r REESE LIBRARY. OK THK, UNIVERSITY OF CALIFORNIA loot No, THE GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS TREATING OP THE FLOW OF WATER THROUGH PIPES, IN CHANNELS AND SEWERS, OVER WEIRS, ETC. BY FREEMAN C. COFFIN, Member of the American Society of Civil Engineers. FIRS T EDITION. FIRST THOUSAND. NEW YORK: JOHN WILEY & SONS. LONDON : CHAPMAN & HALL, LIMITED. 1897. Copyright, 1897, BY FREEMAN C COFFIN. ROBERT DRUMMOND, ELHCTROTYPHR AND PRINTER, NEW YORK. CONTENTS. CHAPTER PACK I. INTRODUCTION i II. UNITS AND SYMBOLS 4 III. FLOW OF WATER IN PIPES ... 6 Formulas 6 Small Pipes, ^-3 Inch 8 Explanation of Diagrams 8 Examples 9 Large Pipes, 4-60 Inch 10 Explanations and Examples 10 Problems Outside of Range of Diagrams n Compound Mains 14 Complex Mains 16 Old or Tuberculated Pipe-lines 24 IV. FLOW OF WATER THROUGH SHORT TUBES 31 Formula and Coefficients 31 Examples 32 V. RECTANGULAR WEIRS 36 Formulas 36 Explanation of Diagrams 37 Examples 38 Wide Crests 40 Velocity of Approach 41 VI. FLOW OF WATER IN CHANNELS 44 Kutter's Formula 44 Values of n , 45 Explanations and Examples 47 iii IV CONTENTS. CHAPTER PAGE VII. PIPE-SEWERS 50 Formula and Coefficient 50 Explanation of Diagram 50 Examples 50 VIII. FIRE-STREAMS AND DISCHARGE OF HOSE-NOZZLES. .. 52 Height of Streams and Head Required 52 Discharge of Hose-nozzles with Head or Pres- sure Indicated at the Hydrant 53 Discharge of Hose-nozzles with Head or Pres- sure Indicated at Base of Play-pipe 53 IX. MISCELLANEOUS PROBLEMS 55 Horse-power of Falling Water, also Horse-power Required to Raise Water 55 Capacity and Size of Pumps. . ; 56 Coal Required for Pumping Water 58 TABLES. Table No. i. Equivalents of U. S. Gallons 60 2. Discharge in Gallons per Minute for Different Velocities in Circular Pipes 62 " 3. n/6 and 6/u Powers of Numbers 65 4. Coefficients of Friction in Old Pipes 66 5- " Discharge in Old Pipes 66 6. " Velocity of Approach for Weirs 67 7. Areas and Values of r in Circular Channels. 68 8. " " " " r in Rectangular Chan- nels 69 9. With Side Slopes of i to i 70 " 10. " " " " 2 to i 71 " " IT. " " " " 3 to i 72 " 12. Equivalents of Pounds Pressure and Feet Head 73 " " 13. Height and Discharge of Fire-streams 74 CONTENTS. DIAGRAMS. Diagram No. I. Flow of Water through Pipes -3 Inches in Diameter. *' " 2-17. Flow of Water through Pipes 4-60 Inches in Diameter. " " 18, 19. Flow through Short Tubes or Entry Head. " " 20. Value of c in V c(rs)\. 21. Weirs with Wide Crests. " " 22-24. " " Thin " . *' " 25,26. Flow of Water in Channels. " 27. " in Pipe-sewers. 14 28. Horse-power of Water. " " 29. Size and Capacity of Pumps. " *' 30. Coal Required in Pumping. " " 3i 32. Discharge of Hose-nozzles. M " A and B. Experiments on Old Pipes. GRAPHICAL SOLUTION .OF HYDRAULIC PROBLEMS, CHAPTER I. INTRODUCTION. IT is not the purpose of this book to discuss the laws governing the flow of water, the hydraulic experiments that have been made, or the formulas derived from them. The object of the book, as conceived by the author, is to provide a convenient instrument or tool for the prac- tising engineer (who is already familiar with hydraulic laws and formulas) with which he can solve quickly and correctly the commonly occurring hydraulic prob- lems by means of diagrams and with a minimum of cal- culation, either mental or written. The diagrams are constructed upon well-known for- mulas, using coefficients that are generally accepted as safe for the conditions to which they are intended to apply. While it was intended to have the computations correctly made and carefully checked, no attempt has been made to secure mathematical refinements, such being in most cases entirely out of place in hydraulic 2 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. problems. The computations for these diagrams were made with the slide-rule and logarithmic cross-section paper.* Such computations are correct to two figures, and sometimes three or four, and for these problems this is a greater degree of accuracy than can usually be secured in all of the data. In regard to the possibility of obtaining sufficiently accurate results from the diagrams themselves, this can be said : That in all problems relating to the flow of water through short tubes, long pipes, sewers, and chan- nels, the result can be read from the diagram with a greater degree of precision than the engineer is warranted in relying upon in practice. The variations in the com- mercial sizes of pipes will probably cause greater errors than those that result from inability to read the diagram closely. Differences that cannot be estimated in the surfaces of pipes, sewers, or channels will produce differ- ences in result beside which any errors from the above sources will be insignificant. Details of actual construc- tion, small variations from the design, differences in the character of surfaces and many other elements, tend to render a refinement in figures absurd. Of the subjects relating to the flow of water treated in this book, it is possible that in weirs alone a few cases will arise where the diagrams will not give results suffi- ciently close ; but such cases are only those in which the weir has been constructed in perfect conformity with the conditions from which the formulas were derived.f * The paper used was that designed by Mr. John R. Freeman. M. Am. Soc. C. E., with four complete squares of lo-inch base on one sheet. f Weirs are excepted from the general statement because the conditions of their construction are more under the control of the engineer than those of the other classes. INTRODUCTION. 3 There are, however, probably few weirs used for ordinary purposes that are so constructed in every detail. If the diagrams are based upon proper formulas, the computa- tions correctly made, and the drawing carefully done, results can be read from them with sufficient precision for all practical purposes. The diagrams are offered to the profession in the hope that they may prove useful, as similar ones have been a valuable aid to the author for several years. Any of the usually occurring hydraulic problems can be solved by them, either in field or office, with no other tables, for- mulas, or information than those contained in this book, and in nearly every case without the use of pencil or paper. CHAPTER II. UNITS AND SYMBOLS. IN choosing the units to be used, preference is given to those in most general use. Measures of quantity of water are always given in U. S. gallons per minute. Whenever possible, a second scale gives cubic feet per second. Table No. i, page 60, gives equivalents of U. S. gal- lons per minute. All linear dimensions are given in feet and decimals, except in the case of diameters of pipes, which are given in inches. SYMBOLS. The following characters will always have the meaning here assigned, unless otherwise stated : a = area in square feet ; _ j coefficient in weir formula, ( coefficient in Chezy formula ; D = diameter in feet ; D' = depth of water in channels in feet ; d = diameter in inches ; g = acceleration of gravity ; h = head in feed for pipes and weirs ; / = length in feet of pipes and weirs ; UNITS AND SYMBOLS. 5 n Kutter's coefficient for character of surface of pipes and channels ; p = wetted perimeter in feet ; Q = cubic feet of water per second ; q = U. S. gallons of water per minute ; R = radius = ; r = hydraulic mean radius of channels and pipes = ; for circular pipes running full r = ; s = hydraulic slope for pipes and channels = ; v = mean velocity in feet per second ; for circular pipes running full v = W '= width of channels. Effective head and loss of head by friction in long pipes is given on diagrams as " friction-head in feet." CHAPTER III. FLOW OF WATER IN PIPES. PROBLEMS connected with the flow of water in pipes are of constant occurrence in the practice of the hy- draulic engineer. They may be divided into three general classes : (I) The size (Z>), length (/), and head (ti) being given to find the quantity (Q) or discharge. (II) The size, length, and quantity being given to find the required head or loss by friction. (III) The quantity, head, and length of pipe line being given to find the size of pipe required. There are many formulas, based upon numerous ex- periments, in use for the solution of these problems. Without discussing the respective merits of these, it may be said that the Chezy formula is used as the basis of computation for the diagrams relating to this subject. This formula was selected on account of its simplicity* and because it seemed possible with a varying coefficient to meet the conditions of different diameters of the pipes and of varying velocities more nearly than by any other form. This formula is expressed as follows : I. v = frs * ; 0$ WATER IN PIPES. 1 With these three forms of the equation, the value of either factor of the problem can be found when the other two are given. Note. When r is known, the diameter of a circular pipe running full can be obtained by multiplying r. by 4. The values of the coefficient c used in these diagrams are those given by Hamilton Smith, Jr., in his work on Hydraulics. These values seem to have been derived with great care from the most reliable hydraulic experi- ments. The reader is referred to the above-named book for further information on this subject.* These values of c are given graphically on diagram No. 30. The full curves are copied from Smith's Hydraulics ; the dotted curves are interpolated by the author. Although the Chezy formula is very simple, there is this practical difficulty in using it in the calculation of individual problems, in the second form in which it is given, i.e., v = c(rs)* : that c cannot be taken correctly until v is known, and an approximation must first be made by taking a value of c according to some assumed or estimated value of v. One or two approximations will generally bring the result. This objection to the formula does not hold, however, in constructing dia- grams, as the first form, or s = , or, reduced to the simpler expression, s = - f-J is the one best adapted for such computations. ACCURACY OF FORMULAS FOR THE FLOW OF WATER IN PIPES. It should be borne in mind that, while it is desirable that formulas should give with accuracy values for cer- * " Hydraulics," by Hamilton Smith, Jr. (John Wiley & Sons). S GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. tain given conditions of pipes, exactly these conditions rarely exist in practice. Therefore too much weight must not be given to mathematical refinements in ordi- nary practice, however desirable these refinements are in experimentation. When, for instance, the value of the coefficient c is in the vicinity of 100, a variation in the coefficient in several formulas of 4 or 5 should not be considered as invalidating either for practical purposes, as perhaps the difference in the commercial sizes of pipes of the same nominal diameter will cause a greater vari- ation. Differences in the interior surfaces of the pipes, that with our present knowledge are impossible of exact estimation, will produce a greater variation than that caused by such differences of the coefficient. Therefore, while it is desirable to use the best formula for clean, straight pipes as a basis, a sufficient margin should be allowed for these differences of condition. The subject of the effect of the age of pipe lines is discussed else- where in this chapter. EXPLANATION OF DIAGRAMS. SMALL PIPES. The diagrams for the flow of water in small pipes (those from three eighths of an inch to three inches in diameter) are constructed for a length of 100 feet. Re- sults for other lengths can be taken proportionately. These sizes are plotted upon one sheet, diagram No. i. The diameter of the pipes is given by curves originat- ing at the upper left-hand corner, the discharge in gal- lons per minute by a vertical scale on the right of sheet, the velocity by dotted curves, and the loss of head from friction by a horizontal scale at the top of the sheet. FLOW OF WATER IN PIPES, 9 Examples of Use. CASE I. Size, length, and discharge given to find head. Example: What is the loss of head in a i-inch pipe 75 ft. long, delivering 16 gallons per minute ? On diagram No. i, at the intersection of the line repre- senting 16 gallons of water with the curve representing i-inch pipe, find the friction in feet, viz , 22 ft.; this is in a length of 100 ft.; therefore 22 X = 16.5 ft., the required answer. CASE II. Size, length, and head given to find discharge. Example: What is the capacity of a 2-inch pipe 350 ft. long with effective head of 40 feet ? To use this diagram the length must be reduced to 100 ft., and the head in like ratio ; thus 40 X =11.4 per 35 100 feet in length. At intersection of 11.4 friction-head and 2-inch curve find capacity in gallons = 70 gallons per minute, flowing with velocity of 7 ft. per second. CASE III. Length, head, and discharge given to find size. Example: What is the size of pipe necessary to dis- charge 175 gallons per minute through 500 ft. in length with a head of 80 ft. ? 80 X = 16 ft. per 100 ft. in length. The intersec- 500 tion of 175 gallons and 16 feet head is between curves of z\ and 3 inches diameter, and of the usual sizes, the latter must be used. 10 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. LARGE PIPES. For pipes larger than 3 inches in diameter there is a diagram for each size. The size is given in the upper right-hand corner of diagram-sheet. On these diagrams the length of pipe is given by a vertical scale on the right- hand side ; the friction-head by a horizontal scale on top and bottom of sheet ; the quantity in gallons per minute by oblique lines radiating from a common origin at the upper left-hand corner. Cubic feet per second are not given on these diagrams, but may be taken from the Table No. i of equivalents of U. S. gallons per minute on page 60. The velocity in feet per second is given by a heavy broken line parallel with the oblique line of quantity. These diagrams are more convenient than those for small pipes, because the solution of all problems within a wide range can be read from them directly for the proper length without multiplication or division. Examples of Use. CASE I. Size, length, and effective head given to find discharge. Example: What is the discharge of a 1 4-inch conduit- line 7500 ft. long with an effective head of 30 ft.? Ans. 1860 gallons per minute; velocity 3.9 ft. per sec- ond. (From diagram No. 7 for 14-inch pipe.) CASE II. Size and length of pipe and discharge given to find loss of head or friction. Example: What is the loss of head in 5500 ft. of 12- inch pipe delivering uoo gallons per minute? The answer will be found at intersection of lines repre- senting 1 100 gallons and 5500 ft. of pipe on diagram No. 6 for i2-inch pipe. It is 18,2 ft. head. The velocity is 3.1 ft. per second. FLOW OF WATER IN PIPES. Il CASE III. Length, head, and required discharge given to find size of pipe. Example: What is the size of pipe necessary to dis- charge 2500 gallons per minute through 7500 ft. of pipe with 30 ft. head ? The diagrams show that a 1 4-inch pipe will discharge 1860 gallons, and a i6-inch pipe will discharge 2675 gal- lons; and of the usual commercial sizes, the latter is the one that must be used. These diagrams cover in their range nearly all of the problems met in practice. If the solution of problems that do not come within their limits is desired, it can be found indirectly as follows. Problems Outside of Limits of Diagrams. When the length of line is greater than that given on the diagram. CASE IV. Size, length, and discharge given to find the friction-head. Divide the given length by some number that will give a quotient within limits of the diagram; find head for this length, and multiply by the number used to divide length. Example: What is the friction-head in 12,000 ft. of 12- inch pumping main, with a delivery of 1000 gallons per minute, or velocity = 2.85 ? 