A HYDRAULIC MOTOR HAVING CENTRIFUGAL POWER
Year: 1876

page_text: PROVISIONAL SPECIFICATION left by Jean Antoine Alexis Cambon de
Lavalette at the Office of the Commissioners of Patents on the 28th June 1876.
JEAN ANTOINE ALEXIS CAMBON DE LAVALETTE, of Paris, in the Republic of
France. "A HYDRAULIC MOTOR HAVING CENTRIFUGAL POWER."
Amongst the prime movers used in manufactures water is one of the most
precious. To use the power which it developes it is generally sought to create
along watercourses falls, the action of which received by a driving engine is
transmitted to the various machines by which it is utilized.
This process has the defect of permitting the establishment of the hydraulic
machines only at certain fixed places, and the object of my Invention is to obviate
this and permit the use of hydraulic machines wherever it is possible to procure a
few cubic meters of water.
Supposing that I have in a tank about 20 cubic meters of water. With this
water I may, for example, create a waterfall of 15 meters in height, giving 800
litres of water per second (using the same water over and over again). I can create
this flow either by raising 800 litres of water to a height of 15 meters, or by forcing
this water into a receiver containing air compressed to one atmosphere and a half,
the water would then be under a pressure of 1k 500.
To raise 800 litres of water per second to a height of 15 meters there is required
10 800 1
Say I lead the water from the fall of 15 meters into a hollow horizontally placed
wheel R (see accompanying Drawings); this wheel has for radii or spokes two
cylinders also hollow of, say, 2m 25 in length, and closed by two valves; apply to
the wheel R a force x capable of causing it to make one revolution per second;
when the wheel turns in order that its speed at its circumference may be 14m 13
(its radius being 2m 25), at what speed must the water leave the cylinders? or
better, what will be the height of the column of water capable of producing such a
speed? When the wheel is at rest the clack valves are closed, and the pressure of
the column of water upon their inner surface will be 1 500 per square centimeter of
[Price 6d.]

page_text: 9 × R¹,
in
surface when the wheel turns, making one revolution per second, this pressure will
Px V2
become nine times greater, as I prove by the following formula, F=
which F represents in kilogrammes the pressure which will be exercised by the
water upon each square centimeter of surface of the valves; P is the weight in
kilogrammes of each square centimeter of the base of the column of water; V is the
speed in meters per second at the circumference of the wheel R.
When the wheel turns the pressure exercised upon each square centimeter of the 1
surface of the valves will be 13k 568. A pressure of 13k 568 per square centi-
meter corresponds to the weight of a column of water of 135m 68 in height. If
the valves are opened the water will escape at the same speed as if it fell from a
height of 135m 68.
The question is to utilize this artificial fall. For this purpose I replace the two 1
clack valves by two turbines (preferably the turbines known as "Fournogron's ").
These two turbines will have a same shaft; the pinion of this shaft will gear
directly with the pinion of another vertical shaft. certainly lose one revolution
As the shaft of the turbines
of the turbine shaft at each revolution of the wheel R.
will make at least 10 revolutions per second I lose 1/10th of the power; suppose even 2
that I lose / I then have a fall of water of 135m in height, giving 800 litres per
second, and upon the yield of this fall I lose 1 of its power. The yield of the
turbine will be 70°; the supply 800 litres per second; the height of the fall 135m.
0-70 x 800 × 135=75,000 k: g. m.
But as I lose 1/8 or 9450 I have
75,600-9540=66,150 k. g. m.=882 horse-power.
Now what power would it have been necessary to apply to the wheel to cause it
to make one revolution per second?
Taking the weight of the wheel with its turbines, the body of water it contains,
as well as the pressure exercised by the column of water, at 16,000 kilogrammes.
The pivot of the wheel has a diameter of 0m 20, its circumference is therefore 0m 62,
we have the formula T=FxPx V.
T is the work absorbed; F is the coefficient of the friction; P, the real pressure,
in kilogrammes; V, the speed of the pivot in meters per second.
T=0.14 x 1600 x 0.62=1388 k. g. m.
there still remains the friction of the shaft of the wheel in the stuffing box; the
circumference of this shaft is 2m 82, the thickness of the stuffing box 0m 10, and the
pressure exercised by the stuffing box upon the cylinder, is about 2k per squaro
centimeter of surface.
T=FxPx V=0·14 × 5640 × 2.82=2226 k. g. m.
The frictions will therefore be 1388 +2226=3614 k. g. m.
Because of certain frictions occasioned by the travelling of the water in the
cylinders I will take the whole of the frictions at 10,000 k. g. m.
Thus to produce a force of 66150 k. g. m. I shall have expended 15600+10,000
k. g. m., together 25600 k. g. m.
It is already a very good result to have been enabled to produce 2, with 1, but
for this purpose it is requisite to have recourse to a foreign motor. My object is to
dispense with this foreign motive power, and to use only the pressure of the water
to create the fall, set in motion the wheel R, and produce a useful working.

