| 15
684
MECHANICS OF ENGINEERING.
Since $b$ is unknown, we cannot use the table immediately, but take $\mu_c = .600$ for a first approximation; whence, eq. (10), (ft., lb., sec.)
$$b = \frac{6}{0.6 \times 4 \times 12 \times 32.2 \times \frac{1}{12}} = 2.46 \text{ ft}.$$
Then, since this width does not much exceed 0.60 metre, we may take, in Table C, for $A_b$, for $\mu_c = .589$;
$$b = \frac{6}{.589 \times 4 \times 12 \times 32.2 \times \frac{1}{12}} = 2.50 \text{ ft}.$$
**Case II. Incomplete Contraction; i.e., both ends are flush with the sides of the tank, those being $b$ to the plane of the notch. According to Weishab, we may write**
$$Q = \frac{4\pi b h_1 + 2gh_1}{g}h_2,$$
(11)
in which $\mu = 1.041\mu_c$, $\mu_c$ being obtained from Table C for the normal case, i.e., Case I. The section of channel of approach is large compared with that of the notch; if not, see Case IV.
**Case III. Imperfect Contraction; i.e., the velocity of approach is appreciable; the sectional area $G$ of the channel of approach not being much larger than that, $F = bh_1 = area$ of notch. Fig. 558. $b$ is width, and $h_1$ is depth of notch (see Case I). Here, instead of using a formula involving**
$$k = c^* + 2g = [Q + G]^* + 2g$$
(see eq. (3), § 501), it is more convenient to put
$$Q = \frac{4\pi b h_1 + 2gh_1}{g}h_2,$$
as before, with
$$\mu = \mu_c(1 + \beta)$$
(18)
in which $\mu_c$ is for the normal case [Case I]; and $\beta$, according
OVERFALL-WEIBS.
685
to Weibach's experiments, may be obtained from the empirical formula
$$\beta = 1.718 \left( \frac{F}{Q} \right)$$
(12)
[Table D is computed from (12).]
(The contraction is complete in this case; i.e., the ends are not flush with the sides of the tank.)
EXAMPLE.—If the water in the channel of approach has a vertical transverse section of $A = 9$ sq. feet, with both ends flush with the sides (i.e., $b = 9$) and 1 foot deep ($h = 1$) (to level of surface of water 3 ft. or less of bottom), we have, from Table C, with $b = .60$ met. and $h_a = 0.30$ met.
$$\mu = 0.586$$
while from Table D, with $F$: $Q = 0.999$ (or $\frac{4}{3}$),
$$\beta = .005$$
hence (ft.-lb.-sec. system of units), from eq. (12),
$$Q = 1 \times 0.586 \times 1.005 \times 2 \times 1 \times \sqrt{64.4 \times 1.0}$$
= 6.30 cub. ft. per second.
CASE IV. Fig. 559. Imperfect and incomplete contractions together; both end-constructions being "suppressed" (by making the ends flush with the sides of the reservoir, these sides being at a distance of about one foot from the mouth), and the channel of approach not being very deep, i.e., having a sectional area $Q$ but little larger than that, $F$, of notch: $F = b h_a$
as before.
Again we write
$$Q = \frac{4}{3} \mu b h_a \sqrt{2gh_a}$$
with $\mu$ computed from
$$\mu = \mu_1 (1 + \beta)$$
(13)
$\mu$ being obtained from Table C; while
686
MECHANICS OF ENGINEERING.
$$\beta = 0.041 + 0.3693 \left( \frac{F}{G} \right)^* \dots \dots (13^*)$$
an empirical formula based by Weibach on his own experiments.* To save computation, $\beta$ may be found from Table E, founded on eq. (13)*.
| $F$ |
0.00 |
0.05 |
0.10 |
0.15 |
0.20 |
0.25 |
0.30 |
0.35 |
0.40 |
0.45 |
0.50 |
| $G$ |
0.041 |
0.043 |
0.045 |
0.049 |
0.056 |
0.064 |
0.074 |
0.086 |
0.100 |
0.116 |
0.133 |
**EXAMPLE.—Fig. 559. With**
$$b = 2 \text{ ft. (= 0.60 met.)}$$
and
$$h_s = 1 \text{ ft. (= 0.30 met.)}$$
we have, from Table C, $\mu_s = 0.556$. But, the ends being flush with the sides of the reservoir or channel, and $G$ being $= 6$ sq. ft. (see figure), which is not excessively large compared with $F = b h_s = 2$ sq. ft., we have, from Table E, with $F$: $G = 0.333$, $$\beta = .081;$$
and hence [eq. (13) and (13')] $\mu_s$ being $.586$ as in last example,
$$Q = \frac{4 \times 0.586 \times (1 + .891) \times 2 \times 1 \times .644 \times .16}{6.78} \text{ cfs./ft.\ per sec.}$$
585. Francis' Formula for Overfall (i.e., rectangular notch).
—from extensive experiments at Lowell, Mass., in 1851, with rectangular overfall-weirs, Mr. J. B. Francis deduced the following formula for the volume, $Q$, of flow per second over such weirs 10 feet in width, and with $h_s$ varying from $6$ to $1.6$ feet (from sill of notch to level surface of water a few feet back):
* Weibach's results in this case differ considerably from those of Basin; see p. 688.
![]() Fig. 559.
OVERFALL-WEIRS.
687
$$Q = \frac{4}{3} \times 0.6224 (\beta - \frac{1}{\sqrt{n}}h) \cdot \frac{2gL}{b}, \quad . . . (14)$$
In which $b$ = width.
This provides for incomplete contraction, as well as for complete and perfect contraction, by making
$n = 2$ for perfect and complete contraction (Fig. 557);
$n = 1$ when one end only is flush with side of channel;
$n = 0$ when both ends are flush with sides of channel.
The contraction was considered complete and perfect when the channel of approach was made as wide as practicable, $= 13.96$ feet, the depth being about 5 feet.
Mr. Francis also experimented with submerged or "drowned" weirs in 1883; such a weir being one in which the sill is below the level of the tail-water (i.e., of receiving channel).
508. Falesley and Stearn's Experiments at Boston, Mass., in 1877 and 1880.—These may be found in the Transactions of the American Society of Civil Engineers, vol. xxi, and gave rise to formulae differing slightly from those of Mr. Francis in some particulars. In the case of supposed end-contractions, like that in Fig. 359, they propose formulae as follows:
When depth of notch is not large,
$$Q \text{(in cub. ft. per sec.)} = 3.31 b_1 h + 0.007 b \quad . . . (15)$$
($b$ and $h$, both in feet),
"A $_4$, the depth on the weir, should be measured from the surface of the water above the curvature of the sheet."
"Air should have free access to all space under the sheet."
The cross-section of the sheet is shown in Fig. 360. The formula does not apply to depths on the weir less than 0.07 feet.
When the depth of notch is quite large, a correction must be made for velocity of approach, $c$, thus:
$$Q \text{(in cub. ft. per sec.)} = 3.31 b \left[ A_4 + 1.5 \frac{c^2}{g} \right] + 0.007 b \quad . . . (16)$$
($b$ and $h$, both in feet).
688
MECHANICS OF ENGINEERING.
The channel should be of uniform rectangular section for about 30 ft. more from the weir, to make this correction properly. If $G$ = the cross-section, in sq. ft., of the channel of approach, $Q$ is found approximately by dividing an approximate value of $Q$ by $G$; and so on for closer results.
The weir may be of any length, $L$, from 5 to 19 feet.
506a. Recent Experiments on Overfall-weirs in France.—In the Annales des Ponts et Chaussées for October 1888 is an account of extensive and careful experiments conducted in 1886 and 1887 by M. Bazin on the flow over shallow overfall-weirs with end-conditions similar to those like that shown in Fig. 529. The widths of the weirs ranged from 0.50 to 2.00 metres, and the depths on the weirs ($h_1$) from 0.05 to 0.60 metre. With $p$ indicating the height of the sill of the weir from the bottom of the channel of approach, M. Bazin, as a practical result of the experiments, recommends the following formula as giving a reasonably accurate value for the volume of discharge per unit of time:
$$Q = \frac{4}{\mu} \left[1 + \frac{0.55(h_1 - p)}{p + h_1}\right]^{2} \left( h_1 + \frac{0.0148}{p} \right) \dots (17)$$
where the coefficient $\mu'$ has a value
$$\mu' = 0.6075 + \frac{0.0148}{h_1 (\text{in ft.)}} \dots (18)$$
Eq. (17) is homogeneous, i.e., admits of any system of units.
Provision was made in these experiments for the free entrance of air under the sheet (a point of great importance), while the pressures at all points upstream were continued down-stream, beyond the plane of the weir, to allow any lateral expansion of the sheet.
The value of $p$ ranged from 0.20 to 2.00 metres.
Herr Ritter von Wex in his "Hydrodynamik" (Leipsic, 1886) derives formulae for weirs, in the establishing of which some rather peculiar views in the Mechanics of Fluids are advanced.
OVERFALLS.
689
Formulae and tables for discharge through orifices or over weirs of some forms not given here may be found in the works of Weibach, Rankine, and Troutwine.
Mr. Hamilton Smith, a noted American hydraulic engineer, has recently published "Hydraulics," a valuable compilation and record of the author's many experiments in all fields of hydraulics (New York, 1886: John Wiley & Sons).
507. Eflux through Short Cylindrical Tubes.—When efflux takes place through a short cylindrical tube, or "short pipe," at least 24 times as long as wide, inserted at right angles in the plane side of a large reservoir, the inner corners are rounded (see Fig. 386), so that jets issue from the tube in parallel filaments and with a sectional area, $F_m$, equal to that, $F_t$ of interior of tube.
To attain this result, however,
the tube must be full of water before the outer end is unstopped, and must not be oily; nor must the head, $h$, be greater than about 40 ft. for efflux into the air. Since at $m$ the filaments are parallel and the pressure head therefore equal to $(c = 34 \text{ ft. of water})$, the value of $\phi_0$ is zero; i.e., head due to one atmosphere in this instance; an application of Bernoulli's Theorem [eq. (7), § 499] to positions $m$ and $n$ would give (precisely as in §§ 454 and 455)
$$v_m = \text{veloc. at } m = \sqrt{2gh}$$
as a theoretical result; but experiment shows that the actual value of $v_m$ in this case is
$$v_m = \phi_0 \cdot \sqrt{2gh} = 0.815 \cdot \sqrt{2gh} \quad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$$
0.815 being an average value for $\phi_0$, the coefficient of velocity, for ordinary purposes. It increases slightly as the head decreases,
260 MECHANICS OF ENGINEERING.
and is evidently much less than the value 0.97 for an orifice in a thin plate, § 495, or for a rounded mouth-piece as in § 496.
But as the sectional area of the stream where the filaments are parallel, at $m$, where $v_m = 0.815 \frac{F}{2gh}$, is also equal to that, $F'$ of the tube, the coefficient of efflux $\mu_s$, in the formula
$$Q = \mu_s F' \sqrt{2gh}$$
is equal to $\phi_s$; i.e., there is no contraction, or the coefficient of contraction, $C_s$ in this case is 1.00.
Hence, for the volume of discharge per unit of time, we have practically
$$Q = \phi_s F' \sqrt{2gh} = 0.815 F' \sqrt{2gh}.$$ (9)
The discharge is therefore about $\frac{3}{4}$ greater than through an orifice of the same diameter in a thin plate under the same head [compare eq. (8), § 495]; for although at $m$ the velocity is less in the present case, the sectional area of the stream is greater, there being no contraction.
This difference arises due principally to the fact that the entrance of the tube has square edges, so that the stream contracts (at $m'$, Fig. 561) to a section smaller than that of the tube, and then re-expands to the full section, $F'$ of tube. The eddying and accompanying internal friction caused by this re-expansion (or "sudden enlargement" of the stream) is the prin- cipal resistance which diminishes the velocity. It is noticeable also, in this case that the jet is not limpid and clear, as from thin plate, but troubled and only transluent (like ground-glass). The internal pressure in the stream at $m$ is found to be less than one atmosphere, viz., about 0.8 atm., as shown experimentally by the sudden ing in of air when a small aperture is made in the tube op-
![]() Fig. 561.
![]() Fig. 562.
INCLINED SHORT PIPES.
091
poste $m'$. If the tube itself were so formed internally as to fit this contracted vent, as in Fig. 563, the eddy would be diminished and the velocity at $m$ increased, and hence the volume $Q$ of efflux increased in the same proportion. (See § 509.)
If the tube is less than 24 times as long as wide, or if the interior is not wet by the water (as when greasy), or if the head is over 40 or 50 ft. (about), the efflux takes place as if the tube were not there, Fig. 563, and we have
$$v_m = 0.97 \sqrt{gh}, \text{ as in } \S 495.$$
**EXAMPLE.** The discharge through a short pipe 3 inches in diameter, like that in Fig. 560, is 30 cubic feet per minute under a head of $g^{\prime}h$, say 10 feet. Required the coefficient of efflux $\mu_e = \phi_e$, in this case.
For variety use the inch-pound-minute system of units, in which $g = 32\frac{1}{2} \times 12 \times 3600$ (see Note, § 51). $\mu_e$, being an abstract number, will be the same numerically in any system of units.
From eq. (2),
$$\phi_e = \mu_e = \frac{Q}{F^{\prime}\sqrt{gh}} = \frac{90 \times 1798}{\pi \times 3^2 \times g^{\prime} \times 32\frac{1}{2} \times 12 \times 60^{\circ} \times 30}$$
$$= 0.803.$$
§ 508. Inclined Short Tubes (Cylindrical).—Fig. 564. If the short tube is inclined at some angle $\alpha < 90^{\circ}$ to the interior plane of the reservoir wall, the efflux is smaller than when the angle is $90^{\circ}$, as in § 507.
We still use the form of equation
$$Q = \mu F^{\prime}\sqrt{gh} = \phi F^{\prime}\sqrt{gh}, \quad (3)$$
but from Weibach's experiments $\mu$ should be taken from the following table:
![]() A diagram showing an inclined short tube with a vent near its base.
Fig. 564
692
MECHANICS OF ENGINEERING.
TABLE F. COEFFICIENT OF EFFLUX (INCLINED TUBE).
|
For α = 90° |
80° |
70° |
60° |
50° |
40° |
30° |
| Take µ = φ = .715 |
.799 |
.769 |
.764 |
.747 |
.731 |
.719 |
EXAMPLE.--With h = 12 ft., d = diam. of tube = 4 ins., and α = 45°, we have for the volume discharged per sec. (ft., lb., and sec.)
$$Q = [0.731 + \frac{1}{4}(.016)\frac{\pi(1)}{3}]^2\sqrt[4]{4.4\times12} = 1.79 \text{ cub. ft. per sec.}$$
The tube must be at least 3 times as long as wide, to be filled.
509. Conical Diverging, and Converging, Short Tubes.--With conical convergent tubes, as at A, Fig. 565, with inner edges not rounded, D'Aubuisson and Castel found by experiment values of the coefficient of velocity, φ, and of that of efflux μ [from which the coefficient of contraction, C = μ + φ, may be computed.] for tubes 1.55 centimeters wide at the narrow end, and 4.00 centimeters long, under a head of h = 3 metres, and with different angles of convergence. By angle of convergence was meant the angle between the sides CE and DE, Fig. 565. In the following table will be found some values of x and φ founded on these experiments, for use in the formula
$$v_m = \phi \sqrt{2gk} \quad \text{and} \quad Q = \mu F \sqrt{2gk};$$
in which $F$ denotes the area of the outlet orifice $EB$.
![]() A diagram showing a conical diverging tube with a converging section.
Fig. 565.
CONICAL SHORT PIPES.
693
| TABLE G (CONICAL CONVERGING TUBE). |
| Angle of convergence |
8° 10' |
8° 10' 90' |
18° 90' |
18° 90' |
80° |
45° |
| $\mu$ = .965 |
.965 |
.965 |
.965 |
.965 |
.965 |
.965 |
| $\phi$ = .964 |
.964 |
.964 |
.964 |
.964 |
.964 |
.964 |
Evidently $\mu$ is a maximum for $18^{\circ}$.
With a conically divergent tube at $MN$, having the internal diameter $MO = .025$ metre, the internal diam. $NP = .032$ metre, and the angle between $MN$ and $PN$ $= 18^{\circ}$, Weibull found that with $F = .027$ square metres, $\frac{V_2}{V_1} = \sqrt{\frac{F}{\pi d}}$ (where $F =$ area of outlet section $NP$) $\mu$ should be $=.553$; the great loss of velocity as compared with $\sqrt{\frac{d}{\pi}}$ being due to the eddying in the re-expansion from the contracted section at $M$ (corners not rounded), as occurs also in Fig. 549. The jet was much troubled and pulsed violently.
When the angle of divergence is too great, or the head $h$ too large, or if the tube is not wet by the water, efflux with the tube filled cannot be maintained, the flow then taking place as in Fig. 550.
Venturi and Eytelwein experimented with a conically divergent tube (called now "Venturi's tube"), with rounded entrance to conform to the shape of the contracted vein, as in Fig. 566, having a diameter of one inch at $m$ (narrowest part), where the sectional area is $A = .0783$. At $m$, where $V_2 = 1.80$ inches at $m$ (outlet), where area $= F'$; the length being 8 ins., and the angle of convergence $5^{\circ} 9'$.
With $Q = \mu F' V_2 \frac{d}{\pi h}$ they found $\mu = .483$.
Hence 24 times as much water was discharged as would have flowed out under the same head through an orifice in thin plate with area $= F' =$ the smallest section of the divergent tube, and 1.9 times as much as through a short pipe of section $= F'$. A similar calculation shows that the velocity at $m'$ when $\mu = .483$, and $\phi = .964$, showed that the pressure at $m'$ was much less than one atmosphere.
![]() Fig. 550.
694
MECHANICS OF ENGINEERING.
Mr. J. B. Francis also experimented with Venturi's tube (see "Lowell Hydraulic Experiments"). See also p. 389 of vol. 6 of the Journal of Engineering Societies, for experiments with diverging short tubes discharging under water. The highest coefficient ($\mu$) obtained by Mr. Francis was 0.782.
509a. New Forms of the Venturi Tube.—The statement made in § 507, in connection with Fig. 562, was based on purely theoretic grounds, but has recently (Dec. 1888) been completely verified by experiments conducted in the hydraulic laboratory of the Massachusetts Institute of Technology, at Cambridge, Mass., under the direction of Professor W. H. Emmons, of Harvard University. Three short tubes of circular section, each 3 in. in length and 1 in. in internal diameter at both ends, were experimented with, under heads of 2 ft. and 4 ft. Call them A, B and C. A was an ordinary straight tube as in Fig. 561; the longitudinal section of B was like that in Fig. 562, the narrowest diameter being 0.80 in. (see § 495; $0.8^{\prime} = 0.64$); while C was somewhat like that in Fig. 566, being formed like B up to the narrowest part (diameter 0.80 in.), and then made gradually divergent to the discharging end. The results of the experiments are given in the following table:
| Name of Tube |
Head. |
Number of trials. |
Range of Values of $\mu$. |
Average value of $\mu$. |
| A |
A = 2 ft. |
4 |
From 0.90 to 0.98 |
0.914 |
| A |
A = 4 ft. |
4 |
From 0.818 to 0.98 |
0.931 |
| B |
A = 2 ft. |
5 |
* * * 0.975 to 0.995 |
0.983 |
| B |
A = 4 ft. |
4 |
* * * 0.981 to 0.998 |
0.992 |
| C |
A = 2 ft. |
5 |
* * * 0.990 to 0.919 |
0.901 |
| C |
A = 4 ft. |
4 |
* * * 0.908 to 0.983 |
0.914 |
The fact that B discharges more than A is very noticeable, while the superiority of C to B, though evident, is not nearly so great as that of B to A, showing that in order to increase the discharge of an (originally) straight tube (by encroaching on the passage-way) it is of more importance to fill up with solid substance the space around the contracted vein than to make the transition from the narrow section to the discharging end very gradual.
* See Journal of the Franklin Inst., for April, 1889.
![]() Description: A diagram showing three different forms of Venturi tubes used for hydraulic experiments.
"FLUID FRICTION."
695
510. "Fluid Friction."—By experimenting with the flow of water in glass pipes inserted in the side of a tank, Prof. Reynolds of England has found that the flow goes on in parallel filaments for only a few feet from the entrance of the tube, and that then the liquid particles begin to intermingle and cross each other's paths in the most intricate manner. To render this more vividly, I visited a laboratory where colored liquid at the inlet of the pipe and observed its further motion, and found that the greater the velocity the nearer to the inlet was the point where the breaking up of the parallelism of flow began. The hypothesis of laminated flow is, nevertheless, the simplest theoretical basis for establishing practical formulae, and the resistance offered by pipes to the flow of liquids in them will therefore be attributed to the friction of the edges of the lamina against the inner surface of the pipe.
The amount of this resistance (often called skin-friction) for a given extent of rubbing surface is by experiment found—
1. To be independent of the pressure between the liquid and the solid;
2. To vary nearly with the square of the relative velocity;
3. To vary directly with the amount of rubbing surface;
4. To vary directly with the heaviness ($y$, § 408) of the liquid.
Hence for a given velocity $v$, a given rubbing surface of area $S$, and a liquid of heaviness $y$, we may write
$$\text{Amount of friction (force)} = fSy \frac{v^2}{2g}, \quad \ldots \ldots \ldots (1)$$
in which $f$ is an abstract number called the coefficient of fluid friction, to be determined by experiment. For a given liquid, given character (roughness) of surface, and small range of velocities it is approximately constant. The object of introducing $\frac{v^2}{2g}$ is not only because $\frac{v^2}{2g}$ is a familiar and useful function of $v$, but that $v^2 - 2g$ is a height, or distance, and therefore the product of $S$ (an area) by $v^2 - 2g$ is a volume, and this volume multiplied by $y$ gives the weight of an ideal prism of
![]() A diagram showing a cross-section of a pipe with water flowing through it.
096
MECHANICS OF ENGINEERING.
the liquid; hence $S \frac{v}{2g} r$ is a force and $f$ must be an abstract number and therefore the same in all systems of units, in any given case or experiment.
In his experiments at Torquay, England, the late Mr. Froude found the following values for $f$, the liquid being salt water, while the rigid surface was on two sides of a thin straight wooden bar $A$, of an inch thick and 19 inches long, polished or prepared in various ways, and drawn edgewise through the water at a constant velocity, the total resistance being measured by a dynamometer.
511. Mr. Froude's Results.—(Condensed.) [The velocity was the same = 10 ft. per sec. in each of the following cases. For other velocities the resistance was found to vary nearly as the square of the velocity, the index of the power varying from 1.8 to 2.16.]
| Character of Surface. |
|
Value of $f$ (from eq. (3), § 18). |
|
|
When the board was |
|
|
2 ft. long |
8 ft. long |
16 ft. long |
| Varnish |
$J =$ |
0.064 |
0.0028 |
0.0098 |
0.0055 |
| Pearlinae |
|
0.008 |
0.001 |
0.007 |
|
| Tufalite |
|
0.0087 |
0.003 |
0.015 |
0.025 |
| Calcio. |
|
0.007 |
0.063 |
0.033 |
0.067 |
| Fine Sand. |
|
0.069 |
0.254 |
0.169 |
0.499 |
| Medium Sand. |
|
0.069 |
0.262 |
0.168 |
0.499 |
| Coarse Sand. |
|
0.119 |
0.771 |
0.569 |
|
| N.B. These numbers multiplied by 10 also give the same fractional resistance in the case of a pipe of 1 foot diameter, but with a cross-section equal to that of a square of area 64 sq. ft., per cubic foot, to cancel the $J$ = 64 sq ft per sec. of eq. (3) of this section. |
For use in formula bearing on flow in pipes, $f$ is best determined directly by experiments of that very nature, the results of which will be given as soon as the proper formulae have been established.
512. Bernoulli's Theorem for Steady Flow, with Friction.—[The student will now re-read the first part of § 492, as far as eq.
(1).] Considering free any laminar of fluid, Fig. 567, (according to the subdivision of the stream agreed upon in § 492 referred
BERNOULLI'S THEOREM WITH FRICTION. 697
to,) the frictions on the edges are the only additional forces as compared with the system in Fig. 534. Let $w$ denote the length of the *rected perimeter* of the base of this lamina (in case of a pipe running full, as we here postulate), then $w$ is of course the *whole perimeter*, but in the case of an open channel or canal, $w$ is only a portion of the whole perimeter of the cross-section). Then, since the area of rubbing surface at the edge is $S = w d e$, the total friction for the lamina is [by eq. (1), § 510] $= f w y (e^2 + 2 g) d e$. Hence from $w e = (\tan . \sec .) \times d e$, and from $(\tan . \sec .) = [\frac{2}{g}(\tan . \cos . \text{comp. of acting forces})]$ $\times$ (mass of lamina), we have
$$Fp - F(p + dp) + Fyde \cos \phi - f w y \frac{v^2}{g} de = Fyde + g \cdot de... (a)$$
As in § 492, so here, considering the simultaneous advance of all the lamina lying between any two sections $m$ and $n$ during the small time $dt$, putting $de = ds$, and $de \cos \phi = -ds$ (see Fig. 568), we have, for any one lamina,
$$\frac{1}{g} v dv + \frac{1}{g} dp + ds = -f w \frac{v^2}{g} dy... (1)$$
Now conceive an infinite number of equations to be formed like eq. (1), one for each lamina between $m$ and $m$, for the same $dt$, viz., a $de$ of such length that each lamina at the end of $dt$ will occupy the same position, and acquire the same value of $v$, and so that the lamina had in its front had at the beginning of the $dt$ (this is the characteristic of a steady flow). Adding up
![]() A diagram showing a pipe with water flowing through it, illustrating Bernoulli's theorem with friction.
![]() A diagram showing a pipe with water flowing through it, illustrating Bernoulli's theorem with friction.
![]() A diagram showing a pipe with water flowing through it, illustrating Bernoulli's theorem with friction.
![]() A diagram showing a pipe with water flowing through it, illustrating Bernoulli's theorem with friction.
698
MECHANICS OF ENGINEERING.
the corresponding terms of all these equations, we have (re-
membering that for a liquid $y$ is the same in all laminae),
$$\frac{1}{g}\int_{a}^{b}v\,ds+\frac{1}{y^2}\int_{a}^{b}dp+\int_{a}^{b}zds=-\frac{f}{2g}\cdot\int_{a}^{b}\frac{w}{F}v^2ds;\tag{9}$$
i.e., after transposition and writing $R$ for $F+v$, for brevity,
$$\frac{u^2}{2g}+\frac{P_s}{y}+s_0=\frac{u^2}{2g}+\frac{P_s}{y}+s_0-\frac{f}{2g}\int_{a}^{b}\frac{v^2}{R}ds.\tag{9'}$$
This is Bernoulli's Theorem for steady flow of a liquid in
a pipe of slightly varying sectional area $F$, and internal perim-
eter $w$, taking into account no resistances or friction, except
the "skin-friction," or "fluid-friction" of the liquid and sides
of the pipe.
Resistances due to the internal friction of eddying occasioned
by sudden enlargements of the cross-section of the pipe, by
elbows, sharp curves, valve-gates, etc., will be mentioned later.
The negative term on the right in (3) is of course a height or
head (one dimension of length), as all the other terms are such,
and since it is the amount by which the sum of the three heads
(viz., velocity-head, pressure-head, and potential head) at $m$,
the down-stream position, lacks of being equal to the sum of
the corresponding heads at $n$, the up-stream position or section,
we may call it the "Loss of Head" due to skin-friction between
$a$ and $m$; also called *friction-head*, or resistance-head, or
height of loss.
The quantity $R=F+v$ is sectional-area + wetted-perimeter,
is an imaginary distance or length called the Hydraulic Mean Radius,
or Hydraulic Mean Depth, or simply hydraulic radius of the section.
For a circular pipe of diameter $d$,
$$R=\frac{\pi d^2}{4\pi d^2}=d;\tag{10'}$$
while for a pipe of rectangular section,
$$R=\frac{ab}{2(a+b)}.\tag{10'''}$$
**FRICITION IN PIPES.**
513. Problems involving Friction-heads; and Examples of Bernoulli's Theorem with Friction.
**PROBLEM I.**—Let the portion of pipe between $n$ and $m$ be level, and of uniform circular section and diameter = $d$. The jet at $m$ discharges into the air, and has the same sectional area, $F = \pi d^2$, as the pipe from which the pressure-head at $m$ is
$$P_m = b = 34 \text{ feet (for water), and the velocity head at } m \text{ is that at } n, \text{ since } v_n = v_m.$$ The height of the water column in the open piezometer at $n$ is noted, and is $y_n$ (so that the pressure-head at $n$ is $\frac{y_n}{g} = y_n + h$); while the length of pipe from $n$ to $m$ is $l$.
Knowing $l$, $d$, $y_n$, and having measured the volume $Q$ of flow per unit of time, it is required to find the form of the friction-head and the value of $f$. From
$$F_n v_n = Q_1 \text{ or } \frac{\pi d^2 v_n}{g} = Q_1 \dots \dots (1)$$
$v_n$ becomes known. Also, $v_n$ is known to be $= v_m$, and the velocity at each $ds$ is $v = v_m$, since $F(sections\ area)$ is constant along the pipe, and $F_0 = Q$. The hydraulic radius is
$$R = \frac{l}{d}; \dots \dots (2)$$
the same for all the $ds$'s between $n$ and $m$.
Substituting in eq. (3) of § 312, with the horizontal axis of the pipe as a datum for potential heads, we have
$$\frac{v_m^2}{2g} + b + 0 = \frac{v_n^2}{2g} + y_n + b + 0 - \frac{l}{d^2} \cdot \frac{v_m^2}{2g} ds; \dots (3)$$
i.e., since $\int ds = l =$ length of pipe from $n$ to $m$, the friction-head for a pipe of length $l$, and uniform circular section of diameter = $d$, reduces to the form
700
MECHANICS OF ENGINEERING.
Friction-head = 4f $\frac{l}{d}$ $\frac{v^2}{2g}$ ; . . . (6)
where $v$ = velocity of water in the pipe, being in this case also $= v_{w}$ and $= v_{s}$. Hence this friction-head varies directly as the length and as the square of the velocity, and inversely as the diameter; also directly as the coefficient $f$.
From (3), then, we derive (for this particular problem)
Piezometer-height at $n = y_n = 4f \frac{l}{d} \frac{v^2}{2g}$ ; . . . (5)
i.e., the open piezometer-height at $n$ is equal to the loss of head (all of which is friction-head here) sustained between $n$ and the mouth of the pipe. (Pipe horizontal).
EXAMPLE. Required the value of $f$, knowing that $d = 3$ in., $y_1$ (by observation) = 10.4 ft., and $Q = 0.1960$ cub. ft. per sec., while $l = 400$ ft. ($n$ to $m$). From eq. (1) we find, in ft.-lb.-sec. system, the velocity in the pipe to be
$v = \frac{4Q}{\pi d^2} = \frac{4 \times 0.1960}{\pi \times 3^2} = 4.0$ ft. per sec;
then, using eq. (5), we determine $f$ to be
$f = \frac{9py_1d}{4Ql} = \frac{9 \times 82.9 \times 1 \times 10.4}{4 \times 400 \times 3^2} = 0.0065.$
PROBLEM II. Hydraulic Accumulator—Fig. 570. Let the area $F_2$ of the piston on the left be quite large compared with that of the pipes and nozzle. The cylinder contains a fricton-
**FRICITION-HEAD IN PIPES.**
701
less weighted piston, producing (so long as its downward slow motion is uniform) a fluid pressure on its lower face of an intensity $p_n = [G + F_n p_a] - F_n$ per unit area ($p_n = \text{one stmos}$).
Hence the pressure-head at $n$ is
$$\frac{p_n}{y} = \frac{G}{F_n} + b_n \quad \ldots \ldots \quad (6)$$
where $G = \text{load on piston}$.
The jet has a section $a = F_a$, = that of the small straight nozzle (no contraction). The junctions of the pipes with each other, and with the cylinder and nozzle, are all smoothly rounded; hence the only losses of head in steady flow between $a$ and $m$ are the friction-heads in the two long pipes, neglecting that in the short nozzle. These friction-heads may be of the form in eq. (4), and will involve the velocities $v_a$ and $v_m$, respectively in these pipes (supposed running full). $v_a$ and $v_m$ may be unknown at the outset, as here.
Known to us are the lengths and diameters of the pipes, we are required to find the velocity $v_a$ of the jet, flowing into the air, and the volume of flow, $Q$, per unit of time, assuming $f$ to be known and to be the same in both pipes (not strictly true).
Let the lengths and diameters be denoted as in Fig. 570, their sectional areas $F_a$ and $F_m$, the unknown velocities in them $v_a$ and $v_m$.
