Question6: What is the neutral wire used for?
Answer6
... The neutral conductor is used to equalize the phase voltages at the load terminals. A \u003d 
and; 
B \u003d b; C \u003d c. In this case, the voltage drops across the load remain equal to the generator phase voltages. If the internal resistance of the generator is negligible (equal to zero), then the voltages at the load remain equal to the phase voltages of the generator, constant and do not depend on the magnitude of the load. ( The current will change, but the voltage across the load will not change).
Question 7: What equations describe the electrical state of a circuit with an unbalanced load?
Answer7
... With an asymmetric phase load and the absence of a neutral wire, phase voltage complexes at the load 
,
,
are related to the corresponding complex source voltages Ů A, Ů V, Ů C by the Kirchhoff equations:
;
;
;
where 
- the complex voltage between the neutral points of the load and the source ( networks).
called the neutral bias voltage.
The neutral bias voltage is calculated using the 2-node method:
where: Ė - complex EMF, 
- complexes of conductivity of load phases.
The currents of the load phases are found according to Ohm's law:
İ a \u003d a / Z a \u003d ( A - )/Z a;
İ b \u003d b / Z b \u003d ( B - )/Z b;
İ a \u003d c / Z c \u003d ( C - )/Z c.
Question 8: How to construct combined vector diagrams of voltages and currents for the investigated modes of a three-phase circuit?
Answer8 .
We begin the construction of vector diagrams with the vectors of line voltages set by the network and not depending on the conditions of the experiment. This is an equilateral triangle formed by line voltage vectors. The vector length corresponds to the line voltage, and the angles between the vectors correspond to the phase shift between the voltage vectors.
Plotting a vector diagram for the case of a uniform load . (symmetrical mode).
1.Select the complex plane (+ 1, j). The real axis +1 is directed vertically upward, the imaginary one along the -X axis. (turn by an angle of + 90 °).
2. Select the voltage scale, for example 1cm → 20V. Vector Ua (to scale) is plotted along the real axis + 1. The end of the vector is denoted by a small letter and.
3.Vectors U b and U c (to scale) draw at angles of + 120 ° and –120 °, respectively. The ends of the vectors are denoted by small letters band crespectively.
4. The point corresponding to the origin is denoted by a small letter n... This is the neutral point of the receiver.
5. We build the vectors of line voltages. To do this, we connect the ends of the phase vectors. We get vectors U a b \u003d U A B, U bc \u003d U BC, U c a \u003d U C A. Note that the receiver line voltages are equal to the generator line voltages.
Dot N in the vector diagram, the corresponding generator neutral point is in the center of the line voltage triangle. In this case, the generator neutral N coincides with the net of the receiver n. In general, the point ncorresponding to the neutral point of the load is found by the serif method. The vectors of currents are plotted in relation to the corresponding vectors of phase voltages, taking into account the phase shift between them.
Below are vector diagrams for different operating modes.
(fig. 8).
Mode 2. Phase loss AND (fig. 9):
In case of phase A breakage and the same load of the other two phases, the neutral point of the receiver nwill move to the middle of the line voltage Ů BC. Z b and Z c will be connected in series and connected to line voltage BC. Voltage drop between points A and nwill increase, and the phase voltages b and c become equal to half of the linear BC.
Mode 3. Phase A short circuit (fig. 9).
When phase A is closed and the other two phases have the same load (that is, when the beginning of the load of phase A is connected to the zero point of the load), point n moves to point A. The phase voltage Ů a becomes equal to zero, the current İ a increases, and the phase voltages b and c become equal to linear.
Resistance, Z a ≠ Z b ≠ Z c, receiver phase voltages a ≠ b ≠ c, the neutral bias voltage appears between the points N and n.
4.1 First, we build a triangle of line voltages.
4.2. Using the serif method (with a compass or a ruler), from each vertex, we postpone the corresponding vectors of the receiver phase voltages. The point of intersection of the arcs will give the neutral point of the receiver n. Generator neutral point Nwe leave in the same place.
4.3 Connect the point nand N... This is the bias voltage vector of the neutral U nN (to scale).
4.4 We build vectors of phase load currents. If the load is light bulbs, which can be represented as active resistances, then there will be no phase shift between the phase voltage and the phase current of the load. Therefore, we postpone the current vectors (to scale) along the corresponding vectors of phase voltages.
***) In the general case, it is necessary to determine the phase shifts between the current and the corresponding phase voltage according to Ohm's law in a complex form and build the current vector using a protractor.
Mode 5... Uneven load with neutral wire (fig. 11).