page_text: To attain this end I have a large cylindrical receiver with very thick sides,
strengthened upon its circumference by rings or hoops, and upon its two plane
faces by ribs. In this receiver I place the wheel R, upon which I replace the clack
valves. In the lower part of the receiver there will be water, and in the upper part
where the wheel is placed there will be air at a compression of 12 atmospheres,
which will press upon the water.
The cylinder which will act as a shaft, and by which the water will enter the
wheel, will be held by friction in a stuffing box. The pivot of the wheel will also
be held in a stuffing box, at the bottom of which there will be a step or socket,
upon which it may pivot; this socket will be open to the air.
As in the receiver there will be a powerful pressure which would act under the
wheel at the point corresponding to the shaft, by which the water will will enter
the wheel; it will be very easy to neutralize this pressure by giving to the pivot a
diameter which would be calculated according to the weight of the wheel and the
pressure of the air.
There only remains to overcome the friction of the shaft and of the pivot as
well as that resulting from the flow of the water in the cylinders, and finally the
resistance of the compressed air.
Taking all these frictions broadly at 14400 k. g. m. in order to drive 800 litres
of water per second into the receiver, there is expended-
1°. 15600 k. g. m. to raise the water to 15 metres.
2°. 14400 k. g. m. to cause the wheel to turn in the receiver with a speed of one
revolution per second.
Together there is expended 30,000 k. g. m. to drive 800 litres of water per
second into the receiver, where it will be at a pressure of 12k 396.
The wheel being at rest the pressure of the water upon the inner surface of the
clack valves would be 1k 500 per square centimeter of surface. The exterior pressure
of the compressed air upon these same clack valves would be 12k 396 per square
centimeter of surface.
Under the effect of the difference of pressure equal to 1k 172 per square centi-
meter of surface the valves would open, and the water would rush into the receiver
with a speed of 15 meters only, the reputed speed of the weight of a column of
water of 11m 72, represented by the pressure of 1m 172, which is the further difference
between the two pressures.
The water would continue to rush into the receiver so long as the pressure of the
compressed air was less than that of the column of water. This difference of
pressure would be maintained because there would issue from the receiver a volume
of water equal to that entering into it, and could be taken to keep the air always
at the same pressure, either by permitting the escape of the excess, or by intro-
ducing more. The water thus driven into the receiver would be at a pressure of
12 396 per square centimeter of surface, it might therefore rise to a height of 123m,
and fall from thence to produce a power.
Instead of causing it to rise to 123m, suppose I give two outlets to the receiver
the one giving out 350 litres, and producing a force of 30130 k. g. m. force which
would be applied to lift the water 15 meters, and to force it into the receiver by
causing the wheel to revolve. The second outlet giving 450 litres, and producing a
force of 38,745 k. g. m. or 5161/, horse-power; this power would be used to perform
the work for which the hydraulic motive-power engine will have been set up. It
will be understood that turbines would be adopted to the above-mentioned outlets.
To start this engine it will only be necessary to force into the receiver a greater
quantity of water than usual, and to further compress the air to the extent of one
atmosphere. This being done the conduit feeding the first turbine will be opened,
that is to say, the turbine, which will raise the water and set the wheel in motion;
directly the wheel is in motion, and as soon as the water rushes into the receiver,
the conduit of the second turbine will be opened, and a useful working will result.
It will then only be necessary to see that the pressure remains constant in the
receiver, and the machine will continue to work.

page_text: These machines might be easily fitted on board ship; they would occupy no more
space than steam engines, give as much power, save fuel, and permit cargo to be
carried in the great space now devoted to carrying fuel.
I may use leather boxed in recesses or other means than stuffing boxes for
preventing the passage of air, and thereby lessen the friction.
The accompanying Two Sheets of Drawing show a centrifugal force hydraulic
engine embodying my Invention.
Figure 1 (Sheet I.) is an elevation of the water wheel with pinion and driving
wheel attached, and Figure 2 is a section thereof drawn to an enlarged scale.
Hare the turbines; B¹, turbine shafts; B¹¹, vertical shaft communicating power
from the turbine shafts; C, bearing for shafts; C, driving wheel; I, strengthening
pieces or ribs; f, joints; B, pinion; g, socket; D, pivot; U, water reservoir.
Sheet II. shows the machine in section through receiver.
A, water tube; B, pinion driven by the toothed wheel C; R, water wheel;
D, pivot; E, receiver containing the water and the compressed air; a, tie-rods;
f, joint; g, socket receiving the pivot D; r, clack valves.