Hello fellow physicists!
I am fond of non-traditional energy, at the link below we have a patent of Markelov V.F.
http://www.macmep.ru/markelov.htm
I studied this topic based on my meager knowledge of physics, but alas, I can’t figure it out to the end.
For this reason, I turn to you - the masters of their craft.
I hope for your help!
Please study the patent first.
I have a question: how legitimate is the power calculation formula, and in particular its two parameters?
N = 9.81 2 Q 0.5 5 H efficiency
where:
9.81 m/s2 - free fall acceleration;
2 - 2 volumes will be displaced through the upper level of the turbine (1-water and 1-air)
Q - water flow in m3/s;
0.5 - density of water-air mixture (0.5 t/m3)
5 - number of impellers;
H - head in m (for a 2-meter turbine = 12m);
Efficiency-type 0.9
The following is confusing:
It is possible to exclude coefficient 2 and density 0.5 from calculations. in the product they give 1. and the impeller is still spinning water, the air will not move it, the water segments will alternately act on each impeller, so the factor two is illegal.
And very embarrassing H in size 12m
Here is a quote from "Markelov":
When calculating the power required for air supply, we took into account atmospheric pressure (1 atmosphere = 10 m of water column), which means that the rising air overcomes the absolute pressure inside the turbine casing, which is the sum of the pressure of the water column in the turbine and atmospheric pressure and is equal to pressure 12 -meter column of water. Absolute pressure inside the turbine housing is neutralized by the buoyancy of the air, but it is present behind the housing and affects the water supply to the turbine. This effect is equivalent to the effect on the water flow of the rarefaction created in the turbine housing by all the volume of air in the turbine (this effect is absent in the hydroturbine) and with the appropriate design of the turbine, we have the right to consider the head as H \u003d H column value + 10 m. (WHO IS FRIENDS WITH PHYSICS - HOW LIMITED IS THIS METHOD OF CALCULATION OF THE HEAD?????????????)
The rest is not in question.
Calculated the power of the turbine with a diameter of 0.2 m and a height of 2 m
The pipe moves 8 bubbles of 4.186 liters. and between them there is the same amount of water, the air segment in height in the pipe occupies 13.3 cm and the water segment, respectively, the same amount.
The calculation used 7 impellers (according to the number of water segments in the pipe at the moment)
In total, there are 33.488 liters in the turbine at the time. air.
Ascent time 5 seconds
33.488 / 5 \u003d 6.69l / s (respectively, here is the pump performance)
6.69*60=401 l/min (pump output per minute)
Well, N \u003d 9-81 * 0.00669 * 7 * 12 * 0.9 \u003d 4.961 kW, but the pressure coefficient is very embarrassing !!! if without it, then 413W.
Well, naturally, from the volume of the pipe, it is necessary to subtract the volume of the shaft with impellers assembled.
The coefficient of air increase during ascent from 2 meters was also not taken into account. it is very small for a given column height
Well, I did not take into account the difference in temperature between air and water.
Thank you. 
The presented method of obtaining energy seems to us the most promising, based on the following considerations:
relatively low manufacturing cost, the possibility of using common improvised materials for the construction of the tank, the possibility of using any air compressor that can be obtained, the relatively small dimensions of the device, which makes it possible to install it in a personal household.
The residence of the author within reach makes it possible to seek advice from him regarding the specific dimensions and shape of the elements of the device.
At the same time, the calculation of capacities by the author makes it not too fundamental the question of exceeding the received power over the spent power by dozens of times, if there is an effect, then it will manifest itself at any ratio of supplied and removed powers.
Moreover, home experiments do not require a powerful material base.
Any home craftsman is able to make a sample using any suitable containers, and adhering to the approximate ratio of dimensions given by the author.
The site administration will be grateful for information about experiments to test and build working samples.
METHOD OF OBTAINING ENERGY
(RF patent N 2059110)
Markelov V.F.,
In 1607, the Danish scientist Cornelius van Drebbel demonstrated to the English king James I a "perpetual" clock, driven, of course, by an equally "perpetual" engine. Drebbel patented them back in 1598. However, unlike numerous other devices with the same name, this engine was indeed "eternal" in a certain sense.
What was the secret of this watch (or rather, their engine)? Drebbel's perpetual clock worked from a drive that, like any other real engine, uses the only possible source of work - non-equilibrium (potential difference) in the external environment.
But the disequilibrium used by Drebbel is of a special kind, although it is also associated with temperature and pressure differences. It can operate in a perfectly equilibrium environment, the temperature and pressure of which are the same at all points. What is the matter here and where does the work come from?
The secret is that potential differences are still present here, but they do not appear in space, but in time.
This can be most clearly illustrated by the example of the atmosphere. Let there be no significant difference in pressure and temperature in the area where the engine is located. But (common at all points) pressure and temperature still change (for example, day and night). These differences can be used to obtain work (in full agreement with the laws of thermodynamics).
In the description of the invention "Method of extracting the energy contained in liquid and gas and converting it into mechanical work" (RF Patent No. 2059110), my version of a pseudo-perpetual and successfully operating solar engine is given. To increase the number of cycles and power, the properties of two media that are not in equilibrium with respect to each other - water and air - are most fully used. Archimedes' law is considered as a consequence of the law of conservation of energy, in which the buoyancy force is linked to the energy costs for the creation of water and air. The amount of this energy also determined such physical properties as, for example, density, heat capacity, and thermal conductivity.
Part of the ratio of energy to density creation is reflected in a disequilibrium factor of 820, and if we could find a way to exploit this disequilibrium to the fullest, we would get an energy gain of 820 times. Disequilibria appear from the moment air is supplied under the water column and increase as it ascends due to an increase in the volume of air and the removal of heat from the water, while the air is supplied with a temperature below the water temperature, because “if, for example, the air pressure is 4 atm (0.4 MPa), and the temperature is +20oC (293 K), then when expanding to atmospheric pressure it will cool down to about -75oC (198 K), i.e. at 95oC". Heat will be removed under conditions close to adiabatic, i.e. with minimal heat loss, because Water is a good heat accumulator, but a poor conductor.
Cooling - water.
CALCULATION OF ENERGY EXTRACTING PNEUMO-HYDRAULIC TURBINE (RF patents N 2120058, N 2170364, N 2024780)
We use a compressor as a source of compressed air. The most suitable are positive displacement and dynamic type compressors. A piston compressor consumes energy several times less than a dynamic one, so we will opt for a positive displacement compressor - a piston one:
Compressed air source - piston compressor VP2-10/9.
We will judge the efficiency of a pneumohydraulic turbine by comparing the consumed and received power, i.e. amount of work per second.
