CA2387387A1

CA2387387A1 - Method of producing energy

Info

Publication number: CA2387387A1
Application number: CA002387387A
Authority: CA
Inventors: Richard Bouthillier
Original assignee: Individual
Current assignee: Individual
Priority date: 2002-05-22
Filing date: 2002-05-22
Publication date: 2003-11-22

Description

METHOD OF PRODUCING ENERGY
The present invention relates to a method for the production of energy and an apparatus therefore.
There are many different types of energy generating methods and apparati. All of the energy generating methods are based on the laws of physics. Thus, hydroelectricity uses the force of gravity wherein water falls in order to turn a turbine.
Internal combustion engines, coal operated power stations and nuclear power plants use the force of expansion to produce a movement. In the case of an internal combustion engine, the expansion of the gases produced by the burning of the gas pushes on the pistons. A coal or nuclear power plant heats a fluid (water).
The vapor thus produced uses a pressure to turn turbines.
It is an object of the present invention to provide a method for generating energy.
It is a further object of the present invention to provide an apparatus suitable for generating energy.
According to the present invention, there is provided a method for generating energy which comprises using a plurality of inflatable balloons or members mounted on a wheel immersed in a liquid.
The invention will best be understood by having reference to the accompanying drawings, in which:
Figure 1 is a schematic view of a conventional waterwheel for generating power using gravity;
Figure 2 is a schematic side elevational view of a power generating device according to the present invention;

Figure 3 is a detailed side view of one of the balloons or compartments thereof;
and Figure 4 is a schematic side elevational view similar to Figure 2 showing calculations for a generation of energy.
Turning initially to Figure 1 which illustrates a well known waterwheel, and which waterwheel is generally designated by reference numeral 10, there are provided a plurality of vanes 12 which define pockets to receive water from an input 14 thereof. A
weight on one side of the waterwheel causes the same to rotate until the water exits at point 16. The rotational movement may be utilized to generate power in a known manner.
Turning to the present invention, and as illustrated in Figure 2, there is provided the energy generating device or apparatus according to the present invention and which will be referred to as a traction wheel and which is generally designated by reference numeral 18.
Traction wheel 18 has a center shaft or hub 22 and is rotatable thereabout.
Traction wheel 18 is shown mounted within a container 20 filled with a liquid such as water.
Traction wheel 18 includes a plurality of expandable pockets or balloons generally designated by reference numeral 24. At the end of each balloon 24, there is provided a wheel or rotatable member 26.
Interconnecting each balloon or compartment 24 are a plurality of ducts 28 as may be best seen in Figure 3. Each duct 28 connects a diametrically opposed compartment to permit the passage of air therethrough.

-2-Each balloon or compartment 24 has a relatively rigid base 30 and which is further defined by a covering which is formed of a flexible but non extensible material.
Thus, each compartment is closed and can only communicate with a diametrically opposed compartment by means of duct 28.
As seen in Figure 2, there is provided a compression guide 32 which is designed to gradually compress each of the balloons or compartments 24 as traction wheel 18 rotates in a clockwise direction.
Figure 1 shows a conventional paddle wheel. Its way of functioning is simple.
The force of this paddle wheel is calculated mainly by the weight of the water volume which is kept in each paddle. Naturally, the rotational force of each paddle is different depending on the position where it is located. This rotational force has two components, one acting downwardly and the other producing the rotating movement. The separation of this force is calculated by using sine and cosine vectors depending on the angle of the paddle vs the center of the rotation. The sine vector is the downward force which is a loss and the cosine is the portion of the force which produces the rotation of the paddle wheel. This movement can then be translated into usable energy. It is possible to increase the force of the paddle wheel by:
- Increasing the weight of the water contained in each paddle by increasing the volume of water contained in each paddle.
- Increasing the number of paddles to the wheel. We thus increase the total weight of the water in the wheel.
- Increasing the diameter of the paddle wheel, which has the effect of lengthening the "arm" of the lever and thus increase the rotation force.

