THE IREIGATION AGE. 



59 



THE PRIMER OF HYDRAULICS* 



By FREDERICK A. SMITH, C. E. 



5. The Chicago Drainage Canal. 



This canal connects Lake Michigan with the Mississippi 

 river via the Chicago, Desplaines and Illinois rivers, and is 

 a splendid example of a successful solution of a great sani- 

 tary and engineering problem, providing at once means for 

 diverting Chicago's great Sewer, the Chicago river, from the 

 same city's great reservoir of drinking water, Lake Michigan, 

 and at the same time introducing sufficient dilution to pre- 

 vent evil effects in its course to the people in the Illinois and 

 Mississippi river valleys ; this on the one hand prevents 

 further pollution of Chicago's water supply and also forms 

 the first and most costly link in a deep water way from the 

 Great Lakes to the Gulf. 



Fig. 84. 



The canal proper starts at Robey street, Chicago, and 

 extends to Summit, 111., a distance of 7.8 miles, having a 

 cross section shown in Fig. 84, 110 ft. wide at the bottom, 

 and side slopes 2 to 1 ; the elevation of the bottom at this 

 point is 24.448 (which means 24.448 feet below lake 

 level). From Summit to Willow Springs, a distance of 5.3 

 miles, the cross section is indicated in Fig. 85, being 202 

 ft. wide on the bottom with side slopes 2 to 1. From Wil- 

 low Springs to Lockport, 14.95 miles, the channel is cut in 

 rock, with a width on the bottom of 160 feet, with sides 

 nearly perpendicular (162 wide 22 ft. above bottom). This 

 section is shown in Fig. 86. The grade through the earth 

 section is 1 in 40,000 ft., or s = .000025; in rock the grade 

 is 1 in 20,000, or j =: .00005. 



At Lockport, where the waters of Lake Michigan join 

 those of the Desplaines river, ample controlling works have 

 been erected to regulate the flow to necessary requirements. 

 While at present the flow is about 400,000 cub. ft. per min., 

 it is expected that this will be gradually increased so that 

 when the population of Chicago becomes 3,000,000, the flow 

 of lake water will reach 600,000 cub. ft. per min., or 10,000 

 cub. ft. per second. Before this latter volume is sent down 

 the canal the earth portion from Robey street to Summit 



Fig. 85. 



will have to be widened out to 202 ft. on the bottom, con- 

 forming to section shown in Fig. 85. The Chicago river con- 

 nects the canal with Lake Michigan and has a length of 6 

 miles from Robey street to Lake street. Its width was quite 

 variable at the time the drainage canal was constructed, 

 but has been improved, and has now almost a uniform width 

 pT of 200 feet, a depth of 26 



-ni_ fffZ r~ ft. for 100 feet in midchan- 



nel, and 16 feet on the dock 

 lines, as shown in Fig. 87. 

 We can now consider 

 several of the problems 



"iso' " 2? 



Fig. 86. 



which present themselves in connection with the "drainage 

 canal. 



6. Problem. 



A flow of 300,000 cub. ft.' per minute stands 22 ft. deep 

 in the narrow channel shown in Fig. 84. What is the coef- 

 ficient of roughness ? 



The cross sectional area of flow is easily computed and 

 is 3,388 sq. ft. ; the flow is 5,000 cubic ft. per second, hence 

 the mean velocity v = 5,000 -=- 3,388 = 1.48 ft. per second. 



The hydraulic radius is next computed : 



r = area DABC H- DA + AB + BC. 



"Copyright, D. H. Anderson. 



r = 3,388 -r- (49.19 + 110 + 49.19). 



r = 3,388j-f- 208.38 16.26. 



Then \,' r = 4.032. 



The slope j = .000025 and V 000025 = .005. 



The data_on hand will give now the factor C since C 

 = "v -r- -\l rs, hence : 



C = 1.48 ~ 4.032 X -005. 



C = 1.48 ~ .02016 = 73.41. 