12,000 ~- 2 = 6000; friction in 6000 ft. = 16.7 ft.; 16.7 X 2 33.4 feet head. CASE V. Size, length, and head given to find discharge. Divide both length and head by the same number, and find discharge for the quotient, which will be the answer. 12 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. Example: What is the discharge through 15,000 ft. of lo-inch pipe with head of 50 ft.? 15,000 -j- 2 = 7500 ft.; 50 -h 2 = 25 ft. head; discharge from diagram for 25 ft. head through 7500 ft. of pipe = 680 gallons; v 2.8. When the length given is less than that for which it is convenient to read results on the diagram. CASE VI. Size, length, and discharge given to find head. Multiply length by any convenient number, and divide the resultant head by same number. Example: What is the loss of head in 36-inch penstock 400 ft. long, discharging 25 cu. ft. per second or 11,200 gallons per minute ? 400 X ioo = 40,000 ft.; 11,200 gallons through 40,000 ft. of 36-inch pipe = 39.2 ft. head; 39.2 -5- ioo = 0.392 head. CASE VII. Size, length, and head given to find dis- charge. Multiply both head and length by any convenient number, and find discharge corresponding to the prod- ucts. Example: What is the discharge of a 3o-inch pipe 200 ft. long with an effective head of 0.4 ft.? 200 X ioo = 20,000 ft. long; 0.4 X ioo = 40 ft. head; the discharge of 3o-inch pipe through 20,000 ft. with 40 ft. head = 10,000 gallons; vel. = 4.5. Note. The entry and velocity heads in the last two examples, an important element, should be considered in all cases of comparatively short lines. This subject is treated in the next chapter. OF WATER IN PIPES. Where the discharge or head is greater than that given on diagram. CASE VIII. Size, length, and discharge given to find head. Divide the given discharge by any number that will bring it within the limits of the diagram. Multiply the friction due to the quotient by the n/6 power of the number used.* A table of n/6 and 6/n powers is given on page 65- Example: What is the loss of head in 500 ft. of 6-inch pipe discharging 1500 gallons per minute? 1500 -r- 2 = 750; 750 gallons through 500 ft. of 6- inch pipe = 24 ft. head; 24 X (2 11 / 6 = 3.56) = 85.5 ft., answer. CASE IX. Size, length, and head given to find dis- charge. Divide the given head by the u/6 power* of any number that will bring head within limits of diagram. On given length find discharge due to this quotient; multiply this discharge by the number taken. * Although the exponent of the formula v = c(rs$ is of the second power, the variation in the coefficient c renders the results practically identical with those of a formula of the n/6th power in which c is constant for all velocities. The reader is referred to a paper in the Journal of Associated Engineering Societies for June, 1894, by Mr. W. E. Foss, in which he shows that a formula of the n/6th power is in practical accord- ance with the results of experiments. He proposes such a formula for the flow of water in pipes and channels which is very simple, and, with the tables given in the paper, renders the computation of such problems much less tedious than by most formulas in use. See page 22 for remarks on the use of the second power instead of the n/6th. 14 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. Example: What is the discharge through 2000 ft. of 8-inch pipe with effective head of 125 ft.? 125 -T- (2 II/6 = 3.56) = 35.1; discharge due to 35.1 ft. head = 925 gallons X 2 = 1850 gallons per minute. Where the discharge or the head is less than can be easily read from the diagram. iL X. Size, length, and discharge given to find head. Multiply the given discharge by a number that will give a product large enough to be easily read from the diagram; divide the head obtained by the n/6 power of that number. Example: What is the head in a i6-inch pipe 5000 ft. long, delivering 100 gallons per minute ? 100 X 10 = 1000; head for 1000 gallons = 3.4ft; 3.4 -f. jo 11 / =0.05 ft. CASE XI. Size, length, and head given to find dis- charge. Multiply the given head by the n/6 power of some number; divide the discharge due to the product by the number. Example: What is the discharge of a i6-inch pipe 6500 ft. long with a head of .25 ft.? 0.25 X io II/6 = 17; discharge due to head of 17 ft. = 2120; 2120 -5- 10 = 212 gallons per minute. COMPOUND MAINS. A compound main is a single line of pipe which is not of uniform diameter for its entire length. It is com- posed of two or more sizes in one line. The friction in such a main can easily be computed when the discharge is given. FLOW OF WATER IN PIPES. lj To find the discharge when the total head is given is more difficult. The finding of both the friction and discharge is much facilitated by the use of these diagrams. CASE I. Size, length, and discharge given to find head. Find head for each size separately, and add results for total head. Example: What is the head in a compound main com- posed of 3000 ft. of i2-inch pipe and 7000 ft. of 16- inch, delivering 1500 gallons per minute? Head in 3000 ft. of 1 2-inch pipe = 17.4 ft. " " 7000 " " 1 6- " " = 9.6 " Total head, 27.0 " Ans. CASE II. Size, length, and head given to find discharge. Assume some quantity for the discharge; find the head due to such discharge in the given line, % the same as in Case I. Divide the correct head by this head, and multiply the assumed discharge by the 6/n power of the quotient. Example i: What is the discharge of a compound main of 4500 ft. of lo-inch pipe and 7000 ft. of i6-inch pipe with a total effective head of 40 ft.? Assume a discharge of 1000 gallons. Head in 4500 ft. of lo-inch pipe = 30.2 ft. " " 7000 " " 16 " " = 4.6 " Total, 34.8 ' \6/n -) = 1000 correct discharge. ^O N6/II Then 1000 X ( ~ 1 = 1000 X 1.075 = IO 75 gallons, l6 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. Check on above calculation : 1075 gals, in 4500 ft. of lo-inch pipe = 34.7 ft. head. " " 7000 " " 16- " " = 5.3 " " Total, 40.0 " " Example 2: What is the discharge of a compound main of 10,000 ft. of i6-inch, 6000 ft. of 1 2-inch, and 2000 ft. of lo-inch pipe with total head of 50 ft.? Assume discharge of 1500 gallons per minute. Head in 10,000 ft. of i6-inch = 13.8 " " 6,000 " " 12 " =34-9 [VIII, page 13. " " 2,000" " 10 " = 28. i, by method of Case Total head for assumed discharge, 76. 8 / c \ 6 /" Then 1500 X Hrs) = Il8 4 gallons, correct dis- charge. * COMPLEX MAINS. The problems involved in the flow of water in complex mains, or a system of mains in which the water flows to the point of discharge through two or more lines of pipe of different sizes and lengths, are of great importance in practice and should receive careful attention. Owing to the immense amount of work required to solve them by the ordinary methods, they are undoubtedly left un- solved in many cases. These problems arise in nearly all cases where an efficient fire system is to be designed for a town or city, or where the efficiency of one already constructed is to be ascertained. A valuable and suggestive paper on this subject was read by Mr. John R. Freeman, M. Am. Soc. C. E., before the New England Water-works Asso- FLOW OF WATER IN PIPES. l^ ciation, and published in the Journal of that Association, Vol. VII. The diagrams for the flow of water in long pipes pro- vide a ready means of solving such problems, and the following examples are given of their use. CASE I. System of piping and discharge given to find head. A 2500'12" B SOOO'lO" C_ Fig- 1 2000 '8" * Example : With reservoir at A and arrangement of piping as shown in Fig. i, what will be the loss of head with a draft of 1000 gallons per minute at C? First, find head in single line AB = 7 ft. The flow will divide at B, part of it going through BC and the remainder through BEDC. It is self-evident that the total friction or loss of head must be the same in one line as in the other, or the flow from the two lines would meet at C under different pres- sures, which is impossible. Assume a loss of head for these lines, and on the dia- grams find the discharge due to this assumed head. Thus assume 20 ft. head; then discharge in BC, 2000 ft. of 10 in. = 1240 gals. " " BEDC, 4000 " " 8 " = 470 " Total discharge for both lines, 1710 " Divide the given discharge by the discharge of the as- sumed head, and multiply the n/6 power of the quotient by the assumed head, and the product will be the true head. I& GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. X 20 = 0.37 X 20 = 7.4, true head. /loooN 11 / 6 \i7ioj Add head in AB = 7 Total loss of head, 14.4 for 1000 gals. (a) If, instead of assuming 20 ft. head, 8 ft. head had been assumed, then discharge in BC = 755 gals. Total assumed discharge, 1035 " 1 / 6 x8 =7 ' 50feet - Add head in AB = 7 Total loss of head, 14.50 " Note. The slight variation in the results of the two examples are due to slight errors in reading diagrams. (b) After the correct head for one rate of discharge is obtained, it is very easy to find it for any other by the following formula : /required discharge y/6 ( hea< ? f r , re ' r. r - 5 ~ X known head = < quired dis- V known discharge / | charge. Thus when draft of 1000 gallons through a given sys- tem of pipes causes a loss of head of 14.50 ft. as above, othe~ drafts would be as follows : /5oo V I/6 - X 14.50= 4.06 ft. iooo/ /I 200V 1 / 6 X 14.50 = 20.3 " \iooo/ \ II/6 X 14.50 = 30.6 " ^ Viooo/ FLOW OF WATER IN PIPES. 10 CASE II. System of piping and head given to find dis- charge. Example: With same arrangement of piping as in Case I, what will be the discharge with total head of 30 ft.? First assume the head in the lines BC and BEDC ; thus assume 15 ft. BC = 2000 ft. of 10 in., discharge = 1060 gals. BEDC = 4000 " " 8 ", " = 400 " Total discharge with 15 ft. head, 1460 " The loss of head in line AB, or 2500 ft. of 12 in., due to a discharge of 1460 gallons = 13.9 ft. Add head in BC and BEDC =15 " Total assumed head, 28.9 " To find the discharge due to the given head use fol- lowing formula: l, given head \ 6/ " j j- i (true dis- -r-r 3 X assumed discharge = j . ssumed head/ ( charge. \ 6/I1 X 1460 =: 1.02 X 1460 = 1488 gallons, ' ^28. 9' the true discharge for 30 ft. head. The most complex problems can be solved by the application of the foregoing principles. Where there are a number of mains, with cross-mains between, it will of course require the exercise of thought and judgment to decide which lines are effective and how they should be classified. The experienced engineer can usually make a very close approximation in such cases without going into the detail of every line, and thus avoid much tedious work. 2O GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. The following case is given as an illustration of a pos- sible treatment of a rather intricate problem. __.500Q-10 : ' ^. A Fig. 2 2000 '12" >/ f 4000'8" E 1 ^^ w 3000'S" y F p^- - X z D With the above arrangement of piping and reservoirs at A, what will be the loss of head caused by a draft of ten i-inch* hydrant streams or 2000 gallons per minute from hydrants on line CD ? As the hydrants will be fed both ways through short lines, the friction in line CD can safely be ignored, although if the location of the hydrants were fixed it could be computed. The cross-lines Bo, woe, yz can also be ignored. The slight effect that they would have would be on the safe side. Assume the loss of head in lines BC, BE, and BmF\ it will be the same in the three.f One of these lines, BmF, is a compound main, and the discharge will be assumed here to start with. Assume 200 gallons; then in line Bm, 2000' of 6", head 8.6 feet " " mF, 3000' " 8", " = 3.2 " Total, 1 1.8 " * See Table No. 12 for definition of fire-steam. f This would only be true in case the pressure in the line CEF was the same at all points ; it is assumed here that it is practi- cally so. FLOW OF WATER IN PIPES. 21 Therefore the discharge with assumed head of n.8 feet will be : in line BC, 5000' of 10", = 562 gals. " " BE, 4000' " 8", - 353 " " " BmF = 200 " Total discharge through BC, BE, and BmF, 1115 " Loss of head in AB due to flow of 1115 gallons : 1115 gallons through 2000' of 12", = 6.8 ft. Add for BC, BE, and BmF, 1 1.8 " Total head for assumed discharge in system ABCF, 18.6 " It is evident that the loss of head in line APD must be equal to that in ABCF. As APD is a compound pipe, assume flow to be 1000 gallons. The head due to this flow is as follows : In AP, 1500' of 12", 4.2 head " PD, 6000' " 14", 7.8 " Total, 12 " Then discharge due to 18.6 feet, or head in ABCF, is /i8.6W I I X 1000 = 1270 gallons per minute. Therefore total discharge with loss of head of 18.6 ft. is : In system ABCF, 1115 gallons per minute " ' " APD, 1270 " Total, 2385 22 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. Finally, to find loss of head due to the required dis- charge of 2000 gallons per minute, use the formula /required discharge V 1 / 6 , ,. , X assumed loss of head ; Vassumed discharge/ discharge or /2000V 1 / 6 X 18.6 = .728 X 18.6 = 13.50 ft. head, ans. If it were desired to find loss of head due to a draft of ten i-J-inch fire-streams or 2500 gallons per minute, or any other quantity, use the same formula. If the discharge for a given loss of head is required in the above system, use the formula / given head W" . [ required . r -H X assumed discharge = < ,. \assumed head / ( discharge. In the above case, suppose that a head of but 15 feet were available, or that it were not desirable to allow of a greater loss of head ; then (ic ^ 6 /" X 2385 = .86 X 2385 = 2050 gallons, the discharge for a loss of head of 15 ft. Note. In all of the cases given in this chapter the 2d power instead of the n/6, and the 1/2 power instead of the 6/1 1, could be used ; and if the assumed quanti- ties closely approximated the real ones, the difference in results would be very small. The following table shows the difference in a few cases. When the friction caused by a draft of 1000 gals, is 14.50 : FLOW OF WATER IN PIPES. 