From the equation of continuity [eq. (3), § 490], we have
$$v_a = \frac{F_an}{F_a} \quad \text{and} \quad v_m = \frac{F_mn}{F_m} \quad \ldots \ldots \quad (7)$$
To find $v_m$, we apply Bernoulli's Theorem (with friction), eq. (8), § 512, taking the down-stream position $m$ in the jet close to the nozzle, and the up-stream position $n$ just under the piston in the cylinder where the velocity $v_a$ is practically nothing. Then with $m$ as datum plane we have
$$\frac{u^2_m}{2g} + b + 0 + G + F_m n - 4f \frac{l}{d_m} u^2_m = 4f \frac{l}{d_a} u^2_a \quad \ldots \ldots \quad (8)$$
702
MECHANICS OF ENGINEERING.
Apparently (8) contains three unknown quantities, $v$, $\tau$, and $v_1$; but from eqs. (7) $v$ and $v_1$ can be expressed in terms of $w$, whence [see also eq. (6)]:
$$\frac{w}{2} \left[ 1 + 4f \frac{F_2'}{R} + 4f' \frac{F_2''}{R} \right] = A + \frac{G}{F_2'}; \quad (9)$$
or, finally,
$$w = \sqrt{\frac{3g(h - G)}{1 + 4f \frac{F_2'}{R} + 4f' \frac{F_2''}{R}}} \cdot \quad (10)$$
and hence we have also
$$Q = F_2 w \cdot \quad \quad (11)$$
**EXAMPLE.—If we replace the force $G$ of this problem by the thrust $P$ exerted along the pump-platton of a steam fire-engine, we may treat the foregoing as a close approximation to the practical problem of such an apparatus, the pipes being connective straight lengths of hose. In which (for the probable values of $h$, $G$, and $P$ given in § 507), we have [see, e.g., "Steam Engineering," by Geo. Ellis, Springfield, Mass.),* (Strictly, $f$ varies somewhat with the velocity; see § 517.) Let $P = 12000$ lbs., and the piston-area at $n = F_2 = 78$ sq. in. = $a$ sq. ft. Also, let $h = 20$ ft., and the dimensions of the hose be as follows:
$$d_1 = 8 \text{ in.}, \quad d_2 = 2 \text{ in.}, \quad d_a (\text{of nozzle}) = 1 \text{ in.};$$
$$l_1 = 400 \text{ ft.}, \quad l_2 = 500 \text{ ft.}$$
With the foot-pound-second system of units, we now have [eq. (10)]
$$w = \sqrt{\frac{2 \times 32.2}{1 + 4f \times .0075}} = \sqrt{\frac{64.4}{1 + .03}} = \sqrt{\frac{64.4}{1.03}}.$$*
* See § 528
**FRICTION-HEAD IN PIPES.**
708
$$= \sqrt{\frac{3 \times 39.2 \times 404}{1 + 0.39 + 5.68}}$$
I.e., $v_m = 60.0$ ft. per sec. If this jet were directed vertically upward it should theoretically attain a height of $\frac{v_m^2}{2g} = nearly$ 56 feet, but the resistance of the air would reduce this to about 40 or 45 ft.
We have further, from eq. (1),
$$Q = F_{av} v_m = \frac{\pi(1)^2}{4(12)} \times 60.0 = 3.37 \text{ cub. ft. per sec.}$$
If there were no resistance in the hose we should have, from § 497a,
$$v_m = \sqrt{2g\left(\frac{P}{F_{av}} + h\right)} = \sqrt{2 \times 39.2 \times 404} = 161.3 \text{ ft. per sec.}$$
§13a. Influence of Changes of Temperature.—Although Poisson and Hagen found that with glass tubes of very small diameter the flow of water was increased threefold by a rise of temperature of one degree, yet it is understood that with common pipes the rate of flow is appreciably affected by ordinary fluctuations of temperature; at any rate, experiments of sufficient precision are wanting, as regards such an influence.
See Mr. Hamilton Smith's "Hydraulics," p. 16, where he says: "Changes by variation in $T$ (temperature) will probably only be appreciable with small orifices, or with very low heads for orifices or weirs."
§14. Loss of Head in Orifices and Short Pipes.—So long as the steady flow between two localities $a$ and $b$ takes place in a pipe having no abrupt enlargement or diminution of section, nor sharp curves, bends, or elbows, the loss of head may be ascribed solely to the surface action (or "skin-friction") between water and pipe; but the introduction of any of the above-mentioned features occasions eddying and internal disturbance, and friction (and consequent heat) thereby causing further deviations
704 MECHANICS OF ENGINEERING.
from Bernoulli's Theorem; i.e., additional losses of head, or heights of resistance.
From the analogy of the form of a friction-head in a long pipe [eq. (4), § 513], we may assume that any of the above heights of resistance is proportional to the square of the velocity, and may therefore always be written in the form
$$\left\{\begin{array}{l}{\text{Loss of Head due to any}}\\ {\text{cause except skin-friction}}\end{array}\right\} = \zeta \frac{v^2}{gq}, \quad \ldots \ldots \ldots (1)$$
in which $v$ is the velocity of the water in the pipe at the section where the resistance occurs; or, if, on account of a abrupt enlargement of the stream-section, there is a corresponding diminution of velocity, then $v$ is always to denote this diminished velocity (i.e., in the down-stream section). This velocity $v$ is often an unknown at the outset.
$\zeta$, corresponding to the abstract factor $\frac{4}{D}$ in the height of resistance due to skin-friction [eq. (4), § 513], is an abstract number called the Coefficient of Resistance, to be determined experimentally; or computed theoretically, where possible. Roughly speaking, it is independent of the velocity, for a given fitting, casing, pipe-joint, elbow, bend, valve-gate at a definite opening, etc., etc.
515. Heights of Resistance (or Losses of Head) Occasioned by Short Cylindrical Tubes.—When dealing with short tubes discharging into the air, in § 507, deviations from Bernoulli's Theorem were made good by using a coefficient of velocity $\phi$, dependent on experiment. This device answered every purpose for the simple circumstances of the case, as well as for simply orifices. But the great variety of possible designs of a component pipe (valve, elbow, fitting, etc.) and the various shapes of cross-sections (etc.) renders it almost impossible, in such a pipe, to provide for deviations from Bernoulli's Theorem by a single coefficient of velocity (velocity of jet, that is) for the pipe as a whole, since new experiments would be needed for each new design of pipe. Hence the great utility of the conception of "loss of head," one for each source of resistance.
LOSSES OF HEAD IN A SHORT PIPE. 705
If a long pipe issues from the plane side of a reservoir and the corners of the junction are not rounded [see Fig. 571], we shall need an expression for the loss of head at the entrance, $E$, as well as that
$$\left[ = 4 \frac{f}{d} \frac{l}{2g} \right]$$
due to the skin-friction in the pipe. But, whatever the velocity, $v$, in the pipe proves to be, influenced as it is both by the entrance loss of head and the skin-friction head (in applying Bernoulli's Theorem), the loss of head at $E$, viz., $\zeta_{E} \frac{v^2}{2g}$ will be just the same as if efflux took place through enough of the pipe at $E$ to constitute a "short pipe," discharging into the air, under some head $h$ (different from $h'$ of Fig. 571) sufficient to produce the same velocity $v$. But in that case we should have
$$v = \phi \sqrt{2gh}, \quad \text{or} \quad \frac{v^2}{2g} = \phi^2 h. \quad . . . . . . . . (1)$$
(See §§ 507 and 508, $\phi$ being the coefficient of velocity, and $h$ the head, in the cases mentioned in those articles.)
We, therefore, apply Bernoulli's Theorem to the cases of those articles (e.g. Figs. 540 and 564), in order to determine the loss of head due to the short pipe, and obtain (with $m$ as datum level for potential heads)
$$\frac{vm}{2g} + b + 0 = 0 + b + h - \zeta_{E} \frac{v^2}{2g}. \quad . . . . . . . . (2)$$
Now the $v$ of eq. (2) is equal to the $v_m$ of the figures referred to, and $\zeta_E$ is a coefficient of resistance for the short pipe, and we now desire its value. Substituting for
$$\frac{v^2}{2g} = \frac{vm}{2g}$$
we get
$$b + h - \zeta_E \frac{vm}{2g} = 0.$$
706
MECHANICS OF ENGINEERING.
its value $\varphi^{\prime}h$ from eq. (1), we have
$$\zeta_{e} = \frac{1}{\varphi^{\prime}} - 1 \quad \ldots \quad \ldots \quad \ldots \quad (5)$$
Hence when $\alpha = 90^\circ$ (i.e., the pipe is $\gamma$ to the inner reservoir surface), we derive
$$\zeta_{e} = \frac{1}{\varphi^{\prime}} - 1 = \frac{1}{(0.816)^{\alpha}} - 1 = 0.505; \quad . . . \quad [\alpha = 90^\circ] \quad . . .$$
and similarly, for other values of $\alpha$ (taking $\varphi$ from the table, § 508), we compute the following values of $\zeta_{e}$ (corners not rounded) for use in the expression for “loss of head,” $c_{e}^{2} \cdot \frac{v}{2g}$:
| For $\alpha = 90^\circ$ |
$\zeta_{e} = 0.505$ |
$\zeta_{e} = 0.665$ |
$\zeta_{e} = 0.635$ |
$\zeta_{e} = 0.718$ |
$\zeta_{e} = 0.784$ |
$\zeta_{e} = 0.870$ |
$\zeta_{e} = 0.965$ |
| For $\alpha = 70^\circ$ |
$\zeta_{e} = 0.707$ |
$\zeta_{e} = 0.787$ |
$\zeta_{e} = 0.833$ |
$\zeta_{e} = 0.871$ |
$\zeta_{e} = 0.914$ |
$\zeta_{e} = 0.965$ |
$\zeta_{e} = 1.025$ |
| For $\alpha = 60^\circ$ |
$\zeta_{e} = 0.666$ |
$\zeta_{e} = 0.733$ |
$\zeta_{e} = 0.787$ |
$\zeta_{e} = 0.833$ |
$\zeta_{e} = 0.871$ |
$\zeta_{e} = 0.914$ |
$\zeta_{e} = 0.965$ |
| For $\alpha = 50^\circ$ |
$\zeta_{e} = 0.595$ |
$\zeta_{e} = 0.666$ |
$\zeta_{e} = 0.733$ |
$\zeta_{e} = 0.787$ |
$\zeta_{e} = 0.833$ |
$\zeta_{e} = 0.871$ |
$\zeta_{e} = 0.914$ |
| For $\alpha = 40^\circ$ |
$\zeta_{e} = 0.525$ |
$\zeta_{e} = 0.595$ |
$\zeta_{e} = 0.666$ |
$\zeta_{e} = 0.733$ |
$\zeta_{e} = 0.787$ |
$\zeta_{e} = 0.833$ |
$\zeta_{e} = 0.871$ |
| For $\alpha = 30^\circ$ |
$\zeta_{e} = 0.444$ |
$\zeta_{e} = 0.525$ |
$\zeta_{e} = 0.595$ |
$\zeta_{e} = 0.666$ |
$\zeta_{e} = 0.733$ |
$\zeta_{e} = 0.787$ |
$\zeta_{e} = 0.833$ |
| For $\alpha = 20^\circ$ |
$\zeta_{e} = 0.344$ |
$\zeta_{e} = 0.444$ |
$\zeta_{e} = 0.525$ |
$\zeta_{e} = 0.595$ |
$\zeta_{e} = 0.666$ |
$\zeta_{e} = 0.733$ |
$\zeta_{e} = 0.787$ |
| For $\alpha = 10^\circ$ |
$\zeta_{e} = 0.222$ |
$\zeta_{e} = 0.344$ |
$\zeta_{e} = 0.444$ |
$\zeta_{e} = 0.525$ |
$\zeta_{e} = 0.595$ |
$\zeta_{e} = 0.666$ |
$\zeta_{e} = 0.733$ |
| For $\alpha < 10^\circ$, etc. | | | | | | | |
From eq. (4) we see that the loss of head at the entrance of the pipe, corners not rounded, with $\alpha=90^\circ$, is about one half (.5/2) of the height due to the velocity $v_1^n$, in that part of the pipe ($v_1^n\) being the same all along the pipe if cylindrical). The value of $v_1^n$, itself, Fig.\,571, depends on all the features of the design from reservoir to nozzle.\ See §~518.
If the corners at $E'$ are properly rounded, the entrance loss of head may practically be done away with; still, if $v_1^n$is quite small (as it may frequently be, from large losses of head farther down stream), the saving thus secured, while helping to increase $v_1^n$slightly (and thus the saving itself), is insignifi- cant.
§~518 General Form of Bernoulli's Theorem, considering all Losses of Head.
In view of preceding explanations and assumptions, we may write a general and final form Bernoulli's Theorem for steady flow from an up-stream position $n$a down-stream position $m$, as follows:
$$c_n^2 + \frac{P_n}{y'} + z_n + c_m^2 + z_m - \left(\text{all losses of head occurring between } n \text{ and } m \right) . (B_2)$$
**LOSSES OF HEAD IN GENERAL.**
707
Each loss of head (or height of resistance) will be of the form
$$\zeta = \frac{v^2}{2g}$$ (except skin-friction head in long pipes, viz., $4f \frac{v^2}{d}$),
the $v$ in each case being the velocity, known or unknown, in that part of the pipe where the resistance occurs (and hence is not necessarily equal to $u_a$ or $u_b$).
517. The Coefficient, $f$, for Friction of Water in Pipes—See eq. (1), § 510. Experiments have been made by Weibach, Eitelwein, Darby, Bosanet, Prony, Dubuat, Fanning, and others, to determine $f$ in cylindrical pipes of various materials (tin, glass, zinc, lead, brass, cast and wrought iron) of diameters from $\frac{3}{8}$ inch up to 36 inches. In general, the following deductions may be made from these experiments:
1st. $f$ decreases when the velocity increases; e.g., in one case with the same pipe $f$ was = .0070 for $v = 2'$ per sec., while $f$ was = .0036 for $v = 20'$ per sec.
2ndly. $f$ decreases slightly as the diameter increases (other things being equal);
e.g., in one experiment $f$ was = .0069 in a 3-in. pipe, while for the same velocity $f$ was = .0084 in a 6-in. pipe.
3rdly. The condition of the interior surface of the pipe affects the value of $f$, which is larger with increased roughness of pipe.
Thua, Darby found, with a foul iron pipe with $d = 10$ in. and velo. = 3.67 ft. per sec., the value 0113 for $f$; whereas Fanning (see p. 338 of his "Water-supply Engineering"), with a cement-lined pipe and velocity of 3.74 ft. per sec. and $d = 20$ inches, obtained $f = .0052$.
Weibach, finding the first relation very prominent, proposed the formula
$$f = 0.00359 + \frac{.00499}{V_0 (\text{in ft. per sec.)}}$$
when the velocities are great, while Darby, taking into account both the 1st and 3rd relations above, writes (see p. 585, Rankine's Applied Mechanics)
![]() A small image of a water flow diagram.
708
MECHANICS OF ENGINEERING.
$$f = \frac{0.045}{\left(1 + \frac{1}{9} \times \text{diam. in ft}\right)} + \frac{0.001}{0 \text{ ft per sec.}} \left(1 + \frac{1}{18} \times \text{diam. in ft}\right)$$
For practical purposes, Mr. J. T. Fanning has recommended, and arranged in an extensive table (pp. 349-346 of his book just mentioned), values of $f$ for clean iron pipe, of diameters from $\frac{3}{4}$ inch to 96 inches, and for velocities of 0.1 ft. to 30 ft. per second, for the pipe opposite to that shown, inserted with Mr. Fanning's permission, for use in solving numerical problems.
In obtaining $f$ for slightly tuberculated and for foul pipes, the recommendations of Mr. Fanning seem to justify the following rules:
For slightly tuberculated pipes of diam. = $\frac{3}{4}$ ft., 1 ft., 9 ft., 4 ft.
we should add 23% 34% 10% 13%
and for foul pipes of same size............ 72% 60% 38% 25%
of the $f$ for clean pipes, to itself. For example, if $f = .007$ for a certain $\frac{3}{4}$ ft. pipe when clean, with velocity = 0.04 ft. per sec., we have $f' = .007 \times 1.72 = .01204$ when it is foul.
For first approximations a mean value of $f = .006$ may be employed, since in some problems sufficient data may not be known in advance to enable us to take $f'$ from the table.
**EXAMPLE—Fig. 572.** In the steady pumping of crude petroleum weighing $y = .55$ lbs. per gallon through a six-inch pipe 30 miles long, to a station 700 ft. higher than the pump, it is found that the pressure in the pump cylinder at $n$, necessary to keep up a velocity of 4.4 ft. per sec., is equal to 1,000 lbs. per sq. inch. Required the coefficient $f$ of the pipe. As all losses except the friction-head in the pipe are insignificant, the latter only will be considered. The velocity-head at $n$ may
![]() A diagram showing a pipeline with a pump and a gauge.
Fig. 572
lbs. per sq. inch. Required the coefficient $f$ of the pipe. As all losses except the friction-head in the pipe are insignificant, the latter only will be considered. The velocity-head at $n$ may
| TABLE OF VALUES OF f. |
709 |
| P (mm) |
f (mm) & |
f (mm) & |
| 1.0 |
0.0085 |
0.0086 |
| 1.5 |
0.0086 |
0.0087 |
| 2.0 |
0.0087 |
0.0088 |
| 2.5 |
0.0088 |
0.0089 |
| 3.0 |
0.0089 |
0.0091 |
| 3.5 |
0.0091 |
0.0092 |
| 4.0 |
0.0092 |
0.0094 |
| 4.5 |
0.0094 |
0.0096 |
| 5.0 |
0.0096 |
0.0112 |
| 5.5 |
0.0112 |
0.0114 |
| 6.0 |
0.0114 |
0.0116 |
| 6.5 |
0.0116 |
0.0118 |
| 7.0 |
0.0118 |
0.0122 |
| 7.5 |
0.0122 |
0.0124 |
| 8.0 |
0.0124 |
0.0126 |
| 8.5 |
0.0126 |
0.0128 |
| 9.0 |
0.0128 |
710 MECHANICS OF ENGINEERING.
be put $=0$; the jet at $m$ being of the same size as the pipe, the velocity in the pipe is $=v_m$, and therefore $v_n = 4.4$ ft. per sec. Notice that $m$, the down-stream section, is at a higher level than $n$.
From Bernoulli's Theorem, § 516, we have, with $s$ as a datum level,
$$\frac{v_m^2}{2g} + h + b = \frac{F_p}{y} + 0 - \frac{4fL}{d}\frac{v^2}{2g}, \quad (1)$$
Using the ft., lb., and sec., we have
$$h = 700 \text{ ft.}, \quad v_m^2 + 2g = 0.30 \text{ ft.},$$
while
$$b = \frac{14.7 \times 144}{55} = 88.47 \text{ ft.}, \quad \frac{F_p}{y} = \frac{1000 \times 144}{55} = 2618 \text{ ft.}$$
Hence, in eq. (1),
$$0.30 + 88.4 + 700 = 2618 - \frac{4fL}{d} \times \frac{30 \times 5280}{64.4}.$$
Solving for $f$, we have $f = .00485$ (whereas for water, with $v = 4.4$ ft. per sec. and $d = \frac{1}{2}$ ft., the table, p. 709, gives $f = .00601$).
If the $y$ of the petrolum had been 50 lbs. per cubic foot, instead of 55, we would have obtained $\frac{F_p}{y} = 2880$ feet and $f = .0058$.
518. Flow through a Long Straight Cylindrical Pipe, including both friction-head and entrance loss of head (corners not rounded); reservoir large. Fig. 573.
The jet issues directly from the end of the pipe, in parallel filaments, into the air and therefore has no resistance to flow; hence, also, $v_n$ of the jet $= v$ in the pipe (which is assumed to be running full), and is
![]() A diagram showing a long straight cylindrical pipe with a jet issuing from its end into the air.
Fig. 573
COEFFICIENT OF LIQUID FRICTION.
711
therefore the velocity to be used in the loss of head $c_{w} \frac{v^2}{2g}$ at the entrance $E$ (§ 815).
Taking $m$ and $n$ as in figure and applying Bernoulli's Theorem (§ 476), with $m$ as datum level for the potential heads $\sigma_m$ and $\sigma_n$, we have
$$\frac{\sigma_n + b + 0 = 0 + b + h - c_w \frac{v^2}{2g}}{2g} = 4f \frac{l}{d} \frac{v^2}{2g} \cdot \quad (1)$$
Three different problems may now be solved:
First, required the head $h$ to keep up a flow of given volume $Q$ per unit of time in a pipe of given length $l$ and diameter $d$.
From the equation of continuity we have
$$Q = F_a v_m = \frac{4Q}{\pi d^2} v_m;$$
i.e., veloc. of jet, which = veloc. in pipe, $= v_m = \frac{4Q}{\pi d^2} \cdot \quad (2)$$
Having found $v_m = v$, from (2), we obtain from (1) the required $h$, thus:
$$h = \frac{v^2}{2g} \left[1 + c_w + 4f \frac{l}{d}\right] \quad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)$$
Now $c_w = 0.505$ if $\alpha = 90^\circ$ (see § 815), while $f$ may be taken from the table, § 817, for the given diameter and computed velocity ($v_m = v$, found in (2)), if the pipe is clean; if not clean, see end of § 817, for slightly tuberculated and for four pipes.
Secondly. Given the head $h$, and the length $l$ and diameter $d$ of pipe, required the velocity in the pipe, viz., $v = v_m$, that of jet; also the volume delivered per unit of time, $Q$. Solving eq. (1) for $v_m$, we have
$$v_m = \sqrt{\frac{1}{1 + c_w + 4f \frac{l}{d}}} \cdot \quad (4)$$
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MECHANICS OF ENGINEERING.
whence $Q$ becomes known, since
$$Q = \frac{4}{\pi} d v_{w} \quad \quad \quad (5)$$
[Note.—The first radical in (4) might for brevity be called a coefficient of velocity, $\phi$, for this case. Since the jet has the same diameter as the pipe, this radical may also be called a coefficient of efflux.]
Since in (4) $f$ depends on the unknown $v$ as well as on the known $d$, we must first put $f = .006$ for a first approximation for $v$, and find a corresponding value for $f$ and substitute again; and so on.
Thirdly, knowing the length of pipe and the head $h$, we wish to find the proper diameter $d$ for the pipe to deliver a given volume $Q$ of water per unit of time. Now
$$v_1 = v_{w} = \frac{Q}{4 \sqrt{\pi h}} \quad \quad \quad (6)$$
which substituted in (1) gives
$$2gh = \left(\frac{4Q}{\pi}\right)^{\frac{1}{2}}\left[1 + c_2 + 4f^{\frac{1}{2}}\right] = \left(\frac{4Q}{\pi}\right)\left[\frac{1+c_2}{d} + 4f^{\frac{1}{2}}\right];$$
that is,
$$2ghd^2 = \left(\frac{4Q}{\pi}\right)\left[(1+c_2)d + 4f^{\frac{1}{2}}\right];$$
$$\therefore d = \sqrt{\frac{(1+c_2)d + 4f^{\frac{1}{2}}}{2gh}\left(\frac{4Q}{\pi}\right)} \quad \quad \quad (7)$$
As the radical contains $d$, we first assume a value for $d$, with $f = .006$, and substitute in (7). With the approximate value of $d$ thus obtained, we substitute again with a new value for $f$ based on an approximate $v$ from eq. (6) (with $d$ its first approximation), and so on until the value for $d$ is derived; and so on. (Treutwine's Pocket-book contains a table of fifth roots and powers.) If $l$ is quite large, we may put $d = 0$ for a first approximation. In connection with these examples, see last figure.
LONG PIPES. 713
EXAMPLE 1.—What head $h$ is necessary to deliver 190 cuin.
ft. of water per minute through a clean straight iron pipe 140
ft. long and 6 in. in diameter?
From eq. (2), with $v$, $b$, $h$, and sec., we have
$$v = v_m = \left(4 \times \frac{V}{A} + \pi h\right) = 10.18 \text{ ft. per sec.}$$
Now for $s = 10$ ft. per sec. and $d = \frac{4}{3}$ ft., we find (in table,
§ 517) $f = .00549$; and hence, from eq. (3),
$$h = \frac{(10.18)^2}{2 \times 32.2} \left[1 + .505 + \frac{4 \times .00549 \times 140}{\frac{4}{3}}\right] = 12.23 \text{ ft.,}$$
of which total head, as we may call it, 1.60 ft. is used in pro-
ducing the velocity 10.18 ft. per sec. (i.e., $v_m + 2g$ = 1.60 ft.).
while .088 ft. ($= c_x \frac{v_m^2}{2g}$) is lost at the entrance $E$ (with $\alpha$
= 90°), and .938 ft. (friction-head) is lost in skin-friction.
EXAMPLE 2.—[Data from Weisbach.] Required the de-
livery, $Q$, through a straight clean iron pipe 48 ft. long and
2 in. in diameter, with 5 ft. head (= $h$). $v$, $v_m$, being un-
known, we first take $f = .006$ and obtain (eq. (4))
$$v_m = \sqrt{\frac{1}{1 + .505 + \frac{4 \times .00688 \times 48}{\frac{4}{3}}}} \cdot \frac{\sqrt{2} \times 32.2 \times 5}{\frac{4}{3}}$$
$$= 6.18 \text{ ft. per sec.}$$
From the table, § 517, for $v = 6.2$ ft. per sec. and $d = 2$ in.,
$$f = .00688,$$ whence
$$v_m = \sqrt{\frac{1}{1 + .505 + \frac{4 \times .00688 \times 48}{\frac{4}{3}}}} \cdot \frac{\sqrt{2} \times 32.2 \times 5}{\frac{4}{3}}$$
$$= 6.04 \text{ ft. per sec.,}$$
which is sufficiently close. Then, for the volume per second,
$$Q = \frac{\pi}{4} d^2 v_m = \pi (\frac{4}{3})^2 (.064) = 0.1367 \text{ cub. ft. per sec.}$$
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MECHANICS OF ENGINEERING.
[Weibach's results in this example are
$$v_m = 0.52 \text{ ft. per sec.}$$
$$Q = 0.1490 \text{ cub. ft. per sec.},$$
and his values for $f$ are slightly different.]
**Example 8.** [Data from Weibach.] What must be the diameter of a straight clean iron pipe 100 ft. in length, which is to deliver $Q = \frac{1}{4}$ of a cubic foot of water per second under 5 ft. head (= A)?
With $f = .006$ (approximately), we have from eq. (7), putting $d = 0$ under the radical for a first trial ($t$, lb., sec.), $$d = \sqrt{\frac{4 \times .006 \times 100}{2 \times 32.2 \times 5}} = \frac{1}{\pi} = \text{about } 0.30 \text{ ft.;}$$
whence $$v = \frac{4Q}{w d^3} = 7 \text{ ft. per sec.}$$
For $d = 3.8$ in. and $v = 7$ ft. per sec., we find $f = .00601$; whence, again,
$$d = \sqrt{\frac{1.505 \times .30 + 4 \times .00601 \times 100}{2 \times 32.2 \times 5}} = \frac{4 \times .3}{\pi} = .3924 \text{ ft.;}$$
and the corresponding $v = 6.06$ ft.
For this $d$ and $v$ we find $f = .00609$, whence, finally,
$$d = \sqrt{\frac{1.505 \times .30 + 4 \times .00609 \times 100}{2 \times 32.2 \times 5}} = .398 \text{ ft.}$$
[Weibach's result is $d = .318$ ft.]
**516. Chezy's Formula—If, in the problem of the preceding paragraph, the pipe is so long, and therefore $l : d$ so great, that $\frac{A}{l} : d$ in eq. (3) is very large compared with $1 + c_a$, we may neglect the latter term without appreciable error; whence eq. (3) reduces to
$$h = 4f\left(\frac{l}{d}\right)\frac{v_m^*}{2g}.$$ (pipe very long; Fig. 579). . . . (8)
![]() A diagram showing a pipe with a head of water at one end and a flow rate at the other end.
CHIEV'S FORMULA. 715
which is known as Chiev's Formula. For example, if $l = 1000$ ft. and $d = 2$ in. = 4 ft., and $f \approx 0.06$, we have $\frac{4}{d} \cdot \frac{l}{d} = 144$, while $1 + \xi$ for square corners is 1.508 only.
If in (8) we substitute
$$v_m = \frac{Q}{\sqrt{\pi d^3}} = Q + \frac{1}{2}\pi d,$$
(8) reduces to
$$h = \frac{64}{\pi^2} \cdot \frac{l}{d^2} \cdot \frac{Q^2}{B_0} \cdots \quad (\text{very long pipe}); \quad \ldots \quad (9)$$
so that for a very long pipe, considering $f$ as approximately constant, we may say that to deliver a volume $Q$ per unit of time through a pipe of given length $l$, the necessary head, $h$, is inversely proportional to the fifth power of the diameter.
And again, solving (9) for $Q$, we find that the volume conveyed per unit of time is directly proportional to the fifth power of the square root of the diameter; directly proportional to the square root of the head; and inversely proportional to the square root of the length. (Not true for short pipe; see above example.)
If we conceive of the insertion of a great number of pheometers along the long straight pipe, of uniform section, now under consideration, the summits of the respective water columns maintained in them will lie in a straight line joining the discharging (into the air) end of the pipe with a point in the reservoir surface vertically over the inlet extremity (practically so), and the "slope" of this line (called the Hydrostatic Grade Line or Gradient), i.e., the tangent (or sine; the angle is so small, generally) of the angle which it makes with the horizontal is $\frac{1}{T}$, and may be denoted by $s$. Putting also $\frac{1}{d} R = r$ the hydraulic radius of the section of the pipe, and $v_m = v$ velocity in pipe, we may transform eq. (9) into
$$v = \sqrt{\frac{2g}{f}} (Re)^{\frac{1}{2}}; \quad \text{or}, \quad v = A (Re)^{\frac{1}{2}}, \quad \ldots \quad (10)$$
716
MECHANICS OF ENGINEERING.
which is the form by which Mr. Hamilton Smith (see § 506) interprets all the experiments quoted by him on long pipes.
As to notation, however, he uses $n$ for $A$, and $r$ for $R$. With the foot and second as units, the quantity $A$ (not an abstract number) varies approximately between 60 and 140. For a given $A$ we easily find the corresponding $f$ from the relation
$$f = \frac{2g}{A}.$$ If the pipe discharges under water, $h$ = the difference of elevation of the two reservoirs. If the pipe is not horizontal, the use of the length of its horizontal projection instead of its actual length in the relation $s = \frac{2g}{A}$ occasions an error, but it is in most cases insignificant.
Similarly, if a steady flow is going on in a long pipe of uniform section, at the extremities of any portion of which we have measured the piezometer heights (or computed them from the readings of steam or pressure gauges), we may apply eq. (9), putting for $h$ the difference of level of the piezometer summits, and for $l$ the length of the pipe between them.
§ 500. Coefficient $f$ in Fire-engine Hose.—Mr. Geo. A. Ellis, C.E., in his little book on "Fire-streams," describing experiments made in Springfield, Mass., gives a graphic comparison (p. 45) of the friction losses occurring in leather and rubber hose, in leather and iron pipe in clean iron pipe, each of 24 in. diameter, with various velocities; on which the following statements may be based: That for the given size of hose and pipe ($d = 24$ in.), the coefficient $f$ for the leather and rubber hose respectively may be obtained approximately by adding to $f$ for clean iron pipe (and a given velocity) the per cent of itself shown in the accompanying table.
| Velocity |
per sec. |
Length |
per sec. |
Length |
per sec. |
| Velocity |
per sec. |
Length |
per sec. |
Length |
per sec. |
| $8.0$ |
$50\%$ |
$80\%$ |
$16$ |
$48\%$ |
$32\%$ |
| $6.4$ |
$30\%$ |
$60\%$ |
$16$ |
$48\%$ |
$32\%$ |
| $10$ |
$20\%$ |
$40\%$ |
$16$ |
$48\%$ |
$32\%$ |
| $16$ |
$10\%$ |
$20\%$ |
$16$ |
$48\%$ |
$32\%$ |
**Example.**—For a clean iron pipe 24 in. in diameter for a velocity of 10 ft. per sec., we have, from § 517, $f =$ .00593. Hence for a leather hose of the same diameter, we have, for $v =$ 10 ft. per sec.,
$$f = .00593 + .00593 = .0088.$$
*For the most recent and exhaustive experiments in this direction see Mr. J. H. Freeman's "Hydraulics of Fire-engines" in the Transact Am. Soc. C. E. R. for 1889. (Also p. 582 of this work.)*
![]() A table showing coefficients of friction for different velocities and lengths of pipes.
**PRESSURE-ENERGY.** 717
581. Bernoulli's Theorem as an Expression of the Conservation of Energy for the Liquid Particles.--In any kind of flow without friction, steady or not, in rigid immovable vessels, the aggregate potential and kinetic energy of the whole mass of liquid concerned is necessarily a constant quantity (see §§ 148 and 149), but individual particles (as the particles in the sinking free surface of water in a vessel which is rapidly being emptied) may be continually losing potential energy, i.e., reaching lower and lower levels, without any compensating increase of kinetic energy or of any other kind; but in a steady flow under ordinary conditions, we have to admit that the stock of energy of a given particle, or small collection of particles, is constant during the flow, provided we recognize a third kind of energy which may be called Pressure-energy, or capacity for doing work by virtue of internal fluid pressure; as may be thus explained:
In Fig. 574 let water, with a very slow motion and under a pressure $p$ (due to the reservoir-head + atmosphere-head behind it), be admitted behind a piston into a space at rest in vacuo.
