Chapter XII: Impact and Reaction of Water (1)

§ 153. When a stream of fluid in steady motion impinges on a solid
surface, it presses on the surface with a force equal and opposite to
that by which the velocity and direction of motion of the fluid are
changed. Generally, in problems on the impact of fluids, it is
necessary to neglect the effect of friction between the fluid and the
surface on which it moves.

_During Impact the Velocity of the Fluid relatively to the Surface on
which it impinges remains unchanged in Magnitude._--Consider a mass of
fluid flowing in contact with a solid surface also in motion, the
motion of both fluid and solid being estimated relatively to the
earth. Then the motion of the fluid may be resolved into two parts,
one a motion equal to that of the solid, and in the same direction,
the other a motion relatively to the solid. The motion which the fluid
has in common with the solid cannot at all be influenced by the
contact. The relative component of the motion of the fluid can only be
altered in direction, but not in magnitude. The fluid moving in
contact with the surface can only have a relative motion parallel to
the surface, while the pressure between the fluid and solid, if
friction is neglected, is normal to the surface. The pressure
therefore can only deviate the fluid, without altering the magnitude
of the relative velocity. The unchanged common component and, combined
with it, the deviated relative component give the resultant final
velocity, which may differ greatly in magnitude and direction from the
initial velocity.

From the principle of momentum, the impulse of any mass of fluid
reaching the surface in any given time is equal to the change of
momentum estimated in the same direction. The pressure between the
fluid and surface, in any direction, is equal to the change of
momentum in that direction of so much fluid as reaches the surface in
one second. If P_a is the pressure in any direction, m the mass of
fluid impinging per second, v_a the change of velocity in the
direction of P_a due to impact, then

P_a = mv_a.

If v1 (fig. 152) is the velocity and direction of motion before
impact, v2 that after impact, then v is the total change of motion due
to impact. The resultant pressure of the fluid on the surface is in
the direction of v, and is equal to v multiplied by the mass impinging
per second. That is, putting P for the resultant pressure,

P = mv.

Let P be resolved into two components, N and T, normal and tangential
to the direction of motion of the solid on which the fluid impinges.
Then N is a lateral force producing a pressure on the supports of the
solid, T is an effort which does work on the solid. If u is the
velocity of the solid, Tu is the work done per second by the fluid in
moving the solid surface.

Let Q be the volume, and GQ the weight of the fluid impinging per
second, and let v1 be the initial velocity of the fluid before
striking the surface. Then GQv1²/2g is the original kinetic energy of
Q cub. ft. of fluid, and the efficiency of the stream considered as an
arrangement for moving the solid surface is

[eta] = Tu/(GQv1²/2g).

§ 154. _Jet deviated entirely in one Direction.--Geometrical Solution_
(fig. 153).--Suppose a jet of water impinges on a surface ac with a
velocity ab, and let it be wholly deviated in planes parallel to the
figure. Also let ae be the velocity and direction of motion of the
surface. Join eb; then the water moves with respect to the surface in
the direction and with the velocity eb. As this relative velocity is
unaltered by contact with the surface, take cd = eb, tangent to the
surface at c, then cd is the relative motion of the water with respect
to the surface at c. Take df equal and parallel to ae. Then fc
(obtained by compounding the relative motion of water to surface and
common velocity of water and surface) is the absolute velocity and
direction of the water leaving the surface. Take ag equal and parallel
to fc. Then, since ab is the initial and ag the final velocity and
direction of motion, gb is the total change of motion of the water.
The resultant pressure on the plane is in the direction gb. Join eg.
In the triangle gae, ae is equal and parallel to df, and ag to fc.
Hence eg is equal and parallel to cd. But cd = eb = relative motion of
water and surface. Hence the change of motion of the water is
represented in magnitude and direction by the third side of an
isosceles triangle, of which the other sides are equal to the relative
velocity of the water and surface, and parallel to the initial and
final directions of relative motion.

SPECIAL CASES

§ 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.

(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).

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.

§ 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].

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]

Three cases may be distinguished:--

(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.

§ 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_.

§ 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.

_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]);

L = R cos ½[phi] = (G/g) bt (v - u)² sin [phi].

