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Lex Vitae; Lex Universalis

 

Antecedents of the Invention:

a) Studies about Wind Turbines

Figure Nē 1: Wind Turbine or Aeolian Turbine

Figure Nē 1: Wind Turbine or Aeolian Turbine

  1. Blades.
  2. Rotor.
  3. Rotor Shaft (low speed shaft)
  4. Gears Box.
  5. Gears.
  6. Brake.
  7. Generator Shaft (high speed shaft)
  8. Electricity Generator.
  9. Cable carrying electricity to substation and transmission wires.
          

The operation of a wind turbine works basically turning the wind force which they rotate the rotor blades in electricity, through a gears box (Figure 1; items 4 and 5) the low speed shaft of the rotor transfers the force multiplied to the high speed shaft of the electricity generator.

The wind turbines are designed to turn the energy of the movement of the wind (kinetic energy) mechanical energy, through the movement of a shaft. Later, in the generators of the turbine, this mechanical energy becomes electricity. The generated electricity can be stored in batteries, or be used directly, sending the energy to the lines of distribution.

Exist a theoretical maximal effectiveness of the wind turbine of 59%. Actually, most of the Aeolian turbines they are much less efficient than this, and diverse types are designed to obtain the maximum possible effectiveness at diverse wind speeds. The best wind turbines have effectivenesses from 35% to 49%.

The gear box is one of the more important main components in a Aeolian turbine (Figure 1, items 4 and 5), located between the main shaft and the generator, its task is to increase the slow rotatory speed of the shaft of rotor to the speed of rotation of the generator of 1000, 1500 or 1800 rpm. Erroneously it is possible to be thought that the gear box could be used to change speeds as if outside a normal box of speeds of a car. Nevertheless, this one is not the case of a gear box of a Aeolian turbine. In this case the gears box always has a constant and a quotient of increase of the speed, if a Aeolian turbine has diverse operational speeds, because it has two types of generators, each one with his own one and is differentiated speed from the rotation (or a generator with two stators).

We analyzed some data provided by General Electric (GE) in its Web Site, doing reference to two types of Aeolian turbines and their corresponding productions of electrical energy.


Website: http://www.gepower.com/...

GE Wind Turbine 1.5 MW operates according to the technical specifications:

Rated Capacity: 1.5 MW (1500 kWh; kilowatts per hour)
Rated Wind Speed: 12 m/s (43.2 km/h)
Annual Energy Yield: 5,800,000 kWh/year.

Details of Rotor
Number of Rotor Blades: 3
Rotor Diameter: 70.5 - 77 m
Rotor Speed (variable): 10.1 – 22.2 rpm


Website: http://www.gepower.com/prod_serv/...

Considering that the speed of the rotor (even the maximal) is not sufficient to rotate the shaft of the generator, through a box of speeds that consists of a series of gears, (Figure 1, items: 4 and 5) the rpm of low speed shaft of the rotor is multiplied, the 22 rpm are converted in 1500 rpm and that force multiplied is transferred to the high speed shaft of the electricity generator.

Deductible, is possible to be inferred:

1500 rpm is equivalent to the 100% of the electricity production.
22 rpm represents 1.47% of the electricity production.

5,800,000 kWh/year is the annual energy yield of this Aeolian turbine. Dividing that value by 365 days, later by 24 hours, this turbine of 1.5 MW is able to generate, theoretically, about 662.10 kWh (kilowatts per hour).

With 22 rpm the calculated values above we can infer that rpm that corresponds to 1.47% of the total energy produced and represents 9.73 kWh (kilowatts per hour).

Considering that the rated capacity of this turbine is 1500 kW, we can calculate that in 24 hours its production will be of 36,000 kWh per day and in a year will be of 13,140,000 kWh, but taking care of that its 5,800,000 kWh/year real production is of concludes that this system has a 44.12% efficiency.

Next we will study another type of turbine, one of most powerful than it has General Electric at the present time.

2) GE Wind Turbine 3.6 MW operates according to the technical specifications:

Rated Capacity: 3.6 MW (3600 kW)
Rated Wind Speed: 14 m/s (50.4 km/h)
Annual Energy Yield: 15,000,000 kWh/year

Details of Rotor
Number of Rotor Blades:
3
Rotor Diameter: 104 m
Rotor Speed (variable): 8.5 – 15.3 rpm

Website: http://www.gepower.com/prod_serv/...

Analogous to the previous process, it is possible to be inferred that through the multiplying effect of the gears box is:

1500 rpm is equivalent to the 100% of the electricity production.
15 rpm represents 1% of the electricity production.

15,000,000 kWh/year is the the electricity production of this Aeolian turbine. Dividing that value by 365 days, later by 24 hours, this turbine of 3.6 MW is able to generate, theoretically, about 1712.32 kWh.