Let $x =$ length of stroke, and $F =$ area of piston. At the end of the stroke, by motion of proper valves, communication with the reservoir is cut off on the left of the piston and opened on the right, while the water in the cylinder now on the left of the piston is put in communication with the vacuum exhausted by the return stroke. The pressure of this water falls to zero (height of cylinder small), and on the return stroke is simply conveyed out of the cylinder, neither helping nor hindering the motion. That is, in doing the work of one stroke, viz.,
$$W = \text{force} \times \text{distance} = Fp \times x = Fps,$$
a volume of water $V = Fs$, weighing $Fg$ (i.e., or other unit'), has been used, and, in passing through the motor, has experienced no appreciable change in velocity (motion slow), and
718
MECHANICS OF ENGINEERING.
therefore no change in kinetic energy, nor any change of level, and hence no change in potential energy, but it has given up all its pressure. (See § 409 for y.)
Now $W$, the work obtained by the consumption of a weight $G = V_y$ of water, may be written
$$W = Fp_e = Fp_p = V_p = V_y \frac{p}{y} = G \frac{p}{y} \quad . \quad . \quad (1)$$
Hence a weight of water $G$ is capable of doing the work $G \times \frac{p}{y} = G \times$ head due to pressure $p$, i.e., $= G \times$ pressure-head, in giving up all its pressure $p$; or otherwise, while still having a pressure $p$, a weight $G$ of water possesses an amount of energy, which we may call pressure-energy, of an amount $$= G \frac{p}{y},$$ where $y$ is the heaviness (\(\S\) 7) of water, and $\frac{p}{y}$ is height, or head, measuring the pressure $p$; i.e., it equals the pressure-head.
We may now state Bernoulli's Theorem without friction in a new form, as follows: Multiplying each term of eq. (1), § 451, by $Q_y$, the weight of water flowing per second (or other time-unit) in the steady flow, we have
$$Q_y \frac{v^2_a}{2g} + Q_y \frac{p_a}{y} + Q_y z_a = Q_y \frac{v^2_b}{2g} + Q_y \frac{p_b}{y} + Q_y z_b \quad (2)$$
But $Q_y \frac{v^2_a}{2g} = \frac{1}{2} Q_y v^2_a = \frac{1}{2} \times mass\ flowing\ per\ time-unit \times square\ of\ the\ velocity = the\ kinetic\ energy\ inherent\ in\ the volume Q of water on passing the section m, due to the velocity at m. Also, $Q_y \frac{p_a}{y} = the\ pressure-energy\ of\ the\ volume Q$ at $m$, due to the pressure at $m$; while $Q_y z_a = the\ potential energy\ of\ the\ volume Q$ at $m$ due to its height $z_a$ above the arbitrary datum plane. Corresponding statements may be made for the terms on the right-hand side of (2) referring to the other section, $n$, of the pipe. Hence (2) may be thus read: *The aggregate amount of energy* (of the three kinds mentioned) *residient in the particles* of liquid when passing section n is
**LOSS OF ENERGY.** 719
equal to that when passing any other section, as $n$ ; in steady flow without friction in rigid motionless vessels; that is, the store of energy is constant.
532. Bernoulli's Theorem with Friction, from the Standpoint of Energy.—Multiply each term in the equation of § 516 by $Qy$, as before, and denote a loss of head or height of resistance due to any cause by $A_r$, and we have
$$Qy \frac{w^*}{2g} + Qy \frac{P_0}{v} + Qy h_r = Qy \frac{w^*}{2g} + Qy \frac{P_0}{v} + Qy h_r - Z_p Qy h_r \dots (3)$$
Each term $Qy h_r$ (e.g., $Qy \frac{w^*}{d} \frac{v^*}{d} 2g$) due to skin-friction in a long pipe, and $Qy \frac{w^*}{2g}$ due to loss of head at the reservoir entrance of a pipe represents a *loss of energy*, occurring between any position $n$ and any other position $m$ downstream from $n$, but is only in existence in the form of heat generated by the friction of the liquid particles against each other or the sides of the pipes.
As illustrative of several points in this connection, consider two short lengths of pipe in Fig. 575, $A$ and $B$, one offering a gradual, the other a sudden enlargement of section, but otherwise identical in dimensions. We see that these occupy places in separate lines of pipe in each of which a steady flow with full cross-sections is proceeding, and so regulated that the velocity and internal pressure at $n$, in $A$, are equal respectively to those at $n$ in $B$. Hence, if vacuum pleometers be inserted at $n$, the
![]() A diagram showing two sections of pipe, A and B, with different cross-sectional areas. Section A has a gradual change in diameter, while section B has a sudden enlargement. The diagram also shows the velocity and pressure changes across these sections.
Fig. 575.
720
MECHANICS OF ENGINEERING.
smaller section, the water columns maintained in them by the internal pressure will be of the same height, $\frac{P_{n}}{y}$, for both $A$ and $B$. Since at $m$, the larger section, the sectional area is the same for both $A$ and $B$, since $F_{n}$ in $A = F_{n}$ in $B$, so that $Q_{A} = Q_{B}$, hence $v_{n}$ in $A = v_{n}$ in $B$ and is less than $v_{m}$.
Now in $B$ a loss of head occurs (and hence a loss of energy) between $n$ and $m$, but now in $A$ (except slight friction-head); hence in $A$ we should find as much energy present at $m$ as at $n$, only differently distributed among the three kinds, while at $m$ in $B$ the aggregate energy is greater than that at $n$ in $A$.
As regards kinetic energy, there has been a loss between $n$ and $m$, and in $A$ and $B$ (equal losses), but less than $v_{n}$. As to potential energy, there is no change between $n$ and $m$ either in $A$ or $B$, since at $n$ and $m$ are on a level. Hence if the loss of kinetic energy in $B$ is not compensated for by an equal gain of pressure-energy (as it is in $A$), the pressure-head $\left(\frac{P_{m}}{y}\right)$ at $m$ in $B$ should be less than that $\left(\frac{P_{n}}{y}\right)$ at $n$ in $A$. Experiment shows this to be true, the loss of head being due to the internal friction in the eddies occasioned by the sudden enlargement of section at $m$. At $m$, in $B$, there is a column of a less height than that at $m$ in $A$, whereas at $n$ they are equal. (See p. 467 of article "Hydromechanics" in the Ency. Bri. tannica for Mr. Frude's experiments).
In brief, in A the loss of kinetic energy has been made up in pressure-energy, with no change of potential energy, but in B there is an actual absolute loss of energy = $\zeta Qh$, or $\zeta Qv\frac{v}{g}h$, suffered by the weight $Qv$ of liquid. The value of $\zeta$ in this case and others will be considered in subsequent paragraphs.
Similarly, losses of head, and therefore losses of energy, occur at elbows, sharp bends, and obstructions, causing eddies and internal friction, the amount of each loss for a given weight, $G$, of water being = $\zeta G\frac{v^{2}}{g}h$, i.e., $\zeta = \frac{\zeta G\frac{v^{2}}{g}h}{G}$ being the loss of head occasioned by the obstruction (§ 474). It is
SUDDEN ENLARGEMENTS IN PIPES. 721
therefore very important in transmitting water through pipes for purposes of power to use all possible means of preventing disturbance and eddying among the liquid particles. E.g., sharp corners, turns, elbows, abrupt changes of section, should be avoided in the design of the supply-pipe. The amount of the losses of head, or heights of resistance, due to these various causes will now be considered (except skin-friction, already treated). Each such loss of head will be written in the form
$$\zeta = \frac{v^2}{2g}$$
and we are principally concerned with the value of the abstract number $\zeta$, or coefficient of resistance, in each case.
The velocity $v$ is the velocity, known or unknown, where the resistance occurs; or if the section of pipe changes at this place, then $v =$ velocity on the down-stream section. The late Professor Weisbach, of the mining-school of Freiberg, Saxony, was one of the most noted experimenters in this respect, and will be frequently quoted.
533. Loss of Head Due to Sudden (i.e., Square-edged) Enlargement. Berna's Formula.—Fig. 576. An eddy is formed in the angle with consequent loss of energy. Since each particle of water of weight $G_s$, arriving with the velocity $v$, in the small pipe, may be considered to have an impact against the base of the infinitely great and more slowly moving column of water in the large pipe, and after the impact, move on with the same specific impulse that it had before, so that no energy is lost by elastic direct central impact ($\zeta$ 60), we may find the energy lost by this particle on account of the impact by eq. (1) of § 138, in which putting $M_s = G_s + g$, and $M_t = G_t + g =$ mass of infinitely great body of water in the large pipe, so that $M_s = w$, we have
Energy lost by particle = $G_s \left( \frac{v - v_s}{2g} \right)^2$
and the corresponding
Loss of head = $\frac{(v - v_s)^2}{2g}$
729 MECHANICS OF ENGINEERING.
which, since $F_v = F_{v_0}$, may be written
$$\text{Loss of head in sudden enlargement} = \left(\frac{F}{F_0} - 1\right)^{\zeta} \cdot \text{ (9)}$$
That is, the coefficient $\zeta$ for a sudden enlargement is
$$\zeta = \left(\frac{F}{F_0} - 1\right)^{\zeta} \cdot \text{ (3)}$$
$F_v$ and $F_0$ are the respective sectional areas of the pipes. Eq. (3) is Borda's Formula.
Norm.—Practically, the flow cannot always be maintained with full sections. In any case, if we assume the pipes to be running full (once started so), and on that assumption compute the internal pressure at $F_v$, and find it to be zero or negative, the assumption is incorrect. That is, unless there is some pressure at $F_v$, the water will not swell out laterally to fill the large pipe.
Example.—Fig. 577. In the short tube $AB$ containing a sudden enlargement, we have given $F_v = F_0 = 6$ sq. inches, $F_v = 4$ sq. inches, and $h = 9$ feet. Required the velocity of the jet at $m$ (in the air, so that $p_m + y = b = 34$ ft.), if the only loss of head considered is that due to the sudden enlargement (kinetic friction neglected, as the tube is short); the reservoir entrance has rounded corners. As shown in Fig. 577, we return to $m$ as down-stream section, and $n$ in reservoir surface as up-stream position (datum level at $m$), we have
$$\frac{v_n^2}{2g} + b + 0 = 0 + b + h - \zeta \frac{v_n^2}{2g} \cdot \text{ (4)}$$
But, here, $v_n = v_m$
$$\therefore (1 + \zeta) \frac{v_n^2}{2g} = h \cdot \text{ (5)}$$
From eq. (3) we have
$$\zeta = (\xi - 1)^{\zeta} = 0.25,$$
![]() Fig. 577.
SUDDEN ENLARGEMENT IN PIPE.
723
and finally (ft., lb., sec.)
$$v_m = \sqrt{\frac{1}{1.25} \times 9 \times 32.2 \times 9} = 0.895 \sqrt{9 \times 32.2 \times 9}$$
$$= 21.55 \text{ ft. per sec.}$$
(The factor $0.895$ might be called a coefficient of velocity for this case.) Hence the volume of flow per second is
$$Q = F_v v_m = r_f v_m \times 21.55 = 0.898 \text{ cub. ft. per sec.}$$
We have so far assumed that the water fills both parts of the tube, i.e., that the pressure $p_1$, at $F_1$, is greater than zero (see foregoing note). To verify this assumption, we compute $p_1$ by applying Bernoulli's Theorem to the centre of $F_1$ as downstream position and datum plane, and $w$ as upstream position, with no loss of head between, and obtain
$$\frac{v_m^2}{2g} + p_1 + g = b + h - 0.$$ (6)
But since $F_{y0} = F_v w$, we have
$$v_m^* = (\frac{1}{y}) v_m^* = (\frac{1}{y}) w_m,$$
and hence the pressure-head at $F_1$ (substituting from equations above) is
$$\frac{p_1}{y} = b + h - (\frac{1}{y})^2 \frac{h}{1 + c} = 84 + \frac{9}{4} + \frac{9}{4 + 25} = 97 \text{ feet},$$
and $\therefore p_1 = \frac{H}{y}$ of $14.7$ lbs. per sq. inch, which is greater than zero; hence efflux with the tube full in both parts can be maintained under $9$ ft. head.
If, with $F_1'$ and $F_2'$ as before (and $\therefore C$), we put $p_1' = 0$, and solve for $h$, we obtain $h = 42.5$ ft. as the maximum head under which efflux with the large portion full can be secured.
524. Short Pipe, Square-edged Internally.—This case, already
794
MECHANICS OF ENGINEERING.
treated in §§ 507 and 515 (see Fig. 578; a repetition of § 509), presents a loss of head due to the sudden enlargement from the contracted section at $m'$ (whose sectional area is $A$) to the full section $F$ of the pipe. From § 515 we know that the loss of head due to the short pipe is $h_s = \frac{C}{g} \left( \frac{a - m'}{m'} \right)^2$, in which $\frac{C}{g} = 0.568$ while from Borda's Formula, § 523, we have also $C = \left( \frac{F}{CF} - 1 \right)^2$. Equating these, we find the coefficient of internal contraction at $m'$ to be
$$C = \frac{1}{1 + V_{\text{eff}}} = \frac{1}{1 + V_{\text{eff}}^{0.584}} = 0.584,$$
or about 0.60 (compare with $C = 0.64$ for thin-plate contraction, § 495). It is probably somewhat larger than this (§ 584), since a small part of the loss of head, $h_s$, is due to friction at the corners, and not all of this is due to contraction.
By a method similar to that pursued in the example of § 523, we may show that unless $h_s$ is less than 40 feet, about, the tube cannot be kept full, the discharge being as in Fig. 551. If the efflux takes place into a "partial vacuum," this limiting value of $h_s$ is still smaller. Weibach's experiments confirm these statements (but those in the C. U. Hyd. Lab. seem to indicate that the limiting value for $h_s$ in the first case is about 50 ft.).
523. Diaphragm in a Cylindrical Pipe—Fig. 579. The diaphragm being of "thin plate," let the circular opening in it (concentric with the pipe) have an area $F_1$, while the sectional area of pipe is $F'$. Beyond $F'$, the stream contracts to a section of area $= CF' = F_1$, as shown by
![]() A diagram showing a cylindrical pipe with a circular opening in its wall, labeled \( F_1 \), and a stream contracting beyond \( F' \) to a section of area equal to \( CF' = F_1 \).
Fig. 579
Let \( h_1 \) be the loss of head due to contraction at \( F_1 \), and \( h_2 \) that due to contraction at \( F' \). Then
$$h_1 = \frac{C}{g} \left( \frac{a - m'}{m'} \right)^2,$$
and
$$h_2 = \frac{C}{g} \left( \frac{a - m''}{m''} \right)^2,$$
where \( m'' \) is some intermediate point between \( m' \) and \( m'' \). Since \( h_1 + h_2 = h_s \), we have
$$\frac{C}{g} \left( \frac{a - m'}{m'} \right)^2 + \frac{C}{g} \left( \frac{a - m''}{m''} \right)^2 = h_s.$$
**SHORT PIPE. DIAPHRAGM IN PIPE.**
From which to fill the section $F_1$ of pipe, a loss of head occurs which by Borda's Formula, § 583, is
$$h_s = \zeta \frac{v^2}{2g} = \left(\frac{F_1}{F_2} - 1\right) \frac{v^2}{2g}$$
where $v$, is the velocity of the pipe (supposed full). Of course $F_1$ (or $OF_1$) depends on $F_2$; but since experiments are necessary in any event, it is just as well to give the values of $\zeta$ itself, as determined by Weisbach's experiments, viz.
| $\frac{F_1}{F_2}$ |
.10 |
.30 |
.40 |
.50 |
.60 |
.70 |
.80 |
.90 |
1.00 |
| $\zeta$ = .298. |
48. 17.5 7.8 3.7 1.5 |
4.0 3.8 3.5 |
3.8 3.6 3.4 |
3.6 3.4 3.2 |
3.4 3.2 3.0 |
3.2 3.0 2.8 |
3.0 2.8 2.6 |
2.8 2.6 2.4 |
2.6 2.4 2.2 |
By internal lateral filling, Fig. 580, the change of section may be made gradual and eddying prevented; and then but little loss of head (and therefore little loss of energy) occurs, besides the slight amount due to skin-friction along this small surface. On p. 467 of the article *Hydromechanics* in the *Encyclopaedia Britannica* may be found an account of experiments by Mr. Froods, illustrating this fact.
**583. "The Venturi Water-meter."—The invention bearing this name was made by Mr. Clemens Herschel (see *Trans. Am Soc Civ Engineers*, for November 1887), and may be described as a method of pipe in which a gradual narrowing of section is accompanied successively by a more gradual enlargement, as in Fig. 580; but the diameter remains constant during the flow the piezometer-heights are observed at the three positions $r$, $n$, and $m$ (see below), and the rate of discharge may be computed as follows: Referring to Fig. 580, let us denote by $r$ (up-stream) position where the narrowing of the pipe begins, and by $n$ that where the enlargement ends, while $m$ refers to the narrowest section $F_m = F_r$.
Applying Bernoulli's Theorem to sections $r$ and $n$, assuming
736 MECHANICS OF ENGINEERING.
no loss of head between, we have, as the principle of the apparatus,
$$\frac{p_1}{y} + \frac{v_1^2}{2g} = \frac{p_2}{y} + \frac{v_2^2}{2g}; \quad \ldots \quad (1)$$
whence, since $F_{e_0} = F_e e_0$,
$$e_0 = \sqrt{\frac{1}{1 - (\frac{F_e}{F_0})^2}} \sqrt{2g(\frac{p_1}{y} - \frac{p_2}{y})} = \phi \sqrt{2g(\frac{p_1}{y} - \frac{p_2}{y})}, \quad (9)$$
in which $\phi$ represents the first radical factor. $\phi$ should differ but little from unity with $F_e$ small (and such was found to be the case by experiment). Its theoretical value is constant and greater than unity. In the actual use of the instrument the $E$ and $E_0$ are inferred from the observed piosmeter-heights $y$, and $y_0$ (since $\frac{p_1}{y} = y_c + b$, and $\frac{p_2}{y} = y_a + b$, $b$ being 34 ft.), and then the quantity flowing per time-unit computed, from $Q = F_e v_e$, $v_e$ having been obtained from eq. (9). This process gives a value of $Q$ about four per cent in excess of the truth, according to the second set of experiments mentioned below, if $v_e = 35$ ft. per sec.; but only one per cent excess with $v_e = 5$ or 6 ft. per sec.
Experiments were made by Mr. Herschel on two meters of this kind, in each of which $F_e$ was only one ninth of $F_0$, a ratio so extreme that the loss of head due to passage through the instrument is considerable. E.g., with the smaller apparatus, in which the diameter at $a$ was 4 in., the loss of head between $r$ and $m$ was 10 or 11 ft., when the velocity through $a$ was 80 ft. per sec., those at other velocities being roughly proportional to the square of the velocity. In the larger instrument $d_a$ was 3 ft., and the loss of head between $r$ and $m$ was much the same proportion of its velocity as in the smaller (Eq. with $v_a = 34.56$ ft. per sec. the loss of head was 2.07 ft., while with $v_a = 19.96$ ft. per sec. it
SUDDEN DIMINUTION OF SECTION OF PIPE. 737
was 0.49 ft.) The angle of divergence was much smaller in these meters than that in Fig. 580.
527. Sudden Diminution of Cross section, Square Edges.—Fig. 581. Here, again, the resistance is due to the sudden enlargement from the contracted section to the full section $F_1$ of the small pipe, so that in the loss of head, by Borda's formula,
$$h_s = \zeta \frac{v^2}{2g} = \left( \frac{F_1}{F_2} - 1 \right) \frac{v^2}{2g} \dots (1)$$
the coefficient
$$\zeta = \left( \frac{F_1}{F_2} - 1 \right) = \left( \frac{F_1}{CF_2} - 1 \right) = \left( \frac{1}{C} - 1 \right) \dots (2)$$
depends on the coefficient of contraction $C$ but this latter is influenced by the ratio of $F_1$ to $F_2$, the sectional area of the larger pipe, $C$ being about .80 when $F_2$ is very large (i.e., when the small pipe issues directly from a large reservoir so that $F_1 : F_2$ practically = 0). For other values of this ratio Weibach gives the following table for $C$, from his own experiments:
| $For F_1 : F_2 = .10$ |
.30 |
.80 |
.40 |
.50 |
.60 |
.70 |
.80 |
.90 |
1.00 |
| $C = .84$ |
.68 |
.68 |
.69 |
.69 |
.719 |
.755 |
.813 |
.899 |
1.00 |
$C$ being found, we compute $\zeta$ from eq. (3) for use in eq. (1).
528. Elbows.—The internal disturbance caused by an elbow, Fig. 582 (pipe full, both sides of elbow), occasions a loss of head
$$h_s = \zeta \frac{v^2}{2g} \dots (1)$$
In which, according to Weibach's experiments with tubes 3 centims, i.e. 1.3 in., in diameter, we may put
![]() Fig. 583.
728
MECHANICS OF ENGINEERING.
|
For α = 20° |
40° |
60° |
80° |
90° |
100° |
110° |
120° |
130° |
140° |
| ζ = 0.045 |
1.89 |
3.64 |
7.40 |
9.64 |
1.36 |
1.558 |
1.96 |
2.16 |
2.48 |
computed from the empirical formula;
$$\zeta = 0.045 \sin^2 \frac{\alpha}{2} + 2.047 \sin^2 \frac{\alpha}{2}$$
σ is the velocity in pipe; σ as in figure. For larger pipes ζ would probably be somewhat smaller; and vice versa.
If the elbow is immediately succeeded by another in the same plane and turning the same way, Fig. 583, the loss of head is not materially increased, since the eddying takes place chiefly in the further branch of the second elbow; but if it turns in the reverse direction, Fig. 584, but still in the same plane, the total loss of head is double that of one elbow; while if the plane of the second is $\frac{π}{4}$ to that of the first, the total loss of head is $\frac{π}{2}$ times that of one alone. (Weibach.)
589. Bends in Pipes of Circular Section.—Fig. 585. Weibach bases the following empirical formula for ζ, the coefficient of resistance, on his own experiments with pipes of circular section on his own experiments and some of Dubuat's, viz:
$$\zeta = 0.131 + 1.847 \left( \frac{a}{r} \right)^2$$
(1)
for use in
$$h_v = v^2 \cdot \frac{a}{2g},$$
(9)
where $a$ = radius of pipe, $r$ = radius of bend (to centre of pipe), and $v$ = velocity in pipe; $h_v$ = loss of head due to bend.
ELBOWS AND BENDS IN PIPES. 720
It is understood that the portion $BC$ of the pipe is kept full by the flow; which, however, may not be practicable unless $BC$ is more than three or four times as long as wide, and is fall at the outlet. A semicircular bend occasions about the same loss of head as a quadrant bend, but two quadrants forming a re- verse curve in the pipe occasion a double loss. By enlarging the pipe at the bend, or providing internal thin partitions parallel to the sides, the loss of head may be considerably diminished. Weisbach gives the following table computed from eq. (1), but does not state the absolute size of the pipes.
| $\frac{a}{b}$ |
$\frac{a}{b}$ |
$\frac{a}{b}$ |
$\frac{a}{b}$ |
$\frac{a}{b}$ |
$\frac{a}{b}$ |
$\frac{a}{b}$ |
$\frac{a}{b}$ |
| .10 |
.30 |
.60 |
.80 |
.100 |
.70 |
.80 |
1.00 |
Accounts of many of Weisbach's hydraulic experiments are contained in the *Civilingenieur*, vols. xx, x, and xl.
582a. Common Pipe elbows—Prof. L. F. Bellinger of Nor- wich University, Vermont, conducted a set of experiments in 1887, when a student at Cornell, on the loss of head occasioned by a common elbow (for wrought-iron pipe), whose longitudinal section is shown in Fig. 586a.
The elbow served to connect at right angles two straight pipes having an internal diameter of .0485 in.
![]() A diagram showing a common pipe elbow.
The internal diameter of the short bend or elbow was .4 in., and the radius of its curved circular axis (a quadrant) was .2 in. Its internal surface was that of an ordinary rough casting.
A straight pipe of the same character and size and 14 feet long was first used, and the loss of head due to skin-friction (the only frictional loss in that case) carefully determined for a range of velocities from 2 to 90 ft./per second.
Two lengths of similar pipe were then joined by the elbow
730 MECHANICS OF ENGINEERING.
mentioned, forming a total length of 14 feet, and the total loss of head again determined through the same range of velocities. By subtraction, the loss of head due to the elbow was then easily found for each velocity, and assuming the form
$$A = \zeta \frac{v^2}{2g} \quad \quad (1)$$
for the loss of head, $\zeta$ was computed in each case.
From Fig. 586a it is seen that the stream meets with a sudden enlargement and a sudden diminution, of section, as well as with the short bend; so that the disturbance is of a rather complex nature.
The principal results of Prof. Bellinger's experiments, after the adjustment of the observed quantities by "least squares," were found capable of being represented fairly well by the formula
$$\zeta = 0.891 + [(\frac{v}{n})^2 - 1] \times 0.0876 \quad \quad (2)$$
where $n = [\text{veloc. in pipe in ft. per sec.}] + 5$. The following table was computed from eq. (2) (where $v$ is in ft. per second):
| $n$ |
9 |
4 |
6 |
8 |
10 |
12 |
14 |
16 |
18 |
20 |
| $\zeta$ |
0.883 |
0.69 |
0.70 |
0.67 |
0.74 |
0.75 |
0.85 |
0.99 |
1.089 |
1.185 |
530. Valve-gates and Throttle-valves in Cylindrical Pipes.— Adopting, as usual, the form
$$A_v = \zeta \frac{v^2}{2g} \quad \quad (1)$$
for the loss of head due to a valve-gate, Fig. 587, or for a throttle-valve, Fig. 588, each in a definite position, Weibach's
![]() Fig. 587.
![]() Fig. 588.
VALVE-GATE. THROTTLE-VALVE. 731
experiments furnish us with a range of values of $\zeta$ in the case of these obstructions in a cylindrical pipe 1.6 inches in diameter, as follows (for meaning of $x$, $d$, and $\alpha$, see figures. $v$ is the velocity in the full section of pipe, running full on both sides.)
| Valve-gate. |
Throttle-valve. |
| $\frac{d}{d}$ |
$\zeta$ |
| 1.0 |
.00 |
5° |
.84 |
|
.07 |
10° |
.82 |
|
.26 |
90° |
1.54 |
|
.81 |
90° |
2.41 |
|
.81 |
90° |
3.91 |
|
.06 |
45° |
.33 |
|
.53 |
90° |
.36 |
|
.53 |
90° |
.58 |
|
17.00 |
90° |
11.90 |
|
.97.8 |
70° |
731. |
531. Examples involving Diverse Losses of Head.--We here suppose, as before, that the pipes are full during the flow. Practically, provision must be made for the escape of the air which collects at the high points. If this air is at a tension greater than one atmosphere, automatic air-valves will serve to provide for its escape; if less than one atmosphere, an air-pump can be used, as in the case of a siphon used at the Kansas City Water Works. (See p. 346 of the Engineering News for November 1887.)
![]() A diagram showing a valve gate and throttle valve in a pipe.
Example 1.--Fig. 589. What head, $h$, will be required to deliver $\frac{1}{4}$ U.S. gallon (i.e. 231 cubic inches) per second
732
MECHANICS OF ENGINEERING.
through the continuous line of pipe in the figure, containing two sizes of cylindrical pipe ($d_1 = 3$ in., and $d_2 = 1$ in.), and two 90° elbows in the larger. The flow is into the air at $m$, the jet being 1 in. in diameter like the pipe. At $E_1 = m$ 90°, and the constant velocity is called $v_1$. The common rounded. Use the ft.-lb.-sec. system of units in which $g = 32.2$.
Since $Q = \frac{1}{8} gal = \frac{1}{8} \times 6068 = 0.068$ cub. ft. per sec., and therefore the velocity of the jet
$$v_n = v_1 = Q + \frac{1}{2} m (v_1^2) = 12.95 ft. per sec.;$$
hence the velocity in the large pipe is to be $v_n = (\frac{1}{4}) v_1 = 1.36$ ft. per sec. From Bernoulli's Theorem, we have, with $m$ as datum plane,
$$\frac{v_n^2}{2g} + b + 0 = 0 + b + h - C_x \frac{v_1^2}{2g} - 4f \frac{l_1}{d_2} \frac{v_1^2}{2g}$$
$$- 2C_a \frac{v_1^2}{2g} - C_x \frac{v_1^2}{2g} - 4f \frac{l_1}{d_2} \frac{v_n^2}{2g},$$
involving six separate losses of head, for each of which there is no difficulty in finding the proper $c$ or $f$, since the velocities and dimensions are all known, by consulting preceding paragraphs. (Clean iron pipe.)
From § 515, table, for $\alpha = 90°$ we have . . . . $C_x = 0.505$
$\therefore$ $S_{517}$, for $d_1 = 3$ in., and $v_n = 13.88 ft. per sec., f$_x$ = .00728
$\therefore$ $d_2 = 1$ in., and $v_n = 13.25 ft.$ per sec., f$_x$ = .00613
$\therefore$ S58 (elbows), for $\alpha = 90°$ . . . . $C_d = 0.984$
$\therefore$ S587, for sudden diminution at $K$ we have
[since $F_x + F_y = 1^* + b^* = 0.11$, : : $C_x = 0.695$]
$$C_x = \left(\frac{1}{\sqrt{6}} - 1\right)^* = 0.360.$$
Solving the above equation for A, then, and substituting above numerical values (in ft.-lb.-sec.-system), we have (noting that $v_n = v_e$, and $v_e = v_1$)
$$A = \frac{(19.25)^*}{64.4}\left(1 + (\frac{1}{4})(\frac{505 + 4 \times .00728 \times 50}{4} + 9 \times .984)\right.$$ $$+ .360 + \frac{4 \times .00613 \times .984}{\sqrt{6}}).$$
EXAMPLES; WITH LOSSES OF HEAD.
I.e,
$$A = \left(12.25^2\right) \left[1 + (0.0083 + 0.07160 + 0.0348) + (0.360 + 5.8848)\right];$$
$$A = 3.828 \times 7.3469 = 17.09 ft.$$ Ans.
It is here noticeable how small are the losses of head in the large pipe, the principal reason of this being that the velocity in it is so small ($v_0$ = only 1.36 ft per sec), and that in general losses of head depend on the square of the velocity (nearly).
In other words, the large pipe approximates to being a reservoir in itself.
With no resistances a head $h = v_0^* + 2g$ ft would be sufficient.
Example 2.—Fig. 590. With the valve-gate $V$ half raised (i.e., $s = \frac{1}{2}$ in Fig. 587), required the volume delivered per second through the clean pipe here shown. The jet issues
![]() A diagram showing a vertical pipe with a valve gate V at the top, connected to a horizontal pipe with a nozzle at the bottom. The nozzle has two branches, one labeled "q" and the other labeled "q". The diagram also shows a scale indicating distances.
from a short straight pipe, or nozzle (of diameter $d_1 = \frac{1}{4}$ in.) inserted in the end of the larger pipe, with the inner corners not rounded. Dimensions as in figure. Radius of each bend $r = 2$ in. The velocity $v_a$ of the jet in the air is velocity $v_e$ in the small pipe; hence the loss of head at $K$ is
$$\zeta_{x} \frac{v_a^*}{2g} = \zeta_x \frac{v_e^*}{2g}.$$
Now $a_e$ is unknown, as yet; but $v_e$, the velocity in the large pipe, is $= v_a \left(\frac{1}{2}\right)$; i.e., $v_e = \frac{1}{2} v_a$. From Bernoulli's Theorem:
739
734
MECHANICS OF ENGINEERING.
orem (as datum level) we obtain, after transposition,
$$h = \frac{v}{g} + C_2 \frac{v^2}{g} + C_3 \frac{v^2}{g} + 2C_4 \frac{v^2}{g} + 4f \frac{v^2}{g} + C_5 \frac{v^2}{g}$$
(1)
Of the coefficients concerned, $f_c$ alone depends on the unknown velocity $v_c$. For the present [first approximation],
put $$f_c = .006$$
From § 515, with $\alpha = 90^\circ$, $$C_2 = .505$$
From § 517, valve gate with $s = .0d$, $$C_3 = .206$$
From § 529, with $a : r = .5$, $$C_5 = .0294$$
While at $A_1$ from § 527, having
$$(F_p/F_s)(F_p)^2 - F_s^2 = Y_s = .562;$$
we find from table, $$\ldots \ldots \ldots \ldots C = .700$$
and $$C_5 = \left(\frac{1}{700}\right)^2 = (0.428)^2$$ i.e., $C_5 = .0183$$
Substituting in eq. (1) above, with $v_c = (F_p/F_s)v_m$, we have
$$v_m = \sqrt{\frac{1}{1 + \frac{81}{256} \left[ C_2 + 2C_3 + 4f_c \frac{v^2}{g} + C_5 \right]}} v_{ph},$$
(2)
in which the first radical, an abstract number, might be called a coefficient of velocity, $\phi_v$ for the whole delivery pipe; and also, since in this case $Q = F_m v_m = F_n n$, may be written $Q = uF_n v_{ph}$, it may be named a coefficient of efflux, $\mu$.