The work done by the jet on the vane is Pu = (G/g)btu(v - u)²(1 - cos
[phi]), which is a maximum when u = 1/3 v. This result can also be
obtained by considering that the work done on the plane must be equal
to the energy lost by the water, when friction is neglected.

If [phi] = 180°, cos [phi] = -1, 1 - cos [phi] = 2; then P =
2(G/g)bt(v - u)², the same result as for a concave cup.

§ 161. _Position which a Movable Plane takes in Flowing Water._--When
a rectangular plane, movable about an axis parallel to one of its
sides, is placed in an indefinite current of fluid, it takes a
position such that the resultant of the normal pressures on the two
sides of the axis passes through the axis. If, therefore, planes
pivoted so that the ratio a/b (fig. 163) is varied are placed in
water, and the angle they make with the direction of the stream is
observed, the position of the resultant of the pressures on the plane
is determined for different angular positions. Experiments of this
kind have been made by Hagen. Some of his results are given in the
following table:--

+-----------+-------------+--------------+
| |Larger plane.|Smaller Plane.|
+-----------+-------------+--------------+
| a/b = 1.0 |[phi] = ... |[phi] = 90° |
| 0.9 | 75° | 72½° |
| 0.8 | 60° | 57° |
| 0.7 | 48° | 43° |
| 0.6 | 25° | 29° |
| 0.5 | 13° | 13° |
| 0.4 | 8° | 6½° |
| 0.3 | 6° | .. |
| 0.2 | 4° | .. |
+-----------+-------------+--------------+

§ 162. _Direct Action distinguished from Reaction_ (Rankine, _Steam
Engine_, § 147).

The pressure which a jet exerts on a vane can be distinguished into
two parts, viz.:--

(1) The pressure arising from changing the direct component of the
velocity of the water into the velocity of the vane. In fig. 153, §
154, ab cos bae is the direct component of the water's velocity, or
component in the direction of motion of vane. This is changed into the
velocity ae of the vane. The pressure due to direct impulse is then

P1 = GQ(ab cos bae - ae)/g.

For a flat vane moving normally, this direct action is the only action
producing pressure on the vane.

(2) The term reaction is applied to the additional action due to the
direction and velocity with which the water glances off the vane. It
is this which is diminished by the friction between the water and the
vane. In Case 2, § 160, the direct pressure is

P1 = Gbt(v - u)²/g.

That due to reaction is

P2 = -Gbt(v - u)² cos [phi]/g.

If [phi] < 90°, the direct component of the water's motion is not
wholly converted into the velocity of the vane, and the whole
pressure due to direct impulse is not obtained. If [phi] > 90°, cos
[phi] is negative and an additional pressure due to reaction is
obtained.

§ 163. _Jet Propeller._--In the case of vessels propelled by a jet of
water (fig. 164), driven sternwards from orifices at the side of the
vessel, the water, originally at rest outside the vessel, is drawn
into the ship and caused to move with the forward velocity V of the
ship. Afterwards it is projected sternwards from the jets with a
velocity v relatively to the ship, or v - V relatively to the earth.
If [Omega] is the total sectional area of the jets, [Omega]v is the
quantity of water discharged per second. The momentum generated per
second in a sternward direction is (G/g)[Omega]v(v - V), and this is
equal to the forward acting reaction P which propels the ship.

The energy carried away by the water

= ½(G/g)[Omega]v (v - V)². (1)

The useful work done on the ship

PV = (G/g)[Omega]v (v - V)V. (2)

Adding (1) and (2), we get the whole work expended on the water,
neglecting friction:--

W = ½(G/g)[Omega]v (v² - V²).

Hence the efficiency of the jet propeller is

PV/W = 2V/(v + V). (3)

This increases towards unity as v approaches V. In other words, the
less the velocity of the jets exceeds that of the ship, and therefore
the greater the area of the orifice of discharge, the greater is the
efficiency of the propeller.

In the "Waterwitch" v was about twice V. Hence in this case the
theoretical efficiency of the propeller, friction neglected, was about
2/3.