With the calculated values above we can deduce that 15 rpm it corresponds to 1% of the total of the produced energy represents 17.12 kWh.

Considering its rated capacity of this turbine is of 3600 kW we can calculate that in 24 hours its production will be of 86,400 kWh per day and in a year will be of 31,536,000 kWh, but taking care of that its real production is 15,000,000 kWh/year, that concludes this system has a 47.56% efficiency.

The values for the calculation of the resulting speeds of the multiplication effect of the gears box are theoretical, the exact number of the relation between these values depends to a large extent on the types and the manufacturers of the Aeolian turbines. These manufacturers do not publish this relation generally to consider like industrial secret.

Be as they are these values, the important thing we can here deduce like mathematical expressions:

 

x = w por z sobre y (x = w * z / y)

n = w sobre y (n = w / y)

w = y por n (w = y * n)



Where x is the value in kWh of the amount that represents rpm of the low speed shaft of the rotor in the total production of electrical energy.

w is the maximum capacity in rpm to rotate the blades of the Aeolian turbine rotor (15 rpm).
y is the resultant of the multiplication of the gears box (1500 rpm).
z is the value of total production of electricity (production by hour; 1712.32 kWh).
n is the factor of relation between w and y (1500/15 = 100).

Deducing, we have:


x = z sobre n (x = z / n)


The result of x is 17,12 kWh.

To construct a perpetual motion machine with the values of x in kWh (kilowatts per hour) whatever is due to calculate would spend an electrical motor to rotate the low speed shaft of the rotor to rpm initial (the value of w). In the case of the studied Aeolian turbines, how many kWh will be necessary to produce 15 or 22 rpm

Is possible to transfer that energy to a feedback system, in other words, the values of x (kWh) could be used to directly produce w in the low speed shaft of the rotor, eliminating therefore the rotor.

Wind Turbine 1.5 MW:
22 rpm = 8.07 kW per hour = 10.82 HP
Real Value: 662.10 kW per hour = 887.89 HP

Wind Turbine 3.6 MW:
15 rpm = 17.12 kW per hour= 22.96 HP
Real Value: 1712.32 kW per hour = 2296.25 HP

Potentially 9.73 kWh can move a rotor at a speed of 22 rpm, in the first case, and in the second case it would be that 17.12 kWh can move a rotor at a speed of 15 rpm through an electrical motor.

Analyzing another situation, all the rotor and the gears box system could be eliminated. Using an electrical motor we can directly connect to the shaft of the generator, and to calculate how much energy in kWh would be needed to produce rpm necessary (the value of y; 1500 rpm) to generate the amount of energy that at first was produced by the wind force which they rotated the blades of the Aeolian turbine rotor. (See Figure Nē 2)

The following extracted data of the Web Site of Procobre Peru indicate the values to us of consumption of electrical energy of a great motor of high efficiency of 200 HP, 1800 rpm (460 volts) that operates almost continuously in an industrial atmosphere to full load:

Efficiency: 96.2%
Energy Out (0.7457 kW/HP): 149.1 kW
Energy In: 155.0 kW
Loss to a load of the 100%:
5.9 kW

A motor industrial but small and, therefore, much less efficient, of 5 HP, 1800 rpm (460 volts)
Efficiency: 89.5 %
Energy Out: 3.73kW
Energy In: 4.17 kW
Loss to a load of the 100%: 0.44 kW

Source: Website of Procobre Perú - Centro de Promociķn del Cobre
http://www.procobre.org/procobre/index.html

Ideally, a system of perpetual motion would have to consume less energy of the one than it produces, resorting for the example of the motor of 200 HP that consumes 155 kWh, this one could be able to directly generate the 1500 rpm in the shaft of the electricity generator. The Figure Nē 2 illustrates the hypothetical example.


Figure Nē 2: Hypothetical illustration of a system of perpetual motion

Figure Nē 2

Figure Nē 2: Hypothetical illustration of a system of perpetual motion:

  1. Electrical Motor: 200 HP, 155 kWh.
  2. Electricity Generator.
  3. Shaft connected directly between the electrical motor and of the generator.
  4. Cable connected to electrical motor to provide electricity from the lines distribution.
  5. Cables that transport the electrical energy produced by the generator to the lines distribution.

The considered values of production of electricity with the generator of a connected Aeolian turbine to an electrical motor of 200 HP would be the following ones:

Generator of Wind Turbine 3.6 MW:

Real energy produced by Wind Turbine 3.6 MW: 15,000,000 kWh/year (1712.32 kW per hour)

1712,32 kWh – 155,0 kWh = 1.557,32 kWh


Here we make calculations where the energy produced by the generator of the Aeolian turbine except the energy spent by the electrical motor of 200 HP. The values indicate the amount of produced energy directly connecting the electrical motor to the shaft of the electricity generator.

