ADIABATIC CHANGES OF MOIST AIR NEUHOFF 441 



indicates the temperature of saturation. This value of 5 can now 

 be obtained directly from a table such as table 3 that contains the 

 appropriate values of 5 computed according to this last equation 

 (6) for all the temperatures likely to come into consideration, i. e., 

 for each degree from + 30 to — 30 and for the values of m I = 

 3.44 to 3.48. For any value of 5 computed from the initial values 

 t and e for which also m l log T can be taken out of table 3, we seek 

 in the table of corresponding temperatures that one which represents 

 the temperature of saturation and which can be obtained by 

 interpolation to the nearest tenth of a degree. 



For example, let us now consider a mass of air expanding adia- 

 batically from the initial pressure p = 760 mm and the initial tem- 

 perature t = 20 C. and the relative humidity 86 per cent. Accord- 

 ing to table 1 for t = 20 the vapor pressure for saturated vapor is 

 e = i7.4 mm and the quantity of moisture at saturation is x m = 14.6 

 grams, hence for relative humidity 86 per cent the vapor pressure 

 will be e a = i5.o mm . The mixing ratio or the weight of aqueous 

 vapor present in (1 + x) kilograms of moist air is obtained from 

 the expression 



x = 622 — ^~- 



P ~ e 



and is 12.5 grams. We may approximately write x/x m = e /e m , 

 whence also follows for x the value 12.5 grams. According to 

 table 4, column 4, there corresponds to this mixing ratio the 

 humidity factor for the dry stage m l = 3.46. If this mass of air 

 expands adiabatically to the point of saturation then for the de- 

 termination of the saturation temperature according to table 3, 

 column 5, we form the product m 1 log T = 8.5355, whence for 

 e = i5.o mm there results 5 = 7.3594. With this value of 5 we 

 enter table 3 under m x = 3.46 in column 11 and find the corre- 

 sponding temperature of saturation^ = 17.0 C. The correspond- 

 ing vapor pressure is e m = i4.4 mm . 



The corresponding pressure is obtained in millimeters at once 

 from the equation (4) and is p = 733. 2 mm , and with this Ave obtain 

 p' =p - e = 7i8.8 mm . If we wish to know the volume (V) of the 

 mixture, or (1 + x) kilograms of air, we shall obtain it from the 

 equation of condition. Thus from table 4, column 3, for x = 12.5 

 grams we obtain Rp = 2.196. Therefore for the initial condition 

 p = 76o mm and T = 293 we have V =0.847 cubic meters and the 

 specific volume v = y/1.0125 = 0.837 cbm. In the saturated con- 



