THE DIAMETER OF PIPES. 197 



3. Solve Ex. 1 by using the value of v as obtained 

 from (8). 



From (8), we have 



170 



v = 8.025\/ 



V 1 . .,5780 



.505 +f 



.5 



Since v is somewhere between 3 and 10, we assume / = 

 .02, and obtain 



v = 8.025\ / 

 V i.i 



170 



.505 -f 231.20 

 = 6.859. 



But v = 6.9 gives more correctly (Art. 105, Sch.) / = 

 .021, and therefore we have 



v = 8.025 A/ 

 V *' 



170 



.505 + 244.265 

 = 6.695, 

 which gives the true value to the first decimal place. 



The discharge, from (4), is 



Q = 2936.86 x 6 2 x 6.7 



= 708370.632 gallons in 24 hours. 



This result is somewhat larger than that obtained from 

 the value of v in (8) of Art. 104. 



107. The Diameter of Pipes. Substituting in (1) 

 of Art. 106 the value of v given in (9) of Art. 105, we have 



