where fi is a constant, the dipole moment. The physical axis of the dipole is directed toward 

 = if fi > 0, toward d = n U ^J. < 0. 



The fluid velocity lies in a plane through the axis of the dipole, which is an axis of 

 symmetry. Its components in the r, 6 directions are, from [118e,f, g], 



?r 



2 fi COS 6 





[119h,i,j] 



P(x,y,2) 



Figure 198 - A dipole at Q or (a;^, yy 2j), 

 with its axis parallel to the a;-axis. 



The stream function i/^, like <5^, is most 

 easily found by differentiating that for a point 

 source, but it can also be found by direct 

 integration. Through any point P draw a 

 circle as before, with the axis of the dipole 

 as its axis; let its radius be &7, and let the 

 distance of its plane from the dipole at Q 

 be h. On the plane of the circle take a 

 ring-shaped element of area centered on the 

 axis; see Figure 197 again, in which there 

 is now a dipole at Q with its axis along 

 QQ\ and again V = 0. The area of the ring 

 will be 2na) ' doJ ' where dco' is its width 

 and cj ' is the radius of either perimeter; and, 



because of the symmetry, the rate of flow across it will be ^nq.co' da' where q. is the com- 

 ponent of the velocity in the direction of the axis. The flow across the entire circle is thus 



•sr 



/ 



q-^ 2n(o 'd(o ' 



Now, by projection, if r\ 0'are the values of r and Q on the elementary ring, 



I?! = q^ cos Q' - qQ sin B' = — (2 cos^ B' - s\r\r 6') 



Zh^ 1 



-— - (3 cos20'-l)=^( 



r'3 \r'5 ;.'3/ 



See Figure 197. Thus the flow across the circle is 



301 



