592 SMITHSONIAN MISCELLANEOUS COLLECTIONS VOL. 5 1 



If this is (Hie to precipitation of aqueous vapor then it corresponds 

 to the con lensation of 4 kilograms of water per square meter or to 

 pth of rainfall of 4 mm over the whole area. 



If instead of the fictitious gas in chamber 2 we had assumed 

 saturated moist air then we could have found the approximate 

 quantity of water condensed and falling away from it as follows: 



By the overturning the lowest layer of 2 changes from T 02 = 303° 

 and p n2 = 76o mm to T' i2 = 297 and p\ = 63o mm . In this lowest 

 layer there is initially 0.02696 kg. vapor associated with each kilo- 

 gram of dry air but in the final stage this is reduced to 0.02271; 

 therefore in this layer there condenses 0.00425 kg. water out of 

 every 1.027 kg. of mixed air and saturated vapor. The highest 

 layer of mass 2 does not expand and contributes no water; we may 

 assume that the whole mass 2 contributes on the average 0.0021 

 kg. water per kg. of its own mass so that the unit column gives up 



260 



— X 10333.0 X 0.0021 = 7.4kg. of water 



760 



Because of the spread of the mass 2 over the whole trough this 

 water is distributed over double the original base giving 3.7 mm 

 depth rainfall in close agreement with the value above given. 



$(39) Second analysis. Approximate method for the computation 

 of the available kinetic energy when the chamber 1 is filled with dry 

 air and chamber 2 with the fictitious gas. 



In this case all the considerations that were made in the analogous 

 second analysis, §25, chapter III, are to be repeated excepting only 

 that X is used instead of « in chamber 2. 



Initially the masses 1 and 2 are each in stable equilibrium within 

 itself; the succession of layers within each mass remains unchanged 

 in the overturning. Hence we obtain p and p and, assuming 

 equal volumes for the two chambers, we find the following approxi- 

 mate values: 



/ K Po- ~ Pi 



r = t ( 1 -- p2 ~ Ph N 

 2 \ 2 ■ p 2 