Compressor performance - the volume of air entering the compressor at atmospheric pressure, i.e. productivity in 0.167 m3/s - the volume of air before entering the compressor and after surfacing in the turbine. When air is supplied under the lower level of the turbine, 0.167 m3 / s of water will be displaced through the upper level and the same amount will flow under the lower level, creating a water-air mixture and its movement inside the turbine housing. The value of 0.167 m3/s corresponds to the water flow when calculating the power of the pneumohydraulic turbine. We will carry out the calculation according to the formula for calculating the power of a hydroturbine:
N=9.81 Q H efficiency,
where 9.81 m/s2 is the free fall acceleration;
Q - water flow in m3/s;
H - head in m;
The efficiency of a real turbine reaches enough high values and under the most favorable regime reaches 0.94–0.95, or 94–95%. We get power in kW. Since the working fluid is a water-air mixture, there is a need to confirm the validity of applying the power calculation formula for a hydraulic turbine. The most efficient mode of operation of the turbine seems to be the mode in which a mixture with a density of 0.5 t/m3 (consisting of 50% water and 50% air) is used. In this mode, the air pressure is slightly higher than the absolute pressure in the turbine casing. The air from the pressure pipe of the compressor exits in separate bubbles at regular intervals, and the volume of the bubbles is equal to the volume of water between them in the turbine housing. The bubble takes the form of a spherical segment and works like a piston in a fixed space, displacing water only upwards, because its downward flow is prevented by a higher pressure, and the flow to the sides is prevented by the incompressibility of water. With a constant supply of 0.167 m3 / s of air, 0.167 m3 / s of water will be displaced, i.e. through the upper level of the turbine, 2 0.167 m3 / s of a water-air mixture with an increased flow rate inside the turbine will be displaced, then
N = 9.81 2 Q 0.5 H efficiency = 9.81 Q H efficiency
Let's take an installation with a water column height of 2 m and determine the required compressor engine power to supply air under this water column, taking into account atmospheric pressure based on the data technical specifications compressor:
At the entire installation height, an upward flow of the water-air mixture will be observed, in which the buoyancy force, independent of the depth of the body immersion, allows at least 5 impellers to be placed. The energy mode of the proposed turbine proceeds in more favorable conditions than in the well-known pump "Airlift", because the water flow occurs below the water level in the turbine, i.e. under conditions close to weightlessness, without a significant rise in water in the turbine housing, for which the main amount of energy is consumed in the pump. Let's take the turbine efficiency equal to 0.9. In this case, the power is:
N = 9.81 0.167 2 5 0.9 = 14.7 kW
Thus, we received energy 13 times higher than the energy spent:
14.7 kW / 1.13 kW = 13
The increase in power due to the placement of additional impellers has been confirmed on experimental models. Indirectly, the performance of the turbine is confirmed by experiments conducted at the St. Petersburg State Technical University. Here is what the doctor of technical sciences, professor, member of the commission on non-
Photo 3, Photo 4
to traditional energy sources under the Government of the Russian Federation, head of the department "Renewable energy sources and hydropower" Elistratov V.V.: "However, based on the hydraulics of hydraulic machines and our numerous experiments on air intake into the impeller of a hydroturbine in order to reduce cavitation erosion, it was shown that that with the improvement of cavitation indicators, energy indicators were significantly reduced. In this case, experiments show that the supplied air creates a counter flow, which, acting on the impeller from below, causes it to rotate in the opposite direction. This is the design of the wheel (Fig. 1). And this action is exerted by a small volume of air in a small area equal to the body of the hydraulic turbine. The proposed installation has the ability to take heat from water and convert it into mechanical energy. Taking into account the temperature difference between water and air, when the water temperature is 80oC ( thermal source, water heated in a solar collector, in the cooling system of turbines, compressors, etc.), and the air temperature is 20oC, the coefficient of increase in air volume, according to Lussac's law, is equal to
1+ (80oС - 20oС) / 273 = 1.2
The power will be
N = 14.7 kW 1.2 = 17.6 kW
Our expectations in terms of energy gain have been confirmed.
17.6 kW / 5 = 3.5 kW 3.5 kW / 1.13 kW = 3.1 times per wheel
When calculating the power required for air supply, we took into account atmospheric pressure (1 atmosphere = 10 m of water column), which means that the rising air overcomes the absolute pressure inside the turbine casing, which is the sum of the pressure of the water column in the turbine and atmospheric pressure and is equal to pressure 12 -meter column of water. The absolute pressure inside the turbine housing is neutralized by the buoyancy of the air, but it is present behind the housing and affects the water supply to the turbine. This effect is equivalent to the effect on the water flow of the rarefaction created in the turbine housing by all the volume of air in the turbine (this effect is absent in the hydroturbine) and, with the appropriate design of the turbine, we have the right to consider the head as H = H w.st. + 10 m. Then the power will be equal to
N = 9.81 0.167 m3/s 12 m 5 1.2 0.9 = 106.14 kW
We received 93 times more energy than we spent.
We will calculate a more powerful power plant capable of providing energy to an average village, military unit, vessel, etc. As a source of compressed air, we take a piston compressor 2VM10 - 63/9 with the following technical characteristics:
Productivity - 1.04 m3/s
Final pressure, MPa - 0.9 (9 Atmospheres)
Compressor shaft power - 332 kW
Water cooling.
We will carry out the calculation for an installation with a water column height of 5 m with 10 impellers placed in it at a depth of 500 mm. The power of the compressor engine for supplying air under a water column of 5 m, taking into account atmospheric pressure, is
5 m (332 kW / 100 m) = 16.6 kW
The power of the installation is
N= 9.81 1.04 m3/s 15 m 10 1.2 0.9 = 1652 kW
Received energy 99 times higher than spent.
Thus, it is possible to obtain any amount of energy while improving the gas composition of water in an environmentally friendly way from an inexhaustible source of energy, using the natural imbalance of water and air in any climatic zone without building expensive dams and lock equipment, without flooding valuable agricultural lands, etc.
CALCULATION OF ENERGY EXTRACTING PNEUMO-HYDRO MOTOR
(RF patents N 2003830, N 2160381)
Compressed air source - piston compressor VP2 - 10/9.
Productivity - 0,167 m3/s
Final pressure, MPa - 0.9 (9 Atmospheres).
Compressor shaft power - 56.5 kW
Water cooling.
We will judge the efficiency of a pneumohydraulic engine by comparing the power expended and received, i.e. the amount of work produced
mine per second. Compressor performance - the amount of air at the compressor inlet, i.e. volume of air at atmospheric pressure. Then 0.167 m3/s is the volume of air at the inlet to the compressor and at the outlet of the upper float of the pneumohydraulic motor shown in Fig. 3. The release of the floats from the air and their filling with water occurs below the water level in the engine housing. With an air pressure of 9 atm, it can be fed under a column of water 90 m high. At an ascent rate of 0.4 m / s, the ascent time will be 225 seconds, while at the entire height of the column, air in motion will be present in the floats. The ascent rate equal to 0.4 m/s was determined as a result of measurements.