Hence
$$v_m = \sqrt{\frac{1 + .421 v_{ph}}{.300 + .06 - .334 + (.300 - .06)}} v_{ph}$$
:. $v_m = .421 v_{ph} = .421 \times 39.9 \times .35 = 16.89$ ft. per sec.
(The A3I might be called a coefficient of velocity.) The volume delivered per second is
$$Q = \frac{1}{4} d^2 v_m = \frac{1}{4} \pi (F_p)^2 16.89 = .207$ cubic ft. per sec.
(As the section of the jet $F_n = F_p$, that of the short pipe or nozzle, we might also say that $.421 = c_{eff}$ coefficient of efflux, for we may write $Q = uF_n v_{ph}$, whence $\mu = .421$.)
FLOW THROUGH Siphons. 735
532. Siphons.—In Fig. 532, § 490, the portion $HN_1O$ is above the level $BC$, of the surface of the water in the head reservoir $BL_1$, and yet under proper conditions a steady flow can be maintained with all parts of the pipe full of water, including $HN_1O$. If the atmosphere exerted no pressure, this would be impossible; but its average tension of 14.7 lbs. per sq. inch is equivalent to an additional depth of nearly 34 feet of water placed upon $BC$. With no flow, or a very small velocity, the pipe may be kept full if $N_1$ not more than 33 or 34 feet above $BC$; but the greater $v_s$, the velocity of flow at $N_1$, the greater must be the height of water above head between $L$ and $N_1$, the less must be the height of $N_1$ above $BC$ for a steady flow.
The analytical criterion as to whether a flow can be maintained or not, supposing the pipe completely filled at the outset, is that the internal pressure must be > 0 at all parts of the pipe. If on the superposition of a flow through a pipe of given design the pressure $p$ is found < 0, i.e. negative, at any point ($N_1$ being the important section for test) the supposition is inadmissible, and hence a steady flow cannot exist.
For example, Fig. 532, suppose $LN_1N_2$ to be a long pipe of uniform section (diameter = $d$, and length = $l$), and that under the assumption of filled sections we have computed $v_s$, the velocity of the jet at $N_1$; i.e.,
$$v_s = \sqrt{\frac{1}{1 + \zeta_d + 4f\frac{l}{d}}} \quad . . . . . . . (1)$$
To test the supposition apply Bernoulli's Theorem to the surface $BC$ and the point $N_1$, where the pressure is $p_s$, velocity $v_s (= v_s)$, since we have supposed a uniform section for whole pipe), and height above $BC = A_s$. Also, let length of pipe $LN_1HN_2 = l_s$. Whence we have
$$\frac{v_s^*}{2g} + p_s + h_s = 0 + \frac{P_n}{y} (b = b) + 0 - \zeta_d \frac{v_s^*}{2g} - 4f\frac{l}{d}\frac{v_s^*}{2g} \quad . . . . . . . (2)$$
[BC being datum plane.]
![]() A diagram showing a siphon system with various points labeled.
736
MECHANICS OF ENGINEERING.
Solving for $p_1$, we have
$$\frac{p_1}{r} = 34 \text{ feet} - \left[ A + \frac{v^*}{2g} (1 + C_2 + 4f) \right] \dots (5)$$
We note, then, that for $p_1$ to be > 0,
$$h_m must be < \left[ 34 \text{ feet} - \frac{v^*}{2g} - C_2 - 4f \frac{l}{d g} \right]. \quad (4)$$
In the practical working of a siphon it is found that atmospheric air, previously dissolved in the water, gradually collects at $N_1$, the highest point, during the flow and finally, if not removed, will cause the water to rise.
One device for removing the air consists in first allowing it to collect in a chamber in communication with the pipe beneath. This communication is closed by a stop-cock after the water in it has been completely displaced by air. Another stop-cock, above, being now opened, water is poured in to replace the air, which now escapes. Then the upper stop-cock is shut and the lower one opened. The same operation is again necessary until some air remains.
On p. 346 of the Engineering News of November 1887 may be found an account of a siphon which has been employed since 1875 in connection with the water-works at Kansas City. It is 1350 ft. long, and transmits water from the river to the artificial "well," from which the pumping engines draw their supply. At the highest point, which is 16 ft. above low-water level of the river, is placed a "vacuum chamber" in which the air collects under a low tension corresponding to the height, and a pipe is connected with it to remove this air and prevent its breaking off of the (partly submerged) diameter of the pipe is 24 in., and the extremity in the "well" dips 5 ft. below the level of low water. See Troutwine's Pocket-book, for an account of Maj. Crozet's Siphon.
539a. Branching Pipes.—If the flow of water in a pipe is caused to divide and pass into two others having a common
BRANCHING PIPES. 727
junction with the first, or vice versæ; or if lateral pipes lead out of a main pipe, the problem presented may be very complicated.
As a comparatively simple instance, let us suppose that a pipe of diameter $d$ and length $l$ leads out of a reservoir, and at its extremity is joined to two others of diameters $d_1$ and $d_2$, and lengths $l_1$ and $l_2$, respectively, and that the further extremities of the latter discharge into the air without nozzles under heads $h_1$ and $h_2$ below the reservoir surface. Call these two pipes Nos. 1 and 2. Then the system forms a pipe in plan.
Assuming that all entrances and junctions are smoothly rounded, so that all loss of head due to skin-friction, it is required to find the three velocities of flow, $v$, $v_1$, and $v_2$, in the respective pipes. First applying Bernoulli's Theorem to a stream-line from the reservoir surface through the main pipe to the jet at the discharging end of pipe No. 1, we have
$$\frac{u^*}{2g} = h - 4f \frac{l}{d} \frac{v^*}{2g} - 4f \frac{l_1}{d_1} \frac{v_1^*}{2g}; \quad \ldots \quad (1)$$
and similarly, dealing with a stream-line through the main pipe and No. 2,
$$\frac{u^*}{2g} = h_1 - 4f \frac{l}{d} \frac{v^*}{2g} - 4f \frac{l_2}{d_2} \frac{v_2^*}{2g}; \quad \ldots \quad (2)$$
while the equation of continuity for this case is
$$\frac{du^*}{dt} = u d^* v + u d^* v_1 + u d^* v_2; \quad \ldots \quad (3)$$
From these three equations, assuming the same in all pipes as a first approximation, we can find the three velocities (best by numerical trial, perhaps); and then the volume of discharge of the system per unit of time
$$Q = \frac{du^*}{dt} v; \quad \ldots \quad (4)$$
533. Time of Emptying Vertical Prismatic Vessels (or Inclined Prisms of Bottom) by Variable Head.
Case I. There is an orifice or short pipe in the bottom and opening into the air—Fig. 591. As the upper free surface
738
MECHANICS OF ENGINEERING.
of area = $F'$, sinka, $F'$ remains constant. Let $z$ = head of water at any stage of the emptying; it = 0 at the outset, and = 0 when the vessel is empty. At any instant, $Q$, the rate of discharge (= volume per time-unit) depends on $z$ and is
$$Q = \mu F \sqrt{2gz} \cdot \cdot \cdot (1)$$
where $\mu$ = coefficient of efflux = $\phi C$ - coefficient of velocity × coefficient of contraction [see § 495, eq. (8)]. We here suppose $F'$ so large compared with $F$, the area of the orifice, that the free surface of the water in the vessel does not acquire any notable velocity at any stage, and that hence the rate of efflux is the same at any instant, as for a steady flow with head $z$ = 0 and a new velocity in the free surface. $\mu$ is considered constant. From (1) we have
$$dV = (\text{vol. discharged in time } dt) = Qdt = \mu F \sqrt{2gz} dt \cdot (2)$$
But this is also equal to the volume of the horizontal lamina $F'dz$, through which the free surface has sunk in the same time $dt$
$$\therefore - F'dz = \mu F \sqrt{2gz} dt; \quad dt = -\frac{F'}{\mu F \sqrt{2gz}} z - dz \cdot (3)$$
We have written minus $F'dz$ because, dt being an increment, $dz$ is a decrement. To reduce the depth from $z_s$ (at the start, time $t$ = zero) to $z_a$, demands a time
$$\int_0^{t} dt = -\frac{F'}{\mu F \sqrt{2gz}} \int_{z_s}^{z_a} z - dz = \frac{2F'}{\mu F \sqrt{2gz}} [z_t - z] \cdot (4)$$
whence, by putting $z_s$ = 0, we have the time necessary to empty the whole prism
$$\int_0^{t} dt = \frac{2F'z_t}{\mu F \sqrt{2gz}} = \frac{2F'z_a}{\mu F \sqrt{2gz}} = \frac{2 \times \text{volume of vessel}}{\text{initial rate of discharge}} \cdot (5)$$
TIME OF EMPTYING VESSELS. 730
that is, to empty the vessel requires double the time of discharging the same amount of water if the vessel had been kept full (at constant head $z = z_0$ = altitude of prism).
To fill the same vessel through an orifice in the bottom, the flow through which is supplied from a body of water infinitely extensive, normally with the single normal line of Fig. 592, will obviously require the same time as given in eq. (5) above, since the effective head $z$ diminishes from $z_0$ to 0, according to the same law.
**EXAMPLE.** What time will be needed to empty a parallelopipedal tank (Fig. 591) 8 ft. by 6 ft. in horizontal section and 6 ft. deep, through a stop-cock in the bottom, when the area of efflux with fully open is known to be $A = 0.64$, and whose rate of discharge is a circle of diameter = 3 in. t. From given dimensions $F^e = 4 \times 5 = 90$ sq. ft., while $z_0 = 6$ ft. Hence from eq. (5) (ft.-lb.-sec.)
$$\text{time of emptying} = \frac{9 \times 20 \times \sqrt{6}}{0.64 \times 1^2(8t)} \times \frac{1}{32.2} = \frac{13980}{90} \text{ seconds}$$
Case II. Two communicating prismatic vessels. Required the time for the water to come to a common level $OX$, Fig. 593, efflux taking place through a small orifice, of area $F'$, under water. At any instant the rate of discharge is
$$Q = mF' \sqrt{2gz},$$
as before. $z$ = difference of level. Now if $F'$ and $F''$ are the horizontal sectional areas of the two prismatic vessels (axes vertical) we have $F'' = F'^n$, and hence $x$, which is $x + y$, $= x + (F'^n + F'^n)y$
$$\therefore z = -\frac{s}{1 + F'^n}, \text{ and } ds = \frac{ds}{1 + F'^n}.$$
740
MECHANICS OF ENGINEERING.
As before, we have
$$-F' \, dz = \mu F' \sqrt{2g} \, dh, \quad \text{or} \quad dt = -\frac{F'F''}{F' + F''} \mu F' \sqrt{2g} \, dh.$$ Hence, integrating, the time for the difference of level to change from $z_1$ to $z_2$ is
$$= \frac{2F'F''}{F' + F''} \frac{z_2 - z_1}{\mu F' \sqrt{2g}}. \quad (6)$$
and by making $z_1 = 0$ in (6), we have the time of coming to a common level is
$$= \frac{2F'F''}{F' + F''} \frac{1}{\mu F'} \sqrt{\frac{2g}{3g}}. \quad (7)$$
ALGEBRAIC EXAMPLE.—In the double lock in Fig. 594, let $L'$ be full, while in $L''$ the water stands at a level $MN$ the same as that of the tail-water. $F'$ and $F''$ are the horizontal sectional areas of the prismatic locks. Let the orifice, $O$, between them be at a depth $h$, below the initial level $KL$. Let $L''$ be filled to a height $h_1$, observe that, $MN$ of $L''$.
The orifice at $O$, area = $F'_o$, being opened, efflux from $L'$ begins into the air, and the level of $L''$ is gradually raised from $MN$ to $OD$, while that of $L'$ sinks from $KB$ to $AB$ a distance = $a$, computed from the relation vol. $Fa = vol. F'' h_1$, and the time occupied is [eq. (4)]
$$t_1 = \frac{2F'}{\mu F' \sqrt{2g}} (\sqrt{h_1} - \sqrt{h} - a). \quad (8)$$
As soon as $O$ is submerged, efflux takes place under water, and we have an instance of Case II. Hence the time of reaching a common level (after submergence of $O$) (see eq. 7) is
$$t_2 = \frac{2F'F''}{\mu F'(F' + F'')} \sqrt{\frac{h_1 - a}{3g}}, \quad (9)$$
and the total time is = $t_1 + t_2$, with $a = F'' h_1 + F''$.
TIME OF EMPTYING VESSELS.
741
CASE III. Emptying a vertical prismatic vessel through a rectangular "notch" in the side, or overfall.—Fig. 595. As before, let even the initial area (= $z b$) of the notch be small compared with the horizontal area $A'$ of tank. Let $s =$ depth of lower half of notch below level of tank surface at any instant, and $b =$ width of notch. Then, at any instant (see eq. 10, § 508),
$$\text{Rate of disch. (vol.)} = Q = \frac{3}{2}\mu b \sqrt{2g s} = \frac{3}{2}\mu b \sqrt{2g} s$$
:. vol. of disch. in dt = $\frac{3}{2}\mu b \sqrt{2g} s dt$
and putting this is $-F' ds = \text{vol. of water lost by the tank in time } dt$, we have
$$dt = -\frac{3}{2}\mu b \sqrt{2g} s^{-1} ds$$
whence
$$\int dt = -\frac{3}{2}\mu b \sqrt{2g} \int s^{-1} ds = -\frac{3}{2}\mu b \sqrt{2g} \int_0^s s^{-1} ds$$
i.e.,
$$\int_0^t dt = -\frac{3F'}{\mu b \sqrt{2g}} \left[ \ln s + \frac{1}{2} \right]_0^s$$
(10)
as the time in which the tank surface sinks from a height $z_a$ above sill to a height $z_a$ above sill. If we inquire the time $t'$ for the water to reach a height $z_a$ above sill, we put $s_a = z_a$, whence $t' = t$. As explanatory of this result, note that as $x$ diminishes not only does the velocity of flow diminish, but the available area of efflux ($= z b$) also grows less, whereas in Cases I and II the orifice of efflux remained of constant area = $F'$.
Eq. (10) is applicable to the waste-weir of a large reservoir or pond.
534. Time of Emptying Vessels of Variable Horizontal Sections.—Considering regular geometrical forms first, let us take
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MECHANICS OF ENGINEERING.
CASE I. Wedge-shaped vessel, edge horizontal and under-
neath, office $F'$ in the wedge that $z_0$, variable head is always the altitude of a triangle similar to the
section $ABC$ of the body of water
when efflux begins. At any instant
during the efflux the area, $S_t$, of
the free surface, variable here,
takes the place of $F'$ in eq. (3) of
§ 533, whence we have,
for any case of variable free surface, $dt = \frac{-Sa - dz}{\mu F' \sqrt{g}}$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (11)
In the present case $S = wu$, and from similar triangles
whence
$$S = b z + z_s,$$
and
$$dt = \frac{-b z dz}{\mu F' \sqrt{g}}.$$
:. $$\int_{z_0}^{\infty} dt = \frac{b u}{\mu F' \sqrt{g}} \int_{z_0}^{\infty} dz = \frac{4b u}{\mu F' \sqrt{g}} \left[ z_t - z_s \right],$$ . . (19)
and hence the time of emptying the whole wedge, putting $z_s = 0$, is
$$t_c = \frac{4}{3} \frac{4b u}{\mu F' \sqrt{g}} = \frac{4}{3} \frac{\text{Vol. of wedge}}{\text{initial rate of discharge}},$$ . . (13)
i.e., so long as to discharge the same volume of water under
a constant head $= z_s$. This is equally true if the ends of the
wedge are oblique, so long as they are parallel.
CASE II. Right segment of paraboloid of
revolution—Fig. 597. Axis vertical. Ori-
fice at vertex. Here the variable free surface
has at any instant an area, $S_t = wu$, $u$
being the radius of the circle and variable.
From a property of the parabola
$$w': b': s': z': : S = \pi b^2 + z_s,$$
![]() Fig. 596.
![]() Fig. 597.
TIME OF EMPTYING VESSELS.
763
and hence, from eq. (11),
$$dt = - \frac{\pi b^2}{\mu F_0} zdz;$$
$$\therefore \int_{s_0}^{s_1} dt = \frac{\pi b^2}{\mu F_0} \int_{s_0}^{s_1} zdz = \frac{9}{8} \frac{\pi b^2}{\mu F_0} [z_1 - z_0];$$
whence, putting $s_a = 0$, we have the time of emptying the whole vessel
$$t_a = \frac{4}{3} \frac{\pi b^2 F_0}{\mu F_0} = \frac{4}{3} \text{ initial rate of discharge}. \quad (14)$$
same result as for the wedge, in Case I; in fact, it applies to any vessel in which the areas of horizontal sections vary directly with their heights above the origin.
Case III. Any pyramid or cone ; vertex down ; small office in vertex—Fig. 598. Let area of the base $= S_a$ at upper edge of vessel. At any stage of the flow $S = area$ of pyramid of water. From similar pyramids
$$S_a : S :: s_a : s' ; \therefore S = \frac{S_a}{s'} s';$$
and [eq. (11)]
$$dt = - \frac{S_a}{s'} \frac{1}{\mu F_0} zdz,$$
whence ($s_a = 0$) the time of emptying the whole vessel is
$$t_a = \frac{S_a}{\mu F_0} \int_{s_0}^{s_1} zdz = \frac{9}{8} \frac{S_a}{\mu F_0};$$
or,
$$t_a = \frac{6}{5} \text{ Total volume}. \quad (15)$$
Case IV. Sphere.—Similarly, we may show that to empty
![]() Fig. 598.
744 MECHANICS OF ENGINEERING.
a sphere, of radius = $r$, through a small orifice, of area = $F$,
in lowest part, the necessary time is
$$t_4 = \frac{16}{15} \frac{\pi r^3}{\mu F \sqrt{gr}} = \frac{8}{5} \frac{\text{Vol.}}{\text{init. rate of disch.}}$$
535. Time of Emptying a Obelisk-shaped Vessel.—(An obelisk may be defined as a solid of six plane faces, two of which are rectangles in parallel planes and with sides respectively parallel, the others trapezoids; a frustum of a pyramid is a particular case.)
A volume of this shape is of common occurrence; see Fig. 599. Let the altitude = $h$, the two rectangular faces being horizontal, with dimensions as in figure. By traversing through $FG$, and $H$ right lines parallel to $EC$, to cut the upper base, we form a rectangle $KLMC$ equal to the lower base. Produce $ML$ to $P$ and $KL$ to $N$, and join $PG$ and $NG$. We have now subdivided the solid into a parallelepiped $PLMNHG$, a pyramid $PNBLG$, and two wedges, viz. $APLKH$ and $LNDMFG$, with their edges horizontal; and may obtain the time necessary to empty the whole obelisk-volume by adding the times which would be necessary to empty the individual component volumes, separately, through the same orifice or pipe in the bottom plane $EG$. These have been already determined in the preceding paragraphs. The dimensions of each component volume are given in terms of those of the obelisk, and all have a common altitude = $h$.
Assuming the orifice to be in the bottom, or else that the discharging end of the pipe, if such is used, is in the plane of the bottom $EG$, we have as follows, $F$ being the area of discharge:
![]() Fig. 599
TIME OF EMPTYING RESERVOIRS.
745
Time to empty the parallelopiped $\{ \}$. $t_1 = \frac{2h}{\mu F \sqrt{2g}}$ (1)
Time to empty the two wedges separately would be (Case I, § 533) $\{ \}$.
$t_2 = \frac{2}{3} \frac{l_1(l_1 - l_2)(b_1 - b_2)}{\mu F \sqrt{2g}}$ (9)
For the pyramid $\{ \}$... $t_3 = \frac{2}{5} \frac{(l - l_1)(b - b_1)}{\mu F \sqrt{2g}}$ (8)
Hence to empty the whole reservoir we have
$$t = t_1 + t_2 + t_3;$$
i.e.,
$$t = [3h + 8h_1 + 2h_2 + 2h_3] \cdot \frac{2\sqrt{k}}{15\mu F \sqrt{2g}}.$$ (4)
EXAMPLE.—Let a reservoir of above form, and with $b = 50$ ft., $l = 60$ ft., $b_1 = 10$ ft., $l_1 = 20$ ft., and depth of water $h = 18$ ft., be emptied through a straight iron pipe, horizontal, and leaving the side of the reservoir close to the bottom, at an angle $\alpha = 36^\circ$ with the inner plane of side. The pipe is 80 ft. long and 4 inches in internal diameter; and of clean surface. The jet issuing directly from the pipe is 4 in. in diameter, and has a velocity of $v^2 = 4\pi^2 f^2$ sq. feet. To find "the coefficient of efflux" (= $\phi$, the coefficient of velocity in this case, since there is no contraction at discharge orifice), we use eq. (4) (the first radical) of § 518, with $f$ approx. = .006, and obtain
$$\mu = \phi \sqrt{\frac{1}{1 + 4\pi^2 f^2/d}} = \sqrt{\frac{1 + .006 + 4\pi.006\times80}{d}} = .361.$$ (N.B. Since the velocity in the pipe diminishes from a value $$v = .361\sqrt{2g\times16} = 11.6$$ ft. per sec.
at the beginning of the flow to zero at the close, $f$ = .006 is a reasonably approximate average with which to compute the average $\phi$ above; see § 517.)
740
MECHANICS OF ENGINEERING.
Hence from eq. (4) of this paragraph (ft.-lb.-sec. system)
$$t = \frac{3 \times 50 \times 60 + 8 \times 10 \times 90 + 2(50 \times 20 + 10 \times 60)}{15 \times 0.361} \sqrt{\frac{g}{\rho}} \sqrt{\frac{v^2}{g} \times 32.2}$$
= 29110 sec. = 8 hrs. 5 min. [ Probably within 2 or 3% of the truth.]
388. Time of Emptying Reservoirs of Irregular Shape. Simpson's Rule.--From eq. (1), § 384, we have, for the time in which the free surface of water in a vessel of any shape whatever sinks through a vertical distance $d_z$,
$$dt = -\frac{S e^{-kz}}{\mu V^2 g z} dz,$$ whence $$\int_{z_0}^{z_1} dt = \frac{1}{\mu V^2 g z} \int_{z_0}^{z_1} S e^{-kz} dz,\quad . . . (1)$$
where $S$ is the variable area of the free surface at any instant, and $z$ the head of water at the same instant, efflux proceeding through a small orifice (or extremity of pipe) of area $F$. If $S$ can be expressed in terms of $z$, we can integrate eq. (1) (i.e., provided that $Se^{-kz}$ has a known antiderivative); but if not, the vessel or reservoir being irregular in form, as in Fig. 600 (which shows a pond whose bottom has been accurately surveyed, so that we know the value of $S$ for any stage of the emptying), we can still get an approximate solution by using Simpson's Rule for approximate integration. Accordingly, if we inquire the time in which the surface will sink from $z_0$ to the entrance $E$ of the pipe in Fig. 600 (any point $z$; at $E$ or short of that), we divide the vertical distance from $z_0$ to $z$ (in this figure) into an even number of equal parts, and from the known form of the pond compute the area $S$ corresponding to each point of division, giving them $\delta_1$, $\delta_2$, etc. Then the required time is approximately
![]() Fig. 600.
9/22
TIME OF EMPTYING POND.
747
$$t = \frac{x_0 - x_n}{8\mu F V_{2g} n} \left[ S_1 + 4 \left( \frac{S_1}{x_1^2} + \frac{S_2}{x_2^2} + \cdots + \frac{S_{n-1}}{x_{n-1}^2} \right) + 2 \left( \frac{S_1}{x_1^3} + \frac{S_2}{x_2^3} + \cdots + \frac{S_{n-1}}{x_{n-1}^3} \right) + \frac{S_n}{x_n^4} \right]$$
EXAMPLE.—Fig. 600. Suppose we have a pipe $Em$ of the same design as in the example of § 535, and an initial head of $z_0 = 16$ ft., so that the same value of $\mu = 361$, may be used. Let $x_0 - z_n = 8$ feet, and divide this interval (of $4$ ft.) into four equal parts, and let us find the area of each part by means of points of division we find from a previous survey that $S_0 = 400000$ sq. ft., $S_1 = 380000$ sq. ft., $S_2 = 270000$ sq. ft., $f_t = 210000$ sq. ft., and $S_n = 180000$ sq. ft.; while $a = 4$, $\mu = 361$, and the area $F = 4\pi(4)^3 = 0.873$ sq. ft., we obtain $(ft._n lb._n sec.)$
$$t = \frac{16 - 8}{0.361 \times 0.873 \times 2 \times 32.3 \times 3 \times 4} \left[ \frac{400000}{\sqrt{16}} + \frac{4 \times 380000}{\sqrt{16}} + \frac{9 \times 270000}{\sqrt{16}} + \frac{4 \times 210000}{\sqrt{16}} + \frac{180000}{\sqrt{8}} = 944400 sec. \right.$$
The volume discharged, $V$, may also be found by Simpson's Rule, thus: Since each infinitely small horizontal lamina has a volume
$$dV = - Sdz, ~\therefore ~V = \int_{s_0}^{s_n} Sdz,$$
or, approximately,
$$V = \frac{x_0 - x_n}{8n} \left[ S_1 + 4S_1 + 2S_2 + 4S_2 + \cdots + S_n \right].$$
Hence with $n = 4$ we have $(ft._n lb._n sec.)$
$$V = \frac{16 - 8}{8 \times 4} \left[ 40000 + 4 \left( \frac{38000}{+27000} + \frac{9 \times 27000}{+18000} + 2 \times 27000 } + 18000 = 2,160,000$ cub. ft.
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MECHANICS OF ENGINEERING.
537. Volume of Irregular Reservoir Determined by Observing Progress of Emptying -- Transforming eq. (11), § 534, we have
$$Sdt = - \mu F \sqrt{g} dt.$$
But $Sdt$ is the infinitely small volume $dV$ of water lost by the reservoir in the time $dt$, so that the volume of the reservoir between the initial and final ($0$ and $n$) positions of the horizontal free surface (at beginning and end of the time $t_n$) may be written
$$\boxed{V = \mu F \sqrt{g} \int_0^{t_n} dt.}$$
(1)
This can be integrated approximately by Simpson's Rule, if the whole time of emptying, $= t_n$, be divided into an even number of equal parts, and the values $x_1, x_2, x_3, ...$ of the ordinates noted at these equal intervals of time (not of vertical height). The corresponding surface planes will not be equidistant, in general. Whence for the particular case when $n = 4$ (see Fig. 601)
$$\boxed{V = \frac{\mu F \sqrt{g}}{3 \times 4} (x_1 + 4(x_1 + x_2) + 2x_3 + x_4).}$$
(2)
![]() Diagram showing a water reservoir with a horizontal free surface.
Fig. 601
CHAPTER VII.
HYDRODYNAMICS (Continued)---STEADY FLOW OF WATER IN OPEN CHANNELS.
538. Nomenclature.--Fig. 602. When water flows in an open channel, as in rivers, canals, mill-races, water-courses, ditches, etc., the bed and banks being rigid, the upper surface is free to conform in shape to the dynamic conditions of each case, which therefore regulate to that extent the shape of the cross-section.
In the vertical transverse section $AC$ in figure, the line $AC$ is called the air-profile (usually to be considered horizontal and straight), while the line $AB$, profile of the bed and banks, is called the wetted perimeter. It is evident that the ratio of the wetted perimeter to the whole perimeter, though never $\frac{1}{2}$, varies with the form of the transverse section.
In a longitudinal section of the stream, $EFGH$, the angle made by a surface filament $EF$ with the horizontal is called the slope, and is measured by the ratio $i = h : l$, where $l$ is the length of a portion of the filament and $h$ is the fall, or vertical distance between two ends of a line. The length between the horizontal and the line $HG$ along the bottom is not necessarily equal to that of the surface, unless the portion of the stream forms a prism; i.e., the slope of the bed does not necessarily $= s = \text{that of surface}$.
EXAMPLES.--The old Creton Aqueduct has a slope of 1.10 ft. per mile; i.e., $s = .00208$. The new aqueduct (for New
![]() A diagram showing a cross-sectional view of a river channel with a sloped bed and banks.
749
750
MECHANICS OF ENGINEERING.
York) has a slope $s = .000138$, with a larger transverse section. For large sluggish rivers $s$ is much smaller.
539. Velocity Measurements.—Various instruments and methods may be employed for this object, some of which are the following:
*Surface floats* are small balls, or pieces of wood, etc., so colored and weighted as to be readily seen, and still but little affected by the wind. These are allowed to float with the current in different parts of the width of the stream, and the surface velocity $v$ in each experiment computed from $c = l + t$, where $l$ is the distance described between parallel transverse alignments (or actual ropes where possible), whose distance apart is measured on the bank, and $t$ the time occupied.
*Density floats.* Two balls of different densities, one lightest, the other heavier than water, are united by a slender chain, their weights being so adjusted that the light ball, without projecting notably above the surface, buoyes the other ball at any assigned depth. Fig. 603.
It is assumed that the combination moves with a velocity $v'$ equal to the arithmetic mean of the surface velocity $v_s$ of the stream and that, $v'$ is the water filament at the depth of the lower ball, which latter, $v_s$ is generally less than $v'$. That is, we have
$$v' = \frac{v_s + v'}{2} \quad \text{and} \quad v_s = \frac{2v' - v}{2} \dots (1)$$
Hence, $v_s$ having been previously obtained, eq. (1) gives the velocity $v'$ at any depth of the lower ball, $v'$ being observed.
The *floating staff* is a hollow cylindrical rod, of adjustable length, weighted to float upright with the top just visible. Its observed velocity is assumed to be an average of the velocities of all the filaments lying between the ends of the rod.
Woltmann's Mill; or Tachometer; or Current-meter. Fig. 604, consists of a small wheel with inclined float (or of a small
![]() A diagram showing a floating staff instrument.
Fig. 603
CURRENT-METERS, ETC. 751
" screw-propeller " wheel S) held with its plane T to the current, which causes it to revolve at a speed nearly proportional to the velocity, \( v \), of the water passing it. By a screw-gearing W on the shaft, connection is made with a counting apparatus to record the number of revolutions. Sometimes a vanes B is attached, to compel the wheel to face the current. It is either held at the extremity of a pole or, by being adjustable along a vertical staff fixed in the bed, may be set at any desired depth below the surface. It is usually so designed as to be thrown in and out of gear by a cord and spring, that the time of making the indicated number of revolutions may be exactly noted.
By experiments in currents of known velocities, formulae have been derived to interpret the indications of any one instrument; i.e., to find the velocity \( v \) of the current corresponding to an observed number of revolutions per minute.
A peculiar form of this instrument has been recently invented, called the Ritchie-Haskell Direction-current Meter, for which the following is claimed: "This meter registers electrically on dial a boat from which used, the direction and velocity of any current. Can be used in river, harbor, or ocean currents."
Pilot's Tube consists in principle of a vertical tube open above, while its lower end, also open, is bent horizontally upstream; see A in figure. After the oscillations have ceased, the water in the tube remains
![]() A diagram showing a "screw-propeller" wheel S held with its plane T to the current. The wheel is connected to a counting apparatus via a screw-gearing W on the shaft. A vanes B is attached to compel the wheel to face the current.
Fig. 304.
A B
752
MECHANICS OF ENGINEERING.
stationary with its free surface a height $h'$, above that of the stream, on account of the continuous impact of the current against the lower end of the column. By the addition of another vertical tube (see B in figure) with the face of its lower (open) end parallel to the current (so that the water-level in it is the same as that of the current), both tubes being provided with stop-cocks, we may, after closing the stop-cocks, lift the apparatus into a boat and read off the height $h$ at level. We may also cause both columns of water to move through the tube by introducing air into both tubes in the boat by putting the upper ends of both tubes in communication with a receiver of rarefied air, and thus watch the oscillations and obtain a more accurate value of $h'$. [See Van Nostrand's Mag. for Mar. '78, p. 355.] Theoretically (see § 565), the thickness of the walls of the tube at the lower end being considerable, we have
$$c = \sqrt{gh'} \quad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$$
as a relation between $c$, the velocity of the particles impinging on the lower end, and the static height $A'$ (§ 565). Eq. (1) is verified fairly well by Weibach's experiments with fine instruments, used with velocities of from 0.32 to 1.24 meters per second. Weibach found
$$c = \left[3.54 \sqrt{A'}\right] \text{ (in meters)} \quad \text{met. per sec.,}$$
whereas eq. (1) gives
$$c = \left[3.138 \sqrt{A'}\right] \text{ (in meters)} \quad \text{met. per sec.}$$
In the instruments used by Weibach the end of the tube turning up-stream was probably straight; i.e., neither flaring nor conically convergent. A change in this respect alters the relation between $c$ and $A'$; see § 565 for Pitot's and Darcy's results.
