THE IRRIGATION AGE. 



141 



THE PRIMER OF HYDRAULICS* 



By FREDERICK A. SMITH, C. E. 



6. The Composite Section. 



To compare the composite section with the true circular 

 one, let us solve the last problem for the composite section. 



Find maximum capacity of 

 composite channel having a di- 

 ameter of 12 ft., using 5 = .000266 

 and n = .015. 



Draw diagram (Fig. 91) show- 

 ing outline of section. 



The maximum capacity of the 

 composite section is when flowing 

 .92 full of circular depth. From 

 table XVI we find opposite .92 cir- 

 cular depth the constant for r = 

 .2908 and for area = .8096; hence 

 r = 12 X .2908 = 3.49. V = 

 1.87, and flow area = .8096 X 12 

 X 12 = 116.58 sq. ft.; from table 

 VI, under slope .0002, we find C opposite \Jr = 1.8 = 120.50, 

 and with the adjustment noted in the preceding paragraph 

 gives us C = 121.9, hence : 



v 121.9 X 1.87 X .0162 = 3.79 sq. ft. per second. 

 The total capacity then will be : 

 116.58 X 3.79 = 444.84 cu. ft. per second. 

 It may be noted that the velocity is the same in the 

 12 ft. circular and 12 ft. composite section, but there is a 

 material gain in capacity, due to the greater flow area. 



Another advantage of the composite section lies in the 

 fact that for dry weather flow it gives much more satis- 

 factory velocities than the circular section ; this will well 

 be illustrated in one of the following applications : 



7. Find flozv area and velocity. 



Suppose the 12 ft. composite sewer is to take care of 

 dry weather sewage flow of 40 cu. ft. per second ; find 

 flow area and velocity, assuming same slope as above. 



From table XVI select a random depth, say .3, which 

 would mean a depth of 12 X .3 = 3.6 ft; the area constant 

 opposite .3 = .090, and hydraulic radius constant = .108 ; 

 hence flow area = .09 X 144 = 12.96 sq. ft., and hydraulic 

 radius = .108 X 12 = 1-296, and Vr~ = 1.14. 



Refer to table VI under slope .0002 and opposite V~ = 

 1.0, we have C = 96.11 ; opposite -\/r = 1.2 we have C= 

 103.98; this shows a difference of 7.87 for .2, or .3935 for 

 1/100; hence for .14 we should add 14 X .3935 = 5.51 to 

 96.11, which gives C = 101.62; hence v = 101.62 X 1.14 X 

 .0162 = 1.88 ft. per second. Multiply this with 12.96 gives 

 = 24.37 cu. ft. flow, which is not enough, as we require a 

 flow of 40 cu. ft. per second. 



Next try a depth of .407; from table XVI we find area 

 constant .1654, and r constant = .1449 ; multiply this by 12, 

 we get : 



R 12 X .1449 = 1.74 and V R 1.32. 



Area == 12 X 12 X .1654 = 23.82 sq. ft. 



Refer to table VI. Under slope .0002 opposite -\JR 

 1.2 we find C_= 103.98. 



Opposite \/R = 1.4 C 110.51. 



Difference = 6.53 for .2; for 1/100 .3265, hence for .12 

 we must add 12 X .3265 = 3.92 = 107.90. 



Now r -= 107.9 X 1.32 X .0162 = 2.306 ft. per second. 



Multiply this by 23.82 gives 54.92 cu. ft. per second. 



This is too much, but we established limits for the flow 

 height. 



Next try in table XVI a depth of .387; the area con- 

 stant is .1497. 



R constant = .13723; multiply by 12 gives R = 1.65 and 



V^ = 1.28. Area = .1497 X 12 X 12 = 21.557 sq. ft. 



From table VI find C for \/R = 1.28, as explained 

 above, gives us C = 106.62. 



v = 106.62 X 1-28 X .0162 = 2.21 ft. per second. 



Multiply 2.21 by 21.557 47.62 cu. ft. per second. 



This is still too .large, but gives a very good idea about 



"Copyright, D. H. Anderson. 



the depth of flow; for accurate work we will try another 

 depth. 



Take .367 in table XVI; area constant .1347, R con- 

 stant .1298; multiply by 12 gives R = 1.56; \/R = 

 1.25; area = .1347 X 12 X 12 = 19.397 sq. ft. 



From table VI find C for \fR = 1.25 = 105.61. 



Then v = 105.61 X 1.25 X .0162 = 2.14 ft. per second. 