Consult table No. 12, (n = .040) and we find opposite 

 4.0 in slope column .000025 the factor C 78.45. Next consult 

 table No. XIII (n = .050), we find opposite 4.0 in slope 

 column .000025 the factor C = 65.02, hence the value of n 

 in this case is greater than .040, and less than .050; by inter- 

 polation we find the difference is 13.43, that is, the factor C 

 decreases 13.43 while the factor n increases from 040 to 050 

 or at the rate of 1.343 for each .001 ; the factor C in question 

 equals 73.41, or 5.04 smaller than the C belonging to 040- 

 hence as often as 1.343 is contained in 5.04, that many times 

 will 001 have to be added to .040 for the required factor 

 n\ this gives 3.7, or (nearest) 4 times .001 added to 040 

 gives .044 as factor n for this section. 



7. Problem. Find the slope s. 



As this volume of the water strikes the earth channel 

 between Summit and Willow Springs where the cross section 

 is considerably larger, it stands to reason that, all other things 

 being equal, there will be a reduction in the slope j on 

 account of the reduced velocity. The area in Fig. 85 equals 

 o,412 sq. ft., and the wetted perimeter equals 300.38 ft hence 



r = 5,412 -=- 300.38 =18.017. 



Then V. = 4.25. 



The velocity v =_ 5,000 -f- 5,412 = .924 ft. 



From tj = C ^\/ rs we have by transposition: 



s = v* -T- CV, in which equation v and r are known. 

 To find C we assume that the factor n is the same as in 

 the preceding case, namely, .044, and again find C by inter- 

 polation, as follows: If were = .040, _C is found from 

 table No. XII (last column), between V r = 4.0 and 5.0; 

 . for 4.0 = 78.45; for 5.0 = 89.28; difference = 10.83- 

 hence difference for .25 = 2.71; add this to 78.45 gives 

 81.16 as C for n = .040; proceed likewise for factor C if n 

 = .050, which produces 2.45 to be added to 65.02 = 67.47 

 Now we have C for .040 = 81.16 and C for .050 = 67.47- 

 hence difference 13.69 or 1.37 for .001; multiply by 4 gives 

 5.48 to be subtracted from 81.16 (because C diminishes as n 

 increases) gives 75.68 as factor in C for the case under con- 

 sideration. 



If we substitute now the known values in : 

 f = if -f- CV, we obtain : 

 J = .924'" -f- 75.6S 3 X 18.017. 

 i = .000008. 



8. Problem. 



We will next consider the flow of 5,000 cub. ft. of water 

 through the rock section shown 

 in Fig. 86. The cross section, 

 of the flow equals 3,542 sq. ft. 

 and the velocity 1.41. As this 

 section eventually will have to 

 accommodate a flow of 10,000 

 cub. ft. per second, the slope 

 of the flow in this case will be only half the final slope of 

 .00005, or s = .000025. and we can now determine the factor 

 n for the rock cut. 



Area of flow = 3,542 sq. ft. 



Wetted perimeter = 160 + 22.02 + 22.02 = 204.04. 



r = 3,542 -i- 204.04 = 17.3596. 



V r = 4.167. 



The factor C can now be determined since 



C = v -r- V rs > namely : 



C = 1.41 -H 4.167 X .005 = 67.1. 



To find coefficient of roughness n proceed as follows : 

 In table No. XII (where n = .040), opposite V r = 4, we 

 find under s = .000025, C = 78.45. In table No. XIII, where 

 n = .050, we find 65.02, and since our factor C = 67.1 lies 

 between the two, it follows that coefficient n is greater than 

 .040 and smaller than .050, and found exactly by interpolation, 

 as explained above ; this brings n .048. 



Right here it may be said that these values of n are 

 taken entirely too high ; the designers of the drainage canal 

 evidently desired to be on the safe side when they computed 

 cross sections and slopes, for it stands to reason that a chan- 

 nel of such regular section should have a much lower factor 

 of roughness than what is indicated above; they should not 



'/no 1 - 

 Fig. 87 



