1. A mountain has an elevation of 19,000 feet. The air at 6,000 feet on the wind
ID: 292280 • Letter: 1
Question
1. A mountain has an elevation of 19,000 feet. The air at 6,000 feet on the windward side has a temperature of 70° in early spring, and condensation takes place at 10,000 feet. Given: Dry adiabatic rate is per 1,000 feet. Wet adiabatic is 3° per 1,000 feet from 10,000 feet to the summit of the mountain. With this information, answer the following questions. a) What is the temperature of the air at 15,000 feet on the windward side? 35 °F b) What is the temperature of the air at the summit of the mountain? c) what is the temperature at the foot of the mountain on the windward side? I??" d) What is the temperature of the air at 15,000 feet on the leeward side? 43 F e) What is the temperature at the bottom of the mountain on the leeward side? 118 F f) Account for the differences in temperatures between the windward and leeward sides? Windward aide moisture and leeward side is 2. A mountain has an elevation of 15,000 feet. The air at 3,000 feet on the windward side has a temperature of 63° in early summer, and condensation takes place at 5,000 feet. Given: Dry adiabatic rate is 5.0° per 1,000 feet. Wet adiabatic is 3.00 per 1,000 feet from 5,000 feet to the summit of the mountain with this information, answer the following questions. a. What is the dew point temperature at 8,000 feet above sea level? b. What is the temperature of the air at the summit of the mountain? c. What is the temperature of the air at 8,000 feet on the leeward side? d. What is the temperature difference between the windward and leeward side of the mountain at 8000 feet above sea level?Explanation / Answer
Given dry adiabatic rate is 5 degree per 1000 ft
wet adiabatic rate is 3 degree per 1000 ft
Condensation takes place at 10000 ft
height of mountain is 19000 ft
A. The temp of air at 12000 ft
When the air parcel rise from 6000 ft at a temp of 70 degree, it loses temp at a dry adiabatic rate till 10000 ft
Thus at 10000 ft temp of air parcel = 70 - (10000-6000)*5/1000= 50 degree
After 10000 ft loses heat at a wet adiabatic rate
thus at 12000 ft temp of air parcel = 50 - (12000-10000)*3/1000= 44 degree F
b. temp at the summit of mountain
height of mountain is 19000 ft
we know that at 10000 ft temp of air parcel
air parcel gets condensed and rise at a wet adiabatic rate
Thus temp of air parcel at 19000 ft = 50 - [(19000 - 10000)*3/1000] = 23 degree F
c. temperature at the foot of the mountain
Given at 6000 ft temprature is 70 degree F
thus at windward side the temperature of parcel increases at a dry adiabatic rate
= 70 + (6000 - 0 ) * 5/1000 = 100 degree F
d. Temp of air at 15000 ft on the leeward side.
Now we have calculate the temp at summit of mountain = 23 degree F
When the air parcel reaches the mountain summit it descend at the dry adiabatic rate
Thus temprature at 15000 ft = [23 + (19000-15000)*5/1000] = 43 degree F
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