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 rightangled 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=ibv2/2g; (6)
where d is the diameter
tea. of the pipe and p the
Valves, Cocks and Sluices.These produce a contraction of the waterstream, 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
d2d(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 increased by 13 %.
(5)
20° 0.046
40°
0.139
0.364
6o°
0.740
8o°
End of Article: COI 

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