# Page:EB1911 - Volume 14.djvu/77

65
HYDRAULICS

Elbojws.-Weisbach considers the loss of head at elbows (fig.91) to be due to a contraction formed by the stream. From experiments 9= 45° 50° 55° 60° 65° 70° 90° with a pipe Ii in. diameter, he found the loss of head I), =g-81,2/zg; (5) § , = 18-7 32-6 58-8 118 256 751 as fe =0'9457 S1H2i4>+2'047 S1114 i4>¢ “ 29° 40° 60° 80° 90° 100° 110° 120° 130° 140° I fe: ' 0.046 I Q139, 0.364 i o.740| o 984| I.26O| 1.5561 1.861 l 2.1581 2.431<| Hence at a right-angled elbow the whole head due to the velocity very nearly is lost. Bands.-Weisbach traces the loss of head at curved bends to a similar cause to that at <~ elbows, f but;tihe coeffi§ cients or ben s are not w very satisfactorily ascer tained. /Veisbach obw tained for the loss of hpad atl a bend in a pipe

i o circu ar section

b*-, ~;='°° f>»=fw2/zg; <6> § fb=o-131+1-847(d/2p)S, { where d is the diameter

of the pipe and p the

FIG. 91. radius of curvature of the bend. The resistance at bends is small and at present very ill determined. Valves, Cocks and Sluices.-These produce a contraction of the - water-stream similar to that for an abrupt 5 loss of head may be taken as before to be =§ ~-2 2g- where 1/ is the xelocity in the pipe beyond the valve and § ', a coefficient determined by experiment. The diminution of section already discussed. The F, fn, /, (7) v l following are Weisbach's results. Sluice in Pipe of Rectangular Section (fi . 2). FIG' 92' Section at s1uice=w1 in pipe=w. g 9 <-11/w= I-0 0-9 0-8 0-7 ~o-6 o-5 0-4 o-3 o-2 0-1 I 4 § '»= 0-oo -o9 -39 -95 2-08 4-02 8-12 17-8 44-5 193 Sluice in Cylindrical Pipe (fig. 93). Ratio of height of opening to diameter of pipe wi/w = 1 0 -3 i % iv § 1.00 0.948 .856 .740 .609 .466 .315 .159 ocoi 0.07 i o26' 0.81 | Loéi 5.52 I 17,0 '978 l § '., = / 3/ ww/4 /' 9 - - »:j; § t nw////////W, ff, ' g~§§ , ,, W”, W, /, W, % // / / / FIG. 93. FIG. 94. Cock in a Cylindrical Pipe (fig. 94). Angle through which cock is turned=0. 0= 5° 10° 15° 20° 25° 30° 35° Ratio of) c r o ss 2 -926 -850 ' -772 -692 -613 -535 -458 sections) § ¢= '05 '29 7| '75 V56 | 3'10 5'47 9°68 6? 409 V450 ' go l 55° 600 650 820 Ratio of c rto s s -385 -315 -250 -190 -137 'OQI o sec ions 5-, = 17-3 3I°2 52-6 IOC) 1-o6 486 so cal Pipe (fig. 95) Throttle Valve in a Cylindri If 0 = 5° 10° 15° I-1 = '24 '52 '90

0 20 25° 50° 35° 40° 1-54 i 2-51 3-91 i 6'22 iw-8 1 § 84. Practical Calculations the following explanations it will be assumed that the pipe is of so great a length that only the loss of head in friction against " on the Flow of Water in Pipes.-In the surface of the pipe needs WW/W H »-'JH / to be considered. In general M /M" /W ”W/W” it is one of the four quantities ' 3 d, i, 'U or Q which requires ' " '”"' ""“” to be determined. For since 'ii the loss of head h is given by /fr -W///2 <f- » '/M//7/7/W” //W the relation h=il, this need not be separately considered. FIG 95 There are then three equations (see eq. 4, § 72, and 9a, § 76) for the solution of such problems as arise:- i'=a(I'l'I/I2d)i (I) where a =o-005 for new and =o-01 for incrusted pipes. W2/22 = idl- (2) Q = i1fd2v~ (3) Problem 1. Given the diameter of the pipe and its virtual slope, to find the discharge Hlld velocity of flow. Here d and i are given, and Q and 'v are required. Find § ' from (I); then v from (2); lastly Q from (3). This case presents no difficulty. By combining equations (1) and (2), v is obtained directly:- U=/ (gdi/25) =/ (g/2¢)/ [di/ll +I/1211) 1- (4) For new pipes / (g/20.) =56-72 For incrusted pipes =4o-13 For pipes not less than 1, or more than 4 ft. in diameter, the mean values of gf' are For new pipes .. .. 0-00526 For incrusted pipes ..... O°01052. Using these values we get the very simple expressions=55-31/ (di) for new pipes 1 (46) =39-11/ (di) for incrusted pipes) Within the limits stated, these are accurate enough for practical purposes, especially as the precise value of the coefficient § ' cannot be known for each special case. Problem 2. Given the diameter of a pipe and the velocity of flow, to find the virtual slope and discharge. The discharge is given by (3); the proper value of of by (1); and the virtual slope by (2). This also presents no special difficulty. Problem 3. Given the diameter of the pipe and the discharge, to find the virtual slope and velocity. Find 11 from (3); § ' from (1); lastly i from (2). If we combine (1) and (2) we get i=§ '(v”/2§) (4/ll) =2¢1{I +I/I2fl}'v'/ed; (5) and, taking the mean values of § ' for pipes from 1 to 4 ft. diameter, given above, the approximate formulae are i=o-ooo3268 112/d for new pipes Q (5a) =o-0006536 vi'/d for incrusted pipes 5 Problem 4. Given the virtual slope and the velocity, to find the diameter of the pipe and the discharge. The diameter is obtained from equations (2) and (1), which give the quadratic expression d”-d(2av2/gi) -1.1212/6gi=0. d=11'v”/gi+/ l(fw”/gi) (<»'v”/gi+I/5)}~ (6) For practical purposes, the approximate equations d=2o.v2/gi+1/12 . (6a) =O'O0O3I oz/i+-083 for new pipes =O'OOO62 of/i-1—083 for incrusted pipes are sufficiently accurate. Problem 5. Given the virtual slope and the discharge, to find the diameter of the pipe and velocity of flow. This case, which often occurs in designing, is the one which is least easy of direct solution. From equations (2) and (3) we getd'=32s“Q”/g1f”i- (7) If now the value of f in (1) is introduced, the equation becomes very cumbrous. Various approximate methods of meeting the difficulty may be used. (a) Taking the mean values of g' given above for pipes of 1 to 4 ft. diameter we get 1l=%/(325/§ 1f“)i/(Qi/i) (8) =O'2216§ / (Q2/i) for new pipes =O'254I Q/ (Q2/i) for incrusted pipes; equations which are interesting as showing that when the value of 5° is doubled the diameter of pipe for a given discharge is only increased

by 13 %.