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
[Omega]/[omega] >= [root]{(h + p_a/G)/h}
= [root]{1 + p_a/Gh};
or, putting p_a/G = 34 ft., if
[Omega]/[omega] >= [root]{(h + 34)/h}.
In practice there will be an interruption of the full bore flow with a
less ratio of [Omega]/[omega], 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 = [omega] [root]{2g(h + p_a/G)};
that is, the discharge is the same as for a well-bell-mouthed
mouthpiece of area [omega], 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.
[Illustration: 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.
[Illustration: FIG. 63.]