626 Scientific Proceedings, Royal Dublin Society. 



VIII. — Statement of Eesults. 



From the figures given in the pievious section it is possible to calculate 

 the rate of solution of the gases dealt with, for any conditions of area 

 exposed, depth, or degree of saturation, provided that the water is kept 

 uniformly mixed. 



The expression can be put either in the form 



dw , 



di = ° - hw 



which gives the rate of solution at any instant, or in the form w - (w - u\) 

 (1 - e~ bt ), which gives the amount dissolved at the end of any given time 

 when iv = saturation value and w 1 = amount of gas in solution initially. 

 For practical purposes it is most convenient to work in percentages of 

 saturation; hence the latter equation becomes w = (100 = w x ) (1 - e~ bt ). 

 and since 



1 = 4 



by substitution 



w = (100 - Wl )(l - e-Jv 1 ) 



as the general equation for any given temperature, and since / varies with 

 temperature according to the equations 



Oxygen / = '0096 (T - 237) 



Nitrogen / = -0103 (T - 240) 



Air / = -0099 (T - 239), 



the corresponding general equation for each gas by substituting these 

 expressions in the formulae is obtained, thus : — 



As an example of the use of these formulae, consider the question of the 

 dissolved oxygen in 1000 c.cs. water, area exposed being 100 sq. cms., 

 temp. 2 - 5° O, and initial gas content = 40 per cent, of saturation. How 

 much gas will be dissolved in an hour ? 

 t = 60 minutes, 



w = 60 (l - eyfir ') = 60 (1 - e- 22z > =.60 (1 - 1991) = 60 x -8009 



= 48 per cent, of saturation. 

 Hence after an hour the water will have risen to 88 per cent, of saturation. 



