360 Lord Kayleigh on the 



A simple example is afforded by the case of a source at A 

 and an equal sink at B, where 6 = n (fig. 2). The fluid enters 



Ffo. 2. 



and leaves the enclosure by two perforations situated at 

 opposite ends of a diameter, the walls being elsewhere impene- 

 trable. The solution may be found independently, or from 

 (19), by changing the sign of cos 6, and adding the equations 

 together. Thus 



+= 



(l-r 4 )rsinfl 

 l-2r 2 cos20 + ; 



-f-tan' 



2r sin 6 

 1-r 2 " 



(20) 



In this case the walls of the enclosure are of necessity 

 stream-lines, the value of ^ being + 1 from to tt, and — 1 

 from to — 7r. 



When e = \ir } that is along OD (fig. 2), 



l7T.lfr = 



1 + 



+ 2 tan 



IT 



dyfr _ (S + i*)(l-i*) 



dr 



(1 + r 2 ) 2 



(21) 



(22) 



From (21) we obtain by interpolation the following 

 corresponding values : — 



ilr 



•00 

 •00 



•25 

 •1330 



•50 

 •2800 



In the neighbourhood of A or B, fig. 

 special form. Thus in the former case, 



'75 1-00 



•4698 1-0000 

 1, (20) assumes a 



l-2^cos2^ + ^=(l-r 2 ) 2 + 4r 2 sin 2 <9 = 4{AM 2 + PM 2 }, 

 (l-?<>sin0 = 4PM.AM, 



tan _1 — — = angle PAO. 



1 — r* 



