30 



Finally I is evaluated by equating [75] and [76]. Then 



1 .-J*- 



x o H +1 



0.V 



1 + 



- 0.0378 log 10 R fi 



R e may be eliminated by use of Equation [58] to give 



H, 



I o ~ H.+1 



1 + 



0.0378 V52.9 log in H n - 4.11 



h: 



[77] 



[78 



The small variation of I Q with Reynolds number R is shown in Figure 

 14. For instance, at R^ = 1500, I = 0.633 and at R e = 100,000, I = 0.554. 

 These values appear reasonable when compared with those based on the Fediaevsky 

 formulations: From Equation [64] for three conditions I = 0.67 (« = for 

 zero pressure gradient) and from Equation [65] for five conditions I = 0.60. 



5 6 7 8 9I0 5 



Figure 14 - Variation of Integral of Shearing Stresses across 

 Boundary Layer with Reynolds Number R a for Flat Plates 



Finally from Equations [71] and [78], ^ is written 



" vu 2 Ah - i I 



• 0378 /52.9 log ln H n - 4.18 



1 + 



[79] 



Using the value of r^ /pU z from [57], derived from the Schoenherr 

 formula, and the value of H Q from [58], the variation of # with Reynolds 

 number is shown graphically in Figure 14. It is seen that ^ o diminishes grad- 

 ually with Reynolds number. 



