THE IRRIGATION AGE 



181 



THE PRIMER OF HYDRAULICS* 



By FREDERICK A. SMITH, C. E. 



V 

 C 



2. Find Velocity in Cast-iron Pipe. 



The pressure in a system of water works is 25 Ibs. per 

 square inch ; what will be the velocity in a 24-in. cast-iron 

 pipe 5,000 ft. long, using the coefficient of roughness .011? 



Solution First find the constant C; the hydraulic radius 

 is 2 -f- 4 = }4 and \/r .tm, or say .71. In Table III in 

 column opposite .7 factor C = 123.11; opposite .8 factor 

 =: 129.77; difference 6.66; hence add .67 to 123.11 = 123.78; 

 being factor C for -\/r = .71. Next substitute values in 



C 

 formula : v - -\/2.3pd -r- /; 



2 



v= 123.78 -r- 2 V2!3 X 25 X 2 -4-5,000; 

 i' = 61.89 V-23; 

 v = 61.89 X .151 = 9.45 ft. per second. 



3. Find Pressure in Cast Iron Pipe. 



Problem \Vhat pressure is necessary to produce a ve- 

 locity of 6 ft. per second in a 36-in. cast-iron pipe 1 mile long, 

 using factor u = .012 ? 



_ Solution Again find factor C first; r = 3 -=- 4 = .75 and 

 V -75 = .866, say .87; in Table IV in ^/V column we find 

 opposite .8, C 117.21; opposite .9, C = 122.68 ; difference, 

 5.47; hence add 7 X .547 = 3.83 to 117.21, making C = 121.04. 



Next apply formula No. 2 in Article xv, substituting 

 given values : 



p = 4 X 5280 X6X6-^2.3X3X 121.04 X 121.04. 



/> = 760320 -=-101. 119 = 7.52 Ibs. per square inch. 



4. Find Diameter of Cast-Iron Pipe. 



Problem It is proposed to supply water to a cast-iron 

 main under a pressure of 12 Ibs. per square inch. What must 

 the diameter be of the pipe if the length of pipe 6,200 ft., the 

 required velocity is 5 ft. per second and factor n = 012 ? 



Solution The selection of the factor C will be difficult, 

 as the diameter of the pipe is not known and a trial factor C 

 must be used. 



In formula 3 in Article xv assume C = 100 then we 

 have : 



d = 4 X 6200 X 5 X 5 -^- 2.3 X 12 X 100 X 100. 



d = 620000 -4- 276000 = 2.246 ft. 



Next we must check back and see what the true factor C 

 is for d 2.246 ; _as follows : 



r=_.561; Vr = .75; looking in Table IV the factor C 

 for V r = .7 = 110.88 jind factor C for V r = .8 = 117.21, 

 hence factor C for V = - 75 ' 114.04. 



If we now use 114 as the factor C in the above composi- 

 tion we obtain : 



d = 620000 -^ 358690 = 1.73. 



This shows that factor C is taken too high and that 1.73 

 is too small. Adding now 1.73 and 2.25 gives 3.98 and divide 

 by 2 as a mean gives very nearly a diameter of 2 ft., which 

 must be again checked up by the true factor C; in Table IV 

 find C for Vr = .71; for .7, C= 110.88; for .8, C= 117.21; 

 difference 6.33, hence add .63 to 110.88, making C = 111.51. 



Now try formula 3 again, using 111 as factor C: 



d = 620000 -r- 340060 = 1.82 ft. 



For general work a 2 ft. diameter pipe would be the prac- 

 tical solution of the problem, though a 22-inch diameter pipe 

 (if the exact size must be had) would undoubtedly be found 

 sufficient ; to try this bear in mind commercial sizes of pipes 

 and try between the limits established. 



