air stability problem

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

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Air stability problem. Surface air = 35 o C temp @ 4000m = -10 o C Dewpoint = +10 o C. 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)?. - PowerPoint PPT Presentation

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Page 1: Air stability problem

Air stability problem

Surface air = 35oCtemp @ 4000m = -10oC

Dewpoint = +10oC

Page 2: Air stability problem

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 (6.5oC and

10oC / 1000m)?

Page 3: Air stability problem

Why is PAR different from DAR

• PAR pseudo-adiabatic lapse rate (wet air, 6.5oC/100m)

• DAR = dry adiabatic lapse rate (10oC/1000m)

• They are different because the cooling rates are different for dry and wet air. The wet air condenses and releases heat.

Page 4: Air stability problem

Figure the Environmental Lapse Rate

• ELR =change in temp / change in altitude * 1000

• (35oC - -10oC) / 4000m *1000 =45oC / 4000m =11.25oC/1000m

• ELR > 10oC/1000m unstable, so this is unstable

Page 5: Air stability problem

Temp of air parcel air at 3500m

• Figure Lifting condensation level (LCL):(Temp at surface – dewpoint) / 10oC * 1000 m(35oC - 10oC) / 10oC * 1000m = 2500meters

• Temp at 2500m is 10oC, the same as the dewpoint

• We know the wet adiabatic lapse rate = 6.5oC/1000m

• 2500m + 1000m = 3500m10oC (dewpoint at 2500m) - 6.5oC = 3.5o

Page 6: Air stability problem

Temp of environmental air at 3500m • We know the environmental lapse rate = 11.25oC/1000m (calculated earlier)

• At 4000m, we’re given the temp = -10oC4000m – 3500m = 500m (i.e. ½ of 1000m)-10oC + ½(11.25oC/1000m) -10oC + 5.625oC= -4.375oC

• 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 = 35oC3500m = 3.5 * 1000m35oC –11.25oC – 11.25oC – 11.25oC – 5.625oC = -4.375oC