82 AGRICULTURAL ENGINEERING 



Assume that an 8-inch tile will be required, then: 



d — diameter of tile, = 8 inches or % foot. 



1000 



h = total head or fall = .4 X = 4 ft. 



100 



I = 1000 feet. 



54d = 54 X % = 36. 



o = area of cross section of tile = M X 3.1416 X (H) 2 =.349 

 square feet. 



V = 48 i/jiiLi = 48 V / -^- = 48 V .002574 

 ^ 1000 + 36 ^ 3108 



= 2.40 feet per second. 

 Q = cubic feet discharged per second, and equals the velocity X 

 area of cross section of tile = 2.4 X -349 = .8376. 



Referring to the preceding table for the discharge per 

 second per acre for the \i-mch. standard, we find .0105. 

 Then the number of acres drained is 



A = = 79.6, or practically 80. 



.0105 



If the answer representing the discharge per second pro- 

 cured in this manner should be too great or too small, the 

 calculation would be made for smaller or larger sizes of tile, 

 as the case may be, and the most practical tile to use chosen. 



To facilitate the use of the formula, a table may be made 

 up from it, showing the number of acres which may be drained 

 with various sizes of tile laid to various grades. The follow- 

 ing is such a table, from Bulletin 68 of the Iowa experiment 

 station. 