For another trial use d = 22" = 1.83 ; r .4575 V r=-68 : 

 C = 103.44 for .6 and 110.88 for .7 ; difference 7.44 ; hence add 

 8 X -74 = 5.92 to 103.44 = 109.36 ; trying again formula 3 using 

 factor C = 109 we get : 620000 -r- 327916 = 1.89 ft. 



This shows that 22 inches is not quite sufficient but is 

 very close. 



Using 23 inches = d; this is 1.917 ft.; r = .4792; V7= 

 .69; C=110. 



Then d = 620000 -=- 333960 = 1.86 ft. 



Reducing this to inches multiply by 12 gives 225^ inches. 



"Copyright, D. H. Anderson. 



0. rind Length of Cast-Iron Pipe. 



A 48-inch cast-iron water main leads from a reservoir, 

 the water level of which is 100 ft. above the supply end of the 

 pipe; what is the length of the pipe if the velocity is 6.5 ft. 

 per second and n = 0.12? 



Solution r = 1.0 and V 7=1.0, so C 127.42. Change 

 the 100 ft. head into pressure by dividing 2.309 into 100 = 

 43.31 Ibs. Use formula four in Article xv. 



/ = 2.3 p d C -=- 4 if. 



1 = 100 X 4 X 127.42 X 127.42 -H 4 X 6.5 X 6.5. 



/ = 38,426 ft. 



6. Find the Factor C for Cast-Iron Pipe. 



Problem Find the Factor C for a 48-inch cast-iron pipe 

 38,426 ft. long having a head of 100 ft. and a mean velocity of 

 flow of 6.5 ft. per second, if n = .012. 



This problem is to check Formula 5 by using data of 

 the preceding problem. 



V 4 X 38426 X 6.5 X 6.5 



=. =127.42. 



100 X 4 



This checks correctly with Table IV. 



It should be observed in the preceding problems that in 

 determining the factor C it was taken from under the column 

 j = .001. This is usually correct for pipes flowing full under 

 pressure, as the equivalent head divided into the length give 

 high slopes ; mostly higher then .001 ; should, however, the 

 pressure become so low as to produce slopes corresponding to 

 .0004, .0002, .0001, etc., then the factor C must be selected 

 accordingly. 



Article XVI. Loss of Head by Enlargement of Channel. 



1. The Principle Involved. 



Whenever the form of a channel is suddenly enlarged, as 

 indicated in figure 92, there occurs a loss of head along AB 



which may be represented by the formula : L 



(% 

 ~7 



Fig. 92. 



in which L is the loss of head in 

 feet, Vi = the velocity in feet per 

 second in the smaller section and v t 

 the velocity in the enlarged section, 

 and g = 32.16. If the change from 

 one section to the other is made 

 gradually as indicated by the dotted 

 lines AC and BD the loss of head is 



very much reduced; if alpha is the angle which AC forms with 



ft v, 2 



(ft v,\ 2 

 - I sin. 

 2 g / 



2. Applied Problem A 12-inch cast-iron water pipe is 

 enlarged to an 18 pipe; if the enlargement is made in 4 ft. and 

 if the velocity in the 12-inch pipe is 6 ft. what is the loss of 

 head? 



S olu ti o n 

 Make sketch as 

 shown in Fig. 93 

 and apply formula 

 by substituting the 

 given quantities : 



Fig. 93. 



AB = V 48' 



+ 3' = V 23 1 5 = 



48.11 ; then sin oc = 



= .0625. 



48.11 



The velocity in the 12" section = 6 ft. = m. 



6 



The velocity in the 18" section = 2.67 = v,. 



2.25 

 (6 2.67)'X.0625 



Then L = = .0118 ft. 



64.32 



If the change had been made suddenly, '. e., if angle 

 <x = 90 then L would be equal to 1.73 ft. 

 1. The Analysis. 



To divide cylindrical channels into equivalent smaller 

 channels of equal diameters : 



Let d = diameter of large conduit. 

 Let x = diameter of small unit. 

 Let n = number of small branches. 



Then x = 





