ADIABATIC CHANGES OF MOIST AIR NEUHOFF 447 



end of the rain stage we should obtain the values a = 40.0 and 

 m n = 3.62. 



By making use of equation (9) we obtain 



log p' - ^ = 2.6008 

 Sy P > 



whence/?' = 482. 6 mm and thence p = 487-2 mm . 



The quantity of vapor present is x = 5.9 grams and the quantity 

 of water is y = 6.6 grams. The volumes are V = 1.2 18 cbm. and 

 v = 1.203 cbm. 



§5. HAIL STAGE 



If fluid water is still present in the air during the freezing stage 

 which begins when the temperature falls to o° C. then the tempera- 

 ture will remain constant and therefore also the quantities that 

 depend upon the temperature, namely, e = 4.6 mm and r = 606.50 

 - 0.695 t - 



The total moisture £ which will be considered as unchanged and 

 which consists of vapor (%), water (y) and ice (z), is therefore still 

 £ = x + y + z. 



The quantity of heat that is needed for isothermal expansion is 

 now made up of the quantity that is needed for the expansion of 

 the air added to that which is needed for the evaporation and freez- 

 ing of water. 



If r e is the latent heat of melting ice then the thermal equation is 



dV 

 dQ = ART - V + rdx - r e dz 



In the case of adiabatic expansion dQ = o and by integration there 

 results 



ART V 

 0= — ik? --log y + r (x - x ) - r e (z - z Q ) 



For the freezing period, which when it occurs is generally of very 

 short duration, we will compute only the condition of the moist 

 air at the end of this stage. 



At the beginning of this hail stage no ice is present, therefore 

 z = o and at the end of the period all the water is frozen, wherefore 

 y = o and therefore z = £ — x, showing that the initial quantity 

 of vapor (x ) has been slightly increased to a new valuer. The con- 

 nection between these quantities is best understood by the use of 

 the scheme given by von Bezold in his lectures: 