23 Gals. Head by n/6 power. Head by 2d power. 1000 14-5 J 4-5 500 4-06 3.62 1200 20.30 21.00 1500 30.60 32.60 2000 51-60 58.00 When a head of 10 feet would cause a discharge of 1000 gallons, the difference for different heads is shown by the following table : Head. Discharge by 6/n power. Discharge by 1/2 power. 10 1000 1000 5 68 S 707 12 noo I0 95 15 1245 1225 20 1458 1415 The advantage of using the 2d power instead of the 1 1/6 is that the formula can be computed on the slide- rule with one setting, which would be as follows : For formula /required discharge \ a , rr-. : X assumed head = true head Vassumed discharge / set assumed discharge on lower slide over required dis- charge on lower scale, and over assumed head on upper slide read true head on upper scale. For formula / given head V/ 2 j- *_ X assumed discharge = true discharge \assumed head/ set assumed head on upper slide under given head on upper scale, and under assumed discharge on lower slide find the true discharge on lower scale, 24 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. OLD OR TUBERCULATED PIPE-LINES. It is now generally recognized that the loss of head caused by the friction in pipes increases with the age of the pipes. There is, however, no generally accepted formula that takes into account the factor of age.* Kutter gives a variable coefficient n for roughness of the sides of channels. In old pipe-lines there are gen- erally no data relating to the condition of these sur- faces, and the age of the line is the only easily deter- mined element. While this is not so satisfactory as the known condition of the pipes in question would be, which is affected by the character of the water and of the coating as well as by the age of the line, there are no other data relating to this condition given in the recorded experiments upon old pipes. It is desirable to have some ready means of approximately estimating the effect of age upon pipe-lines, as the formulas given for clean pipe cannot be safely used for old lines without correction. Although good judgment should always be used, some mathematical assistance is needed. COEFFICIENTS OF LOSS OF HEAD. Wishing to provide some systematic means of correct- ing the results given by the diagrams for clean pipes, so that they may be safely used for old lines, the author made a study of all the experiments upon old pipes that were available to him. As might be expected, the results of these are conflicting and do not furnish a satisfactory basis for a mathematical formula. * Since the above was written a valuable book by Edmund B. Weston, " The Friction of Water in Pipes " (D. Van Nostrand Co ), giving the friction caused by the flow of water in pipes, has been published, which gives coefficients for friction in old lines. FLOW OF WATER IN PIPES. 25 The graphical method seemed to be the most satisfac- tory one for arriving at some conclusion. Accordingly, the results of the experiments expressed as a percentage of the excess of loss of head by friction over the loss under the same condition in new pipes, as computed by the formula adopted in this book, was plotted as shown on diagram A, using the horizontal scale for the percent- age of excess of friction, and the vertical scale for the velocity of flow in feet per second. The result of each experiment of a series is represented by a small circle, and all results of a series are connected by a broken line. By graphical comparison, curves * that seemed to the author to represent the average results of the ex- periments were drawn for different ages of pipe. It may be seen that with a few exceptions the general direction of the broken lines is inclined from the lower left- to the upper right-hand corner. This indicates that the per- centage of increase of friction for any age of pipe is not constant, but increases with the increase in velocity. The rate of this increase (or the inclination of the broken lines) differs greatly in the experiments. The full lines drawn to represent this are inclined in the ratio of one horizontal to two vertical, and represent an increase in the percentage which varies with the square root of the increase in velocity. This rate almost coincides with that of one of the most carefully made experiments, viz., that upon a 48- inch pipe by Mr. Desmond Fitz-Gerald. It is also an approximate average of all the experiments. It is indicated by the experimental results that the excess of friction is greater in small pipes than in large * As the base of the diagram is logarithmic, these curves are straight lines. 26 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. ones, other conditions being the same. There are not sufficient data to make any classification based upon size. In giving the value to the different ages, an endeavor was made to use the approximate average for all sizes. The curve for fifteen years nearly coincides with the above-named experiment on a 48-inch pipe eighteen years old. Table No. 4, giving coefficients by which to multiply the friction loss in new, clean pipe, was compiled from diagram A by adding the percentage of excess of friction- loss, as shown by the curves for several ages of pipe, to unity. For example, the excess of head in a pipe ten years old with a velocity of flow of 2 ft. per second is 30 per cent; adding this to TOO per cent (the friction-loss in clean pipe) makes 130 per cent, or written as a coeffi- cient, 1.30. COEFFICIENTS OF DISCHARGE. Diagram B was designed as a means of deriving coefficients of velocity or discharge from diagram A. This is also constructed upon a logarithmic base. The horizontal scale represents the loss of head by friction ; the vertical scale, velocity in feet per second. Line CD represents the friction in new, clean pipe when a velocity of one foot causes a loss of head of one foot (it is immaterial what value is given to the loss of head caused by a velocity of one foot) as calculated by the formula used in this book. Lines CD', C"Z>", etc., represent the friction in pipes of different ages. The small circles through which these lines are drawn were obtained by multiplying the loss of head given at the intersection of CD with line of velocity upon which FLOW OF WATER IN PIPES. 27 the circles are plotted, by the proper coefficient frorh diagram A. For example, on velocity = 2 the loss of head given at CD is 3.56 ft.; this X (14 per cent from diagram A at v = 2 and age 5 years -f- unity = coef. 1.14) = 4.06 ft., the loss of head in pipes five years old with v = 2. The coefficients of discharge are derived as follows: On vertical lines EF drawn through the intersection of velocity lines with CD, find the velocity at the inter- section of CD', C"D",etc. t as for example: On vertical EF drawn through intersection of v = 3 with CD find the velocity at C" D" = 2.55. This is the velocity in a pipe ten years old under the conditions and with the head that produces a velocity of 3 ft. in new pipes. Then = 0.85 = coefficient of discharge in a pipe ten years old. The coefficients in Table No. 5 were derived in this manner. The velocities in the first column are those that would be produced in new, clean pipes under the conditions of the problem, as given by the diagrams for flow of water in clean pipe. It is not claimed that there is mathematical exactness in these coefficients as applied to any particular case. It is hoped that they represent an approximate average of the results of recorded experiments, and may, when intelligently used, be of some assistance in the solution of problems relating to old lines. Following is a list of the experiments the results of which were used, with references to the sources from which they are taken. 28 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. ] Experi- menter. Place. Diam- eter. Age, Years. Source from which Data were taken. Fitz-Gerald Boston 48" 18 Trans. Am. Soc. C. E., vol. xxxv. Darcy Paris // old ^1 Ehmann .. . Stuttgart... 10" 6 Iben Hamburg.. " n 12 Ganguillet & Kutter, 2 Flow of Water, translated // 15 by Hering and Trautwine. n 6" 25 j Leslie"!!". Edinburgh. 5" 3 I Hamilton Smith's Hydraul- 44 t .... " 6" 8 or 9 ics. Weston ... Providence 6" 4 Simpson . . 2" 7 " 2" Unknown 2" 4 Flow of Water in Pipes. 9" 13 Weston, Trans. Am. Soc. 30" {Heavily C. E., vol. xxn. Green. . Brooklyn. .. 36" tuber- culated Gale Glasgow.. . 48" 8 Forbes Brookline.. *! and 18 [journal N. E. Water-works (16" [ Assn., vol. vi. " " 16" 18 j Hastings... Cambridge. J 3 and 8 Same, vol. vm. ( 36" Coffin Randolph.. 14" 8 See following table. Note. All experiments were upon cast-iron pipe. RESULTS OF EXPERIMENTS UPON A PIPE-LINE EIGHT YEARS OLD AT RANDOLPH, MASS. A 2,7do ft. 14" Pipe. Loss of Head by Friction. Actual. Computed. Per Cent of Computed. Per Cent in Excess. T .80 T.50 2.85 53 2 1.40 6-95 7.90 88 3 1.82 13.80 12.70 1 08 8 4 2.09 18.30 16.40 in II 5 2-34 23.50 20.30 116 16 6 2.50 28.00 22.70 123 23 7 2.80 34-00 28. 121 21 FLOW OP WATER IN PIPES. 29 Examples of Use of Tables Nos. 4 and 5. CASE I. When the discharge is given and the head re- quired in an old line. Example : What is the friction-head in a 20-inch pipe line 10,000 ft. long, 15 years old, with a discharge of 2000 gallons per minute ? Find velocity and friction-head for clean pipe from diagram No. 10: V = 2; friction-head = 7.6. On Table No. 4 find opposite V = 2 and under 15 years :he co- efficient = 1.47; then 7.6 X 1.47 = 1 1. 2 feet-head. (a) What is the head in the above example with a dis- charge of 5500 gallons per minute ? Friction-head for clean pipes from diagram 10 = 49.5; V = 5.6 ft., coefficient fiom Table No. 4 = 1.80; 49.5 X i. 80 = 89 ft.-head. CASE II. When the head is given and the discharge required. Example : What is the discharge of a lo-inch pipe-line one mile long, 20 years old, with head of 40 feet ? From diagram No. 5 find discharge 1070 gallons; v 4.4. On Table No. 5, opposite V = 4 to 5, and under 20 years, find coefficient 0.72; then 1070 X 0.72 = .770 gallons. For compound and complex mains use the same meth- ods, taking an approximate average of the velocities in the different sizes. As an approximate method and for preliminary work the coefficients for an age of 20 years and velocity of 3 feet per second may be memorized and used. They are 1.80 for loss of head and 0.75 for discharge. An age of 20 years and velocity of 3 feet are very close approximations to the values used in ordinary practice. 30 GRAPHICAL SOLUTION O? HYDRAULIC PROBLEMS. Note. Nothing has been said in this chapter about the loss of head caused by bends, branches, and gates in pipe-lines. The author is of the opinion that in lines of new pipes with the ordinary velocities the friction loss as given on the diagrams will include all loss of head in clean-coated pipe-lines as usually laid.* The coefficients for old lines are intended to apply to lines of pipes with the usual special castings and gates. In designing all engineering works it is customary to use a factor of safety. There should be no exception to this custom in the design of pipe lines or systems. It cannot be considered good practice to design a line that will according to the formula used discharge exactly the required amount of water. For instance, if it is abso- lutely necessary that a line shall discharge ten million gallons of water per day, safety would require that a di- ameter be chosen which, when computing the discharge as carefully as possible, will deliver at least eleven mil- lions per day, or an increase of 10 per cent. This is a small factor of safety as compared with that used in other branches of engineering. It is generally the case that the amount supposed to be required is only an approximate estimate of that actu- ally required. In such, as in all cases, the engineer must use his judgment gained from study and experience. No formula, table, or diagram, however correct within their limitations, can alone assure successful design. They are but useful tools for the skilful workman. Note. For sizes of pipes not given on diagram, take area and value of r from table No. 7 and find solution on diagram No. 25 for channels, using n = .on. * See paper on Friction in Several Pumping-mains in the Journal of the N. E. W. W. Assoc., vol. x, No. 4. CHAPTER IV. FLOW OF WATER THROUGH SHORT TUBES. THE flow of water through short tubes is treated in this book on account of its relation to the entry-head of long pipes. By a short tube is understood a pipe that has a length equal to about three times its diameter. The loss of head caused by the flow of water through short tubes includes the loss due to friction at entrance and the head required to generate velocity, and will be designated sim- ply as entry-head hereafter in this book. This loss is constant in any given pipe for any given velocity without regard to the length of the pipe-line. Consequently it must be determined separately from the loss of head caused by friction in a long pipe which is in direct pro- portion to its length. In very long pipe-lines the entry-head is an insignificant element in the problem, but in comparatively short lines, with high velocities, it is an important factor, and should not be ignored. The diagrams for entry-head are constructed from the formula a v = o(2gh)*, or h = with the following value for o as given in Hamilton Smith's Hydraulics: 32 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. Circular pipes: Mouth flush with side of reservoir 0.825 " projecting into reservoir, with square ends -7 I 5 Bell-shaped mouthpiece, small velocities 0.950 large velocities '995 2g was taken as 64.36. There are two diagrams on this subject: No. i for pipe from 4 to 20 inches in diameter ; No, 2 " " " 20 " 60 " The diameter of the pipe is shown by a curve from the upper left-hand corner; the loss of head in feet by hori- zontal scales; the one at the top of the diagram for pipes with the mouth flush with the side of the reservoir; that at the bottom for pipes projecting into the reservoir and with square ends. These are the types most in use. If it is required to find the loss for bell-shaped mouth- pieces, multiply the head for flush pipes, or that given at the top of the diagram, by 0.75. The scale of gallons per minute is given on the right-hand side of the diagram, and the scale of cubic feet per second on the left. Examples. CASE I. Diameter of pipe and discharge given, to find the head. Example : What is the loss of head in a lo-inch pipe dis- charging 4500 gallons per minute or 10 cu. ft. per second when pipe is flush Ans. (from diagram No. 18, top scale), 7.7 ft. OF THE ' UNIVERSITY FLOW OF WATER THROUGH SHORT TUBES. 