Page:EB1911 - Volume 14.djvu/58

Rh But if work is expended in producing irregular eddying motion, the head at the section CD will be diminished.

Suppose the mass ABCD comes in a short time t to A′B′C′D′. The resultant force parallel to the axis of the stream is

p + p0 (1 − ) − p11,

where p0 is put for the unknown pressure on the annular space between AB and EF. The impulse of that force is

{ p + p0 (1 − ) − p11 } t.

The horizontal change of momentum in the same time is the difference of the momenta of CDC′D′ and ABA′B′, because the amount of momentum between A′B′ and CD remains unchanged if the motion is steady. The volume of ABA′B′ or CDC′D′, being the inflow and outflow in the time t, is Qt = vt = 1v1t, and the momentum of these masses is (G / g) Qvt and (G / g) Qv1t. The change of momentum is therefore (G / g) Qt (v1 − v). Equating this to the impulse,

{ p + p0 (1 − ) − p11 } t = (G / g) Qt (v1 − v).

Assume that p0 = p, the pressure at AB extending unchanged through the portions of fluid in contact with AE, BF which lie out of the path of the stream. Then (since Q = 1v1)

(p − p1) = (G / g) v1 (v1 − v);

p / G − p1/G = v1 (v1 − v) / g; (2)

p / G + v2/2g = p1/G + v12/2g + (v − v1)2/2g. (3)

This differs from the expression (1), § 29, obtained for cases where no sensible internal work is done, by the last term on the right. That is, (v − v1)2/2g has to be added to the total head at CD, which is p1/G + v12/2g, to make it equal to the total head at AB, or (v − v1)2/2g is the head lost in shock at the abrupt change of section. But (v − v1) is the relative velocity of the two parts of the stream. Hence, when an abrupt change of section occurs, the head due to the relative velocity is lost in shock, or (v − v1)2/2g foot-pounds of energy is wasted for each pound of fluid. Experiment verifies this result, so that the assumption that p0 = p appears to be admissible.

If there is no shock,

p1/G = p / G + (v2 − v12)/2g.

If there is shock,

p1/G = p / G − v1 (v1 − v) / g.

Hence the pressure head at CD in the second case is less than in the former by the quantity (v − v1)2/2g, or, putting 1v1 = v, by the quantity

(v2/2g) (1 − /1)2. (4)

. THEORY OF THE DISCHARGE FROM ORIFICES AND MOUTHPIECES

§ 37. ''Minimum Coefficient of Contraction. Re-entrant Mouthpiece'' of Borda.—In one special case the coefficient of contraction can be determined theoretically, and, as it is the case where the convergence of the streams approaching the orifice takes place through the greatest possible angle, the coefficient thus determined is the minimum coefficient.

Let fig. 39 represent a vessel with vertical sides, OO being the free water surface, at which the pressure is pa. Suppose the liquid issues by a horizontal mouthpiece, which is re-entrant and of the greatest length which permits the jet to spring clear from the inner end of the orifice, without adhering to its sides. With such an orifice the velocity near the points CD is negligible, and the pressure at those points may be taken equal to the hydrostatic pressure due to the depth from the free surface. Let be the area of the mouthpiece AB,  that of the contracted jet aa Suppose that in a short time t, the mass OOaa comes to the position O′O′ a′a′; the impulse of the horizontal external forces acting on the mass during that time is equal to the horizontal change of momentum.

The pressure on the side OC of the mass will be balanced by the pressure on the opposite side OE, and so for all other portions of the vertical surfaces of the mass, excepting the portion EF opposite the mouthpiece and the surface AaaB of the jet. On EF the pressure is simply the hydrostatic pressure due to the depth, that is, (pa + Gh). On the surface and section AaaB of the jet, the horizontal resultant of the pressure is equal to the atmospheric pressure pa acting on the vertical projection AB of the jet; that is, the resultant pressure is −pa. Hence the resultant horizontal force for the whole mass OOaa is (pa + Gh) − pa = Gh. Its impulse in the time t is Ght. Since the motion is steady there is no change of momentum between O′O′ and aa. The change of horizontal momentum is, therefore, the difference of the horizontal momentum lost in the space OOO′O′ and gained in the space aaa′a′. In the former space there is no horizontal momentum.

The volume of the space aaa′a′ is vt; the mass of liquid in that space is (G / g)vt; its momentum is (G / g)v2t. Equating impulse to momentum gained,

Ght = (G / g) v2t;

∴ / = gh / v2

But

v2 = 2gh, and / = cc;

∴ / = = cc;

a result confirmed by experiment with mouthpieces of this kind. A similar theoretical investigation is not possible for orifices in plane surfaces, because the velocity along the sides of the vessel in the neighbourhood of the orifice is not so small that it can be neglected. The resultant horizontal pressure is therefore greater than Gh, and the contraction is less. The experimental values of the coefficient of discharge for a re-entrant mouthpiece are 0.5149 (Borda), 0.5547 (Bidone), 0.5324 (Weisbach), values which differ little from the theoretical value, 0.5, given above.

§ 38. Velocity of Filaments issuing in a Jet.—A jet is composed of fluid filaments or elementary streams, which start into motion at some point in the interior of the vessel from which the fluid is discharged, and gradually acquire the velocity of the jet. Let Mm, fig. 40 be such a filament, the point M being taken where the velocity is insensibly small, and m at the most contracted section of the jet, where the filaments have become parallel and exercise uniform mutual pressure. Take the free surface AB for datum line, and let p1, v1, h1, be the pressure, velocity and depth below datum at M; p, v, h, the corresponding quantities at m. Then § 29, eq. (3a),

v12/2g + p1/G − h1 = v2/2g + p / G − h (1)

But at M, since the velocity is insensible, the pressure is the hydrostatic pressure due to the depth; that is v1 = 0, p1 = pa + Gh1. At m, p = pa, the atmospheric pressure round the jet. Hence, inserting these values,

0 + pa/G + h1 − h1 = v2/2g + pa/G − h;

That is, neglecting the viscosity of the fluid, the velocity of filaments at the contracted section of the jet is simply the velocity due to the difference of level of the free surface in the reservoir and the orifice. If the orifice is small in dimensions compared with h, the filaments will all have nearly the same velocity, and if h is measured to the centre of the orifice, the equation above gives the mean velocity of the jet.

Case of a Submerged Orifice.—Let the orifice discharge below the level of the tail water. Then using the notation shown in fig. 41, we have at M, v1 = 0, p1 = Gh; + pa at m, p = Gh3 + pa. Inserting these values in (3), § 29,

0 + h1 + pa/G − h1 = v2/2g + h3 − h22 + pa/G;

v2/2g = h2 − h3 = h, (3)