In the presence of a neutral wire, the phase voltages of the receiver become equal to the phase voltages of the source A \u003d and; B \u003d b; C \u003d c:
Neutral in power lines
In power lines of different classes, different kinds neutrals. This is due to the purpose and various equipment for protecting the line against short circuits and leaks. The neutral can be grounded, insulated and effectively grounded.
Deafly grounded neutral
It is used in lines with voltage from 0.4 kV to 35 kV, with a small length of power lines and a large number of consumer connection points. The consumer receives only the phases, the connection of a single-phase load is carried out between the phase and the neutral wire (neutral). The neutral wire of the generator is also grounded.
Isolated neutral
It is used in lines with voltages over 2 kV up to 35 kV, such lines have an average length and a relatively small number of consumer connection points, which are usually transformer substations in residential areas and powerful machines of factories and plants.
Both isolated and effectively grounded neutral can be used in 50 kV lines.
Effectively grounded neutral
It is used on long lines with voltages from 110 kV to 220 kV (p. 1.2.16 of the PUE)
see also
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Notes
Sources
- “Theoretical Foundations of Electrical Engineering. Electrical circuits "Bessonov L. A. Moscow. "Graduate School". 1996 ISBN 5-8297-0159-6
Neutral Wire Excerpt
The cannonade on the left flank will begin as soon as the cannonade of the right wing is heard. The riflemen of Moran's division and the Viceroy's division will open heavy fire when they see the start of an attack from the right wing.The vice king will take possession of the village [Borodino] and cross his three bridges, following at the same height with the divisions of Moran and Gerard, who, under his leadership, will go to the redoubt and enter the line with the rest of the army.
All this must be done in order (le tout se fera avec ordre et methode), keeping troops in reserve whenever possible.
In the imperial camp, near Mozhaisk, September 6, 1812 ".
This disposition, very vaguely and confusedly written, - if you allow yourself to treat Napoleon's orders without religious horror at Napoleon's genius, - contained four points - four orders. None of these orders could be and were not executed.
The disposition says, first: so that the batteries arranged in the place chosen by Napoleon with the guns of Pernetti and Fouche, only one hundred and two guns, which have to align with them, opened fire and bombarded Russian flashes and redoubts with shells. This could not be done, since the shells did not reach the Russian works from the places designated by Napoleon, and these one hundred and two guns fired at an empty one until the nearest commander, contrary to Napoleon's orders, pushed them forward.
The second order was that Ponyatovsky, heading for the village into the forest, bypassed the left wing of the Russians. This could not and was not done because Ponyatovsky, heading to the village in the forest, met Tuchkov, blocking his path, and could not bypass and did not bypass the Russian position.
Third order: General Kompan will move into the forest to capture the first fortification. The Kompana division did not take possession of the first fortification, but was repulsed, because, leaving the forest, it had to be built under canister fire, which Napoleon did not know.
Fourth: The vice king will take possession of the village (Borodino) and cross his three bridges, following at the same height with the divisions of Maran and Frian (about which it is not said: where and when they will move), which, under his leadership, will go to the redoubt and enter the line with other troops.
How much you can understand - if not from the stupid period of this, then from the attempts that were made by the viceroy to carry out the orders given to him - he had to move through Borodino to the left to the redoubt, while the divisions of Moran and Friant had to move simultaneously from the front.
All this, as well as other points of the disposition, was not and could not be fulfilled. Having passed Borodino, the viceroy was repulsed at Koloch and could not go further; the divisions of Moran and Friant did not take the redoubt, but were repulsed, and at the end of the battle the redoubt was captured by the cavalry (probably an unforeseen matter for Napoleon and unheard of). So, none of the orders of the disposition was and could not be executed. But the disposition says that upon entering the battle in this way, orders will be given in accordance with the actions of the enemy, and therefore it might seem that during the battle Napoleon will make all the necessary orders; but this was not and could not be because during the whole time of the battle Napoleon was so far away from him that (as it turned out later) the course of the battle could not be known to him and not a single order of his during the battle could be executed.
Consider the circuit in Fig. 5.12. When Z A ≠ Z B ≠ Z Cunbalanced current system (I А ≠ I B ≠ I C), therefore, in accordance with Fig. 5.5, there is a current in the neutral wire I N \u003d Ia + 1v + Ifrom. This current creates a fall; stresses I N Z Nin the neutral wire.
Due to the voltage drop across the neutral wire
point potentials Nundifferent, therefore the phase voltage of the receiver U "cnot equal to phase voltage
source U c.For these voltages to be equal,
be close to zero resistance of the neutral pro
water.
When Zc decreases to zero (short circuit of the receiver phase), the phase voltage U′c \u003d IcZcwill decrease to zero. A change in the phase resistance of the receiver entails a change in its phase voltage.