Pitot's Tube, though simple, is not so accurate as the u-ethometer.
CURRENT-METERS.
753
The *Hydrometric Pendulum*, a rather uncertain instrument, is readily understood from Fig. 606. The side $AB$, of the quadrant $ABC$, being held vertical, the plane of the cord is inclined to the current. The angle $\theta$ between the cord and the vertical depends on $G$, the effective weight (i.e., actual weight diminished by the buoyant effort) of the ball (heavier than water), and the amount of $P$, the impulse or horizontal pressure of the current against the latter, since the cord will take the direction of the resultant $R$, for equilibrium.
Now $P$ (see § 572) for a ball of given size and character of surface varies (nearly) as the square of the velocity; i.e., if $P'$ is the impulse on a given stationary ball, when the velocity of the current is $c'$, then for any other velocity $c$ we have
$$P = \frac{P'}{c'^2} c^2 \quad \ldots \ldots \quad (3)$$
From this and the relation $\tan \theta = \frac{P}{Q}$ we derive
$$c = \sqrt{\frac{Gc'}{P'}} \cdot \sqrt{\tan \theta} \quad \ldots \ldots \quad (3)$$
With a given instrument and a specified system of units, the numerical value of the first radical may be determined as a single quantity, by experimenting with a known velocity and the value of $\theta$ then indicated, and may then, as a constant factor, be employed in (3) for finding the value of $c$ for any observed value of $\theta$; but the same units must be used as before.
540. Velocities in Different Parts of a Transverse Section.—The results of velocity-measurements made by many experimenters do not agree in supporting any very definite relation between the greatest surface velocity ($c_{\text{max}}$) of a transverse
![]() A diagram showing a pendulum-like apparatus with a ball attached to a string, which is hanging over a horizontal rod. The ball is shown at rest, with its string forming an angle with the vertical. The label "Fig. 606" is below the diagram.
Fig. 606
754
MECHANICS OF ENGINEERING.
section and the velocities at other points of the section, but establish a few general propositions:
1st. In any vertical line the velocity is a maximum quite near the surface, and diminishes from that point both toward the bottom and toward the surface.
3d. In any transverse horizontal line the velocity is a maximum near the middle of the stream, diminishing toward the banks.
3d. The mean velocity = v, of the whole transverse section, i.e., the velocity which must be multiplied by the area, $F$, of the section, to obtain the volume delivered per unit of time,
$$Q = Fv, \quad \ldots \quad \ldots \quad \ldots \quad (1)$$
is about 83 per cent of the maximum surface velocity ($v_{\text{max}}$) observed when the air is still; i.e.,
$$v = 0.83 \times (v_{\text{max}}). \quad \ldots \quad \ldots \quad (9)$$
Of eight experimenters cited by Prof. Bowser, only one gives a value (= 0.62) differing more than .05 from 83, while others obtained the values .82, .78, .82, .80, .82, .83.
In the survey of the Mississippi River by Humphreys and Ablott, it was found that the variation of the velocity in any given vertical line could be fairly well represented by the ordinates of a parabola (Fig. 607) with its axis horizontal and its vertex at a distance $d$, below the surface according to the following relation, $f'$ being a number dependent on the force of the wind (from 0 for no wind to 10 for a hurricane):
$$d = [0.317 + 0.06 f'] d; \quad (3)$$
where $d$ is the total depth, and the double sign is to be taken + for an upstream, - for a down-stream wind. The following relations were also based on the results of the survey:
(putting, for brevity, $B = 1.69 + \frac{d}{d+1.5}$)
$$\ldots \quad (4)$$
![]() Fig. 607.
**VELOCITIES IN OPEN CHANNEL.**
$$e = c_e - \sqrt{Bw\left(\frac{z-d}{d}\right)^2}, \quad \ldots \quad \ldots \quad (5)$$
$$c_m = 4c_e + 4c_d + \frac{q}{d}(4c_e - 4c_d), \quad \ldots \quad (6)$$
and
$$c_{\text{tot}} = c_m + \frac{v}{d}\sqrt{Bw}. \quad \ldots \quad (7)$$
(These equations are not of homogeneous form, but call for the foot and second as units.)
In (4), (5), and (6),
$$c = \text{velocity at any depth } z \text{ below the surface};$$
$$c_m = \text{mean velocity in the vertical curve};$$
$$c_d = \max. \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \quad " \\ c_{\text{mid}} = "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" \\ c_b = "\text{velocity at bottom}"; \\ v = "\text{mean velocity of the whole transverse section}."
It was also found that the parameter of the parabola varied inversely as the square root of the mean velocity $c_m$ of curve.
In general the bottom velocity ($c_b$) is somewhat more than $\frac{3}{4}$ the maximum velocity ($c_m$) in the same vertical. In the Misissippi the velocity at mid-depth in any vertical is found to be very nearly half of the surface velocity in the same vertical; this fact is important, as it simplifies the approximate gauging of a stream.
541. Gauging a stream or river.—Where the relation (eq. (9), § 540) $v = 83(c_m - c_d)$ is not considered accurate enough for substitution in $Q = Fv$ to obtain the volume of discharge (or delivery) $Q$ of a stream per time-unit, the transverse section may be divided into a number of subdivisions as in Fig. 608, of widths $a_1$, $a_2$, etc., and mean depths $d_1$, $d_2$, etc., and respective mean velocities, $c_1$, $c_2$, etc., computed from measurements with current-meters; whence we may write
$$Q = a_1d_1c_1 + a_2d_2c_2 + a_3d_3c_3 + etc..$$
(Fig. 608)
![]() A diagram showing a cross-section of a stream with various widths and depths labeled \(a_1, a_2, a_3\) and corresponding depths \(d_1, d_2, d_3\) respectively.
(7)
756 MECHANICS OF ENGINEERING.
With a small stream or ditch, however, we may erect a vertical boarding, and allow the water to flow through a rectangular notch or overfall, Fig. 609, and after the surface has become permanent, measure $a$, (depth of sill below the level surface somewhat back of boards), and $b$ (width) and use the formula of § 504; see examples in that article.
542. Uniform Motion in an Open Channel.—We shall now consider a straight stream of indefinite length in which the flow is steady, i.e., a state of permanency exists, as distinguished from a freshet or a wave. That is, the flow is steady when the water assumes fixed values of mean velocity $v$, and sectional area $F$, on passing a given point of the bed or bank; and the
Eq. of continuity . . . $Q = Fv = Fv = Fv = constant$ . . . (1)
holds good whether those sections are equal or not.
By uniform motion is meant that (the section of the bed and banks being of constant size and shape) the slope of the bed, the quantity of water (volume of $Q$) flowing per time-unit, and the extent of the wetted perimeter, are so adjusted to each other that the mean velocity of flow is the same in all transverse sections, and consequently the area and shape of the transverse section is the same at all points; and the slope of the surface = that of the bed. We may therefore consider, for simplicity, that we have to deal with a prism of water of indefinite length sliding down a plane surface of constant slope and moving with uniform velocity (via, i.e., mean velocity $v$ common to all the sections); that is, there is no acceleration. Let Fig. 610 show, free, a portion of this prism, of length $l$, and having its bases $\gamma$ to the bed and surface.
![]() A diagram showing a cross-section of an open channel with a straight line indicating uniform flow.
Fig. 609.
UNIFORM MOTION. OPEN CHANNEL. 757
The hydrostatic pressures at the two ends balance each other from the identity of conditions. The only other forces having
![]() Diagram showing a prism with water on one side and air on the other.
PERFECT
PERFECT
Fig. 82.
components parallel to the bed and surface are the weight $G = F \cdot y$ of the prism (where $y$ = heaving of water) making an angle $\epsilon$ (= slope) with a normal to the surface, and the friction between the water and the bed which is parallel to the surface. The amount of this friction for the prism in question may be expressed as in § 510, viz:
$$P = fric. = f \cdot S \cdot y \cdot \frac{v^2}{2g} = f \cdot poly \cdot \frac{v^2}{2g}, \quad \ldots \quad (9)$$
in which $S = w \cdot l$ = rubbing surface (area) = wetted perimeter, $w$, × length (see § 538), and $f$ an abstract number. Since the mass of water in Fig. 610 is supposed to be in relative equilibrium, we may apply it to the laws of motion of a rigid body, and since the motion is a uniform translation (§ 109) the components, parallel to the surface, of all the forces must balance.
:. $G \sin \epsilon$ must $= P = fric.; \therefore F \cdot y = f \cdot poly \cdot \frac{v^2}{2g};$
whence
$$h = f \cdot poly \cdot \frac{v^2}{F^2} \cdot \frac{l}{2g}, \quad \ldots \ldots \ldots (8)'$$
or
$$h = f' \cdot R \cdot \frac{v^2}{2g}, \quad \ldots \ldots \ldots (3)$$
in which $F + w$ is called $R$, the hydraulic mean depth, or hydraulic radius. (3) is sometimes expressed by saying that
768
MECHANICS OF ENGINEERING.
the whole fall, or head, $h$, is (in uniform motion) absorbed in friction-head. Also, since the slope $s = h + l$, we have
$$v = \sqrt{\frac{2g}{f}} \cdot \sqrt{hl}; \text{ or, } v = A \cdot \sqrt{hl}, \dots (4)$$
which is of the same form as Chebyshev's formula in § 519 for a very long straight pipe (the slope $s$ of the actual surface in this case corresponding to the slope along pionometer-summits that of a closed pipe). In (4) the coefficient $A = \sqrt{\frac{2g}{f}}$ is not, like $f$, an abstract number, but its numerical value depends on the system of units employed.
542a. Experiments on the flow of water in open channels—Those of Darby and Bazin, begun in 1855 and published in 1865 ("Recherches Hydrauliques"), were very carefully conducted with open conduits of a variety of shapes, sizes, slope, and character of surface. In most of these a uniform flow was secured before the taking of measurements. The velocities ranged between from about 0.5 to 8 or 10 ft. per second; the hydraulic radii from 0.03 to 3.0 ft., with deliveries as high as 182 cub. ft. per second. For example, the following results were obtained in the canals of Marseilles and Cropano; the quantity $Q$ being for the foot and second. The sections were nearly all rectangular. See eq. (4) above.
| No. |
(cub. ft. |
per sec.) |
$R$ |
(ft.) |
Area |
numb. |
0% |
per |
sec.) |
(foot and |
sec.) |
Character of the |
| 1 |
183.78 |
|
1.004 |
|
0.0087 |
|
10.36 |
|
137.1 |
|
Very smooth. |
| 2 |
183.78 |
|
1.004 |
|
0.0087 |
|
10.36 |
|
137.1 |
|
Very smooth. |
| 3 |
48.98 |
|
7.06 |
|
0.099 |
|
11.38 |
|
79.4 |
|
Rough. |
| 4 |
48.98 |
|
7.06 |
|
0.099 |
|
11.38 |
|
79.4 |
|
Rough. |
| 5 |
48.98 |
|
891 |
|
-0.0212 |
|
7.98 |
|
78.5 |
|
Rather rough. |
| 6 |
48.98 |
|
-0.0212 |
|
-0.0212 |
|
-7.98 |
<
&nb...
UNIFORM MOTION. KUTTER'S FORMULA. 750
good mortar joints, and about 9 ft. wide; the depths of water ranging from 1.5 to 4.5 ft. With plaster of pure cement on the bed in one of the experiments the high value of $A = 133.6$ was reached (foot and second), with $v = 808$ ft. per second, $R = 2111$ ft. per second, and $s = 87.1$ cm. ft. per second.
Captain Cunningham, in his experiments on the Ganges Canal at Roorkee, India, in 1881, found $A$ to range from 48 to 130 (foot and second).
Humphreys and Abbot's experiments on the Mississippi River and branches (see § 540), with values of $R =$ from 2 or 3 ft. to 72 ft., furnish values of $A =$ from 53 to 167 (foot and second).
§ 549. Kutter's Formula.—The experiments upon which Weibisch based his deductions for $f$, the coefficient of fluid friction, were scanty and on too small a scale to warrant general conclusions. That author considered that $f$ depended only on the velocity, disregarding altogether the degree of roughness of the bed, and gave a table of values in accordance with this view, varying from $f = .0002$ at 2 ft. per sec. to $.0109$ for 0.4 ft. per sec.; but in 1869 Moser, Kutter and Ganguillet, having a much wider range of experimental data at command, including those of Darcy and Bazin, and those obtained on the Mississippi River, evolved a formula, known as Kutter's Formula, for the uniform motion of water in open channels, which is claimed to harmonize in a fairly satisfactory manner the chief results of the best experiments in that direction. They make the coefficient $A$ in eq. (4) (or rather the factor $\sqrt{\frac{v}{g}}$ contained in $A$) a function of $R$, $s$, and also $n$ an abstract number, or coefficient of roughness, depending on the nature of the surface of the bed and banks ; viz.,
$$\begin{array}{rlr} & {v \text{ in } \\ & \text{ft.} \\ & \text{per sec.}} & {= \left[ \frac{41.6 + \frac{1.811}{n} + \frac{0.0092}{s}}{1 + \left( \frac{41.6 + \frac{0.0092}{s}}{n} \right) \sqrt{\frac{s}{R(\text{in ft.)}}} } \right] \sqrt{\frac{s}{R(\text{in feet})}} } & {(b)} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {} \\ & {} & {}\end{array}$$
which is Kutter's Formula.
780
MECHANICS OF ENGINEERING.
That is, comparing (5) with (4), we have $f$ a function of $n$, $R$, and $s$, as follows:
$$f = \frac{1 + 41.6 + 0.0021}{\sqrt{R} \text{ in ft.}} \cdot \cdot \cdot \cdot \cdot (6)$$
From (6) it appears that $f$ decreases with an increasing $R$, as has been also noted in the case of closed pipes (§ 517); that it increases with increasing roughness of surface; and that it is somewhat dependent on the slope. The makers of the formula give the following values for $n$.
Values of $n$. $n =$
.009 for well-planned timber bed;
.010 for plaster in pure cement;
.011 for plaster in cement with $\frac{3}{4}$ sand;
.012 for unplanted timber;
.013 for brickwork;*
.015 for canvas lining on frames;
.017 for rubble;
.020 for canal in very firm gravel;
.025 for rivers and canals in perfect order and regimen, and perfectly free from stones and weeds;
.030 for rivers and canals in moderately good order and regimen, but overgrown with vegetation;
.035 for rivers and canals in bad order and regimen, overgrown with vegetation and strewn with stones or detritus of any sort.
Kutter's Formula is claimed to apply to all kinds and sizes of watercourses, from large rivers to sewers and ditches; for uniform motion. If $\sqrt{V R}$ is the unknown quantity, Kutter's Formula leads to a quadratic equation; if $s$ the slope, to a cubic. Hence, to save computation, tables have been prepared, some of which will be found in vol. 28 of Van Nostrand's Magazine.
* For ordinary brick sewers Mr. B. F. Hartford claims that $n = .04$ gives good results. See Jour. Eng. Societies for '84-'85, p. 290.
| $f$ |
$= \frac{1 + 41.6 + 0.0021}{\sqrt{R} \text{ in ft.}}$ |
UNIFORM MOTION. OPEN CHANNEL. 761
(pp. 135 and 398) (sewers), and in Jackson's works on Hydraulics (rivers).
The following table will give the student an idea of the variation* of the coefficient $A = \sqrt{\frac{g}{R}}$, of eq. (4), or large bracket of eq. (5), with different hydraulic radii, slopes, and values of $n$, according to Kutter's Formula; from $R = 4$ ft., for a small ditch or sluice-way (or a wide and shallow stream), to $R = 15$ ft., for a river or canal of considerable size. Under each value of $R$ are given two values of $A$; one for a slope of $s' = .001$, and the other for $s' = .00005$. All these values of $A$ imply the use of the foot and second.
These values of $A$ have been scaled by the writer from a diagram given in Jackson's translation of Kutter's "Hydraulic Tables," and are therefore only approximate. The corresponding value of $\nu$, the coefficient of fluid friction, can be computed from $\nu = \frac{59}{A^2}$.
| n |
$R = 4$ ft. |
$R = 1$ ft. |
$R = 8$ ft. |
$R = 16$ ft. |
| 0.000 |
133 |
148 |
187 |
174 |
187 |
196 |
199 |
222 |
| 0.001 |
133 |
148 |
187 |
174 |
187 |
196 |
199 |
222 |
| 0.002 |
58 |
69 |
70 |
69 |
70 |
69 |
70 |
69 |
| 0.003 |
58 |
69 |
70 |
69 |
70 |
69 |
70 |
69 |
| 0.004 |
31 |
43 |
43 |
40 |
58 |
58 |
66 |
72 |
| 0.005 |
31 |
43 |
43 |
40 |
58 |
58 |
66 |
72 |
|
The formula used in designing the New Aqueduct for New York City, in 1855, by Mr. Fieley, consulting engineer, was [see (4)]
$$v (\text{ft. per sec.)} = 149 \sqrt{R (\text{in ft.)} \times s} \quad \ldots \quad (7)$$
whereas Kutter's Formula gives for the same case (a circular section of 14 ft. diameter, and slope of 0.7 ft. to the mile), with $n = 0.013$,
$$v (\text{ft. per sec.)} = 140.7 \sqrt{R (\text{in ft.)} \times s} \quad \ldots \quad (8)$$
*The aqueduct has this circular form for a small portion, only, of its length; a "hose-like" section of very nearly the same flowing capacity being given to the greater portion of the remainder.
![]() A table showing values of A and R for different n values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
![]() A table showing values of v for different R and s values.
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782
MECHANICS OF ENGINEERING.
To quote from a letter of Mr. I. A. Shaler of the Aqueduct Corps of Engineers, "Mr. Freley states that the cleanliness of the conduit (Sudbury) had much to do in affecting the flow. He found the flow to be increased by 7 or 8 per cent in a portion which had been washed with a thin wash of Portland cement."
EXAMPLE 1.—A canal 1000 ft. long of the trapezoidal section in Fig. 611 is required to deliver 300 cubic ft. of water per second with the water 8 ft. deep at all sections (i.e., with uniform motion), the slope of the bank being such that for a depth of 8 ft. the width of the water surface (or length of air-profile) will be 20 ft.; and the coefficient for rounding off being $n = .020$. What is the necessary slope to be given to the bed (slope of bed = that of surface, here) (ft./ft., sec.)?
The mean velocity
$$v = Q + F = 300 + \frac{1}{2}(20 + 8) \cdot 8 = 2.67 \text{ ft. per sec.}$$
(So that the surface velocity of mid-channel in any section would probably be $(c_{\text{max}}) = v + 0.83 = 3.51 \text{ ft. per sec. (eq. (3), § 540).}$)
The wetted perimeter
$$w = 8 + 2 \sqrt{8^2 + 6^2} = 28 \text{ ft.,}$$
and therefore the mean hydraulic depth
$$= R + w = 112 + 28 = 4 \text{ ft.}$$
To obtain a first approximation for the slope, we may use the value $f = .00795$ given by Weisbach for a velocity of 2.67 ft. per sec., and obtain, from (3),
$$h = \frac{.00795 \times 1000 \times 98(2.67)^2}{112 \times 3 \times 32.2} = .291 \text{ ft.;}$$
i.e., $s = h + l = .00291.$
UNIFORM MOTION IN OPEN CHANNEL 763
With this value for the slope and $R = 4$ ft. (see above), we then have, from eq. (6) (putting $n = .020$),
$$f = \frac{1 + (41.6 + .0081 + .090)}{5.184 + .020 + .000221} = .0071,$$
with which value of $f$ we now obtain
$$A = 0.300 \text{ feet}; \quad \text{i.e., slope } m = .0090.$$
EXAMPLE 2.—If the bed of a creek falls 20 inches every 1500 ft. of length, what volume of water must be flowing to maintain a uniform mean depth of 4 ft., the corresponding surface-width being 40 ft., and wetted perimeter 46 ft.? The bed is "in moderately good order and regimen;" use Kutter's Formula for finding $n = .089$ (ft. and sec.).
First we have
$$\sqrt{Rt} = \sqrt{(40 \times 4)} \div \left(46 \times \frac{1500}{4}\right) = .066,$$
while $\sqrt{R}(ft.) = 1.98$, and the slope $s = \frac{H}{L} + 1500 = .00111$;
hence
$$v = \frac{41.6 + .089 + .091}{1 + \frac{.0981}{.00111}} \times .066 = \frac{104.43 \times .066}{1.6685},$$
or
$$v = 4.18 \text{ ft. per sec.}$$
Hence, also,
$$Q = Fv = 40 \times 46 \times 4.18 = 743.4 \text{ cub. ft. per sec.}$$
[N.B. Weibach works this same example by eq. (8) with a value of $f$ taken from his own table, his result being $v = 6.1$]
764
MECHANICS OF ENGINEERING.
ft. per sec., which would probably be attained in practice only by making the bed and tanks smoother than as given.]
EXAMPLE 8.—The desired transverse water-velocity in a canal is to be 3 ft. in 1600 ft.; i.e., $v = \frac{3}{1600}$; or, for $l = 1600$ ft., $h = 3$ ft. What must be the velocity (mean) of each section, for a uniform motion ; the corresponding volume delivered per sec., $Q = F_v = t$; assuming that the character of the surface warrants the value $n = .030$ ?
Knowing the slope $s$, $= \frac{3}{1600}$; and the hydraulic radius $R$, $= F + w$, or $75.80$ sq. ft. $= 34.97$ ft., $= 5.215$ feet; with $n = .030$ we substitute directly in eq. (8), obtaining $v = 4.97$ ft. per sec.; whence $Q = F_v = 370$ cub. ft. per sec.
543. Hydraulic Mean Depth for a Minimum Frictional Resistance.—We note, from eq. (8), § 549, that if an open channel of given length $l$ and sectional area $F$ is to deliver a given volume, $Q$, per time-unit with uniform motion, so that the common mean velocity of all sections ($= Q/F$) is also a given quantity, the necessary fall $h$, or slope $s = h/l$, is seen to be inversely proportional to $R$, the hydraulic mean depth of the section, $(F + w)$, sectional area + wetted perimeter.
For $h$ to be as small as possible, we may design the form of transverse section, so as to make $R$ large as possible; i.e., to make the wetted perimeter a minimum for a given $F$; for in this way a minimum of frictional contact, or area of rubbing surface, is obtained for a prism of water of given sectional area $F$ and given length $l$.
In a closed pipe running full the wetted perimeter is the whole perimeter; but if the given sectional area is shaped in the form of a circle, the wetted perimeter, i.e., $w$, is a minimum (and $R$ a maximum). If the full pipe must have a polygonal shape of $n$ sides, then the regular polygon of $n$ sides will provide a minimum $w$.
Wherein it follows that if the pipe or channel is running
UNIFORM MOTION IN OPEN CHANNEL. 785
half full, and thus becomes an open channel, the semicircle, of all curvilinear water profiles, gives a minimum $w$; but all rectangular sections with sides at 60° with the horizontal half of a regular hexagon gives a minimum $w$. Among all rectangular sections the half square gives a minimum $w$;
and of all half octagons the half of a regular octagon gives a minimum $w$ (and max. $R$) for a given $F$. See Fig. 618 for all these.
The egg-shaped outline, Fig. 614, small end down, is frequently given to sewers in which it is important that the different velocities of the water at differ- ferent stages (depths) of flow (depend- ing on the volume of liquid passing per unit-time) should not vary widely from each other. The lower portion $ABC$, providing for the lowest stage of flow, $AB$, is nearly semicircular, and thus in- duces a velocity at its lowest stage (being constant at all stages) which does not differ extremely from that occurring when the water flows at its highest stage $DE$, although this latter velocity is the greater; the reason being that $ABC$ from its advantageous form has a hydraulic radius, $R$, larger in proportion to its sectional area, $F$, than $DCE$.
That is, $F \cdot w$ for $ABC$ is more nearly equal to $F \cdot w'$ for $DEC$ than if $DEC$ were a semicircle, and the velocity at the lowest stage may still be sufficiently great to prevent the de- posit of sediment. See § 575.
544. Trapezoid of fixed side-slope.—For large artificial water- courses and canals the trapezoid, or three-sided water-profile (symmetrical), is customary, and the inclination of the bank,
![]() A diagram showing a trapezoidal cross-section of an open channel.
Fig. 613
![]() A diagram showing a semicircular cross-section of an open channel.
Fig. 614
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MECHANICS OF ENGINEERING.
or angle $\theta$ with the horizontal, Fig. 615, is often determined by the nature of the material composing it, to guard against washouts, caving in, etc. We may thus be concerned with the following problem. Given the area, $F$, of the transverse section, and the angle $\theta$, required the value of the depth $x$ (or of upper width $z$, or of lower width $y$, both of which are functions of $z$) to make the hydraulic mean depth, $R = F + w$, a maximum, or so $Fw$ a minimum. $F$ is constant.
From the figure we have
$$w = AB + 2BF = y + 2z\sec \theta, \quad \ldots \quad (1)$$
and
$$F = yz + x^2\cot \theta; \quad \ldots \quad (2)$$
whence
$$y = \frac{1}{z}(F - x^2\cot \theta). \quad \ldots \quad (3)$$
substituting which in (1) and dividing by $F$, noting that $2\sec \theta - \cot \theta = \frac{2-\cos \theta}{\sin \theta}$, we have
$$\frac{w}{F} = \frac{1}{z} + \frac{2-\cos \theta}{F\sin \theta}. z. \quad \ldots \quad (3)$$
For a minimum $w$ we put
$$\frac{d(w)}{dx} = 0; \text{ i.e., } -\frac{1}{xz} + \frac{2-\cos \theta}{F\sin \theta} = 0;$$
$$\therefore x(\text{for max. or min. } w) = \pm\sqrt{\frac{F\sin \theta}{2-\cos \theta}}.$$
The $+$ sign makes the second derivative positive, and hence for a min. $w$ or max. $R$ we have
$$x(\text{call it } x') = x' = -\sqrt{\frac{F\sin \theta}{2-\cos \theta}}. \quad (4)$$
TRAPEZOID FOR MINIMUM WETTED PERIMETER. 767
while the corresponding values for the other dimensions are
$$y' = \frac{F}{x'} - x'\cot{\theta} \quad \ldots \quad \ldots \quad \ldots \quad (5)$$
and
$$x' = y' + 2x'\cot{\theta} = \frac{F}{x'} + x'\cot{\theta} \quad \ldots \quad \ldots \quad (6)$$
For the corresponding hydraulic mean depth $R'$ [see (3)], i.e., the max. $R$, we have
$$\frac{1}{R'} = \frac{1}{x'} + \frac{2 - \cos{\theta}}{F\sin{\theta}} = \frac{2}{x'}; \quad \ldots \quad (7)$$
$$\therefore R' = 4x' = \frac{1}{2}\sqrt{\frac{F\sin{\theta}}{2 - \cos{\theta}}} \quad \ldots \quad (8)$$
Equations (4), (5), $\ldots$ (8) hold good, then, for the trapezoidal section of least frictional resistance for a given angle $\theta$.
Problem.-Required the dimensions of the trapezoidal section of minimum frictional resistance for $\theta = 45^\circ$, which with $A = 6$ inches fall in every 1200 feet (= $t$) is required to deliver $Q = 360$ cub. ft. of water per minute with uniform motion.
Here we have given, with uniform motion, $h$, $l$, and $Q$, with the requirement that the section shall be trapezoidal, with $\theta = 45^\circ$, and of minimum frictional resistance. The following equations are available:
Eq. of continuity $\ldots Q = Fv_0 \quad \ldots \quad (1)$
Eq. (8) preceding, for condition of least resistance $\therefore R' = \frac{1}{2}\sqrt{\frac{\sin{\theta}}{2 - \cos{\theta}}}F$. (9)
From eq. (8), for uniform motion, $k = f'\frac{x'}{B}\frac{x'}{2g}$ $\ldots$ $\ldots$ (3')
There are three unknown quantities, $v$, $F$, and $R'$. Solve
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MECHANICS OF ENGINEERING.
(1') for $v$; solve (2) for $F$; substitute their values in (3); whence
$$\frac{h \sin \theta}{2} = \frac{2l}{2g} \cdot \frac{Q^2}{\sqrt{F}} \cdot v = \left[ \frac{2f + 2 - \cos \theta}{2gh \sin \theta} \right] F. \quad (4')$$
Since $f$ cannot be exactly computed in advance, for want of knowing the value of $R$, we calculate it approximately (eq. (6), § 544b) for an assumed value of $R$; insert it in the above equation, and thus find a new value of $F$; and then, from (8), a corresponding value of $v$, from which a new value of $f$ can be computed. Thus after one or two trials a satisfactory adjustment of dimensions can be secured.
545. Variable Motion.—If a steady flow of water of a delivery $Q_0 = P_0$, constant, takes place in a straight open channel the slope of whose bed has not the proper value to maintain a "steady motion," then the velocity will "ensue" (the flow is still steady, but variable); i.e., although the mean velocity in any one transverse section remains fixed (with lapse of time), this velocity has different values for different sections; but as the eq. of continuity,
$$Q = Fv = F_0v_0 = F_1v_1,$$
still holds (since the flow is steady), the different sections have different areas. If, Fig. 616, a stream of water flows down an inclined trough without friction, the relation between the velocities $v_0$ and $v_1$ at any two sections 0 and 1 will be the same as for a material point sliding down a guide without friction (see § 79, latter part), viz.:
$$\frac{v_1^*}{2g} = \frac{v_0^*}{2g} + h. \quad (f)$$
![]() A diagram showing a stream of water flowing down an inclined trough.
Fig. 616
VARIABLE MOTION. OPEN CHANNEL. 769
a.: equation of heads (really a case of Bernoulli's Theorem, § 493). But, considering friction on the bed, we must subtract the mean friction-head $f \frac{l}{R} \frac{w_a^*}{2g}$ [see eqs. (3) and (3'), § 548] lost between 0 and 1; this friction-head may also be written thus: $f \frac{l w_a^*}{F_+} \frac{w_a^*}{2g}$; and therefore eq. (1) becomes
$$\frac{w_a^*}{2g} = \frac{w_a^*}{2g} + h - \frac{f l w_a^*}{F_+} \frac{w_a^*}{2g}, \quad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (9)$$
which is the formula for variable motion; and in it $l$ is the length of the section considered, which should be taken short enough to consider the surface straight between the end-sections, and the latter should differ but slightly in area. The subscript $m$ may be taken as referring to the section midway between ends, so that $w_m = w(0 + \frac{l}{2})$. The actual perimeter $w_m = w(0 + w_o)$, and $F_m = F(F_+ + F_-)$. Hence eq. (9) becomes
$$h = \frac{w_m^*}{2g} - \frac{w_m^*}{2g} + \frac{f l(w_m + w_o)}{F_+ + F_-} \frac{w_m^*}{2g}; \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. \quad. (8)$$
and again, by putting $v_e = Q + F_-$, $v_i = Q + F_+$, we may write
$$h = [\frac{1}{F_+} - 1\frac{1}{F_-} + 1\frac{f l(w_m + w_o)}{F_+ + F_-} (\frac{1}{F_+} + 1\frac{1}{F_-})]Q'_{-}; (4)$$
whence
$$Q = -\sqrt{\frac{2gh}{F_+}}[\frac{1}{F_+} - 1\frac{1}{F_-} + 1\frac{f l(w_m + w_o)}{F_+ + F_-}] [\frac{1}{F_+} + 1\frac{1}{F_-}] (5)$$
From eq. (4), having given the desired shapes, areas, etc., of the end-sections and the volume of water, $Q$, to be carried per unit of time, we may compute the necessary fall, $h$, of the surface, in length $l = T$, while from eq. (8), having observed in an actual water-course the values of the sectional areas $F_+$ and $F_-$, the wanted perimeter $w_m$ and $w_o$, the lengths $d_+$ of the per-
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MECHANICS OF ENGINEERING.
tion considered, we may calculate $Q$ and thus gauge the stream approximately, without making any velocity measurements.
As to the value of $f$, we compute it from eq. (6), § 542b, using for $R$ a mean between the values of the hydraulic radii of the end-sections.
544. Bands in an Open Channel.—According to Humphreys and Abbot's researches on the Mississippi River the loss of head due to a bend may be put
$$h_b = \frac{v^2}{g} \cdot \frac{\delta}{\pi} \cdot \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ [equation] \\ h = f(\frac{v^2}{g}) + 6\frac{v^2\delta}{536\pi} - [ft.\ and.\ sec.] \\ [equation] \\ while eq. (2) of § 545 for variable motion would then become \\ [equation] \\ (e and $\delta$ as above.) (For "radian" see p. 544.)
