EXERCISE 16. Adisbatic Processes Name Seetion EXERCISE 16 PROBLEMS-PART I (S.I.

Question

EXERCISE 16. Adisbatic Processes Name Seetion EXERCISE 16 PROBLEMS-PART I (S.I. Units) (as shown) Assume that a parcel of air is forced to rise up and over a 4000-meter-high mountain The initial temperature of the parcel at sea level is 30°C, and the lifting condensation level (LCL) of the parcel is 2000 meters. The DAR is 10°C/1000 m and the SAR is 6C/1000 m. Assume that no evaporation takes place as the parcel descends. LCL 30 C 1. Calculate the temperature of the parcel at the following elevations as it rises up ward side of the mountain. (a) 1000 m°C (b) 2000 m C (c) 4000 m eC 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? oC (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)? 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? 3. (a) (b) Why? 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? 95 Copyright O 2017 Pearson Education, Inc.

Explanation / Answer

1)-

a) at 1000 m

DAR= 10c/1000m

temperature at 1000m = 30c - 1000m x 10c/1000m= 20c

b) at 2000 m

DAR= 10c/1000m

temperature at 1000m = 30c - 2000m x 10c/1000m= 10c

c) at 3000 m

SAR= 6c/1000m

temperature at 3000m = 10c - 1000m x 6c/1000m= 4c

d) at 4000 m

SAR= 6c/1000m

temperature at 4000m = 4c - 1000m x 6c/1000m= -2c

2)-

a)-

at leeward side air will go to the adiabatic compression and it has no moisture in it so tempearture increases at rate of DAR = 10c/1000m

so temperature at ground = -2c + 4000m x 10c/1000m = 38c

b)- because air rising at windward side has plenty of moisture with it and air which is drowning leeward side has no moisture in it

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