-3-- Producing a combination of the previous 3 solutions.
The paddle wheel principle utilizes the law of gravity to transform the energy stored in a container or in a flowing current into a useful energy. In order to use the density force, one must reverse the whole process. The wheel will be plunged into a water container and the paddles will be filled with air.
The wheel 18 is now immersed into a liquid container 20 (for example purposes, the liquid will be water).The paddles are no longer open containers containing water, but rather balloons 24 filled with air (or gas) and which are of a circular configuration which fits well the shape of the wheel. The paddles are air tight, supple, non elastic. The top surface of the balloon is formed of a rigid material so as to keep the shape of the balloon and thus facilitate and maximize the expulsion of the air. A compression guide 32 expels the air from the balloon above the wheel so as to push it into the opposed balloon located at the bottom of the traction wheel. Wheels 26 are mounted on each side of the rigid portion of the balloon so as to reduce the rubbing against the compression guide.
The density force is a physics law which is defined as follows ... any body plunged into a liquid gets from the liquid an upward push equal to the weight of the volume of the body moved by this body.
If we take the example of the paddle wheel, if a paddle contains one cubic foot of water, i.e. a weight of 62.23 lbs, and which, by gravity force, makes the wheel turn so as to direct this weight towards the ground. In the present invention, according to the physics law of density, a one cubic foot balloon of air plunged into a water container gets an ascending force equal to the weight of the volume of the liquid moved. The volume moved in our example is of one cubic foot and the liquid is water, so the weight of the

-4-volume of water moved is 62.23 lbs. The force of the paddle wheel and that of the traction wheel are identical. It will be understood that the rotational movement can be translated into usable energy. If not, one will increase the force of the traction wheel by:
- Increasing the volume of water moved by increasing the volume of air contained in each balloon. Weight according to the volume of water moved.
- Adding balloons to the wheel. One increases the total volume of water moved.
- Increasing the diameter of the paddle wheel which has an effect of lengthening the arm of the lever and thus increasing the rotation force.
- Adding on the same drive shaft many traction wheels.
- Producing a combination of the previous 4 solutions.
One may thus observe the similarities between the paddle wheel and the traction wheel at its start position but the movement must be continued.
Figure 3 is an enlarged view of a portion of Figure 2 and shows that the spokes of the traction wheel are ducts 28 which join a balloon with its opposed balloon.
This pipe permits the air to move from one balloon to the other.
The balloon 24 is one of the most important parts of the system. It must be formed of a very flexible, non extensible material. The flexibility is useful during the change of volume that the balloon must undergo during these two phases, either in maximal volume or in minimal volume. It must transform itself with getting damaged.
The most important priority in its construction is that it must not be formed with a stretchy or elastic material so as to maximize the productivity. The importance of the non-extensibility of the balloon will be shown hereinbelow.

-5-In order to best describe the movement, one will consider that the gravity traction wheel in Figure 2 is starting its movement. The air (or gas) pressure in the balloons is of the upmost importance. When a balloon is at its minimum position and maintained thereto, one must inflate the rest of the pipe and of the other balloon at the same pressure than that of the balloons when they are at their deepest position in the water. This pressure will be defined in the calculation section hereinbelow.
Balloon A has just come out of the compression guide. So, it is free to take its maximum volume but will keep its minimum volume. It keeps its minimum volume because its opposed balloon, Balloon B, is not yet in contact with the compression guide and since the balloons are formed of a non-stretch material, nothing pushes the air to move on to the other balloon. The system is in a state of balance. This balance relies on the fact that balloon A gets a water pressure equal to the pressure which is within the balloon. The two balloons have been inflated exactly at that pressure.
The pressure within the two balloons cannot diminish even if the two balloons could take their maximum volumes. The balloons are formed of a non extensible material, so balloon B cannot increase its volume. This would provoke a pressure reduction within the assembly. The pressure of the assembly cannot be reduced by the expansion of balloon B and cannot be done by balloon A even if it were free.
The water pressure forces the balloons to keep the pressure stable within the balloons.
If one forces balloon A to take its maximum volume, as the quantity of air within the balloons did not change, the air pressure will reduce inversely proportionally to its volume increase. We will find ourselves with a balloon which is artificially at its maximum volume. Balloon B will always be at its maximum volume as the pressure in

-6-the two balloons will be inferior to the pressure of balloon A but will always be superior to the air (atmospheric) pressure.
If balloon A is released, the water pressure will force the balloon to reduce its volume down until the pressure within the balloons equals the water pressure.
So, balloon A has just left the compression guide, balloon B does not yet touch the compression guide, the system is in a state of balance. The traction wheel turns and the compression wheel touches the compression guide. While turning, the traction wheel together with the compression wheel and the compression guide, reduces the volume of balloon B and pushes the air towards balloon A. Since the system is in a state of balance, the only restriction to the passage of the air of balloon B toward balloon A is the restriction caused by the friction of the air against the sides of the piping used to this end.
Then, the compression wheel together with the compression guide will maintain the air volume in each balloon assembly on the same side of the traction wheel so as to keep this rotation. The light friction (rubbing) of the compression wheels against the guide will be a loss for the system. The air volume in the pipes does not influence the system. There is the same quantity of pipes on each side of the traction wheel, thus the same quantity of air, on the side of the inflated balloons. This helps the movement but on the side of the guide, it is a disadvantage for the movement. The positive and the negative are equal thus do not change anything to the movement. It has the same effect as for the construction weight of the wheel assembly. It is equally distributed throughout the wheel and is of no consequence to the rotation movement. This weight must be considered only in the construction of the drive shaft.

