Consider an air parcel at 950 hPa with temperature 25C and dewpoint temperature
ID: 159233 • Letter: C
Question
Consider an air parcel at 950 hPa with temperature 25C and dewpoint temperature of 15C.
What is the relative humidity of this air parcel?
ans : RH = (actual) mixing ratio/saturation mixing ratio
actual mixing ratio = 11 g/kg
saturation mixing ratio = 21 g/kg
RH = 11/21 or about 52%
What is the lifted condensation level (LCL) of the air parcel? 830
Lift the parcel (dry) adiabatically to its LCL and moist adiabatically above the LCL to 200 hPa. What is its temperature at this level?
ans : -61.5
Bring the parcel back down to the 950 hPa adiabatically. What is its temperature now?
ans: about 57C (goes off skew-T diagram)Why is this temperature greater than the temperature we started with? Parcel gains latent heat of condensation during ascent, that heat is realized as a temperature increased upon being compressed adiabatically back to 950 hPa.
Could someone explain how we got the answers?
Explanation / Answer
1) we'll find the actual mixing ratio(w) and saturated mxing ratio(ws) from given data formulas are
w = 621.97*e/(Psta-e) = 11
ws = 621.97*es/(Psta-es) = 21 PSTA= 950hpa
then
RH = (w/ws)*100%
= 52.3%
2)
LCL =125(T-Td) T= temperature,td= dew point temperature
= 830
3)