From Section 37 it is seen that these formulas represent a line dipole of moment n, 

 located at (c,0) and having its axis directed at an angle a with the positive a;-axis, together 

 with another of equal moment located at (-c,0) and having its axis inclined at a clockwise 

 angle a from the negative a;-axis. To obtain the second term as here written from [37r], re- 

 place a by n-di&nd note that e"^ = -1. 



The particle velocity is easily found by adding vectorially the velocities due to the 

 two dipoles. 



Along the y-axis, where 6L = r - 0p and r^ - r^, i// = 0. Hence a rigid wall may be in- 

 serted along the y-axis; then either half of the field represents the flow due to a line dipole 

 in the presence of a parallel rigid plane boundary. The other dipole may be regarded as an 

 image of the first in the wall. 



(For notation and method; see Section 34; Reference 2, Section 8.42.) 



51. LINE SOURCE AND CYLINDRICAL BARRIER 



The problem of a uniform line source parallel to a cylindrical barrier of circular cross- 

 section is easily soluble by the method of images. 



Let the source be at P, distant L from the axis of the cylinder, whose radius is a; see 

 Figure 78. Add, outside the cylinder, the flow that would be due to an equal and parallel line 

 source on the inverse line Q in the cylinder, which lies in the plane containing the axis of the 

 cylinder but at a distance h^ = a'^/h^ from the axis. Add also the flow due to a line sink of 

 equal strength located on the axis itself. 



Figure 78 — A line source at P 

 outside a circular cylinder. 



The resultant stream function is from Equation [40c] in Section 40, 

 (x,y) ^- A(d^ + e^-6) 



[51a] 



where -4 is a real constant and 6., d^i 6 are variable angles defined as shown in Figure 78. 

 At any point S on the cylinder the triangles OSQ and OSP are similar; hence angle OSQ equals 

 angle OPS or n-dy, whence d^ = d + n - 6^. Thus at points on the cylinder dy + $2 - d = it 

 and t// = -An and is constant. The streamline for </> = -An thus proceeds from P to the nearest 



117 