§ 164. _Pressure of a Steady Stream in a Uniform Pipe on a Plane
normal to the Direction of Motion._--Let CD (fig. 165) be a plane
placed normally to the stream which, for simplicity, may be supposed
to flow horizontally. The fluid filaments are deviated in front of the
plane, form a contraction at A1A1, and converge again, leaving a mass
of eddying water behind the plane. Suppose the section A0A0 taken at a
point where the parallel motion has not begun to be disturbed, and
A2A2 where the parallel motion is re-established. Then since the same
quantity of water with the same velocity passes A0A0, A2A2 in any
given time, the external forces produce no change of momentum on the
mass A0A0A2A2, and must therefore be in equilibrium. If [Omega] is the
section of the stream at A0A0 or A2A2, and [omega] the area of the
plate CD, the area of the contracted section of the stream at A1A1
will be c_c([Omega] - [omega]), where c_c is the coefficient of
contraction. Hence, if v is the velocity at A0A0 or A2A2, and v1 the
velocity at A1A1,

v[Omega] = c_c v([Omega] - [omega]);

.:v1 = v[Omega]/c_c ([Omega] - [omega]). (1)

Let p0, p1, p2 be the pressures at the three sections. Applying
Bernoulli's theorem to the sections A0A0 and A1A1,

p0 v² p1 v1²
-- + -- = -- + ---.
G 2g G 2g

Also, for the sections A1A1 and A2A2, allowing that the head due to
the relative velocity v1 - v is lost in shock:--

p1 v1² p2 v² (v1 - v)²
-- + --- = -- + -- + ---------;
G 2g G 2g 2g

.: p0 - p2 = G(v1 - v)²/2g; (2)

or, introducing the value in (1),

G / [Omega] \²
p0 - p2 = -- ( ----------------------- - 1 ) v² (3)
2g \c_c ([Omega] - [omega]) /

Now the external forces in the direction of motion acting on the mass
A0A0A2A2 are the pressures p0[Omega]1 - p2[Omega] at the ends, and the
reaction -R of the plane on the water, which is equal and opposite to
the pressure of the water on the plane. As these are in equilibrium,

(p0 - p2)[Omega] - R = 0;

/ [Omega] \² v²
.: R = G[Omega] ( ----------------------- - 1 ) --; (4)
\c_c ([Omega] - [omega]) / 2g

an expression like that for the pressure of an isolated jet on an
indefinitely extended plane, with the addition of the term in
brackets, which depends only on the areas of the stream and the plane.
For a given plane the expression in brackets diminishes as [Omega]
increases. If [Omega]/[omega] = [rho], the equation (4) becomes
_ _
v² | / [rho] \² |
R = G[omega] -- |[rho] ( --------------- - 1 ) |, (4a)
2g |_ \c_c ([rho] - 1) / _|

which is of the form

R = G[omega](v²/2g)K,

where K depends only on the ratio of the sections of the stream and
plane.

For example, let c_c = 0.85, a value which is probable, if we allow
that the sides of the pipe act as internal borders to an orifice. Then

/ [rho] \²
K = [rho] ( 1.176 --------- - 1 ).
\ [rho] - 1 /

[rho] = K =

1 [infinity]
2 3.66
3 1.75
4 1.29
5 1.10
10 .94
50 2.00
100 3.50

The assumption that the coefficient of contraction c_c is constant for
different values of [rho] is probably only true when [rho] is not very
large. Further, the increase of K for large values of [rho] is
contrary to experience, and hence it may be inferred that the
assumption that all the filaments have a common velocity v1 at the
section A1A1 and a common velocity v at the section A2A2 is not true
when the stream is very much larger than the plane. Hence, in the
expression

R = KG[omega]v²/2g,

K must be determined by experiment in each special case. For a
cylindrical body putting [omega] for the section, c_c for the
coefficient of contraction, c_c([Omega] - [omega]) for the area of the
stream at A1A1,

v1 = v[Omega]/c_c([Omega] - [omega]); v2 = v[Omega]/([Omega] - [omega]);

or, putting [rho] = [Omega]/[omega],

v1 = v[rho]/c_c ([rho] - 1), v2 = v[rho]/([rho] - 1).

Then

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.

§ 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.

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;

V = 2[omega] [root](hh1).

We have already, therefore, two conditions to be satisfied by the
pressure curve.

