R = K1G[omega]v²/2g,
where
_ _
| / [rho] \² / 1 \² / [rho] \² |
K1 = [rho] | ( --------- ) ( --- - 1 ) ( --------- - 1 ) |.
|_ \[rho] - 1/ \c_c / \[rho] - 1 / _|
Taking c_c = 0.85 and [rho] = 4, K1 = 0.467, a value less than before.
Hence there is less pressure on the cylinder than on the thin plane.
[Illustration: 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 [omega] is the area of the orifice, the quantity
of water impinging on the plane is obviously
Q = [omega]v1 = [omega] [root](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 = [root](2gh);
where h is the depth of the plane below the free surface. The external
layers of fluid subjected throughout, after leaving the orifice, to
the atmospheric pressure will attain the velocity v, and will flow
away with this velocity unchanged except by friction. The layers
towards the interior of the jet, being subjected to a pressure greater
than atmospheric pressure, will attain a less velocity, and so much
less as they are nearer the centre of the jet. But the pressure can
in no case exceed the pressure v²/2g or h measured in feet of water,
or the direction of motion of the water would be reversed, and there
would be reflux. Hence the maximum intensity of the pressure of the
jet on the plane is h ft. of water. If the pressure curve is drawn
with pressures represented by feet of water, it will touch the free
water surface at the centre of the jet.
[Illustration: FIG. 167.]
Suppose the pressure curve rotated so as to form a solid of
revolution. The weight of water contained in that solid is the total
pressure of the jet on the surface, which has already been determined.
Let V = volume of this solid, then GV is its weight in pounds.
Consequently
GV = (G/g)[omega]v1v;