DPT and Uplift of an Air Mass 16. Our ground level instruments are checked at 10
ID: 114044 • Letter: D
Question
DPT and Uplift of an Air Mass 16. Our ground level instruments are checked at 10:00 am under clear sky conditions. They indcate the following: T 35 C and SH12 gkg. A Thermal Low Pressure system creates uplift of the air mass. Assuming SH remains stable, at what altitude will the air mass reach Lifting Condensation Level (LCL) and Cloud Formation ill occur? Apply the Dry Adiabatic Rate (DAR) to calculate your 17. Consider the same air mass from question 16: the velocity of uplift is measured at 300 meters per hour. At what time can we expect cloud formation? 18,Explanation / Answer
16)Answer: Given SH=12g/kg, mixing ratio= 0.012 =q, say
Relative Humidity RH = 0.263pq [exp(17.67(T-To)/T-19.65)-1 ] where p=101325Pa, To = 273.16 K, T= 35°c = 308.16K
then, RH = 34.7
We know, T-Td = (100 -RH)/5 where Td dew-point tempurature., therefore, T-Td = 13.06
As we know, HLCL = (T-Td)/ R - Rd where R=adiabatic lapse rate = 9.8 °c/Km and Rd= dew point lapse rate= 1.8°c/km
Hence, HLCL = 13.06/8 = 1.6325 km (ans)
17) Answer: velocity = 300 meters per hour. then 1.6325 km= 1632.5 m height will need time= 1632.5/300 = 5.442 Hr
18) dry adiabatic rate = 9.8°c/km, then at LCL, temperature will be = 35-(1.6325*9.8)= 19.0015 °c
After LCL, lapse rate will be moist adiabatic lapse rate = 5.5°c/km, therefore, decrease to T=0°c,
the increase height = 19.0015/5.5 = 3.4548 km,
hence, past LCL, T=0°c at the altitude of 1.6325 + 3.4548 = 5.0873 km (Ans)
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