Some very interesting experiments on the distribution of pressure on a
surface struck by a jet have been made by J. S. Beresford (_Prof.
Papers on Indian Engineering_, No. cccxxii.), with a view to afford
information as to the forces acting on the aprons of weirs.
Cylindrical jets ½ in. to 2 in. diameter, issuing from a vessel in
which the water level was constant, were allowed to fall vertically on
a brass plate 9 in. in diameter. A small hole in the brass plate
communicated by a flexible tube with a vertical pressure column.
Arrangements were made by which this aperture could be moved 1/20
in. at a time across the area struck by the jet. The height of the
pressure column, for each position of the aperture, gave the pressure
at that point of the area struck by the jet. When the aperture was
exactly in the axis of the jet, the pressure column was very nearly
level with the free surface in the reservoir supplying the jet; that
is, the pressure was very nearly v²/2g. As the aperture moved away
from the axis of the jet, the pressure diminished, and it became
insensibly small at a distance from the axis of the jet about equal to
the diameter of the jet. Hence, roughly, the pressure due to the jet
extends over an area about four times the area of section of the jet.

Fig. 168 shows the pressure curves obtained in three experiments with
three jets of the sizes shown, and with the free surface level in the
reservoir at the heights marked.

+------------------------------------------------------+
| Experiment 1. Jet .475 in. diameter. |
+----------------+------------------+------------------+
| Height from | Distance from | |
| Free Surface | Axis of Jet | Pressure in. |
| to Brass Plate | in inches. | inches of Water. |
| in inches. | | |
+----------------+------------------+------------------+
| 43 | 0 | 40.5 |
| " | .05 | 39.40 |
| " | .1 | 37.5-39.5 |
| " | .15 | 35 |
| " | .2 | 33.5-37 |
| " | .25 | 31 |
| " | .3 | 21-27 |
| " | .35 | 21 |
| " | .4 | 14 |
| " | .45 | 8 |
| " | .5 | 3.5 |
| " | .55 | 1 |
| " | .6 | 0.5 |
| " | .65 | 0 |
+----------------+------------------+------------------+
| Experiment 2. Jet .988 in. diameter. |
+----------------+------------------+------------------+
| 42.15 | 0 | 42 |
| " | .05 | 41.9 |
| " | .1 | 41.5-41.8 |
| " | .15 | 41 |
| " | .2 | 40.3 |
| " | .25 | 39.2 |
| " | .3 | 37.5 |
| " | .35 | 34.8 |
| " | .45 | 27 |
| 42.25 | .5 | 23 |
| " | .55 | 18.5 |
| " | .6 | 13 |
| " | .65 | 8.3 |
| " | .7 | 5 |
| " | .75 | 3 |
| " | .8 | 2.2 |
| 42.15 | .85 | 1.6 |
| " | .95 | 1 |
+----------------+------------------+------------------+
| Experiment 3. Jet 19.5 in. diameter. |
+----------------+------------------+------------------+
| 27.15 | 0 | 26.9 |
| " | .08 | 26.9 |
| " | .13 | 26.8 |
| " | .18 | 26.5-26.6 |
| " | .23 | 26.4-26.5 |
| " | .28 | 26.3-26.6 |
| 27 | .33 | 26.2 |
| " | .38 | 25.9 |
| " | .43 | 25.5 |
| " | .48 | 25 |
| " | .53 | 24.5 |
| " | .58 | 24 |
| " | .63 | 23.3 |
| " | .68 | 22.5 |
| " | .73 | 21.8 |
| " | .78 | 21 |
| " | .83 | 20.3 |
| " | .88 | 19.3 |
| " | .93 | 18 |
| " | .98 | 17 |
| 26.5 | 1.13 | 13.5 |
| " | 1.18 | 12.5 |
| " | 1.23 | 10.8 |
| " | 1.28 | 9.5 |
| " | 1.33 | 8 |
| " | 1.38 | 7 |
| " | 1.43 | 6.3 |
| " | 1.48 | 5 |
| " | 1.53 | 4.3 |
| " | 1.58 | 3.5 |
| " | 1.9 | 2 |
+----------------+------------------+------------------+

As the general form of the pressure curve has been already indicated,
it may be assumed that its equation is of the form

y = ab^(-x²).

