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Air stability problem Surface air = 35 o C 4000m = -10 o C Dewpoint = +10 o C.

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Presentation on theme: "Air stability problem Surface air = 35 o C 4000m = -10 o C Dewpoint = +10 o C."— Presentation transcript:

1 Air stability problem Surface air = 35 o C temp @ 4000m = -10 o C Dewpoint = +10 o C

2 questions Is the air stable? What’s the temp of the environmental air at 3500m? What is the temp of the lifted air at 3500m? Why are the DAR and PAR different (6.5 o C and 10 o C / 1000m)?

3 Why is PAR different from DAR PAR pseudo-adiabatic lapse rate (wet air, 6.5 o C/100m) DAR = dry adiabatic lapse rate (10 o C/1000m) They are different because the cooling rates are different for dry and wet air. The wet air condenses and releases heat.

4 Figure the Environmental Lapse Rate ELR = change in temp / change in altitude * 1000 (35 o C - -10 o C) / 4000m *1000 = 45 o C / 4000m = 11.25 o C/1000m ELR > 10 o C/1000m  unstable, so this is unstable

5 Temp of air parcel air at 3500m Figure Lifting condensation level (LCL): (Temp at surface – dewpoint) / 10 o C * 1000 m (35 o C - 10 o C) / 10 o C * 1000m = 2500meters Temp at 2500m is 10 o C, the same as the dewpoint We know the wet adiabatic lapse rate = 6.5 o C/1000m 2500m + 1000m = 3500m 10 o C (dewpoint at 2500m) - 6.5 o C = 3.5 o

6 Temp of environmental air at 3500m We know the environmental lapse rate = 11.25 o C/1000m (calculated earlier) At 4000m, we’re given the temp = -10 o C 4000m – 3500m = 500m (i.e. ½ of 1000m) -10 o C + ½(11.25 o C/1000m) -10 o C + 5.625 o C = -4.375 o C Note the environmental air is colder than the air parcel, which means the parcel is still rising and the system is unstable Alternate calculation going upwards: surface temp = 35 o C 3500m = 3.5 * 1000m 35 o C –11.25 o C – 11.25 o C – 11.25 o C – 5.625 o C = -4.375 o C


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