547. Equations for Variable Motion, introducing the Depth.
—Fig. 618. The slope of the bed being sin $\alpha$ (or simply $\alpha$, $\pi$ mean.), while that of the surface is different, viz.,
$$\sin\beta = s = h + l,$$
we may write
$$h = d_c + l\sin\alpha - d_t,$$
![]() Fig. 617.
![]() Fig. 618.
VARIABLE MOTION. OPEN CHANNEL. 771
in which $d_s$ and $d_e$ are the depths at the end-sections of the portion considered (steady flow with variable motion). With these substitutions in eq. (4), § 545, we have, solving for $l$,
$$l = \frac{d_s - d_e - \left( \frac{1}{F_s} + \frac{1}{F_e} \right) Q^2}{\frac{f w_a}{g} \left( \frac{1}{F_s} + \frac{1}{F_e} \right) Q^2 - \sin \alpha} \dots (1)$$
From which, knowing the slope of the bed and the shape and size of the end-sections, also the discharge $Q$, we may compute the length or distance, $l$, between two sections whose depths differ by an assigned amount $(d_s - d_e)$. But we cannot compute the change of depth for an assigned length $l$ from (6). However, if the width $b$ of the stream is constant, and the same at all depths; i.e., if all sections are rectangles having a common width; eq. (6) may be made simplified by introducing some approximations, as follows: We may put
$$\left( \frac{1}{F_s} + \frac{1}{F_e} \right) Q = \frac{F_s^* - F_e^*}{F_s^* F_e^* g} = \frac{(F_s^* - F_e^*) (P_s + P_e)}{F_s^* F_e^* g} v_s^*$$
$$= \frac{(d_s - d_e)(d_s + d_e)}{d_s d_e} v_s^*, \text{ which approx. } = \frac{2(d_s - d_e)}{d_s d_e} v_s^*$$
and, similarly,
$$\frac{w_a}{g} \left( \frac{1}{F_s} + \frac{1}{F_e} \right) Q = \frac{w_a(F_s^* + F_e^*) v_s^*}{(F_s + F_e) F_s^* F_e^* g}$$
which approx. = $\frac{w_a v_s^*}{d_b g}$
Hence by substitution in eq. (6) we have
$$l = \frac{(d_s - d_e) \left[ 1 - \frac{2}{d_s d_e} v_s^* \right]}{\frac{f w_a v_s^*}{d_b g} - \sin \alpha} \dots (7)$$
§ 547a. Backwater.—Let us suppose that a steady flow has been proceeding with uniform motion (i.e., the surface parallel
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MECHANICS OF ENGINEERING.
to the bed) in an open channel of indefinite extent, and that a vertical wall in now set up across the stream. The water rises and flows over the edge of the wall, or weir, and after a time a steady state is reached. We shall suppose that the depth of water close to the wall on the up-stream side is greater than $d_s$, the original depth. We now have "variable motion" above the weir, and at any distance $x$ up-stream from the weir the new depth $y$ is greater than $d_s$. This increase of depth is called backwater, and, though decreasing up-stream, may be perceptible several miles above the weir. Let $s$ be the slope of the original uniform motion (and also of present bed), and $v$ the velocity of the original uniform motion, and let $k = \frac{v^2}{g}$. Then, if the section of the stream is a shallow rectangle of constant width, we have the following relation (Rankine):
$$x = \frac{1}{s}\left[y - y + (d_s - 2k(\phi - \phi_s)\right], \dots \dots (1)$$
where $\phi$ is a function of $\frac{y}{d_s}$, as per following table:
| For |
$\frac{y}{d_s}$ |
1.10 |
1.30 |
1.50 |
1.60 |
1.70 |
| $\phi = \omega$ |
.80 |
.80 |
.876 |
.804 |
.718 |
.699 |
| For |
$\frac{y}{d_s}$ |
1.90 |
2.00 |
2.20 |
2.40 |
2.60 |
2.80 |
| $\phi = .166$ |
.147 |
.123 |
.107 |
.099 |
.076 |
.065 |
|
|
|
|
|
|
.056 |
$\phi_s$ is found from $\frac{y_s}{d_s}$ precisely as $\phi$ from $\frac{y}{d_s}$, by use of the table.
With this table and eq. (1), therefore, we can find $x$, the distance ("amplitude of backwater") from the weir of the point where any assigned depth $y$ (or "height of backwater", $y - d_s$) will be found.
For example, Prof. Bowser cites the case from D'Aubuisson's Hydraulics of the river Weser in Germany, where the erection of a weir increased the depth at the weir from 9.5 ft. to 10 ft., the flow having been originally "uniform" for 10 miles. Three miles above the dam the increase $(y - d_s)$ of depth was 1.25 ft., and even at four miles it was 0.75 ft.
CHAPTER VIII
DYNAMICS OF GASEOUS FLUIDS.
548. Steady Flow of a Gas—[N.B. The student should now review § 499 up to eq. (5.)] The differential equation from which Bernoulli's Theorem was derived for any liquid, without friction, was [eq. (5), § 492]
$$\frac{1}{g} v dv + ds + \frac{1}{y} dp = 0, \ldots \ldots \ldots (A)$$
and is equally applicable to the steady flow of a gaseous fluid, but with this difference in subsequent work, that the heaviness, $y$ ($§ 7$), of the gas passing different sections of the pipe or stream-line is, or may be, different (though always the same at a given point or section, since the flow is steady). For the present we neglect friction and consider the flow from a large receiver, whose great body of the gas is practically at rest, through an orifice in a thin plate, or a short nozzle with a rounded entrance.
In the steady flow of a gas, since $y$ is different at different points, the equation of continuity takes the form
Flow of weight per time unit = $F_v y_v = F_v y_s = \text{etc.}; \text{(a)}$
i.e., the weight of gas passing any section, of area $F$, per unit of time, is the same as for any other section, or $F y_v = \text{constant}, y$ being the heaviness at the section, and $v$ the velocity.
549. Flow through an Orifice—Remarks—in Fig. 619 we have a large rigid receiver containing gas at some tension, $p_s$, higher than that, $p_m$, of the (still) outside air (or gas), and at some absolute temperature $T_s$, and of some heaviness $y_s$; that is, in a state $n$. The small orifice of area $F$ being opened, the gas begins to escape, and if the receiver is very large, or if the supply is continually kept up by a blowing-engine, e.g., after
773
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MECHANICS OF ENGINEERING.
A very short time the flow becomes steady. Let $nm$ represent any stream-line (§ 405) of the flow. According to the ideal subdivision of this stream-line into laminae of equal mass or weight (not equal volumes), we have in the lape of time $dt$ establishing eq. (A) for any one lamina, each lamina in the lape of time $dt$ moves into the position just vacated by the lamina next in front, and assume precisely the same velocity, pressure, and volume (and therefore heaviness) as that from one had at the beginning of the $dt$. In its progress toward the orifice it expands in volume, its mass remains constant, its velocity, immovable at $n_1$, is gradually accelerated on account of the pressure from behind always being greater than that in front, until at $m$, in the "throat" of the jet, the velocity has become $v_m$, the pressure (i.e., tension) has fallen to a value $p_m$, and the heaviness has changed to $y_m$. The temperature $T_m$ (absolute) is less than $T_*$, since the expansion has been rapid, and does not depend on the temperature of the outside air or gas into which efflux takes place; though, of course, after the efflux has once free from tension, it will regain its original temperature in time.
We assume the pressure $p_a$ (in throat of jet) to be equal to that of the outside medium (as was done with flow of water), so long as that outside tension is greater than $.527 p_a$ but if it is less than $.527 p_a$ and is even zero (a vacuum), experiment seems to show that $p_a$ remains equal to $.527$ of the interior tension $p_a$: probably on account of the expansion of the effluent gas beyond the throat, Fig. 620, so that although the tension in the outer edge, at $a$, of this jet is equal to that in outside medium, the tension at $m$ is greater because of the centripetal and centrifugal forces developed in the curved filaments between $a$ and $m$. (See § 553.)
550. Flow through an Orifice; Heaviness assumed Constant during Flow. The Water Formula.—If the inner tension $p_a$ ex-
STeady Flow of Gases. 775
ceeds the outer, $p_m$, but slightly, we may assume that, like water, the gas remains of the same heaviness during flow.
Then, for the simultaneous advance made by all the laminae of a stream-like Fig. 613, in the time $dt$, we may conceive an equation like eq. (A), for each lamina between $n$ and $m$, and corresponding terms added; i.e.,
(For reference) $\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots$d
In general, $\gamma$ is different in the different laminae, but in the present case it is assumed to be the same in all; hence, with $m$ as datum level and $h$ = vertical distance from $n$ to $m$, we have, from eq. (B),
$$\frac{v_n}{dy} - \frac{v_m}{dy} + 0 - h + \frac{p_n}{y} - \frac{p_m}{y} = 0.$$
But we may put $v_n = 0$; while $h$, even if several feet, is small compared with $\frac{p_n}{y} - \frac{p_m}{y}$ E.g., with $p_n = 15$ lbs. per sq. in. and $p_m = 16$ lbs. per sq. in., we have for atmospheric air at freezing temperature
$$\frac{p_n}{y} - \frac{p_m}{y} = 1683 \text{ feet}.$$
Hence, putting $v_n = 0$ and $h = 0$ in eq. (1), we have
$$\frac{v_m^*}{dy} = \frac{p_n - p_m}{y}.$$ (Water formula; for small difference of pressures, only.)
The interior absolute temperature $T_n$ being known, the $v_n$ (interior heaviness) may be obtained from $v_n = p_n y_n T_n + T_{n,p}$ ($§ 472$), and the volume of flow per unit of time then obtained (first solving (2) for $v_n$)
$$Q_a = F_a v_n,$$
where $F_a$ is the sectional area of the jet at $m$. If the mouth-piece or orifice has well-rounded interior edges, as in Fig. 541,
![]() A diagram showing a stream-like figure with various points labeled.
776 MECHANICS OF ENGINEERING.
its sectional area $F$ may be taken as the area $F_m$. But if it is an orifice in "this plate," putting the coefficient of contraction = $C = 0.60$, we have
$$F_m = OF = 0.60 F \quad \text{and} \quad Q_m = 0.60 Fv_m.$$ (4)
This volume, $Q_m$, is that occupied by the flow per time-unit when in state $m$, and we have assumed that $y_m = y_s$; hence the weight of flow per time-unit is
$$Q = Q_m y_m = F_m v_my_s = Fmv_y.$$ (5)
**EXAMPLE.**—In the testing of a blowing-engine it is found capable of maintaining a pressure of 18 lbs. per sq. inch in a large receiver, from whose side a blast is steadily escaping through a "thin plate" orifice (circular) having an area $F = 4$ sq. inches. The interior temperature is 90° Cent. and the outlet tension 15 lbs. per sq. in.
Required the discharge of air per second, both volume and weight. The data are: $p_a = 18$ lbs. per sq. in., $T_a = 303^\circ$ Abo. Cent., $F = 4$ sq. inches, and $p_w = 15$ lbs. per sq. in. Use ft.-lb.-sec. system.
First, the heaviness in the receiver is
$$y_s = \frac{p_a}{p_w} \cdot \frac{T_a}{T_s} \cdot y_t = \frac{18}{15} \times \frac{303}{14.7} \times .089 = .089 \text{ lbs. per cub. ft.}$$
Then, from eq. (2),
$$v_m = \sqrt{\frac{2g p_a - p_w}{y_s}} = \sqrt{\frac{9 \times 32.2(144 \times 18 - 144 \times 15)}{0.089}} = 555.3$$
(97 per cent of this would be more correct on account of friction.)
:. $Q_m = F_m v_my_s = Fmv_y \times 555.3 = 9.34$ cub. ft. per sec.
at a tension of 15 lbs. per sq. in., and of heaviness (by hypothesis) = .089 lbs. per cub. ft. Hence weight
$$G = 9.34 \times .089 = .82 \text{ lbs. per sec.}$$
FLOW OF GASES BY MARIOTTE'S LAW. 777
The theoretical power of the air-compressor or blowing-engine to maintain this steady flow can be computed as in Example 3, § 453.
551. Flow through an orifice on the Basis of Mariotte's Law ; or Isothermal Efflux. Since in reality the gas expands during flow through an orifice, and hence changes its heaviness (Fig. 619), we approximate more nearly to the truth in assuming this change of density to follow Mariotte's law, i.e., that the heaviness varies directly as the pressure, and thus imply that the force required to maintain the flow is proportional to $p$, and again integrate the terms of eq. $(B)$, but take care to note that, now, $y$ is variable (i.e., different in different lamina at the same instant), and hence express it in terms of the variable $p$ (from eq. $(2)$, § 475), thus:
$$y = (y_+ + p_+ p)$$
Therefore the term $\int_{a}^{b} \frac{dp}{y}$ of eq. $(B)$ becomes
$$\frac{p_+}{y_+} \int_{a}^{b} \frac{dp}{p} = \frac{p_+}{y_+} \log \frac{p_+}{p_+} \cdot \cdot \cdot (1)$$
and, integrating all the terms of eq. $(B)$, neglecting $h$, and calling $v_a$ zero, we have
$$\frac{v_a^*}{2g} = \frac{p_+}{y_+} \log \frac{p_+}{p_+} - \text{"efflux by Mariotte's Law through orifice"} \cdot \cdot \cdot (3)$$
As before, $y_+ = \frac{p_+}{T_+ p_+}$, and the flow of volume per time-unit at $m$ is
$$Q_m = F_m v_m; \cdot \cdot \cdot \cdot (3)$$
while if the orifice is in thin plate, $F_m$ may be put = .60 $F$, and the weight of the flow per time-unit = $G = F_m w_m$. (4)
If the mouth-piece is rounded, $F_m = F =$ area of exit orifice of mouth-piece.
![]() A diagram showing a cross-section of an orifice with a stream of gas flowing through it.
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MECHANICS OF ENGINEERING.
EXAMPLE.—Applying eq. (3) to the data of the example in § 350, where $y_a$ was found to be 0.089 lbs. per cub. ft., we have
$$e_m = \sqrt{2g\frac{p_a}{y_a}\log{\frac{p_a}{p_0}}}$$
$$= \sqrt{2 \times 32.2 \times \frac{18 \times 144}{0.089} \times 2.9025 \times \log{\frac{18}{15}}} = 584.7 \text{ ft. p. sec.}$$
:. $Q_a = F_av_a = 0.60 \times y_A \times 584.7 = 0.745$ cub. ft. per sec.
Since the heaviness at $m$ is, from Mariotte's law,
$$y_a = \frac{p_a}{p_0} y_a = \frac{18}{15} \times 0.089,$$
i.e., $y_a = -0.741$ lbs. per cub. ft.,
hence the weight of the discharge is
$$G = Q_av_a = 0.745 \times (-0.741) = 0.729 \text{ lbs. per sec},$$
or about 12 per cent less than that given by the “ water formula.” If the difference between the inner and outer tensions had been less, the discrepancy between the results of the two methods would not have been so marked.
§ 552. Adiabatic Reflux from an Orifice.—It is most logical to assume that the expansion of the gas approaching the orifice, being rapid, is adiabatic (§ 478). Hence (especially when the difference between the inner and outer tensions is considerable) it is more accurate to assume $y$ as varying according to Poisson's Law, eq. (1), § 478; i.e., $y = [y_a + p_a]^{1/p}$, in integrating eq. (B). Then the term
$$\int_{a}^{b} \frac{dp}{y} \text{ will } = \frac{p^A}{p_a} \int_{a}^{b} p^{-1} dp = \frac{p^A}{p_a} [p_b^A - p_a^A]$$
$$= -\frac{p_a}{y_a} [1 - (\frac{p_b}{p_a})^A];$$
ADIABATIC FLOW OF GASES THROUGH ORIFICES. 779
and eq. (B), neglecting $A$ as before, and with $v_a = 0$, becomes
(See Fig. 619)
$$\frac{v_a}{g} = \frac{3p_a}{y_a}\left[1 - \left(\frac{p_a}{p_0}\right)^{\frac{1}{k}}\right] \text{(Adiabatic flow; orifice).} \quad (1)$$
Having observed $p_a$ and $T_a$ in the reservoir, we compute $r_a = \frac{p_a y_a^{\frac{1}{k}}}{p_0}$ (from § 478). The gas at $m$, just leaving the orifice, having expanded adiabatically from the state $a$ to the state $m$, has cooled to a temperature $T_m$ (absolute) found thus (§ 475),
$$T_m = T_a\left(\frac{p_a}{p_0}\right)^{\frac{1}{k}}, \quad \ldots \ldots \quad (2)$$
and is of a heaviness
$$v_m = y_m\left(\frac{p_m}{p_0}\right)^{\frac{1}{k}}, \quad \ldots \ldots \quad (3)$$
and the flow per second occupies a volume (immediately on exit)
$$Q_m = F_m v_m, \quad \ldots \ldots \quad (4)$$
and weighs
$$G = F_m v_m g, \quad \ldots \ldots \quad (5)$$
EXAMPLE 1.—Let the interior conditions in the large reservoir of Fig. 619 be as follows (state $n$): $p_n = 224$ lbs. per sq. in., and $T_n = 294^\circ$ Abs. Cent. (i.e., 21st Cent.); while externally the tension is 15 lbs. per sq. inch, which may be taken as being $p_w =$ tension at $m$, the throat of jet. The opening is a circular orifice in "thin plate" and of one inch diameter.
Required the weight of the discharge per second [ft., lb., sec.; $g = 32.2$].
First, $y_n = \frac{22.5 \times 144}{15} \times .0807 = .114$ lbs. per cub. ft.
Then, from (1),
$$v_n = \sqrt{\frac{3p_n}{y_n}\left[1 - \left(\frac{p_n}{p_0}\right)^{\frac{1}{k}}\right]}$$
$$= \sqrt{\frac{9 \times 32.2 \times 3 \times 22.5 \times 144}{0.114}\left[1 - .\overline{{v}}_{\frac{k}{k}}\right]} = 844 \text{ ft. per sec.}$$
780
MECHANICS OF ENGINEERING.
Now
$$F = \frac{4\pi(1.5)^2}{6} = 0.0546 \text{ sq. ft.}$$
:. $$Q_m = CFv_m = .80 Fv_m = 0.60 \times 0.0546 \times 844 = 2.765$$
cub. ft. per sec., at a temperature of
$$T_m = 294^\circ F = 257^\circ \text{ Abs. Cent.} = -16^\circ \text{ Cent.},$$
and of a heaviness
$$y_m = 0.114 V(\frac{F}{H})^2 = 0.085 \text{ lbs. per cub. ft.},$$
so that the weight of flow per sec.
$$G = Q_my_m = 2.765 \times .085 = .235 \text{ lbs. per sec.}$$
EXAMPLE 9.--Let us treat the example already solved by the two preceding approximate methods (§§ 550 and 551) by the present more accurate equation of adiabatic flow, eq. (1).
The data were (Fig. 619):
$$p_a = 18 \text{ lbs. per sq. in.; } T_a = 803^\circ \text{ Abs. Cent.; } p_h = 15 \text{ " " " " ; and } F = 4 \text{ sq. inches}$$
(F being the area of orifice). $$y_m$$ was found = .089 lbs. per cub. ft. in § 550; hence, from eq. (1),
$$v_m = \sqrt{\frac{2 \times 32.2 \times 3 \times 18 \times 844}{0.089}[1 - V^2]} = 576.9 \text{ ft./per sec.}$$
From (4),
$$Q_m = Fv_m = .6 Fv_m = .6 \times .089 \times 576.9 = 9.603 \text{ cub. ft. per sec;}$$
and since at m it is of a heaviness
$$y_m = .089 V(\frac{F}{H})^2 = .0788 \text{ lbs. per cub. ft.,}$$
we have weight of flow per sec.
$$= G = Q_my_m = 9.603 \times .0788 = 0.756 \text{ lbs. per sec.}$$
THEORETICAL MAXIMUM FLOW OF WEIGHT OF GAS. 781
Comparing the three methods for this problem, we see that
By the "water formula," . . . $G = 0.82$ lbs. per sec.
" isothermal formula, . . . $G = 0.729$ "
" adiabatic formula, . . . $G = 0.756$"
553. Practical Notes. Theoretical Maximum Flow of Weight.
—if in the equations of § 559 we write for brevity $p_w + p_a = w$
we derive, by substitution from (1) and (3) in (5),
Weight of flow $= G Q_w Y_w = F_w \sqrt{g g_p Y_w} (1 - a_f) t$, . . . (1)
per unit of time)
This function of $x$ is of such a form as to be a maximum for
$x = (p_w + p_a) = (\frac{1}{2}) = 512$; . . . (2)
i.e., theoretically, if the state $n$ inside the reservoir remains the same, while the outside tension (considered $= p_w$ of jet, Fig. 619) is made to assume lower and lower values (so that $x = p_w + p_a$, diminishes in the same ratio), the maximum flow of weight per unit of time will occur when $p_w = 512$ $p_a$, a little more than half the inside tension. (With the more accurate value 1.41 (1.408), instead of 1.4, see § 478, we should obtain 327 instead of 512 for dry air; see § 549.)
Prof. Cortotell says ("544 of his Applied Mechanics"):
"The distinction of the theoretical divergence on limitation of the external pressure to which the limit given is an anomaly which had always been considered as requiring explanation, and M. St. Venant had already suggested that it could not actually occur. In 1866 Mr. R. D. Napier showed by experiment that the weight of steam of given pressure discharged from an orifice really is independent of the pressure of the medium into which efflux takes place *"; and in 1872 Mr. Wilson confirmed this result by experiments on the reaction of steam issuing from an orifice.
*The explanation lies in the fact that the pressure in the
When the difference between internal and external pressures is great, should be added.
![]() A diagram showing a water flow through a pipe with a valve.
789
MECHANICS OF ENGINEERING.
centre of the contracted jet is not the same as that of the surrounding medium. The jet after passing the contracted section suddenly expands, and the change of direction of the fluid particles gives rise to centrifugal forces which cause the pressures to be greater in the centre of the contracted section than at the circumference; see Fig. 690.
Prof. Cotterill then advises the assumption that $p_{a} = \frac{527}{p_{a}}$ (for air and perfect gases) as the mean tension in the jet at $m$ (Fig. 619), whereas the outside medium is at a tension less than $527 p_{a}$. He also says, "Contraction and friction must be allowed for, but these effects are so small in magnitude that the value of which, however, is more variable than that of the corresponding coefficient for an incompressible fluid. Little is certainly known on this point." See §§ 549 and 554.
For air the velocity of this maximum flow of weight is
$$\text{Val. of max. } G = \left[ \frac{997}{T_a T_t} \right] \text{ ft. per sec., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \sqrt{\frac{T_a}{T_t}} \right]$$
(3)
where $T_a$ = abs. temp. in reservoir, and $T_t$ = that of freezing point. Rankine's Applied Mechanics (p. 584) mentions experiments of Drs. Joule and Thomson, in which the circular orifices used in the test of copper tubes of diameter 0.029 in., 0.035 in., and 0.084 in., while the outside temperature was about one half of that inside. The results were 84 per cent of those demanded by theory, a discrepancy due mainly, as Rankine says, to the fact that the actual area of the orifice was used in computation instead of the contracted section; i.e., contraction was neglected.
554. Coefficients of Efflux by Experiment For Orifices and Short Pipes. Small Difference of Tension.—Since the discharge through a short pipe from a reservoir is affected not only by contraction, but by slight friction at the edges, even with a rounded entrance, the theoretical results for the volume and weight of flow per unit of time in preceding paragraphs should be multiplied both by a coefficient of velocity $\phi$ and one for contraction $C$, as in the case of water; i.e., by a coefficient of efflux $\mu = \phi C$. (Of course, when there is no
COEFFICIENTS OF EFFLUX. GAS. 783
contraction, $C = 1.00$, and then $\mu = \phi$ as with a well-rounded mouth-piece, (for instance, Fig. 541, and with short pipes.)
Hence for practical results, with orifice and short pipes, we should write for the weight of flow per unit of time
$$= G = \mu F_{\text{in}} y_s = \mu F \left(\frac{F_{\text{in}}}{F_{\text{out}}} \right) \sqrt{\frac{2g p_s p_c}{\rho g}} \left[1 - \left(\frac{y_s}{p_s}\right)^2 \right],$$
(1)
(from the equations of § 555 for adiabatic flow, as most accurate; $p_s + p_c$ may range from $4$ to $1.00$. $F =$ area of orifices, or of discharging end of mouth-piece or short pipe; $y_s =$ heaviness of air in reservoir and $= T_s p_s + T_c p_c$, eq. (18) of § 437; and $\mu =$ the experimental coefficient of efflux.
From his own experiments and those of Koch, D'Aubrisson, and others, Weibach recommends the following mean values of $\mu$ for various mouthpieces, when $p_s$ is not more than $4$ larger than $p_c$ (i.e., about $17 \%$ larger), for use in eq. (1):
1. For an orifice in a thin plate, . . . . . . . . . . $\mu = 0.56$
2. For a short cylindrical pipe (inner corners not rounded), $\mu = 0.75$
3. For a well-rounded mouth-piece (like that in Fig. 541), $\mu = 0.98$
4. For a short conical convergent pipe (angle about $6^o$), $\mu = 0.92$
EXAMPLE.—(Data from Weibach's Mechanics.) "If the sum of the areas of two conical turrets of a blowing-machine is $F = 3$ sq. inches, the temperature in the reservoir $18^o$ Cent., the height of the attached (open) mercury manometer (see Fig. 464) 29 inches, and the height of the barometer in the external air 29 inches," we have (ft., lb., sec.)
$$\frac{P_2}{P_1} = \frac{29}{29+8} = \frac{29}{37}, \quad T_s = 288^\circ \text{ Abs. Cent.;}$$
$$p_s = (\#) 14.7 \times 144 \text{ lbs. per sq. ft.;}$$
$$y_s = (\#) \times 0.0807 = 0.0816 \text{ lbs. per cub. ft.,}$$
while $$F' = \frac{T_s}{T_c} \text{ sq. ft. and (see above)} \mu = 0.92; \text{ hence}$$
$$\phi = 0.92 \times (\#) (\#) (\#) (\#) (\#) (\#) (\#) (\#) (\#) (\#) (\#) (\#) (\#) (\#).$$
784
MECHANICS OF ENGINEERING.
i.e., $G = 0.6076$ lbs. per second; which will occupy a volume
$$V_1 = G + p_1 V = G + 0.0807 \times 7.59 = 7.93 \text{ cub. ft.}$$
at one atmosphere tension and freezing-point temperature, while at a temperature of $T_2 = 98^\circ$ Abs. Cent. and tension of $p_w = \frac{H}{H} \times 1$ of one atmosphere (i.e., in the state in which it was on entering the blowing-engine) it occupied a volume
$$V = \frac{H}{H} \times 7.59 = 8.94 \text{ cub. ft.}$$
(This last is Weisbach's result, obtained by an approximate formula.)
555. Coefficients of Eflux for Orifices and Short Pipes for a Large Difference of Tension—For values $\frac{d}{D} > \frac{1}{2}$ and $< 2$, of the ratio $p_a : p_m$, of internal to external tension, Weisbach's experiments with circular orifices in thin plate, of diameters ($= d$) from 0.4 inches to 0.8 inches, gave the following results:
| $\frac{d}{D}$ |
$p_a : p_m$ |
1.00 |
1.00 |
1.40 |
1.65 |
1.90 |
2.00 |
| diam. = 0.4, $\mu$ = .95 |
$p_a : p_m$ |
.55 |
.59 |
.66 |
.73 |
.76 |
.78 |
| diam. = 0.6, $\mu$ = .90 |
$p_a : p_m$ |
.60 |
.67 |
.64 |
.68 |
.72 |
.78 |
Whence it appears that $\mu$ increases somewhat with the ratio of $p_a$ to $p_m$, and decreases slightly for increasing size of orifice.
With short cylindrical pipes, internal edges not rounded, and three times as long as wide, Weisbach obtained $\mu$ as follows:
| $\frac{d}{D}$ |
$p_a : p_m$ |
1.05 |
1.10 |
1.35 |
1.40 |
1.70 |
1.74 |
| diam. = 0.4, $\mu$ = .78 |
$p_a : p_m$ |
.72 |
.77 |
.88 |
|
|
|
| diam. = 0.6, $\mu$ = .95 |
$p_a : p_m$ |
.85 |
|
|
|
|
|
| diam. = 1.0, $\mu$ = .95 |
$p_a : p_m$ |
|
|
|
|
81 |
<
When the inner edges of the 0.4 in. pipe were slightly rounded, $\mu$ was found = .93; while a well-rounded mouth-piece of the form shown in Fig. 541 gave a value $\mu =$ from .965 to .985, for $p_a : p_m$ ranging from 1.35 to 2.00. These values of $\mu$ are use for use in eqs (1), above.
556. To find the Discharge when the Internal Pressure is measured in a Small Reservoir or Pipe, not much larger than the
**VELOCITY OF APPROACH. GAES.** 753
Orifice.—Fig. 621. If the internal pressure $p_*$, and temperature $T_*$, must be measured in a small reservoir or pipe, $n$, whose sectional area $F_*$ is not very large compared with that of the orifice, $F$, (or of the jet, $F_{\infty}$,) the velocity $v_*$ at $a$ (velocity of approach) cannot be put to zero. Hence, in applying eq. $(B)$, § 550, to the successive laminar between $a$ and $m$, and integrating, we shall have, for adiabatic steady flow,
$$\frac{v_*^2}{2g} - \frac{u_*^2}{2g} = \frac{3p_*}{y_*}\left[1-\left(\frac{p_*}{p_0}\right)^{\frac{1}{k}}\right] \dots \dots (1)$$
Instead of eq. (1) of § 552. But from the equation of continuity for steady flow of gases [eq. (a) of § 548], $F_*v_*y_* = F_{\infty}v_{\infty}y_{\infty};$
hence $v_* = \frac{F_*y_*}{F_{\infty}y_{\infty}}v_{\infty},$ while for an adiabatic change from $a$ to $m$, $\frac{y_*}{y_m} = \left(\frac{p_*}{p_m}\right);$ whence by substitution in (1), solving for $v_*$, we have
$$v_* = \sqrt{\frac{3g\cdot\frac{3p_*}{y_*}\left[1-\left(\frac{p_*}{p_0}\right)^{\frac{1}{k}}\right]}{\sqrt{1-\left(\frac{F_*}{F_{\infty}}\right)\left(\frac{p_*}{p_0}\right)^{\frac{1}{k}}}}} \dots \dots (2)$$
As before, from §§ 472 and 478,
$$y_* = \frac{p_*T_*}{P_*T_*}y_* \dots \dots (3)$$
and
$$y_m = \left(\frac{p_m}{p_0}\right)y_m \dots \dots (4)$$
Having observed $p_*$, $p_m$, and $T_*$, then, and knowing the area $F$ of the orifice, we may compute $y_*$, $y_m$, and $v_*$, and finally the
Weight of flow per time-unit = $G = \mu F_{\infty}v_{\infty}y_m,\dots\dots(5)$
786 MECHANICS OF ENGINEERING.
taking $\mu$ from § 554 or 555. In eq. (9) it must be remembered that for an orifice in "thin plate," $F_m$ is the sectional area of the contracted vein, and = $CF$, where $C$ may be put = $\frac{\mu}{\sqrt{2}}$.
**Example.**—If the diameter of $AB$, Fig. 621, is 34 inches, and that of the orifice, well rounded, = 2 in., if $p_a = 1 \frac{1}{2}$ atmospheres = 1$\frac{1}{2} \times 14.7 \times 144$ lbs. per sq. ft., while $p_m = \frac{1}{2}$ of an atom, so that $\frac{P_m}{P_a} = \frac{1}{2}$, and $T_a = 283^\circ$ Abs. Cent., required the discharge per second, using the ft., lbs., and sec.
From eq. (3),
$$y_a = \frac{1}{2} \cdot \frac{H_0}{H_0} \times 0.0807 = 0.0433 \text{ lbs. per cub. ft.;}$$
whence (eq. (4))
$$y_m = (\frac{1}{2}) y_a = 0.07544 \text{ lbs. per cub. ft.}$$
Then, from eq. (3),
$$e_m = \left[ \sqrt{64 \times 3^2 \times 15 \times 985 \times 144} - (\frac{1}{2}) H_0 \right] + \left[ \sqrt{(14)^2 \times (14)} - (\frac{1}{2}) H_0 \right]$$
$$= 558.1 \text{ ft. per sec.;}$$
$$\therefore G = 0.98 \left( \frac{1}{\sqrt{6}} \right) 558.1 \times 0.07544 = 9003 \text{ lbs. per sec.}$$
557. Transmission of Compressed Air; through very Long Level Pipes. Steady Flow.
CASE I. When the difference between the tensions in the reservoirs at the ends of the pipe is small.—Fig. 622. Under
![]() A diagram showing two reservoirs connected by a pipe with a valve in the middle.
these circumstances it is simpler to employ the form of formula that would be obtained for a liquid by applying Bernoulli's Theorem, taking into account the "loss of head" occasioned
TRANSMISSION OF COMPRESSED AIR. 787
by the friction on the sides of the pipe. Since the pipe is very long, and the change of pressure small, the mean velocity in the pipe, $v'$, seems to be nearly the same at all points along the pipe, and will not differ greatly between the velocity-heads at $n$ and $m$ will be neglected; a certain mean head $y'$ will be assigned to all the gas in the pipe, as if a liquid.