The pressure measurements will be manometric pressures, not absolute pressures.
At the air (atmospheric) pressure, the manometric pressure is 0 lb/sq in (pound by square inch) or 0 lb/sq ft (pound by square foot). The air pressure is about 14.69 lbs/sq in or ( 14.69 lbs/sq in X 144 sq in/sq ft = 2115.36 lbs/sq ft) 2116.3 lbs/sq ft. I
use 2116.3 lbs/sq ft in the attached conversion table. It changes according to the number of decimals that we keep to make the calculation. When one inflates an automobile tire, one inflates it to 30 lbs/sq in manometric. In absolute pressure, it represents 30 lbs/sq in + 14.69 lbs/sq in = 44.69 lbs/sq in. It is the manometric pressure which will be utilized.
The energy or work provided by the coming back up of the balloon will be:
A) 105 lbs x 6 feet = 630 ft lbs Because the water exerts a net push of 105 lbs on the balloon and the container is 6 ft deep.
The pressure at the bottom of the container will be:
B) 62.32 lbs x 6 feet = 373.92 lbs cult sqft The energy or work necessary to inflate the balloon at the bottom of the container is provided by the following formulae:
C) T = p2u2 - PW
Like:
- The pressure at the bottom of the container is 373.92 lbs/sq ft - The volume of the balloon is 1.77 cubic feet _g_ The energy or work necessary to inflate the balloon at the bottom of the container will be:
373.92 lbs X 1.77 cu ft - 373.92 lbs X 0 cu ft = 661.83 ft lbs sq ft sq ft As 661.83 ft lbs is greater than 630 ft lbs, it is obvious that the system will absorb energy instead of producing it.
To make a relationship with the calculation already established, we will keep the same parameters.
In the example, the calculation presented is based on a balloon (spherical) having a diameter of 1.5 feet. The volume of a sphere is determined as follows:
Volume = 4/3 ~ R3 = 4/3 i ~ (0,75 foot)3 = 1.767 cu ft =~ 1.77 cu ft The voluminal mass of the water is 62.32 lbs/cu ft.
The law of physics indicates that a body plunged into a liquid gets an increasing force equal to the weight of the volume of the liquid moved. The volume moved in our example is 1.77 cu ft.
Forces received by the balloon = 62.32 lbs/cu ft x 1.77 cu ft = 110.31 lbs The number 105 is the result of 62 lbs/cu ft X 1.7 cu ft. It is important in a calculation with multiplications to always keep the same values. If we place the figures in Equation A, the new results will be:
Force X Movement - Work (energy) 110.31 lbs X 6 feet - 661.86 ft lbs In Equation B, we obtain a pressure of 373.92 lbs/sq ft (2.6 lbs/sq in) for 6 ft deep - i.e. 62.32 lbs/sq ft (0.43 lbs/sq in) for each foot of depth. It is precisely this pressure at which the balloons of the traction wheel - i.e. 0.43 lbs/sq is multiplied by the maximum depth at which the balloons go down to.
Equation C is the most important and represents the energy necessary to inflate a balloon of 1.77 cu ft with a pressure of 373.92 lbs/sq ft (2.6 lbs/sq in) and utilizes the formula of the difference between the arrival point and the starting point. At the arrival point, we have 1.77 cu ft of air with a pressure of 373.92 lbs/sq ft (2.6 lbs/sq in) and at the starting point, we had nothing, we had 0 cu ft at 373.92 lbs/sq ft. In manometric pressure, point 0 is the atmospheric pressure. In this example, we do not create air. It is the pumping with a compressor of air at the atmospheric pressure in a balloon of 1.77 cu ft with a pressure of 373.92 lbs/sq ft (2.6 lbs/sq in) which is the pressure generated by water 6 ft high, and the calculation is done considering the pumping mode with an efficiency of 100%. This energy calculation totals 661.83 ft lbs, i.e. about the same energy as the ascension of the balloon to the surface of the water. If we consider the efficiency loss of the compressor to pump the air, we get to the same conclusion. This functioning mode will absorb energy not produce it. Thus, it is imperative to use the traction wheel.
The big difference between the previous example and the traction wheel is the starting point of the inflation of the balloon. In the example, the starting point is the atmospheric pressure; in the traction wheel, the starting point is another balloon of 1.77 cu ft of air with a pressure of 373.92 lbs/sq ft (2.6 lbs/sq in). If we do the calculation of Equation C over, we obtain:

