18 



THE IRRIGATION AGE. 



THE PRIMER OF HYDRAULICS* 



By FREDERICK A. SMITH, C. E. 



Article XIII. Problems. Open Channels. 



A. Triangular Channel. 



1. Find J'elocity When the Other Quantities Are Given. 



In Fig. 81 is shown a problem where a reservoir of water 



a 



A is connected by a timber flume C, made of very smooth 

 planed boards, to a reservoir B whose level is lower than A 

 a distance of 3 feet. The length of flume is 6,600 ft., and 

 cross section of flow is uniform throughout entire length ; 

 the surface levels in A and B are supposed to be constant. 



Find quantity of water discharging per second if the 

 depth of flow GD equals 30 inches and angle EDF 90, 

 the triangle EDF being the cross section of the flow. 



Solution. Assume coefficient of roughness .011. Find the 

 slope .? by dividing 6,600 into 3, which gives .00045. To 

 find r first find area of triangle EFD. which equals GD X EG 

 =2.5 X 2.5 = 6.25 sq. ft. The wetted perimeter equals ED 

 +DF; ED = V 12.50 = 3.535; hence r = 6.25 -=- 7.07. 



r = .8803. 



Take \/ r = V - 88 03 = - 938 - 



Consult "C" table under, n .011 (Table III) ; in column 

 j = .0004 we find C = for .9 = 133.79. 

 for 1.0 138.92. 



Hence, for a one-tenth part of -\/ r C advances 5.13, 

 or .0513 for 1/1,000 part of V ^ so f r - 038 parts the ad- 

 vance is 38 times .0513 = 1.95 ; add this to 133.79 gives 135.74 ; 

 this is the exact factor C for this case. 



The mean velocity v is now determined from the 

 formula : v = C V rs - 



Substitute above values : 

 f = 135.74 X .938 X .021. 



v = 2.67 ft. per second. 



Hence quantity Q in cub. ft. per second. 



Q = 6.25 X 2.67 = 16.687 cub. ft. 



2. Find Size of Flume if the Other Quantities Are Given. 



A variation of the problem may be worked by assuming 

 the slope as before, but require the size of flow for a re- 

 quired capacity. 



Problem : Find depth GD to deliver a flow of 50 cub. 

 ft. per second. 



As stated before, n = .011 and s = .00045. Let GD 

 =.r ; then ;r 2 = area of cross section of flow and 50 -r- .r 2 



rr: v. 



Also r = x -+- 2 -\J~Z. 



Substitute these values in v _=. C \/ rs; we get : 



50 H- x~ C V xs -=- 2 \/2 



Square both sides : 



2,500 C.rs 



x 4 2.828 

 CVi = 2.828 X 2,500. 



Assume C as above at 135 and ^ = .00045, we get : 



X 5 V 2.828 X 2,500 -=- 135 X 135 X .00045. 



.r = 5 V 7,070 -^- 8.2 = 5 V S62.2. 



The 5th root is best extracted by means of logarithms: 



Log. 86:2.2 = 2.93561. 



Divide by 5 = 0.58712. 



Look up number corresponding to this logarithm : 



x = 3.865. 



Hence the depth of flow is 3.865, which will give the flume 

 a capacity of 50 cub. ft. per second. 



To check this back assume depth as 3.865 and ascertain 

 flow: area = 3.865 X 3.865 = 14.938. 



r 3.865 -H 2.828 = 1.367. 



V = 135 \/ 1.367 X .00045. 



V = 135 X .0248. 



V = 3.348 ft. 



Multiply by area 14.938. 



Q = 50.6 cubic ft ; thus it is seen that the problem checks 

 back correctly. 



B. The Rectangular Channel. 



3. Problem. 



A flume of rectangular cross section (see Fig. 

 82) 24x24" is tapping a river at its upper end and is 



i 3mr/<>s :i m il es long, hav- 



A i s_ ing a uniform 



grade of 5 ft. per 

 mile. Find veloc- 

 ity and quantity of 

 water if factor 

 = .012 and if 

 depth of flow = 



Fig. 82. ,, , .. 



Solution. In 



this case n, r and .j are given and the problem can be worked 

 accordingly. 



Area of flow = 2 X l'/3 = 2 2 /3 = 2.67 sq. ft. 



Wetted perimeter = 2'+ l l / 3 + V/ 3 4% = 4.67 ft. 



Hydraulic radius = 2.67 -f- 4.67 = .5717. 



X/.5717 = .756. 



.? = 5^- 5,280 = .00095. 



Consult table No. IV ( = .012) and factor C will be 

 found unde slope .001 (which is nearest to .00095) and 

 opposite V r column, between .5 and .6, or the proportional 

 part for .5717; it will be near enough to compute C for .57; 



C for .50 = 94.52. 



C for .60 = 103.44. 



Difference: 9.12 



This is a variation of .91 for .01 increase in \/ r; thus 

 for .07 it is 7 times .91 = 6.37; add this to 94.52 gives 

 100.89, as C for V = -57. 



In the fundamental formula : 



v = C \l rs. Substitute values : 



V = 100.89 X -57 X V-00095. 



Consult table XIV of square roots; under x find 



V.0009 = .03._ 



Under ^.0010 you find .03162. 

 Difference: .00162. 



Hence \/ .00095 equals .03 plus half of .00162, or V- 00093 

 = .03081. 



v 100.89 X -57 X .03081. 



v = 1.7678 ft. per second. 



The quantity of water per second equals cross section 

 times mean velocity : 



Q = 2.67 X 1.77 = 4.726 cub. ft. per sec. . 



C. Trapezoidal Channel. 



4. Problem. 



A mill race of trapezoidal cross section has a bottom 

 width of 3' 4" and the depth of flow is 2 ft. ; the slope is 

 6 inches in 1,000 ft. ; the side walls of the channel have an 

 inclination of 60 to the horizon, and botttom and walls are 

 constructed of concrete plasterec 1 

 fairly smooth. Find mean veloc- 

 ity of flow. 



Solution. Let Fig. 83 repre- 

 sent the cross section of channel, 

 and AE be depth of flow = 2 ft. ; 

 then AC = BD = 2 -=- sin 60, 

 and CE = FD = 2 -=- cos 60; 

 this makes AC = 2.309 ft. and 



Fig. 83. 



"Copyright by D. H. Anderson. 



CE = 1.1545 ft. The area of flow and length of wetted 

 perimeter can now be readily determined : Area of flow : 

 (AB -f CD) -f- 2 X AE (3.33 + 1.13 + 3.33 + 1.13) -H 

 2X2 = 8.91 sq. ft. ; the wetted perimeter = CA -f- AB + 

 BD = 2.309 + 3.333 + 2.309 = 7.951 ft. 



Hence hydraulic radius = 8.91 : 7.951 = 1.1205. 



And V 1.1205 = 1.059. 



If we now assume the coefficient n = .(n3 we have all 

 the necessary elements to make use of the tables s = 5 -=- 

 1,000 = .0005: table No. V ( = .013), under s.lope .0004 

 (which_Js nearest to our case), we find C opposite \/ r; for 



\/7 = 1.0 C = 115.29 ; for 



V = 1-2 C = 123.06; difference for .2 = 7.77, then 