But it has already been shown that for x = 0, y = h, hence a = h. To
determine the remaining constant, the other condition may be used,
that the solid formed by rotating the pressure curve represents the
total pressure on the plane. The volume of the solid is
_
/[oo]
V = | 2[pi]xy dx
_/0
_
/[oo]
= 2[pi]h | b^(-x²)x dx
_/0
_ _
| |[oo]
= ([pi]h/log_eb) |-b^(-x²)|
|_ _|0

= [pi]h/log_e b.

Using the condition already stated,

2[omega] [root](hh1) = [pi]h/log_e b,

log_e b = ([pi]/2[omega]) [root](h/h1).

Putting the value of b in (2) in eq. (1), and also r for the radius of
the jet at the orifice, so that [omega] = [pi]r², the equation to the
pressure curve is

h x²
y = h[epsilon]^(-½) [root]-- --.
h1 r²

§ 166. _Resistance of a Plane moving through a Fluid, or Pressure of a
Current on a Plane._--When a thin plate moves through the air, or
through an indefinitely large mass of still water, in a direction
normal to its surface, there is an excess of pressure on the anterior
face and a diminution of pressure on the posterior face. Let v be the
relative velocity of the plate and fluid, [Omega] the area of the
plate, G the density of the fluid, h the height due to the velocity,
then the total resistance is expressed by the equation

R = fG[Omega]v²/2g pounds = fG[Omega]h;

where f is a coefficient having about the value 1.3 for a plate moving
in still fluid, and 1.8 for a current impinging on a fixed plane,
whether the fluid is air or water. The difference in the value of the
coefficient in the two cases is perhaps due to errors of experiment.
There is a similar resistance to motion in the case of all bodies of "
_unfair_ " form, that is, in which the surfaces over which the water
slides are not of gradual and continuous curvature.

The stress between the fluid and plate arises chiefly in this way.
The streams of fluid deviated in front of the plate, supposed for
definiteness to be moving through the fluid, receive from it forward
momentum. Portions of this forward moving water are thrown off
laterally at the edges of the plate, and diffused through the
surrounding fluid, instead of falling to their original position
behind the plate. Other portions of comparatively still water are
dragged into motion to fill the space left behind the plate; and there
is thus a pressure less than hydrostatic pressure at the back of the
plate. The whole resistance to the motion of the plate is the sum of
the excess of pressure in front and deficiency of pressure behind.
This resistance is independent of any friction or viscosity in the
fluid, and is due simply to its inertia resisting a sudden change of
direction at the edge of the plate.

Experiments made by a whirling machine, in which the plate is fixed on
a long arm and moved circularly, gave the following values of the
coefficient _f_. The method is not free from objection, as the
centrifugal force causes a flow outwards across the plate.

+---------------+------------------------+
| Approximate | Values of f. |
| Area of Plate +------+-------+---------+
| in sq. ft. |Borda.|Hutton.|Thibault.|
+---------------+------+-------+---------+
| 0.13 | 1.39 | 1.24 | .. |
| 0.25 | 1.49 | 1.43 | 1.525 |
| 0.63 | 1.64 | .. | .. |
| 1.11 | .. | .. | 1.784 |
+---------------+------+-------+---------+

There is a steady increase of resistance with the size of the plate,
in part or wholly due to centrifugal action.

P. L. G. Dubuat (1734-1809) made experiments on a plane 1 ft. square,
moved in a straight line in water at 3 to 6½ ft. per second. Calling m
the coefficient of excess of pressure in front, and n the coefficient
of deficiency of pressure behind, so that f = m + n, he found the
following values:--

m = 1; n = 0.433; f = 1.433.

The pressures were measured by pressure columns. Experiments by A. J.
Morin (1795-1880), G. Piobert (1793-1871) and I. Didion (1798-1878) on
plates of 0.3 to 2.7 sq. ft. area, drawn vertically through water,
gave f = 2.18; but the experiments were made in a reservoir of
comparatively small depth. For similar plates moved through air they
found f = 1.36, a result more in accordance with those which precede.