Work - (Pressure 2 X Volume 2) - (Pressure 1 X Volume 1) - 1.77 cu ft X 373.92 Ibs/sq ft - 1.77 cu ft X 373.92 lbs/sq ft The work utilized to inflate the balloon is zero (0). It is for this reason that it is very important that the balloons be made of a material which is as non-stretch as possible. Because the pressure at the top of the container is lower than that at the bottom of the container, and if the balloon is stretchable, the pressure at the top of the balloon will decrease with the increasing of the volume of the balloon (because of the elasticity of the material) and the traction wheel will have to provide the energy necessary to recuperate this loss of pressure.
But, as explained above, the traction wheel must push the air from the top balloon toward the bottom balloon. In order to do so, it will be necessary to increase the pressure in the top balloon so as to push the air toward the bottom balloon and it is this work that the guide and the compression wheels will provide. This loss of energy will translate into an approximate increase of the maximum pressure of 76.08 Ibs/sq ft (0.5 Ibs/sq in) in the top balloon, that is the loss of the friction of the air against the sides of the pipes.
Compression work Work - (Pressure 2 X Volume 2) - (Pressure 1 X Volume 1 ) W - P2 V2 - P1 Vl - 1.77 cu ft x 373.92 Ibs/sq ft - 1.77 cu ft X 450 lbs/sq ft - 661.83 ft lbs - 796.5 ft Ibs - 134.67 ft lbs In Equation A, we had obtained an energy productivity of 661.81 ft lbs and in order to produce this energy, the estimated loss is 134.67 ft lbs, that is a net increase of 527.14 ft lbs. We keep about 80% of the energy created.

We have calculated the energy produced by a balloon which comes back up to the surface of the water and calculated the loss of energy during the transfer of air from a balloon to another. But the traction wheel has many balloons which produce energy simultaneously. Furthermore, each balloon has an arm lever and a rotational movement, not a translation movement as mentioned above.
Energy calculation of the traction wheel Construction characteristics Diameter in feet 8 Volume in square feet 1.57 Circumference in feet 25.13 Force in lbs 97.89 Number of balloons 16 Rotations per second 2 Length of the balloon 1.57 HP 34.66 in feet Width of the balloon in 2 kWatts 25.86 feet Max. height of the balloon1 Amps at 240 volts gross 107 in feet Amps at 240 volts net 96 (90%) The traction wheel has a diameter of 8 feet, so 25.13 ft of circumference. We choose to install 16 balloons (2513.ft X 16 balloons) the maximum length of each balloon is 1.57 feet. The width of the balloon is 2 feet and the height is 1 foot. The shape of the balloon is a triangle, so the volume occupied by this balloon at its maximum volume is (1.57 X 2 X 1 X'/2) 1.57 sq ft. The force of a balloon is 1.57 cu ft X
voluminal mass of the water 62.32 lbs/cu ft for a total of 97.89 lbs. This density wheel has 16 balloons but only 8 which have their maximum volume. Let's calculate the instantaneous force exerted on the balloons taking into consideration their respective positions on the traction wheel.