33 (a) What would be the loss of head in the above ex- ample, if the pipe projected into reservoir ? Ans. (from lower scale), 10.25 ft. (b) If the mouth of the pipe were bell-shaped ? Ans. 7.7 ft. loss for flush pipe X 0.75 = 5.77 ft. CASE II. Diameter and head given, to find discharge in a short tube or pipe. Example : What will be the discharge through a short tube 12 inches in diameter, flush ends, under head of 4.2 ft.?* Ans. (using upper scale of diagram No. 18), 4850 gallons per minute. If the end projects into reservoir, in above exam- ple, take the head, 4.2 ft. on lower scale, and the dis- charge will be 4190 gallons per minute. CASE III. Available head and required discharge given, to find diameter of pipe. x Example : What size of pipe is required to discharge 2000 gallons per minute with flush ends and head of 2 ft.? Using upper scale of diagram No. 18 for head, the intersection of h = 2 and q = 2000 falls between the curves of 8 and 10 inch pipes ; therefore of the usual commercial sizes a lo-inch pipe would be required. CASE IV. The diameter of a long pipe-line and its dis- charge given, to find total loss of head. Find the loss due to friction in the pipe by the methods of Chapter III, and add to that amount the loss of head caused by entry and velocity found on the diagrams for entry-head, as in Case I : the sum will be the total head. Example : What is the total loss of head in a pipe-line * Head is measured from surface of water to centre of pipe, 34 GRAPHICAL SOLUTION OF HY.DRAULIC PROBLEMS. 14 inches in diameter, 1000 ft. long, flush end, discharg- ing 4000 gallons per minute ? The friction loss, taken from diagram No. 7 for flow of water in long pipes, will be. 16.35 ft- The loss for entry, as shown on diagram No. 18 for entry-head, will be 1.60 " Total, 17.95 " CASE V. The diameter and length of a long pipe and the available head given, to find discharge. This problem cannot be solved directly, as the loss of head caused respectively by the entry and the friction is not known. The methods described in Chapter III, Compound Mains, Case II, may be adopted, and a discharge as- sumed. Example: What will be the discharge of a i6-inch pipe 1000 ft. long, with an effective head of 6 ft., with flush end? Assume a discharge of 3000 gallons per minute ; for this quantity, by diagrams No. 8, the friction-head will be 4.95 ft. No. 18, the entry-head will be 0.5 " Total, 5.45 " Then, following method of Chapter III, Case II, of Compound Main, ( given head ) T / 2 ,. \ , -=-5 , ,. r - [ X assumed discharge ( head due to assumed discharge j / 6 \ I/2 = true discharge, or ( ) X 3000 = 3150* gallons. \5'4S / * See note on page 22. FLOW OF WATER THROUGH SHORT TUBES. 35 The loss of head due to each cause can be taken from the diagram after true discharge is found. In above ex- ample, for a discharge of 3150 gallons, from diagram No. 8, friction-head = 5.40 ft. " " entry-head = .58 " Total, 5.98 "* Note. It will be found advantageous in such cases as the above, where the pipe is comparatively short, to find the friction for a longer length ; in this case the friction- head can be found more closely by finding it for 5000 ft. instead of 1000, and dividing result by 5. The diagrams for entry-head are very convenient for finding loss of head in circular penstocks and other short pipe-lines. In such cases both the friction and the entry-head must be considered. * The entry-head increases as the second power of the increase of velocity, and when it is as great as the friction-head the second power will be as correct as the n/6. CHAPTER V. RECTANGULAR WEIRS. THE diagrams for the discharge of weirs are computed from the well-known Francis' formula, Q = clffl 2 . For heads above 0.5 ft., c = 3.33. For heads below .5, c = the values given in the follow- ing table compiled from the paper of Fteley and Stearns,* which describes their experiments on low heads: VALUE OF C FOR LOW HEADS IN Q = Head. c. Head. c. 0-5' 3-33 0.2' 3.388 0.4 3-337 0.15 3-430 0-3 3-353 O.I 3.528 0.25 3.368 0.06 3-750 END CONTRACTIONS. For each end contraction deduct .ih from the length, or ,2h where there is contraction at both ends. Note. Hamilton Smith, Jr., gives a tables of values for the coefficient c in the formula Q = \c(2g)*lh. \ * Published in vol. xu, Transactions of the Am. Soc. Civil Engineers. f This formula can be reduced to the form of Francis' formula by substituting values of c and 2g. 36 RECTANGULAR WEIRS. 37 Values are given for weirs both with and without end contraction. These coefficients give results slightly different from those obtained by Francis' formula (from one to three per cent), and it is probable that in cases where the weirs are constructed entirely in accordance with the given conditions that the results may be more accurate than with a constant coefficient, even where a correction is made for end contraction, as above. See " Hydraulics," by Hamilton Smith, Jr., for full discussion of the subject and description of experiments. Smith says that no weir measurements of water should be made with h less than 0.2' where accuracy is essential. EXPLANATION OF WEIR DIAGRAMS. Very little explanation is necessary for an understand- ing of these diagrams. They are three in number: No. i gives lengths from o to 4 ft. " heads " o.oi " .5 " discharge " o " 800 gals, per min. " " " o " 1.77 cu. ft. per sec. No. 2 gives lengths from o to 8 ft. " heads " 0.05 " 1.50 ft. " discharge " o " 8000 gals, per min. " " o " 17.7 cu. ft. per sec. No. 3 gives lengths from o to 16 ft. " heads " o.i " 2.5 "* ". discharge " o " 40,000 gals, per min. " o " 89 cu. ft. per sec. * Caution : It is not safe to rely implicitly upon the results with heads greater than 2 feet. Mr. Francis considered the limits of his formula to be from 0.5 to 2 feet. 38 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. The lengths of the weirs are given in feet by a vertical scale on the side of the diagrams ; the horizontal scale at the top gives discharge in gallons per minute ; the scale at the bottom in cubic feet per second, and the oblique lines radiating from the upper left-hand corner give the head in feet. Examples. Example i : What is the discharge of a weir 3 ft. long, with head of 0.26 ft., with contraction at each end ? First find corrected length for end contraction, or 3 .2h = 2.948 ft. On diagram No. 22, at the intersec- tion of lines representing the above values, viz., / = 2.948 and h = 0.26, find discharge = 597 gallons per minute, or 1.325 cu. ft. per second. Example 2: In the above example, if the head is 0.45, what will be the discharge ? Corrected length = 3 (.2h = .09) = 2.91 ft. It will be seen that the intersection of / = 2.91 and h 0.45 does not fall on diagram No. 22; therefore look for it on No. 23. It is 1315 gallons per min., or 2.9 cu. ft. per second. Diagram No. 23 is on a smaller scale than No. 22, and if it is desired to determine the discharge more closely by using No. 22, proceed as follows: Divide the corrected length by 2 or any convenient number; thus, 2.91 -T- 2 = 1.455; find the discharge for this length 655 gallons; multiply this result by 2, or the number used to divide length, 655 X 2 = 1310. On one or other of the three diagrams results for any head up to 1.40 ft. and any length up to 16 ft. can be read directly; also any head up to 2 ft. with lengths under 9.5 ft. RECTANGULAR WEIRS. $$ If it is required to find results where the intersection of / and h does not fall on the diagram that it is desired to use, or if the length is greater than that given on any of the diagrams, advantage can be taken of the principle that the discharge is proportional to the length, and the result found as in Example No. 2 (a). N. B. If there is end contraction, the length must be corrected for it before dividing. Example 3: What is the flow over a weir 20 feet long with contraction at each end and head of 0.2 ft? Corrected length, 20 (.2/1 = .04) = 19.96 ft. To find result on diagram No. 22, where the length is but 4 feet: 19.96 -=- 5 = 3.992 ft. discharge with h = .2 of 554 gallons per min. Multiply this by the divisor used, 554 X 5 = 2770 gallons per min., total discharge. It is often desirable to know how long to construct a weir in order to discharge a certain maximum quantity of water and not exceed a certain head. Example 4: What should be the length of a weir to discharge 2000 gallons per min., h not to exceed 0.6 ? On diagram No. 23 at intersection of h = .6 and q = 2000, find length = 2.96. A weir 3 feet long will meet the re- quirements, including correction for end contraction. All problems connected with rectangular weirs can be solved by these diagrams, and with sufficient accuracy for all except the most careful work. CONDITIONS TO WHICH THE DIAGRAMS APPLY. The conditions of weirs to which the formula and diagrams apply are as follows: The crest must be horizontal, the sides vertical, and the plane of the face of the weir must be approximately at right angles to the line of flow. The edges of the 40 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. weir must be so thin that the escaping water shall only come in contact with the up-stream corner edges. For accurate measurement stiff metallic plates with smooth, straight edges should be used. For weirs where there is no end contraction the sides of the canal should be pro- longed down-stream beyond the crest, but should not ex- tend below its level, as this would prevent the access of air under the escaping vein. For full contraction the water must discharge freely into the air. The air must be given free entrance under the escaping vein. The head should be measured 6 feet up-stream from the weir upon a stake or fixed point set level with its crest. For weirs with end contractions the sides of the weir openings should not be less than twice the maximum head from the sides of the canal, and the same distance from the bottom. The length of the weir should not be less than three times the maximum head. When the sheet falls on an apron, the latter should be not less than the total head below the crest. If the discharge falls into a body of water the latter may be level with the crest.* The error will not exceed one per cent from this cause if the weir is submerged to the extent of 15 per cent of the head on the crest.f WIDE CRESTS. It often happens that measurements are made over weirs with wide crests. Messrs. Fteley and Stearns made a series of experi- ments on the flow over wide crests, which are described * Smith says that if air is admitted freely behind the falling sheet the surface of water may be nearly to level of crest. f See "Submerged Weirs" in the paper of Fteley and Stearns referred to above. RECTANGULAR WEIRS. 41 in the paper already referred to. As the result of their work they give a table of corrections to be applied to the depth or head on a wide crest in order to obtain the depth on a sharp crest which will pass the same quantity of water. The results of their table as worked out are given in the form of a percentage of the discharge over a thin-plate weir on diagram No. 21. To use this diagram take the discharge due to the measured head from the diagrams for thin crests in the usual manner, and multi- ply it by the percentage given on diagram No. 21 for given width of crest and head. Example : What is the discharge of a weir 5 ft. long, without end contraction, and with head of 0.55 over a crest 6 inches wide ? The discharge from the diagram No. 23 for this crest 35 gallons per minute. Multiply this by the per- centage from diagram No. 21 for 6-inch crest with head of 0.55 ft. or 3050 X 0.90 = 2745 gallons per minute. The diagram gives crests from 2 to 24 inches wide, and also gives the proper percentage for overflows of the type of the Lawrence dam worked out from Mr. James B. Francis' formula for that type, Q = 3.0 ^oS//* 1 ^ 3 . This dam is 3 ft. wide on top, with a slope on the up- stream side of about 3 horizontal to i vertical. VELOCITY OF APPROACH. When there is an appreciable current towards the weir the velocity of approach must be considered if accuracy is required. This subject is thoroughly discussed in Smith's Hy- draulics. In the paper of Fteley and Stearns, before re- ferred to, is given a description of a series of experiments 42 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. and their conclusions from them. Any one wishing to make extremely careful measurements is referred to the above, and also to Francis* work on hydraulics, " The Lowell Experiments." For ordinarily careful work Table No. 6 gives correc- tions to add to the measured head before taking results from the diagram. This table is based upon the coeffi- cients given by Fteley and Stearns for weirs with and without end contractions, and for different heads and depths below the crest. The mean velocity of approach is found by dividing the approximate discharge over weir by the area of cross- section of the channel. The head due to this velocity is found by the formula h = when v mean velocity of 2 g approach and zg = 64.36. The coefficients of Fteley and Stearns vary from 1.33, to 1.87 for weirs without, and 1.88 to 2.42 for those with, end contraction, and are used to multiply h as found above, and the product is added to the measured head. The method of using the table is as follows : Find the discharge in cubic feet per second due to the measured head from the diagram ; divide this discharge by the area of the cross-section of the channel, to find mean velocity of approach ;* find the nearest number to ve- locity as found in column i of Table No. 6, and oppo- site it, under the column giving the conditions nearest those of the case, find the correction in feet to be added to the measured head ; finally, find on the diagram the result due to the corrected head. Example : What will be the discharge over a weir 4 ft. long with end contraction and measured head of 0.6 ft. ? * The velocity may also be obtained by observations with floats or a current-meter. RECTANGULAR WEIRS. 