Research website of Vyacheslav Gorchilin
2019-05-16
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Parametric capacity as electrostatic pump
Previously we have reviewed the formulation of the electrostatic pump (EHF) and some necessary data for the planet. Because the surface we will use as a source of electric charges. In this section, we will discuss parametric capacity and ways of its switching to the task.
Consider the Electromechanical version of this capacitance, which is the most simple and straightforward, though not very effective. But we need to solve the problem in the condition which says nothing about efficiency. Looking ahead, we should add that later we will deal with efficiency of real plants.
Parametric capacity
Our parametric capacity shown in figure (1a). It consists of two blades made of conductive materials (A, B), which rotate relative to each other with the help of the motor M. Its axis, and the axis with blades AB, separated by a dielectric to ensure that the supply circuit of the motor could operate independently from the rest of the pump. The blades are electrically connected. During mutual rotation (Fig. 1b) changed their General solitary capacity: in the case when the blade deployed relative to each other, the total capacity of the maximum and is denoted by $$C_0$$, and when the blades are combined, then their minimum total capacity, and is denoted $$C_1$$. The symbol of the two provisions are reflected in the figures (1d) and (1e) respectively. There is also proposed a more sophisticated variant of the nonlinear capacitance composed of several plates that can improve the efficiency of the entire device. Fig.1. Nonlinear capacity. It is possible to design and legend
Perfect for such a condenser would be the placing of the blades under the dome, and even better — with air pumped out (Fig. 1c). It was on such an ideal case we will use in further calculations. Without such a dome, there will be additional losses to ionization of the surrounding air, which we will not be taken into account. It is more important to the principle.
The algorithm of the EHF
Using a symbol (Fig. 1d-1e) try to imagine the process of pumping charges from the Earth over time. To do this, plug the parametric capacitor Cs as shown in figures (2a-2b), and we make synchronization of rotation of the blades and the switch SW1 so that when the maximum capacitance Cs will be closed, the contacts SW1.1 and open — SW1.2, while minimum — on the contrary: closed contacts SW1.2 and open — SW1.1. Grounding on those circuits depicted in the classical way, that is our connection to the source of electric charges. Fig.2. The algorithm of the EHF. Some circuit solutions
The algorithm operation is divided into two half-periods. In the first — the solitary and the maximum capacity Cs is connected to ground (Fig. 2a), so it contains a charge as well as all bodies on the surface of the planet. Obviously, it will be proportional to the area of the plates of the capacitor: $q_0 = k\, \sigma\, S_0 \qquad (2.1)$ where $$k$$ — coefficient of proportionality, $$\sigma$$ is the surface charge density of the Earth according to the formula (1.1) from the previous section, and $$S_0$$ is the maximum area of the plates of Cs. In fact, this expression must be more complex, but given that the maximum capacity of Cs is many orders of magnitude less than the capacity of the Earth, we consider the formula (2.1) is quite acceptable.
In the second half-cycle of the Cs is separated from ground, and the blades of the condenser is rotated, thereby creating a minimal a solitary capacity. From physics we know that a decrease in capacitance, with same charge (and he's hasn't spent), it increases proportionally with voltage. And if so, then between the ground and the blades and Cs is a kind of potential difference which we already can use. But to calculate we will go the other way and imagine what would happen if right now to connect XS1 to the ground? The planet will try to align imbalance and to compensate for the extra for this square plate charge and, therefore, its surplus flows through a jumper in the ground. Overcharging is calculated, obviously, so: $\Delta q = k\, \sigma (S_0 - S_1) \qquad (2.2)$ where $$S_1$$ is the minimum area the plates of Cs. But we let this Delta is not back in the ground, and through the contact SW1.2 in storage capacitor C1, whose capacity should be much greater than the capacitance Cs (Fig. 2b). This is to ensure that all excess charge is poured and accumulated in C1.
In the next half cycle, the contact SW1.2 opens, and Cs through the contact SW1.1, is connected to the ground. Blades Cs again turn around and increase their intimate capacity. Now it turns out that she did not have enough charges (in proportion to its new larger size). Planet again will try to compensate for this non-equilibrium state, but now, in contrast to the previous half-cycle, the charges will flow from the ground to the capacitor Cs. Further, all repeated once again. Thus, we pump electric charge out of the Ground in the storage capacity C1, where the pump acts as a parametric capacitance Cs.
Let's count up how much power we can get from such a setting. To do this, suppose that $$k=1$$, the area of the blades in the two States are respectively 2 and 1 square meter ($$S_0=2 , S_1=1$$), the capacity of the storage capacitor $$C_1=5\cdot 10^{-10}$$ Farad, and losses do not exist. Then the energy added to the cumulative capacity in each period will be so $W = {(\Delta q)^2 \over 2 C_1} = {\left[ k\, \sigma (S_0 - S_1) \right]^2 \over 2 C_1}, \qquad (2.3)$ and is numerically equal to 1.3 nanojoules. This is a very small value. Assuming that blades are rotating with a frequency of 1000 revolutions per second, the resulting output power will be equal to 1.3 microwatt: $P = W\,f \qquad (2.4)$ where: $$P$$ — capacity of the plant, $$f$$ is the rotational speed of the blades Cs. It is obvious that the cost of the rotation of the blades by orders of magnitude exceed the allowance. But we efficiency while not required, we decided the task — developed a pump for electric charges.
To increase the efficiency of EHF
This can be done in several ways: to increase the parametric capacity, to increase the frequency of rotation of the blades or to gather more energy by creating a larger potential difference. The first two methods are very costly and entail unreasonable electrical and mechanical losses at high voltages or square.
The third method assumes that the excess charge according to the formula (2.2) is not simply accumulated, but also participates in the following period (cycle) than creating more and more passionately charges by creating a larger potential difference. This method is shown in figure (2c), where cumulative capacity is included in the circuit ground and now can be comparable in value to a maximum capacitance Cs. To run this scheme requires the initial momentum or charge of the storage capacitor C1. Then, each period will be added to the charge, which will increase the potential difference, which in turn will contribute to more supply charges in the next period. Ie it turns out that the charge here is increasing exponentially, limited only by the parameters of capacitors, mechanical and electrical losses. It is even possible that the ground circuit will need to install current limiting varistors. To remove the excess charge (power output) in the high-resistance load Rn with cumulative capacity using the discharger FV1 or any other threshold device (Fig. 2d).
The following solution is shown in figure (2e), employing two parametric capacitor CS1 and CS2. Here the mechanics of the rotation of the blades of these capacitors are synchronized so that they must work in opposite phase: when the maximum capacity CS1, then CS2 it should be minimal, and Vice versa. L1 and L2 are the two identical Tesla transformer with one difference: the L2 is the reduction ratio of the additional winding. Through it, the load Rn, removed excess capacity. L1 and L2 needs to be inductively linked and work in opposite phase, and resonant frequency should be equal to the rate of turnover of the blades. The principle of this scheme is the same as in the diagram in figure (2c), but unlike it, the charge it collects in the two inductors.