Phase short circuit FROMreceiver neutral point potential pbecomes equal to the potential of the point FROM, and hence the voltage U Aand U "bwill rise to line voltages Ucaand Ubcwhat unacceptable. To protect the receiver from such a mode, for example, fuses are installed in each phase. In case of a short circuit, the fuse link of the fuse burns out, which prevents the transfer of the potential of the point FROMexactly p.
In the presence of a neutral wire, a phase short circuit FROMthe receiver is at the same time a short circuit for the source E C, so the fuse works reliably. In the absence of a neutral wire, the fuse will not work, since the mode
Z C\u003d 0 is not a short circuit for the source E C.
Thus, if the resistance of the neutral wire, called in practice zero wire, significant, then:
1) the receiver phase voltage system is asymmetrical;
2) a change in the load (resistance) of one phase leads to a change in voltage on all phases of the receiver; 3) if the insulation of one phase of the receiver is damaged (short circuit), the receivers of the other two phases may fail due to overvoltages on them; 4) the operation of fuses (or other protective devices) becomes unreliable. Taking this into account, they strive to perform the neutral wire with low resistance.
But what about unexpected breaks in the neutral wire? In this case, it is impossible to operate the circuit due to the danger of failure of the receivers in the event of a short circuit of one of the phases.
More reliable is repeated re-grounding of the neutral wire: at the neutral point of the generator, at the branching points of lines, near public and industrial buildings, at the end of a three-phase line, etc. When the neutral wire breaks, the current passes through the ground.
Note that in order to reduce the asymmetry of the phase voltage of the receivers, in practice, single-phase receivers tend to be distributed evenly over the phases in order to reduce the current of the neutral wire, which is equal to zero with a uniform load.
For the calculation of a three-phase circuit, all methods used to calculate linear circuits are applicable. Usually, the resistances of the wires and the internal resistance of the generator are less than the resistances of the receivers, therefore, to simplify the calculations of such circuits (if greater accuracy is not required), the resistances of the wires can be ignored (Z L \u003d 0, Z N \u003d 0). Then the phase voltages of the receiver U a, U b and U c will be equal, respectively, to the phase voltages of the source of electrical energy (generator or secondary winding of the transformer), i.e. U a \u003d U A; U b \u003d U B; U c \u003d U C. If the total complex resistances of the phases of the receiver are equal Z a \u003d Z b \u003d Z c, then the currents in each phase can be determined by the formulas
İ a \u003d Ú a / Z a; İ b \u003d Ú b / Z b; İ c \u003d Ú c / Z c.
In accordance with the first Kirchhoff's law, the current in the neutral wire
İ N \u003d İ a + İ b + İ c \u003d İ A + İ B + İ C.
Phase voltage - occurs between the beginning and end of any phase. Otherwise, it is also defined as the voltage between one of the phase wires and the neutral wire.
Linear - which is also defined as interphase or between phase - arising between two wires or the same terminals of different phases.
When the power supply is connected with a triangle (Fig.3.12), the end X of one phase is connected to the beginning of the B of the second phase, the end of Y of the second phase - with the beginning of the C of the third phase, the end of the third phase Z - with the beginning of the first phase A. Beginning of A, B and C phases connected with three wires to receivers.
The connection of the phases of the source into a closed triangle is possible with a symmetrical EMF system, since
Ė A + Ė B + Ė C \u003d 0.
If the delta connection is incorrect, i.e. the ends or the beginning of two phases are connected to one point, then the total EMF in the triangle circuit differs from zero and a large current flows through the windings. This is an emergency mode for power supplies and is therefore unacceptable.
The voltage between the end and the beginning of a phase in a delta connection is the voltage between the line wires. Therefore, with a delta connection, the line voltage is equal to the phase voltage.
Neglecting the resistance of the line wires, the line-to-line voltages of the consumer can be equated to the line voltages of the power source: U ab \u003d U AB, U bc \u003d U BC, U ca \u003d U CA. Phase currents İ ab, İ bc and İ ca flow through the phases Z ab, Z bc, Z ca of the receiver. The conditional positive direction of the phase voltages Ú ab, Ú bc and сов ca coincides with the positive direction of the phase currents. The conditional positive direction of the line currents İ A, İ B and İ C is taken from the power sources to the receiver.
In contrast to the star connection, in delta connection, the phase currents are not linear. The currents in the phases of the receiver are determined by the formulas
İ ab \u003d Ú ab / Z ab; İ bc \u003d Ú bc / Z bc; İ ca \u003d Ú ca / Z ca.
