What is the lifting condensation level of this parcel now?
The lifting condensation level of this parcel is now 6000’

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