Energy Produced per year: 13,642,123 kWh (Energy Produced by the hypothetical system of perpetual motion)

Energy Consumed per year: 1,357,800 kWh (Energy Consumed by the hypothetical system of perpetual motion)

Relation between produced energy and consumed energy: 9.05%

As it is observed in the Figure Nē 2, this would hypothetical represent a system of perpetual motion using the generator of a Aeolian turbine and an electrical motor of 200 HP of high efficiency.

The electrical motor (Figure Nē 2; item 1) will directly feed on the energy produced by the electricity generator (Figure Nē 2; item 2). The energy used by the electrical motor would represent 9.05% of the total energy produced by the electricity generator.

The annual amount of energy produced by this hypothetical system of perpetual motion would be sufficient to provide electricity to 1,300 residential users, making a calculation average in each user spends about 10,000 kWh/year (average of cost of residential users in the United States according to the Census of 1997).



b) Studies related about the technology of magnetic levitation developed by Transrapid International

Analyzing the data previous and resorting to the data provided by Transrapid International in its Web Site we found that its system of transport is able to transfer at speeds superior to 400km/h, several tons of payload consuming hardly a few kWh. Transrapid uses the technology of magnetic levitation known as EMS (ElectroMagnetic Suspension)

Website of Transrapid Internacional: http://www.transrapid.de

The consumption of electrical energy to move the sections or wagons of the magnetic levitation trains (MAGLEVs) is very low due to the levitation technology neutralizes the forces of the gravity (weight of the train) and of friction (contract between surfaces) between the train and the rail guides.

According to these data the system of Transrapid spends only 52 Watts per hour (Wh) by each passenger by kilometer to transport it at a speed of 400 km/h.

The following extracted image of the Web site of Transrapid indicates the comparable values to other high-speed railroads.

Specific Energy Consumption: Values in Wh

Specific Energy Consumption:

Values in watts per hour (Wh) per seat-km


The Transrapid MAGLEVs trains count on a minimum of 2 wagons and in average of 90 seats (capacity for passengers).

These data allow to deduce that the energy spent by the system is of:

90 (seats) per 52 Wh would be 4680 Wh per each kilometer.

4680 Wh = 4,7 kWh (kilowatts per hour)


Another relevant data provided by Transrapid is that each section (wagon) of the MAGLEVs train can transport 15 tons of payload. As the propulsion system is located in the guideway or rail guides, the load does not affect the acceleration of the system.

This is an example of the energy consumed by seat/passenger for a route of 300 kilometers with three intermediate stops, depending on a maximum velocity.

200 km/h consumed 32 Wh/km is equivalent to 1.1 l of gasoline per each 100 kilometers.
300 km/h consumed 47 Wh/km is equivalent to 1.6 l of gasoline per each 100 kilometers.
400 km/h consumed 66 Wh/km is equivalent to 2.2 l of gasoline per each 100 kilometers.

The consumption of energy of Transrapid International System is of 3.5 times smaller to the consumption of energy of a vehicle (car) at a same speed and it represents the fourth part of the consumption of energy of an airship of comparable capacity and speed.

This system can arrive at speeds of 1200 km/h, the technology developed by Transrapid International is able to arrive at these values, but they are little practical and unnecessary for the transport of passengers.

Also it is necessary to consider that at speeds near the speed sound (1200 km/h approximately) can be created sonic booms able to break crystals of windows in the neighborhoods of the route of a vehicle that moved at this speed.

Using the technology of magnetic levitation have been created these dynamic systems that use very little amount of energy to make their functions, are the most efficient systems in terms of energy consumption.

The great advantages that the technology of magnetic levitation supposes will serve as it bases for the construction of an ideal perpetual motion machine because the system would spend the possible minimum amount of energy to generate great amounts of energy.

In this system which we will develop with base in the technology of magnetic levitation there are no losses of energy by heat because there are no surfaces in contact, the only energy loss would be the heat produced by the excitation of the electromagnets, but this energy loss irrelevant would be compared to the energy that will produce the system, in addition what we will see next is not a system that depends on a thermoelectrial process, but of mechanical and electromagnetic processes...


Data Update: April 2011

In 2006 I started recording my invention the ElectroMagnetic Levitation and Propulsion Turbine.

The feasibility to build my invention has been shown, in my Antecedents of Invention, I suggested a hypothetical perpetual motion system, connecting an electric motor to an electric generator, an engine that spends far less in relation to what produces the generator, the idealization was not so hypothetical...

An Inventor Colombian is selling the Ucros Kit, which is what I envisioned, the project works this way, having a generator of 3.5 MW, which runs to 3500 rpm, connect an electric motor also works to 3500 rpm. The electric motor replacing work as the driving force of wind, solar energy, hydro or thermal (atomic energy). The electric motor will spend only 1.5 KW of electricity.

Mr. Carlos Ucros Piedrahita has a Facebook Profile, Website and Contact Information.

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