COI  0.1 0.2 0.3 0.t 0.5 o.6 0.7 o.8 0.9 1.0 /w = cc = .624 .632 .643 .659 .68, .712 .755 .813 .892 1.00 c = 22 5.9 47.77 30.83 7.801 1.753 1.796 .797 .290 .060 .000 r • 4 (4) Elbows.-Weisbach considers the loss of

head at elbows (fig.91) to be due to a contraction formed by the stream . From experiments with a pipe i; in. diameter, he found the loss of head = ',v2/2g ; = 0.9457 +2.047 sin' 10 . Hence at a right-angled elbow the whole head due to the velocity very nearly is lost . Bends.-Weisbach traces the loss of head at curved bends to a similar cause to that at elbows, but the coefficients for bends are not very satisfactorily ascertained . Weisbach obtained for the loss of head at a bend in a pipe of circular section fb=i-bv2/2g; (6) where d is the diameter tea. of the pipe and p the Valves, Cocks and Sluices.-These produce a contraction of the water-stream, similar to that for an abrupt diminution of section already discussed . The loss of head may be taken as before to be ~v = 3',v2/2g; (7) where v is the velocity in the pipe beyond the valve and i•,, a coefficient determined by experiment . The following are Weisbach's results . Sluice in Pipe of Rectangular Section (fig . 92) . Section at sluice =w1 in pipe =w . wl/w = I.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 i'e = 0.00 •09 '39 '95 2.08 4.02 8.12 17.8 44'5 193 Sluice in Cylindrical Pipe (fig . 93) .

Ratio of height oft I.o I 1 I } I I } } opening to diameter J} of pipe wtlw = 1.00 0.948 .856 .740 .609 .466 .315 .159 1•n = 0.00 0.07 0.26 0.81 2.06 5.52 17.0 97.8

Cock in a Cylindrical Pipe (fig . 94) . Angle through which cock is turned =B . B = 5° 10° 15° 20° 25° 30° 350 Ratio of •85 0 '772 •692 '613 '535 .458 cross •926 sections j y . _ '05 '29 .75 1.56 3.10 5'47 9.68 B= 40° 45° 5.'° I 55° 6o° 65° 82° Ratio of •385 .315 •250 .190 .137 •091 0 cross sections )} 17.3 31.2 52.6 106 206 486 co - Throttle Valve in a Cylindrical Pipe (fig . 95) f a B = 5° to° 15° 20° 25° 30` 350 1 4G° '24 .52 '90 1 . J4 2.51 3.91 6.22 Io•8 9 = 450 50° 55° 60° 65° 70° 900 - = 18'7 32.6 58.8 118 256 751 co § 84 . Practical Calculations on the Flow of Water in Pipes.-In 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 the surface of the pipe needs to be considered . In general it is one of the four quantities d, i, v or Q which requires to be determined . For since the loss of head h is given by the relation h=il, this need not be separately considered . There are then three equa- tions (see eq . 4, § 72, and 9a, § 76) for the solution of such problems as arise:- ~'=a(I+I/I2d); where a =0.005 for new and =0.01 for incrusted pipes .

02/2g = 4di . (2) Q = . rd2v . (3) Problem I . Given the diameter of the pipe and its virtual slope, to find the

discharge and velocity of flow . Here d and i are given, and Q and v are required . Find from (1) ; then v from (2); lastly Q from (3) . This case presents no difficulty . By combining equations (I) and (2), v is obtained directly:- v=v (gdi/2i-) =v (g/2a)) {di/{1+I/12d} ] . (4) For new pipes . / (g/2a) =56.72 For incrusted pipes . . =40.13 For pipes not less than I, or more than 4 ft. in diameter, the mean values of are For new pipes 0.00526 For incrusted pipes 0.01052 . Using these values we get the very simple expressions- v=55'31,/ (di) for new pipes =39.11V (di) for incrusted pipes S Within the limits stated, these are accurate enough for practical purposes, especially as the precise value of the coefficient i- 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 i- by (I); 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 v from (3); from (I); lastly i from (2) . If we combine (I) and (2) we get =(v2/2g) (4/d) =2a{I+I/12d}v2/gd; (5) and, taking the mean values of i' for pipes from i to 4 ft. diameter, given above, the approximate formulae are i=0.0003268 v2/d for new pipes ] (5a) =0.0006536 v2/d for incrusted pipes 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 (I), which give the quadratic expression d2-d(2 av2/gi) - av2/6gi = o . d =av2/gi+v { (av2/g2) (av2/gi+I/6)} . (6) .

For practical purposes, the approximate equations d = 2av2/gi+ 1 / 12 (6a) =0.00031 v2/i+•083 for new pipes =o 00062 v2/i+.o83 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 get- d' =32'Q2/g7r2i . (7) If now the value of in (I) is introduced, the equation becomes very cumbrous . Various approximate methods of meeting the difficulty may be used . (a) Taking the mean values of given above for pipes of i to 4 ft. diameter we get d=V (32i'/g7r2)J (Q2/i) (8) =0.2216;1 (Q2/i) for new pipes =0•2541V (Q2/i) for incrusted pipes; equations which are interesting as showing that when the value of ° is doubled the diameter of pipe for a given discharge is only in-creased by 13 % .