Gravity-thermal generators. Part 2

2022-12-14

Gravity-thermal generators. Part 0. Armed with the theory and formulas from the of this work, we can proceed to the calculation of some parameters of thermal and gravitational-thermal generators. The main parameter that will interest us is the possible efficiency of real devices, and the conclusion that follows from this is the expediency of their use and production. Generators such devices are named by analogy with electrical machines, because. all devices listed here will generate electricity. From the point of view of thermodynamics, they are called heat pumps. . Generator on two hydraulic accumulators: day-night. This fairly simple power generator is based on the temperature difference in two hydraulic accumulators [1], connected by a relatively thin hose, in the gap of which a reverse pump PM is installed, which generates electricity when a liquid flows through it , or a working fluid . One accumulator is installed outdoors and the other is installed inside. Due to the resulting temperature difference, the flow of the working fluid from one GA to another through the PI pipe begins. We need to calculate the energy that the working fluid will do when flowing through the reverse pump. This mechanical energy will be equal to the electrical energy that the pump will generate, without taking into account its efficiency. .

Since we believe that we know the temperature difference between the internal and external HA, we can immediately apply formula from the previous theoretical part of this work. Only here, for convenience, we write it in a slightly different form: \[A = {p_1 V_1 \over T_1 } \Delta T \tag{2.1} Here: \ -- temperature difference between the internal and external accumulator, \ -- initial values of pressure, air volume and temperature, which can be taken from the internal HA, \ is the adiabatic exponent, which is equal to 0 for air. . Since, as a rule, air is pumped into the GA at an initial pressure of 0 atmospheres, we will take Let's take a GA with a volume of 0 liters, then \, and the temperature in the room where the internal HA is located, we will take equal to +20 ℃, or approximately 0 Kelvin: Then the work done by the air during its compression-expansion will be as follows: \[A \approx 128\cdot \Delta T \tag{2.2} For example, with a temperature difference of 0 degrees, two HA per 0 liters will produce 0 J of energy, which is equivalent to lighting a 0 W lamp for about 0 minutes. Given that such a temperature difference can occur only twice a day: day and night, the expediency of building such a generator is 0. You can slightly improve the parameters of this system, for example, take a HA with a volume of 0 liters, and pump an initial pressure of 0 atmospheres into it , then we will get an energy of approximately 0 J per 0 degrees of temperature difference, which is equivalent to the glow of a 0 W light bulb for 0 hours. As we can see, the efficiency situation has not fundamentally changed, although the output parameters have become more attractive. This generator can be slightly improved by pumping a triatomic gas, for example CO₂, into the GA instead of air. In this case, useful energy will increase by 20%. . Gravity Heat Generator. Ideas about such a generator can be found in publications of the last century, for example, in [2]. The principle of its operation is as follows . Compressor 0 pumps air through pipe 0 into working chamber 0 filled with liquid . Air from the pipe enters containers 0 , partially fills them and displaces liquid from them, forming the vertical force of Archimedes. Since the containers are interconnected by a chain 0 , which rotates on two gears 0, the resulting force rotates this entire structure counterclockwise . The upper gear is connected by a pulley 0 to a generator 0, which generates electrical energy, which is fed further to the control and distribution unit 0. From it, part of the energy goes to the compressor, and part goes to the consumer UG. .

Let's try to calculate such a generator for the ideal case, i.e. when the efficiency of its constituent parts and mechanisms is not taken into account. Then the energy of the compressor spent on pumping the volume of air \ into the working chamber is found according to the formula already known to us: \[A_C = {k\, p_1 V_1 \over k-1} \left[ 0 - \left^{k - 0 \over k} \right] \tag{2.3} Here: \ -- atmospheric pressure, \ -- pressure at the compressor outlet, also equal to the pressure of the height of the water column [3]: \[ p_2 = p_1 + p_w, \quad p_w = \rho_w g H \tag{2.4} where: \ -- liquid density , \ -- gravitational acceleration, for Earth equal to 0 m/s², \ -- working height of the water column . . Now let's calculate a completely different process: the work of the Archimedes force to push this volume out of the working chamber, and then just compare these two energies. We already have a previously obtained formula for describing such a process : \[A = \rho_w\, g \int \limits_{0}^{H} V\, \partial h \tag{2.5} When pushing air containers up, the latter expands as \ rises. To obtain this dependence, we use expression : \[V = V_2 \left} \right)^{1/k} = V_1 \left} \right)^{1/k} \tag{2.6} Here: \ is the volume of air on the surface and at depth \, respectively. Those. as you rise, the pressure of the water column, and hence the air, decreases, and its volume increases. The dependence of pressure on altitude is also obvious: \[p = p_2 - \Delta p\, {h \over H}, \quad \Delta p = p_2 - p_1 \tag{2.7} To simplify further calculations, we denote: \[x = {h \over H}, \quad p = p_2 - \Delta p\, x \tag{2.8} Substituting this into we get: \[V = V_1 \left^{1/k} \tag{2.9} Now we need to substitute this result into : \[A = p_w V_1\, p_1^{1/k} \int \limits_{0}^{1} {\partial x \over ^{1/k}} \tag{2.10} Taking this integral and putting its terms in order, we are surprised to get: \[A = {k\, p_1 V_1 \over k-1} \left[ \left^{k - 0 \over k} \right] \tag{2.11} Yakov Perelman was absolutely right [2] when he once called this generator eternal , because the energy expended by the compressor on compressing air is exactly equal to the work obtained from the Archimedes force , but with the opposite sign. Notice that the initial formulas were taken from different branches of physics, but they completely agreed on the result! It would seem that we are at an impasse and we cannot get any increase from the environment here, but there is a way out. . Here we do not take into account changes in the potential energy of water when the volume of air in the containers changes. it is a periodic process and the more containers will be involved, the less this change will affect the entire process. But we also do not take into account the eddy movement of water entrained by the entire structure, which should reduce the load on the compressor due to the reduced pressure at the air inlet. In the following sections, we will look at other options for increasing COP. . Multistage compressors. The very name of such a compressor suggests that its air compression is divided into several stages. After each stage, the compressed air is cooled, after which it enters the next compression stage. Ideally, such a compressor compresses air in a logarithmic relationship: \[A_C = p_1 V_1 \ln \left \tag{2.12} In reality, the number of its steps is limited , and the dependence is slightly different [4]: \[A_C = n {k\, p_1 V_1 \over k-1} \left[ 0 - \left^{k - 0 \over n \, k} \right] \tag{2.13} where: \ is the number of compressor steps. . The ideal efficiency coefficient of the installation will be as follows: \[ COP = {A \over A_C} = {1 \over n} {1 - ^{k - 0 \over k} \over 0 - ^{ k - 0 \over n\, k}} \tag{2.14} Below is a graph of additional COP increase depending on the number of compressors and installation height .

