Page:EB1911 - Volume 14.djvu/101

pressure due to direct impulse is not obtained. If φ > 90°, cos φ is negative and an additional pressure due to reaction is obtained.

Fig. 164.

§ 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 Ω is the total sectional area of the jets, Ωv is the quantity of water discharged per second. The momentum generated per second in a sternward direction is (G/g) Ωv (v − V), and this is equal to the forward acting reaction P which propels the ship.

The energy carried away by the water

= 1/2 (G/g) Ωv (v − V)2.

(1)

The useful work done on the ship

PV = (G/g) Ωv (v − V) V.

(2)

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

W = 1/2 (G/g) Ωv (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.

Fig. 165.

§ 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 Ω is the section of the stream at A₀A₀ or A₂A₂, and ω the area of the plate CD, the area of the contracted section of the stream at A₁A₁ will be c_c(Ω − ω), where c_c is the coefficient of contraction. Hence, if v is the velocity at A₀A₀ or A₂A₂, and v₁ the velocity at A₁A₁,

vΩ = c_cv (Ω − ω);

∴ v₁ = vΩ / c_c (Ω − ω).

(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),

p0 − p2 =	G	${\Big (}$	Ω	− 1 ${\Big )}$	2	v2
	2g		cc (Ω − ω)

(3)

Now the external forces in the direction of motion acting on the mass A₀A₀A₂A₂ are the pressures p₀Ω₁ − p₂Ω 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) Ω − R = 0;

∴ R = GΩ ${\Big (}$	Ω	− 1 ${\Big )}$	2		v2	;
	cc (Ω − ω)				2g

(4)

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 Ω increases. If Ω/ω = ρ, the equation (4) becomes

R = Gω	v2	${\Big \{}$ ρ ${\Big (}$	ρ	− 1 ${\Big )}$	2	${\Big \}}$,
	2g		cc (ρ − 1)

(4a)

which is of the form

R = Gω (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

K = ρ ${\Big (}$ 1.176ρ/ρ − 1− 1 ${\Big )}$²

ρ =	K =
1	∞
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 ρ is probably only true when ρ is not very large. Further, the increase of K for large values of ρ 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 = KGωv² / 2g,

K must be determined by experiment in each special case. For a cylindrical body putting ω for the section, c_c for the coefficient of contraction, c_c (Ω − ω) for the area of the stream at A₁A₁,

v₁ = vΩ / c_c (Ω − ω); v₂ = vΩ / (Ω −ω);

or, putting ρ = Ω/ω,

v₁ = vρ / c_c (ρ − 1), v₂ = vρ / (ρ − 1).

Then

R = K₁Gωv² / 2g,

where

K1 = ρ ${\Big \{}{\Big (}$

ρ

${\Big )}$

2

${\Big (}$

1

− 1 ${\Big )}$

2

+ ${\Big (}$

ρ

− 1 ${\Big )}$

2

${\Big \}}$.

ρ − 1

cc

ρ − 1

Taking c_c = 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);

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