Figure 100 — Flow net for symmetrical 



flow past a circular cylinder. 



(Copied from Reference 8.) 



qQ = 2U sin d. 



[671] 



The flow net is shown in Figure 100; the points A and B represent the stagnation lines. The 

 streamlines for t/r = follow the plane of the a;-axis to the stagnation line on the cylinder at 

 = 0, divide and proceed around both sides of the cylinder to the other stagnation line, then 

 continue toward negative x. 



If the motion is steady, the Bernoulli equation for the pressure p, assumed zero at in- 

 finity, gives on the cylinder itself 



V = %P{V^- ?2)=i4pCy2(i_4sin2 0), 



[67m] 



Thus p = at (9 = sin"! (±»^) or (9 = ± 30° or ± 150°. At 5 = 0° or 180°, p = p V^/^; at 9 = 90° 

 or 270°, where q = 2\U\ and is a maximum, p = -3p lJ'^/2 and is a minimum. Because of the 

 symmetry, there is no net force on the cylinder. 

 On the a;-axis, where = or 77, 



p = %pU^ 2 ; 



\ 0.2 ^4 / 



[67n] 



on the y-axis, where 6 = 90° or 270°, 



\ y2 ^1 



[67o] 



These formulas for p are plotted in Figure 101. The lower curve shows p along the a!-axis 

 and on the surface of the cylinder, plotted against a;; the upper curve shows p at points on 

 the y-axis outside of the cylinder, plotted horizontally with negative values to the left. 



150 