Linear currents can be determined by phase currents by drawing up equations according to the first Kirchhoff's law for nodes a, b and c (Figure 3.12)
Adding the left and right sides of the system of equations, (3.20), we obtain
İ A + İ B + İ C \u003d 0,
those. the sum of the complexes of line currents is equal to zero for both symmetrical and unbalanced loads.
When connecting the phases of the winding of the generator (or transformer) with a star, their ends X, Y and Z connect to one common point N, called the neutral point (or neutral) (Fig. 3.6). The ends of the phases of receivers ( Z a, Z b, Z c) also connect to one point n... This connection is called a star connection.
Wires A−a, B−b and C−cconnecting the beginning of the phases of the generator and the receiver are called linear, the wire N−nconnecting point N generator with point n the receiver is neutral.
A three-phase circuit with a neutral wire will be four-wire, without a neutral wire three-wire.
In three-phase circuits, phase and line voltages are distinguished. Phase voltage U Ф - voltage between the beginning and end of the phase or between the line wire and neutral ( U A, U B, U C at the source; U a, U b, U c at the receiver). If the resistance of the wires can be neglected, then the phase voltage in the receiver is considered the same as in the source. ( U A=U a, U B=U b, U C=U c). For conventionally positive directions of phase voltages, directions from the beginning to the end of the phases are taken.
Linear voltage ( U L) - voltage between line wires or between the same terminals of different phases ( U AB, U BC, U CA). Conditionally positive directions of line stresses are taken from points corresponding to the first index to points corresponding to the second index (Fig. 3.6).
By analogy with phase and line voltages, phase and line currents are also distinguished:
Phase ( I Ф) are currents in the phases of the generator and receivers.
Linear ( I L) - currents in linear wires.
50. The concept of unbalanced operating modes in three-wire and four-wire circuits. The purpose of the neutral wire.
Three-wire circuit
In general, with an unbalanced load Z ab ≠ Z bc ≠ Z ca. It usually occurs when powered from a three-phase network of single-phase receivers. For example, for a load, fig. 3.15, phase currents, phase angles and phase powers will generally be different.
The vector diagram for the case when there is an active load in the ab phase, active-inductive in the bc phase, and active-capacitive in the ca phase is shown in Fig. 3.16, topographic diagram - in Fig. 3.17.
The construction of the vectors of linear currents is made in accordance with the expressions
İ A \u003d İ ab - İ ca; İ B \u003d İ bc - İ ab; İ C \u003d İ ca - İ bc.
Thus, with an unbalanced load, the symmetry of the phase currents İ ab, İ bc, İ ca is violated, therefore the linear currents İ A, İ B, İ C can be determined only by calculation according to the above equations (3.20) or found graphically from vector diagrams (Fig. 3.16, 3.17).
An important feature of connecting the phases of the receiver with a triangle is that when the resistance of one of the phases changes, the operating mode of the other phases remains unchanged, since the line voltages of the generator are constant. Only the current of this phase and the line currents in the line wires connected to this phase will change. Therefore, the delta connection is widely used to connect unbalanced loads.
When calculating for an unbalanced load, first determine the values \u200b\u200bof the phase currents İ ab, İ bc, İ ca and the corresponding phase shifts φ ab, φ bc, φ ca. Then the linear currents are determined using equations (3.20) in complex form or using vector diagrams
Four-wire circuit
With a symmetrical voltage system and unbalanced load, when Z a ≠ Z b ≠ Z c and φ a ≠ φ b ≠ φ c, the currents in the consumer phases are different and are determined according to Ohm's law
İ a \u003d Ú a / Z a; İ b \u003d Ú b / Z b; İ c \u003d Ú c / Z c.
The current in the neutral wire İ N is equal to the geometric sum of the phase currents
İ N \u003d İ a + İ b + İ c.
The voltages will be U a \u003d U A; U b \u003d U B; U c \u003d U C, U Ф \u003d U Л /, due to the neutral wire at Z N \u003d 0.
Consequently, the neutral conductor ensures the symmetry of the phase voltages of the receiver with an unbalanced load.
Therefore, a four-wire network includes single-phase unbalanced loads, for example, electric incandescent lamps. The mode of operation of each phase of the load, which is under a constant phase voltage of the generator, will not depend on the mode of operation of other phases
It is called zero because in some cases the current in it is zero, and neutral on the basis that it equally belongs to any of the phases.
The purpose of the neutral wire in that it is necessary to equalize the phase load voltages when the resistances of these phases are different, as well as to ground electrical equipment in networks with a solidly grounded neutral.
