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V. THEORY OF THE DISCHARGE FROM ORIFICES AND MOUTHPIECES
| 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 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 Ω be the area of the mouthpiece AB, ω 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 −paΩ. Hence the resultant horizontal force for the whole mass OOaa is (pa + Gh) Ω − paΩ = GhΩ. Its impulse in the time t is GhΩt. 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 ωvt; the mass of liquid in that space is (G/g)ωvt; its momentum is (G/g)ωv2t. Equating impulse to momentum gained,
GhΩt = (G/g) ωv2t;
∴ ω/Ω = gh/v2
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