Its increase or decrease while maintaining the water column and compressor performance is reflected only in the horizontal dimensions of the floats, i.e. on length and width, because the amount of air increases or decreases, which, in turn, increases or decreases the force and does not affect the power of the pneumohydraulic motor. Changing the size of the floats only horizontally allows you to make floats of the required volume while maintaining the water column.
The volume of air at the outlet of the pressure pipe of the compressor at a depth of 90 m, taking into account atmospheric pressure, will be equal to
0.167 (m3/s) / 10 atm = 0.0167 m3/s
because the pressure of 10 m of the water column is 1 atm, and the increase in air volume by the value of the initial volume occurs every 10 m of ascent. If the volume of air did not change, then at the time of ascent it would occupy a volume equal to
0.0167 (m3/s) 225 s = 3.757 m3
Taking into account the increase in air volume during ascent, the volume will be equal to
3.757 m3 10 atm = 37.57 m3
Taking into account the coefficient of thermal expansion, the volume is equal to
37.57 m3 1.2 = 45.084 m3
The buoyancy force of 1 m3 of air is equal to 1000 kg s
This volume of air during ascent will produce
work equal
45.084 tC 0.4 m/s = 18.033 tC m/s
or 18033 kg C m/s
1 kg C m \u003d 9.81 watts, then when recalculating we get:
18033 kg S m / s 9.81 \u003d 176903.73 W or 176.9 kW
Adding to the received power at least 30% of the returned energy due to the reactive force developed when the float is filled with air and water is displaced from it, we get:
176.9 kW + 18 kW = 194 kW
We received 3.4 times more energy than we spent.
The mechanical efficiency of the pneumohydraulic motor will be quite high, because work takes place under conditions of constant lubrication with water, and the floats are mutually balanced. Compressor efficiency is taken into account when considering compressor motor power. The pneumohydraulic motor is equipped with a brake and stops on the move, while air remains in the floats and no energy consumption is required at the next start, because. when the brake is released, the air remaining in the floats will cause the engine to work.
We made a calculation for a commercially available compressor capable of supplying air under a 90 m high water column. This is an option to increase the efficiency of a hydroelectric power plant by placing pneumohydraulic motors on pontoons in reservoirs. Increasing the efficiency of hydroelectric power plants using tailwater is shown in the description of the invention No. 2059110. The design of pneumatic hydraulic motors is characterized by low metal consumption, because consists of lightweight frames. Any river, pond, stream, thermal spring, cooling tower can become a source of energy. At HPPs, due to the mixing of the lower warmer layers of water with the cold upper ones, accompanied by the simultaneous removal of heat, the water temperature will equalize. It is especially important that energy will not have to be saved, because. using natural disequilibrium to obtain it, we do not increase the energy disequilibrium of the Earth, but, on the contrary, return it, removing the consequences of thermal pollution. When it comes to solar energy, we don't use more than we get.
We considered an industrial option for obtaining energy, but there is a huge need for power plants of 3–4 kW. Let's take a look at its size. Let's take the height of the installation with the height of the water column equal to 2 m. Using the same compressor (only for calculation), we determine the power of the compressor motor for supplying air under a water column of 2 m:
N = (2 m 56.5 kW) / (90 m + 10 m) = 1.13 kW
Compressor capacity - 0.167 m3/s
2 m of water column create a pressure of 0.2 atm, then the volume of air at a depth of 2 m, taking into account atmospheric pressure, will be equal to
0.167 (m3/s) / 1.2 atm = 0.139 m3/s
The ascent time from a depth of 2 m is
2 m / 0.4 (m/s) = 5 sec
After 5 seconds, in the floats of the pneumohydraulic motor in the state of motion, taking into account the increase in volume during ascent and the coefficient of thermal expansion, there will be
0.139 (m3/s) 5 sec 1.2 atm 1.2 = 1 m3
When surfacing, work will be done
1000 kgC 0.4 m/s = 400 kgC m/s
Work per second means power.
1 kgC m \u003d 9.81 watts, then the power is
N = 9.81 W 400 = 3924 W = 3.924 kW
Adding 30% of the returned power, we get:
3.924 kW + 0.34 kW = 4.263 kW
With a mechanical efficiency equal to 0.9, we get the power
N = 4.263 kW 0.9 = 3.84 kW
We received 3.4 times more energy than we spent:
3.84 kW / 1.13 kW = 3.4
In order to once again verify the effectiveness of the proposed method of generating energy, let's compare it with the efficiency of a pumped-storage power plant, when water is pumped into a high-level reservoir by a pump or a reversible hydroturbine and used at the lower level in the turbine. In this case, with an efficiency equal to 100%, an amount of energy equal to the spent energy could be obtained. Let us determine the power of the pump motor for supplying water to a height of 90 m with a capacity of 0.167 m3/s:
N \u003d (9.81 0.167 m3 / s 90 m) / 0.75 \u003d 196.5 kW
Let us compare the obtained power with the power of the compressor engine equal to 56.5 kW with a capacity of 0.167 m3 / s of air capable of displacing the same volume of water to a height of 90 m with its supply to the turbine and obtaining 196.5 kW, while spending 3.5 times less energy. In addition, at the entire height of the water column, air remained in motion, which will also do work, which is confirmed by the above calculation. We will additionally consider the possibilities of implementing the proposed method on the graph (Fig. 2)
It follows from the graph that the action of the air buoyancy force immediately begins with the volume Vo. The shaded part is the water column H, which requires compressor energy to overcome its pressure, Vo is the volume of air at depth H, Vk is the volume of air expanded as a result of pressure drop during ascent, Vq is the effective volume of air. The graph shows that for a pneumohydraulic motor, the amount of air in operation is Vq, and for a pneumohydraulic turbine, the air volume equal to Vk is important, because the displaced volume of water works in it, which explains the difference in their efficiency.
Inexhaustible source of energy, absolute ecological cleanliness, active improvement environment, ease of manufacture and quick payback with an increasing need for energy ensure the inexhaustibility of the sales market, and a variety of designs - wide opportunity their applications.
May 1st, 2013
The presented method of obtaining energy seems to us the most promising, based on the following considerations:
relatively low manufacturing cost, the possibility of using common improvised materials for the construction of the tank, the possibility of using any air compressor that can be obtained, the relatively small dimensions of the device, which makes it possible to install it in a personal household.
