Page:EB1911 - Volume 14.djvu/64

Fig. 59.	Fig. 60.

The values of c_v and c_c must here be determined by experiment. The above table gives values sufficient for practical purposes. Since the contraction beyond the mouthpiece increases with the convergence, or, what is the same thing, c_c diminishes, and on the other hand the loss of energy diminishes, so that c_v increases with the convergence, there is an angle for which the product c_c c_v, and consequently the discharge, is a maximum.

§ 51. Divergent Conoidal Mouthpiece.—Suppose a mouthpiece so designed that there is no abrupt change in the section or velocity of the stream passing through it. It may have a form at the inner end approximately the same as that of a simple contracted vein, and may then enlarge gradually, as shown in fig. 60. Suppose that at EF it becomes cylindrical, so that the jet may be taken to be of the diameter EF. Let ω, v, p be the section, velocity and pressure at CD, and Ω, v₁, p₁ the same quantities at EF, p_a being as usual the atmospheric pressure, or pressure on the free surface AB. Then, since there is no loss of energy, except the small frictional resistance of the surface of the mouthpiece,

h + p_a/G = v²/2g + p/G
⁠= v₁²/2g + p₁/G.

If the jet discharges into the air, p₁ = p_a; and

v₁²/2g = h;

v₁ = √(2gh);

or, if a coefficient is introduced to allow for friction,

v₁ = c_v √(2gh);

where c_v is about 0.97 if the mouthpiece is smooth and well formed.

Q = Ωv₁ = c_v Ω √(2gh).

Fig. 61.

Hence the discharge depends on the area of the stream at EF, and not at all on that at CD, and the latter may be made as small as we please without affecting the amount of water discharged.

There is, however, a limit to this. As the velocity at CD is greater than at EF the pressure is less, and therefore less than atmospheric pressure, if the discharge is into the air. If CD is so contracted that p = 0, the continuity of flow is impossible. In fact the stream disengages itself from the mouthpiece for some value of p greater than 0 (fig. 61).

From the equations,

p/G = p_a/G − (v² − v₁²) / 2g.

Let Ω/ω = m. Then

v = v₁m;

p/G = p_a/G − v₁² (m² − 1) / 2g

= p_a/G − (m² − 1)h;⁠

whence we find that p/G will become zero or negative if

Ω/ω ≧ √ {(h + p_a/G) / h }
= √ {1 + p_a/Gh};

or, putting p_a/G = 34 ft., if

Ω/ω ≧ √ { (h + 34)/h}.

In practice there will be an interruption of the full bore flow with a less ratio of Ω/ω, because of the disengagement of air from the water. But, supposing this does not occur, the maximum discharge of a mouthpiece of this kind is

Q = ω √ {2g (h + p_a/G) };

that is, the discharge is the same as for a well-bell-mouthed mouthpiece of area ω, and without the expanding part, discharging into a vacuum.

§ 52. Jet Pump.—A divergent mouthpiece may be arranged to act as a pump, as shown in fig. 62. The water which supplies the energy required for pumping enters at A. The water to be pumped enters at B. The streams combine at DD where the velocity is greatest and the pressure least. Beyond DD the stream enlarges in section, and its pressure increases, till it is sufficient to balance the head due to the height of the lift, and the water flows away by the discharge pipe C.

Fig. 62.

Fig. 63 shows the whole arrangement in a diagrammatic way. A is the reservoir which supplies the water that effects the pumping; B is the reservoir of water to be pumped; C is the reservoir into which the water is pumped.

Fig. 63.

Discharge with Varying Head

§ 53. Flow from a Vessel when the Effective Head varies with the Time.—Various useful problems arise relating to the time of emptying and filling vessels, reservoirs, lock chambers, &c., where the flow is dependent on a head which increases or diminishes during the operation. The simplest of these problems is the case of filling or emptying a vessel of constant horizontal section.

Fig. 64.

Time of Emptying or Filling a Vertical-sided Lock Chamber.—Suppose the lock chamber, which has a water surface of Ω square ft., is emptied through a sluice in the tail gates, of area ω, placed below the tail-water level. Then the effective head producing flow through the sluice is the difference of level in the chamber and tail bay. Let H (fig. 64) be the initial difference of level, h the difference of level after t seconds. Let −dh be the fall of level in the chamber during an interval dt. Then in the time dt the volume in the chamber is altered by the amount −Ωdh, and the outflow from the sluice in the same time is cω √(2gh)dt. Hence the differential equation connecting h and t is

cω √(2gh)dt + Ωh = 0.