43 The channel is 6 ft. wide and 1.5 ft. deep = 9 sq. ft. area, depth below crest i foot. Corrected length = 4 (.2h or .12) = 3.88 ft.; the ap- proximate discharge, from diagram No. 23, due to .6 head is 6 cu. ft. per second. Velocity of approach = 6 -T- 9 = 0.667 f eet P er second. ' On Table No. 6, for weirs with end contraction, opposite v = 0.7, and in column for depth below crest = i foot and head of 0.60, find the correction = 0.013 ; add this to the measured head, = 0.6 + 0.013 0.613. Then, with corrected length = 4 2h = 3.877, the discharge for corrected head of 0.613 6.2 cu. ft. per second or 2780 gallons per minute. Note. Fteley and Stearns give a diagram in the paper referred to from which the head to be added for velocity of approach can be taken directly, without first finding the mean velocity. This diagram only applies to weirs without end contraction. CHAPTER VI. FLOW OF WATER IN CHANNELS. THE engineer meets many problems connected with the flow of water in channels of other than circular sections. There is such variety in the shape of these sections that it is impracticable to construct diagrams for any special forms, ranging as they do from an egg-shaped sewer to a canal of trapezoidal section in earth. Instead of trying to treat these forms (except that of circular pipe-sewers) on special diagrams, a gen- eral diagram is constructed which gives the values of v =i velocity per second, s = slope or fall per 1000, and r = hydraulic mean radius. When any two of these three elements of the problem are known, the third can be read from the diagram. Thus all problems of flow in channels of any section can be readily solved. The basis of these diagrams (Nos. 25 and 26) is the Chezy formula: v = c(rsY' z , as in the case of circular pipes. For channels, however, the value of the coeffi- cient c is found by Kutter's formula, which is 1.811 . . .00281 "1 + 4..6+ \' | .oo28i - all symbols having the meaning given in Chapter II. 44 FLOW OF WATER IN CHANNELS. 45 This formula seems better adapted to channels than any other, on account of the different values that Kut- ter gives to n corresponding with different degrees of roughness of the surface of the channel. These values are given in the following table : VALUES OF IN KUTTER'S FORMULA. Character of Surface. Value of . Well-planed timber 0.009 Plastered with neat cement ) . . . , . > oio Also glazed pipe j Plastered with" mortar composed of one part sand to three of cement on Unplaned timber and uncoated C-I pipe 012 Ashlar and first-class brickwork .013 Second-class brickwork and well-dressed stone ... .015 Rubble in cement in good order 017 " inferior condition, and very firm and regular sides in gravel 020 Rivers and canals in good order, free from vegeta- tion 025 Same, having stones and weeds occasionally 030 " in bad condition, with much vegetation 035 There are two diagrams for channels : No. 25, for those having surfaces corresponding to Kutter's n .on and .013 ; No. 26, for surfaces corresponding to n = .025 and .030. On these diagrams the slope or fall per 1000 is given by curves from the upper left hand ; the hydraulic mean radius, or r, by a vertical scale on each side; and the velocity, or v, by a horizontal scale. 'On diagram No. 25 the upper scale is for ;/ = .on and the lower scale for n = .013 ; on diagram No. 26 the 46 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. upper scale is for n = .025 and the lower one for n = .030. For other values of n the following table gives the percentage to use. These percentages are only a mean of those for different slopes and values of / ; they are usu- ally not more than two or three per cent in error. For n = .015 and .017 and ,035 they are about 10 per cent in error for values of r = . i and 4 ; while for r = i they are about 3 per cent in error for all slopes and values of n. While it may not be considered that the table of percentages gives a very close result in some cases, it must be remembered that this is a subject that contains a very uncertain factor in the value of n. It is probable that the accuracy as given by the percentages is suffi- cient for all practical purposes of design or computation. When greater accuracy is desired, the reader is referred to the table giving values of coefficient c in v c(rs)*/ 2 in Hering and Trautwine's translation of " The Flow of Water," by Ganguillet and Kutter.* This gives the results of Ganguillet and Kutter's formula for said co- efficient, worked out for different slopes and values of r and s. TABLE GIVING PERCENTAGE OF v FOR DIFFERENT VALUES OF . For Diagram No. 25, upper scale When n .009, multiply or divide v by 1.25 n .010, v I.IO " tf=.OII, " " " 1. 00 " n = .012, " ^ tt .90 (t tt " " 9 " .80 " = - OI 5> " " V " .70 " n = .017, " tt v tt .60 John Wiley & Sons. FLOW OF WATER IN CHANNELS. 47 For diagram No. 26, upper scale. When n = .020, multiply or divide v by 1.28 " n = .025, ' ' " v " i.oo " n = .030, " " " v " .80 * = .035, " " " ? " .68 For other values of n than those given on diagrams : When v and s are given, multiply v as found on the diagram by the percentage in the table. When r and v are given, divide v by the percentage before reading result from the diagram. N. B. Always use upper scale of diagram for values of v when using percentages. Explanation and Examples. In using the diagrams, the hydraulic mean radius of the channel, or r, must be calculated. When two of the factors r, v, and s are known, the third can be found by inspection of the diagrams, s is given as the fall in 1000. CASE I. When r and s are given, to find v. Example i: What will be the velocity in a rectangular channel 6 ft. wide, 2 ft. depth of water, fall of 0.25 in 1000, with surface corresponding to n = .on ? Find the value of r : _^__ 6X2 _ f or find area and value p 6 + 2 + 2 ( of r on Table No. 8. On diagram No. 25, at the intersection of r = 1.2 and s = 0.25 per 1000, find the value of v on upper scale = 2.50. If the discharge is required, Q = va = 2.50 X 12 = 30 cu. ft. per second. 48 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. (a) If, in the above example, the surface corresponds with n .013, the velocity will be found on the lower scale, = 2. (/;) If the surface corresponds with n = .015, multiply v, as found in Example i, by the percentage taken from table on page 46, or v = 2.5 X .70 = 1.75. Example 2: What will be the velocity in a canal of trapezoidal cross-section, 10 ft. wide on the bottom, side slopes 3 to i, depth of water 5 ft., s = 0.125 per 1000, with surface corresponding to n = .030 ? Find area and value of r for this section on Table No. ii. Area = 125 ; r = 3.01. At intersection of r = 3.01 and s = 0.125, ^ n( ^ velocity on lower scale, diagram No. 26, = 1.12 per sec. CASE II. When r and v are given, to find s. Example: What is the slope required in a rectangular channel 6 ft. wide, with depth of water of i ft., surface n = .013, to produce a velocity of 3 ft. per second ? On Table No. 8 find area = 6 ft, r 0.75. On diagram No. 25, at the intersection of r 0.75 and v = 3, measured by lower scale, find the value of s = 1.02 per 1000. If in the above example the surface corresponds with n .020, what will be the required slope ? First divide v by the percentage from the table on page 47, for n = .020 = = 2.34. 1.28 At the intersection of r = 0.75 and v = 2.34, on upper scale of diagram No. 26, find value of s = 2.80. CASE III. When the quantity of water to be delivered and the slope is given, to find the size and shape of the channel. The problem is met in this form in designing channels. As v is involved in the quantity and area of the chan- FLOW OF WATER IN CHANNELS. 49 nel, this form of the problem cannot be directly solved by the diagram. Tables Nos. 7, 8, 9, 10, and n give the width and depth of rectangular and trapezoidal channels and diameters of circular ones, with areas and values of r t and will, with the diagrams 25 and 26, facilitate the solution of the above problem. Example : What will be the width, depth, and velocity of a rectangular channel, with slope of 0.5 in 1000 and surfaces corresponding to n = .013, to deliver 30 cu. ft. per second ? It is evident that there are many combinations of width and depth that will meet the conditions. A prefer- ence may be had, however, for a certain approximate width. Assume that to be 6 ft.; there will also probably be a limit to the allowable velocity, say 3 ft. This, with Q = 30, will set the minimum limit of area at IQ ft. Therefore _ = 1.66 deep ; now, with W = 6 and D = 1.66, r = approx. i.i from Table No. 8. With r i.i and s = 0.5, v = 2.68, but with the above area v must = 3. D must be increased ; try D = 1.80 ; then r approx. 1.13, area 10.8, v from diagram = 2.70 ; 2.70 X 10.8 29.2 cu. ft. per second. If a close design is required, a trifling change in D will give the correct result. In this way, by the use of the tables and diagrams, the suitable size and shape of a channel is easily found. CHAPTER VII. PIPE-SEWERS. THE discharge and velocity of pipe-sewers are given on diagram No. 27 for sizes from 6 to 24 inches, when running full ; when running half full, velocity is the same, and the discharge is one-half. This diagram was constructed from Kutter's formula, and gives results for two values of the coefficient n. The scale at the top of the sheet gives the fall or slope in feet per 100 when '=.oii. The scale at the bottom gives slope when n = .013. The diameter of the sewer is given by curves originating at the upper left-hand corner. The discharge in gallons per minute is given by a vertical scale on the right, and cubic feet per second by a vertical scale on the left. The velocity in feet per second is given by the broken curved lines. EXPLANATION OF DIAGRAM. CASE I. Size of pipe and slope per 100 ft. given, to find velocity and discharge. Example : What will be the discharge of a 10-inch pipe-sewer, running full, with a slope of 0.48 per 100, with surface corresponding to Kutter's n = .on ? At the intersection of diameter = 10 inch and slope = 0.48 per 100 taken on the upper scale, find the dis- charge = 790 gallons per minute, or 1.78 cu. ft. per sec- ond, with velocity = 3.25. PIPE-SEWERS. 51 If in the above example the surface corresponds with n = .013, use the lower scale for slope, and the answer would be 650 gallons per minute ; velocity = 2.60. CASE II. The slope per 100 and required velocity or. discharge being given, to find the size of the pipe. Example: What size of pipe is required to secure a velocity of 3 ft. per minute when the slope is 0.32 per- 100 and the surface corresponds with n .on ? At the intersection of velocity = 3, and slope on the upper scale =- 0.32, find next larger size of pipe = 12 inch. If the required discharge is given instead of the ve- locity, the method of solution is the same. CASE III. Size of sewer and required velocity or dis- charge given, to find fall per 100. Example: What is the necessary fall in a 15-inch sewer to produce a velocity of 2.5 ft. per second = dis- charge 1375 gallons per minute, with surface correspond- ing to n = .013 ? At intersection of v 2.5 and diameter = 15 inch, find value of fall read from the lower scale = 0.24. Note. When it is required to find velocity more closely than can be read from velocity-curves, it may be done by reading the discharge instead and taking veloc- ity from Table No. 2, which gives the number of gallons for different velocities in the different diameters. Note. For sizes of sewers, either pipe or brick, not given on diagram No. 27 use table No. 7 to find area and value of /, and solve problem by diagram No. 25 or 26, using such value of n as the case requires. CHAPTER VIII. FIRE-STREAMS AND DISCHARGE OF NOZZLES. IN designing pipe systems for fire-protection, it is desirable to know the pressure or head required at the hydrant to throw a stream to a given height, and the quantity of water required for such stream. Mr. John R. Freeman, M. Am. Soc. C. E., in a valua- ble paper published in the Transactions of the Am. Soc. of Civil Engineers, vol. xxi, describes and gives the results of a great number of careful experiments made by him upon the pressure and discharge of fire-streams.* In this paper are tables giving the height and discharge of fires-treams under various conditions of pressure, length of hose, size and shape of nozzle, etc. Table No. 13 in this book is compiled from Mr. Freeman's tables and gives the head required at the hydrant to deliver fire-streams through lengths of hose varying from 50 to 500 ft.; the height to which 'the streams reach, and the number of gallons per minute required for each. The table gives the results for i-, i-J-, and i^-inch smooth nozzles with the ordinary best quality of 2^- inch rubber-lined hose. * A similar paper by the same author with the same tables was also published in the Journal of the N. E. Water-works Associa- tion, March 1890. 52 FIRE-STREAMS AND DISCHARGE OF NOZZLES. 53 This table is intended for ready reference, and com- prizes those sizes of nozzles in most common use. The reader is referred to the paper mentioned above for more extended tables and a discussion of the subject. Discharge of Hose-nozzles. Diagram No. 31 is con- structed from tables given in the paper referred to. It gives the discharge of hose-nozzles through 50 and 100 ft. of 2^-inch ordinary best quality of rubber-lined hose, and the head or pressure indicated by a gauge at the hydrant. The diagram may be used when the approximate quantity of water delivered by a pump or flowing through a pipe is to be found. Mr. Freeman says that when the hose corresponds with the conditions of his tables the results are close enough for practical purposes, and when careful judgment is used to interpolate between the values of rough and smooth hose, results within 5 per cent can be obtained. The diagram only gives results for ordinary best quality rubber-lined hose with the inside smooth. When testing pumps or lines it is advis- able to use that class of hose. Fifty feet of hose should be used in preference to one hundred. Full curves on diagram represent the size of the nozzles with 50 ft. of hose; the dotted curves with 100 ft. Diagram No. 32 gives the discharge of nozzles and the head or pressure as indicated by a gauge at the base of the play-pipe. In this case the effect of friction in the hose is eliminated, and with smooth, true, carefully- calibred nozzles very accurate results can be obtained.* This diagram is constructed from the formula v o(2gh) T / 2 , using o = 0.99. * See paper on The Nozzle as an Accurate Water-meter, by J. R. Freeman. Trans. Am. Soc. Civil Engineers, vol. xxiv, 1891. 54 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. Mr. Freeman found this coefficient to be 0.995 as tne result of his experiments. On the diagram one curve represents two sizes of nozzles, the larger just double the diameter of the smaller, as f" and i*". The scale of gallons is given for the smaller sizes; it must be multiplied by four for the larger ones. CHAPTER IX. MISCELLANEOUS DIAGRAMS. UNDER this head there are several diagrams that the author has found useful in his practice, and that may be called hydraulic diagrams. HORSE-POWER. Diagram No. 28. This diagram gives the horse-power for falling water, and also the horse-power required to raise water. The vertical scale at the right of the diagram gives the quantity of water in cubic feet per second ; that at the left gives the same in gallons per minute. The horizontal scale at the top gives the horse-power as represented by the total energy or 100$ efficiency; that at the bottom gives the same with an efficiency of 75$, or the amount usually taken as the efficiency of a good water-wheel. CASE I. When the quantity of water and effective head are given to find the power. Example : What is the total horse-power of 8.5 cu. ft. per second with an effective head on wheel of 13 ft. ? At intersection of cu. ft. per sec. = 8.5 and head = 13, find value of horse-power as read on upper scale = 12.6. (a) In the above example what would be the effective power from a wheel with an efficiency of 75$ ? The answer, read from the lower scale, = 9.4 horse-power. 