For a fixed plane in a moving current of water E. Mariotte found f =
1.25. Dubuat, in experiments in a current of water like those
mentioned above, obtained the values m = 1.186; n = 0.670; f = 1.856.
Thibault exposed to wind pressure planes of 1.17 and 2.5 sq. ft. area,
and found f to vary from 1.568 to 2.125, the mean value being f =
1.834, a result agreeing well with Dubuat.

§ 167. _Stanton's Experiments on the Pressure of Air on Surfaces._--At
the National Physical Laboratory, London, T. E. Stanton carried out a
series of experiments on the distribution of pressure on surfaces in a
current of air passing through an air trunk. These were on a small
scale but with exceptionally accurate means of measurement. These
experiments differ from those already given in that the plane is small
relatively to the cross section of the current (_Proc. Inst. Civ.
Eng._ clvi., 1904). Fig. 169 shows the distribution of pressure on a
square plate. ab is the plate in vertical section. acb the
distribution of pressure on the windward and adb that on the leeward
side of the central section. Similarly aeb is the distribution of
pressure on the windward and afb on the leeward side of a diagonal
section. The intensity of pressure at the centre of the plate on the
windward side was in all cases p = Gv²/2g lb. per sq. ft., where G is
the weight of a cubic foot of air and v the velocity of the current in
ft. per sec. On the leeward side the negative pressure is uniform
except near the edges, and its value depends on the form of the plate.
For a circular plate the pressure on the leeward side was 0.48 Gv²/2g
and for a rectangular plate 0.66 Gv²/2g. For circular or square plates
the resultant pressure on the plate was P = 0.00126 v² lb. per sq. ft.
where v is the velocity of the current in ft. per sec. On a long
narrow rectangular plate the resultant pressure was nearly 60% greater
than on a circular plate. In later tests on larger planes in free air,
Stanton found resistances 18% greater than those observed with small
planes in the air trunk.

§ 168. _Case when the Direction of Motion is oblique to the
Plane._--The determination of the pressure between a fluid and surface
in this case is of importance in many practical questions, for
instance, in assigning the load due to wind pressure on sloping and
curved roofs, and experiments have been made by Hutton, Vince, and
Thibault on planes moved circularly through air and water on a
whirling machine.

Let AB (fig. 170) be a plane moving in the direction R making an angle
[phi] with the plane. The resultant pressure between the fluid and the
plane will be a normal pressure N. The component R of this normal
pressure is the resistance to the motion of the plane and the other
component L is a lateral force resisted by the guides which support
the plane. Obviously

R = N sin [phi];

L = N cos [phi].

In the case of wind pressure on a sloping roof surface, R is the
horizontal and L the vertical component of the normal pressure.

In experiments with the whirling machine it is the resistance to
motion, R, which is directly measured. Let P be the pressure on a
plane moved normally through a fluid. Then, for the same plane
inclined at an angle [phi] to its direction of motion, the resistance
was found by Hutton to be

R = P(sin [phi])^{1.842 cos [phi]}.

A simpler and more convenient expression given by Colonel Duchemin is

R = 2P sin² [phi]/(1 + sin² [phi]).

Consequently, the total pressure between the fluid and plane is

N = 2P sin [phi]/(1 + sin² [phi]) = 2P/(cosec [phi] + sin [phi]),

and the lateral force is

L = 2P sin [phi] cos [phi]/(1 + sin² [phi]).

In 1872 some experiments were made for the Aeronautical Society on the
pressure of air on oblique planes. These plates, of 1 to 2 ft. square,
were balanced by ingenious mechanism designed by F. H. Wenham and
Spencer Browning, in such a manner that both the pressure in the
direction of the air current and the lateral force were separately
measured. These planes were placed opposite a blast from a fan issuing
from a wooden pipe 18 in. square. The pressure of the blast varied
from 6/10 to 1 in. of water pressure. The following are the results
given in pounds per square foot of the plane, and a comparison of the
experimental results with the pressures given by Duchemin's rule.
These last values are obtained by taking P = 3.31, the observed
pressure on a normal surface:--

