198 EXAMPLE. 



. . 



- L~ ~~r~ ~J ' 



Or, by logarithms, 



logrf = [2.2450532 + 21og()-{-log(Z + 4.5rf) log A], (2) 

 where d is in inches, and all the other terms are in feet. 



When the pipes are very long, or when d is small as com- 

 pared with I, (2) becomes 



log d = I (2.2450532 + 2 log Q + log I - log h). (3) 



KEM. The value of d can be obtained from (1) only by 

 successive approximations. When considerable accuracy is 

 required, find the value of d from (3), and substitute it in 

 (2), which will give a first approximate value of d ; and 

 this again substituted in (2) will give "a closer approximate 

 value; and so on to any required degree of accuracy. Gen- 

 erally the first approximate value will be found sufficiently 

 accurate for all practical purposes. 



EXAMPLE. 



What is the diameter of a pipe which shall deliver 25000 

 gallons of water per hour, when the length of the pipe is 

 2500 feet, and the head of water 225 feet ? 



Here h = 225 and I = 2500, and the number of cubic 

 feet delivered per second is 



25000 



Substituting in (3), we have 

 log d - (2.2450532 + 2 log 1.1145 + log 2500 - log 225) 



= (1.9870309 + 3.3979400) = .67699; 

 :. d = 4.7533 inches. 