55 56 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. CASE II. Power required to pump water. When the quantity to be pumped and the total head to be pumped against are given to find the effective horse-power required. Example : What is the power required to pump 2000 gallons per minute against a total head on the purnp of 150 ft. ? Answer taken from upper scale = 76 horse-power. DISCHARGE OF SINGLE PUMPS DIAGRAM No. 29. This diagram is designed to give the capacity of recip- rocating pumps or pumping-engines, either single, duplex, or triplex. The horizontal scale at the top and bottom of the sheet gives the capacity in gallons per minute; the horizontal scale between the two parts of the diagram gives the revolutions per minute. The vertical scale on the sides gives the piston-speed in feet per minute. The oblique lines radiating from the upper left-hand corner give the diameters of pistons and plungers in inches. Those from the upper right-hand corner give the length of stroke in inches. The upper part of the diagram is for the smaller sizes of pumps, and the lower part for the larger ones. The diagrams are designed for single pumps with double- acting pistons or plungers. For duplex double-acting pumps multiply the discharge, as taken from the diagram, by 2. For triplex single-acting pumps multiply by i. CASE I. Given the diameter of plungers and piston- speed to find the discharge. Example i: W T hat is the capacity of a single-acting pump with a piston 10" in diameter and a piston-speed of 80 feet per minute, diameter of piston-rod 2 inches ? MISCELLANEOUS DIAGRAMS. 57 At the intersection of lines representing the above values of speed and diameter find the discharge as fol- lows: Discharge of 10" plunger, 80' piston-speed = 326 gallons Less 1/2 discharge of 2" rod at same piston-speed, from diagram, = ^ = 7 " Total, 319 " (a) If the above were a duplex pump, multiply by 2: 319 X 2 = 638 gallons per minute. When the length of stroke and revolutions per minute are given instead of piston-speed, first find, at the inter- section of line representing the number of revolutions with that of length of stroke, the line representing piston- speed, and follow this line to its intersection with that of the given diameter, and read the discharge at that point as before. Example 2: What is the discharge of a duplex pump with i2-inch plungers and 1 8-inch stroke at 35 revolu- tions per minute, diameter of piston-rod 3 inches ? At the intersection of 35 revolutions and i8-inch stroke find piston-speed (107 ft.) ; follow this to intersection of 1 2-inch diameter and there read gallons = 620 gals, per min. Less one half rod same speed = - 4 / = 20 " " " Total, 600 " " " As it is a duplex pump, multiply by 2 = 1200 gallons. CASE II. When the required discharge and piston- speed are known, to find the diameter of plunger. Example i: What size of plunger is necessary in a duplex pump to discharge 4000 gallons per minute; the 58 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. piston-speed not to exceed 130 ft., diameter of rod 3? inches ? As it is a duplex pump, first divide 4000 by 2 = 2000 -f- loss caused by 1/2 rod at given speed, or ^ = 2033 gallons. At the intersection of 2033 gallons and 130 ft. piston-speed find the nearest diameter given on the diagram, which is 20 inches. Then 20-inch diameter delivering 2033 gallons =127 ft. piston-speed. On the line representing this speed find the intersection of the length of stroke desired, and read the number of revolu- tions from the proper scale. In the above case, if the stroke is 30 inches, the number of revolutions will be 25.5. COAL KEQUIRED IN PUMPING ONE MILLION GALLONS OF WATER. DIAGRAM No. 30. This diagram is designed to give the weight of coal consumed in pumping-plants of different duties to raise one million gallons of water to various heads from o to 250. The head in feet is given on the horizontal scale. The duty is given by oblique lines from the upper left- hand corner. The weight of coal in net tons of 2000 Ibs. is given on the vertical scale on the left, and the same in gross tons of 2240 Ibs. on the right. The following examples will be sufficient explanation of the use of the diagram. Example i: What weight of coal will be consumed in a pumping-plant capable of a duty of 60 million ft.-lbs. per 100 Ibs. of coal to raise one million gallons of water against 150 ft. head? At the intersection of 60 millions duty and 150 ft. head find weight of coal = 1.04 net tons or o 92 gross tons. MISCELLANEOUS DIAGRAMS. 59 Example 2: What is the duty of a pumping-plant that raises one million gallons of water for 2.3 net tons of coal consumed against a total head of 200 ft. ? At intersection of 2.3 tons and 200 ft. head find duty = 36 millions approximately. If problems of higher duties than 100 million are to be solved, divide the duty by some number, find answer for the quotient, and divide weight of coal found by same number. Example : How much coal is required to pump one million gallons against a head of 175 feet with machinery capable of a duty of 125 million? Divide 125 million by 5 = 25 million; this with head of 175 will require 2.60 gross tons; this divided by 5 = .52 tons, answer for 125 million duty. Reverse the process to find duty when coal is given. 60 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. TABLE No. 1. TABLE OF U. S. GALLONS PER MINUTE AND THEIR EQUIVALENTS. Gallons per Minute. Gallons per 24 Hours. Cubic Feet per Second. Gallons per Minute. Gallons per 24 Hours. Cubic Feet per Second. I 1440 O.OO2 350 504000 0,780 10 14400 .022 360 518400 .802 20 28800 .044 370 532800 .825 30 43200 .06 7 380 547200 .847 40 57600 .089 390 561600 .869 50 72000 .121 4OO 576000 .892 60 86400 .134 4IO 590400 .914 70 lOOSoo .156 42O 604800 .936 80 II52OO .178 430 619200 .958 9 129600 .200 440 633600 .98! 100 I44OOO .223 450 648000 1.003 no 158400 245 460 662400 -025 120 172800 .268 470 676800 .048 130 187200 .290 480 691200 .069 I4O 2OI6OO 312 490 705600 .091 150 2I6OOO 335 500 72OOOO .112 1 6O 230400 357 510 734400 .136 170 244800 .380 520 748800 159 180 259200 .401 530 763200 .I8l 190 273600 .424 540 777600 .202 200 288000 .446 550 792000 .222 210 302400 .468 560 806400 .248 22Q 316800 .490 570 820800 .269 230 331200 .513 580 835200 .291 240 345600 535 590 849600 312 250 360000 557 600 864000 337 260 374400 579 610 878400 359 27O 388800 .601 620 892800 .381 280 403200 .624 630 907200 .402 290 417600 .647 640 921600 .426 300 432OOO .669 650 936000 449 310 446400 .691 660 950400 .470 320 460800 .713 670 964800 492 330 475200 736 680 979200 515 340 489600 .758 690 993600 538 TABLES. U. S. GALLONS AND THEIR EQUIVALENTS. 61 Gallons per Minute. Gallons per 24 Hours. Cubic Feet per Second. Gallons per Minute. Gallons per 24 Hours. Cubic Feet per Second. 700 IOO8OOO 559 1250 1800000 2.785 710 1022400 .58! I3CO 1872000 2.893 720 1036800 .602 1350 1944000 3-009 730 IO5I2OO .627 1400 2OI6OOO 3.U9 740 1065600 .649 M50 2088000 3.230 750 loSoooo .671 1500 2160000 3-341 7 60 1094400 .692 1550 2232OOO 3-453 770 1108800 715 1600 2304000 3.562 780 1123200 733 1650 2376000 3.676 790 1137600 .760 1700 2448000 3.785 800 1152000 .782 1750 2520000 3-899 810 1166400 .802 1800 2592000 4.010 820 1180800 .827 1850 2664000 4.121 830 1195200 .849 IOXX) 2736000 4-233 840 1209600 .8 7 I 1950 2808000 4-344 850 1224000 .892 2000 2880000 4.456 860 1238400 .918 2050 2952000 4-567 870 1252800 .936 2IOO 3024000 4-683 880 1267200 .960 2150 3096000 4.790 890 1281600 .982 22OO 3168000 4.901 900 1296000 2.005 2250 3240000 5-013 910 1310400 2.027 2300 33I2OOO 5-125 920 1324800 2.048 2350 3384000 5.235 930 1339200 2.073 24OO 3456000 5.347 940 1353600 2.093 2450 3528000 5.458 950 1368000 2.II4 2500 3600000 5-570 960 1382400 2.138 2550 3672000 5.681 970 1396800 2.161 2600 3744000 5-792 980 1411200 2.181 2650 3816000 5-904 990 1425600 2.202 2700 3888000 6.015 IOOO 1440000 2.228 2750 3960000 6.127 1050 1512000 2-339 2800 4032000 6.245 IIOO 1584000 2 450 2850 4104000 6-349 1150 1656000 2 562 2900 4176000 6.464 1200 1728000 2.672 2950 4248000 6-573 3000 4320000 6.684 62 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. TABLE No. 2. GIVING DISCHARGE OF CIRCULAR PIPES RUNNING FULL IN GALLONS PER MINUTE FOR DIFFERENT VELOCITIES. Velocity in Feet per Second. Diameter of Pipe in Inches. 4" 6" 8" 10" 12" 14" .2 8 .18 31 49 70 96 4 16 35 63 98 141 192 .6 24 53 94 147 211 288 .8 3i 70 125 196 282 384 I.O 39 88 157 245 352 480 .2 47 106 188 294 422 576 4 55 123 219 342 493 672 .6 63 141 250 391 563 768 .8 70 159 282 441 634 864 2.0 78 176 313 489 704 960 .2 86 194 345 538 774 1056 4 94 211 376 588 845 1152 .6 IO2 229 407 637 915 1248 .8 ICQ 247 438 686 986 1344 3-o 117 264 470 734 1056 1440 .2 125 282 502 783 1126 1536 4 133 300 532 832 1197 1632 .6 141 317 564 882 1267 1728 .8 149 335 596 930 1338 1824 4.0 156 352 627 979 1408 1920 .2 164 370 658 1027 1478 2016 4 172 388 689 1076 1549 2112 .6 1 80 405 721 1125 1619 2208 .8 188 423 752 1174 1689 2304 5-0 196 441 783 1223 1760 24OO 5 215 485 861 1345 1936 2640 6.0 235 529 940 1468 2112 2880 5 254 573 1017 1590 2288 3120 7.0 274 617 1096 1712 2464 3360 5 293 66 1 H73 1835 2640 3600 8.0 313 705 1252 1957 28l6 3840 5 332 749 1330 2079 2992 4080 9.0 352 793 1410 22OI 3168 4320 5 372 837 1489 2324 3344 4560 IO.O 39i 881 1567 2446 3520 4800 TABLES. DISCHARGE OF CIRCULAR PIPES FOR DIFFERENT VELOCITIES. Continued. Velocity in Feet per Second. Diameter of Pipe in Inches. 15" 16" 18" 20" 24" 30" .2 110 125 159 196 282 440 4 22O 251 318 391 564 880 .6 330 376 476 587 846 1320 .8 441 501 635 782 1128 1760 I.O 551 627 794 978 1410 2 2OO .2 66 1 752 953 1174 1692 2640 4 771 877 III2 1369 1974 3080 .6 882 1003 1270 1565 2256 3520 .8 993 1128 1429 1760 2538 3960 2.0 1103 1253 1588 1956 2820 4400 .2 1213 1379 1747 2152 3102 4840 4 1323 1504 1906 2347 3384 5280 .6 1434 1629 2064 2543 3666 5720 .8 1544 1755 2223 2738 3948 6160 3-o 1653 1880 2382 2934 4230 6600 .2 1763 2005 2541 3130 4512 7040 4 1873 2131 . 2700 3325 4794 748o .6 1984 2256 2858 3521 5076 7920 .8 2094 2381 3017 37i6 5358 8360 4.0 2204 2507 3176 3912 5640 8800 .2 2314 2632 3335 4108 5922 9240 4 2424 2757 3494 4302 6204 9680 .6 2535 2883 3652 4499 6486 IOI20 .8 2645 3008 3811 4694 6768 10560 5-o 2755 3133 3970 4890 7050 IIOOO 5 3031 3447 4367 5379 7755 I2IOO 6.0 3306 3760 4764 5863 8460 13200 5 3582 4073 5161 6357 9165 14300 7.0 3857 4387 5558 6846 9870 15400 5 4133 4700 5955 7335 10575 16500 8.0 4408 5013 6352 7824 11280 17600 5 4684 5326 6749 8313 11985 18700 9.0 4959 5641 7146 8802 12690 19800 5 5235 5954 75-13 9291 13395 2O9OO IO.O 55io 6267 7940 9780 14100 22OOO 64 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. DISCHARGE OF CIRCULAR PIPES FOR DIFFERENT VELOCITIES. Continued Velocity in Feet per Second. Diameter of Pipe in Inches. 36" 42" 4 8" 54" 60" .2 6 3 I 863 1128 1427 1756 4 1262 1727 2256 2854 3512 .6 1893 2590 3384 4281 5269 .8 2524 3454 4512 57o8 7025 I.O 3155 4317 5639 7136 8781 .2 3786 5180 6767 8564 10537 4 4417 6044 7895 9990 12293 6 5048 6907 9023 11417 14049 .8 5679 7771 10151 12844 15807 2.0 6310 8634 11279 14272 17562 .2 6941 9497 12407 15699 I93I9 4 7572 10361 13535 17126 21075 .6 8203 11224 14663 18553 22831 .8 8834 12088 I579I 19981 24588 3-o 9465 12951 16919 21408 26344 .2 10096 13814 18046 22835 28100 4 10727 14678 I9I75 24262 29856 .6 H358 I554I 20303 25689 31613 .8 11989 16405 21431 27116 33369 4.0 12620 17268 22558 28544 35125 .2 I325I 18131 23686 29971 36881 4 13882 18995 24814 31398 38637 .6 T45I3 19858 25942 32825 40394 .8 I5M4 20722 27070 34252 42150 5-o 15775 21585 28198 35679 43906 5 17353 23744 31018 39247 48297 6.0 18930 25902 33838 42815 52687 5 20507 28060 36657 46383 57078 7.0 22085 30219 39474 49951 61468 5 23662 32378 42297 53519 65858 8.0 25240 34536 45H7 57087 70249 5 26817 36694 47937 60655 74639 9.0 28395 38853 50756 64223 79030 5 29973 41012 53576 67791 83420 IO.O 31550 43170 56396 71359 87810 TABLES. TABLE No. 3. TABLE OF 11/6 AND 6/11 POWERS. Num- ber. 1 1/6 Power. 6/1 1 Power. Num her. 1 1/6 Power. 6/1 1 Power. Num- ber. 1 1/6 Power. 6/1 1 Power. 0.50 0.28 0.685 .12 23 .06 74 2.77 35 52 30 .70 .14 27 .07 ./6 2.8 3 36 54 32 .715 .16 31 .08 .78 2.89 37 .56 35 73 .18 .36 .09 .80 2-95 38 08 37 744 .20 .40 .IO .82 3-01 .38 .60 39 .758 .22 44 .11 .84 3-07 39 .62 .42 77 .24 49 .12 .86 3-13 .40 64 .44 .785 .26 53 -13 .88 3-19 41 .66 47 .798 .28 58 .14 .90 3.26 42 .68 49 .81 i -SO .62 15 92 3-32 43 .70 ^52 .824 32 67 .16 94 3.38 43 72 55 .835 i -34 71 17 1.96 3-45 .44 74 .58 .848 .36 76 .18 1.98 3.52 45 76 .61 .86 38 .Si .19 2. 3.56 .46 .78 64 873 .40 .86 .20 2.5O 5-40 .65 .80 .67 855 .42 .91 .21 3-00 7-55 .82 .82 .70 .899 44 .96 .22 3.50 10. 98 .84 73 .91 .46 2.OI 23 4.00 12.75 2.13 .86 76 .92 48 2.06 .24 4.50 15.85 2.27 .88 79 932 50 2. II 25 5-oo 19.20 2.40 .90 .82 944 .52 2.16 25 5 50 22.90 2-53 .92 .86 955 54 2.21 .26 6.00 26.90 2.65 .94 -89. .966 56 2.27 27 6.50 31-20 2.77 .96 93 977 .58 2.32 .28 7. co 35-70 2.88 .98 97 .988 .60 2-37 .29 7 50 40.70 2.99 .00 i. i. .62 2.43 30 8.00 45-6o 3.10 .02 .04 1. 01 .64 2.49 31 8.50 51.00 3.20 .04 .07 1.02 .66 2-54 32 9.00 56.70 3-30 .06 .11 1.0 3 .68 2.60 33 9-50 62.70 3-40 .08 15 1.04 .70 2.65 33 10.00 68.00 3.50 .IO .19 1.05 .72 2.71 34 66 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. TABLE No. 4. COEFFICIENTS FOR LOSS OF HEAD BY FRICTION IN OLD PIPE-LINES. Velocity, Age of Pipe in Years. Feet per Second. / 5 10 15 20 25 30 40 5 I .IO .21 1-33 1.46 1.58 I.7I 1.96 2 23 2 .14 30 47 165 1.82 2.00 2 37 2-74 3 17 37 - -57 i. 80 2.OO 2.22 2.69 3-13 4 .20 43 .66 i-93 2.16 2.41 2.96 3-45 5 .22 .48 74 2.04 2.30 2-57 3-20 3 73 6 .24 52 .81 2.13 2.42 2.72 3.41 3-99 7 .26 .56 .88 2.22 2-53 2.86 3.61 4.22 8 .28 .60 93 2.31 2.63 2-99 3-79 4.44 9 30 .64 99 2-39 2-73 3-n 3-97 4-65 10 32 1.67 2.04 2.46 2.82 3-22 4-13 4-85 TABLE No. 5 COEFFICIENTS OF DISCHARGE IN OLD PIPE-LINES. Velocity, Feet per Second. I 2 3 4 6 8 9 10 Age of Pipe in Years. 5 IO '5 20 25 30 40 50 0-95 0.91 0.86 0.82 0.79 0.77 0.72 0.68 93 .87 .82 .78 75 .72 .66 .62 92 .85 79 75 71 .68 63 59 .91 .84 .78 73 69 .66 .60 56 .90 .83 77 7i .67 .64 .58 54 .89 .82 75 .69 .66 .62 57 53 .89 .80 74 .68 .65 .61 .56 52 .88 79 .72 .67 .63 .60 54 51 .88 79 72 .66 .63 59 53 50 87 78 71 .65 .62 58 53 49 TABLES. 