The residence of the author within reach makes it possible to seek advice from him regarding the specific dimensions and shape of the elements of the device.
At the same time, the calculation of capacities by the author makes it not too fundamental the question of exceeding the received power over the spent power by dozens of times, if there is an effect, then it will manifest itself at any ratio of supplied and removed powers.
Moreover, home experiments do not require a powerful material base.
Any home craftsman is able to make a sample using any suitable containers, and adhering to the approximate ratio of dimensions given by the author.
The site administration will be grateful for information about experiments to test and build working samples.
METHOD OF OBTAINING ENERGY
(RF patent N 2059110)
Markelov V.F.,
In 1607, the Danish scientist Cornelius van Drebbel demonstrated to the English king James I a "perpetual" clock, driven, of course, by an equally "perpetual" engine. Drebbel patented them back in 1598. However, unlike numerous other devices with the same name, this engine was indeed "eternal" in a certain sense.
What was the secret of this watch (or rather, their engine)? Drebbel's perpetual clock worked from a drive that, like any other real engine, uses the only possible source of work - non-equilibrium (potential difference) in the external environment.
But the disequilibrium used by Drebbel is of a special kind, although it is also associated with temperature and pressure differences. It can operate in a perfectly equilibrium environment, the temperature and pressure of which are the same at all points. What is the matter here and where does the work come from?
The secret is that potential differences are still present here, but they do not appear in space, but in time.
This can be most clearly illustrated by the example of the atmosphere. Let there be no significant difference in pressure and temperature in the area where the engine is located. But (common at all points) pressure and temperature still change (for example, day and night). These differences can be used to obtain work (in full agreement with the laws of thermodynamics).
In the description of the invention "Method of extracting the energy contained in liquid and gas and converting it into mechanical work" (RF Patent No. 2059110), my version of a pseudo-perpetual and successfully operating solar engine is given. To increase the number of cycles and power, the properties of two media that are not in equilibrium with respect to each other - water and air - are most fully used. Archimedes' law is considered as a consequence of the law of conservation of energy, in which the buoyancy force is linked to the energy costs for the creation of water and air. The amount of this energy also determined such physical properties as, for example, density, heat capacity, and thermal conductivity.
Part of the ratio of energy to density creation is reflected in a disequilibrium factor of 820, and if we could find a way to exploit this disequilibrium to the fullest, we would get an energy gain of 820 times. Disequilibria appear from the moment air is supplied under the water column and increase as it ascends due to an increase in the volume of air and the removal of heat from the water, while the air is supplied with a temperature below the water temperature, because “If, for example, the air pressure is 4 atm (0.4 MPa), and the temperature is +20oC (293 K), then when it expands to atmospheric pressure, it will cool down to approximately -75oC (198 K), i.e. at 95oC". Heat will be removed under conditions close to adiabatic, i.e. with minimal heat loss, because Water is a good heat accumulator, but a poor conductor.
Cooling - water.
CALCULATION OF ENERGY EXTRACTING PNEUMO-HYDRAULIC TURBINE (RF patents N 2120058, N 2170364, N 2024780)
We use a compressor as a source of compressed air. The most suitable are positive displacement and dynamic type compressors. A reciprocating compressor consumes energy several times less than a dynamic one, so we will opt for a positive displacement compressor - a reciprocating one:
Compressed air source — piston compressor VP2-10/9.
We will judge the efficiency of a pneumohydraulic turbine by comparing the consumed and received power, i.e. amount of work per second.
Compressor performance - the volume of air entering the compressor at atmospheric pressure, i.e. productivity in 0.167 m3/s is the volume of air before entering the compressor and after surfacing in the turbine. When air is supplied under the lower level of the turbine, 0.167 m3 / s of water will be displaced through the upper level and the same amount will flow under the lower level, creating a water-air mixture and its movement inside the turbine housing. The value of 0.167 m3/s corresponds to the water flow when calculating the power of the pneumohydraulic turbine. We will carry out the calculation according to the formula for calculating the power of a hydroturbine:
N=9.81 Q H efficiency,
where 9.81 m/s2 is the free fall acceleration;
Q is the water flow in m3/s;
H is the head in m;
The efficiency of a real turbine reaches quite high values and, under the most favorable conditions, reaches 0.94-0.95, or 94-95%. We get power in kW. Since the working fluid is a water-air mixture, there is a need to confirm the validity of applying the power calculation formula for a hydraulic turbine. The most efficient mode of operation of the turbine seems to be the mode in which a mixture with a density of 0.5 t/m3 (consisting of 50% water and 50% air) is used. In this mode, the air pressure is slightly higher than the absolute pressure in the turbine casing. The air from the pressure pipe of the compressor exits in separate bubbles at regular intervals, and the volume of the bubbles is equal to the volume of water between them in the turbine housing. The bubble takes the form of a spherical segment and works like a piston in a fixed space, displacing water only upwards, because its downward flow is prevented by a higher pressure, and the flow to the sides is prevented by the incompressibility of water. With a constant supply of 0.167 m3 / s of air, 0.167 m3 / s of water will be displaced, i.e. through the upper level of the turbine, 2 0.167 m3 / s of a water-air mixture with an increased flow rate inside the turbine will be displaced, then
N = 9.81 2 Q 0.5 H efficiency = 9.81 Q H efficiency
Let's take an installation with a water column height of 2 m and determine the required compressor engine power to supply air under this water column, taking into account atmospheric pressure, based on the compressor technical characteristics data:
At the entire installation height, an upward flow of the water-air mixture will be observed, in which the buoyancy force, independent of the depth of the body immersion, allows at least 5 impellers to be placed. The energy mode of the proposed turbine proceeds in more favorable conditions than in the well-known pump "Airlift", because the water flow occurs below the water level in the turbine, i.e. under conditions close to weightlessness, without a significant rise in water in the turbine housing, for which the main amount of energy is consumed in the pump. Let's take the turbine efficiency equal to 0.9. In this case, the power is:
N = 9.81 0.167 2 5 0.9 = 14.7 kW
Thus, we received energy 13 times higher than the energy spent:
14.7 kW / 1.13 kW = 13
The increase in power due to the placement of additional impellers has been confirmed on experimental models. Indirectly, the performance of the turbine is confirmed by experiments conducted at the St. Petersburg State Technical University. Here is what the doctor of technical sciences, professor, member of the commission on non-
Photo 3, Photo 4
to traditional energy sources under the Government of the Russian Federation, head of the department "Renewable energy sources and hydropower" Elistratov V.V.: "However, based on the hydraulics of hydraulic machines and our numerous experiments on air intake into the impeller of a hydroturbine in order to reduce cavitation erosion, it was shown that that with the improvement of cavitation indicators, energy indicators were significantly reduced. In this case, experiments show that the supplied air creates a counter flow, which, acting on the impeller from below, causes it to rotate in the opposite direction. This is the design of the wheel (Fig. 1). And this action is exerted by a small volume of air in a small area equal to the body of the hydraulic turbine. The proposed installation has the ability to take heat from water and convert it into mechanical energy. Taking into account the difference in water and air temperatures, when the water temperature is 80oC (thermal source, water heated in a solar collector, in the cooling system of turbines, compressors, etc.), and the air temperature is 20oC, the coefficient of increase in air volume, according to Lussac's law , is equal to
1+ (80oС - 20oС) / 273 = 1.2
The power will be
N = 14.7 kW 1.2 = 17.6 kW
Our expectations in terms of energy gain have been confirmed.
