§ 155. (1) _A Jet impinges on a plane surface at rest, in a direction
normal to the plane_ (fig. 154).--Let a jet whose section is [omega]
impinge with a velocity v on a plane surface at rest, in a direction
normal to the plane. The particles approach the plane, are gradually
deviated, and finally flow away parallel to the plane, having then no
velocity in the original direction of the jet. The quantity of water
impinging per second is [omega]v. The pressure on the plane, which is
equal to the change of momentum per second, is P = (G/g)[omega]v².
[Illustration: FIG. 154.]
(2) _If the plane is moving in the direction of the jet with the
velocity_ ±u, the quantity impinging per second is [omega](v ± u).
The momentum of this quantity before impact is (G/g)[omega](v ± u)v.
After impact, the water still possesses the velocity ±u in the
direction of the jet; and the momentum, in that direction, of so much
water as impinges in one second, after impact, is
±(G/g)[omega](v ± u)u. The pressure on the plane, which is the change
of momentum per second, is the difference of these quantities or P =
(G/g)[omega](v ± u)². This differs from the expression obtained in
the previous case, in that the relative velocity of the water and
plane v ± u is substituted for v. The expression may be written P = 2
× G × [omega](v ± u)²/2g, where the last two terms are the volume of
a prism of water whose section is the area of the jet and whose length
is the head due to the relative velocity. The pressure on the plane is
twice the weight of that prism of water. The work done when the plane
is moving in the same direction as the jet is Pu = (G/g)[omega](v -
u)²u foot-pounds per second. There issue from the jet [omega]v cub.
ft. per second, and the energy of this quantity before impact is
(G/2g)[omega]v³. The efficiency of the jet is therefore [eta] = 2(v -
u)²u/v³. The value of u which makes this a maximum is found by
differentiating and equating the differential coefficient to zero:--
d[eta]/du = 2(v² - 4vu + 3u²)/v³ = 0;
.: u = v or (1/3)v.
The former gives a minimum, the latter a maximum efficiency.
Putting u = (1/3)v in the expression above,
[eta] max. = 8/27.
(3) If, instead of one plane moving before the jet, a series of planes
are introduced at short intervals at the same point, the quantity of
water impinging on the series will be [omega]v instead of [omega](v -
u), and the whole pressure = (G/g)[omega]v(v - u). The work done is
(G/g)[omega]vu(v - u). The efficiency [eta] = (G/g)[omega]vu(v - u) ÷
(G/2g)[omega]v³ = 2u(v - u)/v². This becomes a maximum for d[eta]/du =
2(v - 2u) = 0, or u = ½v, and the [eta] = ½. This result is often used
as an approximate expression for the velocity of greatest efficiency
when a jet of water strikes the floats of a water wheel. The work
wasted in this case is half the whole energy of the jet when the
floats run at the best speed.
§ 156. (4) _Case of a Jet impinging on a Concave Cup Vane_, velocity
of water v, velocity of vane in the same direction u (fig. 155),
weight impinging per second = Gw(v - u).
[Illustration: FIG. 155.]
If the cup is hemispherical, the water leaves the cup in a direction
parallel to the jet. Its relative velocity is v - u when approaching
the cup, and -(v - u) when leaving it. Hence its absolute velocity
when leaving the cup is u - (v - u) = 2u - v. The change of momentum
per second = (G/g)[omega](v - u) {v - (2u - v)} = 2(G/g)[omega](v -
u)². Comparing this with case 2, it is seen that the pressure on a
hemispherical cup is double that on a flat plane. The work done on the
cup = 2(G/g)[omega] (v - u)²u foot-pounds per second. The efficiency
of the jet is greatest when v = 3u; in that case the efficiency =
{16/27}.
