Research website of Vyacheslav Gorchilin
2022-12-12
All articles/Energy ideas
Gravity-thermal generators
"Our world is immersed in a vast ocean of energy, we are flying in endless space at an unfathomable speed. Everything around rotates, moves - everything is energy. We have a daunting task ahead of us - to find ways to extract this energy. Then, extracting it from this inexhaustible source, humanity will move forward with giant steps."
Nikola Tesla, 1891
Introduction
This work was born due to incomprehensibility, inaccuracies and understatements in the field of description of some heat generators. And this is understandable: if you are not a specialist in thermodynamics, then it is very difficult to deal with all the nuances of the behavior of a gas, based on the available literature. Even an ideal gas has at least three changing and mutually dependent parameters: temperature, pressure and density. Changing, for example, the temperature of the air in the room, followed by a change in its pressure and density. But how exactly does each of these parameters change? But it is they who play the main role in the calculation of such generators.
In this work, we will try to obtain the basic regularities and gas formulas for their more practical application, calculate some types of thermal and even gravitational-thermal generators. To do this, we first determine the starting parameters of our calculations. Further, we will consider an ideal gas [1] in an adiabatic process [2]. This means that our gas (and we will mainly use a mixture of several gases - air) consists of very small, elastically colliding particles, and other fields such as gravitational or electric do not act on them. In principle, the results obtained will almost not differ from the operation of a real gas in most applications of such generators. In addition, all gas processes will proceed without exchange with the environment. Let us clarify that the latter applies only to gas, while other media involved in the processes of obtaining energy can carry out such an exchange. Since our readers have different levels of training, we will start with the simplest.
1.1 Heat pump for schoolchildren
At school, they explained to us the principle of operation of a heat pump through the Carnot cycle, mixing isochores, isobrars and other science that was not very clear at that time. The author believes that all these technical terms should be introduced, but only after understanding the very principle of obtaining and returning energy by a heat pump. The very word "thermal" proposes to begin this topic by studying the term "temperature." So what is temperature? The following explanation is not very scientific, but it is very important for understanding some processes.
Imagine that you are walking through a crowd of people. The more people there are and the higher your speed, the more you will sweat and "warm up". If we recall that our ideal gas is elastic balls, then we can immediately introduce an analogy that the temperature depends on only two parameters: the concentration of these balls and their average speed. The greater the speed of these balls (gas molecules) and the greater their concentration, the higher the temperature will be. Two principles of changing the temperature of a gas, or their mixture - air automatically follow from this.
First principle: increasing the speed of gas molecules
It is sold in conventional heaters that are in every home. When this device is heated relative to the surrounding air, its molecules vibrate at a high frequency. And when our ball (gas molecule) flies up to such a molecule, then this energy is transferred to it, and it flies away from the heater at a higher speed. That's it - we heated the room!
Second principle: increase/decrease in the concentration of gas molecules
This principle is applied in air conditioners and refrigerators. Here, too, everything is simple: to increase the temperature, the gas must be concentrated, i.e. compress, and to reduce it - expand. This is what the compressor does, or rather, its piston, on one side of which a higher concentration of gas molecules is formed, and thus a higher temperature relative to the environment, accordingly, on the other hand, their lower concentration and lower temperature are formed. Now it is enough to connect the radiators to the ends of the piston, one of which is put outside, and the second - in the room. If we want to cool the room, then the warm radiator is exposed outside, and the cold one is inside. If we want to heat the room, then do the opposite, which is what is implemented in air conditioners. The refrigerator has the same principle of operation, but it is made a little differently [3]. Note that in this case, the energy is spent only on the operation of the piston, so the received thermal energy here can be several times higher than the costs, which is usually expressed through a special coefficient COP [4].
Our readers will be interested to know about the so-called "Maxwell's Demon" [5]. If it were possible to create a membrane separating a vessel, which would allow fast (hot) gas molecules to fly only from the left side of the vessel to the right, and slow (cold) molecules only from the right side of the vessel to the left, then, after some time, in the right part of the vessel the air would be heated, and in the left part it would be cooled. Such a membrane would make it possible to create heat pumps right in the air!
1.2 Basic laws and some theory
The theory will be very useful to us further, when analyzing some thermal and gravitational-thermal generators. In addition, in this way we will remove many questions about the correctness of the application of formulas in their calculation. To do this, we need only three known gas laws, one gravitational, and one more generalizing law on mechanical work.
1) We write the Mendeleev-Claiperon law as follows [6-7]: \[\rho = {p\, M \over T\, R}, \quad \rho = {m \over V} \tag{1.1}\] Here: \(\rho\) -- gas density (for air, under normal conditions, is 1.2041 kg/m3), \(M\) -- molar mass (for air it is 0.029 kg/mol), \(T\) -- gas temperature (in Kelvin), \(R\) -- universal gas constant equal to 8.314 J/(mol∙K), \(m\) is the mass of the gas, \(V\) is its volume.
2) Poisson's equation for an ideal gas [2,Poisson's Adiabat]: \[p\, V^{k} = const \tag{1.2}\] Here: \(k\) is the adiabatic exponent. For a monatomic ideal gas, it is 5/3, for a diatomic (including air) - 7/5, for a triatomic - 4/3. This equation assumes the constancy of the product indicated there even if the third parameter, the gas temperature, changes.
