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## PROBLEM 1

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**PROBLEM 1**A) The temperature of an air mass at 950 mb is 23.5 ºC. Its relative humidity is 50%. Calculate its dew point temperature and the lifting condensation level. B) Consider an air mass at the same temperature than that on the former paragraph, whose pressure is 850 mb instead of 950 mb, and whose relative humidity is 10% instead of 50%. ¿Which is its dew point temperature? ¿How many g/kg vapor does it contain? PROBLEM 2 An air parcel at 1000 mb and 7 ºC having a wet bulb temperature of 3 ºC raises up to the 625 mb level. During this lifting condensation and partial elimination of liquid water occurs. Later, the air parcel goes down again up to the 1000 mb level. It water loss can be estimated about 1.5 g per kg of dry air. Represente el proceso en un diagrama pseudoadiabático y rellene la tabla adjunta, expresando las unidades adecuadas en cada caso. ¿Where is the lifting condesation level? Find out the final dew point temperature and the final relative humidity.**LCL: 810 mb**13ºC 23.5ºC -9ºC PROBLEM 1 10% 850 = 2.25 gkg-1 850 mb: s = 22.5 gkg-1 50% 950 = 10 gkg-1 950 mb: s = 20 gkg-1 950 = 10 gkg-1 Compare: 850 = 2.25 gkg-1 10-2.25 gkg-1= 7.75 gkg-1**625 mb**-3.5 ºC 7 ºC -13 ºC 12 ºC 3 ºC PROBLEM 2 Sat mixing ratio: 1.0 gkg-1 (1 g vapor + 2 g líquid) Initial sat mixing ratio: 6.5 gkg-1 Initial mixing ratio: 3.0 gkg-1 Sat mixing ratio: 1.5 gkg-1 Initial relative humidity: 3.0/6.5=46% Initial dew point temperature: -3.5 ºC Lifting condensation level: 850 mb Final dew point temperature: -13 ºC Final temperature: 12 ºC Final relative humidity: 1.5/9=17%