1106 



THE IEEIGATION AGE. 



THE PRIMER OF HYDRAULICS* 



By FREDERICK A. SMITH, C. E. 



FIG. 79 



5. The Circular Form. 



Fig. 79 shows the circular form of conduit and while 

 it is a form in very extended use and is easily applied and 



figured when flowing ei- 

 ther half full or full it 

 is very difficult and tedi- 

 ous to determine the hy- 

 draulic radius and condi- 

 tions of flow when the 

 channel flows neither full 

 nor half full. This dif- 

 ficulty becomes apparent 

 if we consider a random 

 height of flow as indi- 

 cated by the line AB in 

 Fig. 79. If the cylinder 

 is just full, or just half 

 full, it is easy to deter- 

 mine the wetted perimeter 

 and area, and consequent- 

 ly the hydraulic radius, in the first case, the full area is 

 divided by full circumference, and, in the second case, it is 

 half area of circle divided by half the circumference. In 

 either case r = d-f-4 if d is diameter of circle; this is easily 

 proven, for the area of circle = ^d* -=- 4 and the circumfer- 

 ence = ird, hence r = (ir</ 2 -:-4) -=- ird. 

 Cancelling ^d we retain : 

 r=d-t-4. 



In the case of the semicircle we have : Area -4- ^d* H- 8 ; 

 circumference ,rd-=-2; hence r = ( w d 2 -f- 8) -4- (wd -r- 2) = 

 d^-4. 



Assume a typical case, taking the surface of fluid at AB 

 with a certain distance below the center; here, if the versed 

 sine MN is given, it will be quite a little problem to deter- 

 mine the area of the segment ANB; it requires still more 

 figuring to get the length of the wetted perimeter ANB; 

 thus, -if d = diameter of circle and m = versed sine, then 

 OM = d/2 m and cos o = (rf/2 w)-r-d/2; this gives 

 the center angle of the sector, as angle AOB = 2a and area 

 of segment ANB can be found by subtracting triangle AOB 

 from sector OANB, thus : 



*P 



Area of sector : 



4X 360 

 Area of triangle: AM X OM 



d 

 AM = rf/2 sin a and OM = -- m, hence area of tri- 



2 



d d Trd'a. 



angle = -sin a ( -- m) and flow area o= 



22 4 X 360 



d d 



sin a ( -- m). 

 2 2 



The circumference or wetted perimeter is found : 



p = Trda +- 360. 



Divide a by p gives r, but it is seen that it requires con- 

 siderable figuring. 



Certain problems in hydraulics require many trial as- 

 sumptions before a satisfactory solution is reached which 

 multiply the work many fold. This the author has sought 

 to avoid by the calculation of a number of tables which 

 are embodied in this book. 



The value of r grows with the depth or flow or the 

 versed sine from zero to a maximum when in (the versed 

 sine) equals ,813d; hence the velocity of flow in a circular 

 channel is greatest when .813 full ; the quantity Q, however, 

 is greatest when m = .949rf, when the product of flow area 

 times velocity reaches a maximum. 



6. Summary Comparisons. 



Summarizing the foregoing deductions may be stated 

 the following facts : 



Copyrighted by D. H. Anderson. 



1. For open channels where there is considerable varia- 

 tions in the volume of flow and where the best velocity is 

 desired for small flow the triangular section is best. 



2. For open channels of nearly constant flow the trape- 

 zoidal section is best, providing height of flow is equal to or 

 greater than width of channel at bottom. 



3. For closed channels flowing full or under pressure 

 the cylindrical section is best. 



7. A New Form of Channel. 



For sewers or other channels which have to accommodate 

 a small dry weather flow and a considerable flow of storm 



water at times a combina- 

 tion section is suggested 

 as indicated in Fig. 80; the 

 lower part being formed 

 by two planes forming a 

 right angle with each other 

 and the upper part being 

 a part of a circle. Let it 

 be a horizontal plane pass- 

 ing through vertex A, then 

 angle o should be 45 and 

 angle BAD = W. The 

 size of such a sewer would 

 be governed by the radius 

 of the cylindrical part, as 

 AB = BC. 



Let R = radius, then 



A 



FIG. 80. 

 area will be: 



a = 



= 3.35627? 2 . 



The wetted perimeter flowing full is p = 



Simplifying this expression we get: 

 87? R 



(6, 



- + ZR. 



4 4 



p 6.71247?. 



Hence : r = a -f- p : substitute values : 



r = 3.3562T? 2 H- 6.71247?. 



r = R -:- 2. 



From this it appears that the hydraulic radius for this 

 combination section is precisely equal to that of the full 

 circle. Consequently when flowing full it develops the same 

 velocity as would be done in a true circle, but its capacity is 

 increased in the ratio of 3.3562 to 3.1416. The main 

 advantage in this shape of a sewer or conduit is the better 

 flow condition it affords in dry weather, when the volume 

 of flow may be as low as one-twelfth of the maximum 

 capacity. The sewer is deeper by EA than the circular sewer, 



CA = ^/ZR 1 .R^Jz and since CE= R 

 AE = R\/~2 R = R (V2 1) 



This makes total height of sewer 2.4147?, as against 27? 

 for true circle. 



This form should prove far more desirable than an 

 elliptical shape, and with the general adaptability of concrete 

 construction now-a-days should be easily built ; it will be 

 referred to later in order to show its superiority over other 

 forms of sewers. 



Article X. The Coefficient of Roughness. 



In the determination of the factor C in Kutter's formula 

 all the quantities entering into consideration are definite 

 and fixed, except the coefficient of roughness . It is there- 

 fore quite important that this value n be properly selected. 

 The quantity n should represent everything which acts as 

 resistance to the flow of water. Many thousands of gaug- 

 ings have been made to determine the value of n in different 

 kinds of channels, and the results obtained are sufficient to 

 act as a guide in the selection of the proper factor for any 

 proposed channel. Some gaugings made show n as low as 

 as .007 and as high as .050, so that all variations in condi- 

 tions are covered by this amplitude. In ordinary practice 

 as low as .007 is impracticable and this book begins with 

 H = .009 as the lowest coefficient of resistance, which applies 

 only to the smoothest glass pipe or polished metal pipes, 

 perfectly straight and uniform. 