17.6 kW / 5 = 3.5 kW 3.5 kW / 1.13 kW = 3.1 times per wheel
When calculating the power required for air supply, we took into account atmospheric pressure (1 atmosphere = 10 m of water column), which means that the rising air overcomes the absolute pressure inside the turbine casing, which is the sum of the pressure of the water column in the turbine and atmospheric pressure and is equal to pressure 12 -meter column of water. The absolute pressure inside the turbine housing is neutralized by the buoyancy of the air, but it is present behind the housing and affects the water supply to the turbine. This effect is equivalent to the effect on the water flow of the rarefaction created in the turbine housing by all the volume of air in the turbine (this effect is absent in the hydroturbine) and, with the appropriate design of the turbine, we have the right to consider the head as H = H w.st. + 10 m. Then the power will be equal to
N = 9.81 0.167 m3/s 12 m 5 1.2 0.9 = 106.14 kW
We received 93 times more energy than we spent.
We will calculate a more powerful power plant capable of providing energy to an average village, military unit, ship, etc. As a source of compressed air, we take a piston compressor 2VM10 - 63/9 with the following technical characteristics:
Productivity - 1.04 m3 / s
Final pressure, MPa - 0.9 (9 Atmospheres)
Compressor shaft power - 332 kW
Water cooling.
We will carry out the calculation for an installation with a water column height of 5 m with 10 impellers placed in it at a depth of 500 mm. The power of the compressor engine for supplying air under a water column of 5 m, taking into account atmospheric pressure, is
5 m (332 kW / 100 m) = 16.6 kW
The power of the installation is
N= 9.81 1.04 m3/s 15 m 10 1.2 0.9 = 1652 kW
Received energy 99 times higher than spent.
Thus, it is possible to obtain any amount of energy while improving the gas composition of water in an environmentally friendly way from an inexhaustible source of energy, using the natural imbalance of water and air in any climatic zone without building expensive dams and lock equipment, without flooding valuable agricultural lands, etc.
CALCULATION OF ENERGY EXTRACTING PNEUMO-HYDRO MOTOR
(RF patents N 2003830, N 2160381)
Compressed air source - piston compressor VP2 - 10/9.
Productivity - 0.167 m3 / s
Final pressure, MPa - 0.9 (9 Atmospheres).
Compressor shaft power - 56.5 kW
Water cooling.
We will judge the efficiency of a pneumohydraulic engine by comparing the power expended and received, i.e. the amount of work produced
mine per second. Compressor capacity - the amount of air at the compressor inlet, i.e. volume of air at atmospheric pressure. Then 0.167 m3/s is the volume of air at the inlet to the compressor and at the outlet of the upper float of the pneumohydraulic motor shown in Fig. 3. The release of the floats from the air and their filling with water occurs below the water level in the engine housing. With an air pressure of 9 atm, it can be fed under a column of water 90 m high. At an ascent rate of 0.4 m / s, the ascent time will be 225 seconds, while at the entire height of the column, air in motion will be present in the floats. The ascent rate equal to 0.4 m/s was determined as a result of measurements.
Its increase or decrease while maintaining the water column and compressor performance is reflected only in the horizontal dimensions of the floats, i.e. on length and width, because the amount of air increases or decreases, which, in turn, increases or decreases the force and does not affect the power of the pneumohydraulic motor. Changing the size of the floats only horizontally allows you to make floats of the required volume while maintaining the water column.
The volume of air at the outlet of the pressure pipe of the compressor at a depth of 90 m, taking into account atmospheric pressure, will be equal to
0.167 (m3/s) / 10 atm = 0.0167 m3/s
because the pressure of 10 m of the water column is 1 atm, and the increase in air volume by the value of the initial volume occurs every 10 m of ascent. If the volume of air did not change, then at the time of ascent it would occupy a volume equal to
0.0167 (m3/s) 225 s = 3.757 m3
Taking into account the increase in air volume during ascent, the volume will be equal to
3.757 m3 10 atm = 37.57 m3
Taking into account the coefficient of thermal expansion, the volume is equal to
37.57 m3 1.2 = 45.084 m3
The buoyancy force of 1 m3 of air is equal to 1000 kg s
This volume of air during ascent will produce
work equal
45.084 tC 0.4 m/s = 18.033 tC m/s
or 18033 kg C m/s
1 kg C m \u003d 9.81 watts, then when recalculating we get:
18033 kg S m / s 9.81 \u003d 176903.73 W or 176.9 kW
Adding to the received power at least 30% of the returned energy due to the reactive force developed when the float is filled with air and water is displaced from it, we get:
176.9 kW + 18 kW = 194 kW
We received 3.4 times more energy than we spent.
The mechanical efficiency of the pneumohydraulic motor will be quite high, because work takes place under conditions of constant lubrication with water, and the floats are mutually balanced. Compressor efficiency is taken into account when considering compressor motor power. The pneumohydraulic motor is equipped with a brake and stops on the move, while air remains in the floats and no energy consumption is required at the next start, because. when the brake is released, the air remaining in the floats will cause the engine to work.
We made a calculation for a commercially available compressor capable of supplying air under a 90 m high water column. This is an option to increase the efficiency of a hydroelectric power plant by placing pneumohydraulic motors on pontoons in reservoirs. Increasing the efficiency of hydroelectric power plants using tailwater is shown in the description of the invention No. 2059110. The design of pneumatic hydraulic motors is characterized by low metal consumption, because consists of lightweight frames. Any river, pond, stream, thermal spring, cooling tower can become a source of energy. At HPPs, due to the mixing of the lower warmer layers of water with the cold upper ones, accompanied by the simultaneous removal of heat, the water temperature will equalize. It is especially important that energy will not have to be saved, because. using natural disequilibrium to obtain it, we do not increase the energy disequilibrium of the Earth, but, on the contrary, return it, removing the consequences of thermal pollution. When it comes to solar energy, we don't use more than we get.