Multiply 19.397 by 2.14 = gives 41.51 cu. ft per second. 



This Answer fulfills the condition; so the Composite 

 Section has a flow depth of .367 X 12 ft. = 4.4 ft., and a 

 velocity of 2.14 ft. per second; which is sufficient to prevent 

 sedimentation. 



S. Comparisons between Circular and Composite Sec- 

 tions: 



For very small quantities the composite section gives 

 better velocities than the circular; to illustrate, find velocity 

 and flow area for a flow of 8 cu. ft. per second in a 

 conduit 12 ft. in diameter. For the composite section, as- 

 sume depth .18; then area cojistant = .0324 and R constant 

 .0648. Then R .78 and \/ R = .88 ; the flow area = 4.67 

 sq. ft. 



From table VI find C for *\/~R~ = .88 = 90.35 and 

 v = 90.35 X .88 X .0162 = 1.29 ft. per second. 



Multiply 4.67 by 1.29 = 6.02 cu. ft. per second, which is 

 not enough. Assume depth Z, area constant -.04, R constant 

 .072; then R = .864 and ^/R =_.93, and flow area 5.76. 



Find C in table VI for VR = .93 = 92.83, and v = 

 92.83 X .93 X .0162 = 1.40 ft. per second. 



Multiply 5.76 by 1.40 = 8.06 cu. ft. per second, which 

 just covers the case, giving a depth of flow of 2.4 ft. and 

 a flow velocity of 1.4 ft. per second. 



To find the depth of flow and velocity in the circular 

 sewer of 12 ft. diameter, we may assume that the velocity 

 will not exceed 1 ft. per second then the flow area will be 

 8 sq. ft. Since the area of a segment equals d ! times area 

 constant, the area constant equals area divided by a*; 

 hence divide 144 into 8, which gives area constant =: .055 ; 

 in table XV we find area constant = .05338 and R con- 

 stant = .0754; then R = .905 and V R = .951; area of seg- 

 ment = .05338 X 144 = 7.69 sq. ft. 



Then in table VI find C under slope .0002 = 91.53 for 

 .90 for V/? 1.0 ; C 96.11, so C for .95 will be half way 

 between 91.53 and 96.11 or 93.82. 



Then r 03.82 X .95 X .0162 = 1.44 ft per second. 

 Multiply this with 7.69 gives 11.07 cu. ft. per second, which 

 is too much. 



The above trial corresponded to a versed sine of .12 

 let us try .09; then the area constant = .03501 and R con- 

 stant = ..0574, hence, 7? = .69, V^ = -83, and area of 

 segment 5.04 sq. ft. 



In table VI we find C for Vrt = .83 = 87.9, and v 

 = 87.9 X .83 X .0162 = 1.18 ft. per second. Multiply 504 

 by 1.18 = 5.95 cu. ft. per second. This is not enough. 

 Let us next try versed sine .1 ; the area constant is .04087, 

 R constant .0635, hence R = .762 and \/R .873, and 

 area of segment 5.885 sq^jft. 



The factor C for \/~R = .87 = 89. 

 V = 89.0 X .87 X -0162 = 1.25. 



Multiply 5.89 by 1.25 gives 7.36 cu. ft. per second. 

 This is not quite enough, but shows that the versed sine 

 is slightly greater, but, we know that the versed sine of .12 

 is too great, and the real one is a very small fraction over 

 .1, perhaps .105. 



The solution shows, however, that while we have a ve- 

 locity of 1.40 ft. per second in the composite section, -we 

 have only a velocity of slightly more than 1.25 ft. per 

 second in the circular conduit. 

 1. General Deductions. 



Whenever flow of water in a pipe is produced by pres- 

 sure the head equivalent to such pressure can be readily com- 

 puted. For instance, where the pressure is given as 20 Ibs. 

 per square inch it is equivalent to the height of a column 

 of water producing 20 Ibs. pressure per square inch. It 

 requires a column of water 2.309 feet high with a cross- 

 section of 1 sq. in. of water to produce a pressure of 1 Ib. 

 on its base ; hence 20 Ibs. pressure would be equivalent to a 

 water column of 20 X 2.309 = 46.18 ft. ; if such a pipe would 

 have a length of 3,000 ft. then the head would be 46.18 ft. in 

 3.000 ft. and the factor ^ = 46.18 -f- 3,000 = .01539. 

 In general let I = length of pipe, 



p = pressure in pounds per square inch, 



