THE IREIGATION AGE. 



99 



THE PRIMER OF HYDRAULICS* 



By FREDERICK A. SMITH, C. E. 



XIV. Closed Channels. 



1. Rectilinear Channels. 



Of the various forms of closed channels, the cylindrical 

 and composite forms will be especially considered, as all 

 the others bounded by right lines are readily solved by the 

 preceding articles ; for instance a closed channel of rectangu- 

 lar section will be considered an 



j) open channel until it flows full ; 



then the hydraulic radius under- 



D, C, goes an immediate change. As 



shown before, the hydraulic radi- 

 us varies directly with the height 

 ts in an open channel ; but if ABCD 

 F j g 89 (Fig. 89) represents a closed 



channel, then when the height of 



flow reaches the top DC, the wetted perimeter is increased 

 to the extent of the face DC, and causes a considerable re- 

 duction in the hydraulic radius ; for instance, let AB 4 ft. 

 and AD = 3 ft; then just before the water reaches DC 

 we have 



r = 3X4H-3 + 4-|-3 = 12-=-10 = 1.2. 

 Then the moment the water reaches DC : 

 r = 3X4-^3 + 4 + 3 + 4 = 12 -=-14 = .857; 

 in case the cross section is a square with the side a, then 

 r = tf -r- 4a = a/4 in the case of a closed channel. 



Similar deductions may be made in regard to closed 

 channels of triangular and trapezoid sections, each of which 

 is considered an open channel until the roof becomes im- 

 mersed, when the reduction of r must be considered. 



2. Cylindrical Channels. 



Channels of circular cross sections are the most gen- 

 erally used and of the greatest industrial importance, as 

 most sewers and water tunnels are at present of that form, 

 and practically all water mains and supply pipes. While 

 most sewers are ordinarily flowing only partially full and 

 only occasionally are completely filled, water pipes as a usual 

 thing are flowing not only full but often under considerable 

 pressure ; therefore the problems of the cylindrical channel 

 comprise three varities : First, channels flowing partially 

 full ; second, those flowing full, and third, those flowing 

 full under a considerable pressure. These various phases 

 are now explained by different applied problems. 



3. Applied Problem. 



A brick sewer 16 ft. diam- 

 eter drains a territory of 4,000 

 acres, and the dry weather flow 

 delivers .011 cu. ft. per second 

 per acre, while the storm flow 

 equals .10 cu. ft. per second per 

 acre; the slope of sewer is .0001 

 and coefficient of roughness 

 .015 ; find depth and velocity of 

 dry weather flow, also depth 

 and velocity of storm flow. 



Solve dry weather flow first. 

 Total flow equals 4,000 X -Oil 

 ; .44.0 cub. ft. per second. To begin, assume a probable 

 hydraulic radius, say 1.44 feet. Draw a diagram as indicated 

 in Fig. 90. in which the line AB represents the height of 

 flow and ED is^the versed line. 



Then \/lA4 = 1.2^ turn to table VI ; we find under 

 slope .0001 opposite Vr = 1.2, so C = 101.77, say 102 for 

 short ; then v = 102 X 1.3 X .01 1.124 ft. per second ; find 

 area corresponding to R = 1.44. in table XV. Divide 1.44 

 by 16 = .09, which is hydraulic radius constant and agrees 

 with area-constant .071 ; multiply .071 by the square of 

 diameter 16 X 16 = 18.18, which is area of segment ; mul- 

 tiply by velocity 1.124, gives the quantity of flow = 20.36 

 cub. ft., which is too small, and hence the value of r must 

 be increased. 



Next assume r = 1.96, hence \/r = 1.4. 

 In table VI under slope .0001 opposite -\/r = 1.4 we find 

 factor C= 10*8.96, say 109. 



*Copyrighted, D. H. Anderson. 



Then v 109 X 1.4 X .01 = 1.526. 



Find area corresponding to R = 1.96 (table XV). 

 Divide 16 into 1.96 = .125, which is hydraulic radius con- 

 stant in table XV, lying about half way between .1206 and 

 .1311; so the corresponding area constant lies between .11182 

 and .12811, or .1199; multiply by 16* = 256 which gives 

 30.48; multiply this by v = 1.526, gives 46.52, cub. ft., which 

 is slightly too great, but near enough for practical purposes; 

 hence the versed sine of flow will be between .20 and .22, 

 say .21, which, multiplied by 16, gives 3.36 ft. for depth of 

 dry weather flow. 



