Research website of Vyacheslav Gorchilin
2020-12-31
5. Calculation method
The calculation will be made for power lines in the form of a coaxial cable with known characteristics. Also, its length should be much less than the wavelength propagated in it.
At the beginning of the calculation, the current (conductance) and voltage are selected, which are maximum for a given cable. Let's choose the brand RK 75-7-12. The diameter of its central core is 1.2 mm, which means that, based on a current density of 4 A / mm2, we choose an effective current value of 4 A. Also, we will select the maximum effective voltage value from the cable passport data. Let it be 1200 V.
Let's choose the cable length, and based on it - the frequency of the installation. For example, if we choose a cable length of 4000 m, then, given that the wavelength should be much shorter, we choose an operating frequency of 9400 Hz.
Considering that the specific capacity of the cable is 67 pF / m, we calculate the capacity of the transmission line along the entire length: $$C = 268 \, nF$$. From here we find the displacement power: $$P_D = 22.8 \, kW$$. We calculate the power according to the formula (1.29), while: $$\omega = 2 \pi f$$, where $$f$$ is the frequency of the sinusoidal signal in the transmission line.
Let's find the active resistance of the center conductor. For cable RK 75-7-12 and length 4000 m this will be: $$R_l = 64 \, Omh$$. Then, the conduction power will be as follows: $$P = UI - I ^ 2 R_l = 4800 - 1000 = 3800 \, W$$, while the power of Joule losses in the transmission line is: $$P_J = 1000 \, W$$. From here you can find the efficiency of the line according to (2.1): $$\eta_l = 0.96$$.
The overall efficiency of the entire system can be found if we assume that the efficiency of the transmitting and receiving parts is 0.95. Then, by (2.2), the efficiency of the entire system is: $$\eta = \eta_ {tx} \eta_l \eta_ {rx} = 0.87$$.
The total portable power for a given transmission line, given the parameters, will be as follows: $$P _ {\Sigma} = \sqrt {P_D ^ 2 + P ^ 2} = 23 \, kW$$
Due to the value of the capacity of the transmission line, it is possible to calculate the inductance of the secondary winding of the transformer TS1 of this installation, based on formula (4.3) and the value of the operating frequency, which will be simultaneously resonant: $$L_1 \approx 1 \, mH$$.
5.1 Power transmission line comparison
Let's compare the power transmission line system calculated above with the same, but without taking into account the bias power. Such a line transmits electrical energy in the classical way - through a conduction current. The voltage in the transmission line, current, its length and other parameters will be the same. The current is chosen the same for reasons of the same heating of the center conductor.
In this case, the total transmitted power will be $$P _ {\Sigma} = P = IU - I ^ 2 R_l = 3.8 \, kW$$, while the Joule losses in the line: $$P_J = 1 \, kW$$. The efficiency of the line in this case will be as follows: $$\eta_l = {P \over P + P_J} = 0.79$$.
The overall efficiency of the entire system can be found if we assume that the efficiency of the transmitting and receiving parts is 0.95. Then: $$\eta = \eta_ {tx} \eta_l \eta_ {rx} = 0.71$$.
Thus, with the same cable parameters (RK 75-7-12), due to the transition from a classical power transmission line to a power transmission line using displacement currents, it is possible to increase the transmitted power 6 times , while increasing the efficiency from 0.79 to 0.87.
Further, examples of reverse calculation of operating devices will be given.
5.2. Installation example for 8 kW
Let's take for the calculation the "Equipment set for a resonant power transmission system with a capacity of 8000 W", developed at the Federal State Budgetary Scientific Institution FNATS VIM, Moscow, Russia. The declared characteristics are as follows [1]:
1. Resonant frequency - 9400 Hz
2. Effective line voltage - 980 V
3. Transmitted power - 7500 W
4. Overall plant efficiency - 0.86
5. Distance (line length) - 1500 m
6. Power transmission lines - coaxial cable RK 75-7-12: specific capacity of the cable - 67 pF / m, nominal diameter of the central core - 1.2 mm, resistance of the central core - 16 Ohm / km.
Let's take the value of the current conduction through the central core of the cable: $$I = 4 \, A$$. In this case, we know the operating voltage in the line from the device parameters: $$U = 980 \, V$$.