We considered the industrial option for obtaining energy, but there is a huge need for power plants of 3-4 kW. Let's take a look at its size. Let's take the height of the installation with the height of the water column equal to 2 m. Using the same compressor (only for calculation), we determine the power of the compressor motor for supplying air under a water column of 2 m:
N = (2 m 56.5 kW) / (90 m + 10 m) = 1.13 kW
Compressor capacity - 0.167 m3/s
2 m of water column create a pressure of 0.2 atm, then the volume of air at a depth of 2 m, taking into account atmospheric pressure, will be equal to
0.167 (m3/s) / 1.2 atm = 0.139 m3/s
The ascent time from a depth of 2 m is
2 m / 0.4 (m/s) = 5 sec
After 5 seconds, in the floats of the pneumohydraulic motor in the state of motion, taking into account the increase in volume during ascent and the coefficient of thermal expansion, there will be
0.139 (m3/s) 5 sec 1.2 atm 1.2 = 1 m3
When surfacing, work will be done
1000 kgC 0.4 m/s = 400 kgC m/s
Work per second means power.
1 kgC m \u003d 9.81 watts, then the power is
N = 9.81 W 400 = 3924 W = 3.924 kW
Adding 30% of the returned power, we get:
3.924 kW + 0.34 kW = 4.263 kW
With a mechanical efficiency equal to 0.9, we get the power
N = 4.263 kW 0.9 = 3.84 kW
We received 3.4 times more energy than we spent:
3.84 kW / 1.13 kW = 3.4
In order to once again verify the effectiveness of the proposed method of generating energy, let's compare it with the efficiency of a pumped-storage power plant, when water is pumped into a high-level reservoir by a pump or a reversible hydroturbine and used at the lower level in the turbine. In this case, with an efficiency equal to 100%, an amount of energy equal to the spent energy could be obtained. Let us determine the power of the pump motor for supplying water to a height of 90 m with a capacity of 0.167 m3/s:
N \u003d (9.81 0.167 m3 / s 90 m) / 0.75 \u003d 196.5 kW
Let us compare the obtained power with the power of the compressor engine equal to 56.5 kW with a capacity of 0.167 m3 / s of air capable of displacing the same volume of water to a height of 90 m with its supply to the turbine and obtaining 196.5 kW, while spending 3.5 times less energy. In addition, at the entire height of the water column, air remained in motion, which will also do work, which is confirmed by the above calculation. We will additionally consider the possibilities of implementing the proposed method on the graph (Fig. 2)
It follows from the graph that the action of the air buoyancy force immediately begins with the volume Vo. The shaded part is the water column H, to overcome the pressure of which the energy of the compressor is consumed, Vo is the volume of air at depth H, Vk is the volume of air expanded as a result of pressure drop during ascent, Vq is the effective volume of air. The graph shows that for a pneumohydraulic motor, the amount of air in operation is Vq, and for a pneumohydraulic turbine, the air volume equal to Vk is important, because the displaced volume of water works in it, which explains the difference in their efficiency.
The inexhaustibility of the energy source, absolute ecological cleanliness, active improvement of the environment, ease of manufacture and quick payback with increasing energy demand ensure the inexhaustibility of the sales market, and the variety of designs - a wide possibility of their application.
The presented method of obtaining energy seems to us the most promising, based on the following considerations:
relatively low manufacturing cost, the possibility of using common improvised materials for the construction of the tank, the possibility of using any air compressor that can be obtained, the relatively small dimensions of the device, which makes it possible to install it in a personal household.
The residence of the author within reach makes it possible to seek advice from him regarding the specific dimensions and shape of the elements of the device.
At the same time, the calculation of capacities by the author makes it not too fundamental the question of exceeding the received power over the spent power by dozens of times, if there is an effect, then it will manifest itself at any ratio of supplied and removed powers.
Moreover, home experiments do not require a powerful material base.
Any home craftsman is able to make a sample using any suitable containers, and adhering to the approximate ratio of dimensions given by the author.
The site administration will be grateful for information about experiments to test and build working samples.
METHOD OF OBTAINING ENERGY
(RF patent N 2059110)
Markelov V.F.,
In 1607, the Danish scientist Cornelius van Drebbel demonstrated to the English king James I a "perpetual" clock, driven, of course, by an equally "perpetual" engine. Drebbel patented them back in 1598. However, unlike numerous other devices with the same name, this engine was indeed "eternal" in a certain sense.
What was the secret of this watch (or rather, their engine)? Drebbel's perpetual clock worked from a drive that, like any other real engine, uses the only possible source of work - non-equilibrium (potential difference) in the external environment.
But the disequilibrium used by Drebbel is of a special kind, although it is also associated with temperature and pressure differences. It can operate in a perfectly equilibrium environment, the temperature and pressure of which are the same at all points. What is the matter here and where does the work come from?
The secret is that potential differences are still present here, but they do not appear in space, but in time.
This can be most clearly illustrated by the example of the atmosphere. Let there be no significant difference in pressure and temperature in the area where the engine is located. But (common at all points) pressure and temperature still change (for example, day and night). These differences can be used to obtain work (in full agreement with the laws of thermodynamics).
In the description of the invention "Method of extracting the energy contained in liquid and gas and converting it into mechanical work" (RF Patent No. 2059110), my version of a pseudo-perpetual and successfully operating solar engine is given. To increase the number of cycles and power, the properties of two media that are not in equilibrium with respect to each other - water and air - are most fully used. Archimedes' law is considered as a consequence of the law of conservation of energy, in which the buoyancy force is linked to the energy costs for the creation of water and air. The amount of this energy also determined such physical properties as, for example, density, heat capacity, and thermal conductivity.
Part of the ratio of energy to density creation is reflected in a disequilibrium factor of 820, and if we could find a way to exploit this disequilibrium to the fullest, we would get an energy gain of 820 times. Disequilibria appear from the moment air is supplied under the water column and increase as it ascends due to an increase in the volume of air and the removal of heat from the water, while the air is supplied with a temperature below the water temperature, because “If, for example, the air pressure is 4 atm (0.4 MPa), and the temperature is +20oC (293 K), then when it expands to atmospheric pressure, it will cool down to approximately -75oC (198 K), i.e. at 95oC". Heat will be removed under conditions close to adiabatic, i.e. with minimal heat loss, because Water is a good heat accumulator, but a poor conductor.
Cooling - water.
CALCULATION OF ENERGY EXTRACTING PNEUMO-HYDRAULIC TURBINE (RF patents N 2120058, N 2170364, N 2024780)
We use a compressor as a source of compressed air. The most suitable are positive displacement and dynamic type compressors. A reciprocating compressor consumes energy several times less than a dynamic one, so we will opt for a positive displacement compressor - a reciprocating one:
Compressed air source — piston compressor VP2-10/9.