Angle position Useful portion to Spokes of the Weight Force Rotation according Wheel Degrees Feet Lbs Ft lbs Col.l Col.2 Col.3 Col.4 Col.2xco1.3xcol.4 0 0 4 97.89 0 22.5 0.3750 4 97.89 146.84 45 0.7070 4 97.89 276.84 67.5 0.9140 4 97.89 357.89 90 1 4 97.89 391.57 112.5 0.9140 4 97.89 357.89 13 5 0.7070 4 97.89 276.84 157.5 0.3750 4 97.89 146.84 180 0 4 97.89 0 Total force 1954.71 The number of rotations per minute of an electric motor is determined by the number of cycles of the electrical source divided by the number of construction poles of the motor. The work that the motor must provide does not affect the rotation of the motor. The gas operated motor, its rotation is determined by the work it provides. When a car goes up a hill, the car slows down; the work required to go up the hill has increased, the speed of the car will decrease so as to find its new equilibrium between the available force (gas consumed by the car) and the work it provides. If we increase the quantity of gas which is consumed by the motor, the car will increase its speed to go up the hill and will find a new balance between the force available (gas consumed by the motor) and the work it provides.
The traction wheel reacts the same as the gas operated motor. Its rotation speed is determined by the force available et the work to provide. I recommend a low rotation speed, approximately 2 rotations per second; an electric motor turns at 1800 rotations per minute (30 rotations per second) most of the time and the gas operated motor turns 3500 rotations per minute (58 rotations per second). The traction wheel is an extremely powerful system in the characteristics shown above, it produces 96 amps with a rotation of 2 rotations per second. It its rotation were that of the electric motor, i.e. 30 rotations per second, it would produce 1440 amps. The higher the rotation speed of the traction wheel, the more sophisticated and costly its construction must be. The wear occurs more rapidly and the efficiency is less. I think that with a rotation inferior to 5 rotations per second, the efficiency will be 90% ( 10% of loss). At a speed of between 5 and rotations per second, the efficiency will be 80%. But it is experience which will really determine these variables. In my example, the rotation speed is 2 rotations per second.
The instantaneous force calculated in our example is 1954.71 ft lbs, the rotation force. The force which the wheel has to maintain its rotation, not the energy which it produces . To calculate the energy which can be produced by the traction wheel, one must find the energy available by taking into consideration the time taken to obtain this energy. The balloon which becomes inflated at the bottom portion of the traction wheel will go back up and occupy each angular position of the circle for a determined time but equal for each position. The energy provided by the balloons in their ascension will become the potential energy that the traction wheel will be able to provide to a generator.
To calculate the useful work accomplished by the ascension of the balloon, it is useful to calculate the useful force of the balloon for each position as well as its vertical movement. The work is determined by the force of the movement, and the power is determined by the time it takes to accomplish this work. See Figure 4.

Useful force portion in each angular position Start angle Arrival angleUseful forceUseful forcePortion of at the startat the arrivalaverage _ ___ force used in this movement_ _ j I80 157.5 __0 0.375 0.1875 157.5 135 0.375 0.707 0.541 135 112.5 0.707 0.914 0.8105 112.5 90 0.914 1 0.957 I

90 67.5 ', 1 0.914 0.957 ' 67.5 45 0.914 0.707 __ 0.8105 45 22.5 0.707 0.375 0.541 i 22.5 0 0.375 0 0.1875 I 0 ~

Vertical movement between each angular position S angle Arrival angleStart depth Arnval depthVertical __ movem 180 157.5 _8 7.71 0.29 157.5 135 7.71 6.83 0.88 -_ 135 112.5 6_.83 5.5 1.33 I I2.5 90 5.5 4 1.5 90 67.5 4 2.5 1.5 67.5 45 , 2.5 1.17 1.33 ' 45 22.5 ~ 1.17 0.29 0.88 22.5 0 I 0.29 0 0.29 0 ~ ~ Total 8 ieds Calculation of a balloon's work Average Force Work Start angle~v~ ogle forced t Vertical i ' portion j in the movemen movement i Feet Pound Ft lbs i1 I 180 157.5 0.1875 0.29 97.8923 5.3596 _ I

157.5 135 0.541 0.88 ~ 97.8923 46.6045 135 112.5 0.8105 97.8923 105.3657 ~ 1.33 l 112.5 90 0.957 1.5 97.8923 140.5243 90 67.5 0.957 1.5 97.8_923 _140.5243 67.5 45 0.8_105 97.8923 105.3657 1.33 --45 22.5 T 97.8923 46.6045 0.541 j ' 0.88 22.5 0 0.1875 0.29 97.8923 5.3596 0 Total 8 ieds Total 595.7084 The total work accomplished by a balloon is 595.7084 ft lbs and within one rotation, there are 16 balloons doing the same work for a total work of 9531.33 ft lbs per rotation. The number of rotations per second is 2, so we obtain an energy (work/second) of 19062.66 ft lbs per second. There are S50 ft lbs/sec in a HP, so our system represents a power of 34.66 HP or 25.86 kWatts (0.746 kWatts per HP). A generator of 25.86 kWatts can produce 107 amps under a voltage of 240 volts. Naturally the conversion of the work of the traction wheel into electricity will be accomplished with a certain loss which we have estimated to 10%. So the estimated amperage production will be of 96 amps.