6 7 TABLE No. 6. GIVING CORRECTIONS IN FEET TO BE ADDED TO HEAD ON THIN-CREST WEIRS FOR VELOCITY OF APPROACH. WITHOUT END CONTRACTIONS. Depth of Bottom of Channel below Crest. Veloc- Depth below Crest = .50. Depth below Crest = i. Depth below Crest = 2.50. Ap- proach. Head on Crest. Head on Crest. Head on Crest. .20 .30t0.50 .4010.60 .80 to i oo 50 I.OO 1.50 2.OO .2 .001 .001 .001 .001 .001 .00: .001 .001 3 .OO2 .002 .002 .002 .002 .002 .002 .002 4 .004 .004 .004 .004 .004 .004 .003 .003 5 .007 .006 .007 .006 .006 .006 .005 .005 .6 .OI .009 .01 .009 .008 .008 .008 .007 7 .012 .013 .OI2 .Oil .OI I .on .01 .8 .cts .017 .016 015 .014 .014 .013 9 .022 .O2I .019 .018 .017 .017 .0 .... .027 .026 .023 .022 .021 .021 .1 .... '033 .031 .028 .02 7 .026 .025 .1 ... 039 037 .033 .032 o->8 .031 .030 4 .051 045 .044 .042 035 .041 5 .... .... .058 .052 05 .048 .047 064. '062* 'o6 3 .8 060 067 9 .08 .077 .075 .... .... .... .083 WITH END CONTRACTIONS. .2 .001 .001 .001 .001 .OOI .OOI .001 .001 3 .003 .003 .003 .003 .003 .003 .003 .003 -4 .006 .005 .006 .005 -005 .005 .005 .005 5 .009 .008 .009 .008 .008 .008 .008 .007 .6 .013 .012 .013 .012 .Oil .Oil .Oil .on 7 .017 .016 .017 .017 .015 .015 .015 .014 .8 .... .021 .023 .022 .02 .02 .019 .019 9 .029 .027 .026 02 S .024 .024 .0 .... .036 0^4 .032 .031 3 .029 .1 .043 .041 .038 037 .036 .036 .2 .051 .048 .046 .044 .043 .042 3 .... .... 057 053 .052 .051 .049 4 .066 .062 .06 59 .058 5 ... .076 .071 .069 .068 .065 .6 .o8l 7 O79 .089 .077 .087 75 .085 .8 9 .... .099 .III .097 .108 .0^4 .105 .... .117 68 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. TABLE No. 7. GIVING AREA AND VALUE OF r OF CIRCULAR CHANNELS RUNNING FULL. Diameter in Ft. and In. e ? < Values of r. Diameter in Ft. and In. .n rt CT # Values of r. ~ ai \r> o u^o^cs in Nor->. in ir> O c^oo ir>r^.o N xr>\o O r^oo O'-i WCM ....? cococOTtTtinoo r^ O O* co Tt mo t^oo OMcoinr^c> CM TtO ooOCMin CMcoOOco O N 00 Tt HHCOTto'oOOd -toco O N Ttco NO Oooo Tt N O nhH^-NNNNcocOTtTtinor^oo N TtO oo m M ci co TtO r^. co coo 6 w N CM M CO CO * M M coTt-min cor^O^i-i comr^O N coo co O^ O >-< ooo r^i^.f^r^cococooococo o O^ N TtO CO 6 Tt mo t^co O* O 70 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. TABLE No. 9. CHANNELS OF TRAPEZOIDAL SECTION. SIDE SLOPES 1 TO 1. GIVING AREA AND VALUES OF r. W- i Ft. W - 2 Ft. W = 3 Ft. arm 4 Ft. Area. r Area. r Area. r Area. r 0-5 0-75 0.31 1-25 0.37 i-75 0.40 2.25 0.42 75 1.32 .42 2.O7 -50 2.82 55 3-57 58 i 2 52 3 .62 4 .69 5 73 1-25 2.82 .62 4.07 -74 5-32 .81 6-57 .87 1.50 3-75 72 5-25 .84 6-75 93 8.25 1.75 4.81 .81 6.56 94 8.31 05 10.06 .12 2 6 .90 8 .04 IO 15 12 .24 2.25 7-35 99 9.60 .14 11.85 .26 14.10 36 2.50 8.75 .08 11.25 .24 13.75 37 16.25 47 2-75 10.30 17 13-05 33 15.80 47 18.55 57 3 12 .27 15 43 18 57 21 .63 3-50 15-75 45 19.25 .62 22.75 77 26.25 .89 4 20 63 24 .80 28 .96 32 2.09 4-50 24-75 .80 29.25 99 33-75 2.14 38.25 2.29 5 30 .98 35 2.17 40 2-33 45 2.48 W = 5 Ft. w- 6 Ft. W = 8 Ft. IT. 10 Ft. Area. r Area. r Area. r Area. r 0-5 2-75 0-43 3-25 0-44 4-25 0.45 5-25 0.46 -75 4-32 .61 5-07 63 6.57 65 8.07 67 i 6 77 7 79 9 -83 II .86 1-25 7.82 .92 9.07 95 n-57 14.07 .04 1.50 9-75 1.05 11-25 i. ii 14-25 .16 I7.25 .21 i-75 ii. 81 1.18 I3-56 1.24 17.06 32 20.56 38 2 14 .31 16 37 20 47 24 53 2.25 16.35 43 18.60 50 23.10 .61 27.60 .68 2-50 18.75 55 21.25 63 26.25 74 31-25 .83 2-75 21.30 .67 24.05 75 29-55 87 35.05 97 3 24 78 27 .88 33 2 39 2. II 29-75 2 33-25 2.09 40-25 2.25 47-25 2.38 4 36 2.21 40 2.31 48 2.48 56 2.6 3 4.50 42-75 2.41 47-25 2.52 56.25 2.71 65-25 2.87 5 50 2.61 55 2-73 65 2-94 75 3-io TABLES. TABLE No. 10. CHANNELS OF TRAPEZOIDAL SECTION. SIDE SLOPES 2 TO GIVING AREA AND VALUES OF r. w = i Ft. w = 2 Ft. W = 3 Ft. w = 4 Ft. Depth. Area. T Area. r Area. r Area. r 0.50 I 0.31 1-50 0-35 2 0.38 2.50 0.40 75 1.88 43 2.6 3 49 3-38 52 4-13 56 i 3 55 4 .62 5 .67 6 71 1.25 4-37 .66 5-62 74 6.87 .80 8.12 .84 1.50 6 78 7-50 .86 9 93 10.50 .98 1-75 7-87 .90 9.62 .98 n-37 -05 13-12 I. II 2 10 .01 12 - 10 14 17 16 1.24 2.25 12.40 .12 14.65 .21 16.90 .29 I9- I 5 .36 2-50 15 23 I7-50 33 20 .41 22.50 I. 4 8 2 75 17-85 34 20.60 44 23-35 53 26.10 .60 3 21 45 24 56 27 .64 30 72 3-50 28 .68 31-50 79 35 .87 38-50 .96 4 36 .90 40 2.OI 44 2. II 48 2.19 4-50 45 2.13 49.50 2.24 54 2-34 58.50 2.42 5 55 2.36 60 2.46 65 2.56 70 2.66 w = 5 Ft. w = 6 Ft. W ' = 8 Ft. w = 10 Ft. Depth. Area. r Area. r Area. T Area. r 0.50 3 0.41 3-50 0.42 4-50 0-44 5-50 o-45 75 4.88 58 5-6 3 .60 7-13 63 8.63 -65 i 7 74 8 . 7 6 IO .80 12 -83 1.25 9-37 .88 10.62 .92 13-12 .96 15.62 1.50 12 1.02 I3-50 .06 16.50 .12 19.50 17 i-75 14.87 I- 15 16.62 .20 20.12 .27 23.62 33 2 18 1.29 20 34 24 -42 28 1.48 2.25 21.40 1.42 23.65 47 28.15 56 32.65 1.63 2.50 25 1-55 27.50 .60 32.50 .69 37-50 -77 2-75 28.85 1.67 31.60 1-73 37.10 83 42.60 .91 3 33 1-79 3 6 1.85 42 .96 48 2.05 3-50 42 2-03 45-50 2.10 52.50 2.22 59-50 2.32 4 52 2.27 56 2-34 6 4 2-47 72 2.58 4-50 63 2.51 67.50 2-59 76.50 2.72 85-50 2.84 5 75 2.74 80 2.82 90 2-97 100 3-09 72 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. TABLE No. 11. CHANNELS OF TRAPEZOIDAL SECTION. SIDE SLOPES 3 TO 1. GIVING AREA AND VALUES OF r. W = i Ft. W=-z Ft. W = 3 Ft. W = 4 Ft. Depth. Area. r Area. r Area. r Area. r 0.50 1.25 0.30 1-75 0-34 2.25 0-37 2-75 0-39 75 2.44 43 3-19 47 3-94 57 4.69 .54 i 4 -55 5 .60 6 -65 7 .68 1.25 5-95 .67 7.20 73 8-45 .78 9.70 .82 1.50 8.25 79 9-75 85 11-25 .90 12-75 95 i-75 10.92 .91 12.67 97 14.42 .02 16. 17 .07 2 14 1.03 16 . 10 18 15 20 20 2.25 17-50 15 19-75 .21 22 .27 24.25 33 2.50 21.25 .26 23-75 .33! 26.25 39 28.75 45 2-75 25-50 -38 28.25 45 31 52 33-75 -57 3 30 50 33 57 36 .64 39 .70 3-50 40.25 74 43-75 1. 8l 47-25 .88 50.75 94 4 52 97 56 2.05 60 2. 12 64 2.18 4-50 65.30 2. 2O 69.80 2.28 74-30 2-35 78.80 2.42 5 30 2-45 85 2-53 90 2.60 95 2.67 W = 5 Ft. W = 6 Ft. W = 8 Ft. W= 10 Ft. Depth. Area. r Area. r Area. r Area. r 0.50 3-25 0.40 3-75 0.41 4-75 0-43 5-75 0.44 75 5-44 55 6. 19 58 7.69 .60 9.19 .62 i 8 70 9 73 n 77 13 .80 1.25 10.95 85 12.20 .88 14.70 93 17.20 .96 1.50 14.25 99 15-75 .02 18-75 07 21-75 1. 12 i-75 17.92 .12 19.67 15 23.17 .21 26.67 1.27 2 22 25 24 .29 28 .36 32 42 2.25 26.50 38 28.75 .42 33-25 49 37-75 -56 2.50 31-25 .50 33-75 55 38.75 63 43-75 70 2.75 36.50 63 39- 2 5 .68 44-75 .76 50-25 83 3 42 i-75 45 .80 5i .89 57 97 3-50 54-25 2 57-75 2.05 64-75 2.15 71-75 2.24 4 68 2.24 72 2.30 80 2.40 88 2-49 4-50 83-30 2. 4 8 87.80 2-59 96.80 2.64 105.80 2-74 5 100 2-73 105 2.80 115 2.91 125 3-oi TABLES. TABLE No. 12. CONVERSION TABLE OF PRESSURE IN POUNDS AND HEAD IN FEET. Pressure in Pounds. a o w F Pressure in Pounds. l ^ Pressure in Pounds. c T3 U rt v & Pressure in Pounds. Head in Feet. Pressure in Pounds. jl I 2.3 41 94.6 66 152.3 91 2IO Il6 267.7 5 ii. 5 i 42 96.9 67 154 6 92 212.3 117 270 10 23-1 i 43 99.2 68 156.9 93 214.6 Il8 272.3 15 34-6 44 101.5 69 159.2 94 2l6.9 119 274.6 20. 46.2 45 103.9 70 161.6 95 219.2 120 276.9 21 48.5 46 106.2 71 163.9 96 221.5 121 279 2 22 50.8 47 108.5 72 166.2 97 223.9 122 281.6 23 53-i 48 no. 8 73 168.5 98 226.2 123 283.9 24 55-4 49 113-1 74 170.8 99 228.5 124 286.2 25 57-7 50 II5-4 75 I73-I 100 230.8 125 288.5 26 60 5i 117.7 76 175.4 101 233-1 130 300 27 62.3 52 . 1 20 77 177-7 102 235-4 135 3H.6 28 64.6 53 122.3 78 1 80 103 237-7 140 323.1 29 66.9 54 124.6 79 182.3 104 240 145 334-6 30 69.2 55 126.9 So 184.6 105 242.3 150 346.2 31 71-5 56V 129.2 81 186.9 106 244.6 155 357-7 32 73-9 57 I3I-5 82 189.2 107 2469 160 369.3 33 76.2 58 133.9 83 191.6 108 249.2 165 380.8 34 78-5 59 136.2 84 193-9 109 251.6 170 392.3 35 80.8 60 138.5 85 196.2 no 253-9 175 403.9 36 83.1 61 140.8 86 198.5 in 256.2 1 80 4I5.4 37 85.4 62 I43.I 87 200.8 112 258.5 185 427 38 87.7 63 145-4 88 203. r H3 260.8 190 438.5 39 90 64 147.7 89 205.4 114 263.1 195 450.1 40 92.3 65 150 90 207.7 US 265.4 200 461.6 74 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS. c ' 1 g * rH 2 | P5 rt H > -S ; fe 1 5. 33 U *&. O 10 VO s " 5- ^r.0 1000 M lOinoO M So-a 8 H H M ~ W cT c?^ M o" S ro *J a"s f T *O ^3 sgs 8 VOC^ 2 2 J'S ? M W ^l^i '7 er/5 H &&& sa ^J-vo OO O , 4j MH r\^ -Ss f T 1 la? o 5~ t^OO OHM 2"^?^ OO O M ro 8 M 5? ajou Sffil O rt 9-j*g gjfgj ^PL^n s^ vo t^ NVO O- O N ^^s; ro -^oo N flj HH -*-i M ffi 2, U5 jj c .. <2 ^ fl3 3 1- 00 *vo t^oo o> o -. w ro TT lovo 1 o ji V > *s.S-a VO ^O rA M oo ^t- o rco O W 1O l^ O* III, &3 $ w *S| t ! 5 S c u iir 'CfcS H) N N o jaqSijq 1U33 jaj 01 UJBSJIS -3JIJ JIBjJ B SB llUIiq UJnUJIXEJ^ K 2 a ^J H rt l o S m !f oo a- o" I?^ O H f> * invo r--oo -^~Y^- V f _/ . C/2 Oa ffi rf O&H u z bi c UVM 1 tn o rt M Q i .g-o a O G jj SH tf? c ^ ^ J^ 3 "S-2 u 3 u Jj g "s 3 PC fe g o 3 5 TABLES. 75 I I ^U&? ! ?Q * 00 M I $ mOO OT ! ~ M M ID t^ O 00 N W T ft ~ O M I ro 4- t^ O O M in O qSiH 1033 jaj 01 JIB j B SB jitnn 30 O O ^ tT% 10VO as 00 ft I 00 10 oo r^ ! >o m M o I * ^ O I -^ m io\o i oo o N aH* Hff N ro -3JIJ JIE,J B SB IJUi M * oo o o 10 t^OO FRICTION HEAD 10 20 IN" FEET IN CLEAN PIPES 30 40 50 \ r art 10 20 30 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No. LIB OF THF UNIVERSITY FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 50 600 fx4 jgsTN &U \ \ \ m 1000 1200 z 1400 ^ 1600 10 20 30 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No. 3 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 50 8 vv> XI 1000 XVv^ &\\ n ,i\ \> 2000 z 3500 4000 10 20 30 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No. 4 FRICTION HEAD IN FEET IN 10 20 30 CLEAN PIPES 40 50 1000 2000 3000 4000 5000 6000 7000 8000 FLOW OF WATER IN LONG PIPES. DIAGRAM No. 5 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 3C \ 40 50 1 1 ^\J ^L^St \ vSrv j lYvsK \ } , V \\K v S 1 n juCC \ A - x : \i V ^ 1 . r| - - - 1000 A V\ i\ ' ' \ \ N v \ '\ ^ ! \; x^ i; \ '' V \ V x" ^JsJ^ > M v\ \ [ \ 'X' ^^ \'i \ A V A ' \ ' -X A. Ss\. - i A A A]\ sf\ Sw 1 ! 3 ! V r ,\- | hlv ^ N \ JxJv ^ ^ i i \ At "VST\ x ^ ^ : - iV \ Y \ \ \ K 5 x \ x x oonn 1 1 \ \ \ l i^ ,'.> A\ A" \ X '' - \. S^x'V. X^^ 1 \ \ \;\\ \\'l \ 3r \ ' X 1 V v V \ ' A \ \ Y , 1\ \^' \^ \' S \ ' '" "^. x^x xN t ; i A L , A KM H\ v v ' *"* S.^\ X X ^^ l~ D A ', ' V V{VX A x \ \ \ ?cc v^ IU :; ; : i. > /,,\\y x - S '-I \ ^ \ \ ^ \ s."^-- Si y~~ i; ! I i \ i ~\ . y \ t Vv\ 5 S l\ i \ \ \ xj \ H. ^ N,. _ ^rt* u. n 1 1 V \ A V * ^ A \ \ rrv .\\ v ^ k \ t> ? "\' \i \\\ V M\\ \ V \ \ \[ "N s+ r\. ^ C "h 1 ', \ - \ \ 1 \ ,\ \ |\ \ . K E ZS vN^ v 3S S "\ iL N S t 3000 H ! l ! \ i l \ \ \ \ \ \ \ \ V f i\ S i ,^ xx ^ ' m r \ \ i\\K\ t N \ K K \^ \ ' ^ s H i 1 p5pp3: \ V xj) s ?p A X, ^ s ^ ^ ' ( m 1 '< '. \\\ \ ,j \ \ y D % , ^TO:O^*5 m MTT \ 1 j \; , \ j 1 - i ^ H i 1 1 I \\\ H\\\ l \^ C S \ S^sS^!: i < \ \ \ i\f\ raj y j\ \ \ X /Lk 'ill 1 J\ f\ w \ \ TSE; \ \ ^ ^ ^ N N ^ f ! ' r ' f t S-^ \ \ K- \ \ k A ^ \\\ x 4000 , M i \ \ \ y > \ a \ i A v \ \ \ ^ *" N. H L dd . \ \ \ > y > v \ A \ \ v X . \ vv "z. H \ ;, V \ v \ \ \^ \ A ^ \ *\ \ ^ KtS J J \ \, \ \ \ j V \ \ v\ \ 1 \ p ., ^ v ^ A \ ; \> / \ \ 1 1 |!\,\ \ v \ \ LJ A N \ \i \^ V- , v_ T h A. J. ^ ^> , 5 1 T \ T \ V i T v L ^ \ ^^S \ 1 V '' \ ^ S V- N \ j I 4! - I v \ !S ! E j 3 "t _ I\ \ ^\ \ \ i ' \ S* \ v_ > ^ ^ ^ L S ^ \ MM' ' - $ ^ ^ \ \ \ \ \ NJ s. \ '' ' \ --A V V ^L i_ ^_ \. ^ \ '\ \ j ' \ \ \ \ y . . \ \ \ fiono j i3 Vi v " i ^ \ i , ill . nT- i V t y y S _. S -r 1 ktjf |H \ \ \ N v ^ S \ L \ ' \ \ \-5----p ^ s X ^ \ \ \ \ \ \ r r \ \ \ \ \ M 1*+ \ \ '. \ L. \ ^ V V U ' \ ^ \ \ \ \ \ \ \ LX 3l\ -i ^ J - - IT ^ y "ry uJLJ \ \ N ~:s__^:70oo : , \ \ \ \ \ \ y. > li 1 r \ Y Y j \ 5 k *y \ ^ H : \ \ > \ \ \ \ \ \^ C" \ \ \ v . \ \ \ \ \ 1 \\\ 44 A-A \ \ 1 \ X i ' A k, ^ S r 1 7 > , ; \ \ \ Y Y \ i \ 1 \ ' \ \ r 5 v A v- !\! 1 \ j i j 5 _i _>____ 8000 10 , 20 3 D 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No.6 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 50 1 I I ~"1 -g V 1 A v| jT ' J^ \YVXK i \ \ \vv}O*vX Hi \A\ V s v^xi^ 1 11 It \ \ \\ \ Ov^^O\>w II 1 \ \ \ \ \ VvV* \ \R S. HI 1 \ A \\ ' \ V\V^ xXiX X -- 2000 |M EEC v^ N ""^^^xj - H^S^V^^?Px$ iiin'V ^ \ NS^v ^^xSv ' ' ^ \ \ \ ' X \ NJ i\ s. ^ ^^X^ ^ III ^A i\ \ V'xl N X. ^ v *X^^S ^ v ^ s^ 11 I 1 ' \ \\\ \\ Nj \ ^ X^ ^ \ S |4V ~^rr \ \ \ M\l ^ - ^ \ - SJX^M ^^"^"^""4" - 4000 1 \ \ \ \ N. \ s v' ; "\ i ^ ^ ^ 'x I'liXVli \N \ \ \. ^ ^ Z s , 1 \l\''\i\\ \. ^ ^ N ^ ^^ x >>. H" 111 1 * '\ r 'V\ : \ i' \ ^ x \ \ N > \ v ;-!3Sj g li ;L.L 'J\\ *\\\ ^ r^ ~ ^ ^ "^ " L -^ -^ "^^^ -rv= -^ ju - | = K| LI! - U. 1 1 1 i ' ll \\'\\ V \^ v ' N. \ x 1 1 ^^ ^\.c 3 iVi '\^l\i \ \ i\ \ i\ ^ N ^ - v S i ^ *^^- /\ '1 fc'^KiYi V \ '\1 '\.i (i i^t ' x i N^ x 3B * 6000 t!tlll\^ ; \\l \ \ \ \ \ 1\. x ^o I* 1 ^\ '\ \\ \\ \ i M \ v i v v [\1 i \^ TV i*S h. X ^ iw. ! 1 ; 1 \ t A 1\\\\ X^'iv* V ' I ^ v "^ k i M'I\P ; I\ \ L V \ X 1 M^\ X j ^ \ rv j* \-\j\ \ 1 \ \ \ Jt^ V, ^\^ X / ^j sfa^ \I \ ^ ' \ t \ Li \ ttL s ^ j ^ ^t (Ilill \ \ \\ \ \ \ '(NL ' i \ \ S X ^ s. 1 \ 1" i \i !*( ' \ i >fe ' \. \ ^ s. ^X . s^ s^ ^ n \ \ ^ 1 T ^ ~^& \ \ " r* i * ^ x \ \ ^&* \ ' 2 \ i . ^ \- ix --t;- ^^--^^- 8000 LH iL 1 o\ ^ ' ^ . 1^^ \ M \ \ fri iu %p ^ ! i % 1^? i \ \ \. v --S <- ^ ^ ^^ En ^5 ^? ii \I!\V' V [I iyi\\iv \ ^ ^_ A j_ ^ ^ \ W V till 1 ! I \ \\ \ \ \ \^ t \ \* X 2v x^ T "~1 i] \ ^ > i \ v\ ^ " n V \ 5 v ^ ii\iil\ \ \ * . , \ s. J S N . . s- iiiiii iii'ii \ v v ^ ^ i_ii\ii\ ' v ' r \ ' \ N v N\i ^ 10000 * t \p| A * \ ^~ V ^ ^ HI ' p itLl 1 n_ ^ t K _S _^ y ^^^ [jj ~j|| MirriH~ \ \ t X~~ \!\ ~* v ' ^ ^--^-A- ~ I i 1 n ! 1* I \ i L. 2 111 III 1 l" \ { ^" V V lltllli [ \ \ \ v_ v ^ V ^ \ IIM \ ^ 1 . ^^ . \ \ , j 1 L "" L. E ^ > L \ f \ ~*\ \ _ v ^ ^ _5 s x 12000 j I I |\ j 1 ,1 "\ ^" |*'JH|\j|\ 1 * >, 1 1 l \ ^ ^. V" \ V j ( M \ 1 1 \ k Y i V ' \ S k 1 T i 1 * 1 \ i \ III 1 Ti ; \ \ " ^ " s " X -r^ 5 \ i 1 i 1 \ 1 \ \ 1 1 i V A ^! \ ' " _S ^ '*^ ~t \ * i 1 1 i i 1 1 V is;. . s 4. . s _ i4ooo I t \ \ I ! \ i \ ^ 1 \ \ " "N \l i I \ I 11 11 u l \ \ ^ I'lllli \ \ \ 1 1 1 V \. \ V "n" ^ \ \ \ \ N 1\ I \ i J ]| 1 |i 1 i i \ ^ ^ 1 1 i y i i \ i \ i \ ^ \_ ^ ^ HOE i i IT 1 ~r\i QL ^ ' " ! - -h*d- -"- -\ - '\ ' " -\ 16000 10 20 30 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No. 7 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 50 [ 1 .. -t* IK V ^ JLvF v V S^ N 1' \ V Sis ! X \ E3| 4J , IV ^xK \l V SEES V' } ^^ \ V v\i ^^31 f\r\f\(\ M ; fv:i\\S 2000 vvAJ r\^ ^v A ' -N \ ) , X- - xK ,i ,\. ,v. N:r.t;v. --^X- lv . , X' SSvv T \ DM ,.,\\\\X \\kv\S 1 l \\v t \ T \- t x^ \ l \\\ HN\^^C^:vv^vs^^^v 1 1 ' i \ v J\K rrrrrrr^^s xs iL jj '. ^ \ \ \ x x x rf rrn V , \ } T\\^ ( \^\ i \^rsN^vs^'s - 4000 PK* \ . > - K\. \' s. S. \ ^ \ ^ \\X\ X v N 1 1 j \ , i\\SAj--C^S5_siSk < Ssi5 \ ; i \ .