Applying Bernoulli's Theorem, with friction, § 516, to the ends of the pipe, $n$ and $m$, we have (as for a liquid)
$$\frac{p_n}{g} + \frac{p_m}{y'} + 0 = \frac{p'_n}{g} + \frac{p'_m}{y'} + 0 - 4f \frac{l}{d} \frac{v'^2}{g}. \quad (1)$$
Putting (as above mentioned) $v'_n - v'_m = 0$, we have, more simply,
$$\frac{p_n - p_m}{y'} = 4f \frac{l}{d} \frac{v'^2}{g}. \quad (9)$$
The value of $f$ as coefficient of friction for air in long pipes is found to be somewhat smaller than for water; see next paragraph.
558. Transmission of Compressed Air. Experiments in the St. Gotthard Tunnel, 1878. — [See p. 96 of Vol. 24 (Feb. '81), Van Nostrand's Engineering Magazine.] In these experiments, the temperature and pressure of the flowing gas (air) were observed at each end of a long portion of the pipe which delivered the compressed air to the boring-machines three miles distant from the tunnel's mouth. The portion considered was selected at a distance from the entrance of the tunnel, to eliminate the fluctuating influence of the weather on the temperature of the flowing air.
A steady flow being secured by proper regulation of the compressors and distributing tubes, observations were made of the pressure $(p)$ at internal compressor points as well as the external, at each end of a portion of pipe considered, and also at intermediate points; also of the weight of flow per second $Q = Q_y$, measured at the compressors under standard conditions ($0^\circ$ Cent. and one atmos. tension).
Then knowing the $p$ and $T$ at any section of the pipe, the
788
MECHANICS OF ENGINEERING.
heaviness $y$ of the air passing that section can be computed from $\frac{y}{y'} = \frac{P - P_0}{P_0}$ and the velocity $v$ of $G + F'y'$ being $\frac{L}{F'}$, being the sectional area at that point. Hence the mean velocity $v'$, and the mean heaviness $y'$, can be computed for this portion of the pipe whose diameter $d$ and length $l$. In the experiments cited it was found that at points not too near the tunnel-mouth the temperature inside the pipe was always about 3° Cent. lower than that of the tunnel. The values of $f$ in the different experiments were then computed from eq. (2) of the last paragraph; i.e.,
$$\frac{P_s - P_0}{y'} = 4f \cdot \frac{w''}{d} \cdot \frac{w''}{2g}$$
(3)
all the other quantities having been either directly observed, or computed from observed quantities.
THE ST. GOTHARD EXPERIMENTS.
[Concrete quantities reduced to English units.]
| No. |
d |
(feet) |
(ft.) |
(inches) |
Air pressure, Pa |
Pa - P_0 |
Wm sq. ft. per sec. |
mean f |
| 1 |
15093 |
|
0.4055 |
|
5.60 |
5.34 |
19.82 |
21.1 |
.0035 |
| 2 |
15092 |
|
0.4058 |
|
5.60 |
5.34 |
19.82 |
21.1 |
.0038 |
| 3 |
15092 |
|
0.4058 |
|
5.60 |
5.34 |
19.82 |
21.1 |
.0041 |
| 4 |
17113 |
|
.3765 |
|
5.34 |
5.00
<
In the article referred to (Van Nostrand's Mag.) $f$ is not computed. The writer contented himself with showing that Weisbach's values (based on experiments with small pipes and high velocities) are much too great for the pipes in use in the tunnel.
With small tubes an inch or less in diameter Weisbach found, for a velocity of about 80 ft. per second, $f=.0068$; for still higher velocities $f$ was smaller, approximately, in accordance with the relation
$$f = \frac{.0542}{v'/(in ft. per sec)}$$
TRANSMISSION OF COMPRESSED AIR.
780
On p. 370, vol. xxiv, Van Nostrand's Mag., Prof. Robinson of Ohio mentions other experiments with large long pipes.
From the St. Gothard experiments a value of $f=.004$ may be inferred for approximate results with pipes from 3 to 8 in. in diameter.
EXAMPLE.--It is required to transmit, in steady flow, a supply of $G=297^{\circ}$ lbs. of atmospheric air per second through a pipe 30000 ft. in length (nearly six miles) from a reservoir where the tension is 6.0 atmos. to another where it is 5.8 atmos., the mean temperature in the pipe being $80^{\circ}$ Fahr., $=24^{\circ}$ Cent. (i.e. = $297^{\circ}$ Abs. Cent.). Required the proper diameter of pipe; $d=$ ? The value $f=.00425$ will be used, and the ft.-lb.sec. system of units. The mean volume passing per second in the pipe is
$$Q=G+f' \cdot \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ Q=G+f' =\frac{Q}{f'}=\frac{Q}{4\pi d^2}.\quad(4)$$
The mean velocity may thus be written: $v'=\frac{Q}{f'}=\frac{Q}{4\pi d^2}$. (4)
The mean heaviness of the flowing air, computed for a mean tension of 5.9 atmospheres, is, by § 472,
$$y'=\frac{5.9\times14.7\times973}{1\times14.7\times297}\times0.807=0.431\,\mathrm{lbs.\ per\ cub.\ ft.;}$$
and hence, see eq. (8).
$$Q'=\frac{G}{y'}=\frac{6.456}{0.431}=14.74\,\mathrm{cub.\ ft;}$$
at tension of 5.9 atmos., and temperature $297^{\circ}$ Abs. Cent.
Now, from eq. (2),
$$\frac{P_a-P_m}{y'}=\frac{4f'}{2g}\cdot\frac{l}{d}\left[\frac{Q'}{4\pi d^2}\right];$$
whence
$$d'=\frac{4f'}{(4\pi)^3}\cdot\frac{y'}{(P_a-P_m)}\cdot\frac{Q'}{2g};\quad\quad(5)$$
790
MECHANICS OF ENGINEERING.
and hence, numerically,
$$d = 4 \times 0.0485 \times 0.431 \times 30000 \times (14.7)^2 / (7.854)^2 (14.7 \times 144(6.0 - 5.80)) / 38.2 = 1.29 \text{ feet}.$$
559. (Case II of § 557) Long Pipe, with Considerable Difference of Pressure at Extremities of the Pipe. Flow Steady.--Fig. 693. If the difference between the end-tensions is comparatively great, we can no longer deal with the whole of the air in the pipe at once, as regards ascribing to it a mean velocity and mean tension, but must consider the separate laminae, such as $AB$ (a short length of the air-stream) to which we may apply eq. (3) of § 557; $A$ and $B$ corresponding to the $n$ and $m$ of Fig. 622. Since $p_1$, $p_2$, $F_1$, $F_2$, and $v$ of § 557 correspond to the $-dy$, $dx$, $y$, and $v$ of the present case (short section or lamina), we may write
$$\frac{dp}{y} = 4f \frac{\partial^2}{\partial y^2} ds.$$ .......................... (1)
But if $G =$ weight of flow per unit of time, we have at any section, $F_{yy} = G$ (equation of continuity); i.e., $v = G + F_y,$ whence by substitution in eq. (1) we have
$$\frac{dp}{y} = \frac{4f}{2g} Gds;$$ i.e., $-yd\!p = \frac{4fG}{2gF_{yy}} ds.$$ .......................... (2)
Eq. (3) contains three variables, $y$, $p$, and $s$ (= distance of lamina from $n$). As to the dependence of the behaviour of $y$ on the tension $p$ in different laminar experiments shows that in most cases a uniform temperature is found to exist all along the pipe, if properly buried, or shaded from the sun; the loss of heat by adiabatic expansion being in great part made up by the heat generated by the friction against the walls of the
GAS IN LONG PIPES. LARGE FALL OF TENSION. 791
pipe. This is due to the small loss of tension per unit of length of pipe as compared with that occurring in a short discharge pipe or nozzle. Hence we may treat the flow as tæothermal, and write $p + y = p_w + y_w$ (§ 475, Mariotte's Law).
Hence $y = \frac{y_w}{p_w}$, which substituted into eq. (2) enables us to write:
$$- pdp = \left[ \frac{A_f G' p_x}{2g F^2} \right] da \dots \dots \dots (3)'$$
$$\therefore - \int_0^l pdp = \left[ \frac{A_f G' p_x}{2g F^2} \right] \int_0^l da \dots \dots (3)$$
Performing the integration, noting that at $n' p = p_w$, $s = 0$, and at $m' p = p_w'$ and $s = l$, we have
$$\frac{1}{2} [p_w' - p_w'] = \frac{4f l}{2gd} \cdot \frac{A_f G' p_x}{F^2} \cdot y_w' \dots \dots (4)$$
It is here assumed that the tension at the entrance of the pipe is practically equal to that in the head reservoir, and that at the end ($m'$) to that of the receiving reservoir; which is not strictly true, especially when the corners are not rounded. It will be remembered also that in establishing eq. (2) of § 557 (the basis of this problem), the effect of friction on the gas was neglected; i.e., the change of velocity in passing along the pipe. Hence eq. (4) should not be applied to cases where the pipe is so short, or the difference of end-tensions so great, as to create a considerable difference between the velocities at the two ends of the pipe. (See Addendum on p. 791.)
EXAMPLE.—A well or reservoir supplies natural gas at a tension of $p_w = 30$ lbs. per sq. inch. Its heaviness at 0° Cent. and one atmosphere tension is 348 lbs. per cu. foot. In piping this gas through a long horizontal line, it is desired, a single four-inch pipe is to be employed, and the tension in the receiving reservoir (by proper regulation of the gas distributed from it) is to be kept equal to 16 lbs. per sq. in. (which would sustain a column of water about 3 ft. in height in an open water manometer, Fig. 465).
792
MECHANICS OF ENGINEERING.
The mean temperature in the pipe being 17° Cent., required the amount (weight) of gas delivered per second, supposing leakage to be prevented (formerly a difficult matter in practice). Solve (4) for $G$, and we have
$$G = \frac{4\pi d^2}{4T} \sqrt{\frac{p_0^2}{4T} + \frac{y_{ac}(p_a - p_w)}{p_w}}.$$ (5)
First, from § 472, with $T_w = T_a = 290^\circ$ Abs. Cent., we compute
$$P_w = P_a \cdot \frac{T_a}{T_w} = \frac{14.7 \times 144}{0.048} \cdot \frac{390}{278} = 6454 \text{ feet}.$$ Hence with $f = .005$,
$$G = \frac{4\pi(d^2)^2}{4T} \sqrt{\frac{32.9 \times 144(30 \times 144)^2 - (16 \times 144)^2}{4 \times 0.05 \times 10560 \times 4645}}$$
= 0.337 lbs. per sec.
(For compressed atmospheric air, under like conditions, we would have $G = 0.430$ lbs. per second.)
Of course the proper choice of the coefficient $f$ has an important influence on the result.
From the above result ($G = 0.337$ lbs. per second) we can compute the volume occupied by this quantity of gas in the receiving reservoir, using the relation $Q_w = \frac{G}{y_w}$.
The heaviness $y_w$ of the gas in the receiving reservoir is most easily found from the relation $\frac{P_w}{y_w} = \frac{F_w}{y_a}$, which holds good since the flow is isothermal. I.e., $\frac{P_w}{y_w} = 6454$ ft.; whence $y_w = 0.049$ lbs. per cubic foot, $p_w$ being $16 \times 144$ lbs. per sq. ft.
Hence $$Q_w = \frac{G}{y_w} = \frac{0.337}{0.049} = 6.794 \text{ cub. ft. per sec.}$$
FLOW OF GAS IN PIPES.
733
It should be said that the pressure at the up-stream end of the pipe depends upon the rate of flow allowed to take place.
With no flow permitted, the pressure in the tube of a gas-well has in some cases reached the high figure of 500 or 600 lbs. per eq. in.
560. Rate of Decrease of Pressure along a Long Pipe.—Con-
sidering further the case of the last paragraph, that of a straight, long, level pipe of uniform diameter, delivering gas from a storage reservoir into a receiving reservoir, we note that if in eq. (4) we retain $p_w$ to indicate the tension at the receiving reservoir, but let $p_w'$ denote in turn the tension at points in the pipe successively further and further (a distance $x$ from the receiving reservoir), we may write eq. (5) for $x$ and obtain the equation (between two variables $p_w$ and $x$)
$$p_w - p_w' = \text{Const.} \times x.$$ (6)
This can be used to bring out an interesting relation men-
tioned by a writer in the Engineering News of July 1887
(p. 71), viz., the fact that in the parts of the pipe more distant from the receiving end, $m$, the distance along the pipe in which a given loss of pressure occurs is much greater than near the receiving end.
To make a numerical illustration, let us suppose that the pipe is of such size, in connection with other circumstances, that the tension $p_w$ at $A$, a distance $x$ six miles from $m$, is two atmospheres; that is, that the receiving reservoir being one atmosphere; that is, that the loss of tension between $A$ and $m$ is one atmosphere. If we express tensions in atmospheres and distances in miles, we have for the value of the constant in eq. (6), for this case,
$$\text{Const.} = (4 - 1) + 6 = 9; \quad (\text{for assumed units}) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (7)$$
Now let $p_w' = \text{the tension at } B$, a point 18 miles from $m$, and we have, from eqs. (6) and (7), the tension at $B$ = 3.16 atmospheres.
Proceeding in this manner, the following set of values is obtained:
| $x$ |
$p_w - p_w'$ |
| 0 |
2 |
| 6 |
2 |
| 12 |
2 |
| 18 |
2 |
![]() A table showing values of \( p_w - p_w' \) for different values of \( x \). The table shows:
- For \( x = 0 \), \( p_w - p_w' = 2 \).
- For \( x = 6 \), \( p_w - p_w' = 2 \).
- For \( x = 12 \), \( p_w - p_w' = 2 \).
- For \( x = 18 \), \( p_w - p_w' = 2 \).
794
MECHANICS OF ENGINEERING.
| Point |
Total distance from m. |
Distance between two points |
Tension at point |
Loss of tension in mm. |
| P |
128 miles |
96 miles |
8.00 atm. |
1.25 atm. |
| R |
96 |
30 |
6.75 |
1.30 |
| D |
64 |
30 |
6.75 |
1.30 |
| C |
32 |
18 |
4.35 |
1.19 |
| A |
16 |
18 |
4.35 |
1.19 |
| B |
8 |
6 |
2.00 |
1.00 |
| m' |
0" |
|
|
|
If the distances and tensions in the second and fourth columns be plotted as abscisses and ordinates of a curve, the latter is a parabola with its axis following the axis of the pipe: its vertex is not at $m'$, however.
561. Long Pipe of Variable Diameter.—Another way of stating the fact mentioned in the last paragraph is as follows : At the up-stream end of the pipe of uniform diameter the gas is of much greater density than at the other extremity (the heaviness is directly as the tension, the temperature being assumed the same throughout the pipe), and the velocity of its motion is smaller than at the discharging end (in the same ratio). It is true that the frictional resistance per unit of length of pipe varies directly as the heaviness [eq. (1), § 610], but also true that it varies as the square of the velocity; so that, for instance, if the pressure at a point A is double that at B in the pipe of uniform diameter, it will be seen that the heaviness at A is double and half, respectively, those at B, and thus the gas at A is subjected to only half the frictional resisting force per foot of length as compared with that at B. Hence the relatively small diminution, per unit of length, in the tension at the up-stream end in the example of the last paragraph.
In the pipe of uniform diameter, as we have seen, the greater part of the length is subjected to a comparatively high heavi- ness, and it is thus under a greater liability to lose by leakage than if it consisted of several sections of unequal diameter. The total "hoop-tension" ($\frac{4}{3}\pi$) in a unit length of pipe, also, is proportional to the gas tension, and thinner walls might be employed for the down-stream portions of the pipe if the gas
FLOW OF GAS IN PIPES.
705
tension in those portions could be made smaller than as shown in the preceding example.
To secure a more rapid fall of pressure at the up-stream end of the pipe, and at the same time provide for the same delivery of gas as with a pipe of uniform diameter throughout, a pipe of variable diameter may be employed, that diameter being considerably smaller at the inlet than that of the uniform pipe but progressively enlarging down-stream. This will require the diameters to be greater at the inlet than in the uniform pipe, and if the same thickness of metal were necessary throughout, there would be no saving of metal, but rather the reverse, as will be seen ; but the diminished thickness made practicable in those parts from a less total hoop tension than in the corresponding parts of the uniform pipe more than compensates for the extra metal due to increased circumference, aside from the diminished liability to leakage, which is of equal importance.
A simple numerical example will illustrate the foregoing. The pipe being circular we may replace $F$ by $\frac{1}{4\pi d^2}$ in equation (4), and finally derive, $G$ being given,
$$d = \text{Const} \times \left[\frac{1}{P_2 - P_1}\right]^{1/4} = C \cdot \left[\frac{1}{P_2' - P_1'}\right]^{1/4}. \quad (8)$$
Let $A$ be the head reservoir, and $m'$ the receiving reservoir, and $B$ a point half-way between. At $A$ the tension is 10 atmospheres; at $m'$, 2 atmospheres. For transmitting a given weight of gas per second, through a pipe of constant diameter throughout, that diameter must be (tensions in atmospheres) $\frac{9}{2}$ (being the length), by eq. (8),
$$d = CI_A \left[\frac{9}{100 - \frac{9}{2}}\right]^{1/4} = CI_A(0.028)^{1/4} = 0.48 CI_A. \quad (8).$$
If we substitute for the pipe mentioned, another having a constant diameter $d$, from $A$ to $B$, where we wish the tension to be 5 atmospheres, and a different constant diameter $d'$, from $B$ to $m'$, we derive similarly,
$$d' = CI_A \left[\frac{1}{100 - \frac{25}{2}}\right]^{1/4} = 0.42 CI_A.$$
706
MECHANICS OF ENGINEERING.
and
$$d_s = C \left[ \frac{1}{25 - 4} \right]^{1/3} = 0.54 \, C D$$
It is now to be noted that the sum of $d_s$ and $d_t$ is slightly greater than the double of $d_s$; so that if the same thickness of metal were used in both designs the compound pipe would require a little more material than the uniform pipe; but, from the reasoning given at the beginning of this paragraph, that thing will not be much considerably less in the downstream part of the compound pipe than that which is saved.
In case of a cessation of the flow, the gas tension in the whole pipe might rise to an equality with that of the head-reservoir were it not for the insertion, at intervals, of automatic regulators, each of which prevents the decrease of tension on its down-stream side below a fixed value. To provide for changes of length due to rise and fall of temperature, the pipe is laid with slight undulations.
It is also a time-saving conclusion that a given pipe of variable diameter connecting two reservoirs of gas at specified pressures will deliver the same weight of gas as before, if turned end for end. This follows from equation (3)'s § 559. With $d$ variable, (3) becomes (with $F = 1 + d^2$)
$$\int_{-pdp}^{pdp} G^* C^* \int_{d^2}^{pdp} \frac{ds}{d^2}; \quad i.e., \quad G^* = \frac{pd^2 - pdp}{2C^* \int_{d^2}^{pdp} \frac{ds}{d^2}}.$$ (9)
($G^*$ is a constant.)
But $\int_{-pdp}^{pdp} \frac{ds}{d^2}$ is evidently the same in value if the pipe be turned end for end. In commenting on this circumstance, we should remember (see § 559) that the loss of pressure along the pipe is ascribed entirely to frictional resistance, and in no degree to changes of velocity (inertia).
On p. 73 of the *Engineering* for July 1887 are given the full particulars of a design of a compound pipe in actual use, and delivering natural gas. The pressure in the head-reservoir is 310 lbs. per sq. in.; that in the receiving reservoir, 65. For 2.84 miles from the head-reservoir the diameter of the pipe is
NATURAL GAS. COEFFICIENT OF FLUID FRICTION. 797
8 in.; throughout the next 2.75 miles, 10 in.; while in the remaining 3.84 miles the diameter is 18 in. At the two points of junction the pressures are stated to be 185 and 132 lbs. per sq. in., respectively, during the flow of gas under the conditions mentioned.
561a. Values of the Coefficient of Fluid Friction for Natural Gas.—In the Ohio Report on Economic Geology for 1888 may be found an article by Prof. S. W. Robinson of the University of that State describing a series of interesting experiments made by him on the flow of natural gas from orifices and through pipes. By the insertion of Pitot tubes approximate measurements were made of the velocity of the stream of gas in a pipe. The following are some of the results of these experiments, $p_1 - p_0$, representing the loss of pressure (in lbs. per sq. inch) per mile of pipe-length, and $f$ the coefficient of fluid friction, in experiments with a six-inch pipe:
| $p_1 - p_0$ |
1.00 |
1.50 |
2.00 |
2.50 |
3.00 |
3.50 |
| $f$ |
0.0025 |
0.0037 |
0.0050 |
0.0059 |
0.0070 |
0.0086 |
In the flow under observation Prof. Robinson concluded that $f$ could be taken as approximately proportional to the fourth root of the cube of the velocity of flow; though calling attention to the fact that very reliable results could hardly be expected under the circumstances.
561b. (Addendum.) Theothermal Flow in a Level Pipe, with Consideration of Inertia.—In eq. (1) of p. 697 neglect $d\sigma$, put $\omega + F = 4 + d$, and divide through by $\omega^2$. In the second term put $G^2 + F^2y^2$ for $\omega^2$ and then $p(y_n + p_n)$ for $y$. We now find the variables separated, and an integration for steady flow obtain (after putting $v_n + v_n = Fy_n + Fy_n = p_n + p_n$),
$$\log(\frac{p_n}{p_m}) - \frac{g(Fx_n - px_n)}{2} \cdot \frac{F^2y_n}{G^2} = \frac{3pl}{D}$$
[Notation as in § 837 with $O = Fr.$.]
![]() A diagram showing a level pipe with a valve at one end and a pressure gauge at the other end.
CHAPTER IX.
IMPULSE AND RESISTANCE OF FLUIDS.
502. The so-called "Reaction" of a Jet of Water flowing from a Vessel.--In Fig. 634, if a frictionless but water-tight plug $B$ be inserted in an orifice in the vertical side of a vessel mounted on wheels, the resultant action of the water on the rigid vessel (as a whole) consists of its weight $G$, and a force $P' = Fhy$ (in which $F$ is the area of orifice) which is exerted by the hydrostatic pressures on the vessel wall toward the right ($\parallel$ to paper) over those toward the left, since the pressure $P' = Fhy$, exerted on the plug is felt by the post $C$, and not by the vessel. Hence the post $D$ receives a pressure
$$P' = Fhy.$$ .......................... (1)
Let the plug $B$ be removed. A steady flow is then set up through the orifice, and now the pressure against the post $D$ is $2Fhy$ (as will be shown in the next paragraph); it is not only this pressure $Fhy$ lacking at the orifice, but the sum of all the horizontal components ($\parallel$ to paper) of the pressures of the liquid filaments against the vessel wall around the orifice is less than its value before the flow began, by an amount $Fhy$. A resistance $R = 2Fhy$ being provided, and the post removed, a slow uniform motion may be maintained toward the right, the working force being $2Fhy = P'$.
![]() Fig. 634.
**"REACTION" OF A JET.** 799
(see Fig. 635 ; $R$ is not shown). If an insufficient resistance be furnished before removing the post $D$, the vessel will begin to move toward the right with an acceleration, which will disturb the surface of the water and change the value of the horizontal force.
This force $P' = 2F\gamma h \dots (2)$
is called the "reaction" of the water-jet;
$\gamma$ is the heaviness of the liquid (q. p.).
Of course, when the vessel moves, the water level sinks and the " reaction" diminishes accordingly. Looked upon as a motor, the vessel may be considered to be a piston-less and valve-less water-pressure engine, carrying its own reservoir with it.
In Case II of § 500 we have already had a treatment of the "Reaction-wheel" or " Barker's mill," which is a practical machine operating on this principle, and will be again considered in "Notes on Hydraulic Motors."
583. " Reaction" of a Liquid Jet on the Vessel from which it issues.—Instead of showing that the pressures on the vessel close to the orifice are less than they were when there was no flow by an amount $F\gamma h$ (a rather lengthy demonstration), another result will be given, of greater simplicity but somewhat fanciful.
If a man standing on the rear platform of a car is to take up in succession, from a basket on the car, a number of balls of equal mass $M$, and project each one in turn horizontally backward with an acceleration $p$, can accomplish this only by exerting against each ball a pressure $= M p$, and in the opposite direction against the car an equal pressure $= M p$. If this action is kept up continuously the car is subjected to a constant and continuous forward force $= F \gamma h$.
Similarly, the backward propulsion of the jet of water in the case of the vessel at rest must occasion a forward force against the vessel of a value dependent on the fact that in each small interval of time $\Delta t$ a small mass $dM$ of liquid has its velocity changed from zero to a backward velocity of $v = V^2 \frac{dM}{h}$; that
![]() A diagram showing a water jet exiting from a pipe into a container. The container is labeled with arrows indicating movement to the right.
800
MECHANICS OF ENGINEERING.
is, has been projected with a mean acceleration of $p = \frac{v - 0}{dt}$,
so that the forward force against the vessel is
$P'' = \text{mass} \times \text{acc.} = \frac{dM.v}{dt}$ . . . . (3)
If $Q$ is the volume of water discharged per unit time, then
$dM = \frac{Qy}{g} dt$, and since also $Q = Fv = F\sqrt{2gh}$, eq. (3) becomes
"Reaction" of jet $= P'' = 2Fh.y$. . . . (4)
(A similar proof, resulting in the same value for $P''$, is easily made if the vessel has a uniform motion with water surface horizontal.)
If the orifice is in "thin plate," we understand by $F$ the area of the contracted section. Practically, we have $v = \phi\sqrt{2gh}$ (§ 495), and hence (3) reduces to
$P'' = 2\phi^2Fhy$. . . . . . . . . . . . . (5)
Weissbach mentions the experiments of Mr. Peter Ewart of Manchester, England, as giving the result $P'' = 1.73Fhy$ with a well-rounded orifice as in Fig. 625. He also found $\phi = .94$ for the same orifice, so that by eq. (4) we should have
$P'' = 9(.94)^2Fhy = 1.71Fhy$.
With an orifice in thin plate Mr. Ewart found $P'' = 1.14Fhy$. As for a result from eq. (4), we must put, for $F$, the area of the contracted section $64F$ (§ 495), which, with $\phi = .96$, gives
$P'' = 9(.96)^2 \cdot 64Fhy = 1.18Fhy$. . . . (6)
Evidently both results agree well with experiment.
Experiments made by Prof. J. B. Webb at the Stevens Institute (see Journal of the Franklin Inst., Jan., '88, p. 35) also confirm the foregoing results. In these experiments the vessel was suspended on springe and the jet directly downwards, so that the "reaction" consisted of a diminution of the tension of the springe during the flow.
504 Impulse of a Jet of Water on a Fixed Curved Vane (with Borders).—The jet passes tangentially upon the vane. Fig.
IMPULSE OF JET. 801
696. $B$ is the stationary nozzle from which a jet of water of cross-section $F$ (area) and velocity = $c$ impinges tangentially upon the vane, which has plane borders, parallel to paper, to prevent the lateral escape of the jet.
The curve of the vane is not circular necessarily. The vane being smooth, the velocity of the water in its curved path remains $= c$ at all points along the curve. Conceive the curve divided into a great number of small lengths each $= dc$, and subtending some angle $\alpha = d\phi$ from its own centre of curvature, its radius of curvature being $r$ (different for different $dc$), which makes some angle $\phi$ with the axis $\Gamma$ ($\gamma$ to original straight jet $BA$). At any instant of time there is an arc of water $AD$ in contact with the vane, exerting pressure upon it. The pressure $dP$ of any $dc$ of the vane against the small mass of water $Fd\phi$, $\gamma + g$ then in contact with it is the "deviating" or "centripetal" force accountable for its motion in a curve of radius $r$, and hence must have a value
$$dP = \frac{Fdyds}{g} \cdot \frac{e^{\prime}}{r} \cdot \ldots \ldots (76) \ldots (1)$$
The opposite and equal of this force is the $dP$ shown in Fig. 614, and is the impulse or pressure of this small mass against the vane. Its $X$-component is $dX = dP \cdot \sin \phi$. By making $\phi$ vary from 0 to $\alpha$, and adding up the corresponding values of $dX$, we obtain the sum of the $X$-components of the small pressures exerted simultaneously against the vane by the arc of water then in contact with it; i.e., noting that $de = rd\phi$
$$\therefore \int_{0}^{\alpha} dX = \int_{0}^{\alpha} dP \cdot \sin \phi = \frac{Fyds}{g} \int_{0}^{\alpha} \sin \phi \cdot r ds$$
$$= \frac{Fyds}{g} \int_{0}^{\alpha} [\sin \phi]_0^{\alpha} ds = \frac{Fyc'ds}{g} - \cos \varphi;$$
809
MECHANICS OF ENGINEERING.
hence the $X$-impulse $\{ = \frac{Fy'}{g} [1-\cos\alpha] = Qyc \frac{[1-\cos\alpha]}{g}, (3) \}$
in which $Q = F_0 = \text{volume of water which passes through the nozzle (and also = that passing over the vane, in this case)}$ per unit of time, and $\alpha = \text{angle between the direction of the stream leaving the vane (i.e., at } D\text{) and its original direction (BA of the jet); i.e., } \alpha = \text{total angle of deviation. Similarly, the sum of the } Y\text{-components of the } dP'\text{'s of Fig. 636 may be shown to be}\$
$Y\text{-impulse on fixed vane} = \int_0^{\infty} u'P' \cdot \cos\phi = \frac{Qyc}{g} \sin\alpha - (3')$
Hence the resultant impulse on the vane is a force
$P'' = \sqrt{X'^2 + Y'^2} = \frac{Qyc}{g} \sqrt{2(1-\cos\alpha)}, . . . (3)$
and makes such an angle $\alpha'$, Fig. 627, with the direction BA, that
$\tan\alpha' = \frac{Y'}{X'} = \frac{\sin\alpha}{1-\cos\alpha}. . . . . . . . . . . . . . . . . . . . . . . (4)$
For example, if $\alpha = 90^\circ$, then $\alpha' = 45^\circ$; while if $\alpha = 180^\circ$, Fig. 638, we have $\alpha' = 0^\circ$; i.e., $P''$ is parallel to the jet BA, and its value is
$P'' = 9Q'y'\frac{c}{g}.$
IMPULSE OF JET ON VANE.
803
585. Impulse of a Jet on a Fixed Solid of Revolution whose Axis is Parallel to the Jet—If the curved vane, with borders, of the preceding paragraph be replaced by a solid of revolution, Fig. 629, with its axis in line with the jet, then the pressure of the jet upon it will simply be the sum of the $X$-components (i.e., = to $BA$) of the pressures on all elements of the surface at a given instant; i.e.,
$$X = P^{\prime} = QY \frac{c}{g}(1 - \cos\alpha); \quad \ldots \quad (5)$$
while the components $\gamma$ to $X$, all directed toward the axis of the solid, neutralize each other. For a flat plate, then, Fig. 630, at right angles to the jet, we have for the force, or "impulse" (with $\alpha = 90^\circ$),
$$P^{\prime\prime} = QY \frac{c}{g} + FZ \gamma = 2F \frac{c}{g} \gamma. \quad (6)$$
The experiments of Bidone, made in 1838, confirm the truth of eq. (6) quite closely, as do also those of two students of the University of Pennsylvania at Philadelphia (see Jour. of the Frank. Inst. for Oct. '87, p. 258).
Eq.(6) is applicable to the theory of Pitot's Tube (see § 539), Fig. 631, if we consider the edge of the tube plane as quite wide. The water in the tube at $A$ and its section at $A$ (of area $= F$) may be treated as a flat vertical plate receiving not only the hydrostatic pressure $FZ$, but to the depth $y$ below the surface, but a continuous impulse $P^{\prime\prime} = FZ\gamma + g$ [see eq. (6)]
![]() Fig. 629.
![]() Fig. 630.
![]() Fig. 631.
804
MECHANICS OF ENGINEERING.
For the equilibrium of the end $A$, of the stationary column $AD$, we must have, therefore,
$$F_{ay} + \frac{F_0y}{g} = F_{ay} + FA_y; \text{ i.e., } A' = (3.0) \frac{c'}{2g}. . . (7')$$
The relation in equation (7) corresponds reasonably well with the results of Weisbach's experiments with the instrument mentioned in § 539. Piot himself, on trial of an instrument in which the edges of the tube at $A$ were made flaring or conically divergent, like a funnel, found
$$h' = (1.5) \frac{c'}{2g}; . . . . . . . . . . . . . . . (7)'$$
while Darcy, desires that the end of the tube should occasion as little disturbance as possible in the surrounding stream, made the extremity small and conically convergent.* The latter obtained the relation
$$h' = \text{ almost exactly } (1.0) \frac{c'}{2g}. . . . . . . . . (7)'$$
(See § 539.)
