where
| K1 = ρ { ( | ρ | ) | 2 | ( | 1 | − 1 ) | 2 | + ( | ρ | − 1 ) | 2 | }. |
| ρ − 1 | cc | ρ − 1 |
Taking cc = 0.85 and ρ = 4, K1 = 0.467, a value less than before. Hence there is less pressure on the cylinder than on the thin plane.
| Fig. 166. |
§ 165. Distribution of Pressure on a Surface on which a Jet impinges normally.—The principle of momentum gives readily enough the total or resultant pressure of a jet impinging on a plane surface, but in some cases it is useful to know the distribution of the pressure. The problem in the case in which the plane is struck normally, and the jet spreads in all directions, is one of great complexity, but even in that case the maximum intensity of the pressure is easily assigned. Each layer of water flowing from an orifice is gradually deviated (fig. 166) by contact with the surface, and during deviation exercises a centrifugal pressure towards the axis of the jet. The force exerted by each small mass of water is normal to its path and inversely as the radius of curvature of the path. Hence the greatest pressure on the plane must be at the axis of the jet, and the pressure must decrease from the axis outwards, in some such way as is shown by the curve of pressure in fig. 167, the branches of the curve being probably asymptotic to the plane.
For simplicity suppose the jet is a vertical one. Let h1 (fig. 167) be the depth of the orifice from the free surface, and v1 the velocity of discharge. Then, if ω is the area of the orifice, the quantity of water impinging on the plane is obviously
Q = ωv1 = ω √ (2gh1);
that is, supposing the orifice rounded, and neglecting the coefficient of discharge.
The velocity with which the fluid reaches the plane is, however, greater than this, and may reach the value
v = √ (2gh);