Taking into account the specific capacity of the cable, we calculate the capacity of the transmission line along the entire length: $$C = 101 \, nF$$ and inductance $$L = 570 \, uH$$. From here we find the displacement cardinality: $$P_D = \omega (C U ^ 2 + L I ^ 2) = 6270 \, W$$.
Let's find the active resistance of the center conductor. For cable RK 75-7-12 and length 1500 m this will be: $$R_l = 24 \, Omh$$. Then, the conductivity power will be as follows: $$P = UI - I ^ 2 R_l = 3920 - 380 = 3540 \, W$$, while the power of Joule losses in the transmission line is: $$P_J = 380 \, W$$. From here you can find the efficiency of the line: $$\eta_l = 0.95$$.
The overall efficiency of the entire system can be found if we assume that the efficiency of the transmitting and receiving parts is 0.95. Then: $$\eta = \eta_ {tx} \eta_l \eta_ {rx} = 0.86$$.
The total transferred power for a given transmission line, with the given parameters, will be as follows: $$P _ {\Sigma} = 7200 \, W$$.
Secondary inductance TS1 : $$L_1 \approx 2.8 \, mH$$.
5.3. Installation example for 20 kW
Let's calculate the power transmission system ROES-20 kW developed by order of Surgutgazprom LLC. The power transmission line was made of an insulated wire, presumably 1 mm in diameter, located directly on the surface of the earth. The declared data of this system are as follows:
1. Electric power on load, kW - 20.5
2. Line voltage, kV - 6.8
3. Line frequency, kHz - 3.4
4. Line length, km - 1.7
In this case, we can only calculate the displacement power, since the material and diameter of the conductor is not exactly known to us. Also, we will not be able to calculate the efficiency of this transmission line.
Taking into account the specific capacity of the cable, which is 12 (pF/m) [2], we calculate the capacity of the transmission line along the entire length: $$C = 20.4 \, nF$$. From here we find the displacement power: $$P_D \approx \omega C U ^ 2 = 20.2 \, kW$$, which, in this case, will approximately coincide with the total power. Despite the lack of many input data, we got a fairly accurate result, from which it can be concluded that in this installation, bias current is mainly used to transfer energy to power lines.
Secondary inductance TS1: $$L_1 \approx 107 \, mH$$.
Conclusions
Based on the formulas obtained, we can conclude that there are two types of energies transferred to power lines: the first one spreads along its cross section, the second one - along the conductor, but not in it. These energies correspond to two known types of currents: conduction current and displacement current, with the help of which these energies are transferred. Electric circuits with conduction current are calculated in the classical way: using Ohm's and Kirchhoff's laws, and to calculate electrical circuits with displacement current, a new mathematical apparatus, part of which is presented in this work.
The resulting formulas correspond to the declared parameters in real ROES devices and confirm the presence of two types of transferred energy in transmission lines. In the transition of telecommunication devices from classical ones, where only conduction current is used, to devices using bias current (or mixed currents), it is possible to achieve a rather tangible saving of power transmission line materials and increase the overall efficiency in ROES. The example shows a comparison of the same transmission line when using only the conduction current there, and when using a mixed current (conduction current and displacement current). In the latter case, it is possible to increase the transmission power by 6 times with a simultaneous increase in the efficiency of the entire ROES from 0.79 to 0.87.
At relatively small distances between the transmitter and receiver (Fig. 2), when the intrinsic capacity of the coil TS1 is comparable to the capacity of the transmission line, obviously, in addition to the line capacitance, the value of the displacement power is also influenced by the coupling capacity [3], which is not taken into account in this work. Such a distance for power lines can be, for example, within one laboratory.

Materials used
1. A set of equipment for a resonant power transmission system with a capacity of 8000 W, FGBNU FNATS VIM, Moscow, Russia. L. Yu. Yuferyov, O. A. Roshchin.
2. Theoretical foundations of electrical engineering: In 3 volumes. Textbook for universities. Volume 3 - 4th ed. / K. S. Demirchyan, L. R. Neiman, N. V. Korovkin, V. L. Chechurin. - SPb .: Peter, 2003 - 377 p .: ill., Ch. 25, Calculation of electrical capacity, pp. 25.1, Capacity of a two-wire transmission line.
3. Communication capacity. Calculation method - http://gorchilin.com/articles/math/capacity