We will judge the efficiency of a pneumohydraulic turbine by comparing the consumed and received power, i.e. amount of work per second.
Compressor performance - the volume of air entering the compressor at atmospheric pressure, i.e. productivity in 0.167 m3/s is the volume of air before entering the compressor and after surfacing in the turbine. When air is supplied under the lower level of the turbine, 0.167 m3 / s of water will be displaced through the upper level and the same amount will flow under the lower level, creating a water-air mixture and its movement inside the turbine housing. The value of 0.167 m3/s corresponds to the water flow when calculating the power of the pneumohydraulic turbine. We will carry out the calculation according to the formula for calculating the power of a hydroturbine:
N=9.81 Q H efficiency,
where 9.81 m/s2 is the free fall acceleration;
Q is the water flow in m3/s;
H is the head in m;
The efficiency of a real turbine reaches quite high values and, under the most favorable conditions, reaches 0.94-0.95, or 94-95%. We get power in kW. Since the working fluid is a water-air mixture, there is a need to confirm the validity of applying the power calculation formula for a hydraulic turbine. The most efficient mode of operation of the turbine seems to be the mode in which a mixture with a density of 0.5 t/m3 (consisting of 50% water and 50% air) is used. In this mode, the air pressure is slightly higher than the absolute pressure in the turbine casing. The air from the pressure pipe of the compressor exits in separate bubbles at regular intervals, and the volume of the bubbles is equal to the volume of water between them in the turbine housing. The bubble takes the form of a spherical segment and works like a piston in a fixed space, displacing water only upwards, because its downward flow is prevented by a higher pressure, and the flow to the sides is prevented by the incompressibility of water. With a constant supply of 0.167 m3 / s of air, 0.167 m3 / s of water will be displaced, i.e. through the upper level of the turbine, 2 0.167 m3 / s of a water-air mixture with an increased flow rate inside the turbine will be displaced, then
N = 9.81 2 Q 0.5 H efficiency = 9.81 Q H efficiency
Let's take an installation with a water column height of 2 m and determine the required compressor engine power to supply air under this water column, taking into account atmospheric pressure, based on the compressor technical characteristics data:
At the entire installation height, an upward flow of the water-air mixture will be observed, in which the buoyancy force, independent of the depth of the body immersion, allows at least 5 impellers to be placed. The energy mode of the proposed turbine proceeds in more favorable conditions than in the well-known pump "Airlift", because the water flow occurs below the water level in the turbine, i.e. under conditions close to weightlessness, without a significant rise in water in the turbine housing, for which the main amount of energy is consumed in the pump. Let's take the turbine efficiency equal to 0.9. In this case, the power is:
N = 9.81 0.167 2 5 0.9 = 14.7 kW
Thus, we received energy 13 times higher than the energy spent:
14.7 kW / 1.13 kW = 13
The increase in power due to the placement of additional impellers has been confirmed on experimental models. Indirectly, the performance of the turbine is confirmed by experiments conducted at the St. Petersburg State Technical University. Here is what the doctor of technical sciences, professor, member of the commission on non-
Photo 3, Photo 4
to traditional energy sources under the Government of the Russian Federation, head of the department "Renewable energy sources and hydropower" Elistratov V.V.: "However, based on the hydraulics of hydraulic machines and our numerous experiments on air intake into the impeller of a hydroturbine in order to reduce cavitation erosion, it was shown that that with the improvement of cavitation indicators, energy indicators were significantly reduced. In this case, experiments show that the supplied air creates a counter flow, which, acting on the impeller from below, causes it to rotate in the opposite direction. This is the design of the wheel (Fig. 1). And this action is exerted by a small volume of air in a small area equal to the body of the hydraulic turbine. The proposed installation has the ability to take heat from water and convert it into mechanical energy. Taking into account the difference in water and air temperatures, when the water temperature is 80oC (thermal source, water heated in a solar collector, in the cooling system of turbines, compressors, etc.), and the air temperature is 20oC, the coefficient of increase in air volume, according to Lussac's law , is equal to
1+ (80oС - 20oС) / 273 = 1.2
The power will be
N = 14.7 kW 1.2 = 17.6 kW
Our expectations in terms of energy gain have been confirmed.
17.6 kW / 5 = 3.5 kW 3.5 kW / 1.13 kW = 3.1 times per wheel
When calculating the power required for air supply, we took into account atmospheric pressure (1 atmosphere = 10 m of water column), which means that the rising air overcomes the absolute pressure inside the turbine casing, which is the sum of the pressure of the water column in the turbine and atmospheric pressure and is equal to pressure 12 -meter column of water. The absolute pressure inside the turbine housing is neutralized by the buoyancy of the air, but it is present behind the housing and affects the water supply to the turbine. This effect is equivalent to the effect on the water flow of the rarefaction created in the turbine housing by all the volume of air in the turbine (this effect is absent in the hydroturbine) and, with the appropriate design of the turbine, we have the right to consider the head as H = H w.st. + 10 m. Then the power will be equal to
N = 9.81 0.167 m3/s 12 m 5 1.2 0.9 = 106.14 kW
We received 93 times more energy than we spent.
We will calculate a more powerful power plant capable of providing energy to an average village, military unit, ship, etc. As a source of compressed air, we take a piston compressor 2VM10 - 63/9 with the following technical characteristics:
Productivity - 1.04 m3 / s
Final pressure, MPa - 0.9 (9 Atmospheres)
Compressor shaft power - 332 kW
Water cooling.
We will carry out the calculation for an installation with a water column height of 5 m with 10 impellers placed in it at a depth of 500 mm. The power of the compressor engine for supplying air under a water column of 5 m, taking into account atmospheric pressure, is
5 m (332 kW / 100 m) = 16.6 kW
The power of the installation is
N= 9.81 1.04 m3/s 15 m 10 1.2 0.9 = 1652 kW
Received energy 99 times higher than spent.
Thus, it is possible to obtain any amount of energy while improving the gas composition of water in an environmentally friendly way from an inexhaustible source of energy, using the natural imbalance of water and air in any climatic zone without building expensive dams and lock equipment, without flooding valuable agricultural lands, etc.
CALCULATION OF ENERGY EXTRACTING PNEUMO-HYDRO MOTOR
(RF patents N 2003830, N 2160381)
Compressed air source - piston compressor VP2 - 10/9.
Productivity - 0.167 m3 / s
Final pressure, MPa - 0.9 (9 Atmospheres).
Compressor shaft power - 56.5 kW
Water cooling.
