2.6 Water Vapor in the Airbr /br /2.6.1 Moisture Parametersbr /br /Mixing ratio: w = m_v/m_d – typically g_H2O/kg_airbr /br /Example 2.7: What is the partial pressure of water vapor, with mixing ratio 5.5g/kg with total pressure 1026.8mb?br /br /e = (m_v/M_w)*p/(m_d/M_d + m_v/M_w) = w/(w+eps) *p = 9mbbr /br /Example 2.8: Calculate the virtual temperature at T=30^oC with w=20g/kgbr /br /T_v = T(eps+w)/(eps+w*eps) = T(1+ 0.61w) = 33.69^oCbr /br /Saturation vapor pressure – equilibrium state between evaporation and condensation if there is enough moisture. The partial pressure of the moisture in this situation is the saturation vapor pressure – depends only on temperature – increases rapidly with increasing tempbr / br /Saturation mixing ratio – the associated mixing ratio:br /br /w_s = rho_vs/rho_d = e_s/(R_vT)/[(p-e_s)/(R_dT)] = 0.622 e_s/(p-e_s) ~0.622 e_s/pbr /br /Dew point – the temperature that a sample of air must be cooled to at constant pressure to arrive at saturation.br /br /Example 2.9br /For air at 1000mb, 18^oC and w = 6g/kg, what is the relative humidity and dew point? Using Pseudo-adiabatic charts – can locate saturation mixing ratio for this point – w_s = 12.9g/kg – so RH = 46.5%. The dew point can be located by finding the temperature at this pressure for which 6g/kg is the saturation value – 6.4^oCbr /br /Lifting condensation level – the level to which a parcel must be lifted adiabatically so that the pressure and temperature drop to the point where the vapor in the parcel is at saturation.br /br /2.6.2 Saturated- and Pseudo-adiabatic processesbr /When lifted, a parcel’s temperature decreases at the dry lapse rate until condensation occurs – at the LCL. Further lifting results in condensation, releasing latent heat and reducing the rate of cooling. If all the condensates are carried by the parcel, the process is still adiabatic and reversible: saturated adiabatic. If the condensates fall out immediately, then the process is irreversible and pseudo-adiabatic. The amount fo heat carried by the condensates is usually small and the pseudo-adiabatic and saturated adiabatic lapse rates are essentially the same.br /br /2.6.3 The Saturated Adiabatic Lapse Rate = consider a lifted parcel undergoing condensation in a saturated-adiabatic manner:br /br /dq = cpdT + gdzbr /br /the heat released by condensation is dq = -Ldw_sbr /br /thus, br /br /dT/dz = -(L/c_p) dw_s/dz – g/c_p = -(L/c_p)(dw_s/dT)(dT/dz) – g/c_pbr /br /so, br /br /Gamma_s == -dT/dz = Gamma_d/[1+(L/c_p)(dw_s/dT)]br /br /This depends upon temperature and pressure and is always less than Gamma_d as the second term in the denominator is always positive.br /br /Example 2.10br /A parcel is intially @ 15^oC, 1000mb and T_dew = 2^oC. What is it’s LCL, and the temperature at that level? If it is lifted a further 200mb, what is its final temperature and how much moisture is condensed out?br /br /Again, this is solved using a pseudo-adiabatic chart. The initial point is located, and the amount of moisture is located from the dew point ~4.4g/kg The intersection of the saturation line from the dew point and the dry adiabat from the actual point determines the LCL at 830mb As we lift further, we follow a saturated adiabat to 630mb, where the temperature of the parcel is -0.5^oC and the saturation mixing ratio is 1.8g/kg – thus 2.6 g/kg of moisture must have condensed during the ascent.br /br /2.6.4 Equivalent potential temperaturebr /br /theta_e = theta exp(Lw_s/c_pT)br /br /the potential temperature of the parcel if all the vapour within is condensed and all the latent heat released go to thermal energy of the dry air. This quantity is conserved in both dry- and saturated- adiabatic motions. br /br /2.6.5 Normand’s rulebr /On a pseudo-adiabatic chart, the LCL of an air parcel is found at the intersection of the potential temperature line which passes through the locus of the parcel and the the pseudo-adiabat that passes through the wet-bulb temperature of the parcel at the same pressure. br /br /2.6.6 Effects of irreversible condensation processesbr /If any of the products of condensation are allowed to fall out as precipitation, the latent heat gained by the parcel will remain when the now drier parcel returns to its original level. The Net effect is increase in T and theta, reduction in w but no change in theta_ebr /br /Example 2.11 – a parcel at 950mb has a temperature of 14^oC and a mixing ratio of 8g/kg. What is the wet-bulb temperature? The air parcel is lifted to 700mb and 70% of the condensate is lost before it returns to 950mb. what is the new temperature, potential temp, wet bulb temp, mixing ratio.br /br /Again, this is solved using the chart.