If a series of cup vanes are introduced in front of the jet, so that
the quantity of water acted upon is [omega]v instead of [omega](v -
u), then the whole pressure on the chain of cups is (G/g)[omega]v{v -
(2u - v)} = 2(G/g)[omega]v(v - u). In this case the efficiency is
greatest when v = 2u, and the maximum efficiency is unity, or all the
energy of the water is expended on the cups.
[Illustration: FIG. 156.]
§ 157. (5) _Case of a Flat Vane oblique to the Jet_ (fig. 156).--This
case presents some difficulty. The water spreading on the plane in all
directions from the point of impact, different particles leave the
plane with different absolute velocities. Let AB = v = velocity of
water, AC = u = velocity of plane. Then, completing the parallelogram,
AD represents in magnitude and direction the relative velocity of
water and plane. Draw AE normal to the plane and DE parallel to the
plane. Then the relative velocity AD may be regarded as consisting of
two components, one AE normal, the other DE parallel to the plane. On
the assumption that friction is insensible, DE is unaffected by
impact, but AE is destroyed. Hence AE represents the entire change of
velocity due to impact and the direction of that change. The pressure
on the plane is in the direction AE, and its amount is = mass of water
impinging per second × AE.
Let DAE = [theta], and let AD = v_r. Then AE = v_r cos [theta]; DE =
v_r sin [theta]. If Q is the volume of water impinging on the plane
per second, the change of momentum is (G/g)Qv_r cos [theta]. Let AC =
u = velocity of the plane, and let AC make the angle CAE = [delta]
with the normal to the plane. The velocity of the plane in the
direction AE = u cos [delta]. The work of the jet on the plane =
(G/g)Qv_r cos [theta] u cos [delta]. The same problem may be thus
treated algebraically (fig. 157). Let BAF = [alpha], and CAF =
[delta]. The velocity v of the water may be decomposed into AF = v cos
[alpha] normal to the plane, and FB = v sin [alpha] parallel to the
plane. Similarly the velocity of the plane = u = AC = BD can be
decomposed into BG = FE = u cos [delta] normal to the plane, and DG =
u sin [delta] parallel to the plane. As friction is neglected, the
velocity of the water parallel to the plane is unaffected by the
impact, but its component v cos [alpha] normal to the plane becomes
after impact the same as that of the plane, that is, u cos [delta].
Hence the change of velocity during impact = AE = v cos [alpha] - u
cos [delta]. The change of momentum per second, and consequently the
normal pressure on the plane is N = (G/g) Q(v cos [alpha] - u cos
[delta]). The pressure in the direction in which the plane is moving
is P = N cos [delta] = (G/g)Q (v cos [alpha] - u cos [delta]) cos
[delta], and the work done on the plane is Pu = (G/g)Q(v cos [alpha] -
u cos [delta]) u cos [delta], which is the same expression as before,
since AE = v_r cos [theta] = v cos [alpha] - u cos [delta].
[Illustration: FIG. 157.]
[Illustration: FIG. 158.]
In one second the plane moves so that the point A (fig. 158) comes to
C, or from the position shown in full lines to the position shown in
dotted lines. If the plane remained stationary, a length AB = v of the
jet would impinge on the plane, but, since the plane moves in the same
direction as the jet, only the length HB = AB - AH impinges on the
plane.
But AH = AC cos [delta]/ cos [alpha] = u cos [delta]/ cos [alpha], and
therefore HB = v - u cos [delta]/ cos [alpha]. Let [omega] = sectional
area of jet; volume impinging on plane per second = Q = [omega](v - u
cos [delta]/cos [alpha]) = [omega](v cos [alpha] - u cos [delta])/ cos
[alpha]. Inserting this in the formulae above, we get
G [omega]
N = --- ----------- (v cos [alpha] - u cos [delta])²; (1)
g cos [alpha]
G [omega] cos [delta]
P = --- ------------------- (v cos [alpha] - u cos [delta])²; (2)
g cos [alpha]
G cos [delta]
Pu = --- [omega]u ----------- (v cos [alpha] - u cos [delta])². (3)
g cos [alpha]