We will judge the efficiency of a pneumohydraulic engine by comparing the power expended and received, i.e. the amount of work produced
mine per second. Compressor capacity - the amount of air at the compressor inlet, i.e. volume of air at atmospheric pressure. Then 0.167 m3/s is the volume of air at the inlet to the compressor and at the outlet of the upper float of the pneumohydraulic motor shown in Fig. 3. The release of the floats from the air and their filling with water occurs below the water level in the engine housing. With an air pressure of 9 atm, it can be fed under a column of water 90 m high. At an ascent rate of 0.4 m / s, the ascent time will be 225 seconds, while at the entire height of the column, air in motion will be present in the floats. The ascent rate equal to 0.4 m/s was determined as a result of measurements.
Its increase or decrease while maintaining the water column and compressor performance is reflected only in the horizontal dimensions of the floats, i.e. on length and width, because the amount of air increases or decreases, which, in turn, increases or decreases the force and does not affect the power of the pneumohydraulic motor. Changing the size of the floats only horizontally allows you to make floats of the required volume while maintaining the water column.
The volume of air at the outlet of the pressure pipe of the compressor at a depth of 90 m, taking into account atmospheric pressure, will be equal to
0.167 (m3/s) / 10 atm = 0.0167 m3/s
because the pressure of 10 m of the water column is 1 atm, and the increase in air volume by the value of the initial volume occurs every 10 m of ascent. If the volume of air did not change, then at the time of ascent it would occupy a volume equal to
0.0167 (m3/s) 225 s = 3.757 m3
Taking into account the increase in air volume during ascent, the volume will be equal to
3.757 m3 10 atm = 37.57 m3
Taking into account the coefficient of thermal expansion, the volume is equal to
37.57 m3 1.2 = 45.084 m3
The buoyancy force of 1 m3 of air is equal to 1000 kg s
This volume of air during ascent will produce
work equal
45.084 tC 0.4 m/s = 18.033 tC m/s
or 18033 kg C m/s
1 kg C m \u003d 9.81 watts, then when recalculating we get:
18033 kg S m / s 9.81 \u003d 176903.73 W or 176.9 kW
Adding to the received power at least 30% of the returned energy due to the reactive force developed when the float is filled with air and water is displaced from it, we get:
176.9 kW + 18 kW = 194 kW
We received 3.4 times more energy than we spent.
The mechanical efficiency of the pneumohydraulic motor will be quite high, because work takes place under conditions of constant lubrication with water, and the floats are mutually balanced. Compressor efficiency is taken into account when considering compressor motor power. The pneumohydraulic motor is equipped with a brake and stops on the move, while air remains in the floats and no energy consumption is required at the next start, because. when the brake is released, the air remaining in the floats will cause the engine to work.
We made a calculation for a commercially available compressor capable of supplying air under a 90 m high water column. This is an option to increase the efficiency of a hydroelectric power plant by placing pneumohydraulic motors on pontoons in reservoirs. Increasing the efficiency of hydroelectric power plants using tailwater is shown in the description of the invention No. 2059110. The design of pneumatic hydraulic motors is characterized by low metal consumption, because consists of lightweight frames. Any river, pond, stream, thermal spring, cooling tower can become a source of energy. At HPPs, due to the mixing of the lower warmer layers of water with the cold upper ones, accompanied by the simultaneous removal of heat, the water temperature will equalize. It is especially important that energy will not have to be saved, because. using natural disequilibrium to obtain it, we do not increase the energy disequilibrium of the Earth, but, on the contrary, return it, removing the consequences of thermal pollution. When it comes to solar energy, we don't use more than we get.
We considered the industrial option for obtaining energy, but there is a huge need for power plants of 3-4 kW. Let's take a look at its size. Let's take the height of the installation with the height of the water column equal to 2 m. Using the same compressor (only for calculation), we determine the power of the compressor motor for supplying air under a water column of 2 m:
N = (2 m 56.5 kW) / (90 m + 10 m) = 1.13 kW
Compressor capacity - 0.167 m3/s
2 m of water column create a pressure of 0.2 atm, then the volume of air at a depth of 2 m, taking into account atmospheric pressure, will be equal to
0.167 (m3/s) / 1.2 atm = 0.139 m3/s
The ascent time from a depth of 2 m is
2 m / 0.4 (m/s) = 5 sec
After 5 seconds, in the floats of the pneumohydraulic motor in the state of motion, taking into account the increase in volume during ascent and the coefficient of thermal expansion, there will be
0.139 (m3/s) 5 sec 1.2 atm 1.2 = 1 m3
When surfacing, work will be done
1000 kgC 0.4 m/s = 400 kgC m/s
Work per second means power.
1 kgC m \u003d 9.81 watts, then the power is
N = 9.81 W 400 = 3924 W = 3.924 kW
Adding 30% of the returned power, we get:
3.924 kW + 0.34 kW = 4.263 kW
With a mechanical efficiency equal to 0.9, we get the power
N = 4.263 kW 0.9 = 3.84 kW
We received 3.4 times more energy than we spent:
3.84 kW / 1.13 kW = 3.4
In order to once again verify the effectiveness of the proposed method of generating energy, let's compare it with the efficiency of a pumped-storage power plant, when water is pumped into a high-level reservoir by a pump or a reversible hydroturbine and used at the lower level in the turbine. In this case, with an efficiency equal to 100%, an amount of energy equal to the spent energy could be obtained. Let us determine the power of the pump motor for supplying water to a height of 90 m with a capacity of 0.167 m3/s:
N \u003d (9.81 0.167 m3 / s 90 m) / 0.75 \u003d 196.5 kW
Let us compare the obtained power with the power of the compressor engine equal to 56.5 kW with a capacity of 0.167 m3 / s of air capable of displacing the same volume of water to a height of 90 m with its supply to the turbine and obtaining 196.5 kW, while spending 3.5 times less energy. In addition, at the entire height of the water column, air remained in motion, which will also do work, which is confirmed by the above calculation. We will additionally consider the possibilities of implementing the proposed method on the graph (Fig. 2)
It follows from the graph that the action of the air buoyancy force immediately begins with the volume Vo. The shaded part is the water column H, to overcome the pressure of which the energy of the compressor is consumed, Vo is the volume of air at depth H, Vk is the volume of air expanded as a result of pressure drop during ascent, Vq is the effective volume of air. The graph shows that for a pneumohydraulic motor, the amount of air in operation is Vq, and for a pneumohydraulic turbine, the air volume equal to Vk is important, because the displaced volume of water works in it, which explains the difference in their efficiency.
The inexhaustibility of the energy source, absolute ecological cleanliness, active improvement of the environment, ease of manufacture and quick payback with increasing energy demand ensure the inexhaustibility of the sales market, and the variety of designs - a wide possibility of their application.
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