+-----------------------------------+-------+-------+-------+------+
| Angle between Plane and Direction | 15° | 20° | 60° | 90° |
| of Blast | | | | |
+-----------------------------------+-------+-------+-------+------+
| Horizontal pressure R | 0.4 | 0.61 | 2.73 | 3.31 |
| Lateral pressure L | 1.6 | 1.96 | 1.26 | .. |
| Normal pressure [root](L² + R²) | 1.65 | 2.05 | 3.01 | 3.31 |
| Normal pressure by Duchemin's rule| 1.605 | 2.027 | 3.276 | 3.31 |
+-----------------------------------+-------+-------+-------+------+

WATER MOTORS

In every system of machinery deriving energy from a natural waterfall there exist the following parts:--

1. A supply channel or head race, leading the water from the highest accessible level to the site of the machine. This may be an open channel of earth, masonry or wood, laid at as small a slope as is consistent with the delivery of the necessary supply of water, or it may be a closed cast or wrought-iron pipe, laid at the natural slope of the ground, and about 3 ft. below the surface. In some cases part of the head race is an open channel, part a closed pipe. The channel often starts from a small storage reservoir, constructed near the stream supplying the water motor, in which the water accumulates when the motor is not working. There are sluices or penstocks by which the supply can be cut off when necessary.

2. Leading from the motor there is a tail race, culvert, or discharge pipe delivering the water after it has done its work at the lowest convenient level.

3. A waste channel, weir, or bye-wash is placed at the origin of the head race, by which surplus water, in floods, escapes.

4. The motor itself, of one of the kinds to be described presently, which either overcomes a useful resistance directly, as in the case of a ram acting on a lift or crane chain, or indirectly by actuating transmissive machinery, as when a turbine drives the shafting, belting and gearing of a mill. With the motor is usually combined regulating machinery for adjusting the power and speed to the work done. This may be controlled in some cases by automatic governing machinery.

§ 169. _Water Motors with Artificial Sources of Energy._--The great convenience and simplicity of water motors has led to their adoption in certain cases, where no natural source of water power is available. In these cases, an artificial source of water power is created by using a steam-engine to pump water to a reservoir at a great elevation, or to pump water into a closed reservoir in which there is great pressure. The water flowing from the reservoir through hydraulic engines gives back the energy expended, less so much as has been wasted by friction. Such arrangements are most useful where a continuously acting steam engine stores up energy by pumping the water, while the work done by the hydraulic engines is done intermittently.

§ 170. _Energy of a Water-fall._--Let H_t be the total fall of level
from the point where the water is taken from a natural stream to the
point where it is discharged into it again. Of this total fall a
portion, which can be estimated independently, is expended in
overcoming the resistances of the head and tail races or the supply
and discharge pipes. Let this portion of head wasted be [h]_r. Then
the available head to work the motor is H = H_t - [h]_r. It is this
available head which should be used in all calculations of the
proportions of the motor. Let Q be the supply of water per second.
Then GQH foot-pounds per second is the gross available work of the
fall. The power of the fall may be utilized in three ways. (a) The GQ
pounds of water may be placed on a machine at the highest level, and
descending in contact with it a distance of H ft., the work done will
be (neglecting losses from friction or leakage) GQH foot-pounds per
second. (b) Or the water may descend in a closed pipe from the higher
to the lower level, in which case, with the same reservation as
before, the pressure at the foot of the pipe will be p = GH pounds per
square foot. If the water with this pressure acts on a movable piston
like that of a steam engine, it will drive the piston so that the
volume described is Q cubic feet per second. Then the work done will
be pQ = GHQ foot-pounds per second as before. (c) Or lastly, the water
may be allowed to acquire the velocity v = [root](2gH) by its descent.
The kinetic energy of Q cubic feet will then be ½GQv²/g = GQH, and if
the water is allowed to impinge on surfaces suitably curved which
bring it finally to rest, it will impart to these the same energy as
in the previous cases. Motors which receive energy mainly in the three
ways described in (a), (b), (c) may be termed gravity, pressure and
inertia motors respectively. Generally, if Q ft. per second of water
act by weight through a distance h1, at a pressure p due to h2 ft. of
fall, and with a velocity v due to h3 ft. of fall, so that h1 + h2 +
h3 = H, then, apart from energy wasted by friction or leakage or
imperfection of the machine, the work done will be

GQh1 + pQ + (G/g) Q (v²/2g) = GQH foot pounds,

the same as if the water acted simply by its weight while descending H
ft.