HYDRAULICS. 99 



vide 13.88 by the diameter of the pipe in inches, and mul- 

 tiply the quotient by the length of the pipe in feet, and 

 the result will be the divisor aforesaid. Divide the first 

 product by this sum, and the square root of the quotient 

 will be the velocity in feet per second of the current in the 

 pipe. 



EXAMPLE. What is the velocity of water in a pipe 5 

 inches diameter and 100 feet long, and under a head of 2 

 feet? 



SOLUTION. 13. 88 -r- 5 =2. 776 x 100=277.6 and 2500x2 

 =5000; then, 5000-7-277.6=18, the sq. root of which is 

 4.24 feet. Ans. 



To find the quantity of water discharged through pipes. 



RULE. Multiply the velocity of the current per second 

 in feet by the area of the transverse section of the pipe in 

 feet, and the product will be the quantity discharged in 

 cubic feet per second. 



EXAMPLE. What quantity of water will a pipe 6 inches 

 diameter and 100 feet long discharge per hour under a 

 head of 2 feet ? 



SOLUTION. By the preceding rule, find the velocity of 

 the current in the pipe, thus : 2500x2 feet, head, =5000, 

 13.88H-6 inches, the diameter of the pipe, = 2.313 x 100 

 feet, length of the pipe, = 231.3, divisor; 50004-231.3= 

 24.34, the square root of which is 4.80 feet, velocity per 

 second. Then, 4.80 x .1963 square feet, area of pipe,= 