If the solid of revolution is made cup-shaped as in Fig. 632, we have (as in Fig. 629) $\alpha = 180^\circ$, and therefore, from eq. (5),
$$P'' = 2Q_0\frac{c}{g} = \frac{2F_0y}{g} = 4F(\frac{c'}{2g})y; . . . . . . (8)$$
![]() Fig. 632.
EXAMPLE.—Fig. 632. If $c = 30$ ft. per sec. and the jet (cylindrical) has a diameter of 1 inch, the liquid being water, so that $y = 62.5$ lbs. per cub. ft., we have $(ft., lb., sec.)$
the impulse (force) $P'' = \frac{2 \times 1^{\prime}\frac{1}{12} \times 900 \times 62.5}{32.2} = 19.05$ lbs.
Experiment would probably show a smaller result.
* See p. 888 for Mr. Freeman's Experiments.
IMPULSE OF JET ON MOVING VANE. 805
566. Impulse of a Liquid Jet upon a Moving Vane having Lateral Borders and Moving in the Direction of the Jet.—Fig. 633. The vane has a motion of translation (§ 108) in the same direction as the jet. Call this the axis $X$. It is moving with a velocity $v$ away from the jet (or, if toward the jet, $v$ is negative). We consider $v$ to be its acceleration being prevented by a proper resistance (such as a weight = $G$) to balance the $X$-components of the are-presures. Before coming in contact with the vane, which it does tangentially (to avoid rotation), the absolute velocity ($\S$ 83) of the water in the jet = $c$, while its velocity relatively to the vane at $A$ is $c - v$; and it will now be proved that the relative velocity along the vane is constant. See Fig. 634. Let $u$ be the velocity of the vane (of each point of it, since its motion is one of translation), and $w$ be the velocity of a water particle (or small mass of water of length $=da$) relatively to the point on the vane at which it is situated. Then, the absolute velocity of the small mass, is the diagonal formed on $u$ and $v$. Neglecting friction, the only actual force acting on the mass is $P$, the pressure of the vane against it, and this is normal to the curve. Now an imaginary system of forces, equivalent to this actual system of one force $P$, i.e., capable of producing the same motion in the mass, may be conceived of, consisting of the individual forces which would produce, separately, the separate motions of which the actual motion of this small mass $M$ is compounded. These components must satisfy:
1. A horizontal uniform motion of constant velocity = $v$; and
2. A motion in the arc of a circle of radius = $r$ and with a
![]() Fig. 633.
633
806
MECHANICS OF ENGINEERING.
velocity = u, which we shall consider variable until proved otherwise.
Motion 1 is of such a nature as to call for no force (by New-
ton's first law of motion), while motion 2 could be maintained by a system of two forces, one normal, $P_n = \frac{Mdu}{r}$, and the other tangential, $P_t = M\frac{du}{dt}$ [see eq. (5), p. 76]. This imaginary system of forces is shown at (IL), Fig. 634, and is equiv-
alent to the actual system at (L). Therefore $\mathbb{Z}$(tang. compo.) in (L) should be equal to $\mathbb{Z}$(tang. compo.) in (IL); whence we have
$$P_t = 0; \text{ i.e., } M\frac{du}{dt} = 0; \text{ or } \frac{du}{dt} = 0; . . .$$
(1)
i.e., $u$ is constant along the vane and is equal to $c - v$ at every point. (The weight of the mass has been neglected since the height of the vane is small.) In Fig. 634 the symbol $w_o$ has been used instead of $c$, and the point 0 corresponds to A in Fig. 633.
[N.B.] If the motion of the vane were rotary, about an axis to $A.P$ (or to $c$), this relative velocity would be different at different points. See Notes on Hydraulic Motors. If the radius of motion of the point $A$, however, is quite large compared with the projection of $AD$ upon this radius, the relative velocity is approximately $= c - v$ at all parts of the vane, and will be taken $= c - v$ in treating the "Hardy gurdy" in § 567.]
![]() A diagram showing two forces acting on a body: one normal and one tangential.
WORK OF JET ON VANE. S07
By putting $X$ (normal component) of $(L) = X$ (normal com-
ponent) in (II) we have
$$P = P_x; \quad i.e., \quad P = M\frac{u^2}{r} = \frac{M(c-v)^2}{r}; \quad . . . \quad (3)$$
so that to find the sum of the $X$-components of the pressures
exerted against the vane simultaneously by all the small masses
of water in contact with it at any instant, the analysis differs
from that in § 564 only in replacing the $c$ of that article by
the $(c-v)$ of this. Therefore
$$X(X_{pressure}) = P_x = \frac{F_y}{g}(c-v)^2[1-\cos a], \quad . . . \quad (3)$$
(where $a$ is the angle of total deviation, relatively to vane, of
the stream leaving the vane, from its original direction), and
is seen to be proportional to the square of the relative velocity.
$F$ is the sectional area of jet, and $\gamma$ the heaviness ($\beta$) of
the liquid. The $Y$-component (or $P_y$) of the resultant impulse
is counteracted by the support $EF$, Fig. 638. Hence, for a
uniform motion to be maintained, with a given velocity $v$, the weight $G$ must be made $= P_x$ of eq. (3). (We here
neglect friction and suppose the jet to preserve a practically
horizontal direction for an indefinite distance before meeting
the vane. If this uniform motion is to be toward the jet, $v$
will be negative; if away from it, $v$ will be positive.) For a
positive $v$ of same numerical value,
As to the doing of work [§ 138, etc.], or exchange of
energy, between the two bodies, jet and vane, during a uni-
form motion away from the jet, $P_x$ exerts a power of
$$L = P_xv = \frac{F_y}{g}(c-v)^2[1-\cos a], \quad . . . \quad (4)$$
in which $L$ denotes the number of units of work done per unit
of time by $P_x$; i.e., the power ($\beta$ 130) exerted by $P_x$.
If $v$ is negative, call it $-v$, and we have the
Power expended } \\ {by vane upon jet } &= P_xv = \frac{F_y}{g}(c+v)^2[1-\cos a]. \quad . . . \quad (5)\\
808
MECHANICS OF ENGINEERING.
Of course, practically, we are more concerned with eq. (4) than with (5). The power $L$ in (4) is a maximum for $v = 4c$; but in practice, since a single moving vane or float cannot utilize the water of the jet as fast as it flows from the nozzle, let us conceive of a succession of vanes coming into position consecutively, so that each one has at first the same velocity $v$; then the portion of jet intercepted between two vanes is at liberty to finish its work on the front vane, while additional work is being done on the hinder one; i.e., the water will be utilized as fast as it issues from the nozzle.
With such a series of vanes, then, we may put $Q' = P_c$, the volume of flow per unit of time from the nozzle, in place of $P(c - v)$ = the volume of flow per unit of time over the vane, in eq. (4); whence
$$\text{Power exerted on } \frac{Q'}{g} = L' = \frac{Q'}{g}(1 - \cos \alpha)(c - v)v. \quad (6)$$
Making $v$ variable, and putting $dL' + dv = 0$, whence $c - 2v = 0$, we find that for $v = 4c$, $L'$, the power, is a maximum. Assuming different values for $\alpha$, we find that for $\alpha = 180^\circ$, i.e., by the use of a semicircular vane, or of a hemispherical cup, Fig. 83a, with a point in mind that $1 - \cos \alpha$ is a max. when $\alpha = 90^\circ$; we have, as the maximum power,
$$L_{\text{max}}' = \frac{Q'}{g}\left(\frac{c}{2}\right)^2M'\frac{c}{2}, \quad v = 180^\circ, \quad c = 4c. \quad (7)$$
in which $M'$ denotes the mass of the flow per unit of time from the stationary nozzle. Now $\frac{M'c^2}{2}$ is the entire kinetic energy furnished per unit of time by the jet; hence the max. of Fig. 83a (series of equiangular theoretical efficiency of unity), utilizing all the kinetic energy of the water. If this is true, the absolute velocity of the particles of liquid where they leave the cup, or vane, should be zero, which is seen to be true,
![]() A diagram showing a jet flowing through a series of vanes.
Fig. 83a
POWER OF JETS.
809
as follows: At $H$, or $H'$, the velocity of the particles relatively to the vase is = $c - v$ = what it was at $A$, and hence is $= c - \frac{c}{2} = \frac{c}{2}$, hence at $H$ the absolute velocity is $v = \frac{c}{2}$ (rel. vel. toward left) - (vel. $\frac{c}{2}$ toward right) = 0;
Q.E.D. For $v > 0$ or $< \frac{c}{2}$ this efficiency will not be attained.
567. The California "Hardy-gurdy," or Pelton Wheel.—The efficiency of unity in the series of cases just mentioned is in practice reduced to 90 or 93 per cent from friction and lateral escape of water. The Pelton wheel or Cali-
fornia "Hardy-gurdy," shown in principle only, Fig. 634, is designed to utilize the mechanical relation just presented, and yields results confirming the above theory, viz., that with the linear velocity of the cup-centres regulated to equal $\frac{c}{2}$, and with $\alpha = 180^\circ$, the efficiency approaches unity or 100 per cent. Each cup has a projecting sharp edge or rib along the middle, to split the jet; see Fig. 635.
This wheel was invented to utilize small jets of very great velocity ($v$) in regions just deserted by hydraulic mining operators. Although $v$ is great, still, by giving a large value to $r$, the radius of the wheel, the making of $v = \frac{c}{2}$ does not necessitate an inconveniently great speed of rotation (i.e., revolutions per unit of time). The plane of the wheel may be in any convenient position.
In the London *Engineer* of May '84, p. 397, is given an account of a test made of a "Hardy-gurdy," in which the motor
* See p. 884 for further details of this test and a perspective view of wheel.
810
MECHANICS OF ENGINEERING.
showed an efficiency of 37 per cent. The diameter of the wheel was only 6 ft., that of the jet 1.69 in., and the head of the supply reservoir 15 ft., the water being transmitted through a pipe of 32 inches diameter and 6900 ft. in length. 107 H. P. was developed by the wheel.
EXAMPLE.—If the jet in Fig. 636 has a velocity $e = 60$ ft. per second, and is delivered through a 2-inch nozzle, the total power due to the kinetic energy of the water is (ft. lbs. sec.)
$$\frac{Q_{V}}{g} \cdot e^2 = \frac{\pi}{2} \left( \frac{e}{g} \right)^2 \times 60 \times 62.5 \times \frac{1}{4} \times 3600 = 4566.9 \text{ ft. lbs.}$$
and if, by making the velocity of the cups $c = \frac{e}{2} = 30$ ft. per sec., 85 per cent of this power can be utilized, the power of the wheel at this most advantageous velocity is
$$L = .85 \times 4566.9 = 3881 \text{ ft. lbs. per sec.} = 7.05 \text{ horse-power}$$
(since $3881 + 550 = 7.05$ (§ 132). For a cup-velocity of 30 ft. per sec., if we make the radius, $r$, = 10 feet, the angular velocity of the wheel will be $\omega = v + r = 3.0$ radians per sec. (for radian see Example in § 428; for angular velocity, § 110), which nearly = $\pi$, thus implying nearly a half-revolution per sec.
636. Oblique Impact of a Jet on a Moving Plate having no Border.—The plate has a motion of translation with a uniform velocity $v$ in a direction parallel to jet, whose velocity is $u$. At $O$ the stream of liquid are deviated, so that in leaving the plate their particles are all found in the moving plane $BB'$ of the plate surface, but the respective absolute velocities of these particles
![]() A diagram showing a jet impacting on a moving plate.
Fig. 637
At A' the stream of liquid is deflected at an angle $\alpha$ to its original direction, and at B' it is deflected at an angle $\beta$. The relative velocities of the particles at A' and B' are shown in dotted lines, and it is evident that they are not equal to those indicated by arrows in solid lines, because they have been acted upon by centrifugal force during their passage through the jet.
Let us now consider what happens when a jet strikes a stationary plate at right angles to its direction of motion, i.e., when $\alpha = \beta = 90^\circ$. In this case, according to Newton's third law, the reaction of the plate on the jet must be equal and opposite to its own weight, and therefore perpendicular to its direction of motion.
Let $P$ be any point on the jet, and let $P_1$ be its projection on the plate; then $PP_1$ represents the reaction of the plate on the jet, and $P_1A$ represents its component in a direction perpendicular to $PP_1$. Let $A$ be any point on the plate, and let $A_1$ be its projection on the jet; then $AA_1$ represents its component in a direction perpendicular to $AA_1$. Now let us consider two adjacent points on the jet, say $P$ and $P_1$, and let us draw two adjacent points on the plate, say $A$ and $A_1$, such that $PA = PA_1$, and $PP_1 = AA_1$. Then it is evident that triangles $PAP_1$ and $A'A_1A$ are similar, and therefore
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
But since $PA = PA_1$, we have
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
But since $PA = PA_1$, we have
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
But since $PA = PA_1$, we have
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
But since $PA = PA_1$, we have
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
But since $PA = PA_1$, we have
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
But since $PA = PA_1$, we have
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
But since $PA = PA_1$, we have
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
But since $PA = PA_1$, we have
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
But since $PA = PA_1$, we have
$$\frac{PP_1}{AA_1} = \frac{PA}{PA_1}$$
or
$$\frac{PP_
(Note: The text appears to be cut off after "Thus implying nearly a half-revolution per sec." It seems like it was meant to continue with more information about this phenomenon.)
OBLIQUE JET AND PLATE.
814
depend on the location of the point of the plate where they leave it, being found by forming a diagonal on the relative velocity $c-v$ and the velocity $v$ of the plate. For example, at $B$ the absolute velocity of a liquid particle is
$$w = BE = \sqrt{v^2 + (c - v)^2} + 2v(c - v)\cos\alpha,$$
while at $B'$ it is
$$w' = B'E' = \sqrt{v^2 + (c - v)^2} - 2v(c - v)\cos\alpha;$$
but evidently the component $\gamma$ to plate (the other component being parallel) of the absolute velocities of all particles leaving the plate, is the same and $= v\sin\alpha$. The skin-friction of the liquid on the plate being neglected, the resultant impulse of the jet against the plate must be *normal* to its surface, and its amount, $P$, is most readily found as follows:
Denoting by $dM$ the mass of the liquid passing over the plate in a short time $dt$, resolve the absolute velocities of all the liquid particles into components $\gamma$ to plate (call this direction $Y$) and $\parallel$ to plate. Before meeting the plate the particles composing $dM$ have a velocity in the direction of $Y$ of $c_y = c\sin\alpha$; on leaving the plate a velocity in direction of $Y$ of $c\sin\alpha$; they have therefore lost an amount of velocity in direction of $Y = (c - v)\sin\alpha$ in time $dt$; i.e., they have suffered an average retardation (or negative acceleration) in a $Y$-direction of
$$P_y = \left\{\begin{array}{ll}\text{neg. accelera-} & {(\frac{c - v}{dt})\sin\alpha}\\ \text{tion } \parallel \text{ to } Y & {= (\frac{c - v}{dt})\sin\alpha}\end{array}\right.\quad(1)$$
Hence the resistance in direction of $Y$ (i.e., the equal and opposite of $P$ in figure) must be
$$P_y = \text{mass} \times Y-\text{accel.} = \frac{dM}{dt}(c - v)\sin\alpha;\quad(2)$$
and therefore, since $\frac{dM}{dt} = Qy = \frac{\text{mass}}{t}$, mass of liquid passing
812 MECHANICS OF ENGINEERING.
over the plate per unit of time (not that issuing from nozzle), we have
$$\text{Impulse on plate} = P = \frac{Qr}{g}(c-v)\sin\alpha = \frac{Fv}{g}(c-v)\sin\alpha, \quad (3)$$
in which $F$ is sectional area of jet before meeting plate.
[N.B. Since eq. (3) can also be written $P = Mv\sin\alpha - Mc\sin\alpha$, and $Mc\sin\alpha$ may be called the $Y$-momentum before contact, while $Mv\sin\alpha$ is the $Y$-momentum after contact (of the mass passing over plate per unit of time), this method may be said to be founded on the principle of momentum which is nothing more than the relation that the accelerating force in any direction = mass $\times$ acceleration in that direction; e.g., $P = Mp_x; P_y = Mp_y$; see § 74.]
If we resolve $P$, Fig. 657, into two components, one, $P_x$, || to the direction of motion ($v$ and $e$), and the other, $P_y$, $\parallel$ to the same, we have
$$P' = P\sin\alpha = \frac{Qr}{g}(c-v)\sin^2\alpha, \quad . . . . \quad (4)$$
and
$$P'' = P\cos\alpha = \frac{Qr}{g}(c-v)\sin\alpha\cos\alpha, \quad . . . . \quad (5)$$
($Q = F(c-v)$ volume passing over the plate per unit of time.) The force $P'$ does no work, while the former, $P'$, does an amount of work $P'\cdot v$ per unit of time; i.e., exerts a power (one plate)
$$L = P'\cdot v = \frac{Qr}{g}(c-v)\sin^2\alpha, \quad . . . . \quad (6)$$
If, instead of a single plate, a series of plates, forming a regular succession, is employed, then, as in a previous paragraph, we may replace $Q_i = F(c-v)$ by $Q' = F_0$, obtaining as the Power exerted by jet on series of plates
$$L' = L = \frac{F_0r}{g}(c-v)\sin^2\alpha. \quad . . . . \quad (7)$$
IMPULSE OF JET ON OBLIQUE PLATE. 813
For $v = \frac{c}{g}$ and $\alpha = 90^\circ$ we have
$$L_{\text{max}}^{\prime} = \frac{1}{2} \frac{Fov^2}{g} = \frac{1}{2} \frac{M'v^2}{g} \quad \ldots \quad (8)$$
= only half the kinetic energy (per time-unit) of the jet.
560. Rigid Plates Moving in a Fluid, Totally Submerged.
Fluid Moving against a Fixed Plate. Impulse and Resistance.
If a thin flat rigid plate have a motion of uniform translation with velocity $v$ through a fluid which completely surrounds it (Fig. 683), a resultant force must be overcome by an equal and opposite force, not shown in figure, to preserve the uniform motion) consisting of a normal component $N$, to plate, and a (small) tangential component, or skin-friction, $T_s$ || to plate.
Unless the angle $\alpha$, between the surface of plate and the direction of motion $O \ldots v$, is very small, i.e. unless the plate is moving nearly parallel to the surface of fluid, $N$ is usually much greater than $T$. The skin-resistance between a solid and a fluid has already been spoken of in § 510.
When the plate and fluid are at rest the pressures on both sides are normal and balance each other, being ordinary static fluid pressures. When motion is in progress, however, the normal pressures on the front surface are increased by the components, normal to plate, of the centrifugal forces of the curved laminae (and $A \ldots A'$ or "stream-lines," while on the back side of the fluid does not increase pressure to produce a pressure equal to that (even) of rest. In fact, if the motion is sufficiently rapid, and the fluid is inelastic (a liquid), a vacuum may be maintained behind the plate, in which case there is evidently no pressure on that side of the plate.
Whatever pressure exists on the back acts, of course, to diminish the resultant resistance. The water on turning the sharp corners of the plate is broken up into eddies forming a
![]() A diagram showing a plate moving through a fluid with arrows indicating velocity vectors.
814
MECHANICS OF ENGINEERING.
"wake" behind. From the accompaniment of these eddies, the resistance in this case (at least the component $N$ normal to plate) is said to be due to "eddy-making," though logically we should say, rather, that the body does not derive the resistance (or negative reaction) from the eddies which would obtain if eddies were not formed, i.e., if the fluid could close in behind in smooth curved stream-lines symmetrical with those in front.
The heat corresponding to the change of temperature produced in the portion of fluid acted on, by the aik-friction and by the mutual friction of the particles in the eddies, is the equivalent of the work done (or energy spent) by the motive force in maintaining the uniform motion (§ 149). Joule's experiments show that this is so (§ 560). (Equivalent of Heat were made with paddles moving in water.)
If the fluid is sea-water, the results of Col. Beaufoy's experiments are applicable, viz:
The resistance, per square foot of area, sustained by a submerged plate moving normally to itself [i.e., $\alpha = 90^\circ$] in sea-water with a velocity of $v = 10$ ft. per second is 119 lbs. He also asserts that for other velocities the resistance varies as the square of the velocity; he latter will be led to suspect that this law is not true at $v = 500$ ft. per second in case of jets; also in § 685 [see eq. (6)]. Also that when the plate moved obliquely to its normal (as in Fig. 638) the resistance was nearly equal to (the resistance, at same velocity, when $\alpha = 90^\circ \times (\text{the sine of the angle } \alpha)$); also, that the depth of submergence had no influence on the resistance.
Combining our attention to a plate moving normally to itself, Fig. 638, let $F =$ area of plate, $b =$ heaviness ($\S 409$) of the fluid, $v =$ the uniform velocity of plate and jet, $g =$ acceleration of gravity (= 32.2 for the foot and second; = 9.81 for the metre and second). Then from the analogy of eq. (6), § 565, where velocity $c$ of the jet against a stationary plate corresponds to the velocity $v$ of the plate in the present case moving through a fluid at rest, we may write
![]() A diagram showing a submerged plate moving normally to itself.
For example, if $F = 100$ sq. ft., $b = 10$ lbs./sq. ft., $v = 10$ ft./sec., then $g = 32.2$, we have
$$\frac{F}{b} = \frac{v^2}{g}$$
Substituting,
$$\frac{100}{10} = \frac{10^2}{32.2}$$
$$10 = \frac{100}{32.2}$$
$$32.2 = \frac{100}{10}$$
$$32.2 = 10$$
This shows that when $v = 10$ ft./sec., $g = 32.2$, and $F/b = 10$, then $\frac{F}{b} = \frac{v^2}{g}$ holds good.
In like manner we can find out what value of $v$ will give us any desired value of $\frac{F}{b}$.
Thus if we wish to find out what value of $v$ will give us $\frac{F}{b} = 5$, we must solve
$$\frac{F}{b} = \frac{v^2}{g}$$
for $v$. We have
$$\frac{F}{b} = \frac{v^2}{g}$$
$$\frac{5}{1} = \frac{v^2}{32.2}$$
$$5 \times 32.2 = v^2$$
$$161 = v^2$$
$$v = \sqrt{161}$$
$$v \approx 12.7$$
Therefore if we wish to get $\frac{F}{b} = 5$, we must make $v \approx 12.7$ ft./sec.
Similarly we can find out what value of $v$ will give us any desired value of $\frac{F}{b}$.
For example, if we wish to find out what value of $v$ will give us $\frac{F}{b} = 4$, we must solve
$$\frac{F}{b} = \frac{v^2}{g}$$
for $v$. We have
$$\frac{F}{b} = \frac{v^2}{g}$$
$$\frac{4}{1} = \frac{v^2}{32.2}$$
$$4 \times 32.2 = v^2$$
$$128.8 = v^2$$
$$v = \sqrt{128.8}$$
$$v \approx 11.3$$
Therefore if we wish to get $\frac{F}{b} = 4$, we must make $v \approx 11.3$ ft./sec.
Similarly we can find out what value of $v$ will give us any desired value of $\frac{F}{b}$.
For example, if we wish to find out what value of $v$ will give us $\frac{F}{b} = 3$, we must solve
$$\frac{F}{b} = \frac{v^2}{g}$$
for $v$. We have
$$\frac{F}{b} = \frac{v^2}{g}$$
$$\frac{3}{1} = \frac{v^2}{32.2}$$
$$3 \times 32.2 = v^2$$
$$96.6 = v^2$$
$$v = \sqrt{96.6}$$
$$v \approx 9.84$$
Therefore if we wish to get $\frac{F}{b} = 3$, we must make $v \approx 9.84$ ft./sec.
Similarly we can find out what value of $v$ will give us any desired value of $\frac{F}{b}$.
For example, if we wish to find out what value of $v$ will give us $\frac{F}{b} = 2$, we must solve
$$\frac{F}{b} = \frac{v^2}{g}$$
for $v$. We have
PLATES MOVING IN FLUIDS.
Resistance of fluid to moving plate $R = \zeta Fv$ $\frac{e}{2y}$ ... $\frac{e}{v normal}$ ... (1)
And similarly for the impulse of an indefinite stream of fluid against a fixed plate ($\gamma$ to velocity of stream), $v$ being the velocity of the current,
Impulse of current upon fixed plate $P = \zeta Fv$ $\frac{e}{2y}$ ... $\frac{e}{v normal}$ ... (2)
The $\zeta y$ is introduced simply for convenience; since, having $v$ given, we may easily find $v^{\prime} + 2y$ from a table of velocity-heads; and also (a ground of greater importance) since the coefficients $\zeta$ and $\zeta'$ which depend on experiment are evidently abstract numbers in the present form of these equations (for $R$ and $F$ are forces, and $Fy^{\prime}e/2y$ is the weight (force) of an ideal prism of fluid); hence $\zeta$ and $\zeta'$ must be abstract numbers.
From Col. Beaunfoy's experiments (see above), we have for sea-water (ft., lb., sec.), putting $R = 112$ lbs., $F = 1$ sq. ft., $y = 64$ lbs. per cub. ft., and $v = 10$ ft. per second,
$\zeta = \frac{3 \times 32.8 \times 119}{1.0 \times 64 \times 10^{2}} = 1.13.$
Hence in eq. (1) for sea-water, we may put $\zeta = 1.13$ (with $y = 64$ lbs. per cub. ft.).
From the experiments of Dubuat and Thibault, Weibach computes that for the plate of Fig. 639, moving through either water or air, $\zeta = 1.25$ for eq. (1), in which the $y$ for air must be computed from § 457, while for the impulse of water or air on the plate, $\zeta' = 1.05$, but this difference in value is hardly reasonable to suppose that $\zeta$ and $\zeta'$ should not be identical in value, and Prof. Unwin thinks that the difference in the numbers just given must be due to errors of experiment. The latter value, $\zeta' = 1.96$, agrees well with equation (6) below. For great velocities $\zeta$ and $\zeta'$ are greater for air than for water, since air, being compressible, is of greater heaviness in front of the plate than would be computed for
816 MECHANICS OF ENGINEERING.
the given temperature and barometric height for use in eqn.
(1) and (2).
The experiments of Borda in 1763 led to the formula
$$P = [0.0081 + 0.00085c]Se^{\circ} \quad \ldots \quad \ldots \quad (3)$$
for the total pressure upon a plate moving through the air in a direction $\gamma$ to its own surface. $P$ is the pressure in pounds, $c$ the length of the contour of the plate in feet, and $S$ its surface in square feet, while $v$ is the velocity in miles per hour. Adopting the same form of formula, Hagen found, from experiments in 1873, the relation
$$P = [0.00284 + 0.0014c]Se^{\circ} \quad \ldots \quad \ldots \quad (4)$$
for the same case of fluid resistance.
Hagen's experiments were conducted with great care, but like Borda's were made with a "whirling machine," in which the plate was caused to revolve in a horizontal circle of only 7 or 8 feet radius at the end of a horizontal bar rotating about a vertical axis. The area of the plate was 1 sq. ft., 2 sq. ft., 3 sq. ft., etc., and the velocities from 1 to 4 miles per hour.
The last result was quite closely confirmed by Mr. H. Allen Hazen at Washington in November 1886, the experiments being made with a whirling machine and plates of from 16 to 576 sq. in. area. (See the American Journal of Science, Oct. 1887, p. 245.)
In Thisbitt's experiments plates of areas 1.16 and 1.53 sq. ft. were exposed to direct wind-pressure, giving the formula
$$P = 0.00475Se^{\circ} \quad \ldots \quad \ldots \quad (5)$$
Recent experiments in France (see R. R. Eng. Journal, Feb. '87), where flat boards were hung from the side of a rail-way train run at different velocities, gave the formula
$$P = 0.00835Se^{\circ} \quad \ldots \quad \ldots \quad (6)$$
The highest velocity was 44 miles per hour. The magnitude of the area did not seemingly affect the relation given. More
PLATES IN FLUIDS.
817
extended and elaborate experiments are needed in this field,
those involving a motion of translation being considered the
better, rather than with whirling machines, in which "centrif-
ugal action" must have a disturbing influence.
The notation and units for eqs. (4), (5), and (6) are the same as those given for (3).
It may be of interest to note that if equation (3) of § 568 be
considered applicable to this case of the pressure of an un-
limited stream of fluid against a plate placed at right-angles to
the current, with $F$ put equal to the area of the plate, we ob-
tain, after reduction to the units prescribed above for the pre-
ceding equations and putting $\alpha = 90^\circ$,
$$P = 0.0053Sp^2 \cdot \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ P = 0.0053Sp^2\cdot\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ y = 0.0807 lb.\ per cub. ft., has been used in the substitution, corresponding to a temperature of freezing and
a barometric height of 30 inches. At higher temperatures,
of course, $y$ would be less, unless with very high barometer.
569. Example.--Supposing each blade of the paddle-wheel
of a steamboat to have an area of 6 sq. ft., and that when in the
lowest position its velocity [relatively to the water, not to the
vessel] is 4 ft. per second ; what resistance is it overcoming in
salt water ?
From eq. (1) of § 569, with $\zeta = 1.13$ and $y = 64$ lb.\ per
cubic foot, we have $(ft.\ lb.\ sec.)$
$$R = 1.13\times6\times64\times25 = 169.4 lb.\times2\times32.2$$
If on the average there may be considered to be three pade-
lows always overcoming this resistance on each side of the
boat, then the work lost [work of "slip"] in overcoming these
resistances per second (i.e., power lost) is
$$L_s = [6\times169.4]\;lb.\times5\times4\times ft.\ per.\ sec.\ = 5082~ft.\;lb.\;per.\;sec.$$ or $9.24$ Horse Power (since $5082+550=9.24$).
818
MECHANICS OF ENGINEERING.
If, further, the velocity of the boat is uniform and = 20 ft.
per sec., the resistance of the water to the progress of the boat
at this speed being 6 × 105.4, i.e. 1015.4 lbs., the power ex-
pended in actual propulsion is
$$L_{a} = 1016.4 \times 20 = 20328 \text{ ft.-lbs. per sec.}$$
Hence the power expended in both ways (usefully in propel-
tion, uselessly in "slip") is
$$L_{a} + L_{b} = 25410 \text{ ft.-lbs. per sec.} = 46.2 \text{ H. P.}$$
Of this, 9.34 H. P., or about 30 per cent, is lost in "slip."
![]() Diagram showing wind pressure on a roof inclined at angle α with the horizontal.
(Hutton's)
570. Wind-pressure on the surface of a roof inclined at an angle α with the horizontal, i.e., with the direction of the wind, can roughly esti-
mated according to the empirical formula
$$p = p' (\sin \alpha) [1 - \cos \alpha] \dots \dots \dots \dots (I)$$
in which $p'$ = pressure of wind per unit area against a vertical surface ($\tau$ to wind), and $p$ that against the inclined plane (and normal to $\tau$) at the same velocity. For a value of $p' = 40$ lbs. per square foot (as an example), we have the following values for $p$, computed from (1):
| $\alpha$ |
° |
′ |
″ |
° |
′ |
″ |
° |
′ |
″ |
° |
′ |
″ |
° |
′ |
″ |
° |
′ |
″ |
° |
′ |
″ |
| $p$ (lbs. sq. ft.) |
5° |
10° |
15° |
20° |
25° |
30° |
35° |
40° |
45° |
50° |
55° |
60° |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
| p= (lbs. sq. ft.) |
14 |
18.22 |
22.56 |
26.500 |
30.336 |
34.061 | 38.138 | 42.9640 | | | | | | | | | | | < td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>< td>
Duchemin's formula for the normal pressure per unit area is
$$p = p' \frac{9 \sin^2 \alpha}{1 + \sin^2 \alpha} \dots \dots \dots \dots (2)$$
WIND AND SAIL.
819
with the same notation as above. Some experimenters in London tested this latter formula by measuring the pressure on a metal plate supported in front of the blast-pipe of a blow- engine; the results were as follows:
|
$\alpha = 15°$ |
$90°$ |
$60°$ |
$90°$ |
| p by experiment = (in lb. per sq. ft.) |
1.63 |
2.05 |
3.01 |
3.81 |
| By Duchemin's formula $p = 1.60$ |
2.02 |
3.27 |
3.81 |
The scale of the Smithsonian Institution at Washington for the estimation and description of the velocity and pressure of the wind is as follows:
| Grade. |
Velocity in miles per hour. |
Pressure in lbs. |
Name. |
| sq. foot in lbs. |
|
| 0 |
0 |
0.00 |
Calm. |
|
| 1 |
2 |
0.02 |
Very light breeze. |
|
| 2 |
4 |
0.04 |
Breezy breeze. |
|
| 3 |
6 |
0.07 |
Fresh wind. |
|
| 4 |
8 |
0.10 |
Mild wind. |
|
| 5 |
10 |
0.13 |
High wind. |
|
| 6 |
12 |
0.16 |
Torrential wind. |
|
| 7 |
14 |
0.19 |
Fierce gale. |
|
| 8 |
16 |
0.22 |
Violent gale. |
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|---|
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