(a) The plane is at rest. Then u = 0, N = (G/g)[omega]v² cos [alpha];
and the work done on the plane and the efficiency of the jet are zero.
(b) The plane moves parallel to the jet. Then [delta] = [alpha], and
Pu = (G/g)[omega]u cos²[alpha](v - u)², which is a maximum when u =
1/3 v.
When u = 1/3 v then Pu max. = 4/27 (G/g)[omega]v³ cos² [alpha], and
the efficiency = [eta] = 4/9 cos² [alpha].
(c) The plane moves perpendicularly to the jet. Then [delta] = 90° -
[alpha]; cos [delta] = sin [alpha]; and Pu = G/g [omega]u (sin
[alpha]/cos [alpha]) (v cos [alpha] - u sin [alpha])². This is a
maximum when u = 1/3 v cos [alpha].
When u = 1/3 v cos [alpha], the maximum work and the efficiency are
the same as in the last case.
[Illustration: FIG. 159.]
§ 158. _Best Form of Vane to receive Water._--When water impinges
normally or obliquely on a plane, it is scattered in all directions
after impact, and the work carried away by the water is then generally
lost, from the impossibility of dealing afterwards with streams of
water deviated in so many directions. By suitably forming the vane,
however, the water may be entirely deviated in one direction, and the
loss of energy from agitation of the water is entirely avoided.
Let AB (fig. 159) be a vane, on which a jet of water impinges at the
point A and in the direction AC. Take AC = v = velocity of water, and
let AD represent in magnitude and direction the velocity of the vane.
Completing the parallelogram, DC or AE represents the direction in
which the water is moving relatively to the vane. If the lip of the
vane at A is tangential to AE, the water will not have its direction
suddenly changed when it impinges on the vane, and will therefore have
no tendency to spread laterally. On the contrary it will be so
gradually deviated that it will glide up the vane in the direction AB.
This is sometimes expressed by saying that the vane _receives the
water without shock_.
[Illustration: FIG. 160.]
§ 159. _Floats of Poncelet Water Wheels._--Let AC (fig. 160) represent
the direction of a thin horizontal stream of water having the velocity
v. Let AB be a curved float moving horizontally with velocity u. The
relative motion of water and float is then initially horizontal, and
equal to v - u.
In order that the float may receive the water without shock, it is
necessary and sufficient that the lip of the float at A should be
tangential to the direction AC of relative motion. At the end of (v -
u)/g seconds the float moving with the velocity u comes to the
position A1B1, and during this time a particle of water received at A
and gliding up the float with the relative velocity v - u, attains a
height DE = (v - u)²/2g. At E the water comes to relative rest. It
then descends along the float, and when after 2(v - u)/g seconds the
float has come to A2B2 the water will again have reached the lip at A2
and will quit it tangentially, that is, in the direction CA2, with a
relative velocity -(v - u) = -[root](2gDE) acquired under the
influence of gravity. The absolute velocity of the water leaving the
float is therefore u - (v - u) = 2u - v. If u = ½v, the water will
drop off the bucket deprived of all energy of motion. The whole of the
work of the jet must therefore have been expended in driving the
float. The water will have been received without shock and discharged
without velocity. This is the principle of the Poncelet wheel, but in
that case the floats move over an arc of a large circle; the stream of
water has considerable thickness (about 8 in.); in order to get the
water into and out of the wheel, it is then necessary that the lip of
the float should make a small angle (about 15°) with the direction of
its motion. The water quits the wheel with a little of its energy of
motion remaining.
§ 160. _Pressure on a Curved Surface when the Water is deviated wholly
in one Direction._--When a jet of water impinges on a curved surface
in such a direction that it is received without shock, the pressure on
the surface is due to its gradual deviation from its first direction.
On any portion of the area the pressure is equal and opposite to the
force required to cause the deviation of so much water as rests on
that surface. In common language, it is equal to the centrifugal force
of that quantity of water.
[Illustration: FIG. 161.]
_Case 1. Surface Cylindrical and Stationary._--Let AB (fig. 161) be
the surface, having its axis at O and its radius = r. Let the water
impinge at A tangentially, and quit the surface tangentially at B.
Since the surface is at rest, v is both the absolute velocity of the
water and the velocity relatively to the surface, and this remains
unchanged during contact with the surface, because the deviating force
is at each point perpendicular to the direction of motion. The water
is deviated through an angle BCD = AOB = [phi]. Each particle of water
of weight p exerts radially a centrifugal force pv²/rg. Let the
thickness of the stream = t ft. Then the weight of water resting on
unit of surface = Gt lb.; and the normal pressure per unit of surface
= n = Gtv²/gr. The resultant of the radial pressures uniformly
distributed from A to B will be a force acting in the direction OC
bisecting AOB, and its magnitude will equal that of a force of
intensity = n, acting on the projection of AB on a plane perpendicular
to the direction OC. The length of the chord AB = 2r sin ½[phi]; let b
= breadth of the surface perpendicular to the plane of the figure. The
resultant pressure on surface
[phi] Gt v² G [phi]
= R = 2rb sin ----- × --.-- = 2--- btv² sin -----,
2 g r g 2
which is independent of the radius of curvature. It may be inferred
that the resultant pressure is the same for any curved surface of the
same projected area, which deviates the water through the same angle.
_Case 2. Cylindrical Surface moving in the Direction AC with
Velocity u._--The relative velocity = v - u. The final velocity BF
(fig. 162) is found by combining the relative velocity BD = v - u
tangential to the surface with the velocity BE = u of the surface. The
intensity of normal pressure, as in the last case, is (G/g)t(v -
u)²/r. The resultant normal pressure R = 2(G/g)bt(v - u)² sin ½[phi].
This resultant pressure may be resolved into two components P and L,
one parallel and the other perpendicular to the direction of the
vane's motion. The former is an effort doing work on the vane. The
latter is a lateral force which does no work.
P = R sin ½[phi] = (G/g) bt (v - u)² (1 - cos [phi]);