3) Thermodynamic work [8]: \[A = \int \limits_{V_1}^{V_2} p\, \partial V \tag{1.3}\] Such work is performed, for example, when gas is compressed or expanded from one volume \(V_1\) to another \(V_2\). The integral form is used when, as the volume of a gas changes, its pressure \(p\) also changes. Interesting is the independence of this formula from temperature changes.
4) Gravity law of Archimedes [9]: \[F_A = \rho_w\, g\, V \tag{1.4}\] where: \(\rho_w\) -- liquid density (for water - 1000 kg/m3), \(g\) -- gravitational acceleration, for Earth equal to 9.81 m/s2, \(V\) -- gas volume (in our case).
5) Mechanical Work Law [10]: \[A = \int \limits_{0}^{H} F_A\, \partial h \tag{1.5}\] Substituting (1.4) into (1.5) we obtain the regularity we need: \[A = \rho_w\, g \int \limits_{0}^{H} V\, \partial h \tag{1.6}\] This law will tell us about the work received as a result of pushing the volume of gas from zero to the height of the water column \(H\). The integral is needed here because the Archimedes force, and hence the volume of the gas, varies with its height. The density of the liquid (water) is assumed to be constant.
The laws presented above will be quite enough to derive the basic relationships that will be needed to calculate energy generators. From them you need to derive simple engineering formulas so that they are "at hand" and they would not need to be deduced each time anew.
1.3 Some practical formulas
We connect the previously obtained laws into simple formulas that can already be used for calculations. And to begin with, we introduce relationships for calculating changes in two parameters, without the explicit presence of the third. Let us apply the law (1.2). For example, how does the pressure of a gas change depending on the change in its volume: \[ {p_2 \over p_1} = \left( {V_1 \over V_2} \right)^{k} \tag{1.7}\] And if we want to see how the temperature of the gas changes depending on the change in its pressure, then (1.7) must be substituted into (1.1), after which we get: \[ {T_2 \over T_1} = \left( {p_2 \over p_1} \right)^{k - 1 \over k} \tag{1.8}\] And the last necessary regularity is how the volume of a gas changes depending on the change in its temperature: \[ {V_2 \over V_1} = \left( {T_1 \over T_2} \right)^{1 \over k - 1} \tag{1.9}\] Now we can already do the simplest calculations. For example, we want to compress or expand a gas from its initial pressure \(p_1\) to its final pressure \(p_2\), and find out how much energy is required for this. Then we find the change in its volume by (1.7) \[ p(V) = p_1 \left( {V_1 \over V} \right)^{k} \tag{1.10}\] and substitute into (1.3): \[A = p_1 V_1^{k} \int \limits_{V_1}^{V_2} {\partial V \over V^{k}} = {p_1 V_1 \over k-1} \left[ 1 - \left({V_1 \over V_2} \right)^{k - 1} \right] \tag{1.11}\] Further, depending on what ratio we know (pressures or temperatures), we need to substitute (1.7) or (1.9). For example, we know the ratio of pressures before and after compression (expansion) of the gas, then the work on compression (expansion) will be as follows: \[A = {p_1 V_1 \over k-1} \left[ 1 - \left({p_2 \over p_1} \right)^{k - 1 \over k} \right] \tag{1.12}\] If we know the ratio of temperatures before and after compression (expansion) of the gas, then substituting (1.9) into (1.11) we get: \[A = {p_1 V_1 \over k-1} \left[ 1 - {T_2 \over T_1} \right] \tag{1.13}\] We will need these formulas in what follows.
Now is the time to clarify the issue of energy gain from the gas compression-expansion cycle. It can be verified that there is no additional energy in such a cycle, and \[{p_1 V_1 \over k-1} \left[ 1 - \left({p_2 \over p_1} \right)^{k - 1 \over k} \right]\ , +\, {p_2 V_2 \over k-1} \left[ 1 - \left({p_1 \over p_2} \right)^{k - 1 \over k} \right] = 0 \tag{1.14}\] An energy increase may appear if a static force field and some tricks are used when expanding the gas, but which we will talk about a little later.
Work on gas compression in a real compressor should be increased by \(k\) times in comparison with (1.12-1.13). The rationale for this is given in [11]: \[A_C = {k\, p_1 V_1 \over k-1} \left[ 1 - \left({p_2 \over p_1} \right)^{k - 1 \over k} \right] \tag{1.15}\] In the next part of this work, we will present some known and yet unknown devices for capturing the free energy around us, and calculate their effectiveness.
 
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Materials used
  1. Wikipedia. Ideal gas.
  2. Wikipedia. Adiabatic Process.
  3. Wikipedia. Refrigerator.
  4. Wikipedia. Coefficient of performance (COP).
  5. Wikipedia. Maxwell's Demon.
  6. Wikipedia. Equation of ideal gas.
  7. Wikipedia. Air Density
  8. Wikipedia. Thermodynamic work.
  9. Wikipedia. Archimedes' principle.
  10. Wikipedia. Mechanical work.
  11. P.A. Trubaev, B.M. Grishko. Technical thermodynamics. Part 2. [PDF]