Thanks to the purpose of the neutral wire the voltage on each phase of the load will be practically the same with an uneven load of the phases. A lighting load switched on by a star always requires a neutral conductor, since a uniform phase load is not guaranteed.
The cross-section of the neutral wire of three-phase lines, in which the neutral wires are not used for grounding (special or reconstructed lighting networks), are taken to be close to half the cross-section of the phase wires.
If, for example, the phase wires have a cross section of 35 mm2, the neutral wire is taken 16 mm2.
The cross-section of the neutral wire of a three-phase system with a solidly grounded neutral, in which the neutral wire is used for grounding, must be at least half the cross-section of the phase wires, and in some cases equal to them.
The neutral wire of overhead lines 320/220 V must have the same brand and cross-section with the phase wires:
in areas made of steel wires, as well as bimetallic and steel-aluminum phase wires, with a cross section of 10 mm2;
if it is impossible to provide by other means the necessary selectivity of protection against short circuits to earth (in this case, it is allowed to take the cross-section of the neutral wires larger than the phase wires).
Since the current of the same magnitude flows in one- and two-phase lines along the neutral and phase wires, then for these lines the cross-sections of the neutral and phase wires are taken the same
51. The causes of transient processes in electrical circuits. Differential equations of the electrical state of circuits and methods for their solution.
Transient processes occur with any changes in the mode of the electrical circuit: when connecting and disconnecting the circuit, when the load changes, when emergency modes occur (short circuit, wire break, etc.). Changes in the electrical circuit can be represented in the form of certain switchings, generally called switching. Physically, transients are processes of transition from an energy state corresponding to a switching mode to an energy state corresponding to a post-switching mode.
Transient processes are usually fast flowing: their duration is tenths, hundredths, and sometimes billionths of a second. Relatively rarely, the duration of transient processes reaches seconds and tens of seconds. Nevertheless, the study of transient processes is very important, since it makes it possible to establish how the signal is deformed in shape and amplitude, to reveal voltage surges in individual sections of the circuit that may turn out to be dangerous for the isolation of the installation, an increase in the amplitudes of currents, which can be tens of times higher than the current amplitude steady-state periodic process, as well as determine the duration of the transient process. On the other hand, the operation of many electrical devices, especially industrial electronics devices, is based on transients. For example, in electric heating furnaces, the quality of the material produced depends on the nature of the transient process. Excessively fast heating can cause rejects, and excessively slow negatively affects the quality of the material and leads to a decrease in productivity.
In the general case, transient processes in an electrical circuit can occur if the circuit contains inductive and capacitive elements that have the ability to accumulate or release the energy of a magnetic or electric field. At the moment of switching, when the transient process begins, energy is redistributed between the inductive, capacitive elements of the circuit and external energy sources connected to the circuit. In this case, part of the energy is irrevocably transformed into other types of energy (for example, into heat energy on an active resistance).
After the end of the transient process, a new steady state is established, which is determined only by external energy sources. When external energy sources are disconnected, the transient process can occur due to the energy of the electromagnetic field accumulated before the onset of the transient regime in the inductive and capacitive elements of the circuit.
52. The laws of commutation and their use in determining the initial conditions.
The first law of commutation is that the current in the branch with the inductive element at the initial moment of time after commutation has the same value as it had immediately before commutation, and then from this value it starts to change smoothly. The above is usually written in the form i L (0 -) \u003d i L (0 +), assuming that the switching occurs instantly at the moment t \u003d 0.
The second law of commutation is that the voltage on the capacitive element at the initial moment after switching has the same value as it had immediately before switching, and then from this value it starts to change smoothly: UC (0 -) \u003d UC (0 +) ...
Therefore, the presence of a branch containing inductance in a circuit switched on under voltage is equivalent to breaking the circuit at this place at the moment of switching, since i L (0 -) \u003d i L (0 +). The presence in the circuit switched on under voltage of a branch containing a discharged capacitor is equivalent to a short circuit in this place at the moment of switching, since U C (0 -) \u003d U C (0 +).
However, voltage surges on inductors and currents on capacitors are possible in the electrical circuit.
In electrical circuits with resistive elements, the energy of the electromagnetic field is not stored, as a result of which transient processes do not occur in them, i.e. in such circuits, stationary modes are established instantly, in a jump.
In fact, any element of the circuit has some kind of resistance r, inductance L and capacitance C, i.e. in real electrical devices, there are heat losses due to the passage of current and the presence of resistance r, as well as magnetic and electric fields.
Transient processes in real electrical devices can be accelerated or slowed down by selecting appropriate parameters of circuit elements, as well as by using special devices.
53. Description of the process of charging and discharging a capacitor connected in series with a resistor. The simplest sawtooth voltage generator.
