8. An air parcel must traverse a 7,000 ft mountain barrier in traveling from Cit
ID: 294351 • Letter: 8
Question
8. An air parcel must traverse a 7,000 ft mountain barrier in traveling from City P to City Q, both of which are at sea level (O feet). The temperature of the parcel at Cty P is 80o F and the dew point (condensation begins at this temperature), is 58° F. Answer the following questions and show the steps in your calculation, (Based on "Adiabatic Processes", Ch.5 in your textbook). (Total 12 pts) 7,000 ft 58° F dew point 80°F (sea level, Oft) P 8a. Initially, the air parcel is not saturated, so it will cool at the dry adiabatic lapse rate (DAR) of 5.50F per 1,000 feet. At what altitude (i.e. Lifting Condensation Level, LCL), will it reach dew point? (3 pts) After it reaches dew point, the rising air parcel will cool at Moist/ Wet Adiabatic Lapse Rate (MAR). (Assume that MAR-3.3°F/1,000 feet). What will be the air temperature at the top of the mountain at 7,000 feet? (3 pts) 8b. City Q is 7,000 feet lower than the top of the mountain. Descending air heats at the Dry Adiabatic Rate of 5.5°F/1,000 feet. What will be the temperature of the air parcel when it descends from the top of the mountain and arrives at City Q (sea level, o ft)? (3 pts) 8c. 8d. As the air parcel reaches city Q(sea level, 0 feet), is its relative humidity higher or lower than whenit was at the summit of the mountain at 10,000 ft? Explain why. (3pts)Explanation / Answer
8a. The initial temperature is 80oF; dew point temperature is 58oF; dry adiabatic lapse rate = 5.5oF per 1000 ft.
The temperature reduction required to reach dew point temperature = 80 - 58 = 22oF
Therefore, LCL = 22*1000/5.5 = 4000 ft
8b. Wet adiabatic lapse rate = 3.3oF per 1000 ft.
After reaching LCL at 4000 ft, the altitude required to reach top of the mountain = 7000 - 4000 = 3000 ft
Therefore, temperature at the top of the mountain = 58 - (3.3*3000/1000) = 48.1oF
8c. The increase in temperature when the air parcel descends 7000 ft at DAR of 5.5oF per 1000 ft is,
= 5.5*7000/1000 = 38.5oF
Therefore, the temperature of the air parcel at city Q, = 48.1 + 38.5 = 86.6oF
8d. The relative humidity (RH) will be lower when it reaches the city Q. The RH reaches the maximum of 100% at LCL when the temperature of the air parcel reaches the dew point temperature. Beyond this, as the parcel rises the RH remains at that level. Therefore, at the summit of the mountain, the RH will be 100%. As the air parcel starts descending, it starts getting warmer and hence the capacity to hold water increases. Therefore, as the air parcel sinks, the RH decreases and when it reaches the city Q which is at the sea level, the RHS will be lower.
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