To find the depth and velocity of the storm flow, pro- 

 ceed in like manner. Calculate the maximum storm flow 

 for 4,000 acres at .10 = 400 cub. ft. water. Assume a prob- 

 able hydraulic radius at 4.0, or sewer flowing half full; 

 then V4.6 = 2.0 A Turn to table VI; we find under slope 

 .0001 opposite V = 2-0 the factor C = 125.13 ; then v 

 125. X 2 X .01 = 2.5 ; find area corresponding to R = 4.0, 

 which is half of area of full 16 ft. circle = 100.53; multiply 

 this by 2.5 gives 251.32 cub. ft. per second ; this isjpo small, 

 so assume a larger hydraulic radius; assume \JT at 2.20; 

 table VI shows C = 125.13, say 125. at -\/r~ = 2.0; at 

 V = 2.5; C = 134, so for V = 2.2 the factor C equals 

 2/5X9 = 33/5, which, added to 125.13, makes 128.73, 

 or about 129.0. 



Then v = 129 X 2.2 X .01 = 2.84 ft. per second. 

 Find area corresponding to R = 4.84 ; divide 16 into 4.84 

 .302 (hydraulic radius constant, table XV) ; this agrees 

 with area constant .63185 and versed sine .75 ; so multiply 

 .63 by 16 2 : - 161.28 sq. ft.; multiply this with v = 2.84, 

 gives 458.03 cub. ft. per second; this indicates that the maxi- 

 mum flow of 400 cub. ft. per second will not reach the 

 .75 X 16 = 12 ft. and r is taken too high, but we have 

 limits established, as^ the depth of flow lies between 8 ft. 

 and 12 ft., and \/r must lie between 2.0 and 2.2; assume 

 V = 2.1 then C 127, and v = 127 X 2.1 X .01 = 2.67 

 ft., r = 4.41 ; divide 16 into 4.41, gives .28 ; in table XV 

 this brings area constant .492 and versed since .60 ; multiply 

 .49 by 16 2 = _ 145.44 sq. ft.; multiply by 2.67, gives 388.32 

 cub. ft. This is very near the requirement, and 'by multiply- 

 ing 16 by .6 we get depth of flow, 9.6 ft. This evidently 

 shows that the size of the 16 ft. sewer is too large for 

 the maximum flow of 400 cub. ft. per second. 



4. Find the Necessary Size. 



Another useful problem to work would be : Find size 

 of a circular sewer to carry a maximum flow of 400 cub. ft. 

 per second when s = .0001 and n = .015. As the maximum 

 flow obtains if height of flow area = .930 we will assume 

 the sewer flowing .93 full at that rate ; the hydraulic radius 

 constant is then .2918, so that in a circle of diameter = 1, 

 the area would be .7612, and r = .292, and wetted perimeter 

 2.61 (see table XV). Assume a sewer 12 ft. in diameter; then 

 at .93d area equals 12 X 12 X .7612 = 109.61 sq. ft.; 

 divide this into 400, gives v = 3.65. _JNow check v as fol- 

 lows : r = 12 X -292 = 3.50 and \/r-= 1.86. In table VI 

 find C = 121; then v = 121 X 1.86 X -01 = 2.25. It is 

 seen that v is too small, hence a larger sewer must be 

 used. 



Try d = 14 ft.; then at .93d area equals 14 X 14 X .76 

 = 148.96 sq. ft. ; divide this into 400 gives v = 2.77 ft. per 

 second. Now r = 14 X .292_= 4.09, and \/V = 2.02 ; 

 in table VI we find C for \/ 1 i r 2.02 = 125; then v = 

 125 X 2.02 X .01 = 2.53 ft. per second, giving a total 

 flow of 348.56 cubic feet. 



This diameter is still too small, so try d = 15 ft.; flow 

 area at .93d = 15 X 15 X .76 = 171.0 sq. ft.; divide into 

 400 gives v 2.34, r = 15 X .292 =4.38 and V = 2.09; 

 C = 127 (table VI)'; then: v. = 127 X 2.09 X .01 = 2.63, 

 which is greater than necessary, as 2.63 X 171 = 449.73 

 cub. ft. 



This size would probably be selected, as there would 

 be an excess in capacity of nearly 50 cub. ft. per second, 

 which would be on the safe side. 



5. Find the necessary slope. 



Another class of problems will present themselves^ Sup- 

 pose it should be desirable, for good reasons, that the di- 

 ameter of the sewer should not exceed 12 ft. ; then it would 

 be necessary to increase the slope s, which factor is now 

 to be developed. 



When flowing .93 full the flow area = 109.6], requiring 

 a velocity of 3.65 ft. per second (see above). On a slope 