\\\\\\\ V^ K"^ \ XJ* H . . \ \ ^Ql u m ign xJL. x V . x \ \ N . \ ^ \ N . - X s ** u m r*v 1\1\ V \ \ \ \ \ \ \ \ \ \ \ S_S ^ ^ liICIUZi lt - ^ K.X ^3/j \ n \ i ' \ \\ \V\ X v - S^ v ,\^ L |X ]x S Zp L \ \ \ ^ \ \ . -. \. , \ v \j \TH, % x . N X >V^ . <;nnn v ls \ \^ 5 ! ^ oOOO i L \ \\N y\ \ X \\\ N \K X x x x \ x>V ]^ ^ ^ s s pffi \f\\^ v\ rp \r.N^\ f \ \ S-A ^ ^ x ^^"3- \ ' 1 1 \ h \ \ ^ ^ v S > \ i > ^ * P a s_uSljLDiEA*i i i X^v^s^"^ N T ^ XX 5<^ M- HV 3 Hf\ 3 t Sl Pr^ \ \\\ NL \ \ ^ ^ % N ^ ^^>v H rm :S::t5:S^:^fE5^s at) :fflLC3:c::!^L:::-:::^. ;_s;.s:^~ v-Sx^.s;.s sooo Fl fe-^ n%Yv A -\\\V ^ ^ . _ _V. ^, X- s -^ ^ x 5I_ \\X.X|\x T 1 \\ \ K } \ \ 5_j_ ___, \ v^ Stj^N* 1 1 i! 1 tn \\N\\.\\\\ ^ X \ N v \ X ^2 N 1 1 ft IL\T \ L^\\l\ i. ;SI_ \ N \ ! \\ x z ' 1 \\ \ \ , N \ \ \ \ \ \ ^ ] tod ujrrn \ \ N N \. \ V \ M \ \ X > ^ \ ^c H \ H \ \ \ v \ \ \ ^ x ^v \ ^ ^ S 1 11 11 i \v\ \ \ \ \ \ ' i \ V ^ \ Xi^ 11 Wt \\\ i'lIS \ \ v , \ --gX --SIOOOO h Iri \ \ L V \ N sT I J 4t^\ V v^s.. L - lAU. i.A -i.-V-V -X ^ -V..v -.27. s ^J.-^.g . ^,,.i!_ 1 11 ' rj n n \ ~ ' \ \ \ \\ 1 I \l \ \ \ \\ \ \ V ^ ^\ \\n\\[\\\\\ \ \ V ^ \ \. ^ 1 J i\, \\\\>A\ ^ X V ^ \ X ^ 1 1 1 1 \ \\\\ :::: \ A > \ r \j d-S3ti CuttAi>-^ ~\ SIISI IS x N \ \\, 4 \ \ \ \ \ i \--v JJI-i^IIS ^-12000 1 ix'' \\\\ \ \\ , i\ \ * \ , V V\ ^ 1 \\\\ \ \\ \ t \ N V \ ^ \ \ ^ !S T- 1 I ] \ \JA \ \ \ \ \ \ \ - -* A ^ 5 ^ * \\\\\\\\ < \ \ : s s s ^ ^ ^ h , [IT" ' \ \ \ , \ \ \ o \\ i ft s ^""^ \ \ S z 1 ! \ \\ \ \ \ \ \ \ \ L. S. U 1 1 \\ \ \ \ \ \ \ \ "^ ^ 3 \- ^ \- J In \ \ V ^ V \ \ \ \ \\ \ \ \ t \ \ . ::$::$ \-~\ ^ - ^'4000 1 i i \ X _ _ [ 111 ! \ \ \ ^ ^ ^ \ sr- 1 \ \ \ \ \ \ ttiiiiiij \ ^ \ \ ~ \ ^S "^ V ' ~^ 1 \ 1 1 I 1 \ V \ 1 r i r v , (T \ \ 11 L J \ \ \ 1 I \ \ \ V L ^ y I rj" t i "j \ i : [] H mj Q j J [ - 3 " : DfEHCHtEdSr 16000 10 20 30 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No.8 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 4000 8000 12000 14000 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No. 9 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 50 r B^ , 4 3 i r ~^^\" r- 1 -} ' t s llu\ v ^ \\ s 4- | -.JK x 2000 , 1 1 < - L I 1 N - N il M ' ' 5 - N jl, , , \ \ U \ - i , : i T * N Af\f\f\ lii v k 4UUU III i -. i KH :. . S -, ru"* - '-'-' -; \ s - L it * * i \ - > " I s U T~I *~t" "* Err _ . _t v .. --.- r - r r K - : -i N ! i t U. , 1 ' .. T . L . . \ * . ^' '-: H4 - - ! \ , s 1 \ 6000 Bj|| | . - | "t -? .- - - \ < * , ; ,, tfflc - . , L - . - ^ B i.i. ',... ^. x v g ^ '^ Li M ' i 1 1 - *-t -, ' . - - N lj: \\.\\ :_^_ t ..i_. j ~y 5 ' ; , V N ^ r } Sfiftfi tii' Hg HmS ! - . ' - ; - | i 5 [ 1 1 N? J 5* i \ . \ , ^ J \ 1 ^t ; '* > !\;\ 1 'i V i j . . . -. i v-.- T' \ ' 1 s , . i z i ! * \ . **" j | , . ' '11 \ j ' - 1 \ -, ' ; : -c ! i .i LLJ _i - ' , \ .- ; ^ ^ 5 10000 L__^ ._ .,_. .. i c 1 ! * rr 1 - - \ i : v - |_T r JO 1 33 L - . [ . _ I K . . If "[I i yn ' \ \ \ \ / ' , 1 ll L ' 1 \ i ^ ^ > j . v ; ' 1 i - - , 1 \ \ ' . 1 i 3 \ N ' ; ! i s . I. j . ' Ji 1 \ 1 1 Q ' j . . 12000 j,. , i ' ^H r l j N -, T~ 1* *~ ' pwc J . X ih-p' n Effii , , - -- r KL 7 i , , i '. ' \1 ' , 1 \ o ! I i R U - 1 ' \ z I i 1 1 1 i , \ s U I 1 1\ \ ! J j j| ! uM i LL _l 1 i ' 1 . ' ; T 1 . \ 14000 | i ' ' ! | 3Tfc \ , 1 1 1 1 1 U 1 :: :: i P \ . j S ( '- ' , \ r 4 ""j \ i Ji j \ ". \ c MIT 1 1 "i ~r i - -H- E I ' j ! : J 16000 10 20 30 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No.10 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 5000 10000 20000 35000 40000 FLOW OF WATER IN LONG PIPES. DIAGRAM No. 11 FRICTION HEAD IN FEET IN CLEAN 10 20 30 PIPES 40 -]=&iin-= sooo 10000 15000 20000 25000 30000 35000 40000 FLOW OF WATER IN LONG PIPES. DIAGRAM No. OF THE UNIVERSITY FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 50 Vl 0000 20000 30000 - i 40000 50000 60000 i \- z u J 70000 I 80000 10 FLOW OF 20 30 40 5 WATER IN LONG PIPES. DIAGRAM No. 13 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 50 10000 20000 30000 40000 50000 60000 70000 80000 30 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No. 14 FRICTION HEAD IN FEET IN CLEAN PIPES 9 K ) 21 3 to 40 ' 50 M ^ i" / Is ; > N . rC > V H XN - 1 \ \ 3 ""Vj \\\ > \ \ V V V vYv X 1 try \ \% X N fUv \A\\V N - 10000 \\VY\\ s x L Is 1 ^- -L->L ran X\\NVS i s^ V 'Y\\Y -. 5 1 NJ ,! \ 1 \ , : . 1' ' \ N . \ v - - Q - \ N - - ^ x . - i^ ' i , \ S . \ . s ^v i \ \ j \ S, V A AA Af> . . vA -: s %i =- X, 5 . *^h -4 = ^ uuuu __i_^_^. > ~ s ~ -" ;^% - j ' , J \ \ ^ j (- . - L bl B ' ^ ; s ^ } -7f r - \ X O . _ ^-_ , . N - Jfc to ^ . \ . S i - 1 s, ] \ \ . sT V \ ^k [^ X 1 _j- \ s k C \^v 1 1 c? * g, ^ \ LS hfc ^c J . \ ^ J ,' 40000 Tg ', :a> \ * ^ ' _ \ \^ ^ J bx JT "~~r~ty - - ~ 3 - S -*V -*T ~~~~ 3 " * ^X"~ * 111 \ \ \ . T \ V ^ x^ \ z 1 \ \ v *~ r I V * 1r t > \ V t S ^ ^ s, Xy -- L ' \ E | \ V t 50000 ' r i j \ f c ^ 1 \ f \ C-^ \ ' N^ ' \ J \ i \ I \ ^ T T * r XT N J rh \ s lie IE I * r 4( i hi -it : ^ ! o \ r \ 50000 II \ I IT j T \ \ - , ti 5 >^ k ~"T"T -I ~F 5 - ~ ~ \ ~ - X " -+- 4 : TT 41 1 \ \ o "!'; " 41 | : TT r N :: TJ \ s TJ j \ \: : 1 \ I |v 7COOO t _r | ' Hi i ; i j c ^ -4-4- : : ~ ' ' V T! : j 11 c - ^ y iiiif t . j \ L J J_ 1 : 1 \ 80000 3 K ) 2( ) 30 4 40 50 FLOW OF WATER IN LONG PIPES. DIAGRAM No. 15 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 50 10000 20000 30000 Is LOOOO -La- 50000 60000 70000 10 30 50 FLOW OP WATER IN LONG PIPES. DIAGRAM No. 16 FRICTION HEAD IN FEET IN CLEAN PIPES 10 20 30 40 50 1[- : 10000 20000 30000 40000 50000 60000 70000 80000 FLOW OF WATER IN LONG PIPES. DIAGRAM No. 17 -*p*^ UNIVERSITY Of HEAD FLOW OF WATER IN SHOE D IN FEET. 8 10 FLUSH ENDS. 12 14 2000 16000 2 > IN FEET. 14 16 18 20 PROJECTING ENDS. HEAD HEA] PLOW OF WATER IN SHOI D IN FEET. ELUSH ENDS. 12 14 20000 40000 60000 c a 80000 100000 en z 3 J O 120000 140000 160000 14 16 18 20 PROJECTING NDS. VELOCITY IN FEET PER. SEC. 2 468 10 I ^ ^ ^ ^ -- ^* ^-* "" ' s "^ ^ ^ ^ ^ "^ \ \ ' ^ ^-^ >^^ \ \ / ^ ^ 1 1 1 J^ ^ x-' ^-- ^-150 I * Y ^^ .^ ^ x ^ **" ' ^^^ 1 C x ' ir x ^^ - ^^-- "*" ! d*2 2^ x x^^ - ' ' "~ ^ ! ^ | j . Lx x ^x " x- *^ ' _***'*'* -- ^ - i t S 7 /' x* |x ' ' ^- ' J ^-^*^' 9% / ' -^* ^- *^^' "*?' ^ y ^ < ^ i - : - ^ X * ^-^-" *** ^ ,/ 22 g- ' ^ ^ x^ '*' J^*- ~~ *^ ^ . " Ijlfi /I/ / j-11.2 -j~*- ^' 140 2131 Z 2 _,.- 1,=" ^ ** ""^ *** ^ ^ y + ^ X * ^-- i?^" -^1^* ^-T^" "" L V _y ' . /' -r^a" --T -^-^*"^ / / / ;j o< r - x r j ^ -- "* / / / S .S ^- ^ ^, -L ' ^ *\s* ( t ^* ^ '""' ** ^ 7 / Z 7 S *- -^ -* , ^" ^ "^ r j 7 * > X ^^ ^ x- ^ ^^. 2Z./Li^ ^ 7 ^^ j?: 3 j- "^ " " i *5n .- 130 /I//,/ y ^ X ^~ . ' *~ ' I t Y j / y 2 x ^ * * -^ ^ ^ //'''//' ^ " "**^ " -W / . / / / ^' ,--1 .. s ; _ ^ _ _: 1 _ _ ST? , / / / / X >^ " 3 tn 1 J> '\ / / / '' H l2^ y x v 7 " x x j il * """"^ " * " " "^^ 1 L/_ 1 y x " ^ r ^* '/ i/ ^ - ^ < ^ - - | -' -' lOrt " *'/""/"" ^ '"'^ -'-' I2 S 1 V S j' r*. ' / - " Z2 Zt/ x- ' x 7 - - " - 5^ '~LL~ Z / , <~ ~t L ^ ^x 7 ^' '' ^J ^ |j y ^ ' y V^> fc ' ^ ^ i / ' / \ '^ ,.-' _-' -- f_ 1 / ' // ' ^i"' ^^ ( v' -''"' ^ / / / x' \ : -- __ _ -t; , m y ./ / /J. v -' ' ^ 7 4t :? IIU o /r/ _ L ll. x :ll S: , - . . ^^ "T, ^- r ? '- x - s ' .-" .---"" U. I ~ ^ " ~r ~ r "i U- / / / ' / ' ' ^ ' i> -r ------ : . .. - - JJ :::-::::":::::::::::;:; g i "~ / ^ - ^ - r : i;2Li;:: - : 7 : --i-a;: : t l I *" " . - , ... , , 1 1 5 S iE* ; , mn ^ ' ' 7 / ' > x .' ' " x '/'/ x ' ' '^^ _ ' /'/'|/ 7 y r ' " ^^ ^ 2 L x ^ " "i f 7^ i - r'" ^ if 1 ', v ' > / ^ '_ - ( ^ii^iiz^z -i;:f :::::::: if :^p- :: = = EE --- ! - :: ! 90 ",''/-// ^ - -i,''-'-/ ^ ' -i- ----" ~ T ""^i; * z! " ^"" - r " '- T--- ^ x ' --- .._.. . " " ' fl fc ' y _ QA / i _ i _ / r ~::z " - 1 it ^ ~L ^ 2 4 6 8 10 COEFFICIENT C. FROM SMITHS HYDRAULICS. DIAGRAM No. 20 es OF THK UNIVERSITY PER. CENT. OF DISCHARGE. 70 80 ^90 100 110 BEE 60 70 80 90 100 110 DISCHARGE OF WIDE CREST WEIRS. DIAGRAM No. 21 GALLONS 200 300 DISCHARGE NS PER MIN. 400 500 600 700 800 .5 1.5 2. - 2.5 3.5 I. ur 2T PER SEC. ECTANGULAR WEIRS. DIAGRAM No. 22 2000 GALLONS 3000 4 6 8 CUBIC FEET DISCHARGE OF EEC NS PER. MIN. 4000 5000 6000 7000 8000 """ "T ' "~ i . i ^*~ h^*J^ -J l^^"-^ "^Ir"'^'"-!^ iT^^S^ '^*^ < ^ v ^si^*" ~-w / A '.. 1 "*^s^ * s ^^ s ^ K ^.I "S^i*"^^! 3 js* 1 -^ -' |^^" ^V^^^" "S^^V ^^ "> _ _ _ _ . u ]^X^ ^ -s^ ^-L^ ; "* *v^ ^*^s^ """i-^ ^^^^ ^jf** . . 2 u EL ^ s " "^^ K " -xT'^'^**^ '"^^ ^* Vvi ^^ ^^t^ *\ * u. x^^^&^c^.^^ ^__^^__ ^xS-w '^v ^^ 4ss s. ^ ^ ' ^^^^ ^^^ ^* ^_ ""*^S"*-^. ^* ^ ^^^r i ^--NP^^-^^- ? '^--^- ^^-^ = ---^---=; ^^Y^- " > v s"f^~^~^ |S^!5Z-Ia^^5I,^Efc ^--^i-A^^ sv ^ ^ V x ^ V . ^^^ ""^ - ^\' "'*%^ ^^ ''x^v s ] X^'^ ' ' ^ x ^ v --x ^ - - x ^ " K ^J^*' t< V'v X. v 'iSc- v *^\^v\ ^ * ^X- X ^ S ^ ^ "^ ^ ^ s l^^X*. ^-X 4 sv ^ v ^5SJ - 1 S^^-^ \^v ~ ^ ^ -^ V "5> ^"> v , v ^ j p V^^^-. ^\^. v"s^ >\^ ^ ^""X. X^xX. Xo v ' ^ ^ s^ ^ ^ 'X^i \N^\ \; ^x ; .{ . >xA >. "N^ s^ xs s x v -, v v ^^> s v v \^'>' j ^^T^^T^^X^I ^^xA ^.x. ^ S ^ S !k S ^^ ^^ ^ Z C\\V v\V\x ^ ^ ^ >o, ^" ^ ^ ~^ s S ^ ^x ^x ^X ^x^ A A A > \v\\\ v^ N, -t ^i' " v \_ ^ x A^v '\ A '\\ X ^N v v A\J " -x v : SA ; : *** ^s; ^^^ ^ ; " XxX^S^'V ^ X S V psp^p^^SSSi^ sp^S:pj|^ffl Sv \^\^ ^^\y N \F*\r^ M^V- x N^ ^ N XK \ "N X>Sk K X XXX 3Z ^ v ^ N ^ ANN \ Xv\K ^ A>\ ^ \\ ^ \\ \\N \ \\ , ' i \A\\ \ N \\ -J \^ X ^ KNik^^^VxXxKSvN f- N^^^ZK^^ 15 S X^ 5 R I ; ^5 ^ \vy j\S ls "w^ xNS $^ 6 2 \ , \\ \\ \ 'AN/' v\\^N^s s IN \ v \ K i CK \ s ^ - t ^ ^ ^ N^ v ^l^l^^sgi 1 3 Sfc^^Ss^^o^^^ ^ ^ v \ \ ^ ^ \| \ x A\ \X N ^^N .\ i^ AN A \ A x A \ \X^ . vA \ ^S^x^^s^^t^^^ \K ^^ \\\^ \\^\^^s ^A \ S 5 \ \ \ x \ N ^ x \ ^% ^^S^^S^xis^^^^^ 7 -^ r ^^^^^ x ^b NX "^^^ x \ \i\ \ ^ \ \ \ \ \ ^ \ \ ^_ \j> v \ N \ \ N \ ^ K Vs <^^ N \^-' % !W-^V^^^^~O X ^ N:S S ||ffi|i|ffl|, 1 ' 1 ' ' ' M ' ' . ' i ' ' ' 1 ' ' ' M ' ' 10 12 E^ET PEJLSEC, IECTANGULAR WEIRS. DIAGRAM No. 23 ' OF THK UNIVERSITY 5000 10000 GALLONS 15000 2C 5 10 15 20 25 30 35 40 45 CUBIC FEET DISCHARGE OF EEC ONS PER. MIN. 20000 25000 30000 35000 40000 ^s ^N ^ ^ X X "^ ^ X, \^ h ^ ^. v,. A y\ "^ s^ -L "^^ " ^ v^ 4 u x, K; '' ^ ^xj ^ ^^ ^ i ^ Urn 5 ^ x^^ X^ ~> v ^- ^-^c ^ 5 i ' ^- ^> ^ ^t .^ "^^-^ / - "^^ s . ^ ^v^ -x^^ ^ / \. x,^ 5 ^^ '^ O -,-X g ; J X 5 ^x 5 k < s X , R ^' "v; ^x; ^ --^^ ^^ ^" ^_ * X , | ^ - x x - "x S 'x 5 6 NX^ 5 N ^ | ^ \ X, X X X 5 ^ , 4 $ X ' -T \ < ! -iX, ^ , <^ . i X X x^^ X 9 ^ ^ \ - t x S ^ X "x W X iS 8 * \ V [ s v ' ' ' - i j \ s :-N \ - - \ x - , . s . ' N X 5 s x \ >. , X, s \0 \ \ ^ \ X .v, \ i - s ^ vj . \ , s , s \2% 3 \ s ^ V ' \ '5 ^ N _ 5 \4 \ I \ \ s \ \ . \ \ V . \^ \ \ . \ > \ -. ^_ \ \ \ c ~ . \ \ ^ \^ , - ^> V . \ j \ ! , x, V \B 45 50 55 60 65 70 75 80 85 r EET PER. SEC. 1ECTANGULAR WEIRS. DIAGRAM No. 1.5 U- o V) _ iil 2. 2.5 f VELOCr FLOW OF WATF CITY IN FEET. 4 5 n^.OII 6 7 8 j . . i 3 j H -JOrr f= ~.}r 3^ - :::: 3 *-U - =4^ 1C ^ ^r r -i . , *=^z i - ^t^~^^- rri |--4 tC^ I ~~- -t^ ~~~T * . d ^"^^-^_i j ^4 -- -*-C^ ~*t -i I I *~~~ --^ 1_ ' - -- = 3 .^^ ^^ ^ ^ N Pr* ^. ~~~~ i _ "~*~J-^L ' --t- _ J r - \ j 5 ^ ^ ** "^'^ J . *~* ~~^ -, j 5^ T-^^ j L -^ J - ^ ^ ^^^ ** . """ ^. B ^"^1^4-^ -, x ^"-^ ^. "^^T^v "- ^, ^ "^ -x ( ^ J ^ . ^^ ^ . "^>-J :: ':3 1 is. *^ , T "^^^j , N 5 ^ j] v, ^ > v - ^ ^, N. . s * "^ ,. > * *> ^ "^ -^ ^ Pv^i N j J N. ^^*^> B X * X |335 h ; 5 . * s s "^ "" i _ _ 1 N H^ [v s ^-^v ^ x > -. N_ "*" ^s^ s * ^ N ' , X x x A ^, ""* ^ "^ s^ " i s. r\ > s x 2 < s s . , - "^ ^ ^ x -^ Vj 1 v ^v N \ V > * x " -- x ^ S S V^N ** ^ v J _ *_ ^ ^ "^ N N ^ < . X^ v ^ ""* s v ^ T ^ ^* N ^ s x X^ i i e V \ N^ ^ "^ _^ \ ^ ^ 1 .0 A ^ \ y ^ \ > \ ^ ^\ ^ r ^ ^ N- ^s x^ , ^ ^ . s N ^ 1 IS N^ . t ^ \ ^ fcu S: H~ ^ ~^- ^ ~S " \r v \- _._ . r " O > \ X ^S X " "\ x . V ^ \ x x ^ \ 1, _ ^ N - A D \ N ^ V \ \ L x t J j \ \ \ \ \* W s "^ \ V V N^ ^ \ C" r j \j N x^ N v r c k N \ N \ \ ^ ^ v ^ s" > x V x V \ \ \ i \ ^ \ \ V "^ \ ; i*-ft V v \ i \ r N \ c"-\1 X i J C * v \^ \ v N \ N \ " ^ r j \ \ s:_: j r V ^ - *- J 1 ^ r~ A. > J ; \ r V N^ N x . r" \ _: E j f -. \ \ \ \ x i \ ^ *~ v 3 jj . v _ L J ' S v A ' r i no.o V ^ \ * r . \ \ j x i \ i \ a \ j T- -\ N "~4 ^ _1 1 , L_ 4 _i 1 L - 1 L j 1_ 5 I_J 1_| 1 1 1 6 OCITY IN FEET. ,TE R IN CHANNELS DIAGRAM No. 25 I = .OI3 (JU O 2. $ 2.5 3.5 I 1.5 VELOCITY FLOW OF WATEl CITY IN FEET. = .D25 2. 2 5 3 3.5 ^ ELK] 1 f"l M. ^ ^ ^u^ era c: l ^ "~*~-*- *_ ~u ^ | f { n q ~ -p i4- 3BE ig *t -Crt""" 1^- L '- ~ T^T - > _^^ ^^ - -~ . M j "+* 5 ', ^^ 1 ^ X. N ^ f 3 E , ^s x ' z "~^-~, ... - I * ^ -- ~^-^^ ^>^ ! ^ -^, x - - x K. -^ ^ 3 "^ X - , s N ^ N, ^ ^ _^ ^-^^^ - * 1 ^- ^ T^*^^: . -. v , "^^ < * = >^I ^ vl -- ^-v^ ^ . s^ T\ -- "^v S X - . ^s, >, ^-^ J f 1 s N -^ . S , s , ^ ] { ; | j ^v^ ^ ! ^^v' - Sj > i *% , N - . , r-k ^ x -.* tv 1 -- -- - ^*s^ i ' V - ' ^ _, \ , j -. M * r " ; ' , o , \ - . ^~ a "^ . 1 N \J . i f 1 \^ K] - s. 1 t i jT j 1 C ^ "V ! "\. ^ \ ; ! Sv I I . 1 ! f^ 1 ' T \| - - ^" Ty \ i " \ V ' 3. \ ' \ ! - -Ai- ^ <; \ 1 ' 4^ 1.5 2. CITY IN FEET. 1 ' ' 1 ' i ' ' ' i ' ri=.030 3 TER IN CHANNELS DIAGRAM No. .2 .3 FALL IN } .6 .7 .8 I. FALL IN FE VELOCITT AND DISCED FEET PER. 1 00 8 9 500 4000 1.2 1.4 L0 2. FEET PER. 1 00 IARGE OF PIPE SEWERS. DIAGRAM No. 27 10. OF THK UNIVERSITY 10 20 SCALE OF FP. |( 30 1000 6000 7000 20 30 SCALE OF FP. 75 HORSE i GIVING HORSE POWER OF FALLING REQUIRED TO PUMP WATER TO DIPl WEIGHT OF WATER TAKEN AT 62 1 00 X- EFFICIENCY. 40 50 3() 40 .75% EFFICIENCY. : POWER ING WATER ALSO OF HORSE POWER DIFFERENT HEIGHTS I 62.4 LBS. PER. CUBIC FOOT. DIAGRAM No. 28 5Q GALLON 60 Q U S 80 100 120 140 CO 160 180 200 300 70 70 500 100 60 60 50 4 REVOLUTIOI 50 4( 1500 2000 2500 GALLONS P DISCHARGE OF S FOR DUPLEX MULTIPLY BY 2. LONS PER. MIN 400 500 600 700 800 j u-^r^^^^ ' ^ ^ ^ -7-X -^^ ^^^^k/r^ *2 7 x _^ -J-T- 60 Tit ^>rK -^ ^C - ^5^ -=tXt2 ZZj ^^^^^ -^^ 4^1 2^2^ _[/ L^^^ j 1 J^ : / 'T~i~~> J/ y^ X ~V-^ - 80 kjMisJ'- ' H^^> X : >V **^ 4- ^ p'^Ty 7^7 - ^^ 100 5P- >< X^'"- S ^ * ^ ~ X X_, , 7^ - X ^X x ^xJ ' 52 / / / 7^-.. >4- 4 ^^^Z N P^^ cc. -j / / :^:z:::i::/:^- -I80UJ ^ ~ / / V ; / U ^ < -V 100 x >> ^ ' X K "^> J / / / '' / '/ / " * -t ">C / j [j Y\ ^-~^ X "- ^ / / / /\ 1 >^*M i'^ ^./ / 7 7 ^r> ^^120 0^ v S^>^/ / ^_i^ ^i ~ BS--B4 jZ_ /^N/ / / / Pfl ' ; 1> ^ \j ' < t- ; id ' S N. Jl /' i/' ' /^V^ / / ^_ 140 r X 1 ; >> Z.NS / / ^A / ::!:E;; % SC^vy^ ^ v : /I I2_ y\. Ou -- / X ^ ^ ' ^ ?^ - "^+ x - " 7 X 160 X ^ ^> ,d. _^_^H^ ' / / N^ ^t ^i I V |_ Z 7 - s ^^' J wi-^ \/ 7 >. >. ,c^ x ,-,_ ISO x \ / - :z. x/ yr ^ x ^ X - ^ ^^ / x \ 2500 3000 3500 4000 4500 2 ONS.PER.MIN. >F SINGLE PUMPS. DIAGRAM No. 29 50 HEAD IN FEET. 100 ISO 200 250 7 50 150 200 250 COAL REQUIRED IN PUMPING. 1000000 GALLONS OF WATER. DIAGRAM No.30 50 HEAD IN FEET. 100 150 200 250 20 40 60 80 PRESSURE IN LBS. DISCHARGE OF NOZZLES. 100 HLL CURVE FOR 50 HOSE. OTTED 100' PRESSURE INDICATED AT HYDRANT. DIAGRAM No. 31 OF THK UNIVERSITY 50 HEAD IN FEET. 100 150 200 250 20 PRESSURE IN LBS. 100 DISCHARGE OF NOZZLES. PRESSURE INDICATED AT BASE OF PLAY PIPE. DIAGRAM FOR LARGE NOZZLES MULTIPLY GALLONS BY 4. No. 32 VELOCITY IN FEET. 10 CD . 01 Ot CO ' > 5* ^fe ^ T < T ^ I UNIVEKSITY OF CALIFORNIA LIBRARY, BERKELEY THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW expiration of loan period. Oft 3 50m-8,'26 YB I 1 044 cf