274 



against llie stream is iinpossihle. we Hiij)|)ose with regard to the fact 

 that the ai-riving gas is free from vapour that c = for ?/ = 0. For 

 the same reason we may further assume that the surface of the 

 liquid extends from // = to // = ao , while for arbitrary y the 

 concentration will not be influenced by the presence of liquid at 

 the boundary z = at greater values of //. As was already said we 

 take foi' c =3 the boundary -c(mdition c = C, while for z =: cc c 

 of coui-se must be zero. 



Problems of this kind may be solved in a general way by making 

 the range in which z may vary finite, further by constructing a 

 solution with (he aid of a series of proper functions, and finally by 

 going to the limit, whereby the range is made infinite. I hope to 

 explain this method at length in my dissertation ; here however it 

 may suffice to give a much simpler treatment, because the purpose 

 is only to find how the quantity of liquid evaporated in unit time 

 depends upon the length of the rectangle, i. e. upon //. 



When we introduce in (III) as a new variable: 



a 

 this equation assumes the form : 



H' = %- ■■■■■■ <"^<" 



The boundary conditions of c are heie : 



c — for .V = 

 c= C „ C = 



C =r ,, S = CO 



The solution of the transformed equation will not contain a or 

 D, because these quantities occur neither in tha differential equation 

 nor in the boundary conditions. 



Therefore is : 



c = i/ {^, ;/) = (/ I z)^^-^, y 



The quantity of the liquid that evaporates in unit time from the 

 part of the surface between // = and y is found by computing 

 the quantity of substance that flows through a plane perpendicular 

 to the axis of y. As the velocity of the gas is az, the quantity of 

 vapour that flows in unit time through a unit surface perpendicular 

 to the axis of //, is azc ■ so the total mass of vapour that flows 

 away per unit breadth in the direction of a*, amounts to: 



