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 = pa/G − (v2 − v12) / 2g.
Let Ω/ω = m. Then
v = v1m;
p/G = pa/G − v12 (m2 − 1) / 2g
= pa/G − (m2 − 1) h;
whence we find that p/G will become zero or negative if
Ω/ω ≥ √ {(h + pa/G) / h } = √ {1 + pa/Gh};