Page:EB1911 - Volume 14.djvu/77

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 11/4 in. diameter, he found the loss of head

ɧe = ζεv2 / 2g;

(5)

ζ_e = 0.9457 sin² 1/2φ + 2.047 sin⁴ 1/2φ.

φ =	20°	40°	60°	80°	90°	100°	110°	120°	130°	140°
ζε =	0.046	0.139	0.364	0.740	0.984	1.260	1.556	1.861	2.158	2.431

Hence at a right-angled elbow the whole head due to the velocity very nearly is lost.

 

Fig. 91.

 

Fig. 92.

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

ɧ_b = ζ_bv² / 2g;

(6)

ζ_b = 0.131 + 1.847 (d/2ρ)^7/2,

where d is the diameter of the pipe and ρ the 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 diminution of section already discussed. The loss of head may be taken as before to be

ɧ_v = ζ_vv² / 2g;

(7)

where v is the velocity in the pipe beyond the valve and ζ_v a coefficient determined by experiment. The following are Weisbach’s results.

Sluice in Pipe of Rectangular Section (fig. 92). Section at sluice = ω₁ in pipe = ω.

ω1/ω =	1.0	0.9	0.8	0.7	0.6	0.5	0.4	0.3	0.2	0.1
ζv =	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 of opening to diameter of pipe${\Big \}}$	1.0	7/8	3/4	5/8	1/2	3/8	1/4	1/5
ω1/ω =	1.00	0.948	.856	.740	.609	.466	.315	.159
ζv =	0.00	0.07	0.26	0.81	2.06	5.52	17.0	97.8

Fig. 93.	Fig. 94.

Cock in a Cylindrical Pipe (fig. 94). Angle through which cock is turned = θ.

θ =	5°	10°	15°	20°	25°	30°	35°
Ratio of cross sections${\Big \}}$	.926	.850	.772	.692	.613	.535	.458
ζv =	.05	.29	.75	1.56	3.10	5.47	9.68

θ =	40°	45°	50°	55°	60°	65°	82°
Ratio of cross sections${\Big \}}$	.385	.315	.250	.190	.137	.091	0
ζv =	17.3	31.2	52.6	106	206	486	∞

Throttle Valve in a Cylindrical Pipe (fig. 95)

θ =	5°	10°	15°	20°	25°	30°	35°	40°
ζv =	.24	.52	.90	1.54	2.51	3.91	6.22	10.8

θ =	45°	50°	55°	60°	65°	70°	90°
ζv =	18.7	32.6	58.8	118	256	751	∞

Fig. 95.

§ 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 equations (see eq. 4, § 72, and 9a, § 76) for the solution of such problems as arise:—

ζ = α (1 + 1/12d);

(1)

where α = 0.005 for new and = 0.01 for incrusted pipes.

ζv² / 2g = 1/4di.

(2)

Q = 1/4πd ²v.

(3)

Problem 1. 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 ζ from (1); then v from (2); lastly Q from (3). This case presents no difficulty.

By combining equations (1) and (2), v is obtained directly:—

v = √ (gdi /2ζ) = √ (g/2α) √ [di / {1 + 1/12d}].

(4)

For new pipes	√ (g/2α) = 56.72
For incrusted pipes	= 40.13

For pipes not less than 1, or more than 4 ft. in diameter, the mean values of ζ 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
⁠= 39.11 √ (di) for incrusted pipes.

(4a)

Within the limits stated, these are accurate enough for practical purposes, especially as the precise value of the coefficient ζ 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 ζ 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 v from (3); ζ from (1); lastly i from (2). If we combine (1) and (2) we get

i = ζ (v²/2g) (4/d) = 2a {1 + 1/12d} v²/gd;

(5)

and, taking the mean values of ζ for pipes from 1 to 4 ft. diameter, given above, the approximate formulae are

i = 0.0003268 v²/d for new pipes
⁠= 0.0006536 v²/d for incrusted pipes.

(5a)

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 ² − d (2αv²/gi) − αv²/6gi = 0.

∴ d = αv²/gi + √ {(αv²/gi) (αv²/gi + 1/6)}.

(6)

For practical purposes, the approximate equations

d = 2αv²/gi + 1/12
  = 0.00031 v²/i + .083 for new pipes
  = 0.00062 v²/i + .083 for incrusted pipes

(6a)

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 5 = 32ζQ2 / gπ2i.

(7)

If now the value of ζ 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 ζ given above for pipes of 1 to 4 ft. diameter we get

d = 5√ (32ζ/gπ2) 5√ (Q2/i)⁠

(8)

= 0.2216 ⁵√ (Q²/i) for new pipes
= 0.2541 ⁵√ (Q²/i) for incrusted pipes;

equations which are interesting as showing that when the value of ζ is doubled the diameter of pipe for a given discharge is only increased by 13%.