[Illustration: FIG. 39.]
§ 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 p_a. 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 [Omega] be the area of the mouthpiece AB,
[omega] 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, (p_a + Gh).
On the surface and section AaaB of the jet, the horizontal resultant
of the pressure is equal to the atmospheric pressure p_a acting on the
vertical projection AB of the jet; that is, the resultant pressure is
-p_a[Omega]. Hence the resultant horizontal force for the whole mass
OOaa is (p_a + Gh)[Omega] - p_a[Omega] = Gh[Omega]. Its impulse in the
time t is Gh[Omega]t. 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 [omega]vt; the mass of liquid in
that space is (G/g)[omega]vt; its momentum is (G/g)[omega]v²t.
Equating impulse to momentum gained,
Gh[Omega] = (G/g)[omega]v²t;
.: [omega]/[Omega] = gh/v²