﻿MO Mr. W. Ellis Williams on the 



by (2) and change to polar coordinates, the equations 

 reduce to 



D ( D -, 1 |>=° • ^ 



where D is the operator 



0?- r-siii0o0\ d#/ 



for steady motion the equation reduces to 



D 2 i|r = (4) 



The boundary conditions require that the velocity should 

 be zero at the surface of the outer sphere and equal to 

 V at the surface of the inner sphere. If a and h be the 

 inner and outer radii respectively, these give : — 



At r = a, 



^|=Va 2 sin0cos0, ^=Vasin 2 . . (5) 



Let us assume ^=(^7 + Br + Cr 2 + DrA sin 2 ; th 



value 



satisfies (4) for all values of the constants A, B, C, D, and 

 on substitution in (5) we get four equations to determine A, 

 B, C, and D. In order to get a solution to compare with the 

 case of a sphere in a cubical vessel given in fig. 3. we put 

 <x = l, 6 = 5*7, making the diameter of the sphere the same 

 as the edge of the cube, and we then obtain A= —'413, 

 B = 1-237,0= -'325, D = -00339, and the stream function is 

 given by 



^= (_— + l-237r--325r 2 +-00339/ W sin 2 0. (6) 



The stream-lines given by this equation have been plotted 

 out. and are shown in fig. 9 and it will be seen that they 

 agree very closely with the experimentally determined lines 

 of fig. 3. A test of the agreement is given by the position 

 of the point of zero velocity. This may be obtained from 



the formula by putting -^ =0, and we thus get r=5'C>, 



while the value measured on the diagram is r = 5*2. Also 

 if the diagram be compared with that of fig. 2 which 

 represents motion in the rectangular trough it will be seen 



