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Air stability problem

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)?.

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Air stability problem

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

  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 (6.5oC and 10oC / 1000m)?

  3. 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.

  4. 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

  5. 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

  6. 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

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