Up to 10% additional addition to can be achieved by applying the shock adiabatic method [5], in which shock waves can propagate from the compressor. But then the whole process of air injection will have to be done in waves, which entails the observance of additional conditions for its occurrence. . No Cost Float Principle. In figure 0, we will present a very unusual principle for obtaining additional energy from the gravity around us. In this figure, the number shows a pool of water, in which a pipe with a height \ is vertically installed, also filled with water. Moreover, the water in such a pipe is kept due to the closed upper end, which is why a strong rarefaction of small volumes of air is formed at its very top. A float with air is lowered into the pool to its height \ and launched into the pipe. Due to the Archimedean force, the float rushes up, and while it rises, we remove useful energy \ from it. We do not discuss the method of energy removal here, but it can be the same as in Figure 0. .

After the bench has risen to the very top, we take it out of the pipe without disturbing the vacuum existing there. The method of such a maneuver is also not discussed here, although it can be cranked using a system of shutters. Then we lower the float down again . The work of lowering the float into the pool to a depth \ is calculated using the already known formula \[A_C = {k\, p_1 V_p \over k-1} \left[ \left^{k - 0 \over k} \right] \tag{2.15} only instead of the upper pressure, the pressure at this depth is substituted here \[ p_p = p_1 + \rho_w g H_p \tag{2.16} Also, the volume of the float is substituted in the formula: . The formula for the work of a rising float also differs from only by some substitutions: \[A = {k\, p_1 V_p \over k-1} \left[ \left^{k - 0 \over k} \right] \tag{2.17} where: \[ p_2 = p_1 + \rho_w g H_2 \tag{2.18} From here you can find the COP of the installation: \[ COP = {A \over A_C} = {^{k - 0 \over k} - 0 \over ^{k - 0 \over k} - 1} x = {\rho_w g H_2 \over p_1}, \quad y = {\rho_w g H_p \over p_1} \tag{2.19} Let's display the obtained regularity in the graph 0. .

The red graph shows us the COP at y=0.01, i.e. when the height of the float is approximately 0 meter. At the same time, the installation height is displayed along the x-axis in meters*10, i.e. the value x=0.1 is approximately equal to 0 meter, and the value x=1 is approximately equal to 0 meters. . As we can see, now the efficiency ratios are quite good. But how to translate such a principle into reality? Any ways to solve this problem "on the forehead" give in the end a very difficult to implement mechanism. Let's try to change its conditions. . Third pressure. The solution of the problem "in the forehead" presented above did not bring high positive results, as is usually the case. Let's try to change its conditions. But what if an ejector [7] is installed between the compressor and the working chamber, and the compressor itself pumps not air, but water ? Then the ejector will suck in air and create a water-air mixture at the entrance to the working chamber, moreover, the formed air bubbles will have a reduced pressure, which we will denote \, and the dynamic pressure of the mixture will be equal to Recall that the total pressure in a horizontal flow is equal to the sum of pressures: static and dynamic [6] \[p_C = p_3 + p_w, \quad p_w = {\rho_w\,v^2 \over 2} \tag{2.20} where: \ is the pressure created by the compressor; \ -- water flow velocity; \ -- dynamic pressure created by the water flow. Our task is to accelerate the flow of water in the ejector to such a speed that \[ p_w \ge p_2 \tag{2.21} In this case, the water-air mixture will flow into the working chamber. The mixture has some inertia and after it enters the working chamber, the pressure in the air bubbles will not immediately increase to This is where the increase in COP occurs: the water column gradually compresses the air bubbles to the pressure \, after which the process occurs in the previously described manner. Note that it is not the compressor that compresses the air here, but gravity!.

It should be noted that in such a scheme it will be necessary to make another circuit, which will again run the excess water from the top of the installation into the compressor, and the compressor itself must have other parameters: be able to pump water well , and create pressure of the order . . . .